Graph Theory & Algorithms - M. Ashraf Iqbal

Graph Theory and Algorithms

M. Ashraf Iqbal

ii

c Copyright 2010 by M Ashraf Iqbal

All rights reserved.

ISBN . . .

. . . Publications

To my grand daughter Nariman

iv

Contents 1 Introduction

1

1.1

Why a new book? . . . . . . . . . . . . . . . . . . . . . . . . .

2

1.2

What do we emphasize? . . . . . . . . . . . . . . . . . . . . .

3

1.3

How the book is organized? . . . . . . . . . . . . . . . . . . .

6

1.4

How is the book designed? . . . . . . . . . . . . . . . . . . . .

9

1.5

Some salient features of the book . . . . . . . . . . . . . . . .

9

1.6

What tools do we use? . . . . . . . . . . . . . . . . . . . . . . 10

1.7

A Possible Sequence . . . . . . . . . . . . . . . . . . . . . . . 10

2 Basic Definitions in Graph Theory and Algorithms

19

3 Problems, Models & Graphs: Why Study Graph Theory?

31

3.1

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

3.2

Reducing One Problem into Another . . . . . . . . . . . . . . 32

3.3

The Satisfiability Problem in Logic Circuits . . . . . . . . . . 34 3.3.1

Reducing a 3-SAT Problem into an Independent Set Problem . . . . . . . . . . . . . . . . . . . . . . . . . . 35

3.3.2

Reducing the 3-CNF Satisfiability Problem into the 3DNF Satisfiability Problem . . . . . . . . . . . . . . . 39

3.3.3

Reducing the 3-CNF Satisfiability Problem into another graph Problem . . . . . . . . . . . . . . . . . . . 41

vi

Contents 3.3.4

Reducing the 2-CNF Satisfiability Problem into a Graph Problem . . . . . . . . . . . . . . . . . . . . . . . . . . 41

3.4

An Activity Scheduling Problem . . . . . . . . . . . . . . . . . 47

3.5

A Dual Machine Serial Processing Environment . . . . . . . . 48

3.6

Sequencing by Hybridization in Computational Biology . . . . 57

3.7

Discussion & Problems . . . . . . . . . . . . . . . . . . . . . . 62

4 Basics of Graph Theory

77

4.1

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

4.2

A Mutual Friendship Graph . . . . . . . . . . . . . . . . . . . 78

4.3

Representation of a Graph . . . . . . . . . . . . . . . . . . . . 79

4.4

Complement of a Graph . . . . . . . . . . . . . . . . . . . . . 82

4.5

Equal Graphs & Isomorphic Graphs . . . . . . . . . . . . . . . 83

4.6

The Degree Sequence . . . . . . . . . . . . . . . . . . . . . . . 86

4.7

Walks, Trails, & Paths . . . . . . . . . . . . . . . . . . . . . . 100

4.8

Multi-graphs and Pseudo-graphs . . . . . . . . . . . . . . . . . 102

4.9

Broad Categories of Graphs & some Special Graphs . . . . . . 102 4.9.1

Tree Graphs . . . . . . . . . . . . . . . . . . . . . . . . 104

4.9.2

Bipartite Graphs . . . . . . . . . . . . . . . . . . . . . 104

4.9.3

Special Graphs . . . . . . . . . . . . . . . . . . . . . . 104

4.10 Integration of Concepts, Properties, and Action Items . . . . . 112 4.11 Self Complementing Graphs . . . . . . . . . . . . . . . . . . . 114 4.11.1 Regular Self Complementing graphs . . . . . . . . . . . 114 4.11.2 Non Regular Self Complementing graphs . . . . . . . . 119 4.11.3 Constructing Self Complementary Graphs . . . . . . . 121 4.11.4 Transforming a SC graph with 4k + 1 vertices into another SC graph with 4k vertices . . . . . . . . . . . . . 124

Contents

vii

4.11.5 Transforming a SC graph with 4k vertices into another SC graph with 4k + 1 vertices . . . . . . . . . . . . . . 129 4.11.6 The Self Complementary problem and Graph Isomorphism . . . . . . . . . . . . . . . . . . . . . . . . . . . 131 4.11.7 A SC graph has diameter 2 or 3 - not less than 2 and not more than 3? . . . . . . . . . . . . . . . . . . . . . 132 4.11.8 Bipartite self complementary graphs . . . . . . . . . . 136 4.11.9 Decomposition of a SC graph G . . . . . . . . . . . . . 140 4.11.10 Permutation, Isomorphism, automorphism & Self Complementing Graphs . . . . . . . . . . . . . . . . . . . . 142 5 Basics of Graph Algorithms 5.1

5.2

5.3

5.4

153

Design of Algorithms . . . . . . . . . . . . . . . . . . . . . . . 156 5.1.1

What is Design? . . . . . . . . . . . . . . . . . . . . . 157

5.1.2

The Moore Method . . . . . . . . . . . . . . . . . . . . 157

The Bucket Algorithm . . . . . . . . . . . . . . . . . . . . . . 157 5.2.1

Understanding the Bucket Algorithm . . . . . . . . . . 159

5.2.2

How does it Work? . . . . . . . . . . . . . . . . . . . . 160

5.2.3

Playing with the Algorithm . . . . . . . . . . . . . . . 161

5.2.4

Solving Other Problems . . . . . . . . . . . . . . . . . 161

5.2.5

The Right Provocation . . . . . . . . . . . . . . . . . . 161

Finding if a Graph is Connected . . . . . . . . . . . . . . . . . 162 5.3.1

The Number of Connected Components

. . . . . . . . 163

5.3.2

Finding a Bridge in a Graph . . . . . . . . . . . . . . . 163

Finding if a Graph is a Tree . . . . . . . . . . . . . . . . . . . 164 5.4.1

Every Edge in a Tree is a Bridge . . . . . . . . . . . . 164

5.4.2

The Number of Edges in a Graph . . . . . . . . . . . . 165

5.4.3

The Spanning Tree of a Tree . . . . . . . . . . . . . . . 165

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Contents 5.4.4 5.5

5.6

5.7

5.8

5.9

A Comparison . . . . . . . . . . . . . . . . . . . . . . . 165

Finding a Spanning Tree of a Graph . . . . . . . . . . . . . . 166 5.5.1

Cutting Edges . . . . . . . . . . . . . . . . . . . . . . . 166

5.5.2

Growing Edges . . . . . . . . . . . . . . . . . . . . . . 166

5.5.3

Selecting Edges . . . . . . . . . . . . . . . . . . . . . . 166

5.5.4

Integrating Concepts and Discovering Algorithms . . . 167

Finding a Minimum Spanning Tree of a Weighted Graph . . . 167 5.6.1

Cutting or Growing Edges: A Krushkal’s like greedy algorithm . . . . . . . . . . . . . . . . . . . . . . . . . 167

5.6.2

Selecting Edges: A Prim’s like greedy Algorithm . . . . 171

5.6.3

A Panoramic Picture of various MST Finding Techniques173

5.6.4

The Maximum Spanning Tree Problem . . . . . . . . . 176

Finding a Path in a Graph . . . . . . . . . . . . . . . . . . . . 177 5.7.1

Cutting Edges . . . . . . . . . . . . . . . . . . . . . . . 177

5.7.2

Selecting Edges . . . . . . . . . . . . . . . . . . . . . . 179

The Shortest Path Problem . . . . . . . . . . . . . . . . . . . 179 5.8.1

Dijkstra’s (like) Algorithm . . . . . . . . . . . . . . . . 181

5.8.2

Discussion on Dijkstra’s (like) Algorithm . . . . . . . . 181

5.8.3

The Shortest Path Problem Redefined: The k-edge Shortest Path Problem . . . . . . . . . . . . . . . . . . 187

5.8.4

The k-edge Longest Path Problem . . . . . . . . . . . . 190

5.8.5

The Shortest Path Problem in Undirected Graphs with Negative Weights . . . . . . . . . . . . . . . . . . . . . 192

Graph Traversal Techniques . . . . . . . . . . . . . . . . . . . 195 5.9.1

Traditional Techniques & the Bucket-Algorithm . . . . 196

5.9.2

The Underlying Data Structure . . . . . . . . . . . . . 196

5.10 Some Graph Theoretic Claims . . . . . . . . . . . . . . . . . . 198 5.11 Shortest Path Algorithms . . . . . . . . . . . . . . . . . . . . 204

Contents

ix

5.11.1 Single Source Shortest Path Algorithms with positive edge weights . . . . . . . . . . . . . . . . . . . . . . . . 205 5.11.2 Single Source Shortest Path Algorithms for Directed Acyclic Graphs . . . . . . . . . . . . . . . . . . . . . . 210 5.11.3 Single Source Shortest Path Algorithms for directed graphs with negative edge weights . . . . . . . . . . . . 218 5.11.4 All Pair Shortest Path Algorithms . . . . . . . . . . . . 227 5.11.5 Johnson’s all Pair Shortest Path Algorithm . . . . . . . 233 5.12 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247 6 Network Flows, Connectivity and Matching Problems

249

6.1

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250

6.2

Definitions & Prior Knowledge . . . . . . . . . . . . . . . . . . 250

6.3

Konig’s Theorem, Menger’s Theorem & Hall’s (Marriage) Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261

6.4

Menger’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . 264 6.4.1

Maximum Edge-Disjoint Paths in Directed Graphs . . 264

6.4.2

The Concept of a Minimum Cut in Directed Graphs . . 271

6.4.3

A Proof of Menger’s Theorem and Finding the Minimum Cut in Directed Graphs . . . . . . . . . . . . . . 275

6.4.4

Finding Maximum Vertex-Disjoint Paths & Minimum Vertex Cut in Directed Graphs . . . . . . . . . . . . . 279

6.4.5

Menger’s Theorem for Undirected Graphs . . . . . . . 282

6.4.6

Edge Connectivity and Vertex Connectivity for Undirected Graphs . . . . . . . . . . . . . . . . . . . . . . . 285

6.5

Konig’s Theorem, Menger’s Theorem & Hall’s (Marriage) Theorem Revisited . . . . . . . . . . . . . . . . . . . . . . . . . . 298

6.6

Network Flows . . . . . . . . . . . . . . . . . . . . . . . . . . 306 6.6.1

Finding Maximum Edge-Disjoint Paths in Multi-Graphs306

6.6.2

The Maximum Flow & the Minimum Cut . . . . . . . 310

x

Contents

6.7

6.8

6.9

7

6.6.3

Algorithmic Issues & Complexity Calculations . . . . . 311

6.6.4

Lower Bounds on Edge Flows and the Max-Cut . . . . 312

The Matching Problem in Bipartite Graphs . . . . . . . . . . 324 6.7.1

Maximum Matching in Un-weighted bipartite graphs . 325

6.7.2

Maximum Matching in Complete (Binary) Weighted Bipartite Graphs . . . . . . . . . . . . . . . . . . . . . 328

6.7.3

Maximum Weighted Matching in Complete Weighted Bipartite Graphs . . . . . . . . . . . . . . . . . . . . . 332

The Max-Flow Min-Cost Problem . . . . . . . . . . . . . . . . 345 6.8.1

Introduction . . . . . . . . . . . . . . . . . . . . . . . . 345

6.8.2

Finding a Maximum Flow or Finding a Shortest Path? 345

6.8.3

Category 3 network flow Problems . . . . . . . . . . . 348

6.8.4

Category 2 (and 1) network flow Problems . . . . . . . 350

6.8.5

A Panoramic Picture of Similar Problems & Solutions (once again) . . . . . . . . . . . . . . . . . . . . . . . . 363

Network flows with lower & upper bounds on flow and the Circulation Problem . . . . . . . . . . . . . . . . . . . . . . . 364 6.9.1

Prior knowledge: . . . . . . . . . . . . . . . . . . . . . 364

6.9.2

New concepts . . . . . . . . . . . . . . . . . . . . . . . 365

6.9.3

New Problems . . . . . . . . . . . . . . . . . . . . . . . 366

6.9.4

Finding a feasible flow in a Circulation graph with one special edge . . . . . . . . . . . . . . . . . . . . . . . . 369

6.9.5

Finding a feasible flow in a network flow graph with one special edge . . . . . . . . . . . . . . . . . . . . . . 370

6.9.6

When upper bound is higher than a non zero lower bound372

6.9.7

Is it possible to solve the Circulation Problem for undirected graphs? . . . . . . . . . . . . . . . . . . . . . 377

Eulerian Graphs & the Chinese Postman Problem

381

Contents

xi

7.1

A Special Class of Graphs . . . . . . . . . . . . . . . . . . . . 382

7.2

Eulerian Circuits and Graphs . . . . . . . . . . . . . . . . . . 388

7.3

Eulerian Trails and Related Problems . . . . . . . . . . . . . . 394

7.4

Eulerian Walk and the Chinese Postman Problem . . . . . . . 397

7.5

The Chinese Postman Problem for Directed Graphs . . . . . . 410

8 Hamiltonian Graphs

423

8.1

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 424

8.2

Prior Knowledge . . . . . . . . . . . . . . . . . . . . . . . . . 424

8.3

8.2.1

Necessary Conditions for a Connected Graph . . . . . . 424

8.2.2

Necessary Conditions for a Hamiltonian Graph . . . . . 424

8.2.3

A Loose Sufficient Condition for a Hamiltonian Graph 425

8.2.4

Sufficient Condition for a Connected Graph . . . . . . 425

8.2.5

A Concept Map . . . . . . . . . . . . . . . . . . . . . . 425

Hamiltonian Graphs . . . . . . . . . . . . . . . . . . . . . . . 428 8.3.1

A Puzzle: . . . . . . . . . . . . . . . . . . . . . . . . . 429

8.3.2

Actually Finding a Hamiltonian Cycle in the Puzzle: . 429

8.3.3

Basic Intuition . . . . . . . . . . . . . . . . . . . . . . 436

8.3.4

Bondy & Chvatal’s Theorem: . . . . . . . . . . . . . . 436

8.3.5

Summary . . . . . . . . . . . . . . . . . . . . . . . . . 438

8.3.6

Closure of a Graph: . . . . . . . . . . . . . . . . . . . . 438

8.3.7

Ore’s Theorem: . . . . . . . . . . . . . . . . . . . . . . 439

8.4

Bipartite Hamiltonian Graphs . . . . . . . . . . . . . . . . . . 446

8.5

Some Theoretical Claims . . . . . . . . . . . . . . . . . . . . . 447

8.6

A Categorization of Hamiltonian Graphs . . . . . . . . . . . . 456

9 Strongly Connected Directed Graphs and Tournaments 9.1

463

Concepts, Properties & Actions . . . . . . . . . . . . . . . . . 464

xii

Contents 9.2

Strongly Connected Directed Graphs . . . . . . . . . . . . . . 471

9.3

Directed Acyclic Graphs (DAG’s) . . . . . . . . . . . . . . . . 476

9.4

Strongly Connected Components . . . . . . . . . . . . . . . . 485

9.5

Tournaments . . . . . . . . . . . . . . . . . . . . . . . . . . . 498

9.6

9.5.1

A Panoramic Picture and a Concept Map . . . . . . . 498

9.5.2

A Hamiltonian Path in a Tournament . . . . . . . . . . 498

9.5.3

A Hamiltonian Cycle in a Strong Tournament . . . . . 503

Unilaterally Connected Directed Graphs: . . . . . . . . . . . . 511 9.6.1

Unilaterally orient-able Un-directed Graphs . . . . . . 518

Chapter 1 Introduction

1.1

Why a new book?

1.2

What do we emphasize?

1.3

How the book is organized?

1.4

How is the book designed?

1.5

Some salient features of the book

1.6

What tools do we use?

1.7

A Possible Sequence

2

Introduction

1.1

Why a new book?

There are a number of excellent books available on topics covering an introductory course on Graph Theory & Graph Algorithms - almost each with a different approach. We shall discuss here why there was a need for another book designed with a different pedagogical structure and suitable for students and practitioners working in diverse fields who intend to use graph theory in their respective fields of study. Before we start making comparison between different pedagogical approaches and talk about the merit of our approach we shall first describe the basic structure of the said course. There are basically three layers on which such a course operates: 1. Definitions (concepts) 2. Implications & theorems (connections between concepts) 3. Algorithms (actions which transform an input into an output) Let us start with a few concepts used in our subject. For example graph, connected graph, tree, vertex, edge, bridge edge, edge weight, path, edge-disjoint paths, & edge cut are all concepts, and a learner needs to become familiar with these concepts. He or she should be able to feel, visualize, and connect these concepts. An implication connects two or more concepts. For example if every edge is a bridge edge in a connected graph G then the graph G is a tree. Menger’s theorem connects maximum edge-disjoint paths to the minimum edge cut in a graph. We need to prove a theorem like this one in a formal but convincing manner. An algorithm converts an input into a useful output. For example an algorithm is to be designed to find maximum edge-disjoint paths as well as the minimum edge cut in a given input graph. Many books tilting towards graph theory do not emphasize graph algorithms. Similarly a number of graph algorithm books ignore the theory part of this course namely graph theorems and their proofs. Unlike these books we operate on all the three layers, thus reinforcing understanding at various levels. In fact we try our best to design constructive proofs which not only prove a connection but also provide an algorithm to find something useful. For example we use constructive techniques to find as well as prove that the maximum number of edge-disjoint paths are equal to the size of the minimum cut in a graph.


3

Most of the other textbooks use a historical perspective of how and when a graph algorithm was discovered. For example Dijkstra designed a greedy algorithm to find shortest path between two vertices in a weighted graph. Similarly Prim designed a greedy algorithm to find a minimum spanning tree in a weighted graph. The historical perspective encourages one to teach these two algorithms in isolation without making any connections between the two algorithms. We think it is important from a learning perspective to integrate as well as differentiate concepts and techniques especially the ones which solve nearly identical problems. We thus encourage our readers to transform one algorithm (Prim’s algorithm) into another (Dijksta’s algorithm). Similarly, it is possible and desirable to transform one proof into another. For example we encourage our readers to transform the techniques used to prove that Dijkstra’s algorithm really finds shortest paths into techniques to prove that Prim’s algorithm really finds a minimum spanning tree in a weighted graph. We think that not only the similarities but also the differences should be highlighted between two almost similar concepts or algorithms. For example Prim’s algorithm can find a minimum spanning tree or a maximum spanning tree (after a minor modification) and can handle positive or negative edge weights while Dijkstra’s algorithm is unable to do so.

1.2


The single most prominent feature which distinguishes this book from other books in this field is an emphasis on transformations (or reductions). We expect our readers to first think of a non graph problem in terms of a graph. (see Fig. 1.2.1). We then encourage our readers to transform that non-graph problem into a graph problem. We also encourage them to transform a graph problem into another graph problem. Similarly we encourage them to transform one theorem into another theorem. We provide platforms where a learner is provoked to transform one proof technique (or an algorithm) into another proof technique (or an algorithm). We think of this course on graph theory & algorithms as a course on (intelligent) transformations. Transforming one concept into another and transforming one graph into another is the driving force behind this exercise (see Fig. 1.2.2).

4

Introduction

Figure 1.2.1: World Cup 2010 round of 16 is represented by a graph. (http://www.fifa.com/worldcup/matches/index.html)


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Figure 1.2.2: A visual depiction of how an algorithm transforms a graph into another graph while trying to find the shortest paths (Chapter 5).

6

1.3

Introduction


The book is organized into eight chapters other than this introduction (see Fig. 1.3.1). The second chapter provides standard definitions. The third chapter describes how a (non-graph) problem is transformed into a graph problem and encourages the readers to think in graph theoretic terms. The non-graph problems come from diverse fields like operations research, civil engineering, digital logic, distributed computing, molecular biology, and computer science. The fourth chapter focuses on basics of graph theory with an introduction to several concepts and properties related to graphs. The fifth chapter handles some important graph algorithms. Here we effectively start using our emphasis on transformations. We start with a stupid algorithm known as the Bucket Algorithm which provides us with a platform where we encourage our learners to modify and mould the Bucket Algorithm to design various useful graph algorithms. We purposely do not use a historical perspective and avoid describing an algorithm in its published (or polished) form. Instead we encourage our learners to devise cruder versions of an algorithm which are relatively easy to discover and appreciate intuitively. We also provide other tools like a colorful visual puzzle which differentiates between and integrates various shortest path algorithms (see Fig. 1.3.2). Chapter 6 is the longest chapter in the book. We discuss concepts related to graph connectivity, network flows, matching problems, minimum cost flows, and the circulation problem. We actively demonstrate how a theorem (and its corresponding proof) like Menger’s theorem is transformed into Hall’s theorem or the Konig’s theorem. We also show how one problem (and its solution), for example, the circulation problem is transformed into another problem like the minimum cost-maximum flow problem. This chapter consists of multiple & diverse topics as described before but a conscious effort has been made to make sure that the number of important milestones or bottlenecks in learning remains very small. For example, all the above mentioned topics depend upon one central and crucial idea, that is, if we need to find multiple edge-disjoint paths in a graph then we should reverse the direction of an already found path before finding another edge-disjoint path. In Chapter 7 we discuss necessary and sufficient conditions for Eulerian graphs and the Chinese Postman problem. All the proofs used here are constructive - we not only prove that an Eulerian Circuit exists in a graph with even degree but we also find that circuit using an algorithm. We show


7

Figure 1.3.1: Organization of our book consisting of nine chapters. The directed acyclic graph illustrates a possible teaching strategy.

8

Introduction

that the Circulation problem (already discussed in Chapter 6) was in fact the Chinese Postman problem in its general form. Both problems and their respective solutions in fact converge into a single problem and a single technique to solve it.

Figure 1.3.2: In shortest path algorithms (described in Chapter 5) we transform one algorithm into another solving the same problem. In Chapter 6, we transform one problem into another and the corresponding algorithm is also transformed in the process. We describe Hamiltonian graphs in Chapter 7. We discuss sufficient conditions for an undirected as well as a directed graph to be Hamiltonian. Hamiltonian directed graphs are discussed again in Chapter 8. We address strongly connected graphs and components, unilaterally connected graphs, and tournaments in this chapter as well. In case of tournament graphs we come back to the problem of finding a Hamiltonian path and a Hamiltonian


9

cycle provided the tournament is strongly connected.

1.4


We have followed a three step design strategy while writing each section and sub-section of this book: 1. Identify a potential bottleneck in learning a specific concept. Identify its nature using theories of learning & pedagogy (this is done through a statistical analysis of student feedback obtained in the last six years of teaching this course at LUMS, Lahore, Pakistan). 2. Remove the bottleneck by introducing a number of bridging concepts and by drawing suitable number of colored diagrams taking advantage of the role of visualization in learning graph theory. A number of research projects undertaken by students registered in a course titled “Problems of Learning & Teaching ” highlighting various problems of learning in the field of graph theory & algorithms have immensely helped us in the design of various sections). Several hundred colored diagrams play a central role in the design of the book. 3. Some text is added where needed in order to supplement diagrams unlike other books where diagrams supplement text.

1.5

Some salient features of the book

1. Discovery based learning is practised by first asking provoking questions before actually describing an algorithm or a theorem. 2. Instead of describing a concept in its finished & sophisticated form we first describe its cruder version which is easy to discover and appreciate 3. Emphasis on prior knowledge, its usefulness and limitations 4. An integrated approach where concepts, algorithms, and theorems reinforce learning and understanding 5. Starting with simple and easy to use building blocks which are used by a learner to construct more sophisticated concepts, algorithms or elaborate proofs

10

Introduction 6. We always make comparisons highlighting similarities as well as differences between concepts, algorithms as well as theorems 7. Constructive proofs with the help of algorithms 8. Complexity of learning under our control because of a conscious effort to keep it under limits. 9. We sometimes encourage our readers to make errors as we think that making an error is a step towards meaningful learning. We then encourage our learners to appreciate by themselves the repercussions of that error and in the process find an alternate path to solve the problem correctly.

10. We use a number of tools from the science of learning, e.g., concept maps.

1.6

What tools do we use?

We occasionally use concept map as a tool for better learning which enable a learner to explicitly make connections between concepts as shown in Fig. 1.6.1. The book has been used several times to teach a graduate level course on Graph Theory & Algorithms at LUMS. It has also been used to teach a similar course at the Virtual University of Pakistan. At the VU, we have used a 3-person drama format instead of a single person monologue. Yasser Hashmi, Komal Syed and I tried to discuss and debate various issues concerning graph theory & algorithms using this book (see Fig. 1.6.3). A lot of technology tools are nowadays available for synchronizing class room video with multimedia slides and lecture notes (synchronizing annotations for educational multimedia). Some of the VU lectures along with Power Point slides are available on Synote, and can be viewed after getting an account on www.synote.com.

1.7

A Possible Sequence

A possible sequence of lectures covering this book is given below in a typical 3-credit course in a semester system at a higher under-graduate or early graduate level.

A Possible Sequence

11

Figure 1.6.1: A concept map showing a number of relevant concepts and a number of theorems which relate different concepts (taken from Chapter 6). We use several visualization tools in order to enable the learner to visualize the working and the subsequent time complexity of an algorithm as shown in Fig. 1.6.2.

12

Introduction

Figure 1.6.2: Visualizing how an algorithm changes a graph and what price is paid in terms of number steps performed. This helps a learner to visualize the time complexity of an algorithm. Each edge in the pink graph exactly corresponds to one step in the algorithm.

A Possible Sequence

13

Figure 1.6.3: A 3-person drama format for teaching Graph Theory & Algorithms at the Virtual University of Pakistan.

14

Introduction

Figure 1.6.4: Virtual University video lectures along with Power Point slides are available on www.synote.com. A learner has the facility of expressing and sharing his or her feedback regarding the understanding of a concept in this software platform.

A Possible Sequence

15

Chapter 3 (1 Lectures) Problems, Models & Graphs: Why Study Graph Theory Reducing one Problem into another, the Satisfiability Problem (in Logic Circuits), An Activity Scheduling Problem (in Civil Works), a Dual Machine Serial Processing Environment (from Distributed Computing), Sequencing by Hybridization (in Computational Biology), discussion & problems (from Operations Research) Chapter 4 (3 Lectures) Basics of Graph Theory A Mutual Friendship Graph, Representation of a Graph, Complement of a Graph, Equal Graphs & Isomorphic Graphs, the Degree Sequence, Walks, Trails & Paths, Multi-graphs and Pseudo-graphs, Broad Categories of Graphs, Integration of Concepts, Properties, and Action Items Chapter 5 (4 Lectures) Basics of Graph Algorithms Design of Algorithms, the Bucket Algorithm, Finding if a Graph is Connected, Finding if a Graph is a Tree, Finding a Spanning Tree of a Graph, Finding a minimum (maximum) spanning (MST) tree, Finding a Path in a Graph, the Shortest Path Problem, Single source & all pair shortest paths, Graph Traversal Techniques, some Graph Theoretic Claims. Chapter 6 (9 Lectures) Connectivity, Matching Problems & Network Flows Definitions, Prior Knowledge, & Warm-up Exercises, Konig’s Theorem, Menger’s Theorem, & Hall’s (Marriage) Theorem, Menger’s Theorem, Finding Maximum Edge-Disjoint Paths, The Concept of a Minimum Cut, a Proof of Menger’s Theorem & Finding the Min-Cut, Finding Maximum Vertex-Disjoint Paths, Menger’s Theorem for Un-directed Graphs, Edge Connectivity & Vertex Connectivity, Konig’s Theorem, Menger’s Theorem, & Hall’s (Marriage) Theorem Re-visited. (3 Lectures) Network Flows: Finding Maximum Edge-Disjoint Paths in Multi-Graphs, the Maximum Flow & the Minimum Cut, Algorithmic Issues & Complexity Calculations, Lower Bounds on Edge Flows: Min-Flow & the Max-Cut, Lower as well as Upper Bounds on Edge Flows, the Matching Problem in Bi-

16

Introduction

partite Graphs, Maximum Matching in Unweighted Bipartite graphs, Maximum Matching in Complete (Binary) Weighted Bipartite Graphs, Maximum Weighted Matching in Complete Weighted Bipartite Graphs. (3 Lectures) The Max Flow Min Cost Problem: Finding a Maximum Flow or Finding a Shortest Path? A Panoramic Picture of Similar Problems & Solutions, Negative Weight Cycles & Improvement in Cost of Flow, Finding Maximum Flow at Minimum Cost (Category 1), a Panoramic Picture of Similar Problems & Solutions (once again), the Circulation Problem, finding a Feasible Flow Under Lower & Upper Bounds, Finding a Minimum Cost Feasible Flow, Connecting the Network Flow Problem with the Circulation Problem. (3 Lectures) Chapter 7 (3 Lectures) Eulerian Graphs & the Chinese Postman Problem A Special Class of Graphs, Eulerian Circuits & Graphs, Eulerian Trails & Related Problems, Eulerian Walk & the Chinese Postman Problem, The Chinese Postman Problem for Directed Graphs Chapter 8 (4 Lectures) Hamiltonian Graphs Prior Knowledge, Hamiltonian Graphs, Bipartite Hamiltonian Graphs, Problems & Claims, discussion & Conclusions Chapter 9 (5 Lectures) Strongly Connected Directed Graphs and Tournaments Concepts, Properties, & Actions, Strongly Connected Directed Graphs, Directed Acyclic Graphs (DAG’s), Strongly Connected Components, Tournaments, Unilaterally Connected Directed Graphs.

A Possible Sequence

Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter

No 1 2 3 4 5 6 7 8 9

A Possible Program of Study in one semester Topic Introduction Definitions Problems, Models, & Graphs Basics of Graph Theory Basics of Graph Algorithms Connectivity, Matching Problems, & Network flows Chinese Postman Problem & Eulerian Graphs Hamiltonian Graphs Strongly Connected Graphs & Tournaments

17

Hours 0 0 2 6 8 18 6 8 10

18

Introduction

Acknowledgements 1. Nooman Nadeem, ”Teaching Science and Mathematics: Discovery Based Learning”, Innovations 2005: World Innovations in Engineering Education and Research, Int. Network for Eng. Ed. and Res. (iNEER), Arlington, VA, pp. 409-418, 2005 2. Alvi Atif, The Magic of Dynamic Programming, Innovations 2004: World Innovations in Engineering Education and Research, Int. Network for Eng. Ed. and Res. (iNEER), Arlington, VA, pp. 409-418, 2004. 3. Alvi Atif, The Magic of Dynamic Programming, Proceedings of the International Conference on Engineering Education, Int. Network for Eng. Ed. and Res. (iNEER), Arlington, VA, Valencia Spain, 2003. 4. Sara Tahir, ”Should We Teach Algorithms?” IJECE, Vol.2, No.2, SummerFall 2003, pp. 134-140. 5. Harika Masood, ”Problems of Learning & Teaching,” proceedings of the Second International Conference on Pedagogies and Learning -2005, Australia. 6. Yasser Hashmi, ”Bridging over Problems of Learning in Finding Strongly Connected Components,” proceedings of the IADIS International Conference on Cognition and Exploratory Learning in Digital Age (CELDA 2005), pages 78 to 87, 2005. 7. Yasser Hashmi, ”Meaningful Learning of Graph Algorithms,” proceedings of the Third International Conference on the Teaching of Mathematics at the undergraduate level, Istanbul, Turkey, 2006.

Chapter 2 Basic Definitions in Graph Theory and Algorithms

20

Basic Definitions in Graph Theory and Algorithms

Graph A graph G is made up of a (non empty) set of objects called vertices or nodes - some of the pair of vertices may be connected by directed or un-directed links which are known as edges. Thus a graph consists of vertices and edges. We may represent the set of vertices by V (G) and the set of edges may be represented by E(G). Please note that the set E is a set of pairs of connected vertices. It is a set of ordered pairs in case of a directed graph and a set of un-ordered pairs in case of an un − directed graph. Usually the size of the vertex set V is represented by p while the size of the edge set E is represented by q. A weighted graph is one where there may be a weight associated with each edge of the graph. (Please see Chapter 4 and 5 for more details)

Multi Graphs We assume that multiple edges (known as parallel edges) between the same two vertices are not allowed in a graph. Similarly an edge coming out of a vertex and terminating at the same vertex (known as a self loop) is not allowed. In M ultiGraphs we allow parallel edges as well as self loops. (Please see Chapter 4 for more details)

Adjacent Vertices A vertex u is said to be adjacent to vertex v if there is an edge {u, v} in graph G. (Please see Chapter 4 for more details)

Degree of a vertex Number of edges connected to a vertex x is known as the degree of vertex x in an un-directed graph G. (Please see Chapter 4 for more details)

Regular graph An un-directed graph G is regular if the degree of each vertex is the same. (Please see Chapter 4 for more details)

21

Star graph A graph of p vertices in which one vertex has degree equal to p−1 while every other vertex has a degree equal to 1. It is known as a star graph because it looks like a star with rays of lighy coming out? (Please see Chapter 4 for more details)

Chain or Cycle graph A connected graph G such that the degree of each vertex is exactly 2. (Please see Chapter 4 for more details)

Complete Graph An un-directed graph G is complete if there is an edge between every pair of vertices of graph G.(Please see Chapter 4 for more details)

Bipartite Graph A graph G is bipartite if the vertex set V (G) can be divided into two subsets (or partites) A and B such that every edge in G connects a vertex in partite A and a vertex in partite B. (Please see Chapter 4 for more details)

Equal Graphs If for every edge {u, v} in a graph G there is an edge {u, v} in graph H and vice versa then the two graphs G and H are equal. (Please see Chapter 4 for more details)

Isomorphic Graphs Two graphs G and H are isomorphic if graph G becomes equal to graph H by some relabeling of vertices of graph H. Please note that two equal graphs are always isomorphic but two isomorphic graphs may not be equal. (Please see Chapter 4 for more details)

22


Complement c(G) of an un-directed graph G The complement c(G) of graph G is a graph with as many vertices as in G and an edge {u, v} in c(G) if and only if there is no edge {u, v} in G. (Please see Chapter 4 for more details)

Transpose of a directed graph G T ranspose T (G) of a directed graph G is obtained by reversing the direction of each edge in the directed graph G. (Please see Chapter 9 for more details)

Self Complementing graph G A graph G is self complementing (SC) if graph G and its complement are isomorphic to each other. (Please see Chapter 4 for more details)

Automorphism If a permutation p of a graph G creates a graph H which is equal to graph G then that permutation is known as an automorphism of graph G. The identity permutation is always an automorphism - it is known as a trivial permutation. We may be more interested in the non trivial permutations of graph G? It may be possible for a certain category of graphs that the only automorphism is the trivial (identity) permutation? Please note that if p is an automorphism of graph G then the permutation p2 is always an automorphism of graph G. (Please see Chapter 4 for more details)

Complementing Permutation If a permutation p of vertices of graph G creates a graph H which is the complement of graph G and at the same time H is isomorhic to G then the permutation p is known as the complementing permutation of graph G and graph G will be a self complementary graph. (Please see Chapter 4 for more details)

Walk You can walk on the edges of graph G edges starting from vertex u and ending at vertex v traversing different edges and vertices. In a walk you

23 can traverse an edge more than once. A walk is open if vertex u and v are different. It is a closed walk if vertex u and v are the same. Please note that in an un-directed graph you can traverse an edge in both directions but in a directed graph you can traverse an edge in only one direction - that is the direction of that edge. (Please see Chapter 4 and 7 for more details)

Trail and Circuit If no edge is repeated in a walk from a vertex u to a vertex v then the walk is known as a trail. A closed trail - that is when vertex u and v are the same then it is known as a circuit. Please note that it is possible to traverse a vertex more than once but an edge should not be traversed more than once in a trail or in a circuit. Thus a trail is always a walk but it is not the other way round. (Please see Chapter 4 and 7 for more details)

Path If neither an edge nor a vertex is repeated in a walk starting from a vertex u and ending at vertex v then the walk is known as a path. It is known as a u − v path. A path is always a trail (or a walk) but it is not the other way round. (Please see Chapter 4 and 5 for more details)

Shortest path Among all paths between vertex u and vertex v, the one with minimum length is known as the shortest path between vertex u and vertex v. In an unweighted graph G the minimum length is measured in terms of number of edges encountered in the u − v path. In a weighted graph G the minimum length is measured in terms of sum of weights of all edges in the u − v path. (Please see Chapter 4 for more details)

Cycle If a path is closed that means you come back to the vertex from where you have started then that path is known as a cycle. A cycle is a circuit but a circuit may not be a cycle as no vertex should be repeated in a cycle. (Please see Chapter 4 and 5 for more details)

24


A Connected Graph An un-directed graph G is connected if there is a path between every pair of vertices of that graph. (Please see Chapter 4 and 5 for more details)

Reachable relation or Transitive closure of a graph The Reachable Relation (or the transitive closure) of a directed graph D is another directed graph in which there is an edge from vertex u to vertex v provided v is reachable from u in D. The Reachable Relation graph of D can be represented by an adjacency matrix A in which A(u, v) = 1 provided there is a directed path from u to v in D, and it is zero otherwise. The Reachable Relation matrix of an undirected connected graph will contain all 1’s. (Please see Chapter 9 for more details)

Square of a directed graph The square of a graph D is another graph (known as) D2 in which there is an edge from vertex u to vertex v provided there is a two edge path from u to v or there is an edge (that is a one edge path) from u to v in D. (Please see Chapter 4 and 8 for more details)

A Strongly Connected Directed Graph A directed graph D is strongly connected if there is a directed path from vertex u to vertex v and a path from vertex v to u for every pair (u, v) in directed graph D. (Please see Chapter 9 for more details)

A Unilaterally Connected Directed Graph A directed graph D is unilaterally connected if there is directed path from vertex u to vertex v or a path from vertex v to u for every pair (u, v) in directed graph D. (Please see Chapter 9 for more details)

Directed Acylic Graph A directed graph D is directedacyclic (or a DAG) if there is directed path from vertex u to vertex v then there is no path from vertex v to u for every pair (u, v) in directed graph D. In other words there are no cycles in a

25 directed acyclic graph. (Please see Chapter 5 and 9 for more details)

A Tournament Graph A directed graph D is a tournament if there is a directed edge from vertex u to vertex v exclusive OR a directed edge from vertex v to u for every pair (u, v) in directed graph D. In other words if we put directions on edges in a completely connected un-directed graph G then the un-directed graph G transforms into a tournament (directed) graph D. It is known as a tournament graph as some (actual) tournaments (like f ootball league are played in the form of a tournament graph? (Please see Chapter 9 for more details)

Disconnected graph An un-directed graph is disconnected if for any pair of vertices u and v there is no u − v path. (Please see Chapter 4 and 5 for more details)

Cyclic graph G A graph G is cyclic if it contains one or more cycles. (Please see Chapter 4 and 5 for more details)

Bridge edge or Cut edge An edge {u, v} is a bridge edge if its removal disconnects an un-directed graph G. (Please see Chapter 4, 5 and 9 for more details)

Acyclic graph or a Tree An un-directed graph G is acyclic if it does not contain a cycle. It is also known as a tree graph provided graph G is connected. In other words a connected graph G is a tree if every edge of G is a bridge edge. (Please see Chapter 4, 5 and 9 for more details)

A Forest A (disconnected)graph G with no cycles. Please note that a tree is a connected graph with no cycles while a forest may be a disconnected graph. (Please see Chapter 4 and 5 for more details)

26


Spanning tree of a graph G A spanning tree (known as ST ) of a connected graph G contains all vertices of G and some edges of G, and it is a tree. In other words if you remove all non bridge edges in graph G then you get a spanning tree of G. (Please see Chapter 5 for more details)

A Binary Tree A binary tree is a tree such that the degree of each vertex is not more than three. (Please see Chapter 4 for more details)

A Path graph A path graph is a tree provided it has two vertices with degree one while all other vertices has degree exactly equal to two. (Please see Chapter 4 for more details)

Hamiltonian Cycle It is a cycle in a graph G which traverses each vertex of G exactly once. A graph which contains a Hamiltonian cycle is known as a Hamiltonian graph. (Please see Chapter 8 for more details)

Eulerian Circuit A circuit in a graph G such that every edge of graph G is traversed exactly once. A graph which contains a Eulerian circuit is known as a Eulerian graph. (Please see Chapter 7 for more details)

Set Cover Given a set of subsets S of the U niversal Set U , what is the smallest subset T of S such that the union of all these sub sets in T covers all elements of U . We have already witnessed the subset sum problem (in previous courses) in which we have to select integers (out of a set of integers) such that the sum of the selected integers is equal to a given constant. (Please see Chapter 6 for more details)

27

Vertex Cover The Universal set U is the set of all edges in a graph. What is the smallest subset of vertices of the graph that covers all edges? (Please see Chapter 6 for more details)

Independent (Vertex) Set It is the largest subset S of vertices of a graph such that no pair of vertices in S has an edge in between. Is there a connection between the vertex cover and the independent set? (Please see Chapter 3 and 6 for more details)

Matching (Independent Edge Set) It is a subset of edges in graph G such that no two edges in the subset has a common vertex in G. The edges in this subset are also known as independent edges. In other words a matching is a set of non-adjacent edges in a graph G; that is, no two edges in the subset share a common vertex. (Please see Chapter 6 for more details)

Maximal Matching This is a matching in which more edges cannot be added in the existing matching to increase the size of this matching. It may be possible to increase the size of the matching by first discarding the initial matching edges. (Please see Chapter 6 for more details)

Maximum Matching This is a matching in a graph with as many edges as possible? Please note that maximum matching is always maximal.Thus it is a matching of maximum size. (Please see Chapter 4, 5 and 6 for more details)

Edge Cover The Universal set is the set of all vertices in a graph. Then the edge cover is the smallest subset of edges, which covers all vertices? How small (or big) can the size of the edge cover become, as compared to the number of vertices in a graph? (Please see Chapter 6 for more details)

28


Perfect Matching A P erf ect Matching is a matching which covers all vertices of the graph. That is, every vertex of the graph is incident to exactly one edge of the matching. Every perfect matching is both maximum and hence maximal. In some literature, the term complete matching is used for it. (Please see Chapter 6 for more details)

Bridge edge or Cut edge An edge {u, v} is a bridge edge if its removal disconnects an un-directed graph G. (Please see Chapter 4, 5 and 6 for more details)

Cut vertex A vertex u is a cut vertex if its removal disconnects an un-directed graph G. (Please see Chapter 6 for more details)

Minimum Edge-Cut or MinCut A set of (minimum number of) edges which if removed will disconnect a special vertex s from another special vertex t in a graph G. (Please see Chapter 6 for more details)

Minimum Vertex-Cut A set of (minimum number) of vertices which if removed will disconnect a special vertex s from another special vertex t in a graph G. (Please see Chapter 6 for more details)

Edge Connectivity It is the minimum number of edges which if removed will disconnect an undirected connected graph G. (Please see Chapter 6 for more details)

Vertex Connectivity It is the minimum number of vertices which if removed will disconnect an un-directed connected graph G. (Please see Chapter 6 for more details)

29

Network Flow problems in a Network Flow graph D It is a directed graph D with two special vertices. One is a source vertex s and the other is a sink vertex t. The source vertex may produce flow while the sink vertex sinks flow. Every edge may have an associated lower bound on flow, an upper bound on flow, and a cost function associated with flow. We may like to find a f easible flow from vertex s to vertex t. We may like to find the maximum or minimum feasible flow or maximum f low at minimum cost.

Feasible Flow in a network Flow Graph A f easible flow in a network flow graph from a source vertex s to a sink vertex t is one in which flow through every vertex (other than the source and the sink vertices) is conserved (that is inflow is equal to out flow) and flow through every edge is within the prescribed upper as well as lower bounds. (Please see Chapter 6 for more details)

The Circulation Problem We consider a network flow graph in which the incoming flow in every vertex should be equal to the outgoing flow in every vertex. The lower bound on flow through every edge is exactly 1. There is a uniform cost of flow through every edge. We need to find a minimum cost f easible f low in the network flow graph (also known as the circulation graph). The problem is solvable in polynomial time for directed graphs. We assume that the Circulation graph is connected if it is un-directed and strongly connected if it is directed. (Please see Chapter 6 and 7 for more details)

The Chinese Postman Problem We are given a graph D which is strongly connected if directed and connected if un-directed. We need to find a closed walk in this graph such that the total distance covered is minimum in terms of the number of edges of the graph. The problem is solvable in polynomial time for directed as well as un-directed graphs. Please note that the Chinese Postman Problem transforms into the Circulation Problem (and vice versa) for directed graphs. (Please see Chapter 6 for more details)

30


Chapter 3 Problems, Models & Graphs: Why Study Graph Theory?

3.1

Introduction

3.2

Reducing One Problem into Another

3.3

The Satisfiability Problem in Logic Circuits

3.4

An Activity Scheduling Problem

3.5

A Dual Machine Serial Processing Environment

3.6

Sequencing by Hybridization in Computational Biology

3.7

Discussion & Problems

32

3.1

Problems, Models & Graphs: Why Study Graph Theory?

Introduction

When we face a real life problem then one possibility is to solve it right from scratch. A more desirable option is to better understand the problem in terms of reducing it into one of the known problems in computer science. There may be good chances that you are able to transform (or reduce) your unknown problem into one of the solvable problems in graph theory. Even if you end up reducing your problem into one of the hard (or unsolvable) problems – you certainly get a better insight. We shall discuss a number of diverse problems in this chapter; the problems are taken from digital logic, civil works, distributed computing, molecular biology, and the field of operations research. Almost all these problems do not seem to have a any relationship with graphs, yet each of them is transformed into a graph problem. As demonstrated in this chapter, it is very much possible (and in fact desirable) to reduce an unknown problem into multiple known problems instead of just one known graph problem. It may happen that out of these multiple known problems, one of the problems may be relatively simple to solve while the other known problem may be a hard one. Once your problem is transformed into a known problem, the complexity of solving your unknown problem will depend upon the complexity of your transformation as well as that of solving the known problem (see Fig. 3.1.1). The real challenge is thus to find an intelligent transformation into a simpler problem.

3.2


A reduction is a transformation of one problem into another problem. Sometimes we reduce a known hard problem into the new problem; thus showing that the new problem is as hard as some of the known hard problems. Usually reductions are used in such a negative context especially in the field of complexity theory. It is, however, possible to make reductions play a positive role: In such cases we transform a problem into one of the solvable problems, thus showing that a new problem is indeed solvable. Transforming one problem into another requires that each instance of the new problem should be transformed into instances of the old problem; we then solve the old problem using a known algorithm and then again transform its results to obtain the final solution of the new problem. We show such a transformation in Fig. 3.1.1. It is important to note the total time complexity of solving


33

Inputs

Outputs

A Known Algorithm for an Old Problem

Algorithm B Algorithm A

Inputs Outputs

A New Problem

Figure 3.1.1: Reducing a new problem into an old problem. The inputs for the new problem should be transformed into the inputs of the old problem. Similarly the outputs should also be transformed. a new problem by reducing it into an old problem will be time complexity of the known algorithm (needed to solve the old problem) plus the time complexities of Algorithm A and that of Algorithm B. We shall categorize a reduction into the following four categories: 1. Reduction of a hard problem (a problem for which) a polynomial time algorithm is not yet designed) into another hard problem. In complexity theory such problems are known as NP-complete problems; 2. Reduction of a problem (with an existing) polynomial time algorithm into another such problem. In complexity theory such problems are known as P problem; 3. Reduction of a P problem into an NP-complete problem; Does it mean that a P problem has become an NP-complete problem? 4. Reduction of an NP-complete problem into a P problem: Does it mean that an NP-complete problem has become a P problem? Why?

34


We shall provide (or discuss) at least one example from each category in this chapter.

3.3


Satisfiability in logic circuits is the problem of finding if we can assign 0 or 1 to the input variables so as to make the output of the logic circuit equal to 1. If no assignment of input variables can make the output 1 then we claim that the logic formula (or the circuit) is not satisfiable. In the 3CNF Satisfiability problem (or the 3-SAT Problem), we are given a boolean expression in 3-conjunctive normal form; in simple words it is the AND output of clauses of OR gates with exactly three inputs (see Fig. 3.3.1). Given such an expression, the 3-Satisfiability problem is to find if it is possible to assign binary values (0 or 1) to its inputs that will make the output equal to 1. In the 3-DNF Satisfiability Problem, we are given a Boolean expression in disjunctive normal form (DNF); in terms of logic circuits it is the OR output of clauses of AND gates with exactly 3 inputs (see Fig. 3.3.2). Again we need to assign input variable such that the output of the circuit is 1.

a

b c

a b c

a b c

a b c

Figure 3.3.1: A logic circuit consisting of OR gates (each with 3 inputs) and one AND gate. The output is 1 for the selected inputs shown in orange color. A lot many of you might have played with logic circuits; have explored assignments of inputs for which the output of the logic circuit is 1 – but many


35

of you may not have realized (or truly appreciated) that the 3-CNF problem is one of the NP-complete problems while the 3-DNF is a solvable problem (in polynomial time). Remember in the 3-CNF problem one has to select inputs such that the output of each OR gate is high so as to satisfy the output. In case of 3-DNF problem the output of the circuit will be high provided the output of any AND gate is high. Surprisingly applying DeMorgans laws it is possible to convert a 3-CNF Boolean expression in terms of a 3-DNF – giving a false impression that we can reduce an NP-complete problem into a P problem. We shall discuss this issue (in detail) in the coming paragraphs. OutPut

a

b

c

a

b c

a b c

a b c

Figure 3.3.2: A logic circuit consisting of AND gates (each with 3 inputs) and one OR gate. The output is 1 for the selected inputs shown in orange color. Problem Set 3.1. Problem 3.1.1. Design an efficient algorithm to solve the 3-DNF Satisfiability problem. Show that it is indeed possible to solve this problem in polynomial time (in fact in linear time). A 3-DNF circuit is shown in Fig. 3.3.2. Problem 3.1.2. Try to use a similar technique to solve the 3-CNF Satisfiability problem. It will not work. The purpose of this problem is to appreciate the inherent hardness of this problem. A 3-CNF circuit is shown in Fig. 3.3.1.

3.3.1

Reducing a 3-SAT Problem into an Independent Set Problem

Let us start with an example from Category 1. We shall reduce the 3-SAT problem into the Independent Set Problem in graphs. Thus in Fig. 3.3.3, the

36


New Problem is the 3-CNF Satisfiability problem: we are given a Boolean expression in 3-conjunctive normal form; in simple words it is the AND output of clauses of OR gates with exactly three inputs (see Fig. 3.3.1). Given such an expression, the 3-Satisfiability problem is to find if it is possible to assign 0 and 1 to its inputs that will make the output equal to 1. The figure also shows a combination of inputs (shown in orange color) for which the output of this logic circuit is 1. Please note that at least one input from each OR gate should be 1 in order to pass the test for Satisfiability. Inputs

Old Problem The Independent Set Problem in a Triangular Graph

Outputs

Algorithm B Algorithm A

Inputs

New Problem The 3-Satisfiability Problem

Outputs

Figure 3.3.3: Reducing the 3-Satisfiability problem into Independent Set problem in a graph. The old problem in Fig. 3.3.3 is the Independent Set problem in a special graph consisting of k triangles with edges connecting certain vertices within different triangles as shown in Fig. 3.3.4. The problem is to find if it is possible to select one vertex from each triangle such that no two selected vertices have an edge in common (they should not be adjacent). In other words we need to find if the size of the Independent Set (a set of vertices with no common edges) in this graph is equal to the number of triangles in the graph. It is obvious that for the size of the Independent Set to be equal to k we have to select exactly one vertex from each triangle such that no two selected vertices are adjacent.


37

Figure 3.3.4: A graph consisting of k triangles where some vertices from different triangles are adjacent. We need to find if the size of the independent Set is equal to k.

Algorithm 1: The 3-SAT Problem: Find for what inputs, the output is high for the given logic circuit? Input : A Boolean formula (or circuit) in 3-CNF. Output: Yes or No; If Yes then a combination of inputs for which the output is high. 1. We consider this as a new problem: You need to transform it into a known problem in graph theory?

Algorithm 2: The Independent Set Problem: Find if the size of the independent set in a given graph is k. Input : A graph consisting of k triangles. Output: Yes/No; If Yes then select a vertex from each triangle such that no two selected vertices is adjacent. 1. We consider this as a known problem in graph theory

38


OutPut AND

c a

¬a ¬b c

a b ¬c

a ¬b c

c

¬c

¬b

a

¬a ¬b ¬c

b

¬a

¬c

¬b

¬a

¬b

Figure 3.3.5: The 3-Satisfiability problem (top) is reduced into the Independent Set problem in the graph shown in the bottom. Orange vertices in the bottom graph provides a solution to the Independent Set problem in the bottom graph while their orange counterparts (input variables to the logic circuit) provides a satisfiable solution to the 3-SAT problem in the top circuit.


39

It is obvious that a possible solution of the Independent Set problem in the graph provides a combination of inputs for which the output will be high in the logic circuit. Thus if the size of the Independent Set in the graph is equal to the number of OR gates in the logic circuit then it is possible to find a combinations of inputs which will make the output high. In fact the reduction goes in both directions; the solution of any one problem implies a possible solution for the other. Unfortunately both these problems belong to the class known as NP-Complete problems. In the next section we shall talk about a Category 2 reduction. Problem Set 3.2. Problem 3.2.1. We have already talked about the 3-Sat problem and its reduction to independent set problem as modeled in Fig. 3.3.3, and depicted in Fig. 3.3.5. Please design Algorithm A and Algorithm B as shown in Fig. 3.3.3. Problem 3.2.2. Try to use a similar reduction to reduce the 2-SAT problem into the independent set problem in a graph. Discuss why this may or may not be possible. Problem 3.2.3. Try to use a similar reduction to reduce the 3-DNF problem into the independent set problem in a graph. Discuss why this may or may not be possible.

3.3.2

Reducing the 3-CNF Satisfiability Problem into the 3-DNF Satisfiability Problem

We know that it is possible to convert a 3-CNF Boolean formula into a 3DNF formula. Such a conversion is done after drawing the truth table or the K-Map of the 3-CNF expression as shown in Fig. 3.3.6. We also know that the 3-CNF Satisfiability problem is NP-Complete while it is possible to solve the 3-DNF Satisfiability problem in polynomial time. Is there a contradiction somewhere? Design the corresponding conversion algorithms (Algorithm No. 1 & 2 of Fig. 3.1.1) and you will be able to resolve this contradiction your self.

40


OutPut AND

Dr aw

OR

OR

OR

K

OR

¬a ¬b c

a b ¬c

¬a ¬b ¬c

c

0

1

00

1

0

01

0

1

11

0

0

10

1

1

ab a ¬b c

Ma p

OutPut OR

AND

AND

AND

AND

a ¬b ¬c

a ¬b c

¬a ¬b ¬c

¬a b c

aw Dr

it rc u Ci

Figure 3.3.6: The 3-CNF Satisfiability problem (top) is reduced into the 3DNF-Satisfiability Problem as shown in the bottom. The truth table, also shown on the right, helps us in this reduction.


3.3.3

41

Reducing the 3-CNF Satisfiability Problem into another graph Problem

Here we intend to discuss a reduction of the 3-SAT problem into the Clique problem in graphs. It is left as an exciting exercise for the reader?

3.3.4

Reducing the 2-CNF Satisfiability Problem into a Graph Problem

We have already talked about reducing a 2-CNF Satisfiability problem into an Independent Set problem in graphs. In fact this problem can be reduced to another graph problem which is solvable in polynomial time. We shall describe this Category 2 reduction briefly in this section. We shall partially justify this reduction and leave the rest of the details as an interesting problem for the reader. The 2-SAT problem is similar to the 3-SAT problem except that now each of the OR gates have two inputs rather than three. We still need to find a combination of inputs for which the output of the circuit is 1. The corresponding graph problem deals with directed graphs; we need to find if any two given vertices, x and y, belong to a single strongly connected component; in other words we need to check if there is a directed path from vertex x to vertex y and from vertex y back to vertex x. This problem and its possible solutions are discussed in detail in Chapter 9. Path finding algorithms are described in Chapter 5. Given the Boolean expression or the logic circuit, we construct a directed graph D according to the following rules: 1. For every variable x in the Boolean expression we create two vertices with labels x and ¬x in the directed graph D. 2. For every OR gate in the Boolean expression with inputs x and y, we add two directed edges: One directed edge from vertex ¬x to vertex y and another directed edge from vertex ¬y to vertex x. Once we have a constructed directed graph D, we claim that the given Boolean expression (or the logic circuit) is not satisfiable (that means for any combination of input variables) if and only if any vertex x and its complement vertex ¬x in graph D belongs to the same strongly connected component.

42


¬a b

OR

¬a ¬b

OR

c

a

¬c

a

b

¬b

¬a

AND OR Convert

¬c

a

OR

c

Figure 3.3.7: The 2-CNF Satisfiability problem is reduced into a graph problem. If any vertex x and its complement in the graph (shown in the right diagram) belongs to the same strongly connected component then the Boolean expression corresponding to the circuit (shown in the left diagram) is not satisfiable. We apply Algorithm 3 for every vertex x (and its complement ¬x) in graph D, and if the answer comes out to be NO in each case then the given Boolean expression is satisfiable otherwise not. If the Boolean expression is satisfiable then we have to find a combination of input variables for which the output of the logic circuit is 1. This requires a deeper understanding and appreciation of different concepts involved in this reduction. We shall provide some hints in this regard and leave the rest as a problem for imagination of the reader. Algorithm 3: Find if two given vertices belong to the same strongly connected component in a directed graph D. Input : A directed graph D, and two vertices x and y. Output: Yes/No; 1. Check if there is a path from vertex x to y and from y to x in the directed graph D. If yes then vertices x and y belong to the same strongly connected component otherwise not.

A 2-CNF expression consisting of a single OR/AND combination is shown in the left diagram of Fig. 3.3.8. It is transformed into a directed graph shown

Tr Lo uth gi Ta c C ble irc o ui f t


¬a b

AND

43

a

a→ b ¬a OR b ¬b → b¬a

0

0

1

1

0

1

1

1

1

1

1

1

1

0

0

0

Im Trut pli h T ca a tio ble n Gr of ap h

For what values of input the output is 1

OR

a

b

¬b

¬a

If a = 1 then b=1 for Output to be 1

Figure 3.3.8: A 2-CNF expression with only one OR/AND gate is reduced into a directed graph. The directed graph is in fact an implication graph with a truth table shown at the top. Note that the implication graph tells us that if a = 1 then b should also be 1 otherwise the output of the logic circuit will be zero.

If a = 0 then Output =1

¬a b

OR AND

¬a ¬b

a

b

¬b

¬a

For what values of input the output is 1

OR If a = 1 then b=1 then ¬a =1

Figure 3.3.9: The 2-CNF Satisfiability problem is reduced into a graph problem. Note that if a = 1 then we face a contradiction in the implication graph. On the other hand if a = 0 then there is no such contradiction and the output of the logic circuit will be 1 for any value of b.

44


in the right diagram of this figure. As you should appreciate this directed graph is not just a directed graph – this is in fact an implication graph. A directed edge in this graph from vertex x to vertex y implies that if the vertex x (or meaningfully) variable x is 1 then y is 1. The truth table of this implication is exactly the same as that of the Boolean expression as shown in this figure. The implication tells us that in order to make the output 1 we shall make b = 1 if a = 1. This makes sense because if both inputs of the OR gate are 0 then the expression will not be satisfiable. The implication graph for a Boolean expression consisting of two OR gates are shown in Fig. 3.3.9. Again the implication graph is telling us how it is possible to make the Boolean expression satisfiable. If a = 1 then b should be 1 so that the output of the top OR gate becomes 1. But if b = 1 then according to the implication ¬a should be 1 so as to make the output of the second OR gate 1 – but that is a contradiction. But if a = 0 then the implication does not dictate any thing – it means that the Boolean expression is satisfiable for any value of b. We show various 2-CNF logic circuits for varying number of OR gates and the corresponding implication graphs in Fig. 3.3.10. It will be interesting to find if the circuit is satisfiable (as we add more OR gates); and in case it is then what inputs should be applied so that the output is 1 for each logic circuit shown in this figure. It should now be possible to design an efficient graph algorithm which operates on the implication graph but which finds the input combination for which the logic circuit output is 1. Please note that in the bottom diagram of Fig. 3.3.10, it is possible to reach from vertex a to vertex ¬a, it is also possible to reach from vertex ¬a to a as shown by the closed path shown in red color. This implies that the logic circuit is not satisfiable. We show the same 2-CNF logic circuit in Fig. 3.3.11. The Satisfiability problem in this logic circuit is transformed into a graph problem in the Implication graph shown in the bottom diagram. The Satisfiability problem is also transformed into an independent set problem where each OR gate in the logic diagram now corresponds to two adjacent vertices (known as an adjacent pair) instead of a triangular graph. If it is possible to select one vertex from each adjacent pair such that no two selected vertices are adjacent then we claim that the logic circuit is satisfiable. This transformation works both ways: if the circuit is satisfiable then we can select an independent set


AND

¬a b

OR

¬a b

OR AND

¬a ¬b

OR

¬a b

OR

¬a ¬b

OR

c

a

¬a ¬b

OR

¬c

a

b

¬b

¬a

a

b

¬b

¬a

a

b

¬b

¬a

Convert

AND

Convert

OR

OR

a

a Convert

¬c

¬a b

c

45

c

¬c

a

b

¬b

¬a

AND OR

OR

Convert

c

Figure 3.3.10: The 2-CNF Satisfiability problem is reduced into a graph problem. In the top three graphs (shown on the right side), vertex a and ¬a do not belong to the same strongly connected component – it means the output of the corresponding logic circuits is satisfiable. In the bottom diagram this is not possible.

46


with size equal to the number of OR gates in the logic circuit. It is now obvious that the independent set problem in the graph (which was derived from a 2-CNF logic circuit) can be transformed into a path finding problem in an Implication graph. Describe an efficient algorithm which finds an independent set in such a graph using an Implication graph. Given a general graph how can you determine that this graph in fact represents a 2-CNF logic circuit? If it does represent such a logic circuit then we can solve this problem after reducing it into an implication graph?

¬a b

OR

¬a ¬b

OR

¬a

b

¬a

¬b

a

c

a

¬c

Transform

AND

a

OR

Tr an sf or m

¬c rm

¬c

OR

a

b

¬b

¬a

an sf o

a

Tr

c

c

Figure 3.3.11: The 2-CNF Satisfiability problem is reduced into an independent set problem shown in the top diagram. The same problem is reduced into a graph problem in an implication graph shown in the bottom diagram. This implies that the independent set problem (in some special graphs) can be reduced into an implication graph problem.


3.4

47


The problem of activity scheduling is described in simple words. In the table below, we have described a number of tasks, their IDs, their duration (in days) and what tasks must be completed before they can begin (prerequisites) (http://www.waa-inc.com/projex/PERT/aoa.htm). We need to find out the minimum possible time to complete a house according to the activities described in the table above. We also need to identify those activities which are critical: that means increasing the time duration of these activities will certainly increase the total completion time. On the other hand there are certain activities which if delayed do not necessarily increase the total completion time. Then there are activities which can run in parallel while some are strictly sequential. How will you model this problem in graph theoretic terms and then solve it? We start with an example of constructing a house as defined in Table 3.1. The problem is then transformed into a graph problem as shown in Fig. 3.4.1. We represent each activity in the table using a vertex, with the duration of the activity as a weight on the vertex. For each task vertex v, we add a directed edge to all the vertices such that the task they represent are dependent on the completion of task represented by v. Then we add a special vertex s and connect it to all such vertices that have no incoming edges (i.e. the tasks they represent have no pre-requisites) as shown in Fig. 3.4.2. Now for each vertex x in Fig. 3.4.2, we assign all the incoming edges to x the same weight as the weight of vertex x in Fig. 3.4.1. A longest path in this graph (Fig. 3.4.3) corresponds to minimum amount of time needed to build the house? Why? The underlying assumption is that we should be able to schedule as many activities in parallel as possible by the pre-requisite relationships. If we are unable to schedule any activity in parallel then the total time needed to construct the house will be the sum total of (the duration of) all activities. So that is an upper limit on time to complete the job; the lower limit will be the length of the longest path in the graph.

48

Problems, Models & Graphs: Why Study Graph Theory? Task ID 1 2 3 4 5 6 7 8 9 10

Task Description Clear land Lay foundations Build walls Electrical wiring Plastering Landscaping Gardening Interior work Roof Handover

Duration 14 28 42 21 21 20 10 35 50 00

Prerequisites none 1 2 3 4, 9 2 6 5, 9 3 7, 8

Table 3.1: We have tabulated a number of tasks, their IDs, their duration (in days) and what tasks must be completed before they can begin (prerequisites). Algorithm 4: Find Minimum Time to Complete the Job. Input : Tasks, Duration, and Pre-requisites (Table 3.1). Output: Minimum Time to complete the job; 1. A new problem: You need to transform it into a known problem in graph theory?

The challenge is to transform one problem (Algorithm 4) into a text book problem (Algorithm 5)

3.5


Chain like program graphs is common in many digital signal processing applications. In such applications each packet or frame of data may be processed through various transforms in a fixed sequence. This kind of computing has a typical serial or chain like structure as shown in Fig. 3.5.1. Each program module in the chain structure may have a very different computation requirement; one module may run faster on one processor while the same module may take much longer time on a different machine in a typical distributed


49

Algorithm 5: Find Longest Path between two given vertices in a directed acyclic graph D Input : A weighted directed acyclic graph D and two vertices Output: The Longest Path between two given vertices in D 1. A text book problem; see Chapter 5

6

7 20

1

2

14

28

10

3

4

42

5

8

21

21

10

35

00

50 9

Figure 3.4.1: Graph representing a specific instance of the activity scheduling problem

10 6

7 00

20

14

1

28

2

42

3

21

4

50

21

5

21

35

8

00

35

9

Figure 3.4.2: Transformation to a new graph

10

50


10 6

7 00

20

14

1

28

2

42

3

21

4

50

21

5

21

35

8

00

10

35

9

Figure 3.4.3: Finding the longest path corresponds to the optimal scheduling of activities heterogeneous computing environment consisting of two processors as shown in the bottom diagram of Fig. 3.5.1. The top diagram of the same figure shows a chain structured modular program consisting of four modules. The execution cost of each module on either processor is indicated below each module. For example the execution cost of module 2 on processor A is 20 while it is 90 on processor B. The communication cost between module 2 and 3 is 50 provided the two modules are placed on different machines; it is zero if the two modules are assigned to the same machine. A module can be processed on either processor A or on processor B but only one processor is active at any time; the total time of computation is the sum of total execution times plus the total communication times. In such a distributed heterogeneous computing environment we should somehow take advantage of the diverse and special characteristics of each machine as assigning all modules of the program to one machine may not be an optimal solution. A straight forward greedy solution would be to assign a module on a processor where it is least costly but then if two modules with a lot of communication traffic in between are assigned to different machines then it will again degrade the over all performance. Let us concentrate on a sub-problem in order to appreciate the intricacies of the problem.


51

Communication Costs

10 Execution Costs Processor B 50 Processor A 70

Processor

50

20

0

90

30

20

20

40

80

Processor

Figure 3.5.1: (Bottom) Processor A and processor B connected with a high speed communication link. The two machines have different capabilities. (Top) A chain structured modular program consisting of four modules (or nodes). The execution cost of each module on either processor is indicated below each module. For example the execution cost of module 2 on processor A is 20 while it is 90 on processor B. The communication cost between module 2 and 3 is 50 provided the two modules are placed on different machines; it is zero if the two modules are assigned to the same machine. There are basically four possibilities for the first pair of adjacent modules to be assigned onto the dual processor system: 1. Module 1 as well as 2 is processed on processor A: Then Cost of Execution will be 70+20 = 90, while there will be no Cost of Communication as the two modules are executed on the same machine. 2. Module 1 as well as 2 is processed on processor B: Then Cost of Execution will be 50+90 = 140, while there will be no cost of communication as the two modules are executed on the same machine. 3. Module 1 is executed on machine A while module 2 is processed on

52

Problems, Models & Graphs: Why Study Graph Theory? B. Cost of Execution will be 70 + 90 = 160, while there will be a Communication Cost = 10, Total Cost = 170. 4. Module 1 is executed on machine B while module 2 is processed on A. Cost of Execution will be 50 + 20 = 70, while there will be a Communication Cost = 10; Total Cost = 80.

The last option provides the optimal solution in terms of minimum completion time for the sub-problem. The challenge is to find the optimal solution for the entire problem efficiently (without enumerating all possibilities as we did for the sub-problem). Please note that the cost consists of two parts, cost of execution of a module on a processor, and the cost of communication between two adjacent modules provided the two modules are assigned to different machines. There is no communication cost if the two (adjacent) modules are assigned to the same machine. The total time of computation is the sum of total execution times plus the total communication times. Problem Set 3.3. Problem 3.3.1. Find the total cost if all modules are assigned to machine A. This is standard sequential processing on machine A; it is not a distributed assignment. Problem 3.3.2. Find the total cost if all modules are assigned to machine B. This is also sequential processing but this time on machine B. Problem 3.3.3. How about if we (initially) ignore communication costs and assign a module on a machine where it is least costly. For example assign module 1 on processor B, module 2 on processor A, and so on. Now calculate the Total Cost of this assignment after taking into account the communication costs. Problem 3.3.4. Find the Optimal Assignment by hit and trial (the size of this problem is small enough)? It should be less than (or equal to) the ones found in 1, 2, and 3. Problem 3.3.5. How to model this problem in graph theoretic terms? Problem 3.3.6. How about if instead of a chain structure, we design Algorithm 2.6 and Algorithm 2.8. Hints are provided in the following figures.


90 50

70

u04

Path B: All Modules on Machine B

Path A: All Modules on Machine A 00

90

70

20

00

20

50

30

53

10+90

10+20

20

40

30

80

20

u04 00

50+30

20+20

20+80

50+40

40

0

00

80

0

Figure 3.5.2: (top) We show the possibility of all modules assigned to either processor A or to processor B. This is equivalent to traversing a path which passes though the top vertices or the bottom vertices in a graph with two dummy vertices. (Bottom) The possibility of some modules assigned to processor A while other modules to processor B is shown with a zigzag path between the two dummy vertices. A shortest path corresponds to an optimal assignment.

54


90 50

10+90

70

10+20

20

50

10

30

20

00

50+30

20+20

20+80

50+40

00

40

90

80

50

20

0

30 0

70

u04

40

20 20

0 80

Figure 3.5.3: (top) Path B corresponds to all modules assigned to processor B while path A corresponds to all modules assigned to processor B. Path X corresponds to some modules assigned to processor A while others are assigned to processor B. (bottom) The module assignment corresponding to path X in the top diagram.


50

90

10

B

50

55

30

20 20

0

A

70

20

40

0 80

B

B

B 90

50 10

30 50

20

70

20 20

0

0

40

80

A

CUT A 70

50

10

90 20

70

40

20

50

30

80

20

40

20

0 80

Cut X 50

90

30

20

CUT B

Figure 3.5.4: (top) A cut in this graph (which disconnects vertex A from vertex B) corresponds to an assignment of modules. Cut A corresponds to all modules assigned to processor A. Cut B corresponds to all modules assigned to processor B. Cut X corresponds to the module assignment as shown in the bottom diagram.

56


Algorithm 6: Find Minimum Cost Assignment Input : Execution & Communication costs of each module on Processors A & B Output: Minimum Time to complete the job & the corresponding module assignment; 1. A new problem: You need to transform it into a known problem in graph theory?

Algorithm 7: Find Shortest Path between two given vertices in a directed acyclic graph Input : A directed acyclic graph D and two given vertices Output: A Shortest Path between two given vertices in D 1. A text book problem; See Chapter 5

Algorithm 8: Find Minimum Cost Assignment Input : Execution & Communication Costs of each module on Processor A & B Output: Minimum Time to complete the job & the corresponding module assignment; 1. A new problem: You need to transform it into another known problem in graph theory?

Algorithm 9: Find Minimum Cut which disconnects vertex s from vertex t in a graph G Input : An un-directed graph G and two vertices (s and t) Output: A Minimum Cut (which disconnect s from t) 1. A text book problem; see Chapter 6


57

The challenge is to design Algorithm 6 and then use the text book Algorithm 7 in order to solve the Minimum Cost Assignment Problem. Another challenge is to design Algorithm 8 and then use the text book Algorithm 9 in order to solve the Minimum Cost Assugnment Problem.

3.6


The Sequencing by Hybridization technique depends on the hybridization of target DNA fragment against a very large array of relatively short probes. With a given unknown DNA sequence (of four letters A, C, G, & T ), an array (also known as the Gene Chip) tells us about all sub-strings of a fixed length that the DNA sequence contains; it does not provide us about the order of the strings or their position in the DNA fragment. For example if the (unknown) sequence s = CAT GAGT then a set of all substrings of length 3 that s contains (known as Spectrum(s, 3)) will be equal to {CAT, AT G, T GA, GAG, AGT }. The elements in the spectrum may not appear in the same order and the challenge is to find the string s, given its spectrum. The sequencing by hybridization problem in molecular biology (or simply the spectrum problem) can be transformed either into a Hamiltonian path problem or an Euler path problem in directed graphs depending upon whether we map every element of the spectrum into a node or an edge of a directed graph respectively. Assume that Spectrum(s, 4) = {AT AC, CAT A, ACAT, ACAG, T ACA, ACAC, CACA, ACAA, CAAC, AACA}. Then we can find a sequence s as shown in Fig. 3.6.1, Fig. 3.6.2, and Fig. 3.6.3. In fact there are multiple strings possible (s1 = ACACAACAT ACAG & s2 = ACAT ACAACACAG) with the same spectrum as illustrated in these diagrams. The challenge is to tranform this problem into a know problem in graph algorithms. The following figures will provide hints to make multiple transformations.

58


Algorithm 10: Find a (Correct) Sequence s given its Spectrum(s, l) Input : Spectrum(s, l) of an unknown s; Output: Sequence s; 1. A new problem: You need to transform it into a known problem in graph theory?

Algorithm 11: Find a Hamiltonian Path in G Input : A graph G Output: A Hamiltonian Path in G 1. A text book problem; see Chapter 8

Algorithm 12: Find an Euler Trail in G Input : A graph G Output: An Euler path in G 1. A text book problem; see Chapter 7

Transform the sequencing problem into Hamiltonian path and then into an Eulerian path problems (see the figures below)


59

Spectrum(s,4) = {ATAC, CATA, ACAT, ACAG, TACA, ACAC, CACA, ACAA, CAAC, AACA}

ACAC ATAC ATAC

ATAC

CATA

ACAT

ACAG

ATAC TACA

TACA

CACA

TACA

CATA

TACA

ACAA

ACAG

ACAT

CAAC AACA

ATAC

CATA

ACAT

ACAG

TACA

Figure 3.6.1: We start with a Spectrum(s, 4) = {AT AC, CAT A, ACAT, ACAG, T ACA, ACAC, CACA, ACAA, CAAC, AACA} and show a directed graph in which every element of the spectrum was mapped onto a vertex. A directed edge exists from one vertex to another provided the last three letters of the first vertex match with the first three letters of the second vertex. That is why there are four directed edges emanating from the vertex T ACA; the edges are going to vertices labeled with ACAA, ACAC, ACAG and ACAT .

60


Start ACAC

ACAC

ATAC

ACAC

ACATAC

ACACA

ATAC

CACA

CACA

ACATA CATA

TACA

ACAA

ACAG

CATA

TACA

ACACAA

ACAT

ACATACA

ACAT

ACACAACAT

CAAC AACA

ACACAAC

ACAA

ACAG

ACAT

Start CAAC AACA

ACACAACA

Figure 3.6.2: (Left) A Hamiltonian path is shown in bold in the graph of Fig. 3.6.1; this Hamiltonian path corresponds to the sequence s1 = ACACAACAT ACAG. (Right) Another Hamiltonian path is shown in the same graph; this Hamiltonian path corresponds to a sequence s2 = ACAT ACAACACAG. Both these diagrams show various stages in the reconstruction of a sequence.


61

Spectrum(s,4) = {ATAC, CATA, ACAT, ACAG, TACA, ACAC, CACA, ACAA, CAAC, AACA}

?

ATAC

?

CATA

?

?

ACAT

?

?

ACAC

ACAG CAG

ACA

ACAT

TACA

ACAA

AACA

TAC

CAC

CACA

AAC

CAA

CAT

CAAC ATA

ATAC

TAC ACA

CAT

CATA

ACAT

CAT

CATA

ATAC

CATA

ATA

ATA ACA

ACAT

TAC CAT

ATAC

ATA

Figure 3.6.3: We start with the same Spectrum(s, 4) = {AT AC, CAT A, ACAT, ACAG, T ACA, ACAC, CACA, ACAA, CAAC, AACA} and this time we map every element of the spectrum into an edge of a directed graph. An edge also specifies its end vertices; the start vertex carries the first three letters of the edge while the end vertex is labeled with the last three letters of the edge. For example the directed edge AACA emanates from CAAC and terminates at AACA.

62


ACA

ACAC

CAC

CACA

ACA

ACAA

CAA

CAAC

AAC

AACA

ACA

ACAT

CAG

ACA

ACAT

CAT

ACAG

CATA

ACA

ATA

TACA

ATAC

TAC

TAC

ATAC

TACA

ATA

ACA

CATA

ACAA

CAT

CAA CAAC

CAG

ACAG

ACA

CACA

CAC

ACAC

ACA

AACA

AAC

Figure 3.6.4: (Top) A Euler path in the graph of Fig. 3.6.1; this Euler path corresponds to the sequence s1 = ACACAACAT ACAG. (Bottom) Another Euler path is shown in the same graph; this Euler path corresponds to the sequence s2 = ACAT ACAACACAG. Please note that in both the Euler paths shown in this diagram there is a vertex which is repeated several times while every edge is traversed exactly once.

3.7


How can we exploit graph theory in order to solve an unknown problem? In other words how can we reduce the unknown problem into a graph problem? Finding the right reduction or transformation is not easy. It is certainly not possible to design an exact algorithm to do this transformation; we can only rely on heuristics. Experience can help you a lot in moving forward. We shall provide you below a certain sequence of questions to guide your search towards a suitable transformation. Note that most of these questions may not have a black and white answer; you should also try to find why or why not. 1. Does your problem have an inherent graphical structure? If yes then the transformation may be a lot simple. For example in the first example


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we have activities and a pre-requisite relationship among themselves; this naturally leads to a directed graph. The activities have weights; these weights can be shifted to vertices if each vertex corresponds to an activity in your graph. 2. Is your unknown problem an optimization problem or otherwise? If it is an optimization problem where you need to maximize (or minimize) some parameter then you need to search for a graph optimization problem. In the first example we need to minimize the total completion time of building a house. Strangely this is equivalent to finding a longest path in a directed acyclic graph as shown in Fig. 3.4.1. In the second example we need to minimize the sum total of execution and communication costs of a modular program. This problem is reduced to finding a shortest path in a directed graph. The last example is not an optimization problem; we need to find an unknown sequence s for which the Spectrum(s, l) is given. This problem is reduced to a decision problem (not an optimization problem) in graph theory whereby we decide if it is possible to find a Hamiltonian Path (or an Euler trail) in a directed graph. 3. Do components of your problem have a sequence (or an order) which can be transformed into a directed graph? The first example certainly implies a directed graph; it is rather impossible to capture its essence without directions. The second example is perhaps more interesting; in its first transformation it is transformed into a directed graph; it is also reduced into a graph problem applicable to an un-directed graph. The last example has an implicit directed structure; only a certain sequence (or order) of letters can correspond to a correct sequence s. Thus this example is also reduced to a directed graph problem. 4. Do you map each element of your unknown problem into a vertex or an edge of a graph? Example 3 is a good illustration of this decision. If you decide to map each element of the Spectrum onto a vertex then an edge in this graph would represent what (and on what basis)? Will this be a directed edge or undirected? 5. Sometimes it is difficult to make a general transformation; it is much easier to make a special case reduction. For example in the second example it is much easier to visualize the special case (in terms of a

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Problems, Models & Graphs: Why Study Graph Theory? graph problem) when all modules are assigned either to processor A or processor B as shown in the top diagram of Fig. 3.5.2. This special case more or less resembles the activity scheduling problem discussed in the first example.

Problem Set 3.4. Problem 3.4.1. The Marriage Problem is a well-known problem in mathematics as well as in any middle class conservative society; we assume that we have a collection of men and an equal number of women; each women knows (or likes) some of the men. The problem is to find out if it is possible to marry each woman to a man she likes or knows. If it is not possible to find such a Perfect Matching then we should perhaps maximize the number of women each of whom is married to man whom she knows. (Please note that here we have not accounted for the amount of liking between a man and a woman – we just claim that a certain woman knows or likes or does not like a certain man. The quantitative aspects of this problem will be discussed in the coming problems.) This problem can be modeled by a graph consisting of each man and woman as a vertex. A liking between a man and a woman is represented by an edge between the two corresponding vertices. The graph is shown below. Please note that the graph consists of two parts – an A part consisting of men, and a B part consisting of women. It is a special graph in the sense that an edge between two vertices belonging to the same part (men or women) is not allowed. Such a graph is known as a bipartite graph. We show two reductions of this problem. One is shown in Fig. 3.7.1, and the other is shown in Fig. 3.7.2. In the first reduction, we reduce the Marriage Problem into a problem of finding maximum edge-disjoint paths between vertex s and vertex t in a graph G. This reduction depends upon the hypothesis that each edge-disjoint path between vertex s and t in graph G corresponds to a matching edge in the bipartite graph. In the second reduction the problem is reduced to finding maximum vertex-disjoint paths between the same two vertices in the graph G. This reduction depends upon the hypothesis that each vertex-disjoint path between vertex s and t in graph G corresponds to a matching edge in the bipartite graph. The problem is to find out which reduction is correct and which one is false. In one case you have to design an informal proof and in the other case you have to design a counter example. Please note that the reduction (which ever is correct) not only tells you if a


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perfect matching is possible, it also tells us which woman to marry whom. In case a perfect matching is not possible, the reduction maximizes the number of marriages taking place. Problem 3.4.2. We consider the same Marriage Problem as described before. We need to find if a Perfect Matching exists in a given bipartite graph C. We also need to identify which woman is marrying whom. If a perfect matching does not exist then we need to maximize the number of marriages as before. Here we start with the adjacency matrix C of the bipartite graph C. We reduce the problem into finding the permanent of the matrix C. The permanent P er(C) of a matrix C is computed like the determinant of C except that the signs of all permutations are positive. For example the P er(C) of the matrix shown in the figure below will be computed as follows:

P er(C) = a11 a22 a33 + a11 a23 a32 + a12 a21 a33 + a12 a23 a31 + a13 a22 a31 + a13 a21 a32 P er(C) = 0 + 1 + 0 + 1 + 1 + 0 = 3 As you can see each individual term in the expression for Permanent is a permutation – there is a one to one correspondence between each non zero term in this expression and a perfect matching in the bipartite graph. Thus if P er(C) is zero then no perfect matching exists. On the other hand if it is non zero then there will be as many perfect matchings as the value of P er(C). The problem of finding a perfect matching is thus reduced to finding the permanent of a matrix. 1. Once we know that a perfect matching exists in the given bipartite graph C, the problem is how to find one? Assume that the only operation that we can perform is to find a Permanent of the matrix after or before removing an edge of the bipartite graph. How can you find a Perfect Matching in this graph? Discuss briefly. 2. An alternate way of finding a perfect matching in a bipartite graph is demonstrated in Fig. 3.7.4. We need to find if an edge (y, z) is part of a perfect matching in the bipartite graph. The said edge will be part of a perfect matching if and only if P er(Cyz) is non zero. On the

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Problems, Models & Graphs: Why Study Graph Theory? basis of this observation, design an efficient algorithm to find all edges belonging to the perfect matching.

3. It is much more complex to find the permanent of a matrix as compared to finding the determinant of a matrix – thus we are tempted to check if the determinant is as helpful as the permanent of a bipartite graph? How about if we find that the determinant of an adjacency matrix of a bipartite graph is non zero? Under such conditions, can we make a claim that a perfect matching exists in the graph? How about if the determinant is zero – can we claim that a perfect matching does not exist in the bipartite graph? Remember that the determinant of a completely connected bipartite graph is zero in spite of the fact that every permutation of the vertices of the bipartite graph is a perfect matching.

4. We know that the determinant of a completely connected bipartite graph is zero, why? How about if we put random weight on each edge of the bipartite graph? The probability of the value of the determinant becoming zero will depend upon the randomness of the weights assigned to the edges of the bipartite graph. So hopefully the determinant of a bipartite graph will not be zero unless there is not a single perfect matching in the bipartite graph. How to find the edges of a perfect matching in this graph? How about using the previous technique?

Problem 3.4.3. Consider the (minimum weight) assignment problem. We are given a complete balanced weighted bipartite graph, and we need to find a minimum weight perfect matching in this graph – that means a perfect matching in which the sum of weights of all (matched) edges is minimum. For the time being assume that the minimum weight perfect matching is unique. We shall try to relax this condition later.We use the same reduction in this problem as was used in the last problem with one slight modification. All edge weights in the adjacency matrix C are raised to the power of 2, and then we take the permanent of matrix C. The highest power of 2 which divides the value of the permanent is the weight of the minimum weight perfect matching.


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P er(C) = 28+1+8 + 28+8+8 + 28+8+8 + 28+8+2 + 23+1+2 + 23+8+8 = 217 + 224 + 224 + 218 + 26 + 219 = 26 (1 + 217 + 224 + 224 + 218 + 219 ) Raising all weights to a power of two and then finding the permanent of the weighted matrix gives us a powerful reduction in which each individual term in the permanent is in fact the weight of a perfect matching raised to power of 2. The value of the permanent will be divisible (with remainder equal to zero) by the weight of minimum weight perfect matching raised to the power of 2 provided we have a unique minimum weight perfect matching in the bipartite graph. 1. Design an efficient algorithm to find edges belonging to the minimum weight perfect matching. The only reduction that you can use is to find permanent of a matrix. Of course you can raise a number to the power of 2 or any other number of your choice. 2. What complications can arise in finding the value of minimum weight perfect matching provided such a matching is not unique? 3. Assume that now there is a possibility that the minimum weight perfect matching is not unique. You need not find the value of the minimum weight perfect matching – just make an efficient check if it is unique or not?

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Figure 3.7.5: A minimum weight assignment problem is reduced to evaluating the permanent of a matrix. Problem 3.4.4. Consider the assignment problem in which we have a complete weighted bipartite graph and we need to find a minimum weight perfect matching. In Chapter 5 we shall reduce this problem into a well known shortest path finding problem in any directed or un-directed graph. Shortest path finding algorithms are relatively simple and are discussed in Chapter 4. Here in this problem we shall discuss a (bizarre) reduction in which we reduce the shortest path problem into the assignment problem. We start with a black box which accepts a complete weighted bipartite graph as input and outputs the minimum weight perfect matching. The output also includes edges (or vertices) belonging to minimum weight perfect matching. We show a weighted graph G in Fig. 3.7.6 where we need to find a shortest path from vertex a to vertex d. The weighted adjacency matrix of this graph is also shown in the top right diagram of this figure. Using this pp matrix we


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generate a weighted (complete) bipartite graph of size (p−1)(p−1) by deleting the column corresponding to vertex a and the row corresponding vertex d as shown in the bottom left diagram. We claim that a minimum weight perfect matching in this bipartite graph gives us a shortest path between vertex a and vertex d as shown in Fig. 3.7.7. 2

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assignment problem. 1. Design an efficient algorithm to transform this problem (of finding a shortest path between two vertices in a weighted graph) into a complete bipartite graph as shown in Fig. 3.7.7. In terms of Fig. 3.1.1, design Algorithm No. 1. 2. Design an efficient algorithm to transform the minimum weight perfect matching into a shortest path between the two given vertices; in terms of Fig. 3.1.1, you will be designing Algorithm No. 2.

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Chapter 4 Basics of Graph Theory

4.1

Introduction

4.2

A Mutual Friendship Graph

4.3

Representation of a Graph

4.4

Complement of a Graph

4.5

Equal Graphs & Isomorphic Graphs

4.6

The Degree Sequence

4.7

Walks, Trails, & Paths

4.8

Multi-graphs and Pseudo-graphs

4.9

Broad Categories of Graphs & some Special Graphs

4.10 Integration of Concepts, Properties, and Action Items 4.11 Self Complementing Graphs

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4.1

Basics of Graph Theory

Introduction

We shall discuss a number of basic definitions in this chapter. We will be solving a couple of puzzles; hopefully the students will realize that in such problems a stage comes when common sense alone is not sufficient to solve the puzzle; we need tools and techniques of graph theory. We shall discuss some necessary conditions for a degree sequence to be graphical. We shall then talk about graph isomorphism and then come back to a discussion of necessary and sufficient conditions for a degree sequence to be graphical. The concept of graph connectedness will also be provided; we shall define a walk, a trail and a path in a connected graph. We shall also talk about some special graphs at the end of this chapter.

4.2

A Mutual Friendship Graph

Assume that we have 5 persons in a hall. Let us name these people as A, B, C, D, & E. Some of these people are friendly with each other while others are not; we assume that their friendship is symmetric; that means if person x knows y then it means that y knows x (for every pair (x, y)). We ask each person with whom he or she is friendly with and we get an answer like A is friendly with B only while B is friendly with A, D, and C. This so called mutual friendship is represented by the top left diagram in Figure 4.2.1. Please note that the double sided arrows emphasize the mutual friendship between two persons; as discussed before we are talking about a symmetric relationship. With each person in this diagram we show the number of people with whom he (or she) is friendly with; thus A is friendly with just one person while B is friendly with three persons. This number in fact represents the amount of popularity a person enjoys; Persons B and D are the most popular persons while persons A and E are the least popular. If we note down this popularity number, and then sort this sequence we get the so called friendship sequence; this sequence is also shown in the top left diagram. The top left diagram in Fig. 4.2.1 is transformed into a graph shown in the top right corner of the same figure. Here each person is represented by a node or a vertex; a double arrow edge is represented by an un-directed edge; the relationship of friendship in the top left diagram is now transformed into an adjacency relationship in the graph shown in the top right diagram. Now instead of a friendship sequence we have a degree sequence; this sequence is


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exactly the same as the friendship sequence and is shown in the top right corner as well.

Figure 4.2.1: We show five persons with a symmetric friendship relationship indicated by lines with double sided arrows in the top left diagram. The relationship is transformed into an un-directed graph as shown in the top right diagram. The friendship sequence as well as the degree sequence is also indicated along with the respective diagrams. The adjacency matrix representing the graph is shown in the bottom diagram.

4.3


The graph shown in the top right corner in Fig. 4.2.1 is modeled by an adjacency matrix data structure as shown in the bottom diagram of Fig. 4.2.1. In this adjacency matrix a 1 represents an adjacency relationship while a

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zero (or an absence of 1) represents no adjacency relationship between the corresponding nodes in the graph. The sum of all one’s in a row corresponds to the degree of a vertex; for example the degree of vertex A is 1 while the degree of vertex B is 3. The degree of each vertex is also shown in the last column of the bottom diagram. When we sort the numbers in the last column then we get the degree sequence or the friend-ship sequence. It will be interesting to explore the (special) structure of the friendship sequence (or the degree sequence); from now onwards we shall use the term “the sequence”. In a group of five persons (as shown in Fig. 4.2.1) the largest number can not be larger than four; the smallest number can not be less than zero. This is because in our understanding of friendship there is no room for self friendship; there is neither any room for negative friendship with some one which means animosity. In addition to these we have other interesting limitations as discussed below. Please note that if these conditions are not met then the sequence will not be graphical; what does that mean? Under such conditions the sequence will neither be a friendship sequence; what does that mean? 1. The sum of all numbers in the sequence will always be an even number. Why? 2. The odd numbers in the sequence will appear even number of times. Why? 3. At least one number will be repeated in the sequence. Why? Problem Set 4.1. Problem 4.1.1. Salma invites three of her office colleagues for dinner. Salma’s husband Aslam (known as A) received the guest when they have arrived while Salma was in the kitchen preparing food. Aslam shook hands with some of the guests. Obviously Salma was curious about whom Aslam shook hands and with whom he did not. She can not ask this question directly so she simple asks every colleague about the number of hands she had shaken; she gets the following answer: B says 3, C says 2 and D says 1. Now Salma asks the same question to her husband; he answers zero (he does not shake hands with females). Is the husband telling the truth assuming the guests told the truth? How about if he answers 3 (he showed no discrimination). Find out the number of times the husband actually shook hands and more importantly with whom he shook hands and whom he has ignored.


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Problem 4.1.2. Salma & Aslam has invited three married couples. Several handshakes took place when the guests have arrived. No one shook hands with one self or with his or her spouse. This time Aslam asked each person including his wife to write on a slip of paper how many hands he or she has shaken. Surprisingly each person gave a different answer but Aslam does not know which answer belongs to whom. Now Aslam is curious about the following: (a) Did a female shake hands with a male? (b) Did Salma shake hands with a male friend? (c) How many times Salma shake hands? (d) How many times Aslam shake hands? (e) Assume that Salma has not shaken hands with any male friend then did Aslam shake hands with a female? Under such conditions was there a husband who refused to shake hands with a female?

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as far as possible; the motivation behind this seating strategy is to encourage strangers to become acquainted with each other.

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Figure 4.3.2: The acquaintance graph of a group of eight people is given. Determine if you can seat these people around a round table so that every two neighbors are acquainted. We have already discussed a number of necessary conditions for a sequence to be graphical; we should be able to draw an actual graph with the same degree sequence. The first problem set (some parts of this at least) also exploits some of the necessary conditions; it will be useful and informative if in each problem you try to actually draw the underlying graph. We shall explore necessary and sufficient conditions for a sequence to be graphical but before we do that let us first discuss some more basic stuff (in general) and graph isomorphism (in particular); we shall need some of its (graph isomorphism) results for formulating necessary and sufficient conditions for a graphical sequence.

4.4

Complement of a Graph

Complement of a graph G is another graph H with the same number of vertices such that there is an edge uv in H if and only if there is no edge uv in G. We show a graph G and its complement H in Fig. 4.4.1 on the left most graph and the middle graph respectively Try to visualize a graph C having same number of vertices as in G or H and having an edge uv provided the edge uv exists either in G or in H. Now try to visualize a graph E having


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same number of vertices as in G or H and having an edge uv provided the edge uv exists in G and in H.

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4.5


We show 4 graphs in Fig. 4.5.1. All these graphs have the same number of vertices, and same number of edges; in fact they all have the same degree sequence as shown in the same figure. So there is a good possibility that they are all equal to each other. Note that vertices with the same degree in two different graphs are colored similarly; the degree of each vertex is also indicated along with each vertex. The graph on the extreme left and the graph on the extreme right are in fact equal; if you draw an adjacency matrix of each graph (after identical labeling) then you will find that the two adjacency matrices are exactly the same. If you closely look at the two middle graphs you realize that they are different graphs meaning that they are neither equal nor isomorphic. This is because in one graph all the four degree vertices are connected in the form of a triangle while in the other graph this is not so. If now you draw the adjacency matrices of the graphs shown on the extreme left and middle left in Fig. 4.5.1 after identical labeling then the two matrices comes out to be different as shown in Fig. 4.5.2. Thus these two graphs are not equal; they may still be isomorphic because we fail to find a visible

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Figure 4.5.1: We show four graphs each having the same degree sequence. Out of these graphs at least two are equal graphs; at least two are unequal graphs (why?); at least two are non isomorphic graphs (why?); and at least two are isomorphic but not equal. Please note that vertices with the same degree are drawn in the same color.

Figure 4.5.2: Graph G shown in the top left corner is not equal to the graph H shown in the top right corner. This is evident from the adjacency matrices of the two graphs shown in the bottom diagram. There is still a possibility that the two graphs are isomorphic to each other as the two graphs have a number of similar qualities and no obvious differences.


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difference like the one that we have found for the two middle graphs of Fig. 4.5.1? These two graphs may still be isomorphic? What does that mean? Two graphs G and H are isomorphic provided they can be drawn with identical graph drawings. If, for example, we rotate graph H of Fig. 4.5.2 by an angle of 180 degrees then we get a drawing which is exactly the same as the graph G; this is shown in Fig. 4.5.3.

Figure 4.5.3: Graph H shown in the top right diagram of Fig. 4.5.2 is redrawn here in the top right corner after rotating it by an angle of 180 degrees. It is quite obvious now that the drawing in the top left diagram is the same as the drawing shown in the top right corner. Let f be an isomorphism from the left graph (G) to the right graph (H). Then the two graphs are isomorphic provided there is an edge between f (u1 ) and f (u2 ) in the right graph if and only if there was an edge between u1 and u2 in the left graph for every two vertices in the two graphs as shown in the bottom diagram of this figure.

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Problem Set 4.2. Problem 4.2.1. Draw as many graphs as possible such that no two of them should be isomorphic to each other and each graph should have a degree sequence 332222. Problem 4.2.2. Draw as many graphs as possible such that no two of them should be isomorphic to each other and each graph should have a degree sequence 4443322. Problem 4.2.3. We show eight graphs with the same degree sequence in Fig. 4.5.4. Find which two of them are equal, which two are isomorphic, and which two are not isomorphic. Find a visible difference incase the two graphs are not isomorphic. Find an isomorphic function from a graph G to graph H in case graphs G and H are isomorphic to each other.

Figure 4.5.4: Eight graphs with same number of vertices, and the same degree sequence. Find which two graphs are equal (and isomorphic), which are isomorphic but not equal and which are not isomorphic (and also not equal).

4.6

The Degree Sequence

Assume that we are given a degree sequence of a graph and we need to find the corresponding graph provided the sequence is graphical. There is a possibility that the degree sequence may not be graphical? Under such conditions

The Degree Sequence

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we claim that it is not possible to draw a graph for that sequence (why it is not possible?). So we need to study necessary and sufficient conditions for a sequence to be graphical; we have already studied some necessary condition at the start of this chapter but those conditions were not sufficient (for a sequence to be graphical). Before moving forward find a sequence which satisfies all necessary conditions that we have discussed (earlier in this chapter) but it is impossible to draw a graph corresponding to this sequence. How about the degree sequence 543211? Let us assume we are given a degree sequence SG equal to 4443322. The maximum degree here is 4 and the total number of integers in the degree sequence is 7. We apply the following procedure on this degree sequence SG and convert it into a new degree sequence SH which is equal to 332222. Algorithm 13: Convert degree sequence SG into SH . Input : Original degree sequence SG (Example 4443322) Output: New degree sequence SH (Example 332222) 1. Remove the maximum degree (which is 4) from the degree sequence SG (thus reducing the length of the sequence from seven to six). The resulting sequence will be 443322 as shown in the top middle diagram of Fig. 4.6.1 2. Subtract 1 from the first 4 (because maximum degree was 4) integers of the remaining sequence as shown in the top right diagram of Fig. 4.6.1. The new sequence SH will become 332222 shown in the same diagram.

Algorithm 13 transforms a degree sequence into another degree sequence; this transformation makes no sense unless we visualize these operations as if they are performed on a graph. Assume that we have a graph G in which the highest degree vertex is known as v and its degree is u. Also assume that (the highest degree) vertex v is connected to the first u vertices (after v) in the degree sequence SG of G as shown in the bottom left diagram of Fig. 4.6.1. Now when we perform the first operation of removing the maximum degree from the degree sequence; that amounts to removing the vertex v from the graph G; the new graph H will have number of vertices one less than G. Please note that removing the vertex v from G means that all

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edges emanating from G will also be removed and that amounts to operation number 2. This is equivalent to reducing the degree by one of all adjacent vertices of v in G. You may have realized that the new graph will have a degree sequence equal to SH . This meaningful interpretation is possible provided we have the crucial assumption: the highest degree vertex v in G is connected to the first u vertices (after v) in the degree sequence SG of G.

Figure 4.6.1: We are given a degree sequence equal to 4443322 shown in the top left diagram. The maximum degree here is 4 and the total number of integers in the degree sequence is 7. We remove the maximum degree from the sequence and subtract 1 from the first 4 integers of the remaining sequence as shown in the top diagrams. The new sequence will be shorter by 1 as compared to the original sequence as shown in the top right diagram. We claim (Havel-Hakimi) that the original sequence SG (corresponding to a graph G) is graphical if and only if the new sequence SH (corresponding to graph H) is graphical. In order to prove this necessary and sufficient condition we have to make and prove two claims as follows: Claim 4.6.1. If the new sequence SH is graphical then the original sequence SG is graphical. Claim 4.6.2. If the original sequence SG is graphical then the new sequence SH is graphical. It is interesting to note that the above claims not only provides us a necessary and sufficient condition for a sequence to be graphical; they also provide us means to draw a graph corresponding to a graphical sequence. So before proving the above claims let first do the more interesting exercise of finding a

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graph. Let us start with the same degree sequence, verify that it is graphical, and then find the actual graph corresponding to this sequence. We start with a p length degree sequence, and apply the following steps which outputs whether the sequence SG is graphical or not. 1. transform the sequence into a new sequence using Algorithm 13 (by removing the first vertex v from the degree sequence with a degree equal to u, and subtracting 1 from the first remaining u degrees of the degree sequence). 2. If you get a number less than zero in the new sequence then the original sequence was not graphical (output “No”, and terminate) otherwise there are two possibilities. Either by carefully looking at the latest new sequence, you can draw the corresponding graph (output “Yes”, and terminate) or if it is not possible then repeat step (1) on the latest new degree sequence (but sort it if it is not already sorted). Each iteration in the above procedure makes the degree sequence smaller in size; a stage comes when it is very much possible to draw the corresponding graph as shown in Fig. 4.6.2. In this figure the seven digit degree sequence is converted into a six digit degree sequence and then ultimately into a four digit degree sequence as shown in the top diagrams. The four digit degree sequence can be recognized to be a graphical sequence. The corresponding 4 vertex graph is shown in the middle right diagram. It is then converted into a 5 vertex graph, and eventually into a 7 vertex graph as shown in the middle left diagram of Fig. 4.6.2. It is very much possible to have two different graphs corresponding to the same degree sequence; this has been illustrated by the bottom diagram of the same figure. The resulting graph shown in the bottom left diagram is different from the middle left graph shown in the same figure. The two graphs are neither equal nor isomorphic (why?).

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Figure 4.6.2: A seven vertex degree sequence is shown in the top left corner. This seven vertex sequence is converted into a six vertex sequence, and then ultimately into a four vertex sequence as shown in the top right corner. This four vertex sequence is graphical as shown by the middle and bottom right diagrams. We get a four vertex graph which is then converted into a five vertex graph and ultimately into a seven vertex graph as shown in the middle and bottom left diagrams. It is very much possible to get two different graphs from the same degree sequence. Let us now take up Claim 4.6.1; it says that if a new sequence SH is graphical then the original sequence SG is also graphical. For example if 332222 is graphical then 4443322 is graphical. Can you design a formal proof for this claim? Is your proof based on induction or is it proof by contradiction? Discuss briefly. Let us design an algorithm to construct a graph G corresponding to an original sequence SG provided the new sequence SH and its corresponding graph H is given.

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Algorithm 14: Find a graph G corresponding to an original degree sequence SG provided the new sequence SH and its corresponding graph H is given. Input : (1) New degree sequence SH (Example 332222), and its graph H is given. (2) Original degree sequence SG is also given (Example: 4443322). Output: Original Graph G corresponding to the degree sequence SG . 1. Note down the highest degree in the degree sequence SG , let it be u. 2. Add a new vertex v in the given graph H, (the degree of v should be u (why?)) 3. Add an edge between the vertex v and the appropriate u vertices in the degree sequence SH (why?).

Let us now take up Claim 4.6.2; it says that if the original sequence SG is graphical then the new sequence SH is graphical. It essentially means that now we are given a graph G corresponding to a degree sequence SG . We are also given the new sequence SH and we need to show that it is graphical; this means that we need to draw the corresponding graph H. If you can prove Claim 4.6.1 then you should be able to prove this claim also; just move backwards in Fig. 4.6.2? Can you design an algorithm similar to (or almost a mirror image of) Algorithm 14. Problem Set 4.3. Problem 4.3.1. Does Algorithm 14 perform the intended function correctly? Problem 4.3.2. Does Algorithm 15 perform the intended function correctly? This is an important question because the answer may be no. The said algorithm provides a correct solution provided we have the crucial assumption: the highest degree vertex v in G is connected to the first u vertices (after v) in the degree sequence SG of G. If the assumption is not true then this algorithm will not provide correct results; this is certainly a serious deficiency of this algorithm? Before correcting this deficiency let us look at it a bit more closely in the next part. Problem 4.3.3. Assume that we are given a degree sequence 4443322 and a graph G as shown in the left diagram of Fig. 4.6.3. Is it possible to apply

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Algorithm 15: Find a graph H corresponding to a new degree sequence SH provided the original sequence SG and its corresponding graph G are given. Input : (1) Original degree sequence SG (Example 4443322), and its graph G is given. (2) New degree sequence SH is also given (Example: 332222). Output: Graph H corresponding to the degree sequence SH . 1. Locate the vertex of highest degree in G; Let this vertex be v and let its degree be equal to u. 2. Remove all edges emanating from the vertex v. 3. Remove the vertex v from the resulting graph and we obtain H (why?). 1

(Note that graph H corresponds to degree sequence SH ?)

Algorithm 15 in order to draw a graph with a degree sequence equal to 332222? Please note that there are three vertices with a highest degree equal to four; two of these vertices are not connected to the first four vertices in the degree sequence of this graph. Problem 4.3.4. Now repeat the above part with the same degree sequences but this time G is as shown in the right diagram of Fig. 4.6.3. What will be the result of our algorithm? Why our algorithm fails this time? Discuss briefly. Please note that there are three vertices in this graph with a highest degree equal to four; none of these vertices are connected to the first four vertices in the degree sequence of this graph. Suppose that we are given a graph F and its degree sequence SF ; the maximum degree vertex v1 has a degree u, and this vertex is not connected with the first u vertices in the degree sequence (after v1 ) of this graph. We need to convert the graph F into another graph G with the same degree sequence but in G the vertex v1 (having highest degree equal to u) is connected to the first u vertices in the degree sequence of this graph. The following algorithm performs this transformation. We claim that such a transformation is always possible. Why? Once it has been established by Algorithm 16 that a given graph F can

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Figure 4.6.3: Two graphs with the same degree sequence 4443322. Please note that the two graphs are not isomorphic. Algorithm 16: Convert graph F with a degree sequence SF into a graph G with the same degree sequence (but with an important difference?) Input : A graph F and its degree sequence SF ; the maximum degree vertex v1 has degree u, and this vertex is not connected with the first u vertices in the degree sequence (after v1 ) of this graph. Output: A graph G with the same degree sequence; the maximum degree vertex v1 has a degree u, and this vertex is connected to the first u vertices in the degree sequence of this graph. 1. Locate vertex vk and vertex vj in the graph F such that v1 is connected to vertex vk and not connected to vertex vj while the degree of vj is larger than that of vk . (If you can not find such vertices then graph F is already transformed into G; there is no need to do any thing else) 2. In the graph H, locate a vertex vn such that vj is connected to vn while vk is not connected to vn . (Why are you guaranteed to find such a vertex vn ? 3. From graph H, remove edges (v1 , vk ) & (vj , vn ), and insert edges (v1 , vj ) & (vk , vn ); graph H is now transformed into graph G?

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always be transformed into another graph G; it gives rises to a number of important conclusions. Let us summarize our recent findings: 1. Given a graph G in which the maximum degree vertex is connected to the first u vertices in the degree sequence of this graph (where u is the degree of the highest degree vertex) it is possible to apply Algorithm 15 in order to transform G into another graph H. 2. Given a graph G in which the maximum degree vertex is not connected to the first u vertices in the degree sequence of this graph (where u is the degree of the highest degree vertex) it is not possible to apply Algorithm 15 in order to transform G into another graph H. Under such conditions we first transform G into another graph G0 using Algorithm 16 and then Algorithm 15 will correctly transform G0 into H. 3. Algorithm 16 has also made it possible to design a constructive proof for Claim Number 2. What is that constructive proof? We show our strategy in handling the necessary & sufficient conditions for a degree sequence to be graphical in Fig. 4.6.5. We show the position of our claims and the respective proofs. Please note that the three algorithms perform a dual purpose; they help us in transforming one graph into another but more importantly they provide crucial insight in designing constructive proofs for the two claims. Problem Set 4.4. Problem 4.4.1. Concentrate on the third line of Algorithm 14: “Add an edge between the vertex v and the appropriate u vertices in the degree sequence SH (why?)”. What does appropriate mean? Carefully read your text book (page 17) and then make your decision appropriately. Remember we have discussed this problem in the class but have not resolved it completely. Problem 4.4.2. We are given a graph G with a degree sequence 432221 as shown in the top left diagram of Fig. 4.6.6. As shown in this figure the highest degree vertex is connected to a vertex of lowest degree while it is not connected to a vertex of a relatively higher degree. It is possible to convert this graph G into another graph with the same degree sequence by using Algorithm 16 as shown in the top right diagram of Fig. 4.6.6. The

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Figure 4.6.4: A seven vertex graph is shown in the left diagram with a degree sequence shown in the top of the graph. There are three vertices in this graph with a highest degree equal to four. All these three vertices (with a degree of 4) are connected to a vertex of minimum degree; the minimum degree in this graph is 2. The graph in the left diagram is converted into a graph shown in the right diagram by deleting two edges and by inserting two edges (shown in bold). The graph in the right diagram has the same degree sequence but there is an important difference. Here there is at least one vertex (v1 ) with a (highest degree equal to four) which is connected to vertices with higher degrees only; v1 is connected to two vertices with degree 4 and two vertices with degree three; it is not connected to a vertex of degree 2.

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Figure 4.6.5: A seven vertex graph is shown in the left diagram with a degree sequence shown in the top of the graph. There are three vertices in this graph with a highest degree equal to four. All these three vertices (with a degree of 4) are connected to a vertex of minimum degree; the minimum degree in this graph is 2.

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problem (that the highest degree vertex is connected to (relatively) lower degree vertices) is still not completely resolved as is evident from the middle left diagram of the same figure. We again apply the same algorithm to resolve the rest of the problem as shown in the middle right diagram. The new graph with the same degree sequence and the problem completely resolved is shown in the bottom diagram. As you may have noticed in this specific problem we have to apply the said algorithm twice to obtain the desired results. Is it possible to do some thing in this specific problem so that the problem is resolved just by applying the said algorithm only once? Can you generalize your findings? How much can you save in time in the worst case analysis?

Figure 4.6.6: A graph with a given degree sequence shown in the top left diagram is converted into another graph with the same degree sequence shown in the bottom diagram. Problem 4.4.3. Let a degree sequence consisting of 11 numbers be SG = a, b, c, d, e, f, g, h, i, j, k and assume that a = 5. Let another sequence be SH = b − 1, c − 1, d − 1, e − 1, f − 1, g, h, i, j, k. Do we know how to draw the graph for SH provided we have a graph for SG ? Discuss briefly. The figure down below may be helpful for your imagination.

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Problem 4.4.4. Let a degree sequence consisting of 11 numbers be SG = a, b, c, d, e, f, g, h, i, j, k and assume that a = 5. Let another sequence be SH = b, c, d, e, f, g − 1, h − 1, i − 1, j − 1, k − 1. Do we know how to draw the graph for SH provided we have a graph for SG ? Discuss briefly. The figure down below may be helpful for your imagination.

Figure 4.6.7: Two graphs are shown in this figure. The graph shown in the left diagram is a connected graph while the graph shown in the right diagram is a disconnected graph. Both these graphs have the same degree sequence which is 54444411111. Problem 4.4.5. We are given a degree sequence SG . We check if the degree sequence is graphical; if it is graphical then we find and draw the actual graph G corresponding to this sequence; In this graph G the highest degree vertex v (with a degree equal to u) will always be connected to the first u vertices in the degree sequence SG . Please verify if this is always right or wrong in general. Discuss briefly. Problem 4.4.6. If the answer to the above problem is no then how can we do some thing to make sure that in the final graph, the highest degree vertex v (with a degree equal to u) will always be connected to the first u vertices in the degree sequence SG ? Problem 4.4.7. We are given a degree sequence SG . We check if the degree sequence is graphical. If the degree sequence is graphical then you are supposed to draw a graph G corresponding to this sequence such that the

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highest degree vertex v (with a degree equal to u) should be connected to the last u vertices in the degree sequence SG . Discuss briefly if this is always possible and how will you do it? Problem 4.4.8. Is there a possibility that a degree sequence SG is graphical but it is impossible to draw a graph corresponding to this sequence such that the highest degree vertex v (with a degree equal to u) in the graph should be connected to the last u vertices in the degree sequence SG . Problem 4.4.9. We are given a graph G (and its degree sequence SG ) in which the highest degree vertex v (with a degree equal to u) is connected to the first u vertices in the degree sequence SG . Is it always possible to convert it into another graph with the same degree sequence but now the highest degree vertex v (with a degree equal to u) is connected to the last u vertices in the degree sequence SG . Either prove or find a counter example. Problem 4.4.10. What are necessary & sufficient conditions for a degree sequence SG to be graphical such that in the resulting graph G the highest degree vertex v (with a degree equal to u) in the graph should be connected to the last u vertices in the degree sequence SG .

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Figure 4.6.8: Two graphs are shown in this figure. The graph shown in the left diagram is a connected graph while the graph shown in the right diagram is a disconnected graph. Both these graphs have the same degree sequence which is 54444411111.

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We show a walk from vertex a to vertex d in a graph shown in Fig. 4.7.1. Please note that in a walk it is possible to traverse an edge (and therefore a vertex) several times. We also show a trail between the same two vertices in the same graph. Remember in a trail it is not allowed to traverse an edge more than once – but it is allowed to traverse a vertex several times. We also show a path between the same two vertices in this graph. You may have realized that in a path it is not allowed to traverse an edge or a vertex more than once. A walk or a trail can always be converted into a path as shown in this figure. A shortest path between the same two vertices is also shown. The path is shortest in terms of number of edges in between the two terminal vertices.


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Figure 4.7.1: A walk from vertex a to vertex d consisting of eight edges as shown in the top left diagram; vertices as well as edges are repeated in the walk. A trail from vertex a to d is shown in the top middle diagram. In this trail only a vertex is repeated; no edge is repeated; the trail consists of six edges. A path is shown from vertex a to vertex d in the top right diagram; the path consists of six edges. The walk in the top left corner (from vertex a to d) is converted into a four edge path as shown in the bottom left diagram. The trail in the top middle diagram is converted into a path from vertex a to d as shown in the bottom middle diagram. The shortest path from vertex a to d is shown in the bottom right diagram.

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4.8


Multi-graphs and Pseudo-graphs

We show a simple graph; it consists of no self loops or parallel edges. This graph can be represented by an adjacency matrix also shown in the same Fig. 4.8.1. We also show graphs with self loops and parallel edges.

Figure 4.8.1: A simple graph containing no self loops and no parallel edges. It can be represented by an adjacency matrix. How about graphs which are not simple?

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We shall talk about very broad categories of graphs and then some special graphs. We categorize graphs into three major categories: (1) Acyclic graphs (or trees), (2) Bipartite graphs, & (3) Cyclic graphs, i.e., graphs containing cycles. Under special graphs we consider: (a) Completely connected graphs, (b) Regular graphs, (c) Cycle graphs, (d) Line graphs, & (e) Star graphs.


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Figure 4.9.1: A cyclic graph which is neither bipartite nor acyclic is shown in the right diagrams. A bipartite graph which is not acyclic is shown in the middle diagrams; an acyclic graph (or a tree), shown in the left diagrams, is also a bipartite graph.

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4.9.1


Tree Graphs

A connected graph G is a tree provided it does not contain any cycles. A tree graph is shown in Fig. 4.9.1 and 4.9.3. We show a number of non isomorphic trees with p larger than 1 and smaller than 6 in Fig. 4.9.2.

4.9.2

Bipartite Graphs

A graph G is bipartite provided it does not contain odd cycles. It may contain even cycles or no cycles at all. If G does not contain a cycle then G is not only bipartite it is also a tree. The vertex set V (G) of a bipartite graph G can be partitioned into two disjoint sets A and B whereas both A as well as B are independent sets. In other words every edge in a bipartite graph connects a vertex from set A to a vertex in Set B. A bipartite graph is shown in the left diagram of Fig. 4.9.1; it is bipartite because it does not contain any cycle at all. Another bipartite graph is shown in the middle diagram of Fig. 4.9.1; it is bipartite because it does not contain any odd cycles. The graph shown in the right diagram of Fig. 4.9.1 is not bipartite as it contains an odd cycle. We show another bipartite graph in Fig. 4.9.3. The A partite as well as the B partite are shown in the middle diagram where the bipartite graph is drawn with a different orientation to highlight the two parts. Please note that partite A is an independent set while partite B is also an independent set but neither of the two is a maximum independent set. The maximum independent set in this graph is shown in the right diagram of this figure. The minimum vertex cover of this graph is also shown in this diagram.

4.9.3

Special Graphs

A graph G is k-regular if the degree of every vertex is exactly equal to k. A graph is a cycle graph if the degree of every vertex is exactly two. A graph G having p vertices is completely connected if the degree of every vertex is p − 1 (please note that the degree of a vertex in a simple graph can not be more than p − 1). A graph is a line graph (or a chain graph) if the degree of every vertex is 2 except for two vertices where the degree is 1. A line graph consisting of two vertices is a special case where the degree of both the vertices is 1.


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Figure 4.9.2: We show a number of non isomorphic trees with p larger than 1 and smaller than 6.

Figure 4.9.3: A graph G is shown in the left diagram. This graph is in fact a bipartite graph as shown in the middle diagram consisting of an A partite and a B partite. The minimum (sized) vertex cover and the maximum (sized) independent set in graph G are shown in the right diagram.

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A graph G is a star graph if the degree of one vertex is p − 1 while the degree of every other vertex is 1. A start graph consisting of two vertices is a special case. A graph G is referred to as a forest if it contains a collection or set of trees.

Figure 4.9.4: The degree of every vertex in a completely connected graph is p − 1. The degree of every node is the same for every vertex in a regular graph. The degree of every vertex is exactly 2 in a cycle graph. The degree of every vertex is 2 except for two vertices where the degree is 1 in a line graph. The degree of every vertex is 1 except for one vertex where the degree is p − 1 in a star graph. It is obvious from Fig. 4.9.4 that a completely connected graph is (p − 1)regular while a k-regular graph may not be completely connected. A cycle graph is also a 2-regular graph as shown in this figure. A line graph is not regular unless it consists of a special case of a connected graph of two vertices. A line graph, being acyclic, is always bipartite. A star graph is not regular unless it has a size equal to 2 when it becomes a line graph which is 1-regular.


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Also a star graph, being acyclic, is always bipartite where one partite consists of size 1 while the other of size p − 1. We show a number of k-regular graphs in Fig. 4.9.5. Please note that all these graphs are bipartite as shown in the bottom diagrams of this figure.

Figure 4.9.5: A 2-regular graph shown in the right diagrams. A 3-regular graph is shown in the middle diagrams. A 4-regular graph is shown in the right diagrams. All these regular graphs are bipartite as shown in the bottom diagrams. We show a number of k-regular graphs in Fig. 4.9.5. It is interesting to compare these graphs with the ones shown in Fig. 4.9.4. These new graphs are regular and not bipartite. They are rather 4-partite graphs meaning that the vertex sets of each of these graphs can be partitioned into four disjoint sets of vertices (or partites); any edge in these graphs connects a vertex in one partite to any vertex in one of the other partites. The four partites are indicated in different colors in the bottom diagrams of this figure. A number of cycle graphs are shown in Fig. 4.9.7. A curious reader might

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appreciate the fact that a cycle graph is bipartite if it consists of even number of vertices, while it is not bipartite if the cycle graph consists of odd number of vertices. What may not be obvious is that an odd cycle graph will always be a 3-partite graph as shown in this figure.

Figure 4.9.6: A 6-regular graph shown in the right diagrams. A 4-regular graph is shown in the middle diagrams. Another 4-regular graph is shown in the right diagrams. All these graphs are 3-partite as shown in the bottom diagrams.


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Figure 4.9.7: Cycle graphs of different sizes are shown. Problem Set 4.5. Problem 4.5.1. The figure below shows a completely connected graph A in the left diagram. It is a 7-regular graph consisting of eight vertices. The middle graph shows graph B which is derived from graph A after deleting or subtracting a number of edges from graph A; graph B is a 6-regular graph. Similarly graph C is derived from graph A by deleting another set of edges from B. Please note that graph C is again a 5-regular graph consisting of 8 vertices. By deleting a different set of edges from graph A, we may obtain a graph which is not isomorphic to graph B. Draw all possible graphs consisting of 8 vertices which are not isomorphic to B but which are 6-regular. Similarly draw all possible non isomorphic graphs of 8 vertices which are 5regular. Problem 4.5.2. The Figure 4.9.9 above shows the same graph A as shown in Fig 4.9.8. Graph E is obtained by deleting a vertex from graph A. Similarly

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Figure 4.9.8: Graphs B and C are derived from graph A by deleting certain edges.

Figure 4.9.9: Graphs E and F are derived from graph A by deleting certain vertices.


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graph F is obtained by deleting another vertex of graph E. Please note that both E and F are completely connected and regular graphs. (a) Is it possible to get a graph which is isomorphic to graph F by deleting any two vertices of graph A? Discuss briefly. (b) Now concentrate on graph E; it is a 6-regular graph; by deleting certain edges transform it into a 5-regular graph. (c) By deleting another set of edges transform it into a 4-regular graph. (d) Draw all non isomorphic graphs which are 4-regular consisting of 7 vertices. Problem 4.5.3. By carefully selecting and deleting certain edges of graph C shown in Fig. 4.9.8, we can obtain a 4-regular graph of eight vertices. How many such non-isomorphic graphs we shall be able to obtain? The figure below shows all non-isomorphic graphs consisting of 8 vertices which are 4regular. Try to match the graphs that you have obtained with the ones shown in the figure below.

Figure 4.9.10: Shows all non-isomorphic graphs consisting of 8 vertices which are 4-regular. Problem 4.5.4. Look at the graphs consisting of 8 vertices and are 4regular as shown in the figure above. (a) Find which of these graphs is a

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bipartite graph? (b) Draw all non-isomorphic bipartite graphs which are connected, and are regular consisting of eight vertices. (c) Draw all nonisomorphic graphs consisting of 12 vertices and are 6-regular. (d) Draw all non-isomorphic graphs consisting of eight vertices which are 3-regular. Out of these graphs indicate which ones are bipartite graphs. Problem 4.5.5. The graph G shown in the figure below is derived from graph A by subtracting certain edges from graph A. Note that graph G is 5-regular. Draw all non-isomorphic graphs consisting of 8 vertices which are 5-regular.

Figure 4.9.11: Graphs G is derived from graph A by deleting certain edges.

4.10

Integration of Concepts, Properties, and Action Items

We have talked about various concepts in graph theory in this chapter. We have mainly confined ourselves to simple graphs in which there are no parallel edges and no self loops. This category of graphs can easily be represented by an adjacency matrix or an adjacency list data structure. We have further limited our study to un-directed graphs in this chapter; we shall study directed graphs in detail in Chapter 8. Un-directed graphs can be further classified into connected graphs and disconnected graphs. Connected graphs can also belong to certain categories like a line graph, a cycle graph, a tree

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graph, a cyclic graph and a completely connected graph. We have also talked about certain properties of graphs, like a Hamiltonian path or an Eulerian path in a graph. We have also talked about a walk, a trail, or a path between two vertices in a graph. It will be interesting if we integrate a couple of concepts with a number of properties.

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4.11


Self Complementing Graphs

A self complementing (SC) graph is a graph G whose complement c(G) is isomorphic to itself. We shall discuss these graphs in some detail as they provide us a platform to connect a number of key concepts in graph theory & algorithms. The following concepts are used in a meaningful manner to advance our discussion on self complementing graphs. 1. Complement c(G) of a graph G 2. Graph isomorphism between two graphs G & H 3. Regular & non regular graphs 4. Degree sequence of a graph 5. Deleting and inserting a vertex in a given graph 6. Eccentricity of a vertex 7. Radius & Diameter of a graph This section is organized as follows. We shall first show some SC graphs in order to give you a feel of such graphs. Some of the necessary conditions of SC graphs will be discussed next. The problem of how to construct a SC graph of fixed vertices will also be discussed. We shall show how one SC graph can be transformed into another SC graph with less or more vertices? Some of the interesting properties of these graphs will also be elaborated. We shall conclude with a number of interesting problems.

4.11.1

Regular Self Complementing graphs

We show a number of non trivial (a graph of one vertex is a trivial example of such graphs) self complementing graphs in the following figure. It can be seen that the left graph is the complement of the right graph as well as isomorphic to the right graph. That means that both these graphs are self complementing. These graphs are also regular as the degree of each vertex is 2. Both of them are also Hamiltonian? Necessary Conditions


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Figure 4.11.1: A graph and its complement which is isomorphic to the original graph.

Does this mean that all regular graphs are self complementing? But a cycle graph of three or six vertices is not self complementing? There must be a class of regular graphs which will be self complementing? The following figure shows the same graph along with the corresponding completely connected graph of five vertices. The complement c(G) of graph G is obtained by deleting all edges of graph G from the corresponding completely connected graph. If c(G) is isomorphic to G then it should have the same number of edges as G - that means the degree of every vertex in a regular self complementing . Then the number of edges in the graph will be exactly graph should be p−1 2 p(p−1) . Does that mean that p or p − 1 in an SC graph should be divisible by 4 4? Are these necessary or sufficient conditions for such a regular graph to be self complementing? It will be rewarding if you draw a couple of regular graphs with the property that the degree of each vertex is p−1 . Can you draw such a graph if p is even? 4 Can you draw such a graph if p is any odd number? It will be essential to answer the above questions before moving forward? If you have tried to draw such graphs then you will soon realize that a nregular graph (where n = p−1 ) is possible if and only if p = 4k + 1 where 4 k is an integer equal to or larger than one. We have already drawn such a graph with p = 5 when k = 1. When k = 2, p will become 9. Let us try to explore graphs with 9 vertices with the degree of each vertex equal to 4. Will all such graphs be isomorphic to each other? Will such graphs (or at

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Figure 4.11.2: A self complementing graph of 5 vertices (left diagram) and a completely connected graph of 5 vertices. least some of them) self complementing? If you take the complement of any such graph it will certainly have the same degree sequence and same number of edges? Again it will be rewarding if you draw some of these graphs before arriving at a conclusion? We show three such graphs in the figure below. Are any two of them isomorphic to each other? Is one complement of the other? Are they self complementing? Please try to answer these questions before moving forward. It will be exciting if you draw the complement of the left graph? You will soon realize that the top right graph is the complement of the top left graph while the bottom graph is isomorphic to the top right graph. What does this example tell us?


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Figure 4.11.3: Top left diagram shows a graph G. Top right diagram shows the complement c(G) of graph G. The bottom graph is equal to c(G).

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If you actually draw all non isomorphic graphs with p = 9 and degree of each vertex equal to 4 you will realize that there are 16 such graphs possible. Most of them transform into one another if you take the complement of one such graph as shown above. A very few of them indeed transform into themselves when you take the complement. One such graph is shown in the figure below. 1

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Figure 4.11.4: A regular self complementing graph of 9 vertices. How about imagining non isomorphic regular graphs having p = 13 and degree of each vertex equal to 6. There are (367860) six-regular non isomorphic graphs with 13 vertices. If we take the complement of one such graph it will transform either into one of such graphs or into itself. One such self complementing graph is shown below. 1

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Figure 4.11.5: A regular self complementing graph of 13 vertices.


4.11.2

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Non Regular Self Complementing graphs

One important question that should agitate you is every self complementing graph a regular graph? Is it possible to have a non regular graph to be self complementing? How about a line graph of four vertices? Its degree sequence is 2211. The degree sequence of a completely connected graph of four vertices is 3333. The degree sequence of the complement of this line graph can be obtained by subtracting 2211 from 3333; it comes out to be the same as the four vertex line graph is self complementary. This provides us a necessary condition for a non regular graph to be self complementary? The degree sequence of c(G) should be such that if we add it (after sorting it in increasing order) with the degree sequence of G (having p vertices) then we should get the degree sequence of a completely connected graph of p vertices.

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Figure 4.11.6: A self complementary line graph of 4 vertices. Necessary Conditions Is it possible to have a non regular graph of five vertices which is self complementary? Let us try out various degree sequences with the above necessary condition? There are only three choices possible under the conditions laid out before: 22222, 33211, & 32221. The first degree sequence corresponds to a cycle graph which was regular and we have already seen it. The other two belongs to non regular graphs and are of interest to us. Now we have to verify if these degree sequences are really graphical, and if they are then do they really belong to graphs which are self complementary? Let us start with the sequence 32221. Can you draw its graph and check out if the graph is self complementary? The sequence is graphical but is itself complementing?

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Now let us check the other sequence which is 33221; this indeed comes out to be graphical as well as self complementary as shown below:

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Figure 4.11.7: A 5 vertex non-regular self complementing graph. Please note that the condition (the degree sequence of c(G) should be such that if we add it (after sorting it in increasing order) with the degree sequence of G (having p vertices) then we should get the degree sequence of a completely connected graph of p vertices) is a necessary condition; it is not sufficient. Thus this condition along with others (for example the number of odd degrees in a degree sequence should be even for a degree sequence to graphical) may narrow down our search for SC graphs but we always have to verify if a given graph G is a SC graph? These simple conditions tells us that an SC graph (which may or may not be regular) can only have p equal to 4k or 4k + 1 vertices where k can be equal to or larger than 1. We have already seen a regular SC graph of 4 + 1 = 5 vertices; we have also seen a non regular SC graph of 5 vertices; we have seen a non regular SC graph of 4 vertices in the last diagram. In fact we have earlier claimed that a regular SC graph can have only 4k + 1 vertices; we can safely say that a SC graph of 4k vertices will always be non regular? Why?


4.11.3

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Constructing Self Complementary Graphs

Let us consider graphs where p = 8 and assume that its degree sequence satisfies the stated necessary conditions for a SC graph. Some of the possible degree sequences are listed here: 66661111, 66443311, 55443322, 44443333, 55552222. It will be interesting if you try to check if a graph corresponding to these degree sequences is indeed self complementary? This will certainly be a tiring process - we shall now discuss ways of constructing large SC graphs bypassing this tiring process? The new method will help us further in making meaningful connection between relevant concepts. Given any graph G we can always find its complement c(G). If we connect graph G and its complement in the following configuration then we claim that the new graph will be a SC graph. Please note that a line between G and c(G) means that each vertex of G is connected to every vertex of c(G); and G may contain more than one vertex. If G contains a single vertex then this graph is certainly SC graph. But if G contains more than 1 vertex then why this composite graph is self complementary?

Figure 4.11.8: Here vertex G is a graph while c(G) is complement of graph G. An edge between G and c(G) means that every vertex G is connected to every vertex of c(G). The resulting super graph is a self complementing graph. Let us actually construct such graph and see how it looks? Assume that G

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is a line graph of two vertices - then c(G) will simply consist of two isolated vertices. How the above configuration would look like when it is actually drawn? If you look at the diagram below you will realize that the graph is indeed self complementary. The degree sequence of this graph is 44443333 it is one of the sequences that we have predicted earlier in our discussion? We can also start with G equal to two isolated vertices - then c(G) will be a line graph of two vertices. Using these building blocks you can also construct another SC graph? If you have the patience of constructing such a graph you will realize that its degree sequence will be 55552222. The number of vertices in the SC graph will still be 4p where G has p vertices. 7 5

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Figure 4.11.9: Here graph G is a line graph of two vertices 1 & 2. Then c(G) will be its complement - and that will be two isolated vertices 3 & 4. Another copy of G will consist of a line graph of two vertices - consisting of vertex 7 and vertex 8. Another way to construct a SC graph is shown below - here also G is any graph and is connected with its neighboring graph in the same fashion? Here x is just one vertex and we claim as before that the resulting graph will be self complementary. Why? The number of vertices in the resulting SC graph will be 4p + 1 where p is the number of vertices in G.


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Figure 4.11.10: Here vertex G is a graph while vertex c(G) is the complement of graph G. An edge between G and c(G) means that every vertex G is connected to every vertex of c(G). Vertex x may be a single vertex graph. The resulting super graph is a self complementing graph. There a number of interesting possibilities in the above configuration. 1. How about if instead of single x vertex we have a graph H connected to its neighbors? In order to make the whole configuration SC should it be any graph H or a special graph H? What special property it should possess? As you can discover yourself H should be a self complementary graph for the above configuration to be self complementary? 2. Is it possible that we insert graph G in the place of x in the above configuration? Please note that G is any graph? 3. How about if we insert a self complementary graph H in the place of x as well as G in the above configuration? 4. What would happen if the self complementary graph H in the above step is regular? How about if it is non regular? 5. Will the above configuration result into a regular or a non regular SC graph? The following diagrams will help you answer these interesting questions. Please find the degree sequence of each of the following graph? Can you figure out how these SC graphs are constructed?

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Figure 4.11.11: A line graph of 4 vertices is used as a building block in the left diagram. A cycle graph of 5 vertices is used as a building block in the right diagram.

4.11.4

Transforming a SC graph with 4k + 1 vertices into another SC graph with 4k vertices

We know that a SC graph should have either 4k or 4k + 1 vertices? This immediately leads us to conjecture that if we have a SC graph of 4k + 1 vertices then by deleting one vertex we can convert it into another SC graph having 4k vertices? Is it straight forward or we need to devise an intelligent algorithm to do so? Similarly if we are given a SC graph having 4k vertices then is it possible to construct a SC graph having 4k + 1 vertices by inserting a new vertex and connecting it to some of the vertices of the original graph? Again do we need some thinking to do so or is it a trivial problem? In order to answer these questions let us start with a regular graph which is SC. It will contain 4k + 1 vertices. By deleting any vertex from this graph is it possible to convert it into another SC graph having 4k vertices? Let us start with a simpler problem of a regular SC graph having 9 vertices as shown below. Can we delete any vertex and the resulting graph would stay a SC graph of 8 vertices? Please note that initially the degree sequence is


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444444444. It becomes 44443333 after deleting that vertex. What about if we delete any vertex from the 25 vertex graph shown earlier. Will the new graph be also self complementary? Why? How about if we have a non regular SC graph of 4k + 1 vertices? Can we delete any vertex to make it another SC graph with 4k vertices? How about this graph? It has a degree sequence 332211. It is no longer a trivial problem and we should devise an algorithm to solve the problem?

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Figure 4.11.12: We remove vertex 1 in the top diagram from a SC graph in the top diagram. We remove vertex 2 from the same SC graph in the bottom diagram.


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Figure 4.11.13: We remove a vertex from the 25 vertex SC graph. Is the resulting graph a SC graph?

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Figure 4.11.14: A SC graph having 5 vertices which is not regular. Here we show another interesting graph which is not regular but is self complementing. Its degree sequence is 774444411. This graph may also provide insights needed to design an intelligent algorithm to solve the above problem.

Figure 4.11.15: A SC graph having 9 vertices. Can we remove any vertex and still it remains a SC graph? Given a degree sequence 774444411 of a self complementing graph - after deleting one vertex the new degree sequence should possess certain properties if it represents a self complementing graph? For example it can be 66443311 or 77443300, etc. The problem is to decide which vertex should be removed?


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Should we always remove the vertex with degree equal to 4? Why? Should that vertex be connected with lower degrees or higher degrees?

4.11.5

Transforming a SC graph with 4k vertices into another SC graph with 4k + 1 vertices

Now let us look into the problem of inserting a vertex in a SC graph consisting of 4k vertices such that the new graph having 4k + 1 vertices is also selfcomplementing. Please note that we can also put an extra constraint that the resulting SC graph should be regular. Again we shall show you a number of graphs and then provoke you to design an efficient algorithm to solve the problem. We show the same graph with a degree sequence 44443333 being converted into a new SC graph with degree sequence 555543333.

Figure 4.11.16: We insert a vertex x in a self complementing graph. Now start with the same original graph (with a degree sequence equal to 44443333) while the new graph should have a degree sequence equal to 444444444. The resulting graph is SC and is also regular - although it has not been drawn in a suitable manner in the top diagram. The graph is redrawn in the bottom diagram to emphasize that it is indeed a beautiful regular graph?

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Figure 4.11.17: After inserting vertex x we have a regular SC graph.


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So the problem is where to connect the inserted vertex and what should be its degree? Perhaps another example may provide you a solution? Please check the degree sequence of the graph before after inserting a new vertex?

Figure 4.11.18: Inserting a vertex x in a SC graph?

4.11.6

The Self Complementary problem and Graph Isomorphism

If we know how to find if G and H are isomorphic then we can always check if G is a self complementary graph? How? Take the complement of G - this is c(G) and now check if G and c(G) are isomorphic. But suppose we know how to check if G is a self complementary graph - then can we solve the graph isomorphism problem? How?

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Figure 4.11.19: We need to check if graph G is isomorphic to graph H? Look at the figure above. We need to check if a given graph G is isomorphic to another given graph H. We claim that if the top (or the bottom) graph is self complementary then graph G is isomorphic to graph H? Why? We can also use the following configuration to check if G and H are isomorphic also by substituting H in an appropriate place in this diagram?

Figure 4.11.20: We need to check if graph G is isomorphic to graph H?

4.11.7

A SC graph has diameter 2 or 3 - not less than 2 and not more than 3?

Instead of proving the above statement let us first do a simple one. Can we prove that for any graph G the following configuration will give us a SC graph with a diameter not more than 3.


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Figure 4.11.21: What is the diameter of this super graph? Again can we prove that for any graph G the following configuration will give a SC graph with diameter not more than 2?

Figure 4.11.22: What is the diameter of this super graph? We shall now attack the more general problem? But before that let us refresh our knowledge regarding some old concepts: the eccentricity of a vertex, and the radius and diameter of a graph. Let us see how the eccentricities of various vertices look like in a line graph of eight and five vertices as shown below. The diameter in the top graph will be 7 while the radius will be 4 in the top diagram. The diameter and radius for the bottom line graph will be respectively 4 and 2. Similarly we show another five vertex graph where the

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Figure 4.11.23: The eccentricities of different vertices in a line graph. eccentricity of each vertex is 1. It is a completely connected graph - the diameter as well as the radius is 1 in this graph. The complement of this graph is also shown in the bottom diagram. Its diameter as well radius will be infinite.

Figure 4.11.24: A completely connected graph G and its complement. Another five vertex graph G and its complement is shown below. Here the eccentricity of each vertex in G is 2 except for vertex 1. Thus its diameter is 2 while its radius is 1. The diameter in the complement of graph G is infinite while its radius is also infinite. The five vertex line graph and its complement are shown in the bottom diagram. The eccentricity of each vertex is indicated in the line graph as well as in its complement. Please note that the diameter in the line graph was 4 while it has reduced to 2 in the complement of the line graph. The diameter as well as the radius in the complement graph is 2. You can well imagine that if the diameter in a graph G is more than 3 then


Figure 4.11.25: A star graph and its complement.

Figure 4.11.26: A line graph and its complement.

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the diameter in its complement is reduced to 2. That means a graph G with a diameter more than 3 cannot be self complementary. Similarly if the radius or the diameter in a graph G is 1 then it cannot be isomorphic to its complement? Why? Does that mean that a SC graph can have diameter 2 or 3 - not more than 3 and not less than 2? How about the radius of a SC graph? Should it be always 2 - not more not less?

4.11.8

Bipartite self complementary graphs

Definition: Assume that the complement of a bipartite graph BP is isomorphic to bipartite graph BP . Let us call these category A SC bipartite graphs. We assume that while taking the complement of graph BP we simply consider it a general graph - thus all the A (and B) partite vertices in BP will become completely connected in the complement of BP . Alternatively in order to find complement of graph BP , we first form a completely connected general graph consisting of all BP vertices - we then remove those edges which are already present in BP - this will give us the complement of graph BP . Please note that bipartite complement of a bipartite graph is different from this complement. Here we first form a completely connected bipartite graph consisting of as many vertices in the A as well as B partites of bipartite graph BP . We then remove existing edges in BP from this completely connected bipartite graph - the resulting graph will be a bipartite complement of bipartite graph BP . Now assume that the bipartite complement of bipartite graph B is isomorphic to bipartite graph B. Let us call these category B SC bipartite graphs. The following graph is a Category A SC graph. Please note that it is not a Category B SC graph? Why? Can you draw another bipartite graph which is a Category A SC graph? Can you prove that such graphs are not possible if any partite contains more than two vertices? We show a category B SC graph BP in the figure below. Please note that this not a Category A SC bipartite graph? Why?


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Figure 4.11.27: A bipartite graph BP shown in the top left diagram. The corresponding completely connected graph is shown in the top right diagram. The complement of graph BP is shown in the bottom diagram.

Figure 4.11.28: A bipartite graph and its bipartite complement.

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The same graph BP is drawn below in a different shape. The corresponding completely connected bipartite graph and the complement of bipartite graph BP is also shown here.

Figure 4.11.29: Top left diagram shows a bipartite graph BP. Top middle diagram shows the corresponding completely connected bipartite graph. Top right diagram shows the bipartite complement of the bipartite graph. The bottom diagrams once again shows the two bipartite graphs shown in the standard form. The degree sequence of the above graph is 22222222. While that of the completely connected bipartite graph is 44444444. As you can see this is a regular SC bipartite graph. The following bipartite graph is a non regular self complementing graph with a degree sequence 33222211. The degree sequence of a corresponding completely connected bipartite graph is also 44444444. Here we show another Category B SC graph with red vertices in one partite


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Figure 4.11.30: A SC bipartite graph which is not regular. and green vertices in another partite.

Figure 4.11.31: A regular SC graph having 12 vertices. From now on wards we shall consider only category B self complementary bipartite graphs - we shall simply refer them as SC bipartite graphs. How many edges are there in a complete SC bipartite graph with m vertices in one partite and n vertices in another partite? That is equal to mn. The number of edges in a SC bipartite graph will be mn/2. Thus either m or n should be even. That is an important necessary condition for a bipartite graph to be self complementary.

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Figure 4.11.32: A SC graph where the size of two partites is different.

4.11.9

Decomposition of a SC graph G

Here we shall talk about the decomposition of a SC graph G into a number of edge-disjoint graphs - one of them is a SC bipartite graph? Initially we shall talk informally and provide some insight and then we shall discuss it more formally. Please note that the following graph is a SC graph with or without the vertex x. Thus even if we remove vertex x the resulting graph G − x will still be a SC graph as shown below - although vertex x provides a useful function of pinpointing certain vertices needed for the decomposition of the graph? Vertices in G − x which are adjacent to x belong to one partite while the rest of the vertices in G − x belong to the other partite. We put vertices in G − x which are adjacent to x in set A while the rest of the vertices in G − x goes in the set B. There will be a bipartite graph having edges going between the set A and the set B. In addition to that there will be a graph inside G consisting of vertices belonging to the set A - we call this the graph G(A). Similarly there is a graph G(B) consisting of vertices in the set B. The original graph can thus be decomposed into edge-disjoint graphs - graph G(A), bipartite graph BP , graph G(B) and the vertex x connected to all


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vertices in graph A.

Figure 4.11.33: A self complementing graph is decomposed into a number of edge-disjoint graphs. It can be observed that the following properties should hold for the original graph G to be SC. 1. The bipartite graph BP (A, B) will also be a SC bipartite graph. The two partites are balanced that is the size of A and B is the same. 2. The graph (x, A) 3. The graph G(A) and graph G(B) - there is some relationship between these two graphs?

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Let us consider another example to confirm/enhance our observations before formally discussing the decomposition of a SC graph.

Figure 4.11.34: A SC graph decomposed into a number of edge-disjoint graphs.

4.11.10

Permutation, Isomorphism, automorphism & Self Complementing Graphs

Any permutation of vertices of any graph G may create a different graph H. This graph H will always be isomorphic to graph G but may not be equal to graph G (as the adjacency matrix may be different). For example consider the permutation p1 equal to (1234)(5678)(9) as shown in Fig. 4.11.35. This is a so called circular permutation in which we map 1 onto 2, 2 onto 3, 3 onto 4, and 4 back to 1. The vertex 9 is mapped onto itself. This permutation permutes the vertices of G and produces a new graph which is not equal but isomorphic to the original graph G. In order to show that this new graph is isomorphic to the original graph G we have to find an isomorphic function


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(or a permutation) which maps the vertices of the new graph back onto the vertices of graph G such that adjacency as well as non adjacency is preserved in the two graphs. In this case this permutation p2 will be (4321)(8765)(9).

Figure 4.11.35: Graph G (top left diagram) and another graph (top right diagram) where the vertices of G are permuted according to the given permutation p1 . The two graphs are not equal but they are isomorphic thus the permutation p1 is not an automorphism of graph G. Please note that graph G is equal to the bottom graph If a permutation p of a graph G creates a graph H which is equal to graph G (that means the adjacency matrix will exactly be the same) then that permutation is known as an automorphism of graph G. The identity permutation is always an automorphism - it is known as a trivial permutation. We may be more interested in the non trivial permutations of graph G? It may be possible for a certain category of graphs that the only automorphism is the trivial identity permutation? Please note that if p is an automorphism of graph G then the permutation p2 is always an automorphism of graph G If a permutation p of vertices of graph G creates a graph H which is the complement of graph G then the permutation p is known as the complementing

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Figure 4.11.36: Graph G (left diagram) and another graph (right diagram) where the vertices of G are permuted according to the given permutation p. The two graphs are not only isomorphic but also equal thus the permutation p is an automorphism of graph G but it is not a complementing permutation of graph G. permutation of graph G and graph G is a SC graph. Any complementing permutation of a graph G is certainly not an automorphism of graph G but the permutation p2 is always an automorphism of graph G if p is a complementing permutation. Thus every self complementing graph G has a complementing permutation p associated with that graph G. Whenever we claim that G is self complementing then we have to find out the self complementing permutation?


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Figure 4.11.37: Graph G (top left diagram) and another graph (top right diagram) where the vertices of G are permuted according to the permutation p. This new graph is also the complement of graph G - that means the permutation p is a complementing permutation of graph G but it is not an automorphism of graph G. The bottom diagram shows another graph where the vertices of p(G) are permuted once again according to the same permutation p. This new graph is equal to graph G - thus p2 (G) is an automorphism of graph G.

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Let us now consider certain self complementary bipartite graphs and the related complementary permutations - two of them are shown in the figures below.

Figure 4.11.38: Bipartite graph BP and a complementing permutation. According to this permutation vertices of one partite are permuted onto vertices of the other partite in the complement of BP . The permutation in Fig. 4.11.38 is a self complementing permutation. The permutation in Fig. 4.11.39 is also a self complementing permutation. But in the top permutation one of the partites is mapped onto the other partite - in the lower permutation it is mapped onto the same partite?


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Figure 4.11.39: Bipartite graph BP and its complement. The complementing permutation p = (1A3A)(2A)(4A)(1B3B)(2B)(4B). According to this complementing permutation vertices of one partite are permuted onto vertices of the same partite. In the decomposition or the synthesis effort to be explained in the coming examples we are more interested in those bipartite graphs which are similar to bipartite graphs similar to the one shown in Fig. 38. Why? In the coming figures you find certain clues of synthesizing self complementing graphs using simple building blocks?

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Figure 4.11.40: A successful attempt to synthesize a self complementing graph. The bottom diagram shows graph G, p(G) and p2 (G) in that order.


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Figure 4.11.43: A failed attempt to synthesize a self complementing graph from individual components. The bottom diagram shows graph G, p(G) and p2 (G) in that order.

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Chapter 5 Basics of Graph Algorithms 5.1

Design of Algorithms

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The Bucket Algorithm

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Finding if a Graph is Connected

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Finding if a Graph is a Tree

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Finding a Spanning Tree of a Graph

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Finding a Minimum Spanning Tree of a Weighted Graph

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Finding a Path in a Graph

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The Shortest Path Problem

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Graph Traversal Techniques

5.10 Some Graph Theoretic Claims 5.11 Shortest Path Algorithms 5.12 Discussion

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Basics of Graph Algorithms

Introduction We have studied a number of concepts related to graph theory in the last chapter. For example, we talked about a bridge edge and a cycle in a graph. We have also studied a number of properties linking different concepts. For example an edge is a bridge edge provided its removal disconnects a graph. We have also studied that a graph is cyclic if it contains a cycle. In this chapter we shall study a number of graph algorithms. For example, how can we check if an edge is a bridge edge; how can we check if a graph is cyclic; how can we find an actual cycle in a cyclic graph (see Concept Map 5.1). Almost all of these algorithms are based on the following easy to understand and friendly to use buliding blocks: 1. A four line procedure known as the Bucket Algorithm which can be molded into a number of useful and powerful graph algorithms based on greedy strategies. 2. Another four line procedure known as 2-edge Shortest Path Algorithm which finds shortest paths of length 2 from a given vertex to every other vertex in a weighted directed graph. Again this building block can be used to design a number of sophisticated shortest path algorithms based on dynamic programming. The initial part of this chapter is based on one of our papers published in IJECE, Vol. 2, No.2, Summer-Fall 2003. The paper was co-authored by Sara Tahir, and its title was Should We Teach Algorithms. The later part of this chapter is based on one of our CS department research reports which was coauthored by Komal Syed, and Yasser Hashmi and its title was Shortest Path Algorithms - Making and Breaking Connections. We also provide a number of powerful learning tools to understand and design various algorithms. The recursiuon tree and the colored puzzle are some of these visual aids which facilitate a learner or a designer in his or her path towards discovery. In this chapter, we provide a detailed study of a number of graph algorithms that have applications in diverse fields like chemistry, biology, mathematics, engineering, social sciences, and ofcourse computer science. The original research papers were about encouraging students to discover and learn (graph algorithms) by themselves with minimal help provided by an instructor in the form of provocative questions.

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Concept Map 5.1. A panoramic picture of some Concepts, Properties, Action items and Graph Algorithms.

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5.1


Design of Algorithms

Teaching the standard course “Analysis & Design of Algorithms” at an undergraduate level in a typical Computer Science program essentially has two objectives. The first objective, dealing with analysis, is to familiarize students with existing algorithms. The second one, which is perhaps far more important, is to equip the students with the necessary tools and techniques, and above all the confidence required in solving a non-textbook problem. This second objective, concerned with the design of algorithms, is essentially a creative effort containing all the ingredients of a thriller: adventure, excitement, challenge, and suspense. There is no guarantee that one who critiques literature can learn to write beautiful poetry. Similarly the ability to understand and analyze algorithms does not guarantee that one could become an efficient algorithm designer. The study of the methods and rules of discovery and invention is a field in its own right. Though there are rules of thumb that can be followed to help an individual design an algorithm, there is no precise algorithm available that can be used to design new algorithms. Despite the fact that one cannot guarantee that a student could become an efficient algorithm designer, we believe that the instructor, following our approach and providing proper guidance, can sow the seeds that could blossom into the genius that produces efficient yet astonishingly simple algorithms. Our experience of teaching algorithms indicates that creativity in algorithm design depends, to a large extent, on how we deal with the analysis phase. We stress that while we are familiarizing students with existing algorithms, we should not formally teach anything. Instead we should encourage, rather incite, students to create algorithms themselves using some very fundamental concepts. The objective is that students should experience the thrill and excitement of discovery even during the initial phases of understanding existing algorithms. Polya [10] remembers the time when he was a student himself: he was always perturbed by the question: ”Yes the solution seems to work, it appears to be correct; but how is it possible to invent such a solution? How could I invent or discover such things by myself ?” We feel that with the availability of some pre-requisite knowledge, timely hints, and stimulating questions posed by the instructor, one can always encourage students to redesign an algorithm right from scratch. It is now


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important to find a good working definition of design (of algorithms).

5.1.1

What is Design?

According to the Webster’s dictionary [1], ”Design is to conceive and plan out in the mind”. In the words of Miller [9], ”Design is the thought process comprising the creation of an entity”. Rine [13] defines design as ”A systematic, directed set of decisions that are introduced, made and deployed, leading to an effective or efficient outcome, solution, or technology”. The last definition suits our discovery based learning approach in which a teacher formulates a directed set of questions and hints in order to help his/her students design algorithms. It is interesting to note that our approach is similar in some aspects with the so-called Moore Method of teaching and learning.

5.1.2

The Moore Method

R. L. Moore was a professor of mathematics at the University of Texas. In the words of Hale [5], ”What was so special about his mode of teaching was that he did not lecture, he did not profess. He sat in the back of the room, mostly quiet, occasionally asking a question, allowing his students to find the answers in their own ways”. Many professors still use his teaching style not only in his subject of specialization (topology), but in analysis, algebra, game theory, and other courses, and have advanced or modified the Moore Method in a number of ways [3], [6]. Taylor [18], while characterizing (his version of) the Moore method of teaching, does not allow collective effort on the part of the students inside or outside of class. He also does not allow the use of any source material. We, on the other hand, encourage lively discussions inside as well as outside the classroom. The teacher, in our model, starts with something (very simple), and then actively guides the students in their path of discovery.

5.2


We start with a simple algorithm which we call the Bucket Algorithm (the bucket symbolizes a friendly container where a child puts every new toy or every new discovery) consisting of just four lines of pseudo code: We shall show how this primitive procedure can be used to reinvent a number

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Algorithm 17: The Bucket-Algorithm input : A Graph G output: A Bucket B 1 Put any vertex x of Graph G in the Bucket B; 2 while there are edges coming out of the Bucket B do 3 Select an edge connecting vertex u in B to v not in B; 4 Put v in B ;

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Figure 5.2.1: Two pictures of what the Bucket B will look like in the initial stages of the Bucket Algorithm.


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of existing powerful algorithms in graph theory ([16], [17], [2], [11], [15], [14], [4]). With some encouragement from the instructor, the students should develop a keen desire and ability to understand the motives behind, and the procedures followed in order to arrive at innovative solutions. They would learn the ways and means of devising their own algorithms. Specifically the Bucket Algorithm would be used to solve the following problems: 1. Find if a given graph is connected. 2. Find the number of connected components of a graph. 3. Find a bridge in a graph. 4. Find if a graph is a tree. 5. Find a spanning tree of a graph. 6. Find a path between two vertices in a graph provided a path exists. 7. Find a minimum-spanning tree of a graph. 8. Rediscover Prim’s Algorithm. 9. Rediscover Kruskal’s Algorithm. 10. Solve the single-source shortest-paths problem: rediscover Dijkstra’s Algorithm. 11. Conduct a breadth first search in a graph. 12. Conduct a depth first search in a graph.

5.2.1

Understanding the Bucket Algorithm

The Bucket Algorithm is simple and straightforward. It is just a 4-line algorithm with a simple while loop (with no conditional statements or recursive calls). We start with something simple but potentially very powerful. Simple, because it is easy to understand and at the same time flexible enough to handle a variety of different problems.

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5.2.2

How does it Work?

We identify a Graph G and a Bucket B (See Figure 5.2.1). Step 1 instructs us to put any node, say node a, of the Graph G into the bucket. Next we choose any edge joining vertex a to any other node, say node b, in the graph (since all other nodes are currently outside of the bucket) and put b in the bucket. See Figure 5.2.1 for a picture of what the Bucket B will look like at this stage. Now we have a set of nodes {a, b} in the bucket giving rise to a set of edges {{a, f }, {a, k}, {b, c}, {b, d}} to choose from in Step 3 as we iterate through the while loop. Figure 5.2.2(middle diagram) shows the bucket B after different iterations through the Bucket Algorithm. Notice that there are two types of vertices: those inside the bucket represented by the set {a, b, c, d, e}, and those outside the bucket, the set {f, g, h, i, j, k}. These two different kinds of vertices give rise to three different kinds of edges. The first is the set of edges connecting vertices inside the bucket with each other: {{a, b}, {b, c}, {b, d}, {d, e}}. The second is the set of edges connecting vertices outside the bucket with each other: {{f, g}, {g, h}, {h, i}, {i, j}, {i, k}}. The third is the set of edges (the “branches coming out of the Bucket B” in Step 2) connecting vertices inside the bucket to vertices outside the bucket (see the middle diagram of Figure 5.2.2). This set of edges is equal to {{a, f }, {a, k}, {c, k}, {d, f }}. An edge belonging to this last set of edges is called a cross edge and is of most interest to us. Depending on the constraint we place on the selection of cross edges in Step 3, we can implement numerous algorithms.


5.2.3

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Playing with the Algorithm

We strongly encourage our readers to play around with the Bucket Algorithm to get comfortable with it. During this activity the instructor should ask thought provoking questions such that the students focus on multiple facets of the algorithm that would later help in designing new algorithms. Such questions could be: Under what conditions would there be no edges coming out of the Bucket? Note that this condition should be met otherwise the algorithm would never terminate. Would all the vertices of the graph move into the bucket after the completion of the algorithm? When would this scenario be true, and when would it be false? Does it make any difference if we have a different starting vertex? Note that there are situations when it really makes a difference. It will be useful at this stage if the students are asked to derive the time complexity of the Bucket Algorithm. Students must come to realize the importance of cross edges: it is because of this cross edge {u, v} we select in Step 3 that we discover the new vertex v.

5.2.4

Solving Other Problems

The above questions would induce a deeper understanding amongst students about how the Bucket Algorithm works under different conditions and give some hints while solving more complex problems. After the students are confident that they understand the idea behind the Bucket Algorithm, the instructor can start asking them to modify it to solve more complex problems. What is the worst-case complexity of this algorithm? It is recommended that the instructor not involve the underlying data structure at this stage in order to tackle the issue of complexity. It may be advisable to include it at a later stage.

5.2.5

The Right Provocation

It is well known that a real understanding of the problem is a necessary condition to solve any problem. According to David [12], “Half way home to solving a problem is a clear understanding of the problem”. Out of a sequence of six questions posed by Skiena [16] in order to guide one to discover the right algorithm, the first question is ”Do I really understand the problem?” Then comes the role of the teacher in terms of how he/she states a problem and provokes (or guides) his/her students to solve it in a specified manner. For

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example if a teacher is talking about Quick Sort, he/she cannot expect his/her students to discover the said algorithm just after understanding the sorting problem. The teacher should first make the students appreciate the need for partitioning the array into halves such that all numbers in the first half are smaller than each number in the second half. Why we should do this and how should we do this are both equally important for designing, discovering (and even understanding) the said sorting algorithm. The understanding of the previous state of an abstract system and the (usefulness of the) final system state after the application of a so called fundamental operation [7] (for example the partitioning procedure in Quick Sort) is crucial in problem solving in computer science as in elsewhere. Please see Concept Map 5.2.

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Assuming that the students know what a connected graph is, the instructor should ask the students: “Can you modify the Bucket Algorithm such that you may be able to determine whether a given graph G is connected?” The emphasis should be on using the existing techniques with minimum modification. The answer is simple: if, after the Bucket Algorithm has been applied to a graph G, there are still any nodes left outside the bucket then the graph is not connected. If, however, all nodes come inside the bucket, then graph G is connected. Notice while students were becoming familiar with the Bucket Algorithm, the instructor asked when there would be nodes left outside the bucket. Brighter students would have been able to identify at that stage that some nodes will be left outside the Bucket B when a graph is not connected since cross edges do not exist connecting them to nodes inside the bucket (Figure 5.3.1). Please see Concept Map 5.2. Not all students may be able to identify this property of the Bucket Algorithm. The instructor in this case will have to make an extra effort to guide such students. Once all students have understood the solution (having arrived at it on their own with well-timed prodding from the instructor) the instructor should start the discussion regarding cost calculation, i.e., the complexity of the modified algorithm.


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Figure 5.3.1: A graph G that is not connected; once the Bucket Algorithm terminates, nodes i and j will be left outside the Bucket B.

5.3.1

The Number of Connected Components

Once we understand how to find if an un-directed graph is connected the above problem becomes simple and very little imagination is needed to answer the above question. Applying the Bucket Algorithm once on a graph with more than one connected component would tell us that the graph is not connected as all the vertices of the graph do not end up in the bucket. The vertices that do end up in the bucket belong to a single connected component. Applying the algorithm again with a new bucket would give us a new connected component, and so on and so forth. The number of times we have to apply the Bucket-Algorithm depends upon the number of connected components, and this would determine the worst-case time complexity.

5.3.2

Finding a Bridge in a Graph

A cut edge or bridge is one whose removal produces a graph with more connected components than the original. There are essentially two different problems here; it is the job of the instructor to at least identify them for those students who cannot visualize the solution immediately. The first problem is to check if a given edge is a bridge. This could be solved if we remove the given edge and then check the number of connected components in the resulting graph. What would be the resulting time complexity of this algorithm? The

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second problem is to find or locate a bridge in a given graph. Once the first problem is solved it should be a simple matter to handle it. How many times the Bucket-Algorithm is applied and what is the resulting worst-case complexity of the algorithm? j

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Figure 5.3.2: A cut edge or bridge is one whose removal produces a graph with more connected components than the original graph.

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The algorithms that solve this problem depend on how we define a tree. This in not only true for this problem but is true for a majority of problems. It highlights the fact that looking at various definitions or properties (which come from a study of graph theory) is sometimes extremely useful and it provides the seed for designing a number of very powerful algorithms. Solving a problem from different angles and then making a comparison is the single most important exercise for a student studying algorithms (Rawlins [11]).

5.4.1

Every Edge in a Tree is a Bridge

We know that a tree having n vertices consists of bare minimum number of edges, which makes it a connected graph. This implies that removing any edge would disconnect a tree. Thus every edge in a tree is a bridge. We already know how to check if a given edge is a bridge in a graph. The problem is thus reduced to repeatedly applying the algorithm designed to


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check if an edge is a bridge. The number of times we would have to do this and finding the resulting complexity is an interesting exercise by itself.

5.4.2

The Number of Edges in a Graph

We can define a tree in a number of ways. In fact, all of these definitions are equivalent implies. For example, a connected graph is a tree provided the number of edges in the graph is exactly equal to one minus the number of vertices in the graph, i.e., p − 1. The catch is that the graph should be connected otherwise the definition would not apply (why?). We know how to find if a given graph is connected using the Bucket Algorithm. So the problem is reduced to counting the number of edges. How complex is this problem? Is it possible to count the number of edges while we are checking if the given graph is connected? Would that perhaps reduce the complexity?

5.4.3

The Spanning Tree of a Tree

We know that a tree has the minimum number of edges required to connect a given number of vertices. A spanning tree of a given graph also satisfies this property, as it is a tree. Thus the spanning tree of a tree would be exactly the same tree. This definition or property can be used to design an algorithm to check if a given graph is a tree.

5.4.4

A Comparison

A comparison of all these algorithms would be extremely beneficial to the students if they are encouraged to work it out independently. Once they have the answers it would again be stimulating for them to compare their findings with their colleagues within the classroom. Encouraging and initiating interesting discussions and even heated debates is one of the most important responsibilities of a teacher: (s)he must simply coordinate and make sure that the interaction is moving in the right direction. Only when the students have gained confidence that they understand the basic problem and can find an efficient solution should we move to more complex problems such as finding whether a given graph is a forest.

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5.5


Finding a Spanning Tree of a Graph

The algorithm that we design to solve this problem depends on how we visualize the development of a spanning tree. We can start with the original graph and start with pruning or removing edges until the graph becomes a tree. Or we can start with no edges and start growing edges until we get a tree. It is also possible to identify some of the so-called cross edges (edges which are coming out of the Bucket), which would constitute the spanning tree. The resulting complexity would change dramatically depending upon the approach used. Each approach has its merits and demerits and the comparison itself is very enlightening especially because each approach has more advanced applications.

5.5.1

Cutting Edges

If we remove all redundant edges from a given graph and just keep edges essential to keep it connected the remaining graph would be a spanning tree of the given graph. This idea would give birth to an algorithm: Remove all edges that do not disconnect the given graph. What would be the worst-case time complexity of this algorithm?

5.5.2

Growing Edges

We start with no edges at all but with p isolated vertices. We add edges out of the edge pool of the graph such that the resulting graph remains a tree. This approach is opposite to the one discussed above: instead of pruning we are growing edges. In the earlier approach we should be careful and should not disconnect the graph. In the second approach we should be careful not to create cycles in the graph. In each case the Bucket Algorithm helps us. How many times we use the Bucket-Algorithm eventually decide the overall worst-case complexity.

5.5.3

Selecting Edges

While running the Bucket Algorithm, we might have noticed that every time we discover a new vertex it is because of a cross edge (step 3), and that the number of such cross edges would be exactly equal to p − 1. If we just keep a record of all such edges we might get the spanning tree of the given


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graph. How efficient would this be if compared with the algorithms described earlier?

5.5.4

Integrating Concepts and Discovering Algorithms

We show a concept map (Concept Map 5.2) which integrates different concepts, properties and the Bucket Algorithm. It is very much possible to discover most of the algorithms (that we have presented in this chapter) in class once we have become comfortable using and manipulating the Bucket Algorithm. We show in Concept Map 5.2, a number of systematic questions that if asked, will provoke the learner to discover a number of interesting techniques, which if refined will lead to a number of important algorithms.

5.6


There could be many non-isomorphic spanning trees possible for a given graph: each approach that we have described for finding a spanning tree of a graph was flexible and there was a lot of maneuvering possible within it, thus giving rise to different spanning trees. What if we find all distinct spanning trees of a given graph using any approach and then select the one with minimum weight? Why is this approach, which looks at all possible solutions and then selects the one of our choice, not feasible (although it is correct)?

5.6.1

Cutting or Growing Edges: A Krushkal’s like greedy algorithm

Each algorithm used to find a spanning tree in the previous section could be used with proper modification to find a minimum-spanning tree of a connected and weighted graph. While cutting edges we select the edge of maximum weight (provided it does not disconnect the graph), having first sorted the edges in descending order of weight. This would give rise to an algorithm very similar to Krushkal’s. See Figure. 5.6.3. Similarly, we can grow edges starting from the edge of minimum weight (making sure no cycle is created). It is very much possible to have multiple non isomorphic minimum spanning trees of a weighted graph but the weight

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Concept Map 5.2. The Concept Map and an iterative sequence (ascending order) of asking questions help student discover or understand a number of useful graph algorithms.

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of each tree would be the same (why?). Please note that here we are not using any fancy data structure since the objective is not to have a complicated design, unlike some textbooks.

5.6.2

Selecting Edges: A Prim’s like greedy Algorithm

While forming a spanning tree we can select any cross edge. In order to form a minimum spanning tree, we should try to include edges of less weight thus excluding those of higher weight. It follows that among all cross edges that we may select we should pick the one of minimum weight. Using this simple technique the Bucket-Algorithm can easily be modified to find a minimum spanning tree of a weighted directed or undirected graph (Figure 5.6.5). It is important that the minimum spanning tree problem is an optimization problem in which we intend to minimize the sum of weights of all edges in the spanning tree. In order to minimize the global sum, we are trying to minimize a local quantity. We are lucky this time: a so-called greedy approach is working optimally and is in fact optimizing the global sum also. However, although greedy approaches are relatively efficient (being based on local conditions only), they are not always optimal. We show here how the simple Bucket Algorithm can be reduced into a minimum spanning tree finding algorithm which resembles Prim’s Algorithm. Algorithm 18: Find Minimum Spanning Tree of a graph G input : A weighted Graph G output: Minimum Spanning Tree (MST) of Graph G 1 2 3

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It would be useful if the students were asked to prove that this greedy approach would actually find a minimum spanning tree. Without reading proofs given in the textbook they should come up with something of their own making. A lively discussion can be initiated to find the merits and demerits of individual work. They should also be asked to derive the time complexity

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of this approach. It would be useful if they compare this approach with Prim’s algorithm. In fact, the two approaches look identical. However the time complexity of Prim’s algorithm (as stated in most textbooks) is better. Why? The reason is in fact more exciting because Prim’s algorithm is not just greedy, there is something else, something magical which cuts down the time complexity for not so obvious reasons. What is that magic? How and why it is working? Can this magic be used elsewhere and under what conditions? We shall discuss it later in this chapter.

5.6.3

A Panoramic Picture of various MST Finding Techniques

It will be useful from a learning perspective if we make a comparison between various minimum spanning tree finding techniques. We should understand both the differences as well as the similarities. We show three different techniques in action on the same weighted graph in Fig. 5.6.6. The top diagrams in this figure show how we grow a minimum spanning tree in a single bucket (this is a Prim’s like algorithm). The middle diagrams show a minimum spanning forest growing up in various buckets using a Kruskal’s like algorithm. In the top diagrams we select an edge of minimum weight out of all edges coming out of the single bucket in which we have our minimum spanning tree growing. In the middle diagrams we can visualize that each vertex is initially in a separate bucket; we select an edge out of all edges (coming out of all buckets) in the graph and thus this technique grows various spanning trees in different buckets. In this algorithm (middle diagram) it is essential to check that the new edge that is selected should not form a cycle with the previously found minimum spanning tree. In the top algorithm there was no explicit need to check that this condition is true; the process of selecting the desired edge out of all edges coming out of the single bucket makes sure that no cycle is generated. How about if we start Prim’s algorithm from every vertex? Thus we may be able to avoid an expensive check (that the resulting graph should not be cyclic) each time we insert a new edge (as in Prim’s algorithm) and at the same time we can exploit the inherent parallelism which was lacking in Prim’s algorithm? We initially put every vertex in a separate bucket; for each bucket we select the edge of minimum weight coming out and thus grow various minimum spanning trees in different buckets. This technique (based

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Figure 5.6.6: Pictures of different Buckets while finding a Minimum Spanning Tree of a weighted graph using different techniques including the Boruvka’s Algorithm.


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on the Boruvka’s algorithm) is shown in action (on the same weighted graph) in the bottom diagrams of Fig. 5.6.6. It is interesting to note the following about the three minimum spanning tree finding algorithms: 1. In case of Prim’s algorithm we need not be worried about the formation of a cycle; the way we move forward makes sure that no cycle is formed. This is because of the fact that we consider edges coming out of the bucket only; if, however we consider all edges incident on a vertex inside a bucket then we do need to be worried about the formation of a cycle. Prim’s algorithm terminates when all vertices come inside the bucket. 2. In case of Kruskal’s algorithm either we are growing a minimum spanning tree (by inserting edges starting from low cost ones) or we are cutting edges starting from high weight edges. In the first case we have to make sure that no cycle is created and in the second case we should be worried about disconnecting the graph. The algorithm terminates when all edges have been considered. 3. In case of Boruvka’s algorithm, we grow a minimum spanning tree in each bucket just like Prim’s algorithm. We try to exploit the inherent parallelism in this scheme where each processor is running a Prim’s like algorithm on each vertex. The big question is do we need to make an expensive cycle check each time we insert an edge? It will be interesting to find under what conditions we need a cycle test and when we do not need such a test? This algorithm terminates when the number of edges selected becomes exactly equal to one less than the number of vertices in the graph. 4. If all edge weights in a graph are unique then we do not need a cycle test in Boruvka’s algorithm. But we do need a cycle test in case the edge weights are not unique. Why? Do we need a cycle test in Kruskal’s algorithm in case all edge weights are unique? We do not need a cycle test in Prim’s algorithm in spite of the fact that edge weights are not unique? Why? 5. It will be useful to formally prove that in case of Prim’s algorithm we do not need a cycle test and still no cycles will be formed. It will also

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5.6.4

The Maximum Spanning Tree Problem

All the minimum spanning tree algorithms described here can easily be modified in order to find the maximum spanning tree of a weighted graph G. A small change in the algorithm can do the job. A curious reader may have realized by now that positive or negative edge weights in a weighted graph does not create any problem while evaluating the minimum or maximum spanning tree of a graph. We shall address these and other related issues in a problem set. Problem Set 5.1. Problem 5.1.1. We can make a small change in one of the Minimum Spanning Tree Algorithms in order to convert it into a Maximum Spanning Tree Algorithm as follows. Apply this algorithm to the graph shown in Fig. 5.6.7, and find the maximum spanning tree while showing the contents of the Bucket B after each step. Algorithm 19: Find Maximum Spanning Tree of a graph G input : A weighted Graph G output: A Maximum Spanning Tree (MaxST) of Graph G 1 2 3

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Problem 5.1.2. Apply the above algorithm to the following graph G consisting of positive as well as negative edge weights. Do you think there will be any complication in finding a maximum spanning tree if the edge weights are negative (or if they are positive). We shall repeat this problem while finding the shortest path (or the longest path) in a graph having negative edge weights. We shall discuss that the shortest path problem becomes complicated if all edge weights are not positive. Do we face a similar problem while finding a minimum spanning tree of a graph having negative edge weights?


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Figure 5.6.7: A Graph G having negative edge weights for problem set.

5.7


It is possible to find a path between two vertices provided the graph is connected. Now instead of checking whether the graph is connected or not, we better check if the two given vertices belong to a single connected component. If we keep moving along the edges connecting one vertex to another within the graph, a time would come when we would reach our destination. What is wrong with this approach? If there are cycles in the graph it is possible that we never reach our destination. What if there are no cycles in the graph – what if we first make a spanning tree of the graph? Even now it would be difficult to find a path, since we might have to do a lot of backtracking.

5.7.1

Cutting Edges

If we remove all redundant edges from a given graph and just keep the edges essential to keep the two vertices connected, the remaining graph would be a “straight forward” path between the two vertices in the given graph (Fig. 5.7.1). What would be the worst-case time complexity of this algorithm? Note that we have used a similar technique to find a spanning tree of a graph. It would be useful to pinpoint the similarities as well as the differences.

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Selecting Edges

Does the problem become simpler if we first find the spanning tree of the given graph? Now if we start moving from the given vertex to the destination vertex, would it be less confusing? Perhaps, but again we may start our journey in the wrong direction and would have to backtrack. Students should experience this confusion and the resultant backtracking. Suppose we apply the Bucket Algorithm starting with the given vertex: the spanning tree thus formed would originate from the given vertex since the given vertex would be the root. We also keep a record of the parent of every vertex in the spanning tree. With this additional information would it be easier to find a path from the given vertex, now the root, to the destination vertex? The answer is still “no” because a parent may have multiple children, and thus there still exist many diversions. However if we start from the destination vertex and keep selecting the parent vertex, we would eventually reach the root without any confusion (Fig. 5.7.2).

5.8


If all edges in the graph were to have the same weight, would the path, found using the algorithms of the previous section, be a shortest path? If not then what should be done to achieve our objective? Note that it is easier to find a shortest path in a graph with uniform edge weights, so first we should solve this problem (which is simpler) before attacking a more complex one. We shall come back to this problem after discussing graph traversal techniques in a later section. Now assume that the edge weights are different. Do we need a different algorithm from the one used to find a shortest path in a graph with uniform edge weights? Why? If somehow we remove edges of higher weights from the graph without disconnecting the two given vertices, would the problem become simpler? What if we first find a minimum spanning tree of a graph and then move backwards from the destination to the source vertex in order to find shortest paths as described earlier? It is obvious from Figure 5.8.1. that a minimum spanning tree of a weighted graph does not always provide shortest distances from a given vertex. It will still be interesting to investigate how a shortest path algorithm resembles and at the same time differs from a minimum spanning tree algorithm. We assume that we are finding shortest paths from a single vertex to all other vertices.

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Figure 5.8.1: The middle diagram shows a minimum spanning tree (MST) of a weighted graph G shown in the left diagram. The right diagram shows a shortest path spanning tree (SST) of the same graph. The shortest distances (as shown in red color) from vertex a provided by the two spanning trees are different as indicated in these diagrams. The weight of the minimum spanning tree is also different from that of the shortest path spanning tree.

5.8.1

Dijkstra’s (like) Algorithm

We produce both the algorithms side by side; it will become obvious that the two algorithms are derived from a common ancestor – the Bucket Algorithm. There is indeed a delicate difference between the two – this difference should make us understand why minimum spanning tree algorithm fails to find a shortest path spanning tree and why a shortest path spanning tree algorithm fails to find a minimum spanning tree of a weighted graph G. It should also be kept in mind that the minimum spanning tree can easily be modified to find a maximum spanning tree of a graph while it is not possible to do so in case of finding a longest path in a graph G (with positive edge weights). The step by step working of the two algorithms is shown in the figure below (Fig. 5.8.2). It is obvious that initially the two algorithms produce similar results but then they depart ultimately producing different results.

5.8.2

Discussion on Dijkstra’s (like) Algorithm

It was interesting to compare the described shortest path algorithm with the corresponding minimum spanning tree algorithm – but at the same time

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Algorithm 20: Find Minimum Spanning Tree of a graph G input : A weighted Graph G output: A Minimum Spanning Tree M ST 1 2 3

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Algorithm 21: Find shortest distance of every vertex from a given vertex x in G & also the shortest path spanning tree of G from x input : A weighted Graph G, a vertex x output: Shortest Distance Dist(i) of every vertex i from x 1

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it will be a learning experience if we look into the working of the shortest path algorithm under conditions when it fails to provide optimal results. But before we do that let us discuss some of the salient features of this algorithm. 1. In the first iteration of the while loop, we consider vertices at a distance of one edge. Out of all such vertices (at a distance of one edge from x), we select a vertex i which is at a shortest distance from x, and put it in Bucket B; the edge (x, i) goes in the shortest path spanning tree. This action is tantamount to a claim that we have found a shortest path from vertex x to vertex i. If all edge weights are positive then this claim will be right – it will, however, be violated if some edge weights in G are negative. The cost of this action is equivalent to making p − 2 comparisons. 2. In the second iteration of the while loop, we consider vertices which are directly connected to x, and which are indirectly connected to x through vertex i. Remember vertex i is the last vertex which went into the Bucket B. Out of all these vertices (some of which are at a distance of one edge and some at a distance of two edges), we select a vertex j which is at a shortest distance from vertex x. The vertex j goes in the Bucket B and the corresponding edge goes in shortest path spanning tree. Again this action means that we have found a shortest path of vertex j from vertex x. The cost of this action is equivalent to making p − 2 + p − 3 = 2p − 5 steps. In order to assert that Dijkstra’s like algorithm is able to find correct results let us first present a scenario where the above mentioned algorithm fails to find shortest paths. Remember that this algorithm works on a greedy strategy – it makes decisions on the basis of local conditions – but produces optimal results on a global basis provided all edge weights in a graph are positive. We show a directed graph in the Figure 5.8.3 where some edge weights are negative. It is obvious from this figure that Dijsktra’s algorithm fails to produce correct results. The same weighted graph and its (correct) shortest path spanning tree is shown in Fig. 5.8.4. We show another weighted graph in the right diagram of this figure. This graph is different from the one shown in the left (and middle) diagram in the following ways:

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Figure 5.8.2: Pictures of the Bucket B while finding a Minimum Spanning Tree (left diagrams) as compared to the pictures while finding a Shortest Path Spanning Tree (right) from vertex a.


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Figure 5.8.3: Pictures of the Bucket B while finding a shortest path spanning tree using a Dijsktra’s like algorithm. The final answer as shown in the bottom right diagram is incorrect.

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1. The shortest distance path from vertex a to any other vertex is a simple path – no edge or vertex is repeated in this path. The magnitude of this shortest path is finite. In other words the shortest distance between any two vertices does not reduce if we move in a cyclic path. The above observation is true for the left or the middle graph in the figure below (Fig. 5.8.4). Please note that there are negative weight edges in this graph but no negative weight cycles: a negative weight cycle is a cycle in a graph where the net sum of the weight of the edges in the cycle is less than zero. 2. In the graph shown in the right diagram (Fig. 5.8.4) the above observations are not true. The shortest distances of the red vertices keep on decreasing if we move in a cyclic path as shown in red color with the result that the shortest distance of such vertices ultimately approaches minus infinity. This happens in any graph where there are negative weight cycles.

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Figure 5.8.4: The left diagram shows an (incorrect) shortest path spanning tree produced by a Dijkstra’s like algorithm. The middle diagram shows the correct shortest path spanning tree for the same graph. The right diagram shows a graph where shortest distances of red vertices from vertex a keep on decreasing if we move in a cyclic path as shown in red.


5.8.3

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The Shortest Path Problem Redefined: The kedge Shortest Path Problem

In order to overcome the above mentioned complication we redefine the shortest path problem as follows. Instead of finding the shortest path between two vertices we intend to find k-edge shortest paths from a given vertex to every other vertex in a directed graph containing negative edges – the graph may even contain negative weight cycles. A k-edge shortest path (between two given vertices) requires that the path should be shortest but it should not consist of more than k edges. The weighted adjacency matrix of the given directed graph provides all one edge paths from vertex x. If there are no parallel edges in the graph then these paths will also be the shortest paths from vertex x to every vertex adjacent to x. Such one edge paths are shown in the top left diagram in Figure 5.8.5. Once we have the 1-edge shortest paths – we can convert them into 2-edge shortest paths and then into 3-edge shortest paths according to the algorithm described below. It is evident from the figure below – as we move from k-edge to (k + 1)-edge shortest paths there is some improvement in the length of a shortest path; if no improvement takes place in any path then we stop; now we claim that we have found the shortest distance of every vertex from x. For the graph shown in the Fig. 5.8.5, this happens when k = 5; now if we increase k to 6 there is no improvement in any shortest distance. Algorithm 22: Find (k+1)-edge shortest distance of every vertex from a given vertex x in a weighted directed graph G. input : A weighted directed Graph G; a vertex x; k-edge shortest distance of every vertex i from vertex x denoted by Distk (i) output: (k + 1)-edge Shortest distance of vertex i from vertex x, it is denoted by Distk+1 (i) 1 2 3

for i = 1 to p do for j = 1 to p do Distk+1 (i) = min{Distk (i), w(i, j) + Distk (j)};

Let us explore the consequences of the presence of negative weight cycles in a graph like the one shown in the right diagram of Fig. 5.8.4. This graph is reproduced in Fig. 5.8.6 where we find different edge path from vertex

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Figure 5.8.6: We show k-edge shortest paths from vertex a to every other vertex in graph G. When k changes from 1 to 6, the shortest paths of some vertices change. But when k goes beyond six then we are caught in a negative cycle – the shortest path of three vertices (belonging to a negative weight cycle and shown in red color) keeps going down, and this situation never becomes stable.

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x. When k increases the shortest path distances go down. Thus k keeps increasing and the distances keep going down. This situation never becomes stable because of the presence of a negative weight cycle comprising three vertices shown in red color. The shortest distances of the rest of the vertices become stable as soon as k approaches 5.

An Important Conclusion If there are no negative weight cycles in a directed graph consisting of some negative weight edges then we can use our k-edge shortest path technique to find shortest paths. As soon as all the k-edge shortest paths become stable with increase in k, we claim that we have found the shortest paths. Please note that the resulting shortest paths will be simple paths; no edge or vertex is repeated in these paths. It means that k would be less than or equal to p − 1 in case there are no negative weight cycles in the graph.

5.8.4

The k-edge Longest Path Problem

The k-edge longest path problem is quite similar to the k-edge shortest path problem; it is just like maximum spanning tree versus minimum spanning tree problem. As we increase k the longest path of certain vertices increases; it may become stable after some time as shown in Fig. 5.8.7 where we operate on the same graph of Fig. 5.8.6. If, however, there are positive weight cycles in the graph then we shall be caught in an infinite loop and the longest distances (of at least some vertices) will keep increasing with increase in k. This is quite expected in view of our prior experience: if there are negative weight cycles in a graph then the shortest path (of certain vertices) keep increasing with k. Problem Set 5.2. Problem 5.2.1. Please recall the weighted graph of Fig. 5.8.5, the same weighted graph is reproduced below in the left diagram of Fig. 5.8.8. Remember if we increase k beyond 5 while finding k-edge shortest paths then no (shortest) distance (with respect to vertex a) changes with k. We now change the magnitude of the weight associated with the edge (f, g) from 4 to 40 as shown in the middle diagram of this figure. Find k-edge shortest paths with respect to vertex a while k changes from 1 to 10. Find how and at what value of k, the shortest paths becomes stable.

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Problem 5.2.2. We show another weighted graph as shown in the right diagram of Figure 5.8.8. We need to explore if shortest distances of all vertices with respect to vertex a become stable when k increases while finding k-edge shortest paths. Please note that there is a big negative weight equal to -40 associated with the edge (d, b). Find k-edge shortest distances of all vertices with respect to vertex a; vary k from 1 to 11. Find the value of k at which the shortest distances become stable, i.e., if we further increase k then there is no change in the shortest distance of any vertex with respect to vertex a. Problem 5.2.3. In spite of a large negative weight in the graph shown in the right diagram of Fig. 5.8.8, the shortest distances become stable at a finite value of k? Why? What do you think are necessary and sufficient conditions for the shortest distances to become stable for a finite value of k in a weighted graph? Problem 5.2.4. Describe an efficient algorithm (based on finding k-edge shortest distances) to find if a directed graph contains negative weight cycles. Problem 5.2.5. Is it possible to use the above algorithm to find if an undirected graph contains negative weight cycles? Problem 5.2.6. Find k-edge longest paths for each vertex with respect to vertex a in the graph shown in the left diagram of Fig. 5.8.8. Find if the longest paths become stable with increasing value of k. Problem 5.2.7. Repeat the above problem for the graph shown in the right diagram of Fig. 5.8.8. Problem 5.2.8. What do you think are necessary and sufficient conditions for the longest distances to become stable for a finite value of k in a weighted graph? Problem 5.2.9. Comment on the claim that the longest path problem is a hard problem while the shortest path problem is a solvable problem.

5.8.5

The Shortest Path Problem in Undirected Graphs with Negative Weights

We consider the shortest path problem in undirected graphs with negative edge weights but no negative weight cycles. We already know that if there


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are no negative weights then a simple greedy strategy (like that of Dijskrta’s Algorithm) will solve the shortest path problem in undirected graphs. With negative edge weights (but no negative weight cycles) we expect that our kedge shortest path technique will work with the hope that the k-edge shortest paths will stabilize for a finite value of k and we shall get the optimal answer. Remember that the so called technique was designed for directed graphs and in order to apply it to undirected graphs we have to first convert the undirected graph into a directed one. We show an undirected graph in left diagram of Fig. 5.8.9. The correct shortest paths & distances with respect to vertex a are indicated in the middle diagram. The undirected graph is converted into a directed graph as shown in the right diagram of this figure. 0

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Figure 5.8.9: We show an undirected graph with negative edge weights but no negative weight cycles in the left diagram. The shortest distances with respect to vertex a are indicated in the middle diagram. The undirected graph is converted into a directed graph as shown in the right diagram. When we apply our k-edge shortest path algorithm to the directed graph (shown in the right diagram of Fig. 5.8.9) we run into a complication; we have created a two-edge negative weight cycle in the directed graph. The working of the k-edge shortest path technique on this graph is shown in Fig. 5.8.10. It is quite evident that k-edge shortest distances do not stabilize in this graph with increase in k, and we get incorrect answers for shortest distances. The challenge is how to handle this complication? Please understand that we are limiting our focus on a directed graph which is derived from an undirected graph as shown in Fig. 5.8.9. Let us represent a k-edge shortest path of any vertex j from vertex x by SP athk (j). Each such shortest path starts from vertex x and terminates

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at vertex j; assume that the last vertex in this shortest path is yj before terminating at j. Now we are in a position to find (k + 1)-edge shortest path of a vertex i from vertex x in a graph using the following algorithm. Algorithm 23: Find (k + 1)-edge shortest path & distance of every vertex from a given vertex x in a weighted directed graph G. input : A weighted directed Graph G; a vertex x; k-edge shortest path of every vertex j from vertex x denoted by SP athk (j); Its weight is represented by Distk (j); The last vertex in a shortest path SP athk (j) is vertex yj before terminating at j. output: (k + 1)-edge Shortest path of vertex i from vertex x, it is denoted by SP athk+1 (i); its weight is equal to Distk+1 (i) 1

We need to find SP athk+1 (i): We consider k-edge shortest path for every vertex j in the graph except for j where yj = i. Out of the remaining paths we claim that Distk+1 (i) = min{Distk (i), w(i, j) + Distk (j)} ;

Problem Set 5.3. Problem 5.3.1. Apply the above Algorithm to the graph shown in Fig. 5.8.9 and verify that it provides correct results for values of k in the range of 1 to 5. Check if k-edge shortest paths stabilize for a finite value of k. Compare your results with the correct shortest paths given in the middle diagram of Fig. 5.8.9. Problem 5.3.2. Apply the above Algorithm to the graph shown in Fig. 5.8.11 for values of k in the range from 1 to 8. Please note that this graph contains negative weight cycles. Problem 5.3.3. Consider an undirected graph containing negative weight cycles. Will it be possible to provide a warning that the given graph contains negative weight cycles? What modifications are needed in our existing algorithms to solve this problem?

5.9


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Figure 5.8.11: We show an un-directed graph (left diagram) with a negative weight cycle. The undirected graph is converted into a directed graph as shown in the right diagram. move in a systematic manner to ensure that we do not miss out any vertex belonging to the same connected component [3]. You might have noticed that the Bucket-Algorithm is essentially a graph traversal algorithm. However, in most of the current textbooks, you would only find some very specific techniques like the Breadth and Depth First Search traversal algorithms – more specialized and less flexible than our Bucket Algorithm.

5.9.1

Traditional Techniques & the Bucket-Algorithm

It is interesting to note that the Breadth as well as Depth First Searches are two different implementations of the Bucket-Algorithm (Fig. 5.9.1). While introducing this algorithm we purposely did not disclose the implementation details ignoring the underlying data structure required to program the algorithm. The objective was to highlight the basic idea and initially suppress the programming details. Baase [2] uses JAVA to describe algorithms and this may be one reason why the book is relatively difficult to read even if students have prior knowledge of the language. Cormen [17] and Skiena [16] use a pseudo programming language and operate at a slightly higher level. The Bucket Algorithm is simple because it is more abstract and flexible.

5.9.2

The Underlying Data Structure

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5.10

Some Graph Theoretic Claims

We make the following claims about the BFS traversal in an un-directed and connected graph G: Claim 5.10.1. In the first iteration of the BFS traversal, all vertices at a distance of one edge from the starting vertex are selected (or goes in the Bucket); in the k th iteration of the BFS traversal, all vertices at a distance of exactly k edges from the starting vertex will be discovered. Please see the Figure 5.10.1 below. Claim 5.10.2. BFS spanning tree of a graph G is a minimum distance spanning tree in terms of number of edges between the starting vertex and any other vertex in graph G.


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Claim 5.10.3. If there is no edge in G other than the ones in the BFS spanning tree of G then G is a tree provided G is a connected graph. Claim 5.10.4. If there is an edge in G other than the ones in the BFS spanning tree of G then this edge will connect two vertices x and y in the BF S spanning tree such that Dist(x) − Dist(y) is either zero or one whereas Dist(x) is the distance of vertex x from the starting vertex in the BFS spanning tree. Please see Fig. 5.10.2 Claim 5.10.5. If there is an edge in G other than the ones in the BFS spanning tree of G connecting vertex x and vertex y in the BFS spanning tree such that Dist(x) − Dist(y) is zero then there will be an odd cycle in graph G. (Remember Dist(x) is the distance of vertex x from the starting vertex in the BFS spanning tree). Please see Fig. 5.10.2. j

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Figure 5.10.2: BFS spanning tree of a graph G is shown. The edge between vertex h and vertex j creates an odd cycle in the graph as both these vertices are at the same distance from vertex a. As we know the distance of vertex h from vertex a, we also know a path between vertex h and vertex a as shown in green color in the top right diagram. We also know a path between vertex j and vertex a; these two paths along with edge j, h makes a cycle. It will be an odd cycle as vertices h and j are at the same distance from vertex a. Claim 5.10.6. If there is an edge in G other than the ones in the BFS spanning tree of G connecting vertex x and vertex y in the BFS spanning


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5.11

Shortest Path Algorithms

We have already done the following in earlier sections of this chapter. 1. Modifying the Bucket algorithm to find a minimum spanning tree (MST). 2. Modifying the Bucket algorithm to find a shortest path spanning tree (SST) or shortest distances in a weighted graph from a given vertex. 3. Finding k-edge shortest or longest paths in a weighted graph. On the basis of that precious prior knowledge we can design interesting shortest path algorithms. The interesting thing about this style of design is that we shall be using a single building block (sometimes the Bucket Algorithm and some times a 2-edge Shortest path Algorithm) to design or describe an algorithm. We intend to do the following in this section: 1. Analyze the existing minimum spanning tree & shortest path finding algorithms and improve their efficiency as far as possible. This is about single source shortest path algorithms assuming that all edge weights are positive in the given weighted graph. We shall further modify the Bucket algorithm to achieve our objectives. 2. Design of a shortest path algorithm for directed acyclic graphs with negative as well positive edge weights. We shall again modify the Bucket algorithm to achieve our objective. 3. Design of single source shortest path algorithms based on our prior knowledge of finding k-edge shortest paths from a given vertex. Here we assume that there may be negative edge weights in the given weighted directed graph. We also assume that the directed graph may have cycles. 4. Design of all pair shortest path algorithms including the slow all pair, faster all pair, Floyd-Warshall and Johnsons shortest path algorithms.


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We copy here the modified Bucket algorithm designed in the last section to find shortest paths from a given start vertex. We shall study this algorithm using different tools before improving its time complexity. We introduce a visual tool known as ”opening up the graph” as shown in Fig. 5.11.1. We apply the shortest path algorithm and find the shortest distances from vertex a in this graph where a = 1 also shown in Fig. 5.11.1. The visual tool demonstrates at what stage and when a vertex enters the Bucket while the shortest path algorithm moves forward. As soon as a vertex enters the bucket its color changes from blue to orange and then its distance from the start vertex cannot change - it is fixed and finalized. The shortest distances from vertex 1 as found by this algorithm are also indicated in the bottom diagram. Algorithm 24: (Crude-Dijkstra): Find shortest distance of every vertex from a given vertex a in G & also the shortest path spanning tree (SST ) of G from vertex a input : A weighted Graph G, a vertex a output: Shortest distance Dist(k) of every vertex k from vertex a 1 2

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Put vertex a in Bucket B; Initialize Dist(a) = 0, and Dist(k) of every other vertex k from vertex a equal to ∞; while there are edges coming out of the Bucket B do Select the edge for which Dist(j) + w(j, k) is minimum where vertex j is in B and vertex k is outside the Bucket; Dist(k) = Dist(j) + w(j, k); Put vertex k in B; Put edge (j, k) in SST ;

Food for thought: The graph shown in Fig. 5.11.1 has negative edge weights. That is why the shortest paths found by the crude Dijkstra’s algorithm are not correct. We have already witenessed that this algorithm does not always provide correct results for graphs with negative weight edges. But we also know that this very algorithm provides correct results when all edge weights are positive.

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Figure 5.11.1: We use a new tool ”Opening up the graph” as shown here. It shows at what stage (edge distance) a vertex enters the Bucket while executing crude shortest path algorithm. It also shows how and when the distance of a vertex changes from the start vertex. As soon as a vertex enters the bucket its color changes from blue to orange and then its distance from the start vertex cannot change - it is fixed and finalized.


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Figure 5.11.2: The number of edges coming out of the Bucket B at any time will be proportional to p2 under worst case conditions. In fact there are two complications with this algorithm: 1. It does not provide correct results for negative edge weights. Although it provides correct results for positive edge weights. 2. Its time complexity is too high. We shall reduce the time complexity in the following refined version of shortest path algorithm. The issue of negative edge weights will be handled in the next sub-section. The number of edges coming out of the Bucket B at any time is proportional to p2 as illustrated in Fig. 5.11.2. This requires as many comparison steps to move forward in the while loop. As this loop runs as many times as p so the overall time complexity will be p3 of this algorithm. Somehow we should reduce the number of relevant edges coming out of the Bucket to as small as p - this will automatically cut down the time complexity of this single source shortest path algorithm to p2 . Surprisingly this can be done with a slight modification in the crude algorithm as shown below. What we essentially do is to change the graph itself at each step as shown in Fig. 5.11.3. Now we consider the edges coming out of the Bucket from (only) vertex a instead of edges coming out of the Bucket from all vertices in the Bucket. The outcome of this step is that the number of edges (which really matters) coming out of the bucket B from vertex a is limited by p and not p2 . It is interesting to note how the shortest distances are provided in the output in this new algorithm. The shortest paths, however, could not be found

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Algorithm 25: (Refined-Dijkstra):Find shortest distance of every vertex from a given vertex a in weighted graph G input : A weighted Graph G, a vertex a output: Modified adjacency matrix of graph G in which the row corresponding to vertex a gives shortest distances of every vertex from vertex a. 1 2 3

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without an extra effort using this refined algorithm. It is important to note that we may be able to make similar modifications in the crude minimum spanning tree algorithm to make it more efficient. The refined version of that algorithm is given below - its time complexity also reduces from O(p3 ) to O(p2 ). Again it will be interesting to understand that this algorithm will provide the weight of the minimum spanning tree of a weighted graph - not the minimum spanning tree itself as shown in the diagrams below. Algorithm 26: Find weight of MST of a weighted graph G input : Adjacency matrix of a weighted Graph G output: Modified adjacency matrix of graph G in which the row corresponding to vertex a gives weight of MST 1 2 3

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We can recover shortest paths by adding a parent table as shown in Fig.


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Figure 5.11.4: We modify the graph as we move forward in the minimum spanning tree algorithm. 5.11.6 & 5.11.7. You can yourself make suitable modifications in the algorithm according to these modifications. Food for thought: Given a parent table and shortest distances as shown in Fig. 5.11.7 determining the actual shortest paths in a directed graph is an interesting problem. We have provided a hint in the same figure. See how by increasing memory or space requirements we can reduce the time complexity of an algorithm. If we can recover shortest paths then we should also be able to recover a minimum spanning tree using the refined minimum spanning tree algorithm.

5.11.2

Single Source Shortest Path Algorithms for Directed Acyclic Graphs

We know that by definition a directed acyclic graph contains no cycles. We can find shortest paths in such graphs in the presence of negative edge weights and we can also find longest paths in the presence of positive edge weights. We can even solve the Hamiltonian Path problem in this very restricted class


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Figure 5.11.5: The original graph is shown in the top diagram. Shortest distances with respect to vertex a are indicated in the bottom left graph. Weight of the minimum spanning tree is indicated in the bottom right graph. Please note that the shortest paths and minimum spanning tree are not easily available here.

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Figure 5.11.6: We need to add a parent table array in order to find shortest paths in addition to shortest distances.

Figure 5.11.7: How to recover shortest paths with the help of a parent table?


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of graphs. The Bucket Algorithm (described earlier) can easily be modified to create an algorithm which can find shortest paths from any given vertex in a very efficient manner. The intuition of this algorithm comes from the observation that we can always arrange the vertices of a DAG such that all edges in the DAG move from left to right as shown in the figure below. This observation is explained in detail in Chapter 8 while discussing directed acyclic graphs. If we need to find shortest paths from a source vertex a then we should put that vertex in the bucket first. The crucial question is which will be the next vertex to go in the bucket and on what basis. An answer to this question will not only determine the character of this algorithm - it will also decide its time complexity. Recall how we select the next candidate vertex which enters into the bucket in case of Dijkstra like algorithm or in case of Prim’s like algorithm. In both these algorithms we made certain comparisons to select the next entrant into the bucket. Please see the concept map in Fig. 5.11.9.

Figure 5.11.8: A directed acyclic graph D shown in the top diagram. In the bottom diagram we arrange its vertices such that all edges move from left to right. The numbers inside each vertex is the start time and finish times obtained during a depth first search of the directed graph. So coming back to the important question: which vertex (and on what basis)

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Figure 5.11.9: A concept map depicting which vertex should next enter the bucket in different algorithms. should enter the bucket after putting the start vertex in the bucket? The interesting observation is that after arranging vertices of a DAG (such that all edges move from left to right) the next vertex to enter the bucket has already been decided. It is the next left vertex in the new arrangement of the vertices of the graph. We need no comparisons or extra steps to make this decision. The corresponding algorithm is described below. Its working is shown in Fig. 5.11.10. It will be an interesting challenge to derive the time complexity of this elegant shortest path algorithm.


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Algorithm 27: Find shortest distance of every vertex from a given vertex a in a directed acyclic graph D input : A directed acyclic and weighted Graph D, a vertex a output: Modified graph D in which the weighted edges coming out of vertex a provides shortest distances from this vertex. 1 2 3 4 5 6

Arrange vertices of the DAG such that all edges move from left to right Put given vertex a in Bucket B while there is an edge going out of the Bucket B do Select the next right vertex j; Put vertex j in B for every edge coming out of B from vertex j do assign min{w(a, k), w(a, j) + w(j, k)} to w(a, k)

Food for Thought 1. Consider the directed acyclic graph shown in Fig. 11.10. We need to find shortest paths from vertex a in this graph. The graph is already drawn such that all edges are going from left to right. The next vertex which goes in the bucket is vertex b in spite of the fact that the edge joining vertex a with vertex b is the heaviest edge in this graph. If we had applied Dijkstra’s algorithm to this graph then it would have selected vertex c (instead of vertex b) as the edge joining vertex a with vertex c having the minimum weight. Do you think Dijkstra’s like algorithm will also find correct shortest paths in a DAG? And at what cost? 2. Assume that we have a DAG with multiple source vertices. Can we still apply this algorithm without any modification to find shortest paths? How about if we need to find shortest paths from a vertex other than the source vertex in a DAG? 3. How about if there are negative weight edges in the graph? Would this algorithm still provide correct results? 4. How about if we need to find longest paths instead of shortest paths from a given vertex in this or any other directed acyclic graph? Which vertex will next end up in the bucket? What changes are needed in this algorithm to find longest paths? Please see figure 5.11.11.

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5. Can we use a similar algorithm to find a Hamiltonian Path in a directed acyclic graph provided it exists?

Figure 5.11.10: We need to find shortest distances from vertex a in this graph. The vertices of the DAG have been arranged such that all edges move from left to right. Now the next vertex to move in the bucket will always be the next left vertex. Weight of every edge not indicated in the diagram is equal to 1.


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Figure 5.11.11: We need to find longest distances from vertex a. The vertices of the DAG have been arranged such that all edges move from left to right. Now the next vertex to move in the bucket will always be the next left vertex. Weight of every edge not indicated in the diagram is equal to 1.

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Single Source Shortest Path Algorithms for directed graphs with negative edge weights

We have discussed directed acyclic graphs and the ease with which we can find shortest paths in such graphs in the last section. If a graph is cyclic then we cannot use the simplicity and elegance of this algorithm. Then we need to revert back to algorithms discussed earlier. Remember we have already reduced the time complexity of our modified shortest path algorithm from p3 to p2 . That was Dijkstra’s like algorithm which can handle positive edge weights in a graph with cycles but does not provide us with correct results if there are negative edge weights in the graph. Now is the time to get rid of the bucket as it is hindering our way to handle negative edge weights. Let us recall the (i + 1)-edge shortest path algorithm described earlier in this chapter. It has been copied below. Algorithm 28: (Bellman-Ford Building Block-A): Find (i + 1)-edge shortest distance of every vertex k from a given vertex a in a weighted directed graph G. input : A weighted directed Graph G; a vertex a; i-edge shortest distance of every vertex k from vertex a denoted by Disti (k) output: (i + 1)-edge Shortest distance of vertex k from vertex a, it is denoted by Disti+1 (k) 1 2 3

for each (terminating) vertex k do for each (intermediate) vertex j do Disti+1 (k) = min{Disti (k), w(j, k) + Disti (j)}

A 2-edge shortest path algorithm Let us make this building block consistent with our earlier policy of modifying the graph as we move forward in the algorithm. This will make it friendlier to use as a building block. In this building block we just convert the one edge distances into 2-edge shortest distances as given in the following algorithm. We assume that all distances are to be measured and minimized with respect to a start vertex a. The terminating vertex is k and the intermediate vertex is indicated by vertex j. We shall follow this terminology throughout this section. Here w(a, k) represents the initial one edge path of vertex k from


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Algorithm 29: (Bellman-Ford Building Block-B): Find 2-edge shortest distance of every vertex k from a given vertex a in a weighted directed graph G. input : A weighted directed Graph G; a vertex a; 1-edge shortest distance of every vertex k from vertex a provided by the row corresponding to vertex a in the adjacency matrix G output: Modified graph G in which row corresponding to vertex a provides 2-edge shortest distance of every vertex k from vertex a 1 2 3

for each (terminating) vertex k of graph G do for each (intermediate) vertex j of graph G do Find a 2-edge path from vertex a to k passing through j, and assign min{w(a, k), w(a, j) + w(j, k)} to w(a, k)

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4 5 ∞ 1 2-edge Shortest paths?

Figure 5.11.12: Execution of the 2-edge shortest path algorithm from start vertex a. Edge weights not shown are equal to 1. Please see the accompanying figure (Fig. 5.11.12). Here a = 1. We vary j and k for the entire vertex range and find the 2-edge shortest distances of every vertex from vertex a. The 2-edge shortest distances are indicated in the modified graph shown in the bottom right corner.


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Figure 5.11.13: We execute the k-loop first and then the j-loop in the 2-edge shortest path algorithm. Here intermediate vertex j is fixed at 4 and the destination vertex k is varied first. Please note that it is immaterial whether you execute the j-loop first and the k-loop later or vice versa. The end result remains the same. This switching of two loops is an exciting idea in the development of shortest path algorithms - it also becomes a powerful tool for the algorithm designer (Fig. 5.11.13).

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Figure 5.11.14: Execution of the 2-edge shortest path algorithm is illustrated. A recursion tree corresponding to the equation w(a, k) = min{w(a, k), w(a, j) + w(j, k)} is also shown in the bottom diagram. The number of edges in the recursion tree corresponds exactly to the number of steps performed by the two edge shortest distance algorithm.

The Recursion Tree We illustrate the 2-edge shortest path algorithm by one additional tool known as the recursion tree (although strictly speaking it is a DAG) as shown in Fig. 5.11.14. Please note that the number of edges in the recursion tree is exactly equal to the number of steps performed by the algorithm. A vertex (13) in the bottom row of the recursion tree corresponds to one edge distance of vertex 3 from vertex 1 in original graph. While vertex (13) in the top row of the recursion tree corresponds to the two edge shortest distance from vertex 1 to vertex 3 in the original graph. As you can understand this 2-edge shortest path algorithm becomes a building block for so many shortest path algorithms.


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Bellman-Ford like Shortest Path Algorithm In order to find a shortest distance in a weighted graph with negative edge weights we just have to run the 2-edge shortest path building block sufficient number of times as shown in the following algorithm. Algorithm 30: (Bellman-Ford1): Find shortest distance of every vertex k from a given vertex a in G input : Adjacency matrix of weighted Graph G and a start vertex a output: Modified adjacency matrix G in which row corresponding to vertex a provides shortest distance of every vertex k from vertex a in G 1 2 3 4

for i=1 to p-1 do for each (intermediate) vertex j of graph G do for each (terminating) vertex k of graph G do Find a two edge path from vertex a to vertex k in G passing through an intermediate vertex j, and assign min{w(a, k), w(a, j) + w(j, k)} to w(a, k)

This algorithm resembles the so called Bellman-Ford algorithm with a time complexity of p3 as we have three nested for loops. Here we introduce another tool to study an algorithm.

The Colored Puzzle The colored puzzle highlights the fact that Algorithm 29 can be used to become a building block for Bellmam-Ford Algorithm. Algorithm 29 has two f or loops. One is the terminating vertex k loop represented by brown color in the colored puzzle. The other is the intermediate vertex j loop represented by orange color in the puzzle. The outer most loop in the Bellman-Ford Algorithm (known as Algorithm 30) is represented by blue color in Fig. Fig. 5.11.15. This outer loop executes Algorithm 29 p − 1 number of times and minimizes the value of w(a, k) according the following equation: w(a, k) := min{w(a, k), w(a, j) + w(j, k)}. See without this blue box the remaining two boxes (brown and orange) represents the 2-edge shortest path algorithm known as Algorithm 29. Please note that the nesting of f or loops in these algorithm can be nicely

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captured by nested colored boxes in the colored puzzle. The positioning of the j and k loops in the Bellman-Ford Algorithm can be interchanged without affecting the outcome of this algorithm as shown in Fig. 5.11.15. The position of the blue loop, however, cannot be changed without adversely affecting the performance of the algorithm. The recursion tree corresponding to the Bellman-Ford Algorithm is shown in Fig. 5.11.16 along with the corresponding colored puzzles. Other shortest path algorithms can also be represented by this puzzle with the addition of the start vertex a loop shown by green color in coming figures. It is interesting to note that each such algorithm appears in pairs - thanks to the colored puzzle which provokes a learner to note this interesting property. for i =1 to p-1 for each vertex k for each vertex j

for i =1 to p-1 for each vertex j for each vertex k

Figure 5.11.15: A colored puzzle depicting the positioning of the different for loops in the shortest distance finding algorithm. It is possible to run the intermediate vertex j-loop (orange color) first and then the destination vertex k-loop (brown color) or vice versa. The outer most loop in the algorithm is represented by blue color.


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Figure 5.11.16: The recursion tree corresponding to Bellman-Ford like algorithm. The complexity of algorithm is equal to the number of edges in this recursion tree - which is equal to p3 . The colored puzzle corresponding to this algorithm is again shown in the bottom diagram. Time Complexity of Bellman-Ford shortest path algorithm We have witnessed that the time complexity of Bellman-Ford1 shortest path algorithm is O(p3 ) with an adjacency matrix data structure. How about a sparse graph and if we represent it using an adjacency list representation? It may be a good idea to see if the time complexity can be reduced to at least O(pq) which will be less than O(p3 ) for a sufficiently sparse graph. It is also possible to rephrase this algorithm so that we do not need any

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extra intelligence (in its implementation) to make it a O(pq) algorithm. The rephrased algorithm is shown below. Algorithm 31: (Bellman-Ford2): Find shortest distance of every vertex k from a given vertex a in G input : Directed and weighted graph D. Distance Dist(a) of vertex a from itself is zero and from any other vertex it is infinite. output: Distance array Dist(k) storing minimum distances of every vertex k from vertex a. 1 2 3

for i=1 to p-1 do for every directed edge (j, k) in graph D do Dist(k) = min{Dist(k), Dist(j) + w(j, k)}

Dist(k) = min{ Dist(k), Dist(j) + w(j, k)} a Dist(k) Dist(j)

j a

w(j,k)

k a Dist(k) = 6

Dist(k) = 6 Dist(j) = 4 Dist(j) = 4

j

w(j,k) = 3

k

New Dist(k) = 6

j

w(j,k) = 1

k

New Dist(k) = 5

Figure 5.11.17: How and when the (shortest) distance Dist(k), with respect to vertex a changes when we consider the directed edge (j, k). Line 3 is the basic building block of this algorithm. Its working is explained in the figure below. In this version of Bellman-Ford the time complexity will always be pq with an adjacency list data structure and without an extra intelligence in its implementation.


5.11.4

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All Pair Shortest Path Algorithms

We shall describe now three all pair shortest path algorithms. They will all use a single building block which is the 2-edge shortest path algorithm described earlier. The slow all pair shortest path algorithm has a O(p4 ) complexity, faster all pair shortest path algorithm has a O(p3 log2 p) complexity while Floyd-Warshall all pair shortest path algorithm has O(p3 ) time complexity under worst case conditions. The Slow All Pair shortest Path Algorithm The Bellman-Ford like algorithm finds shortest distances from a fixed start vertex in a graph. If we run this algorithm for every vertex in the graph then we end up with the so called slow all pair shortest path algorithm with a time complexity of O(p4 ). This algorithm is depicted pictorially by the colored puzzle shown in the left diagram of the figure below. Algorithm 32: Slow All Pair: Find shortest distance of every vertex from every vertex in G input : Adjacency matrix of weighted Graph G output: Shortest distance of every vertex from every vertex in G 1 2 3 4 5

for each (start) vertex a of graph G do for i=1 to p-1 do for each (intermediate) vertex j of graph G do for each (terminating) vertex k of graph G do Find a two edge path from vertex a to vertex k in G passing through an intermediate vertex j, and assign min{w(a, k), w(a, j) + w(j, k)} to w(a, k)

The Faster All Pair shortest Path Algorithm It is interesting to note that if we switch the blue p-loop with the green a-loop in the colored puzzle as shown in the diagram below then we end up with another all pair shortest path algorithm, known as faster all pair shortest path algorithm. It will be interesting to derive its time complexity and compare it with that of slow all pair shortest path algorithm. The working of this algorithm is shown in the diagram below.

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for each vertex a for i =1 to p-1 for each vertex j for each vertex k

for i =1 to p-1 for each vertex a for each vertex j for each vertex k

w(a,k) = min{w(a,k), w(a, j)+w(j, k)} Figure 5.11.18: The start vertex loop is represented by green color. The intermediate vertex j loop is represented by orange color and the terminating vertex k loop is represented by brown color. There is another p-loop - it is represented by blue color. The diagram shows the effect of switching between the blue and the green loops. The left diagram depicts the slow all pair shortest path algorithm while the right diagram represents the faster all pair shortest path algorithm.

Algorithm 33: Faster All Pair: Find shortest distance of every vertex from every vertex in G input : Adjacency matrix of weighted Graph G output: Shortest distance of every vertex from every vertex in G 1 2 3 4 5

for i=1 to p-1 do for each (start) vertex a of graph G do for each (intermediate) vertex j of graph G do for each (terminating) vertex k of graph G do Find a two edge path from vertex a to vertex k in G passing through an intermediate vertex j, and assign min{w(a, k), w(a, j) + w(j, k)} to w(a, k)


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Figure 5.11.19: Working of the faster all pair shortest path algorithm is shown. Please note that from two edge all pair shortest distances we jump to 4-edge and then to 8-edge shortest distances.

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Figure 5.11.20: The recursion tree and the colored puzzle corresponding to faster all pair shortest path algorithm. Here it is also obvious that from 2-edge shortest distances we directly jump to 4-edge shortest distances.


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Food for thought 1. Is the time complexity of this algorithm any better than that of slow all pair shortest path algorithm? If yes then why? 2. You may have noticed that in this algorithm (FAster All Pair) the number of edges in shortest paths jumps in the powers of 2 as shown in Figure 5.11.20. What does that mean? The outer most loop (the blue colored loop in the colored puzzle) should run from 1 to p or from 1 to log2 p? How will it affect the time complexity of this so called faster all pair shortest path algorithm. 3. In the slow all pair shortest path algorithm the blue loop should run from 1 to p and not from 1 to log2 p? Why? All Pair (Floyd-Warshall) Shortest Path Algorithm Consider the faster all pair shortest path algorithm. We know it can be represented by the colored puzzle as shown in Fig. 5.11.21. The recursive equation used as a building block is also indicated in this figure. There are essentially four for loops in this algorithm represented by four nested rectangles in the multi colored puzzle. If the intermediate vertex j loop (also known as the orange loop in the colored puzzle) becomes the outermost loop while the source vertex a loop (green) and destination vertex k loop (brown) are inner loops (in any order) then (surprisingly) we end up with one of the most efficient all pair shortest path algorithms as shown in Fig. 5.11.22. Please note that now we have only three loops and there is in fact no need to have the fourth loop - the so called blue loop in the colored puzzle. The time complexity will now be O(p3 ) as there are only three loops. Its time complexity is as good (or as bad) as that of Bellman-Ford algorithm (which is a single source shortest path algorithm) for non sparse graphs. The basic building block of this algorithm is the same recursive equation that we used in other algorithms. See Fig. 5.11.22 & 5.11.23.

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w(a,k) = min{w(a,k), w(a, j)+w(j, k)} for i =1 to log2p for each vertex a for each vertex j for each vertex k

for i =1 to log2p for each vertex a for each vertex k for each vertex j

Figure 5.11.21: The colored puzzles corresponding to faster all pair shortest path algorithm.

Algorithm 34: (Floyd-Warshall): Find shortest distance of every vertex from every vertex in G input : Adjacency matrix of weighted Graph G output: Shortest distance of every vertex from every vertex in G 1 2 3 4

for each (intermediate) vertex j of graph G do for each (start) vertex a of graph G do for each (terminating) vertex k of graph G do Find a two edge path from vertex a to vertex k in G passing through an intermediate vertex j, and assign min{w(a, k), w(a, j) + w(j, k)} to w(a, k)

Shortest Path Algorithms for vertex i =1 toj p-1 for each for each vertex k for each vertex a

233 for each vertex j for each vertex a for each vertex k

Figure 5.11.22: We show the possibility of having the intermediate vertex j loop become the outer most loop. There is no fourth loop in this diagram and surprisingly there is no need for it. Once the j-loop becomes the outer most loop then it does not matter if the order of the green loop and that of the brown loop is interchanged. Food for Thought: What is the trick or intuition behind this algorithm? Is this not surprising that without a fourth loop we can design an all pair shortest path algorithm? A spectrum of single source and all pair shortest path algorithms We show a spectrum of shortest path algorithms in Fig. 5.11.24. We also show connections or links between different algorithms. Note that the Bucket algorithm is the ancestor of most of these algorithms. This panorama of shortest path algorithms is also depicted by the colored puzzle shown in Fig. 5.11.25. Please appreciate the fact that for the entire galaxy of such algorithms we use a single building block - the 2-edge shortest path algorithm.

5.11.5

Johnson’s all Pair Shortest Path Algorithm

We have studied in the last section that all pair shortest path algorithm (Floyd-Warshall) has a worst case time complexity of O(p3 ). This is the best performance seen so far for an all pair shortest path algorithm. In this subsection we shall study another all pair shortest path algorithm which works faster than O(p3 ) for graphs which are sufficiently sparse. Required Prior Knowledge: First we shall talk about the prior knowledge required to understand this

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Figure 5.11.23: The recursion tree corresponding to Floyd-Warshall all pair shortest path algorithm. The number of edges in the recursion tree is equal to the number of steps performed by the said algorithm - and this is equal to O(p3 ).


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Figure 5.11.24: We show a concept map of various single source and all pair shortest path algorithms.

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w(a,k) = min{w(a,k), w(a, j)+w(j, k)}

2-edge shortest path algorithm

Bellman-Ford (single source)

Slow All Pair

Faster All Pair

Floyd-Warshall (all pair)

Figure 5.11.25: We show the galaxy of single pair and all pair shortest path algorithms. Each algorithm can be represented by a different color arrangement in the rectangular puzzle. It is interesting to note that each shortest path algorithm has at least one dual with the same performance and output.

Figure 5.11.26: It will be interesting to see if any of these colored arrangements represents one of the already discussed shortest path algorithms.


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algorithm. If that becomes clear then it is almost trivial to appreciate the innovation behind this algorithm. It is interesting to note that this is not entirely a new algorithm - it innovatively combines two shortest path algorithms (Dijkstra + Bellman-Ford) and creates an all pair shortest path algorithm such that the overall time complexity becomes better than that of best known shortest path algorithm - Floyd-Warshall (p3 ) under certain conditions. 1. We know that Dijkstra’s shortest path algorithm finds correct shortest paths from a single vertex provided all edge weights are positive. We also know that its time complexity is O(p2 ) if we use an adjacency matrix as a data structure. Its time complexity can be improved with an adjacency list data structure provided we use a minimum heap to locate the next vertex which goes in the bucket. The improved time complexity is O(qlogp). This can further be improved to O(plogp + q) if we use a Fibonacci heap to implement the minimum priority queue. If all edge weights are positive then we can apply Dijkstra’s algorithm p times to find all pair shortest paths in O(p2 logp + pq). This time complexity is better than O(p3 ) (Floyd-Warshall) in sufficiently spare graphs. But if there are negative edge weights then we shall get incorrect results. 2. We know that Bellman-Ford algorithm can find shortest paths from a single source vertex in time O(p3 ) even if there are negative edge weights in a directed graph. The time complexity of this algorithm is O(pq) provided we use an adjacency list as a data structure to represent the input graph. Again this is an improvement over O(p3 ) provided we have a sufficiently sparse graph. If we convert this O(pq) Bellman-Ford algorithm into an all pair shortest path algorithm then its time complexity would become O(p2 q) which is worse than O(p3 ) for Floyd-Warshall. So we need to do something more? Something very innovative? Johnsons’Algorithm = Bellman-Ford + Innovation + Dijkstra Johnson’s algorithm first uses Bellman-Ford to check if there are any negative weight cycles. If there are no negative cycles then this algorithm somehow uses results of Bellman-Ford algorithm to convert negative edge weights into positive without disturbing the relative path lengths. This requires time proportional to pq. Once all edge weights are made positive we can use

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the improved Dijkstra’s algorithm to find all pair shortest paths in time O(p2 logp + pq). Consider the Bellman-Ford algorithm described earlier and reproduced here. It was already discussed that its worst case time complexity will be O(pq). All distances are measured with respect to vertex a. We initialize Dist(a) = 0 and set Dist(x) of every other vertex x from vertex a equal to infinity. Algorithm 35: Find shortest distances of every vertex k from vertex a in D input : Directed and weighted graph D. Distance Dist(a) of vertex a from itself is zero. output: Distance array Dist(k) storing minimum distances of every vertex k from vertex a. 1 2 3

for i = 1 to p − 1 do for every directed edge (j, k) in the graph D do Dist(k) = min{Dist(k), Dist(j) + w(j, k)};

Checking Negative Weight Cycles in a directed graph: The above algorithm finds shortest paths correctly in case there are no negative weight cycles in the directed graph. If there are negative weight cycles then complications arise as already discussed. Under such conditions this algorithm should at least inform us that in the given graph there are negative weight cycles. What modification is needed in this algorithm for this extra intelligence? The required modification is simple and elegant? We run the outer most loop p − 1 times and store distance of each vertex x from the start vertex a. We then run this loop one more time to check if any distance changes. If it does then it means we have negative weight cycles in the directed graph reachable from vertex a as shown in the figure below. If it does not then there are no negative cycles reachable from vertex a in the directed graph. Please see Fig. 5.11.27 . Food for Thought 1. How about if the distance of no vertex changes after an initial iteration i when i < p − 1. Does that mean there are no negative weight cycles and should we stop without further iterating? See Fig. 5.11.28.


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Figure 5.11.27: If we find i-edge distances in this graph from vertex a then we observe that some distances will change when i goes from 4 to 5. This does not confirm that there is a negative weight cycle. But when a distance of a vertex from vertex a changes when i goes from 5 to 6 then that is a confirmation that there is indeed a negative weight cycle in this graph. 2. How about if we apply Bellman-Ford at a vertex which is not reachable to a negative cycle. See Fig. 5.11.29. Does that mean we have to apply Bellman-Ford algorithm at each vertex to find if there are any negative weight cycles in the graph? But that will be very costly?

Figure 5.11.28: After finding 3-edge shortest paths in this graph from vertex a there will be no change in distance calculations. Does that mean that we should stop here and declare that there are no negative weight cycles in this graph? Applying Bellman-Ford Algorithm once to determine Negative Weight Cycles The problem is how can we apply Bellman-Ford algorithm just once from a (special) vertex and check if there are any negative weight cycles in the di-

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Figure 5.11.29: If we apply Bellman-Ford algorithm in this graph to find shortest distances from vertex e then we should never be able to confirm that there will be negative weight cycles in this graph. If, however, we apply this algorithm from vertex a then it is possible to verify that indeed there are negative weight cycles in this graph. rected graph? The answer to this problem is given in elegant transformation shown in Fig. 5.11.30. The application of Bellman-Ford algorithm in the transformed graph just once (from the newly added vertex x) provides us the following information: 1. Whether there are any negative weight cycles in the graph. If there are any negative weight cycle then the algorithm should not move forward and should terminate. 2. If there are no negative weight cycles then what are the shortest distances of each vertex from the newly added vertex x. This information will further be used to convert negative edge weights into positive edge weights. If the graph does not contain any negative weight cycles but contains negative weight edges then we should somehow try to make the negative weight edges positive and then apply Dijsksta’s algorithm from each vertex as already planned. How about adding a big positive number in each edge weight such that every (edge) weight becomes positive? See what complication would arise if we do so as shown in Fig. 5.11.31. Thus we cannot add an arbitrary positive number in each edge weight as it will disturb the relative path lengths in the new graph as shown in the following figure. Converting negative weight edges into positive weight edges Please note that we need to simultaneously fulfill the following two objectives:


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Figure 5.11.30: The transformation in this diagram allows us to apply Bellman-Ford algorithm just once and verify if there are any negative weight cycles in the graph. All edges coming from vertex x have a zero weight. We apply Bellman-Ford algorithm to this transformed graph and find shortest distances from vertex x.

Figure 5.11.31: We can always convert negative edge weights by adding a positive number in each weight so that we can apply a more efficient algorithm to find shortest paths in a graph with positive edge weights. But adding a constant in each edge weight disturbs the relative weights of different paths and leads to a wrong answer.

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1. All negative edge weights should be converted into positive weights. 2. All relative distances between two vertices should remain the same in the modified graph with positive edge weights. First we shall devise a simple scheme to alter edge weights such that relative distances between any two vertices do not change in the given graph. Then we shall modify this scheme so as to convert all negative edge weights into positive edge weights thus fulfilling both the above objectives. Consider a directed edge (j, k) with a weight equal to w. How about associating any arbitrary number Label(j) with a vertex j and another arbitrary number Label(k) with vertex k. Now the edge weight w(j, k) is changed according to a formula where wnew = wold +Label(j)−Label(k). If we associate an arbitrary number with each vertex of the graph and change edge weights according to the formula described then we claim that relative path lengths between any two vertices will be the same in the new graph as compared to the old graph. See Fig. 5.11.32. In fact the new path length from a vertex a to a vertex b will be the old path length plus Label(a) − Label(b). Putting arbitrary labels solves only one problem, namely the relative distances between two vertices remain the same in the new graph as compared to the old graph. The other problem (converting all negative edge weights into positive) remains to be solved. It requires that vertices should not be labeled arbitrariliy but with some intelligence as described below. So now the problem is reduced to finding an appropriate number Label(j) to be associated with a vertex j and another number Label(k) to be associated with vertex k (assuming that there is an edge from j to k with a weight w(j, k)) such that if w(j, k)old is negative then w(j, k)new = w(j, k)old + Label(j) − Label(k) becomes positive.This would require that wjk (old) + Label(j)−Label(k) is greater than or equal to zero. It means that wjk (old)+ Label(j) is not less than Label)k). In other words Label(k) should be at least equal or more negative than wjk (old)+Label(j) assuming that wjk (old), Label(j), and Label(k) are all negative. Before finding a systematic scheme of providing labels to each vertex let us try to work out a simple example. What numbers are desirable and should be associated with two adjacent vertices j and k are indicated in Fig. 5.11.33. Here vertex x is a vertex added to the given directed graph just like the one shown in Fig. 5.11.30. In the top diagram we try to initially label k such that


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Figure 5.11.32: How relative distances remain the same if we associate an arbitrary number with a vertex and then add a number in the edge weight of each edge according to a fixed formula.

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weight of edge (j1 , k) becomes positive. In the bottom diagram we intend to re-label k such that weight of edge (j3 , k) also becomes positive. Please note that the new label of k (equal to -80) will make sure that all edges incident to vertex k have positive edge weights. Now coming back to the general question: What systematic scheme should be applied which guarantee the allocation of desirable numbers to vertices of a graph such that all edge weights become positive while relative distances between any two vertices remain the same? After looking at the last figure you must have some idea of what is going on or what should be done. If you are still undecided then read the following paragraphs and look at the coming figure. Let us recall the basic building block of Bellman-Ford algorithm and how the shortest distance of a vertex k from a source vertex x is updated. See Fig. 5.11.34 for a demonstration of this updating. Please see once again Fig. 5.11.33 where the label of a vertex is updated and now loook at Fig. 5.11.34 where the shortest distance of vertex j is updated. It should become quite evident now that if Label(j) is the shortest distance of vertex j from vertex x and Label(k) is the shortest distance of vertex k from vertex x then the required inequality would be satisfied and the edge weight for edge (j, k) will eventually become positive if it was initially negative. See Fig. 5.11.35. The interesting thing is that we need not spend extra time in finding these labels as they have already been found while checking if the given graph has negative weight cycles.

Summary So now we are in a position to describe Johnson’s algorithm in the following meaningful manner as applied on a directed graph D. 1. Check if there are any negative weight cycles in the graph D using Bellman-Ford algorithm applied to the graph after inserting a source vertex x as shown in Fig. 5.11.35. If there are no negative weight cycles in graph D then move to step 2 otherwise terminate. 2. In step 1 we have already found shortest distance Dist(k) of each vertex k from newly inserted vertex x. That shortest distance Dist(k) become Label(k) of vertex k. Now draw another copy of graph D with all edge


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Figure 5.11.33: We assume that vertices j1 , j2 , and j3 are already labelled while we need to label k so as to make edge weights positive. In the top diagram we try to initially label k such that weight of edge (j1 , k) becomes positive. In the bottom diagram we intend to re-label k such that weight of edge (j3 , k) also becomes positive. Please note that the new label of k (equal to -80) will make sure that all edges incident to vertex k have positive edge weights.

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Dist(k) = min{ Dist(k), Dist(j) + w(j, k)} x Dist(k) Dist(j)

j

w(j,k)

k

x

x Dist(k) = -70

Dist(k) = -10

j

Dist(j) = -20

k

w = -40

j

New Dist(k) = -60

k

w = -40

Dist(j) = -20

New Dist(k) = -70

Figure 5.11.34: While executing Bellman-Ford algorithm we use a basic building block as shown above. The bottom diagrams show how and when the shortest distance of vertex k from a source vertex x is updated.

x Transform

0

b 0

b -4

8 -2

-4

a

c

1 -2

2

d 0

3

-2 a

-4 c

1+0 -2+2

2+4

e 0

-4+4

8-2

d 0

3+0

e 0

Figure 5.11.35: How to determine which labels to associate with each vertex.

Discussion

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weights modified. All negative edges become positive as shown in Fig. 5.11.35. 3. Now apply Dijksta’s shortest path algorithm at every vertex in the modified graph (with positive edge weights) to find all pair shortest distances and paths.

5.12

Discussion

The most important task of a teacher should enable the students to discover and acquire experience of independent work. According to Polya [10]: If the student is left alone with his problem without any help or with insufficient help, he may make no progress at all. If the teacher helps too much, nothing is left to the students. The teacher should help, but not too much and not too little, so that the student shall have a reasonable share of the work. In this chapter we have demonstrated how a teacher can help students discover a number of graph algorithms with some initial help, starting with something seemingly simplistic yet capable of being transformed into a number of powerful algorithms with minor modifications. We have shown that by asking thought provoking questions it becomes possible for the teacher to guide the students while solving complex problems. We have also shown that making comparisons between various techniques and solutions provides a deep insight which itself is very useful in solving otherwise difficult problems [8]. At times it is almost impossible to solve a given problem while it is easy to solve a related problem (the shortest path problem is solvable while the longest path problem is unsolvable). Similarly it is difficult to solve a problem in its original form while it is easier to solve it while placing certain restrictions (the graph isomorphism problem is solvable for trees but is difficult to solve in general). It is extremely useful to find why a certain technique works under certain conditions and why it fails in others (greedy methods provide optimal solutions in finding the shortest path but fail to find the longest path). The theory of NP-Completeness connects all problems that are NP-Complete: it is also possible to find a useful relationship among solvable problems and this is what we have attempted to do in this chapter. According to Hale [5]: “There are different kinds of learning, but I refer here to the intellectual kind. To learn means to cause your mind to function in a different way: new memories are created and/or new connections are forged.”

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These relationships provide the algorithm designer a perspective that proves invaluable when solving new problems and analyzing old one’s.

Acknowledgement We are thankful to the Department of Computer Science, Lahore University of Management Sciences for providing support for this research. We wish to specially thank R. Khan for providing motivation as well as inspiration for this project. We also wish to thank S. Skiena, S. Baase, S. Mahkari, H. Mian, A. Alvi, K. Fahd, J. Ikram, M. Maud, T. Jadoon for their help and encouragement.

Chapter 6 Network Flows, Connectivity and Matching Problems 6.1

Introduction

6.2

Definitions & Prior Knowledge

6.3

Konig’s Theorem, Menger’s Theorem & Hall’s (Marriage) Theorem

6.4

Menger’s Theorem

6.5

Konig’s Theorem, Menger’s Theorem & Hall’s (Marriage) Theorem Revisited

6.6

Network Flows

6.7

The Matching Problem in Bipartite Graphs

6.8

The Max-Flow Min-Cost Problem

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6.1

Network Flows, Connectivity and Matching Problems

Introduction

We shall first address the problem of vertex or edge connectivity in general graphs. We shall also discuss the matching problem in bipartite graphs. Specifically we shall be describing Menger’s Theorem which relates maximum number of vertex-disjoint (or edge-disjoint) paths with minimum number of vertices (or edges), which if removed will disconnect a special node from another special node in a graph. We shall also discuss Konig’s Theorem which relates the size of the vertex cover to the size of maximum matching in a bipartite graph. The Marriage (Hall’s) theorem provides necessary and sufficient conditions for a bipartite graph to have a perfect matching. Our intention (and desire) in this chapter, will be to integrate concepts so that the enabled learner is able to appreciate the bigger picture where one relationship implies another and one theorem can be used to prove the other. That is why we first provide a unified picture and then go deeper in order to analyze each area in detail. In addition to making formal proofs for a number of theorems, we shall be designing algorithms to solve a number of related problems. We shall also be discussing the network flow problem, the maximum flow at minimum cost problem, and last but not the least the Circulation problem. We use a single building block in this entire chapter for designing almost every algorithm.

6.2


Set Cover: Given a set of subsets S of a Universal Set U , what is the smallest subset T of S such that the union of all these sub sets in T covers all elements of U (it means that every element in U is contained in any of the subsets in T and the set T has the smallest possible size). We have witnessed the subset sum problem in which we have to select integers (out of a set of integers) such that the sum of the selected integers is equal to a given constant. Vertex Cover: The Universal set U is the set of all edges in a graph. What is the smallest subset of vertices of the graph that covers all edges in the graph? Independent (Vertex) Set: What is the largest subset S of vertices of a graph such that no pair of vertices in S has an edge in between? Is there a


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Figure 6.2.1: A perfect binary tree is shown. The root vertex is shown at the top of the diagram. Each leaf vertex is at a distance of 3 from the root vertex. Any independent set will include leaf vertex u, while any vertex cover will include non leaf vertex v. Every edge connecting a leaf vertex u and a non leaf vertex v will always be part of the edge cover. Note that every non leaf vertex is a cut vertex and every edge is a cut edge or a bridge in a tree graph. Matching (Independent Edge Set): A sub-graph of G, where the maximum degree of each vertex is one, i.e. no edge has a common end point. These edges are also known as independent edges. Perhaps a more meaningful definition will be a set A of non-adjacent edges; that is, no two edges in the set A share a common vertex. The size of a matching is equal to the size of set A. Maximal Matching: This is a matching in which more edges cannot be added to increase the size of this matching. Find a simple algorithm to find a maximal matching in a line graph. (A line graph is a tree where the degree of each node is not more than 2) Maximum Matching: It is a matching in a graph with maximum possible size? How bad can a maximal matching become as compared to maximum matching? (It means how small the size of the maximal matching can become as compared to the size of the maximum matching) Edge Cover: The Universal set is the set of all vertices in a graph. What is the smallest subset of edges, which covers all vertices? How small (or big) can the size of the edge cover become, as compared to the number of vertices

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in a graph? A Perfect Matching is a matching which covers all vertices of the graph. That is, every vertex of the graph is incident to exactly one edge of the matching. Every perfect matching is both maximum and hence maximal. In some literature, the term complete matching is used for perfect matching. A Perfect Binary Tree: We know that a tree is a connected graph where each edge is a bridge edge. A binary tree is a tree where the degree of each vertex is less than or equal to 3. Here we define a perfect binary tree. A perfect binary tree (shown in Fig. 6.2.1) has the following features: 1. It has a single vertex with degree equal to 2. Let us call it the root vertex. The root vertex is shown as the top most vertex in Fig. 6.2.1. 2. It has exactly 2h+1 − 1 vertices where h is the path length (in terms of number of edges) between the root vertex and any leaf vertex in the perfect binary tree. The perfect binary tree shown in Fig. 6.2.1 has h equal to 3 and it has 15 vertices. 3. It has exactly 2h leaf vertices. As described before each leaf vertex in a perfect binary tree has a path length equal to h from the root vertex. The perfect binary tree shown in Fig. 6.2.1 has 8 leaf vertices. 4. It has exactly 2h − 1 non leaf vertices including the root vertex. All non leaf vertices other than the root vertex has a degree exactly equal to 3. There are 7 non leaf vertices in tree shown in Fig. 6.2.1. Problem Set 6.1. We are given a perfect binary tree G (see Fig. 6.2.1) and we intend to solve a number of problems related to connectivity and matching. We know that in a tree every edge is a cut edge or bridge while every (non leaf) vertex is a cut vertex. We also know that there is always a unique path between every two vertices in a tree. We shall observe later in this chapter that there is indeed a relationship between graph connectivity and number of paths in a graph (Menger’s Theorem). In a perfect binary tree graph G, a leaf vertex u will always part of an independent set (why?). While a non leaf vertex v which is adjacent to a leaf vertex will always be part of a vertex cover (why?). Using this logic all leaf vertices will be part of the independent set and will not be part of the vertex cover. Independent set and vertex cover vertices are shown in Fig. 6.2.2 for the perfect binary


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Figure 6.2.2: The top left diagram shows the vertices belonging to the vertex cover highlighted by double circled vertices. The rest of the vertices belong to the independent set in this diagram as shown in the top right corner. The bottom left diagram highlights, in brown bold, the maximum matching edges in the given tree. The bottom right diagram shows the edges belonging to the edge cover; these edges are shown in bold in green. You may have noticed that the size of the maximum matching is equal to the size of the vertex cover in this graph. The maximum matching in this graph is not a perfect matching, as all vertices here are not matched as shown by black circled vertices. You may have also noticed some relationship between the size of the vertex cover and the size of the independent set?

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tree graph of Fig. 6.2.1. In a perfect binary tree graph an edge connecting a leaf vertex u with a non leaf vertex v (see Fig. 6.2.1) will always be part of the edge cover. Edges belonging to the edge cover are shown in bold in the bottom right diagram of Fig. 6.2.2. We shall observe later that in a bipartite graph (such as a tree) the size of a maximum matching is equal to the size of vertex cover (Konig’s Theorem). It may thus be possible to find a maximum matching given a vertex cover in a tree graph. The bottom left diagram of Fig. 6.2.2 shows a maximum matching in the perfect binary tree graph of Fig. 6.2.1. It also indicates matched and unmatched vertices and edges. Based upon the above observations it is possible to design efficient algorithms to solve the following problems in a tree graph. Please note that a perfect binary tree graph is a fairly restricted structure and our existing prior knowledge of graph theory and algorithms is sufficient to solve these problems. Problem 6.1.1. How can we efficiently find an edge cover in a perfect binary tree? Problem 6.1.2. How can we efficiently find a vertex cover in a perfect binary tree? Problem 6.1.3. How can we find an independent (vertex) set in a perfect binary tree? Problem 6.1.4. How can we find an independent (edge) set, meaning a maximum matching in a perfect binary tree? Problem 6.1.5. Does a perfect matching exist in a perfect binary tree? Discuss briefly. Problem 6.1.6. For a perfect matching to exist in a tree graph, what conditions are necessary and which are sufficient? Problem 6.1.7. We know that a tree is a (restricted) bipartite graph consisting of two halves. A perfect matching may exist for the tree graph if the two halves have the same size. If the two halves are not of the same size, then a perfect matching cannot exist, however a near perfect matching may exist. A near perfect matching requires that all vertices belonging to the smaller half are matched to vertices in the larger half in a bipartite graph. Check Fig. 6.2.2 to see if a near perfect matching exists in a perfect binary tree.


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Concept Map 6.1. A concept map showing a number of relevant concepts and a number of theorems which relate different concepts.

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Figure 6.2.3: There are several paths between vertices s and t in this graph. By removing certain edges (or vertices) it is possible to remove all these paths between the two vertices; the resulting graph will be a disconnected graph. Problem 6.1.8. What are the necessary and sufficient conditions for a near perfect matching to exist in a tree graph? Edge Connectivity λ(G) of a Graph G is the minimum number of edges which if removed will disconnect the graph G (see Disconnected Graph). A disconnected graph G has λ(G) = 0, while a tree graph T has λ(T ) = 1. The edge connectivity for a completely connected graph (having p vertices) is equal to p−1. Sometimes we not only want the graph G to be disconnected but also want to make sure that a special vertex s is separated from another special vertex t, where both s and t belonged to G. So our new requirement is that when G is disconnected (by removing certain edges) then vertices s and t should belong to different connected components of G. Thus λ(s, t) is the minimum number of edges which if removed will destroy all paths between vertices s and t in G. Vertex Connectivity κ(G) of a Graph G is the minimum number of vertices which if removed will disconnect the graph G (see Disconnected Graph). Similarly we define κ(s, t) as the minimum number vertices which if removed from G, will disconnect it so that all paths between vertices s and t are destroyed, vertices s and t of G would now belong to different connected components.


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Edge-disjoint paths do not share any edge, while vertex-disjoint paths do not share any vertex except the terminal vertices. Vertex-disjoint paths are also edge-disjoint but it may not be true the other way round. Disconnected Graph: A graph G may simply be disconnected into two or more connected components, that is, it may be broken up into more connected components or we may require that G would be disconnected such that two special vertices s and t of G should lie in separate connected components. Problem Set 6.2. We are given a graph G with two special vertices s and t as shown in Fig. 6.2.3. We intend to explore different paths between these two vertices and see how we can destroy these paths in G. There is only one bridge edge in this graph, if we remove this edge {s, a}, then the graph G is disconnected but vertex s is still accessible to vertex t. There is no cut vertex in this graph. We have to remove a lot more than one vertex in order to disconnect G so as to destroy all paths between vertex s and t. Problem 6.2.1. Draw all 4-edge paths between vertex s and t. See Fig. 6.2.4. Problem 6.2.2. Draw all 5- and 6- edge paths between the two vertices. See Fig. 6.2.4. Problem 6.2.3. Out of all paths that you have listed in Problem 6.2.1 and Problem 6.2.2 shortlist the ones that are edge-disjoint. What is the maximum number of such paths? Problem 6.2.4. Out of all paths that you have listed in Problem 6.2.1 and Problem 6.2.2 short list the ones that are vertex-disjoint. What is the maximum number of such paths? Problem 6.2.5. Draw a 4-edge path between the two vertices such that not more than one additional edge-disjoint path is possible between the two vertices. Problem 6.2.6. Draw more than one edge-disjoint paths between the two vertices such that at least one path should be of length six.

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Figure 6.2.6: A 6-edge path and a couple of 4-edge disjoint paths. Problem 6.2.7. Draw a 6-edge path between the two vertices such that no additional edge-disjoint (or vertex-disjoint) path is possible between the two vertices. Problem 6.2.8. The left diagram of Fig. 6.2.7 shows a 6-edge path (shown in bold) between vertices s and t in the same graph. Note that if we remove all edges in this 6-edge path, then all the paths between vertex s and t are destroyed. We know that there existed more than one path between the two vertices, but only because we have selected one wrong path initially, that other possible paths have disappeared. Why? What are its ramifications in designing an algorithm to find all possible edge-disjoint paths in a graph? Problem 6.2.9. In fact it is not necessary to remove all the edges in the single 6-edge path as shown in Fig. 6.2.7 in order to destroy connectivity between s and t. A minimal subset of these edges shown by a cut is sufficient to do the job. List down these edges. Do these three edges corresponds to a minimum number of edges which if removed will destroy all paths between vertices s and t. Can you find another subset of three edges which when removed will disconnect s and t? What is λ(s, t) for this graph? Problem 6.2.10. The right diagram of Fig. 6.2.7 shows two edge disjoint paths of length 4 (shown in bold) between the vertices s and t. Once these paths are selected, it will no longer be possible to find an extra edge-disjoint path between the two vertices. You can remove all edges belonging to the

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paths and see for yourself that the connectivity between s and t is destroyed. What is a minimal subset of these edges which will do the same job of destroying the connectivity between s and t? One such subset is shown in the right diagram of Fig. 6.2.7. This subset consists of four edges unlike three in the last part?

Problem 6.2.11. The left diagram of Fig. 6.2.8 shows a minimum sized subset of edges, which if removed will destroy the connectivity between vertices s and t. The right diagram shows a minimum sized subset of vertices which if removed will disconnect s and t from each other. This subset consists of two vertices, e and f ; the diagram also shows, in bold, two vertex-disjoint paths between s and t. Is this a coincidence that the (maximum) number of vertex-disjoint paths is exactly equal to the (minimum) number of vertices which if removed will disconnect s and t? The left diagram shows an edge cut of size three. Does this mean that there will be three edge-disjoint paths in this graph? Find these three paths. Be careful as this is the same cut shown in the left diagram of Fig. 6.2.7 where there existed a single path of six edges (with no room for additional paths). Now we need to find three edge-disjoint paths between the same vertices.

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6.3


Before discussing connectivity, matching, and the above theorems in detail in subsequent sections, we shall first provide a panoramic picture of how the different concepts are interrelated. How one theorem implies another and can be derived from each other. The right diagram of Fig. 6.2.8 is an illustration of Menger’s Theorem. It states that the maximum number of vertex-disjoint paths between vertices s and t are equal to minimum number of vertices which, if removed from the graph will disconnect s and t (see Concept Map 6.1). The left diagram of Fig. 6.3.1 shows a bipartite graph with partite A and partite B; it also shows a maximum matching with the maximum matching edges shown in bold. The vertex cover of this bipartite graph is also indicated in this figure. Konig’s Theorem tells us that the cardinality of a maximum matching in a bipartite graph is equal to the (minimum) size of the vertex cover (see Concept Map 6.1). The right diagram of Fig. 6.3.1 shows the same bipartite graph with two dummy nodes designated as s and t. All vertices in the first partition of the bipartite graph are connected to s while all vertices in the second partition are connected to the dummy vertex t as shown in the right diagram of Fig. 6.3.1. The vertex cover in the bipartite graph (shown

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in the left diagram) becomes the minimum sized subset of vertices which, if removed will destroy all paths between vertex s and t in the right diagram (why?). Similarly each matched edge (in the left diagram) will correspond to a vertex-disjoint path between vertex s and t in the right diagram (why?). A relationship between (minimum) vertex cover and maximum matching in a bipartite graph (Konig’s Theorem) is transformed into a relationship between κ(s, t) and (maximum) number of vertex-disjoint paths between vertex s and t in the right diagram of Fig. 6.3.1 (Menger’s Theorem). Note that all vertices in the set A are not matched to all vertices in the set B although the size of A is equal to size of B in the given bipartite graph. In other words a perfect matching does not exist in this bipartite graph. Had a perfect matching existed then the size of the vertex cover would have been equal to the size of A (or B). What should be the size of the neighborhood N (S) for every subset S of A which will guarantee a perfect matching (or a vertex cover equal to the size of A) is the subject of Marriage (Hall’s) Theorem. It states that a bipartite graph (of equal halves) has a perfect matching if and only if |N (S)| ≥ |S| for every S ⊆ A. For a better understanding of the Marriage Theorem a few neighborhood subsets are indicated below for the bipartite graph of Fig. 6.3.1. N ({a1 }, {a2 }) = {b1 , b2 } N ({a1 }, {a2 }, {a3 }) = {b1 , b2 } N ({a2 }, {a3 }) = {b2 } N ({a2 }, {a3 }, {a4 }) = {b2 , b3 , b4 } N ({a1 }, {a2 }, {a3 }, {a4 }) = {b1 , b2 , b3 , b4 } Out of the five neighborhood subsets N (S), only three satisfy the condition that |N (S)| ≥ |S|, while in the remaining two the said condition is violated. It is interesting to note that if we remove vertex a3 from A and vertex b3 from B then the condition |N (S)| ≥ |S| is true for every S ⊆ A. Under such conditions the size of the maximum matching and that of the vertex cover will be equal to the size of remaining set A which is equal to three. You may have realized yourself that this necessary and sufficient condition (applicable to the bipartite graph shown on the left of Fig. 6.3.1) implies that the (maximum) number of vertex-disjoint paths (and κ(s, t)) in the graph shown on the right side of Fig. 6.3.1 would be equal to the size of partition A (and B).


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Menger’s Theorem

Connectivity of an un-directed graph is expressed in terms of minimum number of edges or minimum number of vertices which if removed will disconnect the graph G. (see Concept map 6.1). Menger’s Theorem equates the maximum number of edge disjoint (vertex disjoint) paths between vertices s and t to minimum number of edges (vertices) which if removed will disconnect s from t. Connectivity of a directed graph D also poses similar problems. We shall start this section with directed graphs and then generalize our results for undirected graphs. We shall also address the following algorithmic problems: 1. Given a directed graph D containing two special vertices s and t, how can we find the maximum edge-disjoint (vertex-disjoint) paths from s to t in a directed graph D? 2. Given a directed graph D how can we find the minimum number of edges (vertices) which if removed will destroy all paths from vertex s to t in D? 3. Given an undirected graph how can we find its connectivity: the minimum number of edges (belonging to Edge Cutset) or minimum number of vertices (belonging to Vertex Cutset) which if removed will disconnect the graph into two or more connected components? The graph theoretic aspects of the above problems and a proof for Menger’s Theorem will be presented after a better appreciation of the issues involved.

6.4.1

Maximum Edge-Disjoint Paths in Directed Graphs

Given a directed graph D and two specific vertices s and t, how can we efficiently find the maximum number of edge-disjoint paths from vertex s to vertex t? Let U be the minimum sized subset of the edge set of D, such that D − U does not contain any directed path from vertex s to t. The size of U will obviously be represented by λ(s, t). Then according to Menger’s Theorem, the maximum number of edge-disjoint paths from vertex s to t will be equal to λ(s, t). Thus the problem of finding maximum number of paths from s to t is, in fact, equivalent to finding a minimum sized edge set which if

Menger’s Theorem

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Concept Map 6.2. A concept map showing a number of concepts related to vertex connectivity and edge connectivity and some important relationships that we shall explore in this section.

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removed from D will disconnect t from s. How about devising the following common sense algorithm to find the maximum number of edge-disjoint paths from vertex s to t in a directed graph D? Algorithm 36: Find Maximum edge-disjoint paths from s to t in D. input : Directed graph D, and vertices s & t. output: (Maximum) edge-disjoint paths from s to t in D. 1

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Let us apply this simple algorithm to solve the problem in the graph of Fig. 6.4.1. Using any path finding algorithm, we can find a directed path P1 , from vertex s to t in D. Let us look at what are different possibilities for P1 . It may be (s → a → c → t), (s → b → d → t), (s → a → d → t), or (s → d → t). It cannot be (s → d → a → c → t) or (s → b → d → a → c → t) because on directed edge (a, d) we can move from ‘a’ to ‘d’ but not from ‘d’ to ‘a’. Once we have selected an initial path P1 , we can move forward in Algorithm 36. But before finding another edge-disjoint path we should first remove the edges of P1 (why?) and then find another path P2 . If we are lucky, we shall be able to find the two edge-disjoint paths; one being (s → a → c → t), and the other (s → b → d → t or s → d → t). But if we are unlucky, we shall land in a difficult situation as depicted in Fig. 6.4.1. If path P1 = (s → a → d → t) (as shown by bold lines in Fig. 6.4.1) then we are blocked; it has now become impossible to find another path in this graph. Thus an initial wrong choice will make things hard for us. We are in dire need of some innovation? There are potentially two ways to fix this problem: 1. Pick the initial path more intelligently - but this may not be an efficient solution, why? 2. Do not delete the edges of the chosen path - and do something else with these edges. What is that something else? Before we actually try something new consider the following:

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Figure 6.4.1: If we initially select a path as shown by bold red lines (left diagram) and remove it from the graph (right diagram) then it becomes impossible to find another path in this graph. 1. Algorithm 36 fails to find the maximum number of edge-disjoint paths in a graph but it still finds maximal number of edge-disjoint paths in a graph. 2. Is this complication, (where an initial wrong choice will make things hard) a general phenomenon in almost all graphs or is this a problem in a certain class of graphs? 3. Is there a class of directed or undirected graphs where an initial wrong selection would not create any complication? How is the class of graphs (where an initial wrong selection really matters) structurally different from the other class (where an initial wrong selection does not matter)? 4. Is it possible to convert one class of graphs into another? If this is possible then after the conversion we can use our (stupid) Algorithm 36 to find out maximum edge-disjoint paths in a graph? 5. We are given a directed acyclic graph D with a source vertex (a vertex with zero in-degree) s and a sink vertex (a vertex with zero out-degree) t. The in-degree is equal to the out-degree for every vertex i of D other than the vertices s and t. Show that if you apply Algorithm 36 to this graph then you will be able to find the maximum number of edgedisjoint paths without any complication. How about if the directed graph D is cyclic but the in-degree as well as the out-degree of every node are equal? You should consider both cases: when vertices s and t are not part of any cycle and when they are also part of some cycles. See Fig. 6.4.2 before arriving at a conclusion.

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Figure 6.4.2: A directed graph D in which the in-degree of every vertex is equal to the out-degree of every vertex including vertices s and t. This diagram is provided by Khawaja Fahd. Finding edge-disjoint paths in a general directed graph may be hard to find; expecting an innovation to achieve this objective may be unrealistic; let us try to solve a simpler problem (finding pseudo edge-disjoint paths) and then use our newly found experience and confidence to solve the harder problem (finding edge-disjoint paths). In edge-disjoint paths an edge (x, y) cannot be shared by more than one path; in case of pseudo edge-disjoint paths we follow the following rules of the game: 1. A directed edge (x, y), not already in use by a path, can be traversed (or used) by a path from x to y and not from y to x. The status of such an edge when traversed by a single path from x to y will be “used”. 2. A directed edge (x, y), already traversed by a path from x to y, can be traversed by a new path from y to x and not from x to y. The status of such an edge, which is traversed by two paths in opposite directions will be “not used by any path” or “unused”. Please note that rule number 1 is common in pseudo edge-disjoint as well as edge-disjoint paths. It is rule number 2, which makes the two categories different. Finding maximum number of pseudo edge-disjoint paths is a problem which can be solved using the exciting process of discovery based learning. We shall later show in this section that the number of pseudo edge-disjoint paths between two vertices in a graph is exactly equal to the number of edgedisjoint paths, in fact, given a set of pseudo edge-disjoint paths in a graph it is possible to find an equal sized corresponding set of edge disjoint paths in the graph.

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So the current problem is how to maximize the number of pseudo edgedisjoint paths in D? Instead of deleting all edges in a (recently found) path (as is done in step number 2 of Algorithm 36) we reverse the direction of each edge in the path. Algorithm 37 is based on this innovative idea. Let us carefully look at this new algorithm which claims to find the maximum number of pseudo edge-disjoint paths from s to t in a directed graph D. It also outputs the Status(e) of every edge e of D. Initially no edge of D, is occupied by any path and thus the Status of every edge is unused. In step number 2 we find a path from vertex s to t in D using any path finding algorithm. As soon as an unused edge is occupied by a path its Status is changed from unused to used in step number 4. In step number 4 (instead of removing every edge in the path as was done in Algorithm 36) we reverse the direction of every edge in the path found in step number 1, and then go back to step number 2. Note that Algorithm 37 (unlike Algorithm 36) allows an edge to be used by two different edge disjoint paths moving in opposite directions. Algorithm 37: Find Maximum pseudo edge-disjoint paths from s to t in directed graph D. input : Directed graph D, and vertices s & t, Status(e) = unused for every directed edge e in D. output: Maximum pseudo edge-disjoint paths from s to t in D and a Status(e) for every directed edge e of D. 1 2

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Example 6.4.1. Let us execute this simple four line algorithm on the graph of Fig. 6.4.3. We apply the above algorithms on the graph D shown in the same figure. We first find maximum pseudo edge-disjoint paths from vertex s to vertex t in this graph. In this process (Algorithm 37) we convert graph

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Algorithm 38: Find Minimum Cut in a graph D. input : Directed graph D, and vertices s & t. output: Edges of D that belong to the Minimum Cut. 1 2 3

Convert D into F using Algorithm 37; Let P be the set of vertices (of F ) reachable from vertex s in F ; Minimum Cut will consist of edges (x, y) of D Where x belongs to P and y belongs to V (D) − P ;

Figure 6.4.3: Original Graph D

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D into F . Graph F helps us identify the set P using Algorithm 38. Using this set P we identify edges of D which belong to the minimum cut. The paths P1 (green) and P2 (blue) that we have found during the execution of this algorithm are not edge-disjoint paths (see Fig. 6.4.4), yet this algorithm tells us the maximum number of possible edge-disjoint paths in D (why and how?). The most useful result of this algorithm is the Status of every edge e in D after its termination. The diagram on the bottom right of Fig. 6.4.4 shows the original graph D with used edges only. After deleting unused edges (from D) we end up with this new graph; this new graph has some very desirable properties which we shall elaborate later. The Minimum Cut is found in Fig. 6.4.5. Before jumping to any conclusions and in order to gain more confidence let us solve another example. Example 6.4.2. We execute Algorithm 37 once again on a directed graph D shown in Fig. 6.4.6. The top left diagram shows the directed graph D with a selected path P1 from vertex s to vertex t shown in bold. The top right diagram shows the edges of another path P2 shown in bold. The bottom left diagram shows the edges of the path P3 shown in bold. The original graph D is shown in the bottom right diagram with used edges shown in bold. Again Algorithm 37 provides the maximum number of edge-disjoint paths present in this graph although the paths P1 , P2 & P3 found by this algorithm are not edge-disjoint as shown in Fig. 6.4.6. Let us again consider graph D (of Example 6.4.1) minus the unused edges as shown in the bottom of Fig. 6.4.4. Let us also consider the graph D (of Example 6.4.2) minus the unused edges shown in Fig. 6.4.6. Both these graphs share something in common; they both belong to a (desirable) class? You can apply (the very naive) Algorithm 36 to this class of graphs and you will find maximum sized subset of edge-disjoint paths without any difficulty.

6.4.2

The Concept of a Minimum Cut in Directed Graphs

Let us define the concept of a cut more formally. Let S be the set of vertices of D containing s but not t and let the subset T = V (D) − S. Then (S, T ) is a set of directed edges from a vertex in S to a vertex in T . The set (S, T ) will be a subset of E(D) and is known as a cut in the graph D (a cut signifies edges which if removed from a graph will destroy all paths from vertex s to

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vertex t). Out of many possible cuts we show only three cuts in the directed graph shown in the top diagram of Fig. 6.4.7. Please note that it is the same directed graph for which we have found maximum number of edge-disjoint paths as shown in Fig. 6.4.1. For Cut A : SA = {s, a, b, d}; TA = {c, t}; (SA , TA ) = {(a, c), (d, t)} For Cut B : SB = {s, a, b}; TB = {c, d, t}; (SB , TB ) = {(s, d), (a, c), (a, d), (b, d)} For Cut C : SC = {s, b, d}; TC = {a, c, t}; (SC , TC ) = {(s, a), (d, t)} It is interesting to note that the edge (a, d) is physically cut by Cut B as well as Cut C but it is only included in the Cut B and not in Cut C. This is because in Cut B vertex ‘a’ belongs to SB while vertex d belongs to TB as shown in the bottom left diagram of Fig. 6.4.7. While in Cut C, vertex ‘a’ belongs to TC and vertex d belongs to SC . Thus the edge (a, d) in the Cut C does not belong to the set (SC , TC ) as shown in the bottom right diagram of Fig. 6.4.7. Looking at this problem from a different angle: In Cut B you have to remove all four edges present in this cut in order to destroy all paths from vertex s to t in D. While in Cut C you have to remove only two edges present in this cut and vertex s is disconnected from t; edge (a, d) is not contributing to a path from s to t in D and is therefore not required to be removed. The size of Cut B is larger then the size of Cut A and that of Cut C. By a little inspection in this graph it can easily be verified that Cut A is indeed one of the minimum cuts, (SA , TA ) is equal to U (which is the minimum sized subset of edges which, if removed will destroy all paths from s to t in D). Thus λ(s, t) = |(SA , TA )| = 2. According to Menger’s Theorem the maximum number of edge-disjoint paths in this graph will equal λ(s, t), which is 2. How can we find the minimum cut systematically? We shall discuss it in a proof of Menger’s Theorem.

6.4.3

A Proof of Menger’s Theorem and Finding the Minimum Cut in Directed Graphs

We are given a directed graph D and two special nodes s and t. We intend to prove Menger’s Theorem in terms of edge connectivity of a directed graph. The theorem states that the maximum number of edge-disjoint paths from vertex s to t is equal to the minimum number of edges of D which if removed will destroy all paths from s to t in D. We shall provide a constructive proof.

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Using this proof one can also find the maximum number of edge-disjoint paths as well as the minimum cut. This proof technique can easily be adapted for undirected graphs and can also be used to prove the vertex form of Menger’s Theorem which states that, the minimum number of vertex-disjoint paths from s to t is equal to the cardinality of minimum sized vertex cut-set which will disconnect t from s in D. We know that number of edge-disjoint paths from vertex s to t in a graph cannot exceed the size of a cut (SX , TX ) where vertex s belongs to the subset SX while vertex t belongs to TX as defined earlier in this section. As we have seen before there may be different cuts possible with different sizes in the same graph as shown in Fig. 6.4.7. If we try to maximize the number of paths from s to t they will only be limited by the bottleneck in the graph which will be the minimum cut. That means the maximum number of edgedisjoint paths from s to t in D will not exceed |(S, T )|min where the minimum is taken over all cuts (S, T ) in D. Menger’s Theorem tells us that the two quantities (maximum number of paths and size of the minimum cut) are exactly equal. We shall make the following claims here: Claim 6.4.1. Algorithm 37 finds k pseudo edge-disjoint paths from s to t in D before it terminates; there can not be more than k pseudo edge-disjoint paths in D. Claim 6.4.2. The maximum number of pseudo edge-disjoint paths from vertex s to t is equal to the minimum number of edges of D which if removed will destroy all paths from s to t in D, i.e., the minimum cut. Claim 6.4.3. The number of real (not pseudo) edge-disjoint paths from s to t in D can not exceed the number of (pseudo) edge-disjoint paths from s to t in D. The maximum number of the two paths will be exactly equal (given one, we should be able to find the other). Proof for Claim 6.4.1 & 6.4.2: In step number 2 of Algorithm 37 we find a path from vertex s to t in D using any path finding algorithm. We then reverse the direction of each edge of the path just found, and go back to step number 2 to find a new path (see Fig. 6.4.4). Please note that each (pseudo) path found in step number 2, may consist of unused edges (not used by any path so far) or used edges (occupied by a previous path in the

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opposite direction) as shown in the top right corner of Fig. 6.4.6. We keep on doing this until we are no longer able to find a path from s to t in D and then the algorithm terminates. It is obvious that if there is a (pseudo) path left (in addition to the one’s that are already found) then Algorithm 37 will find it before its exits. a

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Figure 6.4.8: The bottom left diagram of Fig. 6.4.5 is duplicated in the top left corner; it shows the number of vertices (enclosed in the shaded area) which are reachable from s in D after the termination of Algorithm 37. The maximum number of pseudo edge-disjoint paths is shown in bold in the top right diagram. The bottom diagram shows the modified graph D0 with used edges shown in bold. We have seen that the maximum number of pseudo edge-disjoint paths cannot exceed the size of a minimum cut; what we need to show now is that there will actually be a cut whose size will be equal to the maximum number of paths from s to t in D. Let us assume that while executing this algorithm we have reached a stage where we are no longer able to find a new path and the Algorithm 37 terminates. Although no (additional) path from vertex s to t exists anymore, but it still may be possible to reach a number of vertices in the modified graph D from the vertex s. Let P be a nonempty set (a subset of V (D)) containing vertex s and all other vertices which are reachable from s. We define subset Q to be equal to V (D) − P . It is important to note that the vertex t will belong to Q otherwise it will be in P and then we can find an extra path from vertex s to vertex t in D. The Cut (P, Q) will consist of edges such that each edge which is part of this cut will be occupied by a unique

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path from vertex s to t in D. The number of paths cannot be larger than the size of this cut as discussed before. Can the number of paths be smaller than the size of this cut? If the number of paths is smaller than the size of the cut then you can identify at least one edge (x, y) which is part of the cut but not part of any path from s to t in D. Under such conditions vertex y will also be reachable from vertex s which contradicts our initial assumption. Thus each edge of the Cut (P, Q) will be part of exactly one pseudo path from vertex s to t in D. This situation is depicted in Fig. 6.4.5. When Algorithm 37 is unable to find an extra path it terminates as shown in the bottom left corner of Fig. 6.4.5. The set of vertices which are reachable from s (known as P ) are shown shaded in this diagram. The Cut (P, Q) consists of edges all of which are already occupied by existing paths as shown in the bottom right corner of Fig. 6.4.5. The Cut (P, Q) is in fact the Minimum Cut equal to the maximum number of edge-disjoint paths from vertex s to t in D. Proof for Claim 6.4.3: We have seen that the Algorithm 37 not only provides maximum number of pseudo edge-disjoint paths (see Fig. 6.4.4 & Fig. 6.4.6.) but also the status of every edge is provided by it. If we remove all edges of D which have a Status = unused from D then the resulting graph D0 would have the following properties: 1. It will be a directed acyclic graph. 2. The in-degree of every node other than the vertices s and t will be equal to its out-degree. 3. The out-degree of s will be equal to the in-degree of t in D. The above three properties guarantee that the modified graph D0 has as many (real) edge-disjoint paths from s to t as the number of pseudo edgedisjoint paths in the original graph D. In fact it is possible to find the (real) edge-disjoint paths from the modified graph D0 (as shown in the bottom of Fig. 6.4.4) using (the very simple) Algorithm 36.

6.4.4

Finding Maximum Vertex-Disjoint Paths & Minimum Vertex Cut in Directed Graphs

We have shown earlier that the maximum number of edge-disjoint paths (from vertex s to vertex t) in a directed graph is equal to the minimum

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number of edges (which if removed will destroy all paths from vertex s to t). The vertex form of Menger’s Theorem equates maximum number of vertexdisjoint paths (from vertex s to vertex t) to minimum number of vertices which if removed will destroy all paths from vertex s to t in a directed graph. The Fig. 6.4.9 shown below shows a directed graph H in the top left diagram. It also shows how this graph would look like if we remove a number of vertices. Please note that removal of certain vertices does not destroy all paths from vertex s to vertex t in this graph (see the top right and bottom left diagrams) while the removal of some other vertices does indeed destroy the s − t connectivity in this graph.

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Figure 6.4.9: Various vertices are removed to show that the graph does not necessarily become disconnected by their removal and only specific ones make the graph disconnected. There are basically two issues that we would like to tackle: 1. An efficient algorithm to find maximum number of vertex-disjoint paths from vertex s to t and a minimum cut in terms of vertex cut-set. 2. A proof that maximum number of vertex-disjoint paths (from vertex s to vertex t) is equal to the minimum number of vertices which if removed will destroy all paths from vertex s to t in a directed graph.

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Interestingly both these problems can be resolved using our prior experience provided we make a couple of transformations on a given directed graph H (left diagram) and convert it into another directed graph D (right diagram) as shown in Fig. 6.4.10 using Algorithm 39. We split each vertex x (excluding vertex s and t) of directed graph H into x1 and x2 in D; each directed edge (x, y) in H is transformed into a directed edge (x2 , y1 ) in D. We insert an extra edge between x1 and x2 in D for every vertex x in H except for s and t as shown here. For every edge (s, x) in H insert a directed edge from s to x1 in D. Similarly for every edge (x, t) in H insert a directed edge from x2 to t in D. Thus for every edge in H, there will be a corresponding edge in D as shown in brown color in the right diagram of Fig. 6.4.10. Let us call these edges external edges. In addition to these edges we have edges of the form (x1 , x2 ); we call these edges internal edges. Thus for the new graph D we have V (D) = 2V (H) − 2 and E(D) = E(H) + V (H) − 2. Algorithm 39: Transform directed graph H into directed graph D. input : Directed graph H with special vertices s & t. output: Directed graph D with internal & external edges marked 1

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1. There is a path from vertex s to t in D corresponding to any path from s to t in H. The corresponding path in D passes through external as well as internal edges as shown in Fig. 6.4.11 (top left corner). 2. Every path going from vertex s to t in H has to pass through a number of k intermediate nodes (nodes other than s and t), the corresponding path in D will pass through k internal edges, and thus any number of edge-disjoint paths in D will correspond to the same number of vertexdisjoint paths in the directed graph H. 3. A Minimum (edge) Cut in graph D will correspond to a Minimum (vertex) Cut in graph H provided the Minimum (edge) Cut in D passes through internal edges only. The above observations provide us enough insight to find maximum edgedisjoint paths in the directed graph D which will correspond to maximum vertex-disjoint paths in H. We can construct (an almost) similar constructive proof that the maximum number of vertex-disjoint paths from s to t in H is exactly equal to the minimum number of vertices which if removed from H will destroy all paths from s to t in H. We can use Algorithm 39 to convert graph H into graph D and then use Algorithm 37 without any change to find the maximum edge-disjoint paths and the Minimum Cut, the working of this approach on the directed graph D of Fig. 6.4.10 is shown in Fig. 6.4.11. There is just one problem to be resolved, the Minimum Cut as found by the original Algorithm 37, may pass through some of the external edges as shown in the middle right diagram in Fig. 6.4.11. It is important to appreciate that the minimum cut should (be forced to) pass through internal edges only. If it passes through (some of the) external edges then the minimum cut will not correspond to a minimum vertex cut. In order to do so we make a one line modification in our earlier approach in terms of Algorithm 40 as described below. We first add the external edges of graph D in graph F and then find the appropriate cut as shown in the bottom diagrams of Fig. 6.4.11 (thereby forcing the entire cut to pass through internal edges alone).

6.4.5

Menger’s Theorem for Undirected Graphs

We shall consider Menger’s Theorem in terms of edge connectivity (Menger’s Theorem in terms of vertex connectivity can also be derived from in a similar

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Algorithm 40: Find a Minimum Cut in a directed graph D passing through internal edges only. input : Directed graph D with special vertices s & t and internal & external edges marked output: Internal Edges of D that belong to the Minimum Cut. 1 2 3 4

Convert D into F using Algorithm 37; Add all the external edges of Graph D to F ; Let P be the set of vertices (of F ) reachable from vertex s in F ; Minimum Cut will consist of edges (x, y) of D Where x belongs to P and y belongs to V (D) − P ;

fashion). It states that the maximum number of edge-disjoint paths between two specific nodes in an un-directed graph G is equal to minimum number of edges needed to destroy all paths between the two specific vertices. An un-directed graph G is shown in Fig. 6.4.12 along with two vertices s and t and a path from s to t shown by bold lines. The undirected graph G of Fig. 6.4.12 is transformed into a directed graph D as shown in Fig. 6.4.13. The path between s and t in G (Fig. 6.4.12) is also shown in bold in the top left diagram of Fig. 6.4.13. Each undirected edge {x, y} of G is thus transformed into two directed edges (x, y) and (y, x) in D. It is reasonable to assume that any directed path from s to t in D will consume either (x, y) or (y, x) but not both. For every un-directed path between vertex s and t in G there is a corresponding directed path from s to t (or from t to s) in D. The maximum number of directed edge-disjoint paths from s to t in D will be equal to the maximum number of un-directed edge-disjoint paths between s and t in G. We can in fact use Algorithm 37 without any changes to find maximum number of edge-disjoint paths from s to t as shown in Fig. 6.4.13. The minimum cut corresponding to minimum number of edges required for destroying all paths between s and t can also be found using similar techniques as shown in the bottom left diagram of Fig. 6.4.13.

Maximum Vertex-disjoint Paths & Minimum Vertex Cut in un-directed graphs Now considering Menger’s Theorem in terms of vertex connectivity of a graph where s and t are already given. Menger’s Theorem states that the maximum

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Figure 6.4.12: This is Graph G (left diagram); if we initially select a path as shown by bold red lines and then remove its edges from G then it becomes impossible to find another path in the remaining un-directed graph (right diagram). number of vertex-disjoint paths between two specific nodes in an un-directed graph G is equal to minimum number of vertices needed to destroy all paths between the two specific vertices. An un-directed graph G is shown at the top of Fig. 6.4.14 along with two vertices s and t. The un-directed graph G of Fig. 6.4.14 is transformed into a directed graph D as shown in the middle diagram of Fig. 6.4.14. Thus each un-directed edge {x, y} of G is transformed into two directed edges (x, y) and (y, x) in D. Then all the vertices except for s and t are split into two as shown in the bottom diagram of Fig. 6.4.14; for every vertex x there is a corresponding two vertices {x1 , x2 }. Thus the total number of vertices and edges are almost doubled. |V (D)| = 2|V (G)| − 2 and |E(D)| = 2|E(G)|. We then apply the standard algorithm for finding edgedisjoint paths from vertex s to vertex t as shown in the top-right diagram of Fig. 6.4.14 and reverse the edges. The only difference from the previous procedure is that we apply Algorithm 39 to find the Minimum Cut for vertexdisjoint paths. This way in the bottom-left diagram of Fig. 6.4.15 the external edges of Graph D of Fig. 6.4.14 are added where they do not exist in graph F (not shown in the Figure) and the minimum cut is found and applied to Graph G.

6.4.6

Edge Connectivity and Vertex Connectivity for Undirected Graphs

Edge Connectivity λ(G) of an undirected graph G is the minimum number of edges which if removed will disconnect graph G. Vertex Connectivity κ(G) of a Graph G is the minimum number of vertices which if removed will

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Figure 6.4.15: The top-left diagram shows the initial graph F , where path P1 (pink) is found and the edges of the path are reversed as shown in the topright diagram. The middle diagram shows another path P2 (blue) found and its edges are reversed in the middle-right diagram, as can be seen no more paths can be found. Hence we apply Algorithm 39 and find the minimum cut for the vertex-disjoint paths in the bottom-left diagram and that cut is applied to original Graph G of Fig. 6.4.14.

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disconnect graph G. The problem is to efficiently find both edge connectivity and the vertex connectivity of an undirected graph G. The diagrams in Fig. 6.4.16 show two undirected graphs. The top graph shows its edge connectivity. The bottom diagram shows the vertex connectivity of the bottom graph.

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Figure 6.4.16: The top diagram shows the edge connectivity of the top graph and the bottom one shows the vertex connectivity of the bottom graph. Our prior knowledge and expertise tells us that given an undirected graph G and two already selected special vertices s and t we can efficiently find λ(s, t) which is equal to minimum number of edges which if removed will disconnect vertex s from vertex t in G. (We actually do it by finding maximum number of edge-disjoint paths from vertex s to t in a graph G). Is it possible to arbitrarily select s and t in the graphs below, find λ(s, t) and claim that λ(s, t) is equal to the edge connectivity of graph G, that is, λ(s, t) = λ(G)? The problem may not be that simple as it is evident from the diagram below (Fig. 6.4.17). There is a possibility that a wrong choice for vertex s and vertex t may give you an incorrect result as shown in the following diagrams. What is then the way out? Perhaps we should consider all possible (s − t) pairs in the graph G? As the graph G consists of p vertices, there could be O(p2 ) distinct pairs of vertices. For each such pair (we call it an s−t pair) we find the maximum number of edge-disjoint paths from vertex s to vertex t in the undirected graph G. We know that the maximum number of edge disjoint

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Figure 6.4.17: Various Min-cuts in the same graph. paths is equal to λ(s, t). Out of all the O(p2 ) values for λ(s, t) (because of O(p2 ) distinct pairs of s−t vertices) we select the one with the minimum value and that will be the edge connectivity λ(G) of the graph. It will be a useful experience to derive the overall time complexity of this simple algorithm. The next logical step should be to find ways to reduce this complexity. You may have realized that there is no need to consider all O(p2 ) distinct pairs of vertices; only O(p) pairs will be sufficient for calculating edge connectivity as shown in Fig. 6.4.18. The problem of efficiently finding vertex connectivity is slightly more complex as we shall explain in the coming section. Edge Connectivity of an undirected Graph We have already hinted before that in order to efficiently compute edge connectivity of an undirected graph, we may fix s arbitrarily but then compute λ(s, t) for every different value of t in the undirected graph as shown in the figure below. We thus have to apply the maximum edge-disjoint paths finding algorithm for all possible p − 1 pairs where s is fixed while t takes on every possible value as shown in Fig. 6.4.19. This intelligent observation will certainly cut down the time complexity of our earlier technique as described in the algorithm given below. We select a vertex s and keep it fixed throughout the working of the algorithm. We select a different vertex t in each of the p − 1 graphs and in each such graph we find the maximum number of edge-disjoint paths from vertex s to vertex t in time equal to O(pq). We repeat this process as many number of times as the number of graphs. Out of all results we select the minimum.

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Algorithm 41: Find edge connectivity of an un-directed graph G. input : Undirected Graph G with vertices numbered from 1 to p. output: Edge Connectivity of graph G. 1

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Figure 6.4.19: When is vertex s fixed and when does vertex t change its position? The total time complexity will thus be O(p2 q). Problem Set 6.3. Problem 6.3.1. We now know how to find the magnitude of λ(G). How about finding the actual edges belonging to the minimum sized set of edges which if removed will disconnect graph G? Problem 6.3.2. Assume that instead of finding edge connectivity (which is an optimization problem) we intend to solve the corresponding decision problem: Is the edge connectivity of graph G less than or equal to k where k is an arbitrary number and is always less than p (why?)? Of course you can use the edge-connectivity finding algorithm to solve this problem but then it will be an overkill. So let us design an efficient algorithm to solve this problem. Once we have solved the decision problem it becomes almost trivial to solve the corresponding optimization problem of finding the (minimum) edge connectivity of a graph. In fact the problem becomes a search problem in a finite search space. You have the option of making a linear search or a more efficient binary search in order to find the edge connectivity of a graph. Try to derive the time complexity of this algorithm. Let us now try to make the previous algorithm more efficient. We make p − 1 copies of graph D numbered 2 to p and then fix vertex s and t in each copy. In the j th copy of graph G we select vertex j as t where 2 ≤ j ≤ p. In every

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Figure 6.4.20: We make p − 1 copies of graph D numbered 2 to p and then fix vertex s and t in each copy. Now instead of finding Max-Paths or the Min-Cut in each graph we find k edge-disjoint paths in each copy of graph G starting from k equal to 1. If we can find k edge-disjoint paths in all copies then we move forward otherwise we exit out with the edge connectivity of the graph G.

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Algorithm 42: Find if the edge connectivity λ(G) of a graph G less than or equal to k? input : An un-directed graph G output: Yes/No 1

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copy of graph G we select vertex 1 as s. Now instead of finding Max-Paths or the Min-Cut in each graph we find k edge-disjoint paths in each copy of graph G starting from k equal to 1. If we can find k edge-disjoint paths in all copies then we move forward otherwise we exit out with the edge connectivity of the graph G. The modified Algorithm 43 is described below. Try to derive the time complexity of this algorithm. Algorithm 43: Find edge connectivity λ(G) of graph G. input : Un-directed Graph G with vertices numbered from 1 to p. output: Edge Connectivity λ(G) of graph G. 1

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Vertex Connectivity of an undirected Graph We know how to find κ(s, t), that is the minimum number of vertices which if removed will destroy all paths between vertex s and t in a given undirected graph G. If we find a κ(s, t) for every possible pair of vertices in the graph G and then select the minimum (out of all O(p2 ) possible values), we shall get the vertex connectivity κ(G) for graph G. How about if we use the same complexity cutting strategy used in finding the edge connectivity; select s arbitrarily but allow t to have all possible values? Please concentrate on the following algorithm (Algorithm 44)

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Algorithm 44: Find vertex connectivity κ(G) of graph G. input : Undirected Graph G with vertices numbered from 1 to p. output: Vertex Connectivity κ(G) of graph G? 1

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Assume that the algorithm (Algorithm 44) is applied to the graph shown below. The vertex connectivity κ(G) of this graph is also indicated here. The outcome of the above algorithm when applied to this graph (Fig. 6.4.21) is shown in Fig. 6.4.22. a

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We have earlier proved Menger’s Theorem which states that the maximum number of edge-disjoint (vertex-disjoint) paths from vertex s to t in a directed graph D is equal to the minimum number of edges (vertices) of G which need to be removed in order to disconnect vertex t from s. This theorem is applicable to any graph while the other two theorems (Hall’s Theorem and Konig’s Theorem) are applicable to bipartite graphs only. Let us now start with an (undirected) bipartite graph B. We transform this graph into a directed graph D after adding vertices s and t as shown in the left diagrams in Fig. 6.5.1. We find maximum edge-disjoint paths from s to t in the graph D as shown in the bottom right diagram of this figure. Each path from vertex s to vertex t in this diagram corresponds to a matching edge in the bipartite graph B as shown in the top right diagram of this figure. Thus maximizing number of paths in D is equivalent to finding maximum matching in B. The directed graph D with maximum number of edge-disjoint paths is transformed into a directed graph F after the direction of each path in D is reversed as shown in the top left diagram of Fig. 6.5.2. The following observations will help us in proving the remaining two theorems, namely Konig’s and Hall’s Theorem. 1. Every edge-disjoint path from vertex s to t in the graph D is in fact a vertex-disjoint path between the same two vertices. We shall, therefore, refer to these paths as paths only and not vertex-disjoint or edgedisjoint paths. 2. There will be a path between vertex s and t in D corresponding to every matching edge in the bipartite graph B. Thus the maximum number of paths between vertex s and t in G will be equal to the size of the maximum matching in the bipartite graph. If you cannot find an additional path from vertex s to vertex t in graph D then it means that you can not improve the size of the existing matching in the bipartite graph. 3. The minimum cut in terms of number of edges will be equal to the minimum cut in terms of minimum number of vertices.


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Figure 6.5.1: The top left diagram shows a bipartite graph B. The bipartite graph is transformed into a directed graph D with additional vertices s and t as shown in the bottom left diagram. We find maximum number of edgedisjoint (or vertex-disjoint) paths in graph D as shown in the bottom right diagram. Corresponding to each edge-disjoint path from vertex s to vertex t in the directed graph D, there is a matching edge in the bipartite graph as shown in the top right diagram. Maximizing number of paths in D is equivalent to finding maximum matching in B.

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Figure 6.5.2: The top left diagram shows directed graph D where it is no longer possible to find an additional path from vertex s to t. However it is still possible to reach some vertices from s; these vertices form the set P . The set X contains those vertices which are common in the set P and the partite A. The middle right diagram shows the minimum cut. The bottom left diagram shows the set Y containing (minimum number of) vertices which if removed will destroy all paths from vertex s to t in the directed graph. The set Y is in fact the vertex cover in the bipartite graph shown in the bottom right diagram.


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4. The minimum sized vertex-disconnecting set in the graph D will correspond to the vertex cover in the bipartite graph B. 5. Let P represent the vertices which are reachable from vertex s once we have found maximum number of paths and it is no longer possible to find an additional path from s to t in D as shown in the top left diagram of Fig. 6.5.2. Obviously P will not contain t (why?). Let X represents vertices which are common between P and the partition A as shown in the middle left diagram of the same figure. The neighborhood N (X) of X is shown in the middle right diagram of this figure. 6. The minimum cut in terms of edge-disconnecting set will be edges from s to A − X and from N (X) to t as shown in the middle right diagram of Fig. 6.5.2. The size of the minimum cut will thus be equal to the size of A − X and N (X). 7. The size of the minimum cut in terms of vertex-disconnecting set will be kA − Xk + kN (X)k. This set (A-X +N(X)) is denoted by Y in the bottom left diagram of Fig. 6.5.2. The same set Y is in fact the vertex cover in the bipartite graph in the bottom right diagram. 8. If the set P does not contain any vertex of A then it implies that the number of paths from s to t is equal to the degree of node s in the graph. It also implies that every vertex belonging to A is matched to a vertex in B in the corresponding bipartite graph. 9. If the set P contains a vertex x belonging to A then obviously vertex x is not matched to a vertex in the B in the maximum matching of the bipartite graph. Under such conditions the size of the maximum matching will be (at least one) less than the size of partition A. There are two major cases to be considered as shown in Fig. 6.5.3. Problem Set 6.4. Problem 6.4.1. The Fig. 6.5.4 below shows a bipartite graph B in the top left diagram. It is first transformed into a directed graph D as described earlier, we find maximum edge-disjoint paths in D, then reverse the direction of each path, the resulting directed graph F is shown in the top right diagram of this figure. Each matching edge in the bipartite graph corresponds to a

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Problem 6.4.2. The figure shown above (Fig. 6.5.4) shows two different maximum matchings of the same bipartite graph B in the bottom diagrams. For each bipartite graph (with a maximum matching indicated) draw the corresponding graph F . You know that in graph F , there is a directed path from vertex t to vertex s corresponding to each matching edge in B. Find the set P, X, A − X, and N (X). Also find the minimum cut in terms of vertices as well as edges.

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Problem 6.4.3. The left diagram of Fig. 6.5.5 shows a bipartite graph with maximum matching edges indicated in different colors. It is obvious that the maximum matching in this bipartite graph is not a perfect matching – all vertices of partition A are not matched. Under such conditions vertex s will be part of a directed cycle, such that the number of vertices of partition A in the cycle will always be one larger than the number of vertices of partition B belonging to the same cycle. Can we prove this observation in general or is this localized to this graph only? What are the implications of this observation? Is this something to do with the proof of Hall’s Theorem?

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Figure 6.5.5: Problem 6.4.4. We show a bipartite graph with two different maximum matchings in Fig. 6.5.6. The corresponding graph F showing reversed paths is shown on the right side of each bipartite graph. The set of vertices reachable from vertex s, known as the set P is also indicated in each graph F . It is quite evident from these diagrams that for different maximum matchings in the same bipartite graph we get the same set P . Is this a coincidence in this bipartite graph or will it always be true. Prove or give a counter example. Problem 6.4.5. The figure shown below (Fig. 6.5.7) shows a bipartite graph B with maximum matching edges shown in different colors. The right diagram shows the same bipartite graph with two extra vertices s and t added to it. Note that the bipartite graph B in this problem is transformed into an un-directed graph G instead of a directed graph D. It is obvious that corresponding to every matching edge in this bipartite graph, there is a path from


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between vertex s and vertex t in the graph G (Is this one to one correspondence between a matching edge in the bipartite graph and an edge-disjoint path in graph G a general phenomena or is it restricted to this bipartite graph? We shall address this issue in the next problem). Find the minimum edge cut & minimum vertex cut in graph G and show that it is equal to the size of the vertex cover in this bipartite graph B.

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Figure 6.5.7: Problem 6.4.6. Assume that we convert a bipartite graph (any bipartite graph – not necessarily the one shown in Fig. 6.5.7) into an undirected graph G as shown in the figure above. We find maximum edge-disjoint (or vertexdisjoint) paths between vertex s and vertex t in graph G. Do you think that corresponding to every edge-disjoint path between vertex s and t in this undirected graph, there will be a corresponding matching edge in the bipartite graph B? Either prove or give a counter example.

6.6 6.6.1

Network Flows Finding Maximum Edge-Disjoint Paths in MultiGraphs

We consider here the problem of finding maximum number of edge-disjoint paths (and a minimum cut in terms of edges) from a vertex s to a vertex

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t in a directed multi-graph. We assume that there are no self loops but parallel edges are allowed in the directed multi-graph. We can use our earlier algorithms to solve this problem; calculating the resulting time complexity would be interesting? We can certainly make adjustments in order to increase the efficiency of our earlier approach. This may be possible without damaging the original character of our algorithm. We reproduce an earlier algorithm below for a ready reference. We designed this algorithm to find maximum edge-disjoint paths in a directed graph. Algorithm 45: Find Maximum pseudo edge-disjoint paths from s to t in directed graph D. input : Directed graph D, and vertices s & t, Status(e) = unused for every directed edge e in D. output: Maximum pseudo edge-disjoint paths from s to t in D and a Status(e) for every directed edge e of D. 1 2

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Whenever we find a path P from vertex s to a vertex t according to step number 2 of this algorithm we find the vertices through which P passes; all paths parallel to P (that means passing through the same set of vertices) can be discovered right away. For example if P = (s → a → d → t) as shown in the top left diagram of Fig. 6.6.1 then we have not one but three edgedisjoint paths passing through the same vertices. Instead of finding these paths sequentially (strictly according to this algorithm) we should be able to do it in one go as shown in the same figure. How this will cut down the time complexity of our modified approach? We have asked you to derive the time complexity of this algorithm (with or without modification) when applied to Multi-graphs. It will be useful if we spend some time on the selection of a suitable data structure to represent

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a multi-graph. The left diagram of Fig. 6.6.2 reproduces the multi-directed graph of Fig. 6.6.1. Please note that it is an un-weighted graph. The right diagram shows how it can be represented by a weighted graph. The weight of each edge (x, y) in this weighted graph corresponds to the number of edges from vertex x to vertex y in the un-weighted multi-directed graph shown in the left diagram. Note that this graph (shown in the right diagram) is a simple graph with no parallel edges, and can conveniently be represented by a weighted adjacency list or adjacency matrix data structure. How can we exploit this data structure in order to use Algorithm 45 (efficiently) to solve the edge connectivity problem? When you derive the time complexity of Algorithm 45 (or its modified version) you may realize that the complexity expression may depend upon the graph edge weights in addition to the size of the problem (that means number of vertices and edges). That is in fact a more serious problem (than having a less efficient algorithm); it implies that your algorithm is no longer an algorithm but a technique? Remember that in an algorithm the complexity depends upon the number of vertices and edges of a graph and does not depend upon the edge weights in the graph while in a technique it may depend upon edge weights as well. Why has this happened? How can we over come this shortcoming?

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The Maximum Flow & the Minimum Cut

We are given a network which is essentially a weighted directed graph with two special vertices (a source vertex) s and (a sink vertex) t as shown in Fig. 6.6.3. Each directed edge is weighted with a positive integer; the weight of the edge is known as the capacity of the edge. If (u, v) is a directed edge then the capacity of this edge is denoted by c(u, v). The network flow problem is to find the maximum flow which can take place from the source vertex to the sink vertex such that the flow f (u, v) taking place in any edge (u, v) does not exceed the capacity of the edge, that is, c(u, v). We assume that flow can originate from the source vertex and can be consumed by the sink vertex. All vertices other than the source or the sink (known as intermediate vertices) can neither generate any flow nor consume any flow. It means that the incoming flow through an intermediate vertex is exactly equal to the outgoing flow through that vertex. The net flow coimng out of the source verex s is greater than zero while the netflow sinking in the sink vertex t is also finite. There are a number of following related problems and issues with network flows: 1. The network flow f (N ) taking place in a network graph N is equal to the net flow coming out of the source vertex s or the net flow consumed in the sink t. 2. The net flow coming in or going out of a vertex other than s to t is zero, which means that whatever flow goes into an intermediate vertex is equal to the flow coming out of that vertex and no new flow is generated by the vertex itself, whereas in the case of s and t the flow going out of s is absorbed by the flow going into t. 3. Let us assume that (S, T ) is a cut in the network such that the source vertex s belongs to set S while the sink vertex t belongs to set T . Then the flow taking place on the cut from S to T will be donated by (S, T ) while the flow taking place from T to Swill be represented by (T < S). Thus the network flow f (N ) is given by the equation: f (N ) = f (S, T ) − f (T, S) . Hence every cut will have some flow in the direction from s to t represented by f (S, T ) and some flow in the direction from t to s which is represented by f (T, S); the network flow f (N ) is the difference between the two. The maximum flow in a

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network is achieved through maximizing the flow in one direction (that is f (S, T )), and minimizing the flow in the opposite direction (that is f (T, S)). 4. The flow f (N ) in the network is bounded by the expression: f (N ) ≤ min{c(S, T )} where the minimum is taken over all cuts (S, T ) in N . 5. In every network the value of the maximum flow is equal to the capacity of a minimum cut. This relationship is described by the famous MinCut-MaxFlow Theorem. 6. Given a network it is possible to efficiently find the maximum flow and the minimum cut in the network.

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6.6.3

Algorithmic Issues & Complexity Calculations

Although we have not formally described an algorithm to find a maximum flow in a network in the last section, the curious reader may have realized that the algorithm for finding maximum edge-disjoint paths for multi-graphs can be used to find maximum flow as well the minimum cut in the network. There is however, one serious problem regarding (the complexity of) this algorithm, we have briefly talked about this issue before and shall try to settle it now. The Fig. 6.6.4 shows a network graph with upper bounds on edge flow. Here we apply our earlier techniques of finding maximum edge-disjoint paths in a

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directed graph with just one important difference – whenever we find a path from vertex s to vertex t, we use breadth first search – thus ensuring that we find a shortest path in terms of number of edges. In the shortest path we move from vertex s to 1 and then to t – thus ensuring a network flow of 3 units from the source to the sink as shown in the top right diagram of this figure. We find another shortest path from s to 2 to t in the second step – again a flow of 3 units is possible in this path. The maximum flow and the minimum cut are found after applying BFS twice in this graph. Please note that if the upper bound on flow in every edge is 3000 instead of 3 – even then we have to apply the same number of steps to find the maximum flow. In other words the time complexity does not depend upon the magnitude of the upper bound on flow at least in this example. In fact, it can be proved that application of BFS (a minor change in our algorithm) ensures (something really big) that the time complexity of the resulting algorithm will not depend upon the magnitudes of the upper bounds on flow – it will only depend upon the size of the problem. Unfortunately the said proof and the resulting time complexity calculations are beyond the scope of this book – we, therefore rely on our earlier techniques. Let us first find out what extra price we have to pay if we do not use BFS. The Fig. 6.6.5 below shows the same network graph. We apply our earlier technique of finding any path (not necessarily shortest) from vertex s to vertex t in this network. Fig. 6.6.5 shows various stages of the working of our algorithm while it is trying to find a maximum flow in the network. This time we take a longer path from vertex s to vertex t. The path goes from s to 1 to 2 and then to t. We reverse its edges and then take another path – this time it goes from vertex s to 2 to 1 to t. The algorithm finally converges and we get the correct answer but after passing through a number of iterations proportional to k where k is an upper bound on flow in the network. Here the time complexity has become dependent not only on the size of the problem but the magnitude of the numbers involved.

6.6.4

Lower Bounds on Edge Flows and the Max-Cut

We show a directed graph with two special vertices known as s and t in the Fig. 6.6.6 below. We also show a number of cuts in this graph, every cut cuts a number of edges which if removed will destroy all paths from vertex s to vertex t. The Min-Cut passes through minimum number of edges while the

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Max-Cut passes through maximum number of edges. The size of the MinCut specifies the maximum number of edge-disjoint paths between s and t in a graph (according to Menger’s Theorem). The size of the Min-Cut in this graph is only two and thus there are only two edge-disjoint paths in this graph (between s and t) which are also indicated in the right diagram shown below (Fig. 6.6.6). Please note that we have already described an efficient algorithm to find a Min-Cut or maximum edge-disjoint paths in a graph.

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Figure 6.6.6: The left diagram shows various cuts which disconnect graph and the right diagram shows the Min-Cut for this graph. Interestingly the Max-Cut in a graph ia also important. The Max-Cut is indicated in the diagram above and is replicated again in Fig. 6.6.7. As it is clear from its name Max-Cut passes through maximum number of edges which if removed will destroy all paths between vertex s and t in the graph. The (size of the) Max-Cut specifies the minimum number of paths from vertex s to t where each path passes through every edge of the graph at least once (see the right diagram of Fig. 6.6.7). Remember the (size of the) Min-Cut specifies the maximum number of (edge-disjoint) paths where each edge is traversed at most once. Food for thought If we intend to find a Max-Cut in a network flow graph then why not remove all edges of the graph - the source vertex will certainly be separated from the sink vertex and the size of the cut will be maximum? Please think about this strategy and find what is wrong with this otherwise simple approach? We now address the problem of how to find Max-Cut and minimum number

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Figure 6.6.7: The left diagram shows the Max-Cut for the graph and the right diagram shows the minimum number of paths where each path passes through every edge of the graph at least once. of paths between vertex s and vertex t in a given directed acyclic graph, where each edge is traversed at least once. Instead of devising an entirely new algorithm let us explore if we can solve the problem using existing or modified algorithms. The intellectual exercise of reducing one problem into another is always an exciting venture – especially when the similarity between the two problems is not so obvious. We have earlier discussed the network flow problem where each edge has an upper bound on the amount of the flow that can take place, and we need to maximize the total flow taking place in the entire network. The Fig. 6.6.8 shows a network graph D with the maximum capacity of each edge indicated. The maximum flow and the corresponding minimum cut are shown in the right diagram of the same figure. We show the same network graph with the same edge capacities in Fig. 6.6.9 except that each weight associated with an edge in this network graph signifies not the upper bound but a lower bound on the flow that can take place through that edge. The problem is to find a minimum flow in this graph such that the flow taking place through any edge does not go below the lower bound of that edge. In other words we need to find the maximum cut in the network graph D that will disconnect vertex t from vertex s in D. The minimum flow and the corresponding maximum cut are indicated in the right diagram of Fig. 6.6.9. It may have become obvious now that if we can solve the problem of finding a Max-Cut and Min-Flow in a network graph with lower bounds on edge

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capacities then we can also solve the problem of finding minimum number of paths between vertex s and vertex t in a given directed graph where each edge is traversed at least once. Remember when we have (only) an upper bound on flow that can take place through any edge in a network then we start with a small acceptable flow – so small that it can take place through every edge without violating any bounds. A zero flow through every edge is a possible answer. Then we try to increase and maximize the flow from vertex s to vertex t in the network. Now when we have lower bounds on flow that can take place through any edge then we should start with a large acceptable flow – large enough that the lower bound (or limit) on flow through any edge is not violated. We should be careful about one thing – the conservation of the flow taking place in the network. It requires that the total flow coming towards a vertex should be exactly equal to the total flow going out of that vertex (except for vertices s and t in the network). Algorithm 46 finds an acceptable flow through the network D. The input as well as the output networks of this algorithm is shown in Fig. 6.6.10; we assume that the lower bound on flow through each edge is equal to 1. Please note that an acceptable (or legal flow) shown in the right diagram of Fig. 6.6.10 is not a minimum flow taking place from vertex s to t in the network. Algorithm 46: Find an acceptable (not necessarily minimum) flow from vertex s to vertex t in a given network graph D. input : A weighted directed network graph D with vertices s and t. The weight of any edge signifies the lower bound of flow that can take place through that edge. output: A weighted network graph D. The weight of each edge signifies an actual and acceptable flow taking place through that edge. 1 2

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Figure 6.6.10: The left diagram shows the input graph to Algorithm 46 and the right diagram shows one of the possible output for the algorithm. After we have found an acceptable flow from vertex s to vertex t in the network graph D, now we need to minimize it. Instead of minimizing this flow directly we do so indirectly – by pushing an opposite flow taking place from vertex t to vertex s in D as shown in the Fig. 6.6.11 below. We start with an acceptable flow in the network graph D as shown in the top left diagram of Fig. 6.6.11. The weight of each edge in this diagram shows how much actual flow is taking place in this network without violating any lower bound on flow taking place in any edge. The network graph D is converted into a directed graph F as follows: 1. We copy graph D into graph F without any edge weights. It means that all vertices as well as directed edges of D are copied in directed graph F . 2. For every edge (x, y) in D with a weight equal to w, there is an additional directed edge (y, x) in graph F with a weight equal to w −c(x, y), where c(x, y) is the lower bound on flow that can take place in the edge (x, y) in the graph D. The resulting graph F is shown in the top right diagram of Fig. 6.6.11. We find the maximum flow in graph F taking place from vertex t to vertex s as shown in the middle left diagram. The maximum flow taking place through any edge (y, x) in this graph is represented by m(y, x). The Min-Cut in graph F from vertex t to vertex s is shown in the middle right diagram. The

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corresponding minimum flow in the graph D is shown in the bottom left diagram. Here the weight of an edge (x, y) is equal to w(x, y) − m(y, x). The Max-Cut in the graph D from vertex s to vertex t is shown in the bottom right diagram. Please note that the Max-Cut in D from s to t cuts the same edges as the Min-Cut in graph F from vertex t to vertex s, the size of the two cuts are however different as is evident from Fig. 6.6.11.

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Problem Set 6.5. Problem 6.5.1. We are given a directed network graph D with special vertices s and t. Each edge in this network has a lower as well as an upper bound on edge flow; we assume that the lower bound as well as upper bound is the same for each edge although the two bounds may be different from each other. (a) Design an efficient algorithm to find if an acceptable flow is possible in this network. (b) If an acceptable flow is possible and is provided to you then find a maximum flow in this network. (c) If an acceptable flow is possible then find a minimum flow in this network. Problem 6.5.2. How about if the lower bound is a positive number (it may be different for different edges) while the upper bound on flow is the same for all edges? Repeat the earlier problem for this new situation. Problem 6.5.3. Musharraf designed the following intelligent algorithm to find an acceptable flow provided it exists in the network graph of the above problem: (a) Initially ignore the upper bound on each edge and find a minimum flow in the network (from vertex s to vertex t) keeping into account the lower bounds on flow through each edge; (b) An acceptable flow exists if and only if the flow through any edge does not exceed its upper bound. Once we find an acceptable flow – it can always be maximized or minimized, prove that this algorithm always finds the correct result or find a counter example. Problem 6.5.4. Kashif finds the following counter example for the above problem as shown in the figure below (Fig. 6.6.12. Now Musharraf insists that his algorithm will find an optimal answer for the first problem in this problem set. Either prove that Musharraf is right or find a counter example. Problem 6.5.5. We are given a network graph D with special vertices s and t. We need to find minimum number of paths from vertex s to vertex t such that each vertex of D is traversed at least once by any of the s − t paths. Problem 6.5.6. We are given a network graph D with special vertices s and t as in Fig. 6.6.12. We need to make sure that a same fixed amount of flow should take place through every directed edge in that graph. How can you design an efficient algorithm which will output either yes or no; yes in case an acceptable flow is possible and no in case it is not possible? A special case of this problem is when the fixed flow through every edge is exactly one. Such graphs (where an acceptable flow) is possible have a special structure as we shall study in coming chapters.

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D without the unused edges. The two graphs are reproduced in the figure below for ready reference. We claim that in any network graph D without the unused edges, it is possible to push a flow of exactly one unit in every directed edge, either prove this or give a counter example. Problem 6.5.9. A directed network graph D without unused edges has two important properties: (a) every edge is part of a directed path from vertex s to vertex t in D, & (b) We can use (the so called stupid) Algorithm 36 to find maximum edge-disjoint paths in this graph. Can you think of a directed graph where property (a) is applicable but (b) is not? Can you think of a graph where (b) is applicable but (a) is not? Problem 6.5.10. Consider a directed graph D where the in-degree of every vertex (other than s and t) is equal to the corresponding out-degree. (a) Is it possible to push a flow of exactly one unit through every directed edge in this graph? (b) If it is possible then does it mean that every edge in this graph will be part of a directed path from vertex s to vertex t in this graph? (c) If (a) is possible then can we find maximum edge-disjoint paths from s to t in this graph using the stupid Algorithm 36?

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bipartite graphs. We shall devise efficient algorithms to find maximum matching in un-weighted bipartite graphs and weighted maximum matching in a weighted bipartite graph.

6.7.1

Maximum Matching in Un-weighted bipartite graphs

Given a maximal matching, how can we find a maximum matching in a bipartite graph? A maximal matching in a bipartite graph is shown by bold lines in the top left corner of Fig. 6.7.1. The matching problem is converted into a connectivity problem by adding two dummy vertices s and t as shown in the top right corner of Fig. 6.7.1. We add directed edges from vertex s to every vertex in partite A; We convert undirected edges of the bipartite graph (top left diagram) into directed edges as shown in the top right corner; please note that the direction of each such edge is from the partition A to partition B. We also add directed edges from every vertex of the partition B to vertex t; this completes the transformation from a bipartite graph of the top left corner into the directed graph shown in the top right diagram. As explained before each matched edge in the bipartite graph (top left corner) corresponds to a path in the directed graph (top right corner), in fact any two matched edges in the bipartite graph will correspond to two edge-disjoint directed paths from vertex s to vertex t in the corresponding directed graph. The problem of finding maximum matching in a bipartite graph is thus transformed into the problem of finding maximum number of edge-disjoint (or vertex-disjoint) paths in the directed graph. Once we find the maximum number of paths in the directed graph (see the bottom left diagram of Fig. 6.7.1) we can find the corresponding matched edges in the bipartite graph as shown in the bottom right diagram of the same figure. The algorithm described above works well for finding a maximum matching in an unweighted bipartite graph. Unfortunately it is difficult to adapt this algorithm to find a maximum weighted matching in a weighted bipartite graph. We, therefore, present (in the coming sub-section) a slightly different version of this algorithm to find maximum matching in an unweighted bipartite graph? Problem Set 6.6. We are given a bipartite graph having a partition B (consisting of a number of boys) and a partition G (consisting of a number of girls). An edge between a boy and a girl shows a degree of compatibility. We need to find maximum pairs of boys and girls such that the boy and girl

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in every pair are compatible. This problem is also known as the Marriage Problem, it can be modeled by an un-weighted bipartite graph as shown below (Fig. 6.7.2). The problem can be solved by techniques similar to the ones that we have just studied.

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Figure 6.7.2: A bipartite graph showing a partition B consisting of a set of boys, a partition G showing, a set of girls. Compatibility between a boy and a girl is indicated by an edge between the corresponding vertices. Problem 6.6.1. Assume that we are required to solve the marriage problem: namely we intend to marry maximum number of boy; each boy can marry a single girl and each girl can marry a single boy. How will you model this problem in terms of a known graph problem? Under what conditions it will be possible? Problem 6.6.2. Assume that we are required to solve the marriage problem: namely we intend to marry maximum number of girls; each girl can marry a single boy. How will you model this problem in terms of a known graph problem? Under what conditions it will be possible? Problem 6.6.3. Assume that we intend to solve the decision problem in which we intend to find a yes/no answer corresponding to the question: Is it possible to marry all boys? (Or is it possible to marry all girls?)

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Problem 6.6.4. Assume that a boy can marry four girls and each girl can marry a single boy; we intend to maximize the number of marriages taking place. How will you model this problem in terms of a known graph problem? Under what conditions will it be possible? Problem 6.6.5. Assume that a boy can marry four girls; we intend to maximize the number of girls who are married. How will you model this problem in terms of a known graph problem? Under what conditions will it be possible? Problem 6.6.6. Assume that a boy can marry four girls; we intend to maximize the number of boys who are married. How will you model this problem in terms of a known graph problem? Under what conditions will it be possible? Problem 6.6.7. We intend to maximize the number of married boys and girls subject to the condition that a boy can marry four girls while a girl can marry a single boy. Problem Set 6.7. Problem 6.7.1. The maximum matching in a bipartite graph is equal to the maximum number of edge-disjoint paths in a directed graph D. In other words the problem of finding maximum matching in a bipartite graph is transformed into the problem of finding maximum number of edge-disjoint (or vertex-disjoint) paths in a directed graph D. How about if we remove all directions from the directed graph D, thus converting it into an undirected graph. Can we still claim that the maximum matching in the bipartite graph is equal to the maximum number of edge-disjoint paths between vertex s and vertex t in the undirected graph D. Discuss with the help of an example. Problem 6.7.2. We are given a bipartite graph with edge weights equal to either zero or one. We need to find maximum weighted matching in this weighted bipartite graph. Discuss how you will solve this problem efficiently using similar techniques.

6.7.2

Maximum Matching in Complete (Binary) Weighted Bipartite Graphs

We are given an unweighted balanced bipartite graph as shown in left diagram of Fig. 6.7.3. A maximum matching in this graph is also shown in this


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diagram. It can be found using our expertise gained in the last section. Here we shall describe another algorithm to find the maximum mathing this new algorithm may not be as efficient as the one described before but it has the added advantage of being flexible. It can handle (with or without minor modification) the more general problem of finding maximum weighted matching in a bipartite graph. As shown in Fig. 6.7.3 we convert this bipartite graph into a complete weighted bipartite graph - see the right diagram of this figure. An edge in the original bipartite graph has a weight of 1 in the complete bipartite graph while every other edge has a weight equal to zero. Thus the complete bipartite graph has binary weights. A maximum weighted matching in the right graph of this figure will be a maximum matching in the original bipartite graph. We describe in the following paragraphs a useful algorithm to solve the maximum weighted matching problem in a complete bipartite graph with binary weights. The results of this algorithms can be used to find maximum matching in a bipartite graph. The algorithm can also be used to find maximum weighted in a complete bipartite graph with non binary weights as described in the next section. a1

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efficient but useful algorithm) which works on the directed graph D and outputs the maximum weighted matching in the bipartite graph. Algorithm 47: Find a maximum weighted matching in a complete balanced bipartite graph G. input : A complete balanced (binary) weighted bipartite graph. output: A maximum weighted matching in G. 1

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Problem Set 6.8. Problem 6.8.1. What is the time complexity of Algorithm 47 and the previous algorithm described for unweighted graphs in the last section? Note that we are applying a simple path finding algorithm for un-weighted edges in the previous algorithm of the last section and its time complexity was O(p + q), also it is applied p (number of vertices) times on the graph until all the paths have been found. Whereas in the case of Algorithm 47 for a complete weighted bipartite graph, the Bellman-Ford algorithm is applied and its time complexity is O(p3 ) and again it is applied p number of times on the given network graph. Problem 6.8.2. Would we be able to get the maximum matching in a bipartite graph if the graph is neither complete nor balanced by applying Algorithm 47? Problem 6.8.3. How is Algorithm 47 applicable to find maximum matching in a general weighted graph D with weights greater than 1? Discuss the possible modification that need to made to the general graph D before we can use it as input for the Algorithm, the modifications should be such that the maximum matching in the new bipartite is equal to the maximum matching in the original graph D.


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6.7.3


Maximum Weighted Matching in Complete Weighted Bipartite Graphs

Problem: We need to find a maximum, weighted, perfect matching in a weighted complete and balanced bipartite graph G. All edge weights are positive. Input: A weighted complete & balanced bipartite graph G with positive edge weights, assume that vertices in the A partite are numbered as a1 , a2 , a3 . . . ak , while vertices in the B partite are numbered as b1 , b2 , b3 . . . bk . We define layers in the graph G: layer 1 contains a1 and b1 ; layer x contains vertices ax and bx . The graph consisting of first x layers contains first x vertices from both the partitions A & B, and all edges between these vertices as shown in Fig. 1. Output: A maximum weighted perfect matching. See Fig. 6.7.5. Graph consisting of First 2 Layers 13

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Figure 6.7.5: A complete balanced bipartite graph G is shown with positive edge weights. The partition A as well as the partition B is indicated in the left diagram. Maximum weighted matching is shown in red color. The weight of the maximum matching is equal to 15 + 8 + 16 = 39. The first two layers of bipartite graph G are shown in the right diagram. A Comparison of Two Algorithms: We shall describe two algorithms to solve the above problem. The algorithms, that we describe, are similar in some respects and different in others; it will be interesting to make a


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comparison between the two. Algorithm 48: Find maximum weighted matching in a complete bipartite graph G input : A weighted complete bipartite graph G output: maximum weighted matching 1 2

for x = 1 to k do Find maximum weighted matching in the first x layers of bipartite graph G;

Algorithm 49: Find maximum weighted matching in a complete bipartite graph G input : A weighted complete bipartite graph G output: maximum weighted matching 1 2

for x = 1 to k do Find maximum weighted matching of size x in bipartite graph G;

The working of the two algorithms is shown in Fig. 6.7.6. It is obvious from these diagrams that the intermediate results may be different but the end results are same for the two algorithms. Do not move forward before understanding Fig. 6.7.6. Algorithm No. 48 requires us to find the maximum weighted matching in the first x layers. In order to implement this algorithm efficiently we need to design an efficient procedure which performs the following function: Procedure No. 1: Given a maximum weighted matching in the first x layers of G it finds the maximum weighted matching in the first x + 1 layers of G. Similarly in order to implement Algorithm No. 49 efficiently we need to design an efficient procedure which performs the following function: Procedure No. 2: Given a maximum weighted matching of size x in G it finds the maximum weighted matching of size x + 1 in G. Before discussing the details of the two procedures we first need to transform the weighted bipartite undirected graph G into a directed graph D with

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Maximum Matching in the first layer: Value = 13 13

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Figure 6.7.6: The left diagrams show the working of Algorithm No. 48 when applied to a bipartite graph G. The right diagrams show the working of Algorithm No. 49 when applied to the same bipartite graph G. Please note that intermediate results may be different but the end result is the same for the two algorithms.


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additional vertices s and t according to the following rules as shown in the diagram below. 1. Add vertex s to G such that in the resulting graph D there is a directed edge from vertex s to every unmatched vertex in A with a weight equal to 0. 2. Now add another vertex t such that there is a directed edge from every unmatched vertex in B to vertex t with a weight equal to 0. 3. For every matched edge (a, b) in bipartite graph G add a direction going from vertex b to a in the directed graph D. For every unmatched edge (a, b) in G add a direction going from vertex a to b in the directed graph D. Note that vertex a belongs to A while b belongs to B in the bipartite graph G. 4. The sign of every weight w for every matched edge in G is changed to a minus sign in the directed graph D. All other edge weights retain their original signs in D as there are in G. We shall now discuss the details of Procedure No. 1 & 2. Algorithm 50: Procedure 1: Transform maximum weighted matching in the first x layers into a maximum weighted matching in the first x+1 layers in a bipartite graph input : Maximum weighted matching in the first x layers of bipartite G output: Maximum weighted matching in the first x + 1 layers of bipartite G 1 2

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Transform the weighted bipartite graph into a directed graph D; Find a longest path from vertex s to t passing through the first x + 1 layers of D and now reverse the edges in this path; Every directed edge from B partite to A partite in D corresponds to a maximum weighted matching in the first x + 1 layers of G;

Claim 6.7.1. The weight of the longest path from vertex s to t in D is equal to the gain in the weight of maximum matching when the size of matching is increased from x to x + 1 in the bipartite graph G.

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Figure 6.7.7: The diagrams show the working of Procedure No. 1. Maximum weighted matching in the first two layers in a bipartite graph G is converted into a maximum weighted matching in the first three layers of the same bipartite graph.


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Algorithm 51: Procedure 2: Transform maximum weighted matching of size x into a maximum weighted matching of size x + 1 in a bipartite graph input : Maximum weighted matching of size x in bipartite graph G output: Maximum weighted matching of size x + 1 in bipartite graph G 1 2

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Transform the weighted bipartite graph into a directed graph D; Find a longest path from vertex s to t in D and now reverse the edges in this path; Every directed edge from B partite to A partite in D corresponds to a maximum weighted matching of size x + 1 in G;

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Figure 6.7.8: The diagrams show the working of Procedure No. 2. Maximum weighted matching of size 1 in a bipartite graph G is converted into a maximum weighted matching of size 2 in the same bipartite graph.

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Figure 6.7.9: The diagrams illustrates that every path from s to t in the directed graph D passing through the first three layers corresponds to a matching in the first three layers of the bipartite graph G. The longest path corresponds to a perfect matching of maximum weight.


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Claim 6.7.2. Assume that we have already found a maximum matching in the first x layers of graph G, we now need to extend this maximum matching in the first x + 1 layers of G as shown in Fig. 6.7.9. We transform the bipartite graph into a directed graph also shown in Fig. 6.7.9. Suppose we have found two paths (one longest and one relatively shorter) from vertex s to vertex t, Path 1 is the longest path and has a weight equal to y and Path 2 is a relatively shorter path and has a weight equal to z. If we select Path 1 (as dictated by Procedure No. 1) then we need to reverse edges belonging to Path 1. Then the reversed edges of Path 1 plus edges belonging to Path 2 will form a cycle in the graph with a net negative value at most equal to y − z. If on the other hand we reverse edges in Path 2 then a cycle with a net positive value at most equal to y − z will be formed as shown in Fig. 6.7.10.

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Claim 6.7.3. We shall never encounter a situation where there will be a positive weight cycle in D (provided we follow steps given in Procedure No.1 or Procedure No. 2). Negative weight cycles will, however, be present in the directed graph D. Problem Set 6.9. Problem 6.9.1. Assume that we are given a complete balanced weighted bipartite graph with positive edge weights. Describe an efficient algorithm to find a maximum matching in the very first layer of the bipartite graph. Carefully derive the time complexity of this algorithm. Problem 6.9.2. In Procedure No. 1 (and Procedure No. 2), we need to find a longest path from vertex s to vertex t. Someone thinks that the longest path problem is NP-Complete. Please comment. Problem 6.9.3. The left diagram of Fig. 6.7.13 shows a directed graph in which there is only one edge going from a vertex b to a vertex a. We need to find what will be the maximum length (in terms of the number of edges) of the longest path passing through the first two layers of the graph. Please note that there are 2k + 2 vertices in this directed graph and your answer should be independent of edge weights in the graph. Problem 6.9.4. The right diagram of Fig. 6.7.13 shows a directed graph in which there are x edges going from a b vertex to an a vertex. We need to find what will be the maximum length (in terms of number of edges) of the longest path passing through the first x + 1 layers of the graph. Please note that there are 2k + 2 vertices in this directed graph and your answer should be independent of edge weights in the graph. We are working under the assumption that there are no positive weight cycles in the directed graph D. Problem 6.9.5. Assume that we have found a 3-edge longest path from vertex s to vertex t in a directed graph consisting of the first x + 1 layers of graph D (see right diagram of Fig. 6.7.13). Now we need to find a 5-edge longest path from s to t in the same graph consisting of first x + 1 layers of graph D. Design an efficient algorithm to solve this problem. Carefully derive the time complexity of this algorithm in terms of the size of the problem. Discuss briefly if your algorithm is a greedy algorithm or does it use dynamic programming.


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Finding Maximum Matching of size one

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The Matching Problem in Bipartite Graphs Only 1 edge is reversed

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Figure 6.7.13: The left diagram shows a directed graph D in which there is only one edge going from a vertex b to a vertex a. The right diagram shows another directed graph D where there are x edges going from a vertex b to a vertex a. Problem 6.9.6. Is it possible to use an existing textbook algorithm (without any modification) in order to solve the previous problem? Discuss briefly. Problem 6.9.7. Carefully derive the time complexity of Procedure No. 1 and then the time complexity of Algorithm No. 48 (see Fig. 6.7.11). Problem 6.9.8. Assume that the time complexity of Procedure No. 1 is bounded by O(x · k 2 ). Now derive the time complexity of Algorithm No. 48. Problem 6.9.9. Find the actual longest path in each of the directed graphs shown in Fig. 6.7.14. Problem 6.9.10. Assume that we are given a complete balanced weighted bipartite graph with positive edge weights. Describe an efficient algorithm to find a maximum matching of size 1 in this bipartite graph. Carefully derive the time complexity of this algorithm. Problem 6.9.11. Consider the working of Procedure No. 2. What would be the maximum value of the length (in terms of number of edges) of the longest path from vertex s to vertex t in D? Discuss briefly. Problem 6.9.12. Carefully derive the time complexity of Procedure No. 2 and then the time complexity of Algorithm No. 49 (see Fig. 6.7.12).

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Network Flows, Connectivity and Matching Problems 13

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Figure 6.7.14: We need to find longest path from vertex s to t in the directed graphs shown here. Problem 6.9.13. Although the worst case time complexity of Algorithm No.48 and 49 may be the same in terms of the Big O notation yet one algorithm is considerably faster than the other. Discuss briefly. Problem 6.9.14. Suppose we need to find a best possible matching of size x in a given complete balanced weighted bipartite graph G of size k and x is much smaller than k. Discuss how you will solve this problem and carefully derive the time complexity of your algorithm in terms of x and k. Problem 6.9.15. Suppose we need to find a best possible matching for the first x members of partite A in a given complete balanced weighted bipartite graph G. Discuss if we can use Algorithm No. 48 or 49 to solve this problem? How can you design a better algorithm to solve this problem? Discuss briefly. Problem 6.9.16. Suppose we need to find a best possible matching for the first x members of partite B in a given complete balanced weighted bipartite graph G. Discuss if we can use Algorithm No. 48 or 49 to solve this problem? How can you design a better algorithm to solve this problem? Discuss briefly.


6.8 6.8.1

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The Max-Flow Min-Cost Problem Introduction

We consider the Maximum Flow at Minimum Cost problem in a network graph as shown in Fig. 6.8.1. We assume that the capacity as well cost per unit flow through every edge is a positive integer; it is indicated with each edge in the figure below. We need to push the maximum flow in this network graph (starting from vertex s and ending at vertex t) at a minimum cost. A possible solution of this network graph is shown in the right diagram of the same figure; we are pushing a maximum flow of 3 units at a total cost of 19. The problem is to efficiently find maximum flow at minimum cost in a given network graph. 1/8 2/8

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Figure 6.8.1: We show a directed graph D (left diagram) having two special vertices s and t. The capacity/cost of an edge is shown along with each edge. We need to find maximum flow coming out of vertex s and being absorbed by vertex t. We also need to make sure that the maximum flow is taking place at minimum cost. The maximum flow at minimum cost is indicated by colored lines in the right diagram.

6.8.2

Finding a Maximum Flow or Finding a Shortest Path?

Before answering the above question it is important to revise the relevant prior knowledge, and also remind ourselves of some of the powerful techniques which will be useful in solving this problem. It is interesting to note

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that this challenging problem (which we call a Category 1 problem) has two requirements: 1. We need to maximize the flow and we already know how to do it (but we should also remember our shortcomings and limitations). 2. We need to minimize the cost of (unit) flow starting from vertex s and ending at vertex t, and this somehow looks like finding a shortest path from vertex s to t (we are aware of a number of simple and efficient algorithms to solve shortest path problems). We know how to solve the above two problems in isolation but how can we fulfill the two requirements simultaneously? If we start finding flows without looking at costs then we may end up with a maximum flow but at higher cost. If we start minimizing the cost by finding shortest paths, then how will we be able to tackle the problem of maximizing the flow? Different edges have different capacities; finding shortest paths without considering capacities will create complications. It is certainly a very exciting mixture of two important problems. Before we solve this general problem we shall try to reflect on similar problems we have already solved and what are some of the special cases of this general problem which can be resolved using our prior knowledge? We shall then extend or modify specialized solutions to solve this general problem. So let us list down the general problem (once again) and its special cases: Category 1: Given a network flow graph as shown in Fig. 6.8.1 how can we find maximum flow at minimum cost from a source vertex s to a sink vertex t? We assume that edge capacities are integers while per unit cost of flow through any edge may be any non-negative real number. Category 2: Given a network flow graph as shown in Fig. 6.8.2 how can we find the maximum flow at minimum cost from a source vertex s to a sink vertex t? We assume that the capacity of each edge is exactly 1 while per unit cost of flow through any edge may be a real number. Please note that this problem is equivalent to finding the maximum edge-disjoint paths from vertex s to vertex t at minimum cost, i.e. the sum of edge costs in all edge-disjoint (shortest) paths should be as small as possible. Category 3: The network flow graph is derived from a complete balanced and weighted bipartite graph. Here we add a source vertex s to the A and


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a sink vertex t to B as shown in Fig. 6.8.3. We assume that the capacity of each edge is one in the network flow graph. Here the problem is to find maximum flow at minimum cost in the network flow graph shown in the right diagram. It is possible to recognize that finding maximum flow (or maximum edge-disjoint paths) at minimum cost in the right diagram is equivalent to finding a minimum cost perfect matching in the bipartite graph shown in the left diagram. A related problem in a bipartite graph would be to find a maximum cost perfect matching - that would require us to find maximum flow (or maximum edge-disjoint paths) at maximum cost. Both problems have their applications in graph theory and elsewhere. 13

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Figure 6.8.3: In a Category 3 problem a network flow graph is derived from a weighted bipartite graph by inserting a source vertex and a sink vertex. All edge weights associated with source vertex s and sink vertex t are zero

6.8.3

Category 3 network flow Problems

We shall now try attacking these problems starting from Category 3. We need to recall our expertise of finding maximum cost perfect matching in a weighted bipartite graph discussed in earlier sections. A curious reader might have noticed that while solving the maximum cost perfect matching problem in the bipartite graph shown in Fig. 6.8.3 we were essentially solving the maximum flow at maximum cost problem as shown in a network flow graph in Fig. 6.8.4. Similarly while we were solving the minimum weighted perfect matching


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Figure 6.8.4: A maximum weighted perfect matching in bipartite graph (left diagram) corresponds to a maximum flow at maximum cost in the right diagram.

problem in a bipartite graph we are essentially solving the minimum cost maximum flow problem. We have in fact solved the Category 3 network flow problem without explicitly saying so as our primary objective was to find a perfect matching of maximum (or minimum) cost. Maximum Flow at Maximum Cost in Category 3 problems Please recall Algorithm 47 which was designed to find a maximum weighted matching in a complete bipartite graph. It is slightly modified as shown below (Algorithm 52). As we shall show later this algorithm is powerful enough to handle Category 2 and 1 network flow problems. What we essentially do here is to find a longest path P from vertex s to vertex t in the given graph. We reverse the direction of each edge in path P and also multiply weight of each edge in the path by negative 1. We then again find a longest path from the source vertex to the sink vertex in the modified graph. We apply this algorithm to the network flow graph shown in the right diagram of Fig. 6.8.3. The step by step working of this algorithm is shown in Fig. 6.8.5. The algorithm terminates when it is no longer possible to find a path from the source to the sink as shown in the bottom right diagram of Fig. 6.8.5. The problem is how to recover maximum flow at minimum cost from this (final) graph F .

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Algorithm 52: Find a Maximum Flow at maximum cost from vertex s to vertex t in a directed network graph D belonging to Category 3 input : Directed & Weighted graph D, and vertices s & t, output: Maximum Flow at minimum cost from s to t in D 1 2

3 4 5

Copy Graph D into F ; Find a flow of one unit through a longest path P (in terms of edge costs) from vertex s to t in F . If you are successful in finding a flow then keep a record of the path found otherwise exit with graph F as output; for each edge (x, y) of path P in graph F do Reverse the direction of edge (x, y) Multiply the weight of edge (x, y) with a negative sign and go back to Step 2

The solution to this problem is again consistent to our earlier approach of deleting unused edges from the network flow graph. The unused edges in final graph F are those which have positive weights as shown in Fig. 6.8.6. By deleting these edges it is possible to find maximum flow at maximum cost in a Category 3 problem as shown in Fig. 6.8.6. One would like to compare this answer with the one obtained while finding a maximum weighted perfect matching in a bipartite graph shown in Fig. 6.8.4. Maximum Flow at Minimum Cost in Category 3 Problems The above algorithm can easily be adopted to find maximum flow at minimum cost or in other words maximum edge-disjoint paths at minimum cost in a Category 3 network flow graph. The only change that we need to do is to replace longest path in line 2 of this algorithm by shortest path as shown in the following algorithm (Algorithm 53).

6.8.4

Category 2 (and 1) network flow Problems

Here we shall consider Category 2 (and Category 1) network flow problems. We intend to solve the maximum flow at minimum cost problem using multiple algorithms in order to provide a better insight to the problem and its possible solutions. We shall first show that Algorithm 53 can be used solve these problems without any modification. But in order to do that we need


Find a longest path from s to t 13

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Figure 6.8.5: The step by step working of Algorithm 52 is shown here on the network graph of Fig. 6.8.3. The final graph F is shown in the bottom right corner. Please note that in this graph it is no longer possible to find another path from the source vertex to the sink vertex.

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Network Flows, Connectivity and Matching Problems 13

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Re gra mov ph e +i F f ve rom ed gra ges i ph n D

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Maximum Flow at Maximum Cost in the apposite direction

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Maximum Flow at Maximum Cost in the normal direction

Figure 6.8.6: Final graph F taken from the last figure is shown in the top left diagram. We can recover maximum flow at maximum cost by removing edges with positive weights in this final graph F . Algorithm 53: Find a Maximum Flow at minimum cost from vertex s to vertex t in a directed network graph D belonging to Category 3 input : Directed & Weighted graph D, and vertices s & t, output: Maximum Flow at minimum cost from source vertex s to sink vertex t in D 1 2

3 4 5

Copy Graph D into F ; Find a flow of one unit through a shortest path P (in terms of edge costs) from vertex s to t in F . If you are successful in finding a flow then keep a record of the path found otherwise exit with graph F as output; for each edge (x, y) of path P in graph F do Reverse the direction of edge (x, y) Multiply the weight of edge (x, y) with a negative sign and go back to Step 2


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to do some serious graph theoretic work in terms of claims and some hints for their proofs. Claim 6.8.1. We are given a directed & weighted graph D. The graph D may contain cycles but it does not contain any negative weight edge. It contains a source vertex s and a sink vertex t. We copy this graph in graph F . We find a shortest path P from source vertex s to the sink vertex t in directed graph F . We now reverse the direction of each edge in path P in graph F . We also multiply the weight of each edge in path P by negative one in graph F . Now we claim that as directed graph D does not contain any negative edges therefore graph F will not contain any negative weight cycle. Claim 6.8.2. We are given a directed & weighted graph D with no negative weight cycles but it may contain negative weight edges. It contains a source vertex s and a sink vertex t. We copy this graph in graph F . We find a shortest path P from source vertex s to the sink vertex t in directed graph F . We now reverse the direction of each edge in path P in graph F . We also multiply the weight of each edge in path P by minus one in graph F . Now we claim that as directed graph D does not contain any negative weight cycle then graph F will also not contain any negative weight cycle. Negative Weight Cycles & Improvement in Cost of Flow We show a flow of one unit in the network D shown in the top left diagram of Fig. 6.8.7. Please note that this flow of one unit is not taking place on a shortest path from vertex s to vertex t. We reverse the direction of the edges used by the flow and multiply the cost of these edges by negative 1 as shown in graph F (see top right diagram). We then find a negative weight cycle in this new graph F highlighted by orange color. The value of net weight in this cycle is -2. By redirecting the flow in the negative weight cycle it is possible to reduce the cost of flow by an amount exactly equal to 2 as shown in the bottom diagram of this figure. If on the other hand there are no negative weight cycles then it is not possible to reduce the cost of existing flow. Such a scenario is shown in Fig. 6.8.8. Claim 6.8.3. Given a network flow graph D. We find a flow of one unit from vertex s to vertex t. If there is a negative weight cycle in graph F with a net weight equal to -k (graph F is derived from a network graph D after reversing the edges in the direction of the flow and multiplying the costs in

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An Improvement is possible as there is a negative cycle

One edge-disjoint path or a flow of 1 unit in a Category 1 Problem

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Reverse the edges in the path shown earlier There is a negative cycle of value = -2

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Figure 6.8.7: We show a flow of one unit in the network shown in the top left diagram. If there is a negative weight cycle (after reversing the edges in the direction of the flow) then the cost of flow can be further reduced.


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these edges by minus 1) then it is possible to reduce the cost of the existing flow by k by re-adjusting the flow in the direction of the negative cycle. Claim 6.8.4. If there is no negative weight cycle in graph F (derived from a network graph D after reversing the edges in the direction of the flow and multiplying the costs in these edges by negative 1) then it is not possible to reduce the cost of the existing flow by re-adjusting the flow in any direction. 2/8

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Figure 6.8.8: We show a flow of two units in the network shown in the top left diagram. If (after reversing the edges in the direction of the flow) there are no negative weight cycles then the cost of flow can not be further reduced. Claim 6.8.5. Given a graph D and a finite flow taking place from vertex s to vertex t. We know the path taken by each unit of flow from vertex s to vertex t in D; we keep a record of the length of each such path and call these paths existing paths in D. If there is no improvement possible in the cost of existing flow then there will be no negative weight cycle in graph F (derived from a network graph D after reversing the edges in the direction of the flow and multiplying the costs in these edges by negative 1). Please see Fig. 6.8.9.

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Graph F: we have reversed the edges in the direction of the flow in graph D and have multiplied the edge costs by -1

If there are no improvement possible in existing flow in graph D then there will be no negative cycles in F

Figure 6.8.9: We show two existing paths from s to t in graph D (top left diagram). The paths are reversed in graph F as shown in the top right diagram.


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Finding Maximum Flow at Minimum Cost In view of the above claims it is obvious that Algorithm 53 can be used to find maximum flow at minimum cost in Category 2 as well as Category 1 network flow problems. It will be interesting to derive the time complexity of that algorithm for both these categories. Algorithm 54: Find Maximum Flow at Minimum Cost from s to t in a directed graph D in a Category 1 problem input : Directed & Weighted graph D, and vertices s & t, output: Maximum Flow from s to t in D at Minimum Cost 1 2

3

4

Copy graph D into a graph F ; Find Maximum Flow from vertex s to t in F (ignoring costs). Reverse the edges in the path of every flow; multiply the costs corresponding to these edges with negative 1; Find if there is a negative weight cycle in graph F. If you find one then go to step 4 else output the maximum flow at minimum cost and exit; Remove the negative weight cycle by adjusting the flow accordingly, now go to step 3;

We now describe an alternate algorithm (Algorithm 54) to find a maximum flow at minimum cost for Category 1 problems. Our last three claims support the argument that the following algorithm would be able to correctly solve maximum flow at minimum cost problem in Category 1 network flow graphs. The step by step detailed working of the above algorithm on a Category 1 problem is shown in Fig. 6.8.10. First we find maximum flow; we then reverse the edges in the direction of the flow; multiply the cost of these edges with minus one. If there are any negative weight cycles in the graph then we have to remove every negative weight cycle; the removal of a negative weight cycle will certainly reduce the cost of flow by at least one unit, but it may give rise to another negative weight cycle. The process is repeated until there are no more negative weight cycles left in the graph. The process of finding maximum flow at minimum cost may be accelerated if we some how find a negative weight cycle of higher value and remove it as shown in Fig. 6.8.11. Problem Set 6.10. Problem 6.10.1. We have demonstrated the working of an algorithm (to find maximum flow at minimum cost in Category 3 problems) in Fig. 6.8.4.

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Find the Max Flow in D

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Figure 6.8.10: We show different steps in finding maximum flow at minimum cost in a Category 1 problem. We first find maximum flow ignoring costs (cost comes out to be = 27). We then remove negative weight cycles and subsequently reduce cost.

The Max-Flow Min-Cost Problem 2/-8

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Maximum Flow at Minimum Cost (8+5+4+2) in the original Graph D

Figure 6.8.11: We show different steps in finding maximum flow at minimum cost in a Category 1 problem. A maximum flow of 3 units can pushed in this network at an optimal cost of 8+4+5+2 = 19. The cost is thus reduced from 27 to 19 for the same amount of flow.

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Figure 6.8.12: Removing a negative weight cycle of a relatively higher value may speed up the process of finding a maximum flow at minimum cost.


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Figure 6.8.13: We show different steps in finding maximum flow at minimum cost in a Category 2 problem by removing negative weight cycles. Please note that in this category all edge capacities are equal to one

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Describe this algorithm (known as Algorithm 53) in your own words and carefully derive its time complexity. Is this possible to use this technique (without any appreciable change) to find the minimum cost for a fixed flowthe amount of fixed flow may not be the maximum flow in the network. Discuss briefly. Problem 6.10.2. We show an interesting special case of Category 4 in Fig. 6.8.14. In this special category of graphs the sum of capacities of all edges coming out of s is equal to the sum of capacities of all edges going into vertex t. The cost of every edge coming out of s and going into t is zero. The capacity of every edge in the D − s − t graph is infinite while the cost is a positive integer indicated in the diagram. We need to find maximum flow at minimum cost using an efficient algorithm. Find if you can use any earlier techniques (Algorithm 53) to solve this interesting problem. You may design a better algorithm if you want and if you can?

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Figure 6.8.14: We show a special network graph D. Here edge capacities in graph D − s − t are all infinite. The sum of capacities of all edges coming out of source s is exactly equal to the sum of capacities of edges going in sink t while the costs of these edges are zero. Problem 6.10.3. Algorithm 54 can be used to find maximum flow at minimum cost in Category 1 problems. The step by step working of this algorithm is demonstrated in Fig. 6.8.10. Carefully derive the time complexity of this algorithm. Is it possible to solve problems belonging to this category using Algorithm 53 even if the cost of a unit flow in an edge is a real number? Discuss briefly.


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Problem 6.10.4. We have applied Algorithm 54 to solve a Category 2 problem as shown in Fig. 6.8.12. In this category all edge capacities are equal to 1, thus maximum flow here corresponds to maximum edge-disjoint paths as shown in Fig. 6.8.6, applying Algorithm 54 seems to be overkill. Describe an efficient algorithm to handle Category 2 problems. Now consider a special case of this category where costs are expressed by binary numbers, this special case is shown in the bottom diagram of Fig. 6.8.6. Design a (very) efficient algorithm to find a maximum flow at minimum cost in this special category. Because of its restricted nature, we expect to solve it using a simple and a very efficient algorithm.

6.8.5

A Panoramic Picture of Similar Problems & Solutions (once again)

Please note that Category 1 problems are the most general. We have not yet devised an exact algorithm to find maximum flow (even if we ignore costs) in such problems. On the other extreme Category 2 & 3 problems can be solved using very efficient algorithms. In between these two extremes there is an exciting range of problems to be explored. Some of these problems were addressed in the last problem set. 1/7

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Category 3 The Max flow Min-Cost Problem

Category 3 The Max flow Min-Cost Problem

Figure 6.8.15: We show that a Category 3 problem (where edges adjacent to s and t have non-zero cost) is as difficult (or as easy) to solve as a Category 3 problem where the s and t edges have zero costs (top diagram). The bottom diagram shows an interesting variation of Category 3 problem where we need to minimize cost not for a maximum flow but for a fixed flow in the network.

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6.9


We shall discuss the Circulation Problem and the related problems of finding feasible flow in network flow graphs. We start this section with the required prior knowledge which is essential to understand the theory and the practice described here. We shall also list down the specific problems that we shall address here. It is interesting to note that this section relies on old concepts like network flows, maximum flow and minimum cut, maximum flow at minimum cost, etc. We shall introduce one new concept and that is of a Circulation graph. We provide a systematic and step by step treatment of a number of theorems and algorithms. These theorems as well as algorithms depend upon a number of powerful transformations. Understanding of these transformations is a must for appreciating the new knowledge described here.

6.9.1

Prior knowledge:

1. Given a network flow graph with zero lower bound and a finite upper bound on flow in each edge, how can we find a maximum flow from a source vertex to a sink vertex? The maximum flow as well as the minimum cut is indicated in the graph shown below (Fig. 6.9.1). 4

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Figure 6.9.1: Left diagram shows a network flow graph with upper bound on flow in each edge. The lower bound on flow through every edge is zero. The right diagram shows the maximum flow and the minimum cut in the network shown in the left diagram. Please note that the flow is conserved at every vertex except the source and the sink vertex. 2. We are given a network flow graph with zero lower bound, a finite


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upper bound on flow in each edge. We are also given a cost per unit flow in each edge of the network graph, and now we need to find a maximum (or a fixed) flow at minimum cost from a source vertex to a sink vertex in the network. For example in the following network flow graph (Fig. 6.9.2) the minimum cost of a flow of one unit from the source to the sink vertex is 4. 2/8

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Figure 6.9.2: A network flow graph with upper bound on flow in each edge as well as per unit cost of flow through that edge. The lower bound on flow through every edge is zero. The problem is to find maximum flow at minimum cost. Please note that the flow is conserved at every vertex except the source and the sink vertex. 3. Using our prior knowledge, we should also be able to solve a slightly different problem in which we can find in terms of yes or no if a fixed flow of k units can be pushed in a network flow graph from a source vertex and taken back into a sink vertex from the network. For example in the graph, shown in Fig. 6.9.1, it is possible to push a flow of 2 units from the source vertex to the sink vertex.

6.9.2

New concepts

In a network flow graph we assume that there is a single source vertex and a single sink vertex. For the rest of the vertices the law of conservation of flow holds - that means the actual flow entering a vertex is equal to the actual flow coming out of it. In a Circulation Graph the law of conservation of flow should hold for every vertex. In a network flow graph, in contrast, the source vertex has the capability to produce an infinite flow while the sink vertex has

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the capability to sink an infinite flow. Thus in a circulation graph there is no source and no sink vertex. Every edge in a circulation graph may have a nonzero lower bound and an upper bound on flow through this edge. There may also be per-unit cost associated with each edge.

6.9.3

New Problems

1. We need to know how to find a feasible flow in a Circulation Graph where every edge has a lower bound as well as an upper bound on flow. For example a feasible flow of 5 is possible in the circulation graph shown below (Fig. 6.9.3).

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Figure 6.9.3: A Circulation graph with lower as well as upper bound on flow in each edge. A feasible flow in the graph is also indicated. Please note that the flow is conserved at every vertex without any exception. 2. We need to find a minimum cost feasible flow in a Circulation graph. Every edge has an associated lower bound, an upper bound, and perunit cost on flow through that edge. Usually this problem is known as the minimum cost Circulation Problem or the Circulation Problem. Please see the solution of the Circulation problem in the following diagram (Fig. 6.9.4). 3. We need to find a feasible flow in a network graph with a source and a sink vertex. Every edge has a lower bound as well as an upper bound on flow (see Fig. 6.9.5). Please note that if the lower bound on flow in each edge is zero then a zero flow will always be a feasible flow from the


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Figure 6.9.4: A Circulation graph (left diagram) with lower bound equal to 1 and upper bound equal to five on flow in each edge. The cost of a unit flow through every edge is equal to one in this graph. A feasible flow in the circulation graph is indicated in the middle diagram. A minimum cost feasible flow is indicated in the right diagram. Please note that the flow is conserved at every vertex without any exception. source to the sink. If, however, the lower bound on flow in any edge is nonzero then finding a feasible flow is not a trivial problem as the zero flow is not a correct answer. 4. We need to find a maximum or a fixed flow at minimum cost in a network graph with a source and a sink vertex. Every edge has an associated lower bound, an upper bound, and per unit cost on flow through this edge. Please see the solution of this problem in the diagram below (Fig. 6.9.6).

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Figure 6.9.5: A network flow graph with lower bound, upper bound, and actual flow in each edge is indicated in the respective order. Please note that the flow is conserved at every vertex except the source and the sink vertex. Also note that the feasible flow taking place may not be a maximum flow (or a minimum flow) from the source to the sink vertex.

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Figure 6.9.6: Left graph is a network flow with a flow of 4 units taking place from the source to the sink vertex. Each edge has an associated lower bound, an upper bound, per unit cost on flow, and actual flow taking place in that edge. A flow of 4 units is a feasible flow but it is taking place at a higher cost in the left diagram. The right diagram shows the same amount of flow - that is 4 units - taking place at minimum cost.


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Before finding feasible flow in a Network graph and in a Circulation graph we should again note that in a Circulation graph the flow is conserved at every vertex while in a network flow graph it is not conserved at the source as well as the sink vertex. Thus it may be possible that a feasible flow does not exist in a Circulation graph (see the right diagram of Fig. 6.9.7) while a feasible flow exists in the same graph having a source and sink vertex as shown below in the left diagram of Fig. 6.9.7.

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Figure 6.9.7: Left graph is a network flow graph with a source and sink vertex; here a feasible flow exists. Right graph is a circulation graph; here a feasible flow does not exist because of obvious reasons.

6.9.4

Finding a feasible flow in a Circulation graph with one special edge

Consider the graph shown in the left diagram in Fig. 6.9.8. Here all edges except one have a lower bound equal to zero. The edge (s, a) has a lower bound equal to the upper bound on flow and they are both equal to 2. We transform the left circulation graph into the right network flow graph and then claim that a feasible flow exists in the circulation graph if and only if we can push a specified flow (equal to 2 in this case) from the source vertex x and pull the same amount of flow from vertex y in the network flow graph. It will be interesting to formally prove this claim. Obviously a feasible flow does not exist in this example because of obvious reasons.

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Figure 6.9.8: Left graph is a Circulation graph with associated lower bound and upper bound on flow in each edge. Please note that except for one edge all edges have lower bounds equal to zero. The edge (s, a) has a lower bound equal to upper bound equal to 2. The Circulation graph is converted into a network flow graph shown in the right diagram. Here the source vertex x is trying to push a flow of 2 units in the network graph while the sink vertex y is trying to pull 2 units from the network graph.

6.9.5

Finding a feasible flow in a network flow graph with one special edge

Consider the network flow graph shown in the left diagram with a designated source and a sink vertex (Fig. 6.9.9). We transform this network flow graph into a circulation graph as shown in the middle diagram. We claim that a feasible flow exists in the network flow graph (left diagram) if and only if a feasible flow exists in the circulation graph shown in the middle. It will be interesting to formally prove this claim. Once this relationship is established then we can use earlier results to find a feasible flow provided it exists as shown in Fig. 6.9.10.


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Figure 6.9.9: Left diagram is a network flow graph with associated lower bound and upper bound on flow in each edge. First it is converted into a Circulation graph (middle diagram) and then we find a feasible flow in the circulation graph.

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Figure 6.9.10: Top left diagram is a network flow graph with associated lower bound and upper bound on flow in each edge. First it is converted into a Circulation graph (top right diagram). We find a feasible flow in the Circulation graph and then convert it into a feasible flow in the original network flow graph shown in the bottom diagram.

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6.9.6


When upper bound is higher than a non zero lower bound

We again consider a circulation graph (shown in the left diagram) where every edge has a lower bound equal to zero except for one special edge - but now the upper bound is higher than a non zero lower bound on flow through this edge. We transform this circulation graph (left diagram) into another circulation graph shown in the middle diagram and claim that a feasible flow in the left circulation graph exists if and only if a feasible flow exists in the middle circulation graph. It will be interesting to formally prove this claim. Once this relationship is established then we can use earlier results to find a feasible flow provided it exists as shown in Fig. 6.9.11.

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Figure 6.9.11: Left graph is a Circulation graph with associated lower bound and upper bound on flow in each edge. Please note that except for one edge all edges have lower bounds equal to zero. The edge (s, a) has a lower bound equal to 2 and an upper bound equal to 7. This edge is split into two edges as shown in the middle graph. The Circulation graph is converted into a network flow graph shown in the right diagram. Here the source vertex x is trying to push a flow of 2 units in the network graph while the sink vertex y is trying to pull 2 units from the network graph.


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Finding feasible flow in a general Circulation graph Here we consider a circulation graph where each edge may have a nonzero lower bound and a different upper bound on flow through this edge. Using earlier transformations we convert the circulation graph shown in the left diagram into a network flow graph shown in the middle diagram. The middle network flow diagram is in turn transformed into another simplified network flow diagram shown in the right diagram. We claim that a feasible flow in the circulation graph exists if and only if we can push a specified amount of flow from vertex x in the middle or the right diagram.

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Figure 6.9.12: Left graph is a Circulation graph with associated lower bound and upper bound on flow in each edge. The Circulation graph is converted into a network flow graph shown in the middle diagram. Here the source vertex x is trying to push a flow of number of units in the network graph while the sink vertex y is trying to pull the same amount of units from the network graph.

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Figure 6.9.13: Left graph is a Circulation graph with associated lower bound and upper bound on flow in each edge. The right diagram shows a feasible flow in the circulation graph. The middle diagram shows an intermediate stage. How to solve the minimum cost Circulation Problem Consider the following circulation problem where the lower bound for each edge is 1 while the upper bound on flow through each edge is 3. In addition to lower and upper bounds we have a cost associated with each edge. This is per unit flow cost for each edge - in this example it is equal to 1. We need to find a feasible flow of minimum cost in this circulation.


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Figure 6.9.14: The left diagram is a Circulation graph with associated lower bound and upper bound on flow in each edge. The cost of per-unit flow through each edge is equal to 1. The Circulation graph is converted into a network flow graph shown in the right diagram. Here the source vertex x is trying to push a flow of 2 units in the network graph while the sink vertex y is trying to pull the same amount of units from the network graph. If we just want to find a feasible flow (not the minimum cost feasible flow) then we know what to do. We convert the circulation (left diagram) into a network flow graph as shown in the right diagram. We try to push 2 units of flow from vertex x. If we are successful then it means that a feasible flow exists in the circulation. But we know that this feasible flow may not be the minimum cost feasible flow. Such a feasible flow is shown in the left diagram of Fig. 6.9.15.

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Figure 6.9.15: The left diagram shows that it is possible to push 2 units of flow from the source vertex x at some cost. The right diagram shows that it is possible to push the same amount of flow from vertex x at a lower cost in fact at a minimum cost. In order to find a feasible flow at minimum cost in the circulation graph of Fig. 6.9.14, we should push 2 units of flow from vertex x at a minimum cost as shown in the right diagram of Fig. 6.9.15. This flow of 2 units at minimum cost is then translated into a feasible flow at minimum cost as shown in the right diagram of Fig. 6.9.16. This provides us a solution to the (minimum cost) Circulation Problem.


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Figure 6.9.16: The left diagram shows the source vertex x pushing two units of flow at minimum cost. The right diagram shows the resulting solution of the minimum cost Circulation Problem. Here the flow in each edge is 1 except for the red bold edges where the flow is 2 units. Our Strategy of Solving the Circulation & other related problems Our strategy of solving the Circulation and other related problems are summarized in the following four figures which are self explainatory.

6.9.7

Is it possible to solve the Circulation Problem for un-directed graphs?

We need to find a (or size of) maximum cut in an undirected network flow graph having vertices s and t. We call this Problem 1. The cut should separate vertex s from vertex t. But we know that this problem is a hard problem? Assume that we need to find minimum flow from vertex s to vertex t in a network flow graph which is un-directed. We call this Problem 2. We know that minimum flow from vertex s to vertex t is equal to max cut in a network

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Network Flow Problem (Zero lower bounds)

Figure 6.9.17: Finding a feasible flow in a Circulation with non zero lower bounds.

Circulation Problem?

Network Flow at Min Cost

(non zero lower bounds)

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Find Feasible Flow at Min Cost

Figure 6.9.18: Finding a minimum cost feasible flow in a Circulation.

Find Feasible Flow

Network Flow Problem? (non zero lower bounds)

Find Feasible Flow

Circulation Problem? (non zero lower bounds)

Network Flow Problem (Zero lower bounds)

Figure 6.9.19: Finding feasible flow in a network with non zero lower bounds.


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flow graph? Suppose we can solve the circulation problem (Problem 3) at minimum cost in an un-directed graph? Then we can use the solution of Problem 3 in order to solve Problem 2? We can then use this solution to find a solution to Problem 1? 5,7

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Figure 6.9.20: The network flow problem is first converted into a circulation problem. The top figure shows that if we need to find max cut in this directed graph (Problem 1) then we should find the min flow from vertex s to vertex t (Problem 2). In order to find minimum flow we need to make it a circulation as shown above and then find a minimum cost circulation (Problem 3) in this graph - that will give us a minimum flow from s to t in the network flow graph as shown below. The minimum flow is 5 and thus the max cut will also have same size. Thus we moved in the following fashion to find the max cut in a network flow graph? In case of un-directed graphs we have the same sort of strategy - but instead a negative result.

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Figure 6.9.21: Finding minimum flow at minimum cost in a network flow problem. The maximum cut can now be found.

Solve Problem 3

Solve Problem 2

Solve Problem 1

Figure 6.9.22: Solving one problem solves another. It essentially means that we cannot find a minimum cost circulation in an un-directed graph while we can solve this problem in a directed graph?

Chapter 7 Eulerian Graphs & the Chinese Postman Problem 7.1

A Special Class of Graphs

7.2

Eulerian Circuits and Graphs

7.3

Eulerian Trails and Related Problems

7.4

Eulerian Walk and the Chinese Postman Problem

7.5

The Chinese Postman Problem for Directed Graphs

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Introduction We shall start with a special category of graphs which was earlier discovered in the last chapter. Exploiting our prior knowledge about this category we define a number of new categories of graphs which are to some extent similar and at the same time different from the graphs we have seen previously. Eulerian graphs belong to one of these categories which we discuss in detail. We make a number of inter-related claims about such graphs and then show how the proof of one claim can lead to the proof of another. We then come back to our categories of graphs and look at these in the light of our newly acquired experience about Eulerian graphs. Thus we start with a panoramic picture, move in depth with one category, and then come back to the panoramic picture with new tools and techniques. We end this chapter with a detailed study of the Chinese Postman problem for both directed as well as undirected graphs.

7.1


We have studied the problem of finding the maximum number of edge-disjoint paths from a source vertex to a sink vertex in a graph in the last chapter. We have claimed earlier that there is a class of graphs where even an unintelligent algorithm (like Algorithm 36) can efficiently find the maximum edge-disjoint paths. We shall study this and similar classes of graphs in detail in this chapter. Please recall Algorithm 36 which is reproduced below. We shall consider connected undirected graphs; for a directed graph we shall assume that the underlying undirected graph is connected. Algorithm 55: Find Maximum edge-disjoint paths from s to t in D input : Directed graph D, and vertices s & t. output: (Maximum) edge-disjoint paths from s to t in D 1

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Find a directed path P from vertex s to vertex t in D. If you are successful in finding a path then keep a record of this path otherwise exit the algorithm; Remove all edges in the path P and go to step 1;

Consider the directed graph D shown in Fig. 7.1.1. In this graph vertex a is a source vertex having only out-degree while the vertex d is a sink vertex


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having only in-degree; for the rest of the vertices of D, the in-degree is equal to the corresponding out-degree. In this special class of graphs the in-degree of vertex d is always equal to the out-degree of vertex a (why?). All directed graphs fulfilling the above properties are known as Class A graphs.

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Figure 7.1.1: A directed graph D with two special nodes a and d. The indegree is equal to the out-degree for every node in this graph except a and d. We can find maximum edge-disjoint paths from vertex a to d in this class of directed graphs using Algorithm 55. We can use Algorithm 55 to find maximum number of edge-disjoint paths from vertex a to d in this graph (there is no need to use the more sophisticated Algorithm 37). Please note that we can use any traversal algorithm (Breadth First Search, Depth First Search, etc) in order to execute the first step of this algorithm. The maximum (number of) edge-disjoint paths in the directed graph D are shown in the left diagram of Fig. 7.1.2. The right diagram shows the same graph with one trail and one path from vertex a to vertex d in the given graph. Please note that each edge of this graph is covered either by the trail or by the path as shown in this diagram. In fact all class A directed graphs possess the following properties: 1. The maximum number of edge-disjoint paths from the source vertex to the sink vertex is equal to the out-degree of the source vertex or the in-degree of the sink vertex. 2. Every edge of the graph is covered by one of the edge-disjoint trails (or paths) from the source vertex to the sink vertex.

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3. The maximum number of edge-disjoint paths from the source vertex to the sink vertex can be found by Algorithm 55. We shall prove these properties later in this chapter. z

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Figure 7.1.2: We show a directed graph of Class A: Vertex a has only outdegree while vertex d has only in-degree. For every other node in this graph the in-degree is equal to the corresponding out-degree. There are two edgedisjoint paths from vertex a to vertex d in the graph D as shown in the left diagram. There are two edge-disjoint trails in this graph such that each edge of the graph is covered exactly once by either of the two trails as shown in the right diagram. Let us now define a Class B category of graphs: In this category of directed graphs we again have two special nodes a and d. There is, however, an important difference this time: Vertex a has an out-degree but also an indegree while the vertex d has an in-degree but also an out-degree. We also assume that vertex a has an out-degree larger than the in-degree, while the vertex d has an in-degree larger than the out-degree. Thus the vertex a resembles a source vertex while vertex d resembles a sink vertex. For the rest of the vertices in D the in-degree is equal to the respective out-degree. How is this class B different from class A and in what respect are they similar? Try to answer this question before moving forward. Again we shall discuss the proofs later. Class C deals with directed graphs in which there are no special vertices; the in-degree of every vertex is equal to the corresponding out-degree. Class C may also include undirected graphs where the degree of every vertex is even. Class E deals with undirected graphs where the degree of every vertex is even except for two special vertices a and d where the degree is an odd number. Class F deals with a more general category of undirected graphs in which


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Concept Map 7.1. A concept map showing various classes of some special graphs.

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every graph G (having p vertices) has 2k nodes with an odd degree; the rest of the p − 2k vertices have an even degree. Remember that the number of vertices having odd degree in a graph can not be odd (why?). We show a Class B directed graph in Fig. 7.1.3. The out-degree of vertex f is 3 while the in-degree is equal to 1; the out-degree of vertex x is 1 while its in-degree is equal to 3; for the rest of the nodes, the in-degree is equal to the corresponding out-degree. We can use Algorithm 55 to find the two edge-disjoint paths from vertex f to vertex x in D; the two paths are shown in Fig. 7.1.3. Can we also find trails in this graph such that each edge of this graph is covered exactly once by one of the trails? We show a Class C directed graph in the left diagram of Fig. 7.1.4. There is no special vertex in this graph; the in-degree of every vertex is equal to the corresponding out-degree. An undirected graph belonging to the same category is shown in the left diagram of this figure. The degree of each node in this undirected graph is even. It is interesting to note that a directed graph (where the in-degree of each node is equal to the corresponding out-degree) belongs to the same class as an undirected graph where the degree of each vertex is even (why?). As you should discover yourself Class C directed and undirected graphs have some special and interesting properties. One such property is shown in Fig. 7.1.4. Both the graphs shown in this diagram can be partitioned into edge disjoint cycles (or circuits) shown in different colors. We can use any traversal algorithm to find a cycle in such graphs; if we now remove edges of this cycle from the original graph then the new graph will also belong to the same class (why?). If however we remove (all edges in) a path from any vertex x to a vertex y in a Class C graph then the new graph will not be a Class C graph any more; it will become a Class A or Class B graph. Similarly if we add a path between vertex a and d (these are the only two vertices having an odd degree) in a Class E graph then it will be transformed into a Class C category. How other properties of a graph change (or do not change) after such a transformation is interesting to explore. Before moving forward, here are a few questions that you should be able to answer by yourself: 1. What are the necessary and sufficient conditions for a (connected) graph to be cyclic?


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Figure 7.1.3: A directed graph of Class B: The out-degree of node f is larger than the in-degree while the in-degree of x is larger than its out-degree. For every other node in this graph the in-degree is equal to the corresponding out-degree.

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Figure 7.1.4: Every node in the directed graph has an in-degree equal to the corresponding out-degree as shown in the left diagram. An un-directed graph where the degree of each vertex is even is shown in the right diagram. Both these graphs belong to our Class C category. The edge set of these graphs can be partitioned into edge-disjoint cycles (shown by different colors) which if combined together will create a circuit consisting of all edges of the graph.

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2. Let us consider a special case of Class C graphs where the degree of each vertex is not only even it is exactly two. Try to visualize such a graph. What can you say about other properties of this special graph? 3. Can we claim that in such a graph every vertex will lie on some cycle? 4. Can we make the above claim if the degree of every vertex is even? 5. What about if each vertex in a graph has a degree equal to at least two? Remember we are considering connected graphs only as mentioned earlier in this section.

7.2


A graph G is Eulerian provided it contains an Eulerian circuit; a circuit which contains every edge of G. Please remember that in a circuit it is possible to visit a vertex several times but we are allowed to traverse an edge only once. Now we make a number of claims (it will become evident from these claims that an Eulerian graph belongs to our Class C category): 1. A graph G is Eulerian if and only if every edge of G lies on an odd number of cycles. 2. A graph G is Eulerian if and only if the degree of each vertex is even. 3. A graph G is Eulerian if and only if the edge set of G can be decomposed into edge-disjoint cycles. 4. The degree of each vertex in a graph G is even if and only if the edge set of G can be decomposed into edge-disjoint cycles. We shall start with proving Claim Number 4 and then work backwards in order to prove earlier claims. Assume that a graph G can be decomposed into edge-disjoint cycles; then we should be able to prove that the degree of each vertex of G is even (you should be able to do it easily). So we assume that the degree of each vertex is even and now we should be able to prove that the graph G can be decomposed into edge-disjoint cycles. Let us do


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it now. As the degree of each vertex in G is even we can find a cycle C in G using any traversal algorithm. If we remove all the edges belonging to C from G then in the resulting graph the degree of each vertex will again be even (why?) but this new graph will have fewer edges as compared to the original graph G. It is possible to build logic on this observation in order to design a formal proof. Let us now tackle Claim Number 3. If the edge set of a graph can be partitioned into edge-disjoint cycles then we can always combine these cycles to create a circuit which will cover every edge of the graph exactly once; so the graph will be Eulerian. This part of the proof is done and let us attempt the other part: if a graph is Eulerian then (we shall be able to prove that) it can be partitioned into edge-disjoint cycles. If the graph is Eulerian then there will be an Eulerian Circuit inside that graph as shown in left diagram of Fig. 7.2.1. If this circuit is a cycle then the proof is complete otherwise it will consist of several cycles. Remember in a circuit a vertex may be repeated but an edge cannot be repeated. Thus the cycles forming the Eulerian circuit will be edge-disjoint as shown in the left diagram of Fig. 7.2.1.

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Figure 7.2.1: An Eulerian graph containing an Eulerian circuit is shown in the left diagram. The circuit can be decomposed into edge-disjoint cycles as shown in the right diagram. We have earlier proved in Claim Number 4 that a graph can be partitioned into edge-disjoint cycles if and only if the degree of each vertex in the graph is even. In Claim Number 3 we have proved that a graph G is Eulerian if and only if the edge set of G can be decomposed into edge-disjoint cycles. Combining the two we can prove that a Graph is Eulerian if and only if the degree of each node in the graph is even.

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Figure 7.2.2: An Eulerian graph is shown. Consider vertex a which is adjacent to vertex b in this graph. We intend to prove that the edge ab is part of an odd number of cycles in this graph. Please note that if there is a counter clockwise cycle from vertex f to vertex y to vertex z and back to f then there is clockwise cycle from vertex f to z to y and back to vertex f .

Let us now concentrate on Claim Number 1. First assume that every edge (m, n) in graph G is part of an odd number of cycles. It means that the degree of node m as well as that of n will be even; thus G will be an Eulerian graph. Now assume that the graph G is Eulerian; we have to prove that every edge (m, n) in G will be part of an odd number of cycles as shown in Fig. 7.2.2. As the graph G is Eulerian thus the degree of each node will be even; thus the degree of node m as well as that of node n will be even. Proving that edge (m, n) will be part of an odd number of cycles is equivalent to proving that there are an odd number of paths between vertex m and vertex n (why?). Proving that there is an odd number of paths between vertex m and vertex n is in fact equivalent to proving that there is an odd number of trails from vertex m to vertex n. This may not be very obvious so we shall prove this after first finding the number of distinct trails between the two vertices. Please remember that in a trail we may repeat vertices but we cannot repeat edges while in a path neither vertices nor edges can be repeated. We start from a (in Fig. 7.2.2) and arrive at the adjacent vertex e; as the degree of every vertex in G is even so if you can enter a vertex then you can leave it as well; in fact there will always be an odd number of ways out (why?). This means that if you draw a tree of all possible trails then the out-degree of every node in this tree will be an odd number as shown in Fig. 7.2.3. The total number of trails in the graph of Fig. 7.2.2 will be equal to


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the number of leaf vertices in the tree of Fig. 7.2.3 which will be an odd number (why?). Every cycle in the graph contributes to two trails as shown in Fig. 7.2.2 and Fig. 7.2.3. Thus the total number of paths between vertex m and vertex n will be an odd number (why?). There are a number of algorithmic issues apart from the above (theoretical) claims and their respective proofs: 1. How can you efficiently check if a graph G is Eulerian using different (necessary & sufficient) conditions for a graph to be Eulerian? 2. How can you efficiently find an Eulerian circuit in an Eulerian graph? 3. How can you efficiently find edge-disjoint cycles in an Eulerian graph? There are a number of exciting theoretical problems which you should attempt before moving forward: 1. How can you generalize the four claims that we have made for multigraphs; in multi-graphs we allow parallel edges and self loops? (Hint: Can you convert a multi-graph into a simple graph?) 2. How can you generalize (or modify) the four claims in case of directed graphs?


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Problem Set 7.1. Please read the following algorithm which is primarily designed to find edge-disjoint cycles in a graph G where the degree of every vertex is even. Algorithm 56: Find Cycles in an un-directed graph G where degree of every vertex is even input : Undirected graph G where degree of every vertex is even output: Edge-disjoint cycles in graph G 1 2

Find a Cycle C in graph G, and keep a record of it; Remove all edges in the Cycle C from graph G. If some edges are still left in G then go to step 1 otherwise exit;

Problem 7.1.1. Apply the above algorithm on the graph G as shown in Fig. 7.2.4, and specify each cycle output by the algorithm. e

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Figure 7.2.4: Problem 7.1.2. Some one claims that the algorithm outputs edge disjoint cycles. Discuss why or why not. Problem 7.1.3. Someone claims that the algorithm outputs maximal number of edge-disjoint cycles in a graph G. Prove the claim or give a counter example. Problem 7.1.4. Someone claims that the algorithm outputs maximum number of edge-disjoint cycles in a graph G. Prove the claim or give a counter example.

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Problem 7.1.5. Assume that we apply the above algorithm to graph H which is different from graph G. How would the algorithm behave in this type of graph? What will be the output of the algorithm? How will the output be different in this case? Discuss briefly. Problem 7.1.6. It will be useful to design the following algorithms for an undirected graph G.

Algorithm 57: Find graph G given a set of edge-disjoint cycles of G input : Edge-disjoint cycles in an unknown graph G output: Un-directed graph G (degree of every vertex in G will be even)

Algorithm 58: Find an Eulerian circuit in graph G given a set of edge-disjoint cycles of G input : Edge-disjoint cycles in an unknown graph G output: An Eulerian circuit in G

Algorithm 59: Find a set of edge-disjoint cycles of G given an Eulerian circuit in a graph G input : An Eulerian circuit in G output: Edge-disjoint cycles in an unknown graph G

Problem 7.1.7. How about if we have a directed graph in which the indegree of every vertex is equal to its out-degree. How will the solution of earlier problems be affected?

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Let us now consider Class E undirected graphs where the degree of every vertex is even except for two special vertices f and x where the degree is an odd number (will the degree of the two odd vertices be the same? Why?).


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Figure 7.2.5: A graph (with every vertex having even degree) can be split up into edge-disjoint cycles. Several sets of edge-disjoint cycles are shown here, some having a larger size than others. Note that every edge is part of an edge-disjoint cycle. We can use Algorithm 56 to find out efficiently one such set of edge-disjoint cycles.

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Figure 7.2.6: We have seen that if the degree of every vertex is even then the graph can be split into edge-disjoint cycles again shown in the top diagram. On the other hand if a graph can be split into edge-disjoint cycles or if the edge-disjoint cycles of a graph are given then we can reconstruct the original graph as shown in the middle diagram. The original graph will have all nodes with even degree. The reconstruction algorithm will help you reconcile with this claim. There will be an Eulerian circuit in the reconstructed graph.


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Such a graph is shown in the left diagram of Fig. 7.3.1. Its counterpart in directed graphs is shown in the right diagram of Fig. 7.3.1; here the outdegree of one special vertex is larger than its in-degree by one, while it is the other way round for the other special vertex; for the rest of the vertices the in-degree is equal to the corresponding out-degree. In both these graphs it is possible to find an Eulerian trail from vertex f to vertex x; a trail in which every edge of the graph is covered (exactly once). How can we prove this and how can we find an Eulerian trail? Perhaps we can find a constructive proof which will solve both the problems. Again we should appreciate that we can use Algorithm 55 to find maximum number of edge-disjoint paths from a source to a sink vertex? We have earlier presented Class B and Class A category directed graphs in Fig. 7.1.2 and 7.1.3. These graphs are reproduced in Fig. 7.3.2 for comparison with graphs in Fig. 7.3.1. We have made certain claims about Class A graphs earlier in this chapter. These claims can be generalized with some interesting modifications for Class B directed graphs and then proved using our newly acquired experience of Eulerian Graphs? We now present one last claim which is applicable to Class F undirected graphs. This class comprises of undirected graphs having 2k nodes with an odd degree; the rest of the p − 2k vertices have an even degree. The claim for such a graph G is that the edge set of G can be partitioned into k trails where each trail is connecting two odd vertices. Such a graph is shown in the left diagram of Fig. 7.3.3. with odd vertices shown in bold. The right diagram shows the same graph where we add an extra edge between two odd vertices converting this graph into a Class C category where the degree of each vertex is even. It will then become possible to find an Eulerian circuit in the resulting graph which is also shown in the right diagram of Fig. 7.3.3. This observation should lead you to design a formal proof for the above claim.

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We know that if the degree of every vertex is even in an undirected graph G then we can find an Eulerian Circuit in G by traversing each edge exactly once. If, however, the degree of every vertex is not even then the problem is to traverse each edge at least once (not exactly once) and making sure

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Figure 7.3.1: An un-directed graph shown in the left diagram; the degree of vertex f and x is odd while the degree of every other vertex is even; this is a Class E un-directed graph. A directed graph D with two special nodes f and x is shown in the right diagram; the in-degree is equal to the out-degree for every node in this graph except f and x; the out-degree of vertex f is one larger than its in-degree and it is the other way round for vertex x in this directed graph.

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Figure 7.3.2: A directed graph of Class B category is shown in the left diagram; the out degree of vertex f is three while its in-degree is 1; it is the other way round for vertex x. A directed graph of Class A category is shown in the right diagram; the in-degree is equal to the out-degree for every node in this graph except a and d; vertex a has only out-degree while vertex d has only in-degree.


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Figure 7.3.3: An undirected graph of Class F shown in the left diagram; the degree of vertex f , b, y and x is odd while the degree of every other vertex is even. By adding two edges between odd vertices we can convert this graph into a Class C category where the degree of every vertex is even.

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that the number of edges, traversed more than once, are minimized. This problem is faced by any post man delivering letters in houses along lanes or a sweeper who is sweeping roads; he would certainly like to traverse each lane at least once while making sure that the traversals of the same lane should be minimized. This problem is formally defined as the Chinese Postman Problem: We need to find a shortest closed walk in a graph G which passes through every edge of G at least once. Such a walk (walk because some edges will be traversed more than once) is also known as an Eulerian walk in the graph G. In the next figure, we show graphs where there are two vertices of odd degree. In the left diagram of this figure we have an un-weighted graph while the right diagram shows a weighted graph. As you can well imagine, in a weighted graph, we need to minimize the total sum of edge weights in a closed walk which covers every edge of the graph at least once. There are basically two problems that we intend to solve simultaneously but first we need an understanding of these problems in isolation. 1. We need to convert a given graph G into an Eulerian graph H by duplicating some existing edges of G. Now an Eulerian circuit in H will correspond to a closed Eulerian Walk in G; in graph G we shall be traversing an edge twice when in graph H we shall be traversing a duplicated edge. 2. We need to make sure that the Eulerian trail that we have created in the first part is shortest in terms of number of edges involved or in terms of edge weights. In other words the Eulerian graph H should be of minimum size in terms of number of edges or in terms of sum of edge weights of H. We show a simple technique of converting a graph with odd vertices into a graph with all vertices even in Fig. 7.4.2. We find a path from one odd vertex to another odd vertex and duplicate every edge encountered in that path. The resulting graph will be Eulerian as shown in the right diagram of this figure. Please note that the degree of the two terminal odd vertices in the path will become even while the degree of an even vertex in the middle of the path will stay even. Indeed there could be paths of different edge lengths between the two odd vertices in a graph as shown in Fig. 7.4.3. All these possible paths gives rise to different number of duplicated edges required to convert a graph into an Eulerian graph. Each resulting Eulerian graph H will


Two vertices with odd degree (red color) in an unweighted graph

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Two vertices with odd degree (red color) in a weighted graph

Figure 7.4.1: An undirected graph and un-weighted graph (left diagram). The right diagram shows a weighted graph. Both graphs have two vertices with an odd degree. have different number of edges; we need to select the one with the minimum number. The Chinese Postman Problem for an un-weighted graph can thus be rephrased: We need to convert a graph G (having some odd vertices) into an Eulerian graph H (having all vertices with even degree) by duplicating minimum number of existing edges of graph G. For a weighted graph we need to convert G into an Eulerian graph by duplicating certain edges such that the sum total of edge weighted corresponding to duplicated edges is minimized. We show different Eulerian graphs corresponding to a weighted graph G in Fig. 7.4.4. It is interesting to note that optimal graph, shown in the right diagram, consists of more edges than the graph shown in the middle diagram.

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Figure 7.4.2: Shows how can we convert a graph G having two odd vertices into a graph H where the degree of each vertex is even. The duplicated edges are shown in red color in the graph H. An Eulerian Circuit in H is equal to an Eulerian Walk in G.

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The duplicated edges are shown in red; the graph becomes Eulerian

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Figure 7.4.3: It is possible to convert a graph having two odd vertices into different Eulerian graphs with varying number of edges. The Eulerian circuit in each graph is also indicated.

Eulerian Walk and the Chinese Postman Problem 2

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Figure 7.4.4: It is possible to convert a graph having two odd vertices into different Eulerian graphs with varying number of edges. The Eulerian circuit in each graph is also indicated. It is quite obvious now that a shortest path between the two odd vertices provides us an optimal solution to the Chinese Postman Problem given that the graph has only two odd vertices. What about if a graph G consists of more than 2 odd vertices as shown in Fig. 7.4.5? How about if we apply the same trick of finding a shortest path between vertices belonging to different pairs of odd vertices? (Remember the total number of odd vertices in any graph will always be even). Let us start with an arbitrary selection of vertices in the three pairs as shown in Fig. 7.4.6. The resulting Eulerian graph is shown in the right diagram of the same figure. The total number of edges duplicated is also indicated in this diagram. Finding shortest paths for a fixed set of pairs of odd vertices certainly helps to reduce the cost of making the degree of each vertex even - but a different set of pair of odd vertices may help us in further reducing this cost as shown in Fig. 7.4.7. Thus the problem is reduced to finding the set of pairs of odd vertices which minimizes the cost of duplicating the edges. Do we have to enumerate all possible sets of pairs of odd vertices in the graph G? How to do that systematically and estimate what is the total number of possibilities? Is a brute force approach suitable or should we find a way to solve this problem in a more efficient manner? Given 2k items (corresponding to 2k odd vertices) numbered from 1 to 2k; let us draw a completely connected graph consisting of 2k = 6 vertices (of G) as shown in Fig. 7.4.8. A perfect matching in this graph provides the desired set of all pairs of odd vertices. It is possible to put weights on edges of the completely connected graph as shown in Fig. 7.4.9. A minimum weight

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Figure 7.4.6: We find shortest paths between a fixed set of odd vertices in graph G. The resulting Eulerian graph H is shown in the right diagram.


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Figure 7.4.7: We find shortest paths between a different set of odd vertices in graph G. The resulting Eulerian graph H is shown in the right diagram. perfect matching in the completely connected weighted graph will provide an efficient solution to our problem as shown in the following figures.

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Figure 7.4.8: A perfect matching in a completely connected graph K of odd vertices of graph G. An efficient solution of the Chinese Postman Problem in an un-directed graph G is thus reduced to the following steps: 1. Identify odd vertices in graph G. 2. Create a completely connected graph K; every odd vertex of G is a vertex of this completely connected graph K. The weight of an edge between two vertices in graph K is equal to the weight of the shortest path between the corresponding odd vertices in graph G. 3. A minimum cost perfect matching in the completely connected graph K provides us the desired pairs of vertices (x, y) in graph K. For every such pair (x, y), we duplicate edges along the shortest path between vertex x and vertex y in G.


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Figure 7.4.10: We enumerate all distinct perfect matching in a completely connected graph consisting of 6 odd vertices of G. Please note that looking at all enumerations will be very costly; how do we find a better solution?


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7.5



We discuss here the Chinese Postman Problem for a directed graph. If a directed graph G have all vertices where the in-degree is equal to the corresponding out-degree then the graph G is Eulerian. If, however, some vertices have out-degree larger or smaller than the in-degree then we need to convert graph G into an Eulerian graph H by duplicating minimum number of existing edges of G. We show an undirected graph G in the top left corner of Fig. 7.5.1. By putting directions on edges of G, this graph is converted into a directed graph as shown in the top right diagram of the same figure. The undirected graph G has even as well as odd vertices - that is why graph G is not Eulerian. In contrast, the directed graph D has three types of vertices as described below. 1. The out-degree is equal to the corresponding in-degree. The difference ∆ (equal to out-degree minus in-degree) for any vertex is zero for such vertices. Two such vertices exist in the directed graph D and are shown in green color. 2. The out-degree is larger than the corresponding in-degree; hence ∆ is positive. There are three such vertices in graph D, all shown in red. 3. The out-degree is less than the corresponding in-degree; hence ∆ is negative. There are four such vertices in graph D, all shown in orange. If all vertices in a directed graph D have ∆ equal to zero then graph D is Eulerian. If, however there are vertices with +ive and/or -ive ∆ then we face the challenge of solving the Chinese Postman Problem. A solution to this problem for an undirected graph G is not helpful even if the directed graph is derived from the same undirected graph as shown in the bottom left and right diagrams of Fig. 7.5.1. We show a directed graph G where there is only one vertex with positive ∆ as shown in Fig. 7.5.2. Here the problem is simple: Finding a path from each of the three orange vertices to the only red vertex and duplicating edges in these paths will convert the graph G into an Eulerian graph. Finding shortest paths from each orange vertex (with ∆ negative) to the only red vertex (with ∆ positive) will provide us the optimal solution to


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Figure 7.5.1: We again show an undirected graph G in the top left diagram. It is converted into a directed graph D as shown in the top right diagram. It is obvious that graph G is not Eulerian. After duplicating certain edges in graph G it is transformed into an Eulerian graph as shown in the bottom left diagram. An Eulerian walk in G is shown in the bottom right diagram. The Eulerian walk in G is not an Eulerian walk in the directed graph D.

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Figure 7.5.2: We show a directed graph G where there is only one vertex (shown in red) where the out-degree is larger than the in-degree. All other vertices have either out-degree equal to in-degree (shown in green) or less than in-degree (shown in orange).


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the Chinese Postman problem as shown in the same figure. Please note that all the three shortest paths originating from vertices with ∆ equal to minus 1 are terminating at a single vertex; that single vertex has ∆ exactly equal to +3; this will make the new directed graph Eulerian; duplicating edges on shortest paths ensures that the number of extra edges (added) is minimized. We show a directed graph in the top diagram of Fig. 7.5.3. There are two vertices (4 & 10) with ∆ negative shown in orange color. Vertex 4 has ∆ equal to −3 while vertex 10 has ∆ equal to −1. Similarly there are three vertices with ∆ positive, all shown in red color; vertex 9 has ∆ equal to +2. Please note that there should be three paths coming out of vertex 4 in order to increase its ∆ from −3 to zero. Similarly two paths should be terminating at vertex 9 to convert its ∆ from +2 to zero (please see Fig. 7.5.3). In order to make ∆ of every vertex zero (thus converting the graph G into an Eulerian graph) we should find out which path originating from an orange vertex should terminate at which red vertex. We create a bipartite graph B consisting of an A partite (consisting of all orange vertices) and a B partite (consisting of red vertices) as shown in the middle left diagram of Fig. 7.5.3. Please note that this will be a complete bipartite graph; there will be an edge from every orange vertex to every red vertex in B signifying that there will be a path from every orange vertex to every red vertex in graph G. Further adding vertices s and t and edge capacities (middle right diagram) the problem is converted into a flow problem. Finding a maximum flow efficiently (in polynomial time) in this (special) network graph N is by itself an interesting problem. The maximum flow in this graph (as shown in the bottom diagram) will provide the required information. The graph G will become Eulerian but the cost in terms of number of edges of G that are duplicated may be high; the problem is to minimize the cost also. The solution is simple: we create a weighted bipartite graph B where the weight of an edge from an orange vertex x to a red vertex y in B signifies the weight of the shortest path between the orange vertex x to red vertex y in graph G. Finding a maximum flow with minimum cost in a network graph N will ensure that the graph becomes Eulerian and the number of edges (which are) duplicated are minimized at the same time. This has been illustrated in Fig. 7.5.4. An efficient solution of the Chinese Postman Problem in a directed graph G is thus reduced to the following steps:

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Figure 7.5.3: A maximum flow in a network graph N (middle right diagram) helps us find how to convert a directed graph into an Eulerian graph.


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1. Identify orange and red vertices in G. 2. Create a weighted bipartite graph B consisting of orange vertices as an A partite and red vertices as the B partite. The weight of an edge from an orange vertex x to a red vertex y in B signifies the weight of the shortest path between the orange vertex x to red vertex y in graph G. 3. Convert bipartite graph B into a network flow graph after adding vertices s & t. The capacity of an edge from s to an orange vertex x is equal to ∆(x). The capacity of an edge from a red vertex y to t will be equal to ∆(y). The costs of these edges will be zero. The cost of any other edge in N will be the corresponding weight in the bipartite graph; the capacity of these edges will be infinite. 4. A minimum cost maximum flow in the network graph N will provide us the desired pairs of orange/red vertices in graph G. Corresponding to every such (x, y) pair (x belongs to partite A consisting of orange vertices, while y belongs to partite B of red vertices), duplicate edges in the shortest path from vertex x to vertex y in graph G. It will be interesting to derive the time complexity of the above algorithm and compare it with that of the algorithm used to solve the Chinese Postman Problem for undirected graphs. Problem Set 7.2. Problem 7.2.1. The examples of directed graphs that we have considered in this section have one thing in common: If we sum ∆ of all vertices in a graph it comes out to be zero. Is this a coincidence or will every directed graph possess this property? Discuss briefly. Problem 7.2.2. Show that a necessary & sufficient condition for a directed graph D to have a solution to the Chinese Postman Problem is that graph D should be strongly connected; it means that there should be a path between every pair of vertices in graph D. Problem 7.2.3. We show a directed graph in the top left diagram of Fig. 7.5.5. Here there are four vertices with ∆ negative (shown in orange color) while

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The Chinese Postman Problem for Directed Graphs 8

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Figure 7.5.5: We show a directed graph G where there are three vertices, shown in orange color, where the out-degree is less than the corresponding in-degree. Vertices where the out-degree is larger than the in-degree are shown in red color. Green vertices have out-degree equal to the in-degree. We describe how we can convert a directed graph G into an Eulerian graph H by duplicating minimum number of edges of G.

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there are three red vertices with ∆ positive. We demonstrate an alternate scheme to solve the Chinese Postman problem. Describe an algorithm behind this demonstration and calculate the time complexity of this algorithm in comparison with that of the algorithm discussed in the text (based on minimum cost maximum flow algorithm). Problem 7.2.4. We show two directed graphs in Fig. 7.5.6. Solve the Chinese Postman problem for both these graphs using an efficient algorithm. 8

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Figure 7.5.6: We show two directed graphs which are not Eulerian. We need to convert these graphs into Eulerian graphs by duplicating minimum number of existing edges of these graphs. Problem 7.2.5. Derive the time complexity of the algorithm (that we have described in the text) which can be used to solve the Chinese Postman prob-


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lem for directed graphs. Compare its value with the time complexity of the algorithm used for undirected graphs. Problem 7.2.6. We show a directed graph in the left diagram of Fig. 7.5.7; please note that the in-degree of every vertex in this diagram is exactly equal to its corresponding out-degree; thus the graph is Eulerian. The Chinese Postman is supposed to distribute mail on streets shown in red color only. We need to find an Eulerian circuit consisting of edges corresponding to red streets alone; if that is not possible then we should allow him (the Chinese Postman) to traverse the red edges at least once and come back after traversing a minimum number of edges in this graph. Please describe an efficient algorithm to solve this problem; show the detailed working of this algorithm on this graph. Problem 7.2.7. We show another directed graph in the right diagram of Fig. 7.5.7. This graph is not Eulerian; it is neither strongly connected yet it may be possible for the Chinese Postman to distribute mail in (only red) streets while traversing minimum number of extra graph edges.

Figure 7.5.7: We show two directed graphs; the graph shown in the left diagram is Eulerian while the one shown in the right diagram is neither Eulerian nor it is strongly connected. The Chinese Postman is supposed to deliver mail in red streets only. Problem 7.2.8. We show a directed graph G in the left diagram of Fig. 7.5.8. The Chinese Postman is supposed to start delivering his mail from a specific

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point and end his delivery job at another point as shown in the figure. The postman is supposed to traverse every edge of this graph exactly once. If this is not possible then he should traverse each edge at least once and also minimize the number of edges which are traversed more than once. The right diagram of this figure shows a transformation of graph G into another graph where certain edges of G are duplicated. Show that an Euler trail is possible from vertex 8 to vertex 3 in this graph. Design an efficient algorithm to efficiently solve the above problem. Does the right diagram of Fig. 7.5.8 provide an optimal solution to our problem for the graph shown in the left diagram? Discuss briefly? 5

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Figure 7.5.8: We show a directed graph G in the left diagram; the red vertices have in-degree smaller than the corresponding out-degree while the orange vertices have in-degree larger than the out-degree; the green vertices have out-degree equal to the corresponding out-degree. The right diagram shows a transformation of graph G into another graph where it is possible to find an Eulerian Path from vertex 8 to vertex 3.


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Chapter 8 Hamiltonian Graphs 8.1

Introduction

8.2

Prior Knowledge

8.3

Hamiltonian Graphs

8.4

Bipartite Hamiltonian Graphs

8.5

Some Theoretical Claims

8.6

A Categorization of Hamiltonian Graphs

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8.1

Hamiltonian Graphs

Introduction

In this chapter, we shall address the problem of finding a Hamiltonian cycle and a Hamiltonian Path in a graph. We shall also be considering certain conditions (necessary and then sufficient) for a graph to be Hamiltonian. We begin by using constructive proof techniques for certain sufficient conditions; the same techniques can be used to find a Hamiltonian Cycle in a graph where a Hamiltonian Cycle is guaranteed (by the sufficient conditions) to exist. Students of this topic sometimes get confused between necessary and sufficient conditions. In case of Eulerian Graphs where we have a neat classification, we can easily differentiate between necessary and sufficient conditions. In Hamiltonian Graphs however, we are unaware of such a characterization so we shall be talking about necessary and sufficient conditions separately.

8.2 8.2.1

Prior Knowledge Necessary Conditions for a Connected Graph

Let us start with a familiar and simple example of a connected graph. For a graph (with p vertices) to be connected, there should be at least p − 1 edges otherwise the graph will be guaranteed to be disconnected. Such a graph is known as a tree where each vertex is a cut vertex and each edge is a bridge edge. But then there are graphs where the number of edges is more than this critical number but still these graphs are not connected. That means this (q ≥ p − 1) will be a necessary condition for a connected graph. Having edges less than this limit will be a sufficient condition for a graph to be disconnected. It will be interesting to find sufficient conditions for a graph to be connected, but we shall discuss this issue later in this topic.

8.2.2

Necessary Conditions for a Hamiltonian Graph

A graph can not be Hamiltonian unless it is connected. It is thus a necessary condition, although a loose necessary condition. A tree graph is connected but it can not be Hamiltonian as it has no cycles (it can, however, contain a Hamiltonian Path). Thus a Hamiltonian graph should be more than a tree graph. It should be connected and there should be a cycle spanning all vertices. This means that no vertex should be a cut vertex and no edge is a

Prior Knowledge

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bridge edge in a Hamiltonian graph. Again these will be necessary conditions as there are graphs which satisfy these conditions but are not Hamiltonian. On the other hand if a graph contains a bridge edge (or a cut vertex) then it is certainly not Hamiltonian. We can certainly find sharper necessary conditions as are explained in most other text books.

8.2.3

A Loose Sufficient Condition for a Hamiltonian Graph

Let us think of a familiar graph where a Hamiltonian Cycle is guaranteed to exist and we can find this cycle using a trivial algorithm. A tree graph is just connected and belongs to one side of the extreme of connected graphs. The other side of the spectrum is a completely connected graph where every vertex is connected to every other vertex. Any permutation of the vertices of this graph will give you a Hamiltonian Cycle (see Fig. 8.2.1). Thus the completely connected property is a sufficient condition for a graph to be Hamiltonian, although it is perhaps an overkill and thus not a sharp sufficient condition. This loose sufficient condition will be used in later pages to derive sharper sufficient conditions. It is interesting to note that complete connectedness is not a sufficient condition for a graph to be Eulerian.

8.2.4

Sufficient Condition for a Connected Graph

Before finding a tighter sufficient condition for a Hamiltonian graph, let us solve a much familiar and also simpler problem of finding a sufficient condition for a graph to be connected. Obviously a completely connected condition would be just too loose for a graph to be just connected. It will be useful if a learner solves this problem by himself or herself. The minimum degree of for a graph to be connected. any vertex in a graph should be at least (p−1) 2 It can also be shown that if the maximum degree of any vertex in a graph is equal to (p−2) then you can always draw a disconnected graph. (How?) This 2 implies that the condition (minimum degree ≥ (p−1) ) is certainly a sharp 2 condition for connectedness.

8.2.5

A Concept Map

We show a concept map indicating some necessary and sufficient conditions for a graph to be connected, Eulerian, and Hamiltonian (see Concept Map

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1). In case of Eulerian graphs we can find a nice characterization that is necessary and sufficient for a graph to be Eulerian. Subsequently we can design efficient algorithms to actually find an Euler Cycle in a graph provided the graph satisfies the given conditions. In case of connected or Hamiltonian graphs it is not possible to find necessary and sufficient conditions. Thus for a general graph which does not satisfy the sufficient conditions for a Hamiltonian Cycle it is extremely hard to design an algorithm which can find a Hamiltonian Cycle (provided such a cycle exists) or out puts in negative if a Hamiltonian Cycle does not exist. In case of connected property again it is not possible to devise necessary and sufficient conditions, however it is possible to design efficient algorithms which can determine if a given graph is connected or not. A simple Breadth First Search (or any traversal algorithm) can solve this problem efficiently. u5

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Figure 8.2.1: A completely connected graph of 6 vertices. Two Hamiltonian Cycles are shown in the completely connected graph; each cycle corresponds to a different permutation of vertices of the graph. Summary It will be interesting to summarize that what is possible (or not possible) and at what cost. If the number of edges is less than p − 1 then the graph will certainly be disconnected. We can still draw a disconnected graph if the . If the minimum degree of a node in the graph is less than or equal to (p−2) 2 (p−1) minimum degree in a graph is equal to or more than 2 then certainly the graph will be connected. The graph may still not be Hamiltonian; in fact we can draw a graph with the above property which is clearly not Hamiltonian. Interestingly a minimum degree equal to (p−1) is a sufficient condition for 2 graph to contain a Hamiltonian Path. And a minimum degree slightly larger than (p−1) is a sufficient condition for a graph to be Hamiltonian. We shall 2

Prior Knowledge

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Concept Map 8.1. A concept map of certain necessary and sufficient conditions for a graph to be connected, Eulerian, and Hamiltonian.

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Hamiltonian Graphs

now provide constructive proof techniques to prove a number of sufficient conditions for a graph to be Hamiltonian.

8.3

Hamiltonian Graphs

We shall make a drastically different start. Instead of simply describing the algorithms, we shall make you solve a number of related but simple problems. You just need some prior knowledge in addition to some common sense. It is important for the learner to gain confidence: You can discover and create new knowledge. We shall also encourage you to solve a puzzle; the experience that you will gain will provide you powerful tools that we shall use in this chapter in solving various problems. Deciding when a specific tool can be used and where it can not be used is certainly a valuable learning experience. Problem Set 8.1. Let us start with simpler problems. For each of the problem you are supposed to design a formal proof; Problem 8.1.1. Suppose G is a star graph then G2 has a Hamiltonian Cycle. Please note that the diameter of a star graph is equal to 2 irrespective of the order of the graph. Problem 8.1.2. Suppose G is a line graph then G2 has a Hamiltonian Cycle. Please note that the diameter of a line graph is p − 1, and as it is obvious it is a function of the order of the graph. No graph (other than a line graph) can have a diameter as large as p − 1. For each of the above two problems it will be useful if the learner actually draws such graphs and then discovers the answer himself or herself. The k th power Gk of a graph G is a graph with same number of vertices as in G in which two vertices are adjacent if and only if they are at most d distance apart from each other in G. If the diameter of a graph G is k then Gk will be a completely connected graph and that is why G2 of a star graph is Hamiltonian. The diameter of a line graph is proportional to the size of the graph but still G2 is Hamiltonian. It will be interesting to explore if this is a more general result: G2 of any tree is Hamiltonian? Either show it or find a counter example. Let us now try to solve a problem which belongs to the so called critical activity section. You must solve this problem before moving forward. Please see Fig. 8.3.1 and Fig. 8.3.2.

Hamiltonian Graphs

8.3.1

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A Puzzle:

We are given a line graph with p nodes, the starting and end node of the line graph are designated as u1 and up respectively. The intermediate nodes are labeled as ui , where 2 ≥ i ≥ p − 1 as shown in Fig. 8.3.1. You are allowed to insert new edges between u1 and any of the intermediate nodes. Similarly you can insert edges between up and any of the intermediate nodes. While inserting edges you should keep the degree of u1 and up exactly the same. You can not insert an edge between the two terminal nodes otherwise a Hamiltonian Cycle will immediately be formed (we already have a Hamiltonian Path between the two terminal vertices). While inserting an edge from u1 to an intermediate vertex (and then an edge from up to an intermediate vertex) try your best that a Hamiltonian Cycle is not formed. The problem is to find out (by an actual trial and error drawing) the minimum degree of the two terminal vertices when it will be impossible to resist a Hamiltonian Cycle. That will be a sufficient condition for this graph to be Hamiltonian. If, instead of resisting a Hamiltonian Cycle, you are adamant to form one as soon as the first opportunity arises then what will be the minimum degree of the two terminal vertices which will guarantee a Hamiltonian cycle? Now come back to the previous problem where we try to resist a Hamiltonian Cycle as far as possible. But let us relax the condition that the degree of the two terminal vertices should be the same. Now the degree of one terminal vertex will be larger and the other smaller. Again find what will be the minimum sum of the two degrees (of the terminal vertices) when it will be impossible to resist a Hamiltonian Cycle. Again this will be a sufficient condition for the modified graph to be Hamiltonian. You may like to solve the puzzle for the graph shown in Fig. 8.3.2

8.3.2

Actually Finding a Hamiltonian Cycle in the Puzzle:

Let us look at the same problem from an algorithmic point of view: You are given a graph G which contains a Hamiltonian path between two non adjacent vertices u and v. Assume that the sum of the degrees of u and v is equal to or larger than p. Also assume that the Hamiltonian Path between u and v is already provided as an input. You are supposed to design an efficient algorithm which will output the actual Hamiltonian Cycle in G.

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Figure 8.3.1: A line graph with six nodes. We should add extra edges in this graph but do our best to avoid a Hamiltonian Cycle.

Algorithm 60: Find a Hamiltonian Cycle in a graph G where deg(u)+ deg(v) ≥ p for a pair of non adjacent vertices u & v in G input : A Graph G; A Hamiltonian Path between vertex u and v. output: A Hamiltonian Cycle in G 1

Use the knowledge and expertise that you have gained while solving the puzzle;

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Figure 8.3.3: We are given three graphs where a Hamiltonian path exists between two special vertices. The Hamiltonian path between the two vertices is also indicated by shaded lines in each graph. In the top graph the degree sum of the two special vertices is equal to the number of nodes in the graph; in the bottom graphs the degree sum of the two special vertices is less than the number of nodes in the graph. Please note that a Hamiltonian Cycle exists in each of these graphs.

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Apply your algorithm on the graphs given in Fig. 8.3.3 in order to find a Hamiltonian Cycle in each graph; please note that each of the graphs does indeed contain a Hamiltonian Cycle (which can easily be found by hit and trial method). Let us see where and why your algorithm fails and where it does find a Hamiltonian Cycle. Is this not strange that a graph contains a Hamiltonian Cycle but your algorithm can not find it? Let us now describe the details of the algorithm which solves the problem outlined above. The working of the algorithm is illustrated in Fig. 8.3.4. Algorithm 61: Find a Hamiltonian Cycle in an un-directed graph G where deg(u) + deg(v) ≥ p for a pair of non adjacent vertices u & v in G; a Hamiltonian path between u and v is also given. input : A Hamiltonian Path between vertices u and v where deg(u) + deg(v) ≥ p in an un-directed graph G. output: A Hamiltonian Cycle in G 1

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Let us index the vertices of the Hamiltonian path like 1, 2, 3, . . . p. The index of u will be 1 while that of v will be p; Find a vertex x such that vertex 1 is adjacent to vertex x + 1 while vertex p is adjacent with vertex x in G; The Hamiltonian Cycle will be 1, 2, . . . x, p, p − 1, p − 2, . . . x + 1, 1;

It is interesting to note that our algorithm also has a serious short coming: a graph may have a Hamiltonian Cycle but we can not find it (as shown in Fig. 8.3.5). Interestingly there may be graphs where the condition (that deg(u) + deg(v) ≥ p for a pair of non adjacent vertices u & v in G) is not met yet our algorithm will be able to find a Hamiltonian cycle as shown in the same figure.

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Figure 8.3.4: Intermediate stages of how Algorithm 61 works to find a Hamiltonian Cycle in a graph G where the condition (that deg(u) + deg(v) ≥ p for a pair of non adjacent vertices u & v in G) is met and where there is a Hamiltonian path between vertices u and v in G.

Hamiltonian Graphs

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Degree = 2

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Figure 8.3.5: A graph G shown (left diagram) where the condition (that deg(u) + deg(v) ≥ p for a pair of non adjacent vertices u & v in G) is not met and the Algorithm 61 will NOT be able to find a Hamiltonian cycle; but a Hamiltonian Cycle exists in this graph as shown by blue lines. A graph G shown where the condition (that deg(u) + deg(v) ≥ p for a pair of non adjacent vertices u & v in G) is not met yet the Algorithm 61 will be able to find a Hamiltonian cycle as shown in the right diagram.

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8.3.3

Hamiltonian Graphs

Basic Intuition

Let us carefully look at the ramifications of Algorithm 61. It provides a couple of very powerful theoretical results and useful algorithmic tools which can be used to find a Hamiltonian cycle in a certain category of graphs. Assume that a graph G has two vertices where the degree sum of these two vertices is more than p as shown in the figure below. We insert an edge between these two vertices and assume that in the resulting graph there is a Hamiltonian Cycle as shown in the top left diagram of Fig. 8.3.6. This essentially means that there will be a Hamiltonian Path starting from vertex 1 and terminating at vertex p in the original graph G. Then, we claim, that there will be a Hamiltonian Cycle in graph G (without that extra edge) as shown in the right diagrams of the Fig. 8.3.6. Now assume that after inserting that extra edge between the two vertices the resulting graph does not have a Hamiltonian Cycle. Then, we claim, that there will not be a Hamiltonian Cycle in the graph G (without that extra edge) as shown in the left diagrams of the same figure. Consider a graph G where the degree sum of any pair of vertices in G is equal to or larger than p as shown in Fig. 8.3.7. Now visualize a hypothetical Hamiltonian Cycle in this graph starting from 1, passing through 2, 3, 4, . . . p and then going back to 1. If this hypothetical Hamiltonian Cycle does not pass through original graph edges then insert Extra edges as shown in (red) in this figure to create a cycle consisting of some original and some Extra edges. You know that we can always delete Extra edges one by one (using the techniques that we have recently acquired) forcing the Hamiltonian Cycle to divert through only graph edges. This intuition provides a powerful technique which can be used in most of Hamiltonian finding algorithms in Hamiltonian graphs.

8.3.4

Bondy & Chvatal’s Theorem:

Now let us come back to a text book theorem which says that a graph G is Hamiltonian if and only if G + uv is Hamiltonian provided vertices u and v are non adjacent and deg(u) + deg(v) ≥ p. Before attempting to prove this theorem let us look at the problem more closely and try to relate it to the previous knowledge that we have acquired. G+uv is Hamiltonian means that there is a Hamiltonian Cycle in the graph G+uv. There are two possibilities:

Hamiltonian Graphs

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We insert a direct edge from u1 to up Assume that now a HAM Cycle does Not exist in the resulting graph

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Now we remove the direct edge from u1 to up A HAM Cycle will NOT exist in graph G?

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Now we remove the direct edge from u1 to up A HAM Cycle will still exist in graph G?

Figure 8.3.6: We make a number of claims in this diagram: A graph G is Hamiltonian if and only if the graph G plus an edge between two non adjacent vertices of G is Hamiltonian provided certain conditions are met. u1

up

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Figure 8.3.7: We make a number of claims in this diagram: A graph G is Hamiltonian if and only if the graph G plus an edge between two non adjacent vertices of G is Hamiltonian provided certain conditions are met.

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Hamiltonian Graphs

1. The Hamiltonian Cycle passes through the edge uv. Then there will certainly be a Hamiltonian Path between the vertex u and vertex v. Under such conditions we can map this problem to the last problem that we have discussed except that the actual Hamiltonian Path between u and v is not provided here. Such a possibility is indicated in the left diagram of Fig. 8.3.8. Again notice the figure of eight in this diagram. 2. The Hamiltonian Cycle does not pass through the edge uv. Under such conditions a Hamiltonian Path between u and v may not exist. It will be instructive to actually draw such a graph. As the edge uv is not part of the Hamiltonian Cycle so it is possible to remove this edge and still a Hamiltonian Cycle will exist in G. Such a possibility is shown in the right diagram of Fig. 8.3.8. How about if we are supposed to design an algorithm to find a Hamiltonian Cycle in the graph G provided the above stated conditions are true. Again it will be interesting to relate this algorithmic problem with the last such problem that we have discussed.

8.3.5

Summary

Let us look at what we have really understood so far. If there is a graph G in which there are two adjacent nodes u and v such that deg(u) + deg(v) ≥ p then G is Hamiltonian if and only if G + uv is Hamiltonian. This means the possibility, that the graph G is not Hamiltonian while the graph G + uv is Hamiltonian, does not exist. This also means that if we remove the edge uv even then there will be a Hamiltonian Cycle in G provided there was a Hamiltonian Cycle in G + uv. The edge uv does not make any difference; its inclusion does not convert a non Hamiltonian graph into Hamiltonian and its removal does not convert a Hamiltonian graph into a non Hamiltonian graph provided other conditions are also met.

8.3.6

Closure of a Graph:

A closure c(G) of a graph G of order p is a graph obtained from G by recursively joining pairs of non adjacent vertices u and v whose degree sum is more than or equal to p until no such pair remains in G. It can now be proved that a graph G is Hamiltonian if and only if its closure c(G) is Hamiltonian.

Hamiltonian Graphs u5

439 u4

u6

u5

u3

u1

u2

u4

u6

u3

u1

u2

Figure 8.3.8: A graph containing a Hamiltonian Cycle which passes through the edge between vertex u1 and u2 is shown in the left diagram. The right diagram shows a graph where the Hamiltonian Cycle does not pass through the edge between vertex u1 and u2 . You can verify that a Hamiltonian path does not exist between these two vertices in the right diagram.

8.3.7

Ore’s Theorem:

The usefulness of the above result becomes obvious when in a graph G every pair of non adjacent vertices satisfies the above mentioned condition, that is, degu + v ≥ p, then the closure c(G) will be obviously be a completely connected graph. We have already stated that complete connectedness is a very loose sufficient condition for a graph to be Hamiltonian. We have also stated that for a completely connected graph, finding a Hamiltonian Cycle is a trivial problem. But remember the Hamiltonian Cycle that we have found belonged to the closure of G; the same Hamiltonian Cycle may not exist in the actual graph G? This is because of the fact that c(G) consists of two types of edges: 1. Edges which were originally present in G, that means E(G). 2. Extra edges inserted between very pair of non adjacent vertices u and v where deg(u) + (v) ≥ p. These edges were not part of G. We call these edges extra edges. The Hamiltonian Cycle in c(G) will pass through some of the original edges of the graph. But it may also pass through some of the extra edges belonging to the second type. How to find a Hamiltonian Cycle which passes through E(G) alone is an interesting problem. The following algorithm can be used to find a Hamiltonian Cycle in graph G. It can also be used constructively

440

Hamiltonian Graphs

to prove that if deg(u) + deg(u)v ≥ p for every non adjacent pair of vertices in a graph G then G is Hamiltonian. This is certainly a tighter sufficient condition for a graph G to be Hamiltonian as compared to the complete connectedness of a graph. This sufficient condition was originally discovered by Ore. Algorithm 62: Find a Hamiltonian Cycle in a graph G where deg(u)+ deg(v) ≥ p for very pair of non adjacent vertices u & v in G. input : An un-directed Graph G where deg(u) + deg(v) ≥ p for very pair of non adjacent vertices u & v in G. output: A Hamiltonian Cycle in G 1

2 3

4

We find a closure of graph G known as c(G) - this closure is our starting graph H. We find a Hamiltonian Cycle in graph H. This cycle may pass through some of the graph G original edges and may also pass through some of the edges not present in G (but are present in H) known as extra edges; while there are extra edges in the Hamiltonian Cycle in graph H do Remove an extra edge uv from the cycle and from graph H thereby creating a Hamiltonian Path between two vertices u and v in the new graph H; We know that deg(u) + deg(v) ≥ p, thus if there is a Hamiltonian Path between u and v in H then we can find a new Hamiltonian cycle in H which does not pass through the edge uv (use Algorithm 61);

Problem Set 8.2. Problem 8.2.1. Draw three different graphs; in the first graph G the degree sum of each pair of non adjacent vertices should be at least p. In the second graph H the degree sum of each pair of non adjacent vertices is not p but the closure of H is a completely connected graph. In the third graph J, the degree sum of each pair of non adjacent vertices is not equal to p. Discuss briefly where it is possible to use Algorithm 62 without any modification to find a Hamiltonian Cycle. Also pinpoint the graph where it may not be easy to find a Hamiltonian Cycle using the knowledge that we have acquired until now, and last but not the least discuss where Algorithm 62 have to be modified in order to find a Hamiltonian Cycle.

Hamiltonian Graphs

441

Dirac’s Theorem: A graph G where the degree of each node is more than or equal to p/2 is Hamiltonian. A formal proof of this theorem can easily be derived from the above discussion. A constructive proof can also be designed on similar lines. Example 1: We show a graph G where the degree of each node is equal to p/2 in the left diagram of Fig. 8.3.9. The closure of this graph is a completely connected graph as shown in the right diagram of the same figure. The Hamiltonian Cycle is also indicated in this diagram; interestingly the Hamiltonian Cycle passes entirely through extra edges. We apply Algorithm 62 to this graph and show a number of intermediate results in Fig. 8.3.10. We remove extra edges one by one and each time find a new Hamiltonian Cycle; the extra edges removed are indicated by dotted lines while a Hamiltonian Cycle is shown by bold lines in Fig. 8.3.10. Please note that the order in which extra edges are removed is not very important in this example. Why? The order will, however, become important in the next example. u5

u4

u6

u5

u3

u1

u2

u4

u6

u3

u1

u2

Figure 8.3.9: A graph G where the degree sum of each pair of non adjacent vertices is equal to p; the graph is shown in the left diagram. The closure of this graph is shown in the right diagram. Note that the closure is a completely connected graph containing a Hamiltonian Cycle also shown in the right diagram. Note that the Hamiltonian Cycles passes entirely through extra edges. Example 2: We show a graph G in the top left diagram of Fig. 8.3.12; the degree of some of the nodes is also indicated in this diagram; the degree sum of a pair of non adjacent vertices is less than 6 in this graph. The closure of this graph is, however, a completely connected graph as shown in the top right diagram. The closure of this graph thus contains a Hamiltonian

442

Hamiltonian Graphs Remove One of the Extra edges shown in orange

u5

Find Hamiltonian cycle (blue color) in the remaining graph

u5

u4

u6

u6

u3

u3

u1

u2

u2

w

th e

gr

ap

h

u1

u4

u1

ra


R

ed

Remove One more Extra edge

u4

u6

u1

u3

u6

u3

u2

u5

u2

w

th e

gr

ap

h

u5

u4

u4

ra


R

ed


u1

u6

u4

u3

u5

u6

u2

h ap gr th e ed

ra

u3

R

u4

u5

u2

u2


w

u1

u6

u3

u5


u3

u1

u1

u6

u4

u5

u2

Figure 8.3.10: We start with the closure of the graph shown in the last figure. We apply Algorithm 62 to this graph; remove an extra edge and find a new Hamiltonian Cycle. We repeat this process until all extra edges are removed. All extra edges which are removed are shown by dotted lines.

Hamiltonian Graphs

u5

443

u4

u5

u6

u3

u1

u6

u2

u4

u3

u5

u2

u6

u3

u4

u5

u2

Redraw the graph and remove the last Extra Edge

u1

u1

u2

Hamiltonian Cycle shown in blue. Another Extra Edge can be removed

u1

u2

u3

u5

u5

u4

u5

u4

u6

u2

u6

u2

Redraw the graph and remove another Extra Edge

Redraw the graph and remove another Extra Edge

u1

u3

u5

u3

u5

Hamiltonian Cycle shown in blue. Another Extra Edge can NOT be removed

u3

u2

u6

u4

u6

Hamiltonian Cycle shown in blue. Another Extra Edge can be removed

u1

u6

u1

u3

u1

Extra edges are shown in red; We remove one Extra Edge

u4

u4

u4

u6

u3

u1

u2

Hamiltonian Cycle, shown in blue, passes through No Extra Edge

Figure 8.3.11: It is possible to remove more than one extra edge (in a single step) from the graph. Initially the Hamiltonian Cycle passes through all the extra edges as shown in the top left diagram. At the end it passes through the graph edges as shown in the bottom right diagram.

444

Hamiltonian Graphs

4 u5

u4 2

u5

u4

u5

u4

3

u6

u3

u6

u2 4

u1

u1

We start with a graph where degree sum of a few pairs of vertices is not p

u5

u5

u1

u3

u1

u4

u5

Redraw the graph and try to remove the last Extra Edge through which the Hamiltonian Cycle passes NOT Possible?

u1

u3

u2

Delete an Extra Edge

u5

u2

u4

u6

u2

u4

u3

u2

Add Extra Edges where the degree sum condition is recently satisfied. The closure is complete

Find a Hamiltonian Cycle in the remaining graph as shown in blue color

?

u3

u1

u1

u3

The Hamiltonian Cycle does not pass through any Extra Edge

u6

u2

u6

u2

u5

u6

Add Extra Edges where the degree sum condition is satisfied. This will increase the degree?

u4

u6

u3

u4

u6

u1

u3

u2

Find a different Hamiltonian Cycle in the remaining graph as shown in blue color. It passes through one Extra Edge; remove other Extra Edges

Figure 8.3.12: While removing Extra edges, one should be very careful about the order in which they are removed otherwise one may fall into a trap as shown in the bottom diagrams.

Hamiltonian Graphs

445

A given graph where degree sum condition for every pair of vertices is satisfied

We add Extra Edges so that a Hamiltonian Cycle is formed; the closure is not yet complete

1

1

6

2

5

3

6

2

5

3

4

4

1 2

6

It is possible to add just one Extra Edge so as to get a Hamiltonian Cycle

5

3 4

Figure 8.3.13: There is no need to make the closure of a graph complete; we can add just enough extra edges such that a Hamiltonian Cycle becomes possible in the graph.

446

Hamiltonian Graphs

Cycle. If we now remove the extra edges in the wrong order then we may end up with a difficult situation as depicted in the bottom right diagram of Fig. 8.3.12. Why this order has suddenly become important and why it was not important in the graph of Fig. 8.3.9? It is essential for you to answer these questions before moving forward. 4

u5

u4

u5

2

u4 4 4

3 u6

u3

u1

u5

u2

4

u4

u6

u3

u1

u6

u2

u3

u1

u2

u5

u4

u6

u3

u1

u2

Figure 8.3.14: The top left diagram shows a graph in which the degree sum of every pair of non adjacent vertices is not equal to (or larger than) the number of nodes in the graph. The closure of this graph is still a completely connected graph as shown in the bottom left diagram. The bottom right diagram shows that a difficult situation arises if we remove extra edges in the wrong order.

8.4

Bipartite Hamiltonian Graphs

We are now in a position to design or discover similar sufficient conditions for a bipartite graph to be Hamiltonian (It will also be exciting to discover some of the necessary conditions for a bipartite graph to be Hamiltonian). Let us solve puzzle No. 8.3.1 once again. Be careful this time; while inserting edges you should keep in mind that in a bipartite graph, an edge uv exists provided


447

vertex u and vertex v belong to different partites. You should also keep the degree of u1 and up exactly the same while inserting new edges. You can not insert an edge between the two terminal nodes otherwise a Hamiltonian Cycle will immediately be formed (we already have a Hamiltonian Path between the two terminal vertices). While inserting an edge from u1 to an intermediate vertex (and then an edge from up to again an intermediate vertex) try your best that a Hamiltonian Cycle is not formed. The problem is to find out (by an actual trial and error drawing) the minimum degree of the two terminal vertices when it will be impossible to resist a Hamiltonian Cycle. That will be a sufficient condition for this graph to be Hamiltonian. Please see Fig. 8.4.1, Fig. 8.4.2 and Fig. 8.4.3 u1

u2

u3

u4

u5

u6

u7

u8

Figure 8.4.1: We show a bipartite graph consisting of eight vertices containing a Hamiltonian Path between two end vertices. If we insert an edge between the two end vertices then a Hamiltonian Cycle will immediately be formed.

8.5


We shall describe a number of special Hamiltonian graphs and then present a number of graph theoretical claims. Hamiltonian Graphs: p-closure is complete in G: A Hamiltonian Cycle exists in G. Hamiltonian Connected: (p+1)-Closure is complete: A Hamiltonian Path passes through every pair of vertices. It is always Moorish Connected.

448

Hamiltonian Graphs

up

u1

u7

u2

u6

u3 u4

u5

up

u1

u7

u2

u6

u3 u4

u5

Figure 8.4.2: Here we show a Hamiltonian Path between two vertices in the bipartite graph shown in Fig. 8.4.1. Notice the bi-colored vertices; similar color vertices belong to one partite while all the other color vertices are in the second partite. In a bipartite graph, an edge uv exists provided u and v belong to different partite. Edges not allowed in a bipartite graph are indicated in both the diagrams in the form of dotted lines.


u

u

449

v

v

u

v

u

v

Figure 8.4.3: A bipartite graph of sizes 8, 10, and 12. You can easily think of sufficient condition for a bipartite graph to be Hamiltonian.

450

Hamiltonian Graphs

Moorish Connected: A Hamiltonian Path passes through every pair of adjacent vertices. It may or may not be Hamiltonian Connected. Claim No. 1: Given a graph G (having p vertices), we add a vertex x to G and connect it to vertex u and vertex v of G, the new graph is known as graph H as shown in the figure below. Now if p + 1 closure of H (having p + 1 vertices) is complete then there will be a HAM path in G with end points u and v in G. If vertex u and v are adjacent in graph G then there will be a HAM cycle in G passing through u and then v.

k

u

k

G k

Degree of every vertex larger than or equal to k

k+1

k

2

k

x

k

k+1

k

k

v

u

k

k

If p+1 closure is complete in graph G+x then there will be a HAM cycle in the graph G+x.

G k

v

k

k

Then there will be a HAM cycle passing through vertex u & v provided u and v are adjacent in G

Figure 8.5.1: We show a graph G; we add and connect a vertex x to vertices u and v of G as shown in the top diagrams. If the closure of G+x is complete then there will be a HAM cycle in G + x and a HAM path in G between vertex u and vertex v of G.


451

Example No. 1: We show a graph G in Fig. 8.5.2 and we need to check if our newly acquired knowledge can really confirm if there is a HAM path between two vertices 4 & 5 in graph G. We know before hand that a HAM path exists between these vertices but we also know that there is a possibility that a HAM path exists in a graph but we are unable to confirm or find it. We are lucky in this example as we do find a HAM path between the two given vertices. It is somewhat surprising to note that a p closure of graph G is not complete yet a p + 1 closure of graph G + x is complete. Claim No. 2: Assume that in a graph G, the p + 1 closure is complete then there will be a HAM path between every pair u, v of G. It follows that such a graph G is Hamiltonian, in fact it is more than that: every edge u, v will be part of some HAM cycle. Such a graph G (where a HAM path exists between every pair of vertices in G) is known as Hamiltonian Connected. 3

1

4

2

x 2

2

1

2

x

5

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3

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4

2

x 2

5

4

3

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2

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4

4

1

2

5 5

3

3

3

2

Connect x to 4 and 5. Degree sum of 4 and 5 becomes 6

2

1

3

2

Connect 4 & 5. Degree sum of other pairs becomes 6

1

4

5

4

5

2

2

5

4

2

Connect vertices where degree sum is equal to 6

1 5

2 5

5 5

5

3

2

There is a HAM path between vertex 4 and 5

x

5

4

4

2

x 2

4

5 5

4

3 All vertices in graph G+x are now completely connected

3 All vertices in G are completely connected

Figure 8.5.2: We show a graph G; we add and connect a vertex x to vertices 4 and 5 of G as shown in the top diagram. The closure of G + x is complete and there will be a HAM cycle in G + x and a HAM path in G between vertex 4 and 5 of G.

452

Hamiltonian Graphs

Claim No. 3: Given a graph G (having p vertices), we add a new vertex x to G and connect it to every vertex of G, the new graph is known as the graph H as shown in Fig. 8.5.3. Now if p + 1 closure of H (having p + 1 vertices) is complete then there will be a HAM path in G.

k+1

k

k+1

k

G

k

p

k


k+1

x k

k+1 k+1 AH

AM

pat h in

G

If p+1 closure of G+x is complete then there will be a HAM cycle in G+x

Figure 8.5.3: We show a graph G; we add and connect a vertex x to every vertex of G as shown in the top diagrams. If the closure of G + x is complete then there will be a HAM path in G. Example No. 2: We show a chain graph G in Fig. 8.5.4 with end vertices 3 and 5. We know that a HAM path exists between vertex 3 and 5 in this graph. We add a vertex x to G and connect it to vertex 3 and 5 as shown in the top left diagram. The p + 1 closure of graph G + x (having p + 1 vertices) is not complete; thus we can not confirm that a HAM path exists between vertex 3 and 5 in G. Then we connect vertex x to every vertex in G; now the p + 1 closure of G + x becomes complete as shown in the rest of the diagrams; thus confirming that a HAM path do exist in this graph. Claim No. 4: Assume that in a graph G, the p − 1 closure is complete then there will be a HAM path in G. Claim No. 5: Given a graph G (having p vertices) with three vertices u, v, and w such that u is adjacent to v and v is adjacent to w. We add a vertex x and a vertex y to G such that vertex x is connected to vertices u and v of G,


1

2

2

2

4

5

2

2

453

1

3

3

2

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3

x

1

3

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4

2

2

x

2

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3

3

3 We add vertex x to G but now connect x it to every vertex of graph G

A HAM path exists between 3 and 5 but the 6 closure of this graph is not complete

1

5

5

5

4

5

5

x

We connect 2 and 4 as the degree sum is 6

1

3

5

3

4

5

5

2

x

2

5

3

3

3

The closure is a completely connected graph.

x

Now we can connect 2 and 5 and 3 & 4 as the degree sum is 6

Figure 8.5.4: The p + 1 closure of the graph G + x (having p + 1 vertices) is not complete as shown in the top left diagram. The p + 1 closure of graph G + x is complete and thus a HAM path exists in the graph G as shown in the rest of the diagrams.

454

Hamiltonian Graphs

and vertex y is connected to vertex v and w of G, the new graph G + x + y is known as graph H as shown in the figure below. Now if p + 2 closure of H (having p+2 vertices) is complete then there will be a HAM cycle in G which will be passing from vertex u to v and then to vertex w (see Fig. 8.5.5).

u 2

k k

G

k

x k+2

k


v

k

y k

k

k+1

2

k k+1

k

If p+2 closure is complete in G+x+y then there will be a HAM cycle passing from u to x to v to y and then to w, and so on.

u k

k

v

k

G k k

w

w

There will be a HAM cycle passing through u then v and then w in G provided u is adjacent to v and v is adjacent to w.

Figure 8.5.5: If the p + 2 closure of graph G + x + y (having p + 2 vertices) is complete then there will be a HAM cycle in G passing from vertex u to vertex v and then to vertex w of G. Claim No. 6: Assume that in a graph G, the p + 2 closure is complete then some HAM cycle will pass through every pair of adjacent edges {u, v} and {v, w} in G. Claim No. 7: In a regular graph G where the degree of every vertex is p/2, a HAM path exists between vertex u and vertex v of G if and only if p + 1


455

closure of G + x is complete (please see Fig. 8.5.6 in this regard). Claim No. 8: In a regular graph G where the degree of each vertex is p/2, every edge {u, v} will be part of a HAM cycle. Claim No. 9: A regular graph G, where the degree of each vertex is p/2, is Hamiltonian Connected (that means there is a HAM Path between every pair of vertices in G) if and only if p + 1 closure of G + x is complete for every pair u and v of G (please see Fig. 8.5.6).

u

p/2

p/2

u

p/2

p/2

G Degree of every vertex is exactly p/2

v

p/2

2

p/2

x

p/2

p/2

v

p/2

There will be a HAM path passing through vertex u & v in G

G

p/2

If and only if

p+1 closure is complete in graph G+x

Figure 8.5.6: We show a regular graph G; the degree of every vertex is p/2. We add a vertex x to G and connect x to vertex u and v in G as shown in the right diagram. Problem Set 8.3. Problem 8.3.1. How can you modify the algorithm (which was earlier used to find a Hamiltonian Cycle in a Hamiltonian graph) which makes sure that the resulting Hamiltonian Cycle passes through a given edge. Problem 8.3.2. Look at the graphs in Fig. 8.5.7. You can use our past experience of finding a Hamiltonian cycle in these graphs. If some of the graphs do not satisfy sufficient conditions for Hamiltonian graphs then there is a possibility that a Hamiltonian Cycle exists but our expertise is not able to find it.

456

Hamiltonian Graphs

1. For every graph in Fig. 8.5.7, check if there is a Hamiltonian cycle and also check if our expertise can find it or can not find it. 2. Design an efficient algorithm to find a Hamiltonian path in these graphs between any two vertices. 3. Design an efficient algorithm to find a Hamiltonian path in these graphs between two given (not any two but two special) vertices. Problem 8.3.3. Draw a graph which is Hamiltonian (that means it satisfies the sufficient conditions) but in which there is not a Hamiltonian path between every pair of non adjacent vertices. We call it category A graphs. Problem 8.3.4. Draw a graph which is Hamiltonian (that means it satisfies the sufficient conditions) but in which there is a Hamiltonian path between every pair of non adjacent vertices. We call it category B graphs. Problem 8.3.5. Draw a graph which is Hamiltonian (that means it satisfies the sufficient conditions) but in which there is a Hamiltonian path between every pair of vertices. We call it category C graphs. Problem 8.3.6. Draw a graph which is not Hamiltonian (that means it does not satisfy the sufficient conditions) but in which there is a Hamiltonian path between every pair of non adjacent vertices. We call it category D graphs. Problem 8.3.7. Draw a graph which is not Hamiltonian (that means it does not satisfy the sufficient conditions) but in which there is a Hamiltonian path between every pair of vertices. We call it category E graphs.

8.6


Category A: There is not a Hamiltonian path between every pair of non adjacent vertices in G and G is Hamiltonian (that means it satisfies the sufficient conditions). Category B: There is a Hamiltonian path between every pair of non adjacent vertices in G and G is Hamiltonian (that means it satisfies the sufficient conditions) and it is not Hamiltonian Connected (B1). Also consider the option when it is Hamiltonian Connected (B2).


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Figure 8.5.7: You may find some graphs here which you were supposed to draw in earlier problems.

458

Hamiltonian Graphs

u5

u4

u6

u3

u1

u2

Figure 8.5.8: A panoramic hierarchy of graphs; some containing a Hamiltonian Path, some containing a Hamiltonian Cycle, etc.


459

Category C: G is Hamiltonian (that means it satisfies the sufficient conditions) & Hamiltonian Connected (C1). Also consider the option when G is not Hamiltonian Connected (C2). Category D: There is a Hamiltonian path between every pair of non adjacent vertices in G and G is not Hamiltonian (that means it does not satisfy the sufficient conditions). Category E: G is Hamiltonian Connected but is not Hamiltonian. Problem 8.3.8. Try to draw a graph G where G2 is not Hamiltonian while G3 is Hamiltonian. Problem 8.3.9. Try to draw a graph G where G2 is not Hamiltonian while G3 is Hamiltonian Connected. Problem 8.3.10. Try to draw a graph G where G2 is not Hamiltonian while G3 is not Hamiltonian Connected. Problem 8.3.11. Try to draw a connected graph G where G3 is not Hamiltonian Connected. Problem 8.3.12. Prove that G3 of a graph G is always Hamiltonian. Problem 8.3.13. A graph G where the minimum degree is k is given as shown below. We need to determine conditions under which there will be a Hamiltonian Path in G (see the top diagrams in Fig. 8.6.1). Problem 8.3.14. We need to determine k such that a Hamiltonian Path (or a cycle) passes through every two vertices u and v in G (see the bottom left diagram in Fig. 8.6.1). Problem 8.3.15. We need to determine k such that a Hamiltonian Path (or a cycle) passes through three vertices u, v and w in G (see the bottom right diagram in Fig. 8.6.1). Problem 8.3.16. Assume that we are given a graph G with number of vertices equal to p where p is even, and the degree of every vertex is larger than or equal to p/2. Initially you can assume that G is a regular graph and thus the degree of every vertex is the same. You can relax this condition later and cater to the condition when the degree of every vertex may be larger than p/2. We need to find (that is design an efficient algorithm to solve this

460

Hamiltonian Graphs

k

k+1

k

k+1

k

p

G

k+1

x

k k

k+1

u

k+1

k

x

u

k+1

k+1

k

2 2

v

k

x

k+1

2 k+1

v

k

y

k+1

w

Figure 8.6.1: Figures for the last problem set

k


461

problem) if there is a Hamiltonian path between two given vertices. If there is really a Hamiltonian path then our algorithm should be able to output this path (we should provide a proof for this). Consider two options separately: (a) when the two given vertices are adjacent, and (b) when they are not adjacent. The following graphs in Fig. 8.6.2 may provide some food for thought.

462

Hamiltonian Graphs

Figure 8.6.2: We show here a number of regular graphs where the degree of each vertex is exactly p/2 and p is even.

Chapter 9 Strongly Connected Directed Graphs and Tournaments 9.1

Concepts, Properties & Actions

9.2

Strongly Connected Directed Graphs

9.3

Directed Acyclic Graphs (DAG’s)

9.4

Strongly Connected Components

9.5

Tournaments

9.6

Unilaterally Connected Directed Graphs:

464

Strongly Connected Directed Graphs and Tournaments

Introduction We shall talk about some specific features of directed graphs in this chapter. As with un-directed graphs, directed graphs can also be classified into different categories. We shall be talking about strongly connected directed graphs, directed acyclic graphs, unilaterally connected directed graphs and tournaments. An undirected graph is known as strongly orient-able if it has the potential of becoming a strongly connected directed graph. We shall discuss and prove (again using constructive proof strategies) necessary and sufficient conditions for an undirected graph to be strongly orient-able. We shall also describe efficient algorithms to find if a directed graph is strongly connected, and if not then what are its strongly connected components, and how they are connected.

9.1


We have earlier classified an un-directed graph on the basis of connectedness. We had an un-directed graph which was not connected, and then we had a just barely connected graph known as a tree. We have also discussed a connected graph which was not a tree that means a graph which was cyclic. On the other extreme we have seen the completely connected graph. We have also talked about certain actions on graphs like the square of a graph or the complement of a graph. These actions on a graph G transform G into another graph H. Graphs are also judged by different properties that they possess. For example a line graph which is acyclic and is not Hamiltonian. A completely connected graph is always Hamiltonian. When we make a transformation on a graph we create a graph with new properties. We have seen in the last chapter that if we take the square of a line graph or a star graph then the resulting graph is Hamiltonian. Some of the properties of graphs, some action items (or so called transformations), and a classification of un-directed and directed graphs are shown in a concept map (see Concept map 1). This is how we shall study directed graphs in this chapter. We shall start with different categories (shown as concepts in Concept map 1) of directed graphs. Each category of directed graph possesses some specific properties. It will be interesting to examine how different properties of a graph change (or do not change) if we transform a graph into a new directed graph using


465

Concept Map 9.1. A concept map showing some of the properties of graphs, some action items or transformations, and a classification of undirected and directed graphs. Please note that the action items are represented by elliptical objects while the properties are shown by square boxes.

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some of the action items shown in Concept map 1.

Concepts As you can see in Concept map 1, directed graphs can also be classified on the basis of connectedness. A directed graph where you can reach any vertex from any other vertex is known as a strongly connected graph. A directed graph having no cycles is called an directed acyclic graph or a DAG. If we remove directions from a directed graph D then the resulting graph will become undirected and is known as the underlying undirected graph. If the underlying undirected graph of D is connected then D is known as a weekly connected directed graph. A directed graph D is unilaterally connected provided for every pair of vertices u and v, there is a path from u to v or a path from v to u. In a strongly connected directed graph there is a path from u to v and a path from v to u. That means a strongly connected graph is always a unilaterally connected graph but not vise versa. Both strongly connected as well as unilaterally connected directed graphs are always weakly connected but again it may not be true the other way round. Fig. 9.1.1 shows three different directed graphs. One of them is strongly connected. The other is unilaterally connected but not strongly connected. The third is neither unilaterally connected nor strongly connected. One of these graphs is acyclic while the rest are cyclic. Please try to pinpoint which one is which. When we put directions in an un-directed (or bi-directed) graph G then it becomes a directed graph D. Not all un-directed graphs can be converted into a strongly connected directed graph. Those which can be converted into a strongly connected graph are known as strongly orient-able un-directed graphs. Un-directed graphs which can be converted into a unilaterally connected directed graph are known as unilaterally orient-able undirected graphs. Completely connected un-directed graphs are both unilaterally orient-able and strongly orient-able. This means that if we intelligently put directions in a completely connected un-directed graph then the resulting directed graph will be strongly connected (or unilaterally connected). It is also possible to put directions in a completely connected graph in such a manner that the resulting directed graph is acyclic, and is therefore not strongly connected. A directed graph formed by putting arbitrary directions in a completely connected undirected graph has a special name that is a tournament. It is obvious that a tournament directed graph may be acyclic


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Figure 9.1.1: Three different directed graphs. One of them is strongly connected. or may contain a cycle, in fact it may contain a Hamiltonian Cycle, as we shall discuss in detail in this chapter. Problem Set 9.1. A graph consisting of six vertices is shown in Fig. 9.1.2. Only two edges in this graph are directed while the rest are left un-directed or bi-directed. You are supposed to put directions in these undirected edges in order to fulfill some objective as defined below. After putting appropriate directions draw the directed graph D. Problem 9.1.1. The resulting graph should be acyclic and not unilaterally connected. Problem 9.1.2. The resulting graph should be acyclic and unilaterally connected. Problem 9.1.3. The resulting graph should be cyclic and not unilaterally connected. Problem 9.1.4. We have not asked you to draw a graph which should be acyclic and strongly connected. Why?

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Problem 9.1.5. The resulting directed graph should be cyclic and strongly connected. Problem 9.1.6. The resulting directed graph should be cyclic but not strongly connected. Problem 9.1.7. The resulting graph should be unilaterally connected but not strongly connected. Problem 9.1.8. The resulting graph should be unilaterally connected, not strongly connected but cyclic.

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Figure 9.1.2: A graph in which only two edges are directed while the rest are left un-directed. You are supposed to make it a directed graph by putting directions on un-directed edges in order to fulfill certain objectives Problem Set 9.2. Draw a directed graph D in order to fulfill the following objectives: Problem 9.2.1. A directed graph in which every node lies on a directed cycle but the graph is not strongly connected. The directed graph may not be weekly connected. Problem 9.2.2. A directed graph in which every node lies on a directed cycle but the graph is not strongly connected. The directed graph should be weekly connected. Problem 9.2.3. A directed graph in which if there is a path from u to v then there is a path from v to u for every pair of vertices u and v in D, and D is not strong. Will the resulting graph always be unilaterally connected? Discuss briefly.


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Problem 9.2.4. A weekly connected directed graph in which for every pair of vertices u and v, if there is a path from u to v then there is no path from v to u. Will the resulting graph be strongly connected, unilaterally connected, or acyclic? Discuss briefly.

Action Items Let us now discuss some of the action items or so called transformations. The square of a graph D is another graph (known as) D2 in which there is an edge from vertex u to vertex v provided there is a two edge path from u to v or there is an edge (that is a one edge path) from u to v in D. This definition is almost the same as given in the last chapter for un-directed graphs. The Reachable Relation (or the transitive closure) of a directed graph D is another directed graph in which there is an edge from vertex u to vertex v provided v is reachable from u in D. The Reachable Relation graph of D can be represented by an adjacency matrix A in which A(u, v) = 1 provided there is a directed path from u to v in D, and it is zero otherwise. The Reachable Relation matrix of an undirected connected graph will contain all 1’s. Similarly the Reachable Relation matrix of a strongly connected directed graph will also contain all 1’s. The transpose of a directed graph D is another directed graph E in which there is an edge from u to v if and only if there is an edge from v to u in D. We shall introduce more action items as the need arises.

Properties A directed graph D is Hamiltonian provided there is a spanning cycle in the graph. We know that a spanning cycle passes through every vertex exactly once. A Hamiltonian Path or a spanning path spans every vertex only once but is not closed so you can not come back from where you have started. An Eulerian Trail is a closed trail which spans all edges exactly once. An Eulerian Trail may pass through a vertex several times but it should pass through every edge exactly once. A closed spanning walk in a directed graph D passes through every node of D and comes back from where it has started. As we know in a walk a node as well as an edge may be traversed several times. A directed graph D is n − cyclic if it contains a directed cycle consisting of n nodes where n ≤ p. If n = p then the graph D is Hamiltonian. A directed graph D is transitive if and only if there is an edge from u to v provided

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there is an edge from u to any vertex w and an edge from w to v for every pair of vertices u and v in D. Problem Set 9.3. For the directed graph D shown in Fig. answer the following:

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Problem 9.3.1. Find a Reachable Relation Matrix for this graph. Problem 9.3.2. Is this graph strongly connected? Why? Problem 9.3.3. Draw the transpose of the graph. Is the transpose also strongly connected? Why? Problem 9.3.4. Is this graph Hamiltonian? Discuss briefly. Problem 9.3.5. Find a closed spanning walk in this graph. Problem 9.3.6. Draw D2 . Problem 9.3.7. Is D2 Hamiltonian? Discuss briefly. Problem 9.3.8. Is D or D2 transitive? Discuss briefly.

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Let us focus on a strongly connected graph (we consider it as the current concept). Keep in mind a number of properties a directed graph can possess. These properties are: 1. Hamiltonian Cycle 2. Eulerian Trail 3. Closed spanning walk 4. n-cyclic 5. Transitive. Let me now pick one action item, that is, the Reachable Relationship of a directed graph. When a graph is strongly connected then it is possible to reach every vertex from every vertex in that graph. This directly implies that the Reachable Relation matrix of a strongly connected graph will contain all 1’s (see Concept map 9.2). Does it also imply that if the Reachable Relation matrix of a graph contains all 1’s then the graph is strongly connected? I think it is quite obvious that the answer is yes. A directed graph which is Hamiltonian contains a Hamiltonian Cycle; this implies that every vertex is reachable on this cycle, and thus such a graph will be strongly connected. Is it the other way round also, that is every strongly connected graph is Hamiltonian? I think you should be able to find a counter example quite easily. Please do it before moving forward. A strongly connected graph may not be Hamiltonian but it perhaps should always contain a cycle of length n where n < p. Why it can not be an acyclic graph? Because otherwise it will not be possible to reach from v to u if it is possible to reach from u to v. Thus a strongly connected graph should never be acyclic. Again it is not the other way round that is every cyclic graph may not be strongly connected (we have already witnessed this in a problem set). Let us now focus on Eulerian directed graphs. In such graphs it is possible to find a Eulerian trail; a trail passes through every edge exactly once but may pass through a vertex several times. As every vertex is visited at least once and we come back from where we have started thus every Eulerian graph

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will be strongly connected but not vise versa. After looking at Hamiltonian or Eulerian graphs which are quite restrictive in nature, let us look at a relatively more general graph D which contains a closed spanning walk. We know that it is possible to traverse a node or even an edge several times in a walk. We also know that in a strongly connected graph it is possible to move from any vertex to any other vertex; it means that you can always come back to a vertex from where you have started; that will in fact be a closed walk, it may, however, be not a closed spanning walk. Let us call this a closed walk U . As U does not span all vertices thus there will be a vertex u in U which will be connected to a vertex v of D which is not part of U (why?). We can start from u, go to vertex v; it will still be possible to come back to vertex u; This will result in another closed walk W which when merged in U gives us a closed walk with more vertices that in U . We repeat this process until the closed walk becomes a closed spanning walk (see Fig. 9.2.1). Let us summarize of what we have gained so far. A graph D is strongly connected if and only if we can find a closed spanning walk in D or D is strongly connected if and only if the Reachable Relationship of D contains all 1’s. 1

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Figure 9.2.1: A closed walk U is indicated in a directed graph D. Another closed walk W is also shown in the same directed graph. When we merge U and W we shall get a closed spanning walk in the graph D. We are now in a position to design algorithms to solve a number of related problems. Algorithm 63 answers if a given directed graph is strongly connected (or not) in a polynomial time algorithm. It uses the Reachable Relation of a graph as an intermediate or a so called bridging concept. Once we have the Reachable Relation we have some information about the nature of the directed graph.


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Algorithm 63: Check if directed graph D is strongly connected input : Directed graph D output: Graph D is Strongly Connected or not 1

Transform the Directed Graph D into a Reachable Relation graph by applying any traversal algorithm on every vertex of D. If the Reachable Relation matrix contains all 1’s then the graph D is strongly connected otherwise not;

Concept Map 9.2. A map showing the concept of a strongly connected graph, a couple of properties and an action item which is the Reachable Relation. It is interesting to note that the Reachable Relation graph of a Hamiltonian, Eulerian, and strongly connected directed graph is exactly the same.

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If the relation matrix contains all 1’s then the graph D is strongly connected otherwise it is not.

Khawaja Fahd says that we should just check the diagonal boxes in the Reachable Matrix and there is no need to check any other box. What do you think?

It is interesting to note that it is possible to design a much more efficient algorithm to solve the same problem but then we need some more concepts and a deeper insight into the problem. We shall address this problem once we acquire the required knowledge. Algorithm 64: Find a cycle in a directed graph D input : A directed graph D which is not acyclic output: A cycle in D 1 2

Find a closed walk u in D; Convert it into a cycle;

Algorithm 64 finds a cycle in a directed graph D. The given directed graph D may not be strongly connected but it should contain a cycle. If D is strongly connected then it will certainly contain a cycle. In this algorithm first we find a closed walk in graph D using any traversal algorithm. A closed walk can always be converted into a cycle. It will be interesting to derive the time complexity of this algorithm and to make it more efficient if possible. Algorithm 65: Find a closed spanning walk in a strong graph D input : A strongly connected directed graph D output: A closed spanning walk in D 1 2

Find a closed walk u in D; Increase its size until it becomes a closed spanning walk as previously explained in this section;

Note that this algorithm is directly related to the (constructive) proof which proves that a directed graph D is strongly connected if and only if D contains


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a closed spanning walk. The proof was earlier presented informally in this section. It will again be interesting to derive the time complexity of this algorithm and to make it more efficient if possible. Let us discuss strongly connected graphs in terms of one more property (transitive) and one more transformation (transpose). Is a strongly connected graph D transitive? Should it never be transitive? If it is transitive then will it be a special strongly connected graph? If a strongly connected graph D is not transitive then will D2 be transitive? Under what conditions D2 be transitive? We know that in a strongly connected graph D it is possible to reach any vertex from any vertex in D. Now assume that we reverse the direction of each edge in graph D, that is, we have taken the transpose of D. Will it be still possible to reach any vertex from any vertex in this graph? If yes then the transpose of D will always be strongly connected provided D was strongly connected. We next consider a class of un-directed graphs (known as strongly orient-able) where if we put a direction intelligently on each edge then the resulting directed graph will be strongly connected. On the other hand there are un-directed graphs where it is impossible to convert them into strongly connected directed graphs. It is interesting (or shocking) to note that the underlying un-directed graph of a strongly connected graph may not be strongly orient-able (why?).

Strongly orient-able Un-directed Graphs We need to discover necessary and sufficient conditions for an undirected graph G to be transformed into a strongly connected graph. It is obvious that the un-directed G should be connected. It should not be acyclic otherwise it will be impossible to create a directed cyclic graph out of it which is strongly connected. We should remember that the desired directed graph should not only be cyclic; every directed edge should be part of a directed cycle otherwise it will be impossible to ensure reach ability from any vertex to every other vertex. This implies that a graph G should not contain any bridge edge for G to be strongly orient-able. One of the necessary conditions for a graph to be Hamiltonian is that there should be no cut vertex. Is this also a necessary condition for a graph G to be strongly orient-able? The following algorithm converts an un-directed graph with no bridge edges into a strongly connected directed graph by putting appropriate direction on each edge (see Fig. 9.2.1). The algorithm is quite similar to the algorithm

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used to find a closed spanning walk in a strongly connected directed graph. Algorithm 66: Convert an un-directed G into a strongly connected D input : An un-directed graph G output: A strongly connected directed graph D 1 2 3

Find a closed spanning walk u in G; Put appropriate directions; Increase the size of the spanning walk until it becomes a closed spanning walk as previously explained;

The algorithm can also be used as a constructive proof to prove that if there are no bridge edges in an un-directed graph G then it is possible to convert G into a strongly connected directed graph D. We have yet to prove that if there is a bridge edge in a graph G then it is impossible to convert G into a strongly connected graph D. That will in fact complete the proof of a theorem which states that a connected graph G is strongly orient-able if and only if G contains no bridge edges. We have seen in the last chapter that G2 of a line graph is Hamiltonian and is therefore strongly orient-able. We have also seen that G2 of a star graph is also Hamiltonian. It will be worthwhile to conjecture that if G is connected but not strongly orient-able then G2 is certainly strongly orient-able. You should either prove this conjecture or find a counter example.

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In a strongly connected directed graph every vertex is reachable from every other vertex, that is, a vertex u is reachable from v and the vertex v is reachable from u for every vertex u and v in D. As opposed to a strongly connected graph, a DAG is quite the apposite; here if a vertex v is reachable from u then the vertex u is not reachable from v for every pair of vertices u and v in a DAG. In other words there is no cycle in a DAG and that is why the name: directed acyclic graph. As it should be evident; a Hamiltonian Cycle, an Eulerian Trail, an n-Cycle, or a closed walk can not exist in a DAG. Thus a DAG can never be strongly connected. Can a DAG be unilaterally connected? Can a DAG be not unilaterally connected? Should a DAG be always transitive or always not transitive? These are some of the questions


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which should trouble your mind. Answer these questions before moving forward. There are certain nodes in a DAG which have some special features not noticed in other directed graphs. For example every DAG will have at least one source node and at least one sink node in addition to other ordinary (sometimes called intermediate nodes). A source node is defined as a node with no in degree while a sink node has no out degree. An intermediate node has both in degree as well as out degree. It is interesting to note that the underlying undirected graph of a DAG will be an undirected graph. Looking from the other way round you can start from any undirected graph G and put a direction on each edge such that the resulting directed graph becomes a DAG (in fact it can be converted into several different DAG’s). If the undirected graph is a tree then we may put an arbitrary direction on each edge and the resulting directed graph will guaranteed to be a DAG. On the other hand if an un-directed graph is completely connected then we have to be careful in selecting the direction of an edge in order to transform G into a DAG. Try your luck and design an algorithm which can transform an un-directed completely connected graph into a DAG (see Fig. 9.4.4). We shall shortly provide an efficient algorithm to solve this problem; as this DAG is derived from a completely connected un-directed graph so it is known as a tournament DAG. A tournament DAG is always transitive (why?). Perhaps you may like to find a non tournament DAG which is also transitive? Let us look at a DAG along with an action item, the transpose of a DAG. Will the transpose of a DAG always a DAG? Why? It will obviously contain the same number of vertices and exactly the same number of edges, the direction of each edge is reversed only. But then a DAG consists of a source vertex, a sink vertex, and a number of intermediate vertices. What happens to these vertices when we take a transform of a DAG? A source node has no in-degree but when we reverse directions it will have no out-degree so it will be transformed into a sink as shown in Concept map 9.3. Similarly the sink of the original DAG will be transformed into a source in the transpose of the DAG. How about if we take the square of a DAG? Will the resulting graph also a DAG? How the roles of different vertices change or remain the same in the square graph? Will the resulting graph be transitive? If we draw a Reachable Relation matrix A for a DAG, how the matrix will tell us that it represents the reachable relation for a directed acyclic graph?

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If we take the transpose of a DAG then obviously the Reachable Relation matrix for the transpose will be different from that of the original graph; it should still tell us somehow that it represents a directed acyclic graph? Let us now introduce a new action item; you must have become familiar with Depth First Search of a graph while studying algorithms. The output of the DFS are: 1. DFS spanning tree consisting of tree edges 2. Forward edges 3. Cross edges 4. Back edges 5. Finishing times of all vertices (please see Concept map 9.3). We know that a directed graph is acyclic if and only if a DFS of the directed graph does not produce a back edge. We also know that if a directed graph is a DAG then the vertex with maximum finishing time will always be (one of) the source vertices of the DAG. More importantly it is beneficial to sort the vertices of the DAG on the basis of decreasing finishing times. In this so called topological sort, the vertices of the DAG can be so arranged on a horizontal line such that all directed edges go from left to right. The source vertex will always be on the extreme left while the sink will be on the extreme right on the horizontal line (see Fig. 9.3.2). This visualization of a DAG may encourage us in devising a scheme to convert an un-directed graph into an acyclic directed graph; the directed edges should always go only from left to right. It is interesting to note that the vertices of a cyclic graph can not be so arranged on a horizontal line such that all edges go in only one direction. Problem Set 9.4. For the directed acyclic graph D shown in Fig. 9.3.1, do the following: Problem 9.4.1. Perform a DFS from any vertex u in D; identify tree edges, back edges, forward edges, and cross edges if any. Problem 9.4.2. Find the vertex of maximum finishing time. Problem 9.4.3. Start a DFS from a vertex other than u and again locate a vertex with the maximum finishing time.


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Concept Map 9.3. A concept map showing the concept of a directed acyclic graph, a couple of properties and two action items which are Reachable Relation and transpose of a DAG.

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Problem 9.4.4. Repeat the above three steps (1, 2, & 3) on the same graph with one change; reverse the direction of the edge between vertex 2 and vertex 4. Discuss the outcome?

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Figure 9.3.1: A directed graph D for Problem Set. We are now in a position to address a number of algorithmic issues. Let us try to address the following problems: Problem Set 9.5. Design an efficient algorithm to solve the following: Problem 9.5.1. Find if a given directed graph is acyclic. Problem 9.5.2. Find the source and the sink node in a DAG. Problem 9.5.3. Find if a given DAG has a node u such that you can reach every other vertex of the DAG from u. Problem 9.5.4. Find if a DAG has a node u such that you can reach u from every other vertex of the DAG. Problem 9.5.5. Find if a given directed graph (not necessarily a DAG) has a node u such that you can reach every vertex from u. Problem 9.5.6. Convert a completely connected un-directed graph into a DAG. Problem 9.5.7. Convert a connected graph into a DAG.


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Algorithm 67: Check if D is a DAG; find the source and the sink. input : A directed graph D output: Yes/No; if yes then the source and the sink vertices 1 2 3

Apply DFS on graph D. If no back edge then D is a DAG; Source vertex will be the one with maximum finishing time; Sink vertex: You find out;

You should try to solve these problems yourself before moving forward. Algorithm 67 used DFS as a bridge to solve a couple of problems. The source vertex can be found directly by identifying the vertex with maximum finishing time. It is not possible to find the sink node directly. Perhaps you should take the transpose of D and then again perform a DFS now on the transpose of D. Look at Concept map 9.3 for further hints. Algorithm 68: Check if D is a DAG; find the source and the sink. input : A directed graph D output: Yes/No; if yes then the source and the sink vertices 1

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Find a Reachable Relation R for the given graph D. If all diagonal entries in R are zero then D is a DAG; Source: Vertex with maximum reachability in R?; Sink: Vertex with zero reachabiliy?;

You might have noticed that Algorithm 68 is much more costly than Algorithm 67. This is because the Reachable Relation Matrix R is expensive as compared to a single DFS. But then the Reachable Relation provides much more information about a graph. Using a Reachable Relation Matrix we can easily find if there is a node u in D from where it is possible to reach every vertex of D. In fact it can answer this question even if the graph D is not acyclic. Algorithm 67 can also be slightly modified to locate a vertex u in a special DAG from where it is possible to reach every other vertex. First we can find the vertex with maximum finishing time; this will be a source vertex. Now we can start a new BFS from this very vertex; if all vertices of D are spanned by the BFS then the DAG possesses the desired characteristics (Does it mean that D is strongly connected?). How about if instead of a DAG we have an ordinary directed graph which may contain cycles? There

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may still be a node u in D from where it is possible to reach every where. This node u is not a conventional source vertex as it may have an in-degree. How about if we perform a DFS in this directed graph, will the maximum finishing time be able to locate this u in D? It is important for you to verify or contradict this conjecture at this very stage before moving forward. Algorithm 69: Convert a completely connected graph into a DAG input : An undirected completely connected graph G output: A DAG D (with same number of nodes and edges) 1 2 3 4

Find a Hamiltonian Path in G; Arrange the Hamiltonian Path as a Horizontal line; Put directions on this line going from left to right only; Put back the remaining edges of G with directions going from left to right only;

Algorithm 69 can be used to put appropriate directions on a completely connected un-directed graph in order to convert it into a DAG (see Fig. 9.4.4). It is left as an exercise for the learner to design an algorithm in order to convert an ordinary connected graph into a DAG. Please note that if the undirected graph G is not completely connected then it will be hard to find a Hamiltonian Path in G (as is done in Algorithm 69. Perhaps in an ordinary graph, we should first put directions arbitrarily and then remove cycles by reversing back edges? Or put edge directions such that no cycle is created in the first place? Or why not convert the given un-directed graph into a completely connected un-directed graph and then use Algorithm 69. But then we have to remove the extra edges that we have inserted into the original un-directed graph? So what? It is interesting to note that finding a Hamiltonian Path in a DAG is not a hard problem as it is in other graphs. We can easily design an efficient algorithm to check if such a path exists, and if it does then the algorithm outputs the actual Hamiltonian Path. The first step should be to do a topological sort on the vertices of the given directed graph; now the vertices of the DAG can be so arranged on a horizontal line such that the source vertex will always be on the extreme left while the sink will be on the extreme right on the horizontal line (see Fig. 9.3.2). All edges of the graph will go from left to right in this arrangement otherwise it will not be a directed acyclic graph. For a Hamiltonian path to exist in this graph there should be a directed edge


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(from left to right) between every two consecutive vertices; a Hamiltonian path in D will essentially be a path between topological sorted vertices in D (in the topological sorted order). Fig. 9.3.2 shows a directed acyclic graph D. We start a Depth First Search in this graph starting from a vertex designated as a start vertex in the top diagram. The resulting start / finishing times are indicated along with each vertex in this diagram. The vertices of D are topologically sorted on the basis of decreasing finishing times as shown in the bottom diagram. As one can see all edges of this graph are going from left to right; this observation confirms that D is a directed acyclic graph. A directed edge between two consecutive vertices (8/9 & 1/6) is missing; thus it is not possible to find a directed path passing through all the topologically sorted vertices. It implies that a Hamiltonian path does not exist in this directed acyclic graph. Algorithm 70: Find a Hamiltonian path in a DAG input : A directed acyclic graph D output: A Hamiltonian path in D provided it exists 1 2

Perform a topological sort on vertices of D; A path passing through all topological sorted vertices means a Hamiltonian path exists otherwise not.;

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Figure 9.3.2: A depth first search is conducted on the directed acyclic graph D starting from a vertex labeled as start node as shown in the top diagram. The start/finishing times of each vertex is also indicated here. Vertices of D are topologically sorted on the basis of decreasing finishing times as shown in the bottom diagram. The directed graph D is acyclic thus all edges move from left to right in the bottom diagram. A direct edge between two consecutive vertices is missing in the bottom diagram thus a Hamiltonian path is not possible in this graph.


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We know that in a strongly connected graph D, a vertex u is reachable to vertex v and the vertex v is reachable to vertex u for every pair of vertices u, v in D. Let us now consider a (related) class of directed graphs which we call independently connected directed graphs. In such a graph every vertex u in D may not be reachable to every vertex v in D but if u is reachable to v then v is reachable to u for every pair of vertices u, v in D. An independently connected graph may not be strongly connected but a strongly connected directed graph is always independently connected (see Concept map 9.4). An independently connected graph D which is not strongly connected is shown in the top of Fig. 9.4.1. It is quite evident from this figure that graph D consists of two sub-graphs E and F ; both of these sub-graphs are strongly connected while graph D as a whole is not strongly connected. It is now natural to define a new concept before moving forward: a strongly connected component in a directed graph D is a maximal set of vertices of a directed graph in which vertices u and v are reachable from each other for every pair of vertices u and v in that maximal set. The two sub-graphs in Fig. 9.4.1 are in fact two strongly connected components; in the top diagram there is no edge between component E and F . In the bottom diagram of Fig. 9.4.1, we add an edge between component E and F ; the resulting graph is no longer independently connected. The problem that we intend to address in this section is to understand how we can efficiently find the following: 1. Nodes belonging to a strongly connected component, & 2. The number of strongly connected components in a directed graph. The concept of independently connected graphs is introduced as it is much simpler to solve the above problem in such graphs. It is important to appreciate that in an independently connected directed graph there are no edges among strongly connected components (see top diagram of Fig. 9.4.1 and Concept map 9.4). Thus if we start a BFS from any vertex belonging to any strongly connected component in an independently connected graph then the search is confined to that strongly connected component; it can not enter into another strongly connected component; it is in fact an efficient way of determining vertices belonging to a strongly connected

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Figure 9.4.1: Sub-graph E and sub-graph F are both strongly connected components. In the top diagram, there is no edge between the two components and graph D is independently connected. In the bottom diagram there is an edge from E to F and the graph D is neither strongly connected nor independently connected. component (it is as simple as determining nodes belonging to a connected component in an un-directed graph). In a directed graph which is neither strongly connected nor independently connected there are edges between different strongly connected components; note the directed edge from component E to component F in the directed graph D shown in the bottom of Fig. 9.4.1 If you start a BFS in this graph then the search may not be confined to only one strongly connected component; it may also enter another strongly connected component. For example if you start a BFS from vertex 1 belonging to component E then the search will traverse vertices belonging to component E as well as F (see Fig. 9.4.1) Thus the problem of determining nodes belonging to a strongly connected component becomes harder if a directed graph is neither strongly connected nor independently connected. If some how we can remove edges connecting different strongly connected components in a directed graph (like the edge (3,5) in D shown in the bottom of Fig. 9.4.1) thereby converting it into an independently connected graph then the problem of determining nodes belonging to a strongly connected component is much simplified. The


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question is how to locate and then remove these edges? It is important at this learning stage to apply the Reachable transformation to a number of independently connected directed graphs and draw the corresponding Reachable Matrices. We shall certainly encourage you to draw the Reachable matrix for the independently connected graph shown at the top of Fig. 9.4.1. If you carefully look at the resulting Reachable matrix you will notice that the relationship it depicts on this graph is reflexive, symmetric as well as transitive; it is an equivalence relationship on the vertices of the independently connected graph. The corresponding equivalence classes (that the relationship generates) are in fact the strongly connected components E and F of the independently connected graph D. In simple words it means that in an independently connected graph a vertex is reachable only to vertices belonging to the same strongly connected component; it is not reachable to vertices belonging to other strongly connected components. Once we have the Reachable matrix of an independently connected graph, it becomes trivial to pinpoint vertices belonging to a strongly connected component. The Reachable Relation Matrix for the independently connected graph of Fig. 9.4.1 is shown by the bold 1’s in any of the table shown in Fig. 9.4.2.

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The picture is not so rosy for a directed graph which is not independently connected. For example if you draw the Reachable matrix of the graph shown in the bottom of Fig. 9.4.1, you will realize the inherent complication; you can reach all vertices belonging to component E as well as component F from vertex 1, although vertex 1 belongs to strongly connected component E (please see Reachable Relation Matrix A in the table shown in Fig. 9.4.2). The Reachable relation is not symmetric; hence it is not an equivalence relationship. It is thus no longer trivial to determine vertices in a strongly connected component in such directed graphs? If somehow we can convert it back into a symmetric relation without disturbing the connectivity inside any strongly connected component and without changing the number of strongly connected components then the problem will become much simpler? This means that some how we should be able to remove the small 1’s in the Reachable Relation Matrix A of table in Fig. 9.4.2. You might have noticed that removing those small ones in the table shown in Fig. 9.4.2 is again equivalent to removing the edge (3,5) in the bottom diagram of Fig. 9.4.1 thereby converting it again into an independently connected directed graph. The question is again how to do it? Let us again consider an independently connected directed D and assume that we take a transpose of D. It is interesting to note that the graph D is different from its transpose but the Reachable relation of the two graphs will be exactly the same as demonstrated by bold 1’s in the two tables of Table shown in Fig. 9.4.2. If, however, we take the transpose of a directed graph which is not independently connected then things will be different. We show the same directed graph D as depicted in the bottom of Fig. 9.4.1 as well as its transpose on the top of Table in Fig. 9.4.2. The Reachable Relation Matrix A of D and the Reachable Relation Matrix B of the transpose of D are also shown in this table. It is quite obvious in the tables that the reachability within a strongly connected component is not affected by the transpose transformation (see the bold 1’s in the two tables in Fig. 9.4.2). The reachability across strongly connected components is certainly affected by taking the transpose of a directed graph; this is indicated by the small 1’s in the two tables given in the figure. The problem of determining strongly connected components in a directed graph is thus reduced to recovering the bold 1’s (or throwing away the small 1’s) in the Reachable Relation Matrices of Fig. 9.4.2. A careful look at the two tables will provide the required answer.


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A problem For each of the graphs shown in Fig. 9.4.3, find the Reachable Relation Matrix and compare this matrix with the ones given in Table in Fig. 9.4.2 (this comparison will certainly be a useful learning experience). Please note that: 1. Graph H is the same as graph D in Fig. 9.4.2 except that the edge (3,5) in D is reversed. It is very different from the transpose of D in Fig. 9.4.2 except for one similar edge and that is (5,3). 2. Graph I is the same as the transpose of D in Fig. 9.4.2 except that the edge (5,3) is reversed and another edge (6,2) is added. Again it is quite different from graph D in Fig. 9.4.2 except for one common edge, that is (3,5). 3. Graph J is almost the same as graph D in Fig. 9.4.2 except the connectivity inside strongly connected component E in D is altered. 4. Graph K is almost the same as graph I except that the edge (2,6) in I is reversed in K thus creating a new cycle in the directed graph.

Algorithm 71: Find strongly connected components of graph D input : A directed graph D output: Strongly connected components of graph D 1 2 3 4

Find Reachable Relation Matrix A of D; Find Transpose T (D) of D; Find Reachable Relation Matrix B of T (D); Take AND of Matrix A & B;

Algorithm 71 is a straight forward algorithm to find all strongly connected components in a directed graph as already illustrated in the Table shown in Fig. 9.4.2. It may not be very efficient (still a polynomial time algorithm) but certainly a good start for a new learner in this field. It uses two transformations: Reachable Relation as well as the Transpose as intermediate building blocks of this algorithm. (please see Fig. 9.4.2). We shall describe hints for designing a more efficient algorithm to solve a similar problem in

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the coming paragraphs. Let us for the time being restrict ourselves to find one strongly connected component of a directed graph. Again consider an independently connected directed graph D consisting of K strongly connected components. We know there are no edges between strongly connected components in an independently connected graph. You are supposed to insert extra edges between the k strongly connected components such that the new directed graph should still have as many as k strongly connected components. When you insert edges you will soon realize that in order to achieve the above mentioned objective you should be careful not to create a cycle otherwise all strongly connected components within that cycle will collapse into one strongly connected component and the above condition will be violated. Please recall graph I & K in Fig. 9.4.3. Graph I has two strongly connected components in spite of the extra edges (2,6) while Graph K is reduced into one strongly connected component because of a cycle created by the edge (6,2). This immediately suggests that strongly connected components in a directed graph are always connected in the form of a directed acyclic graph (see Concept map 9.2). This further implies that there will be at least one source and one sink strongly connected component. In the top left diagram of Fig. 9.4.2, the component E was a source while the component F was a sink. While in the top right diagram of the same figure, component F was a source and E was a sink strongly connected component.

Algorithm 72: Find a strongly connected components of graph D input : A directed graph D output: One Strongly Connected Component of D 1 2 3 4

Call DFS on D; find vertex u of maximum finishing time; Find Transpose T of D; Call BFS on T starting from u.; Output all vertices traversed by the BFS;

Before discussing this algorithm (Algorithm 72), let us describe another algorithm (Algorithm 73) which will also output a strongly connected component of F . This looks very similar but there are interesting differences between the two algorithms. A comparison of the two will certainly be an exciting learning experience.


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Concept Map 9.4. A map showing the concept of a strongly connected graph, a couple of properties and an action item which is the Reachable Relation. It is interesting to note that the Reachable Relation graph of a Hamiltonian, Eulerian, and strongly connected directed graph is exactly the same.

Algorithm 73: Find a strongly connected component of graph D input : A directed graph D output: One Strongly Connected component of D 1 2 3 4

Find Transpose T of D; Call DFS on T : find vertex u of maximum finishing time; Call BFS on D starting from u; Output all vertices traversed by the BFS;

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Figure 9.4.2: The Reachable Relation Matrix A for D is shown on the left. A similar matrix B for the Transpose of D is shown on the right. Reachability within a strongly connected component is shown by a bold 1’s.


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Figure 9.4.3: Directed graphs for the problem set As you may have noticed that Algorithm 72 locates a vertex u of maximum finishing time after doing a DFS on D. Vertex u will belong to the source strongly connected component of directed graph D. If we start a traversal from this node in the graph F then the traversal will not only traverse vertices belonging to the source strongly connected component; it will also enter and traverse vertices belonging to other strongly connected components. We therefore take the transpose of D and then start a traversal from vertex u. The source strongly connected component in D will become a sink in the transpose of D; this time a traversal will be contained in the sink component of the transpose of D; the output will thus be all vertices belonging to the source component in D as shown in Concept map 9.5. Algorithm 73 first takes the transpose of D and then locates the vertex u of maximum finishing time. This vertex will belong to the source component of the transpose of D. The output of this algorithm will thus be the source or the sink strongly connected component of directed graph D (see Concept map 9.5)? Please recall the algorithm in which we have tried to solve the problem of determining if a given directed graph is strongly connected. What does that mean in terms of our newly acquired knowledge of strongly connected components? A strongly connected graph is just one strongly connected component. But how to determine if a directed graph possesses this property? How about if the sink strongly connected component consists of all vertices of the directed graph? We can determine this in O(p + q) instead of O(p3 )

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Concept Map 9.5. A directed graph F consists of a DAG of its strongly connected components. Transpose of F also consists of a (different) DAG of strongly connected components. The source strongly connected component belonging to F is the same as the sink strongly connected component of the transpose of F .


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for Algorithm 63. What we need to do is to slightly modify the last line of Algorithm 72 or 73. It should read; if BFS traverses all vertices of the directed graph D then D is strongly connected. Problem Set 9.6. Fig. 9.4.4 (top) shows a completely connected undirected graph. The bottom of this figure shows two directed graph derived from the top graph by putting a directions on edges of the un-directed complete graph (such a directed graph is known as a tournament). It is interesting to note that the directed graph (bottom right) is a directed acyclic graph as all edges are going from left to right (see Algorithm 69). The bottom left diagram of this figure is one strongly connected directed graph or component. Please note that all edges in the bottom left diagram are not directed. You may put any directions on these edges (indicated by thin lines); the graph still remains a strongly connected graph. Please solve the following:

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Figure 9.4.4: A completely connected undirected graph of five vertices is shown in the top diagram. Directions are added in this undirected graph in order to convert it into an acyclic directed graph shown in the bottom right. Different directions are added in the completely connected un-directed graph to convert it into a strongly connected graph shown in the bottom left. Problem 9.6.1. For the two bottom graphs in Fig. 9.4.4 determine the corresponding strongly connected components and the underlying DAG connecting the strongly connected components.

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Problem 9.6.2. Convert the completely connected un-directed graph in Fig. 9.4.4 into a directed graph such that the resulting directed graph (which will be another tournament graph) is neither strongly connected nor acyclic. Find its strongly connected components and determine the DAG connecting these components. Problem 9.6.3. Find and draw all non isomorphic directed graphs derived from the completely connected un-directed graph shown in the figure. Draw the corresponding DAG of strongly connected components for each of the tournament graph that you have drawn. Problem 9.6.4. Is it possible to convert the top un-directed graph Fig. 9.4.4 into a directed graph consisting of two strongly connected components such that one strongly connected component should consist of two nodes and the other will consist of the remaining three nodes? Discuss briefly how this is possible or why it is not possible. Problem 9.6.5. Is it possible to draw a tournament of 5 nodes which consists of 4 strongly connected components? Discuss briefly. Problem Set 9.7. Fig. 9.4.5 show three non-isomorphic directed graphs (tournaments) derived by putting directions on edges of the completely connected un-directed graph of Fig. 9.4.4. The corresponding DAG of strongly connected components of each tournament graph is drawn in front of it. Note that the top two directed graphs are not isomorphic to each other yet the DAG’s of strongly connected components are isomorphic to each other. The DAG of strongly connected components of the bottom graph is not isomorphic to any of the two. Problem 9.7.1. Carefully examine each of non-isomorphic tournaments that you have drawn in the last problem set, and the respective DAG of strongly connected components of each tournament. Group all tournaments which correspond to the same DAG; please see Fig. 9.4.5; it shows two tournaments (top and middle) with the same DAG and one with a different DAG (bottom). Problem 9.7.2. How many non-isomorphic DAG’s are possible corresponding to all possible tournaments of 5 vertices. Draw all these DAGs. What are their respective sizes in terms of number of nodes? Please remember that each node in a DAG corresponds to a strongly connected component.


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Figure 9.4.5: We show here three non-isomorphic directed graphs derived by putting directions on each edge of a completely connected graph of Fig. 9.4.4. The corresponding DAG of strongly connected components of each tournament graph is drawn in front of it. Note that all edges are not directed as shown by thin lines in the bottom graph; you may put any directions on these edges, the strongly connected components and the corresponding DAG will still remain the same.

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Problem 9.7.3. Is it possible to draw a DAG of 4 strongly connected components corresponding to a tournament of size 4? Problem 9.7.4. Is the DAG of strongly connected components of a tournament also a tournament? Why? Problem 9.7.5. Derive an exact expression for the number of non isomorphic DAGs of strongly connected components of tournaments of size p.

9.5 9.5.1

Tournaments A Panoramic Picture and a Concept Map

A directed graph obtained by putting directions on edges in a completely connected un-directed graph is known as a tournament (see Fig. 9.5.1). The directed graphs that you have studied in Fig. 9.5.1 are in fact all tournaments. While playing with tournaments in previous problems, you might have noticed that a tournament can be strongly connected while another tournament may be a directed acyclic graph again shown in Fig. 9.5.1 . Within these two extremes there is a lot of variety of possible tournaments consisting of various strongly connected components (see Concept map 9.6). There are several non-isomorphic tournaments possible which are not acyclic (Fig. 9.5.2) but there is always a unique acyclic tournament for a fixed number of vertices (why?). Similarly a tournament is transitive if and only if it is acyclic (why?). We shall study these and many other interesting properties of tournaments in this section. Let us start with some very simple properties; a tournament is always unilaterally connected. Why? Only because it is a tournament? Recall how a tournament is constructed from a completely connected un-directed graph and also recall the definition of a unilaterally connected directed graph?

9.5.2

A Hamiltonian Path in a Tournament

Every tournament has a Hamiltonian path; we can design an efficient algorithm to find that path. Assume that we have a tournament of p nodes and we have already found a path P of length k in this tournament. The goal is to find a Hamiltonian path in the tournament which will be a simple path of length p − 1. We can take any node u in T which is not part of P . We know

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Concept Map 9.6. An un-directed and completely connected graph can be converted into a directed graph by putting directions on each edge in an arbitrary fashion. Such a directed graph is known as a Tournament; it possesses some very special properties as shown in this concept map.

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Figure 9.5.3: A directed path P of length k is shown in a tournament. Consider any vertex u in T which is not included in P. Vertex u will be connected to every vertex in P; you are at liberty to put any directions in the top diagram you will always end up with an extended path P of length k+1. Any exercise to put directions so that the path length does not increase will fail as shown in the bottom diagrams. The extended Hamiltonian path is shown in bold in the bottom diagrams.

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that u is connected to all nodes in T (why?) as shown in Fig. 9.5.3 (top); you may put any directions on these edges, you are bound to get an increase in the path length of P from k to k + 1. It will be an interesting (but futile) exercise to put directions such that the path length does not increase by one as is demonstrated in Fig. 9.5.3 (bottom). Algorithm 74: Extend path length in P from k to k + 1 input : A Tournament T , and a simple path P of length k in T output: A simple path P of length k + 1 in T 1 2

Consider any node u in T not already in the path P ; Include u in P and output the new path P ;

Using Algorithm 74 it is possible to find a Hamiltonian path in a tournament. The algorithm also serves as a constructive proof that there is always a Hamiltonian path in any tournament.

9.5.3

A Hamiltonian Cycle in a Strong Tournament

We know that a tournament can be acyclic; thus it may not be strongly connected; it will not contain a Hamiltonian cycle. We now prove that if a tournament is strongly connected then it will always be Hamiltonian. Before proving that general result, we first prove a simpler hypothesis: if there is a cycle C of length k in a tournament then it is possible to find a cycle of length k +1 in that tournament. How about if we use a similar approach used earlier for extending the path length. Try your luck and check if it is possible. See if you can find the Basic Idea and Time Complexity of Algorithm 75 as shown below. Algorithm 75: Extend cycle length in T from k to k + 1 input : A strong Tournament T, and a cycle C of length k in T output: A cycle of length k + 1 in T

Before providing an answer to the above problem, let us look at a related problem: Assume that we are given a Hamiltonian cycle in a tournament T of size k. Also assume that now we add another vertex u to T such that T +u is strongly connected and is also a tournament. We will now show that T + u

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is Hamiltonian. As T + u is a tournament thus u will be connected to every vertex of T; the vertex u will have a finite in-degree and a finite out-degree as T + u is strongly connected (see Fig. 9.5.4). Thus it will always be possible to find a Hamiltonian cycle in T + u as shown in Fig. 9.5.4. It means that if T is Hamiltonian then T + u will also be Hamiltonian provided T + u is a strong tournament. 3

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Figure 9.5.4: A Hamiltonian cycle in a tournament T of size k is shown. A new vertex u is added to T such that T + u is a tournament and is also strongly connected. The vertex u will have at least one incoming edge and one outgoing edge (right). It is thus always possible to find a Hamiltonian cycle in T + u as shown in bold in the left diagram. How the above hypothesis will help us in finding a Hamiltonian cycle in a strongly connected tournament or in proving that a strongly connected tournament is Hamiltonian? How about finding a cycle of length 3 in a strong tournament; that is certainly not a hard problem? Then we should locate a vertex u such that the cycle length increases from 3 to 4, and so on until we find a Hamiltonian cycle. This type of approach will be very similar to Algorithm 74 where we extend the path length incrementally and finally output the Hamiltonian path. Is it feasible or not? Carefully look at Fig. 9.5.5 before making up your mind. As you can see there is a serious complication here; it is possible that a tournament T is not strongly connected but T + u is a strongly connected tournament. Then how can we show that T + u will be Hamiltonian? Let us look at the extreme; how about if T is a directed acyclic tournament but T +u is a strongly connected. In fact this special case is simple to solve as there is only one way to convert a directed acyclic tournament graph (we already

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know that it will have a Hamiltonian path inside it) into a strongly connected graph is to insert the extra node u is to connect it with the source node of the DAG and then connect the sink with node u; this will certainly result in a Hamiltonian graph (why?). We have already considered the possibility when T is strongly connected; we have also considered the possibility when T is a directed acyclic graph; but there is a possibility when T is neither strongly connected nor acyclic. How to cater to that class of graph; it will be interesting for you to explore?

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Figure 9.5.5: A directed cycle C of length 3 is shown (enclosed in a dotted circle) in a strongly connected tournament of size 5 (top). Consider vertices 4 and 5 in this figure; both these vertices are outside the cycle C and are double circled. Please note that vertex 4 has all incoming edges from vertices belonging to the cycle C while vertex 5 has all out going edges to vertices belonging to the cycle C. A cycle of length 5 is shown in bottom left diagram. A cycle of length 4 is shown in the bottom right. Let us come back to the problem of designing Algorithm 75 or proving that if a cycle C of length k exists in a strong tournament then a cycle of length k + 1 also exists in the strong tournament. Before proving this hypothesis let us look at its repercussions. If it is right then the tournament will have a cycle of every possible length until we have a Hamiltonian cycle; thus a strongly connected tournament is Hamiltonian, in fact it is more than that;

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a cycle of length k is possible in a strongly connected tournament where k is equal to or more than 3. Let us now come to the proof of the above hypothesis. Assume that we have found a cycle C of length k in a strongly connected tournament. Now we intend to find a cycle of length k + 1 in the same tournament. There are essentially two possibilities; 1. You can find a vertex u such that it has at least one incoming and one out going edge connecting u to vertices already in the given cycle C. This possibility is quite similar to the one illustrated in Fig. 9.5.4. You can certainly find a cycle of length K +1 in this situation by extending the cycle length of C from k to k + 1. 2. If you can not find a vertex u with the above property then you will certainly find a vertex u and a vertex v such that (a) all edges to u from every vertex of C are incoming towards u, (b) all edges from v to every vertex of the cycle C are out going, and (c) there is an edge from u to v. Note that if (a) or (b) is not true then we shall end up with the first possibility. If (c) is not true then the tournament T will not be strongly connected. This scenario is depicted in Fig. 9.5.5. Now it is possible to find a cycle of length k + 2 (see the bottom left of Fig. 9.5.5); it is possible to convert this cycle into a cycle of length k + 1 by a simple manipulation (see the bottom right of Fig. 9.5.5and also Fig. 9.5.6 ). This completes the constructive proof that you can extend the cycle length in a strong tournament of p nodes from 3 to p−1. This also provides an efficient algorithm to find a cycle from length 3 to all the way to a Hamiltonian cycle of length p − 1.

The Concept of a Rip Vertex Let us define a new term before designing an alternate strategy to find a Hamiltonian cycle in a strong tournament. You are already familiar with the concept of a cut vertex in an un-directed graph (a vertex v in an undirected connected graph G is a cut vertex provided G − v is not connected). Corresponding to a cut vertex in a connected un-directed graph there is a counter part concept in a (strongly) connected directed graph: a vertex v is a rip vertex (in a strongly connected directed graph D) provided D − v is not

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Figure 9.5.6: Extending a 3-cycle into a 5-cycle in a strongly connected tournament.

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(strongly) connected. Now assume that we are given a strongly connected tournament T of size p. We first locate and then remove a non rip vertex u. The resulting graph T − u will be a tournament (why?) and a strong tournament (because u was not a rip vertex) of size p − 1. We can do this step of removing a non rip vertex recursively until the original tournament is reduced to such a small sized tournament where it will be trivial to find a Hamiltonian cycle. Now we can start inserting back the removed vertices one by one in the last removed first inserted order; the Hamiltonian cycle will grow incrementally (with the graph) as shown in Fig. 9.5.8, until you find the Hamiltonian cycle in the original tournament T . On the basis of these ideas it is possible to construct an alternate algorithm to find a Hamiltonian cycle in a strong tournament. It will be a learning experience to compare the working of the two algorithms that we have described to find a Hamiltonian cycle in a strong tournament. In the earlier algorithm we grow a cycle within the original tournament from a small size to p. In the later algorithm we remove non rip vertices from a tournament one by one until it is possible to find a Hamiltonian cycle in the reduced size tournament; we then insert the removed vertices one by one and Hamiltonian cycle also increases incrementally with the size of the graph. It will be interesting to derive and compare the time complexities of the two algorithms. Problem Set 9.8. It is important for you to answer the following in order to meaningfully understand the last algorithm that we have described to find a Hamiltonian cycle in a strong tournament. Problem 9.8.1. Indicate which vertices are rip vertices and which vertices are not rip vertices in Fig. 9.5.5. Problem 9.8.2. Draw a strong tournament in which no vertex is a rip vertex. Problem 9.8.3. Draw a strong tournament in which there are at least 3 rip vertices in a tournament of size 5 or 6 . Problem 9.8.4. Prove that there always exists at least one non rip vertex in a strong tournament. Problem 9.8.5. How can you efficiently find a non rip vertex in a strong tournament?

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Figure 9.5.8: Extending a 3-cycle into a 5-cycle in a strongly connected tournament.


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Problem 9.8.6. What will be the overall time complexity of finding a Hamiltonian cycle in a strong tournament using the above algorithm. Problem 9.8.7. Why we insist that a non rip vertex should be removed instead of a rip vertex? Problem 9.8.8. If instead of removing a non rip vertex in a strong tournament, we remove a rip vertex then what will be the complication in reconstructing the Hamiltonian cycle in T ? Problem 9.8.9. Can you visualize a tournament graph where there is no rip vertex? Please see Fig. 9.5.9

9.6


A directed graph D is unilaterally connected provided for every pair of vertices u and v, there is a path from u to v or a path from v to u. In a strongly connected directed graph there is a path from u to v and a path from v to u. That means a strongly connected graph is always a unilaterally connected graph but not vise versa. We have also seen that a tournament graph is also unilaterally connected. In this section we shall discuss various properties of unilaterally connected directed graphs. Let us first address the problem of how a Reachable Relation Matrix A of a unilaterally connected graph would look like. It is clear from the definition of such directed graphs that either A[i, j] = 1 or A[j, i] = 1 for every pair of vertices i, j in D. If instead of one both A[i, j] and A[j, i] are equal to 1 for every pair of vertices i, j in D then D will be more than unilaterally connected; in fact it will be strongly connected. Thus a directed graph D will be unilaterally connected if A[i, j] = 1 for every pair of vertices i, j in D if j > i. This means that the Reachable Relation Matrix A for a unilaterally connected directed graph will have all 1’s either on upper side of the diagonal (or on the lower side of the diagonal) as shown in Table. 9.6.1. Each of the matrices, shown in Table 9.6.1 corresponds to a unilaterally connected graph which is also directed acyclic (why?) assuming that all other entries in the matrix are equal to zero. A variety of non-isomorphic directed acyclic graphs are possible which will produce such a Reachable Relation matrix. It will be useful at this stage to draw a couple of such

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Figure 9.5.9: The effect of deleting a vertex in a Special Tournament


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Concept Map 9.7. A map showing the concept of a unilaterally connected directed graph, a couple of properties and an action item which is the Reachable Relation. An un-directed connected graph which can be converted into a unilaterally connected directed graph is known as a unilaterally orient-able graph. A tree graph is unilaterally orient-able provided it is a path graph.

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Figure 9.6.1: The Reachable Relation Matrix A for a unilaterally connected directed graph should have all 1’s either on the upper side of the diagonal (left) or on the lower side as shown in the right table.

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Figure 9.6.2: A unilaterally connected directed graph D having minimum number of edges with all 1’s in the upper side of the diagonal and 0’s else where as shown in the left diagram. The right diagram shows such a graph with maximum number of edges. Both theses graphs are directed acyclic.


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graphs with the above property and the assumption that all other entries in the matrix are zero; we show in Fig. 9.6.2, such a unilaterally connected graphs, one having minimum number of edges (left) and the other having maximum number of edges as shown on the right of this diagram. We shall soon prove that a unilaterally connected directed acyclic graph D will always contain a Hamiltonian path inside D. But before that let us try to imagine what will happen if there are more 1’s then are absolutely essential for a graph to be unilaterally connected. Assume that in addition to all 1’s on the upper side of the diagonal there are some additional 1’s as shown (in bold) in the top diagram of Fig. 9.6.3. Under such conditions one would expect that the resulting directed graph will be unilateral, as well as cyclic containing a Hamiltonian path (as before) as shown in the middle diagram of Fig. 9.6.3. What is quite unexpected is that the Hamiltonian path may disappear because of the extra 1’s that we have added in the Reachable Relation matrix as shown in the bottom diagram of Fig. 9.6.3. Let us summarize our observations before making a number of formal proofs. We have observed that a unilaterally connected graph contains a Hamiltonian path if it is acyclic (see Fig. 9.6.2). If there are cycles in the unilaterally connected graph then there may or may not be a Hamiltonian path inside the graph (see Fig. 9.6.3). What we have not observed until now is that whether a unilaterally connected directed graph contains a Hamiltonian path or not it will certainly contain an open spanning walk. In fact we shall prove that a directed graph is unilaterally connected if and only if it contains an open spanning walk. Let us first find a constructive proof that a unilaterally connected directed acyclic graph will always contain a Hamiltonian path inside it. We are dealing with a directed acyclic graph D; thus if we do a topological sort on vertices of D then the ordering of the vertices of the graph D along a horizontal line is such that all directed edges will be going from left to right as shown in Fig. 9.6.2. There will not be any edge going from right to left because otherwise graph D will be cyclic. We are dealing with a unilaterally connected directed graph D; thus for every consecutive vertex pair {u, v} in D either u is reachable to v or v is reachable to u; assume that u is on the left of v in linear ordering then there must be a direct edge from u to v otherwise the condition stated above will be violated or there will be an edge going from right to left. This means there will be a direct edge between every two consecutive vertices (again from left to right); this would imply a Hamiltonian path in D (it will

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Figure 9.6.3: A Reachable Relation matrix A for a unilaterally connected directed graph with some additional 1’s in the matrix (top). The extra 1’s (shown in bold) will make the unilaterally connected directed graph cyclic but there is still a Hamiltonian path inside the graph as shown in the middle diagram; what is quite unexpected is that the Hamiltonian path may disappear from the resulting directed graph because of the extra 1’s as shown in the bottom diagram.


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essentially be a path between topological sorted vertices in D).

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Figure 9.6.4: A unilaterally connected directed graph D having 5 strongly connected components. Note that inside each strongly connected component there is a closed walk. Also note that the directed acyclic graph connecting all strongly connecting components should be a line graph otherwise D will not be a unilaterally oriented directed graph. Let us now consider the possibility when a unilaterally connected directed graph is not directed acyclic (see Fig. 9.6.3) and so it may not have a Hamiltonian path; now we claim that a directed graph D is unilaterally connected if and only if it contains an open spanning walk. Let us first handle the hypothesis that if D is unilaterally connected then there will be an open spanning walk inside D. We have already observed that if D is a unilaterally connected directed acyclic graph then there will be a Hamiltonian path inside D; thus there will be (not only a spanning walk) but a spanning path inside D. If, however, D contains cycles then we can always find strongly connected components of D and we know that the graph interconnecting its strongly connected components will be a directed acyclic graph. We can again make a topological sort on strongly connected components of D and argue that there will be an edge going from left to right between every two consecutive strongly connected components of D. Thus there will be a Hamiltonian path passing through all strongly connected components of D as shown in Fig. 9.6.4. We also know that there is a closed walk spanning all vertices inside each strongly connected component; thus there will be an open spanning

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walk inside D.

9.6.1

Unilaterally orient-able Un-directed Graphs

We know that if we put directions on edges in an un-directed acyclic connected graph G (that means a tree) then no matter what is our direction scheme the resulting graph D would be a directed acyclic graph. We also know that a directed acyclic unilaterally connected graph D always contain a Hamiltonian path. Thus when we put directions on edges in a tree graph in order to make it unilaterally connected then the resulting directed graph should also have a Hamiltonian path inside it. This would require that there should be an un-directed Hamiltonian path inside the tree graph G in the first place otherwise it would have been impossible to convert G into D with a directed Hamiltonian path inside it. The only tree graph which contains a Hamiltonian path is in fact a line graph. Thus a tree graph is unilaterally orient-able if and only if it is a line graph (or a path graph). Now let us find necessary and sufficient conditions for a general connected graph G (G is no longer restricted to be a tree and may contain cycles) to be unilaterally orient-able. There are essentially three possibilities: 1. Every edge of G is a bridge edge. 2. Every edge of G is a non-bridge edge. 3. Some edges are non bridge edges while some are bridges. We have already considered the first possibility when G is a tree and every edge of G is a bridge edge. Let us consider the second possibility when every edge of G is a non bridge edge. If you recall this is a sufficient condition for a graph to be strongly orient-able. If G is strongly orient-able then it will also be unilaterally orient-able. Now we shall consider the third possibility when some edges of G are bridge edges while some other edges are non bridges as shown in Fig. 9.6.5. If the graph G contains K bridge edges and if we remove all these K edges then G will be decomposed into k + 1 connected components. Consider each connected component as a single (super) vertex. If we put back the bridge edges then these (super) vertices will be connected in the form of a tree (why?). A tree of ordinary vertices can be unilaterally


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Figure 9.6.5: An un-directed connected graph G with 4 bridge edges shown in bold in the top diagram. If we remove all bridge edges then the un-directed graph is disconnected into 5 connected components shown in shaded circles. The connected components are always connected in the form of a tree as shown in the bottom diagram.

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orient-able if and only if the tree is a line graph. A tree of super vertices can be unilaterally orient-able provided: (1) The tree is a line graph and (2) We can find a closed spanning walk inside each connected component (why?). So once again concentrate on a single connected component; no edge inside the component will be a bridge edge (why?). Thus each connected component is strongly orient-able (and will subsequently become a strongly connected component in the directed graph); there will be a closed spanning walk inside each component. This completes the proof that if a graph G is unilaterally orient-able then it will contain an open spanning walk. You may prove the converse as an exercise?

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Graph Theory & Algorithms - M. Ashraf Iqbal

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