Math776: Graph Theory (I) Fall, 2013 Homework 1 solution 1. [page [page 30, #2 ] Determine the average degree, number of edges, diameter, girth, and circumferenc cumferencee of the hypercube graph Q d . Solution by Shuliang Bai:
• The average degree is d, d , according to its definition. • Number of edges is 2 − × d. According to it definition, thevertex is coded by binary strings of length d length d,, so |V ( V (Q )| = 2 . Then |E (Q )| = ∈ d(v) = × 2 × d = − × d. 2 • Because the distance between any two vertices depends on the number of different d 1
d
d
d
1 2
d 1
1 2
v V
d
bits, so the diameter is d, d , i.e. diam G = G = d d..
• Girth(Q Girth(Q ) = ∞, this is because there are no cycles on hypercube graph Q . Girth(Q Girth(Q ) = ≥ × 4, where d where d 2, this is because Q because Q = K = K Q − . • Circumference of Q of Q is 2 . 1
d
1
d
d
2
d 1
d
2. [page [page 30, #3 ] Let G Let G be be a graph containing containing a cycle C cycle C ,, and assume assume that G that G contains contains a path of length at least k least k between two vertices of C . C . Show that G that G contains contains a cycle of length at least k .
√
Solution Solution by Melissa Bechard: Bechard: Let C = (V c , E c ) denote the cycle in G, and let P = (V p , E p ) denote the path of length at least k in G. Notice Notice,, if V p V c = V p then we have a path of length k with distinct edges, and we can then connect the end points of this path by traversing C to C to result in a cycle of length greater than k k. Assume this is not the case. Let V p V c = n < k . Notice, if n n k, then could take the path along C along C connecting connecting these k distinct vertices, and then traverse C and C and the result would be a cycle of length at least k . Also, between each of the n the n vertices of P that P that lie on C on C there exist some subpath of P . P . Let P 0 P be P be a subpath of P with P with the property that both of P 0 ’s endpoints lie on C and P and P 0 is the maximum length such subpath of P of P .. Let ℓ(P 0 ) = m. m . Additionally, any other subpath of P has P has length less than or equal to m. m . We then have the following inequality:
| √ ∩ | √
√
≥ √
| | ∩ | | | ≥ √
⊆
k
≤ ℓ(P ) P ) ≤ √ nm ≤ km
(as defined) (since (since each path path between between the vertices vertices in the intersecti intersection on has length m)
≤
√ ≤ m, hence k ≤ m. Thus, we have a path, P , of length at √ least k , whose end points lie on C on C .. So, we can traverse C traverse C to to connect these end points √ and result with a cycle of length at least k .
k This inequality gives us √ k
0
3. [page [page 30, #8 ] Show that every connected graph G contains a path of length at least min 2δ (G), G 1 .
{
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1
Solution by Nicholas Stiffler: Let P = x0 x1 . . . xm be a path in graph G of maximal length. We will denote the length of P by l. If l G 1, we are done. Otherwise, the set O = V (G) V (P ) is nonempty, and since the graph is connected, there exists a V (P ) O path, P ′ = y 0 y1 . . . yk that is non-trivial.
≥ | |−
\
−
Proposition If l < 2δ (G) then there is a cycle spanning V (P ). Proof Observe that N (x0 ) P and N (xm ) P , because if either endpoint of P is adjacent to a vertex outside of P , then the path can be extended and is not maximal. If x0 xi+1 and x m xi are both edges in P , then there is a cycle C = x0 . . . xi xm . . . xi+1 x0 .
⊂
⊂
The occurrence of these two edges can be shown by the Pigeonhole Principle. A special case occurs where there is an edge x0 xm , since the other edge is given in the path. The vertex x0 has at least δ (G) 1 neighbors out of x2 . . . xm−1 because it is ad jacent to x1 and not adjacent to x m (would create an obvious cycle). For each neighbor xi there is a corresponding vertex xi−1 to which x m is not adjacent. So, x m must have at least δ (G) 1 neighbors out of x1 . . . xm−2 , and of those δ (G) 1 are forbidden. Since m < 2δ (G), there are fewer than 2 δ (G) 2 possible neighbors. So by the Pigeonhole Principle one of the neighbors is forbidden, so there is a cycle.
−
−
{
{
−
}
}
−
By deleting an edge of the cycle spanning V (P ) incident with y0 you can extend the remaining path with P ′ , forming a path longer than P , which is a contradiction.
x0
xi+1
xm
xi
x0
subgraph
x0
xi+1 xi P
xi+1 xi C spanning V (P )
Figure 1: Visualization of proof
2
xm
xm
4. [page 30, #9 ] Show that a connected graph of diameter k and minimum degree d has at least about kd/3 vertices but need not have substantially more. Solution by Robert Wilcox: We want to show that a connected graph of diameter k and minimum degree d has at least about kd vertices but need not have substantially 3 more. Let x0 and xk be vertices such that the shortest path, P, between the two has length k, the diameter. Let v be a vertex not on P that is adjacent to a vertex on P. Let i be the smallest integer such that xi is adjacent to v. If j > i + 2 then xj cannot be adjacent to v or x 0 P xi vxj P xk would be a path from x 0 to xk of length less than k, a contradiction. Thus any vertex off of P can only be adjacent to at most 3 vertices on P. Since there are k vertices on P, each with at least d neighbors and only 3 vertices can share a neighbor there must be at least kd/3 vertices in G. However constructing such a path with each vertex off of the path adjacent to exactly 3 vertices of P will yield a graph with not substantially more than kd vertices. 3 5. [page 30, #12 ] Determine κ(G) and λ(G) for G = P m , C n , K n , K m,n , and Q d ; d, m, n 3.
≥
Proof by Garner Cochran: κ(P m ) = 1 If you remove an interior vertex of P m , you disconnect graph. λ(P m ) = 1 If you delete any edge, the graph becomes disconnected. κ(C n ) = 2 If you delete any vertex of C n you are left with P n−1 , therefore you must delete one more vertex to disconnect it, so κ(C n ) = 2. λ(C n ) = 2 If you delete any edge of C n you are left with P n , therefore you must delete one more vertex to disconnect it, so λ(C n ) = 2. κ(K n ) = n
−1
You must delete all but one vertex to disconnect the graph because every vertex is has an edge to every other vertex. λ(K n ) = n
−1
Again since every vertex has an edge to every other vertex, the way to disconnect K n by deleting the least number of edges is deleting all of the edges incident to one vertex. κ(K m,n ) = min m, n
{
}
The minimum number of vertices to disconnect the graph K m,n is by deleting whichever side is smaller, if you delete any less than that, you will still have vertices on each side, and since there are edges connecting every vertex on one side of the graph to the other side, the graph will still be connected. λ(K m,n ) = min m, n
{
}
Let m n, you must delete all the edges incident to a vertex on the side with n vertices. This will allow you to disconnect a vertex from the graph. If you try to do it by deleting less than m edges, you will be able to find some vertex on the n side which is still
≤
3
completely connected. From the neighborhood of this vertex, we note that since none of the vertices on the m side are disconnected, we may find a path from these vertices to those on the n side. κ(Qd ) = d We must delete d vertices to disconnect the graph. If we delete all vertices with only one 1 in the string, then there is no way to get from the string of all zeroes to the string of all ones. If we try to delete only d 1 vertices, there will always be some bit it is possible to change to continue exploring the graph.
−
λ(Qd ) = d If we delete all edges incident to one vertex we disconnect the graph, and since d(v) = d for all v G then it is possible to disconnect the graph by deleting d edges.
∈
Again, by deleting only d 1 edges, we have not disconnected any individual vertex, and since we are always able to explore the graph further, for the same reason as above, d 1 edges will not disconnect the graph.
−
−
6. [page 31, #18 ] Show that a tree without a vertex of degree 2 has more leaves than other vertices. Can you find a very short proof that does not use induction? Proof by Rade Musulin: Let G be a tree with no vertex of degree 2. Looking at the average degree: 2 E 2( V 1) v ∈V d(v) = = < 2. V V V
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Let L be the set of leaves in G and let O be the set of vertices which are not leaves in G. Note that the minimum degree of an element of O is 3 because no vertex has degree 2. So, v∈V d(v) v ∈L d(v) + v ∈O d(v) 2 > = V V L +3 O V + 2 O 2 O v ∈L 1 + v ∈O 3 = = =1 + . V V V V
| |
≥
So,
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| | | | | | | | | | | | | | 2O , V
| | | | |V | > 2|O|, |L| + |O| > 2|O|, |L| > |O|. 1 >
4
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