Encrypting and Decrypting

Outline 1 2

CSc 466/566 Computer Security 3 4

7 : Cryptography — Public Key Version: 2012/02/15 16:15:24

Department of Computer Science University of Arizona

5 [email protected] c 2012 Christian Collberg Copyright

Christian Collberg

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History of Public Key Cryptography

Introduction RSA Algorithm Example Correctness Security GPG Elgamal Algorithm Example Correctness Security Diffie-Hellman Key Exchange Diffie-Hellman Key Exchange Example Correctness Security Summary

Introduction

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Public-key Algorithms

Definition (Public-key Algorithms) RSA Conference 2011-Opening-Giants Among Us:

Public-key cryptographic algorithms use different keys for encryption and decryption.

http://www.youtube.com/watch?v=mvOsb9vNIWM&feature=related

Rivest, Shamir, Adleman - The RSA Algorithm Explained: http://www.youtube.com/watch?v=b57zGAkNKIc

Bob’s public key: PB

Bruce Schneier - Who are Alice & Bob?:

Bob’s secret key: SB

http://www.youtube.com/watch?v=BuUSi_QvFLY&feature=related

Adventures of Alice & Bob - Alice Gets Lost:

EPB (M) = C

http://www.youtube.com/watch?v=nULAC_g22So http://www.youtube.com/watch?v=nJB7a79ahGM

DSB (C ) = M DSB (EPB (M)) = M

Introduction

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Introduction

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Public Key Protocol

Public Key Encryption Protocol. . .

Key-management is the main problem with symmetric algorithms – Bob and Alice have to somehow agree on a key to use.

Bob

Alice

In public key cryptosystems there are two keys, a public one used for encryption and and private one for decryption. plaintext

1

Alice and Bob agree on a public key cryptosystem.

2

Bob sends Alice his public key, or Alice gets it from a public database.

3

Alice encrypts her plaintext using Bob’s public key and sends it to Bob.

4

Bob decrypts the message using his private key.

Introduction

PB

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Public Key Encryption: Key Distribution

Alice SA , PA

PA , PC

PA , PB

PB , PC PA , PD

Carol SC , PC

Eve

decrypt

plaintext

SB

Introduction

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In practice, public key cryptosystems are not used to encrypt messages – they are simply too slow.

Bob SB , PB

Instead, public key cryptosystems are used to encrypt keys for symmetric cryptosystems . These are called session keys , and are discarded once the communication session is over.

PB , PD 1

Bob sends Alice his public key.

2

Alice generates a session key K , encrypts it with Bob’s public key, and sends it to Bob.

3

Bob decrypts the message using his private key to get the session key K .

4

Both Alice and Bob communicate by encrypting their messages using K .

SD , PD

Advantages : n key pairs to communicate between n parties. Disadvantages : Ciphers (RSA,. . . ) are slow; keys are large Introduction

ciphertext

A Hybrid Protocol

Dave PC , PD

encrypt

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Introduction

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Hybrid Encryption Protocol. . .

Outline 1 2

Bob

Alice

3 4

K

encrypt

EPB (K )

PB

decrypt

K

SB 5

M

encrypt K

EK (M)

decrypt

M

K 6

Introduction

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RSA


RSA

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RSA: Algorithm

Bob (Key generation): Generate two large random primes p and q. Compute n = pq. Select a small odd integer e relatively prime with φ(n). 4 Compute φ(n) = (p − 1)(q − 1). 5 Compute d = e −1 mod φ(n). 1 2 3

RSA is the best know public-key cryptosystem. Its security is based on the (believed) difficulty of factoring large numbers. Plaintexts and ciphertexts are large numbers (1000s of bits).

PB = (e, n) is Bob’s RSA public key. SB = (d, n) is Bob’ RSA private key.

Encryption and decryption is done using modular exponentiation.

Alice (encrypt and send a message M to Bob): 1 2

Get Bob’s public key PB = (e, n). Compute C = M e mod n.

Bob (decrypt a message C received from Alice): 1

RSA

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RSA

Compute M = C d mod n. 12/83

RSA: Algorithm Notes

RSA Example: Key Generations

How should we choose e? It doesn’t matter for security; everybody could use the same e. It matters for performance: 3, 17, or 65537 are good choices.

n is referred to as the modulus , since it’s the n of mod n.

1

Select two primes: p = 47 and q = 71.

2

Compute n = pq = 3337.

3

Compute φ(n) = (p − 1)(q − 1) = 3220.

4

Select e = 79.

5

Compute

You can only encrypt messages M < n. Thus, to encrypt larger messages you need to break them into pieces, each < n.

d

Throw away p, q, and φ(n) after the key generation stage.

= 79−1 mod 3220

Encrypting and decrypting requires a single modular exponentiation.

= 1019

RSA

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RSA Example: Encryption 1 2

6

P = (79, 3337) is the RSA public key.

7

S = (1019, 3337) is the RSA private key.

RSA

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RSA Example: Decryption

Encrypt M = 6882326879666683. Break up M into 3-digit blocks: m = h688, 232, 687, 966, 668, 003i

3

= e −1 mod φ(n)

1

Note the padding at the end. Encrypt each block:

Decrypt each block: m1 = c1d mod n = 15701019 mod 3337

c1 = m1e mod n

= 688

= 68879 mod 3337 = 1570 We get: c = h1570, 2756, 2091, 2276, 2423, 158i

RSA

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RSA

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In-Class Exercise: Goodrich & Tamassia R-8.18

In-Class Exercise: Goodrich & Tamassia R-8.20

Alice is telling Bob that he should use a pair of the form (3, n) Show the result of encrypting M = 4 using the public key (e, n) = (3, 77) in the RSA cryptosystem.

or (16385, n) as his RSA public key if he wants people to encrypt messages for him from their cell phones. As usual, n = pq, for two large primes, p and q. What is the justification for Alice’s advice?

RSA

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In-Class Exercise: Stallings pp. 270-271

RSA

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RSA Correctness We have C

= M e mod n

M = C d mod n. 1

Generate an RSA key-pair using p = 17, q = 11, e = 7.

2

Encrypt M = 88.

3

Decrypt the result from 2.

To show correctness we have to show that decryption of the ciphertext actually gets the plaintext back, i.e that, for all M
RSA

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RSA

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RSA Correctness: Case 1

RSA Correctness: Case 1. . .

From the key generation step we have d

= e −1 mod φ(n)

from which we can conclude that M φ(n) mod n = 1 follows from Euler’s theorem.

ed mod φ(n) = 1 ed

Theorem (Euler)

= kφ(n) + 1

Let x be any positive integer that’s relatively prime to the integer n > 0, then x φ(n) mod n = 1

Case 1, M is relatively prime to n: C d mod n = M ed mod n = M kφ(n)+1 mod n = M · (M φ(n) )k mod n = M · 1k mod n = M mod n = M RSA

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RSA Correctness: Case 2

RSA

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RSA Correctness: Case 2. . . We have that φ(n) = φ(pq) = φ(p)φ(q) By Euler’s theorem we have that

Assume that M is not relatively prime to n, i.e. M has some factor in common with n, since M < n. There are two cases: 1 2

M kφ(n) mod q = M kφ(p)φ(q) mod q = (M kφ(p) )φ(q) mod q = 1

M is relatively prime with q and M = ip, or M is relatively prime with p and M = iq.

Thus, for some integer h

We consider only the first case, the second is similar.

M kφ(n) = 1 + hq Multiply both sides by M M · M kφ(n) = M(1 + hq) M kφ(n)+1 = M + Mhq

RSA

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RSA

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RSA Correctness: Case 2. . .

RSA Security

We can now prove Case 2, for M = ip: d

C mod n = M = M

ed

Summary: Compute n = pq, p and q prime. Select a small odd integer e relatively prime with φ(n). Compute φ(n) = (p − 1)(q − 1). 4 Compute d = e −1 mod φ(n). 5 PB = (e, n) is Bob’s RSA public key. 6 SB = (d, n) is Bob’ RSA private key.

mod n

kφ(n)+1

1 2 3

mod n

= (M + Mhq) mod n = (M + (ip)hq) mod n = (M + (ih)pq) mod n

Since Alice knows Bob’s PB , she knows e and n.

= (M + (ih)n) mod n

If she can compute d from e and n, she has Bob’s private key.

= (M mod n) + ((ih)n mod n)

If she knew φ(n) = (p − 1)(q − 1) she could compute d = e −1 mod φ(n) using Euclid’s algorithm.

= M mod n = M

RSA

If she could factor n, she’d get p and q!

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Security of Cryptosystems by Failed Cryptanalysis

1

Propose a cryptographic scheme.

2

If an attack is found, patch the scheme. GOTO 2.

3

If enough time has passed ⇒ The scheme is secure!

RSA

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RSA Security. . .

If we can factor n, we can find p and q and the scheme is broken. As far as we know, factoring is hard.

How long is enough? 1 2

RSA

We need n to be large enough, 2,048 bits.

It took 5 years to break the Merkle-Hellman cryptosystem. It took 10 years to break the Chor-Rivest cryptosystem.

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RSA

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RSA Factoring Challenge

RSA Factoring Challenge. . . ✞

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☎

Name : RSA−576 Digits : 174 188198812 92 0 60 7 96 3 83 8 69 7 23 9 46 1 65 0 43 9 8 07 1 63 5 63 3 79 4 17 3 82 7 00 7 63 3 56 4 22 988859715 23 4 66 5 48 5 31 9 06 0 60 6 50 4 74 3 04 5 3 17 3 88 0 11 3 03 3 96 7 16 1 99 6 92 3 21 2 05 7340318795 50 6 56 9 96 22 1 30 5 16 87 5 93 0 76 5 02 57 0 59

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The factoring research team of F. Bahr, M. Boehm, J. Franke, T. Kleinjung continued its productivity with a successful factorization of the challenge number RSA-640, reported on November 2, 2005. The factors are:

On December 3, 2003, a team of researchers in Germany and several other countries reported a successful factorization of the challenge number RSA-576. The factors are

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☎

1900871281 66 4 82 2 11 3 12 6 85 15 7 39 3 54 1 39 7 54 7 18 9 67 8 99 68 5154936666 38 5 39 0 88 0 27 10 3 80 2 10 4 49 8 95 7 19 1 26 14 6 55 7 1

47277214610 7 43 5 30 2 53 6 22 30 7 19 7 30 4 82 24 6 32 9 14 6 95 30209711645 9 85 2 17 11 3 05 2 07 11 2 56 3 63 5 90 39 7 52 7

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RSA

RSA Factoring Challenge. . .

RSA

☎ ✞

☎

Name : RSA−1536 Digits : 463 184769970 32 1 17 4 14 7 43 0 68 3 56 2 0 20 0 16 4 40 3 01 8 54 9 33 8 66 3 4 1 0 1 71 4 71 7 85 7 74 9 10 6 5 1 696711161 24 9 85 9 33 7 68 4 30 5 43 5 7 44 5 85 6 16 0 61 5 44 5 71 7 94 0 5 2 2 2 97 1 77 3 25 2 46 6 09 6 0 6 469460712 49 6 23 7 20 4 42 0 22 2 69 7 5 67 5 66 8 73 7 84 2 75 6 23 8 95 0 8 7 6 4 67 8 44 0 93 3 28 5 15 7 4 9 657884341 50 8 84 7 55 2 82 9 81 8 67 2 6 45 1 33 9 86 3 36 4 93 1 90 8 08 4 6 7 1 9 90 4 31 8 74 3 81 2 83 3 6 3 502795470 28 2 65 3 29 7 80 2 93 4 91 6 1 55 8 11 8 81 0 49 8 44 9 08 3 19 5 4 5 0 0 98 4 83 9 37 7 52 2 72 5 7 0 525785919 44 9 93 8 70 0 73 6 95 7 55 6 8 84 3 69 3 38 1 27 7 96 1 30 8 92 3 0 3 9 2 56 9 69 5 25 3 26 1 62 0 8 2 3676490316 03 6 55 1 37 14 4 79 1 39 3 23 47 1 69 5 66 9 88 06 9

Name : RSA−768 Digits : 232 123018668 45 3 01 1 77 5 51 3 04 9 49 5 8 38 4 96 2 72 0 77 2 85 3 56 9 59 5 33 4 7 92 1 97 3 22 4 52 1 51 7 2 640050726 36 5 75 1 87 4 52 0 21 9 97 8 6 46 9 38 9 95 6 47 4 94 2 77 4 06 3 84 5 9 25 1 92 5 57 3 26 3 03 4 5 373154826 85 0 79 1 70 2 61 2 21 4 29 1 3 46 1 67 0 42 9 21 4 31 1 60 2 22 1 2 4 0 4 79 2 74 7 37 7 94 0 80 6 6 5 351419597459 85 69 0 21 43 41 3

Name : RSA−2048 Digits : 617 251959084 75 6 57 8 93 4 94 0 27 1 83 2 4 00 4 83 9 85 7 14 2 92 8 21 2 62 0 4 0 3 2 02 7 77 7 13 7 83 6 04 3 6 6 202070759 55 5 62 6 40 1 85 2 58 8 07 8 4 40 6 91 8 29 0 64 1 24 9 51 5 08 2 1 8 9 2 98 5 59 1 49 1 76 1 84 5 0 2 808489120 07 2 84 4 99 2 68 7 39 2 80 7 2 87 7 76 7 35 9 71 4 18 3 47 2 70 2 6 1 8 9 63 7 50 1 49 7 18 2 46 9 1 1 650776133 79 8 59 0 95 7 00 0 97 3 30 4 5 97 4 88 0 84 2 84 0 17 9 74 2 91 0 0 6 4 2 45 8 69 1 81 7 19 5 11 8 7 4 612151517 26 5 46 3 22 8 22 1 68 6 99 8 7 54 9 18 2 42 2 43 3 63 7 25 9 08 5 1 4 1 8 65 4 62 0 43 5 76 7 98 4 2 3 387184774 44 7 92 0 73 9 93 4 23 6 58 4 8 23 8 24 2 81 1 98 1 63 8 15 0 10 6 7 4 8 1 04 5 16 6 03 7 73 0 60 5 6 2 016196762 56 1 33 8 44 1 43 6 03 8 33 9 0 44 1 49 5 26 3 44 3 21 9 01 1 46 5 7 5 4 4 45 4 17 8 42 4 02 0 92 4 6 1 651572335 07 7 87 0 77 4 98 1 71 2 57 7 2 46 7 96 2 92 6 38 6 35 6 37 3 28 9 9 1 2 1 54 8 31 4 38 1 67 8 99 8 8 5 0404453640 2 35 2 73 8 19 5 13 7 86 3 65 6 43 9 12 1 20 1 03 9 71 2 28 2 21 2 0 7 2 03 5 7

Name : RSA−896 Digits : 270 412023436 98 6 65 9 54 3 85 5 53 1 36 5 3 32 5 75 9 48 1 79 8 11 6 99 8 44 3 2 7 9 8 28 4 54 5 56 2 64 3 38 7 6 4 455652484 26 1 98 0 98 8 70 4 23 1 61 8 4 18 7 92 6 14 2 02 4 71 8 88 6 94 9 2 5 6 0 93 1 77 6 37 5 03 3 42 1 1 3 098239748 51 5 09 4 49 0 91 0 69 1 02 6 9 86 1 03 1 86 2 70 4 11 4 88 0 86 6 9 7 0 5 64 9 02 9 03 6 53 6 58 8 6 7 4337317208 1 31 0 41 0 51 9 08 6 4 25 4 79 3 28 2 60 1 39 1 25 7 62 4 03 3 94 6 37 3 26 9 39 1 Name : RSA−1024 Digits : 309 135066410 86 5 99 5 22 3 34 9 60 3 21 6 2 78 8 05 9 69 9 38 8 81 4 75 6 05 6 6 7 0 2 75 2 44 8 51 4 38 5 15 2 6 5 106048595 33 8 33 9 40 2 87 1 50 5 71 9 0 94 4 17 9 82 0 72 8 21 6 44 7 15 5 1 3 7 3 68 0 41 9 70 3 96 4 19 1 7 4 304649658 92 7 42 5 62 3 93 4 10 2 08 6 4 38 3 20 2 11 0 37 2 95 8 72 5 76 2 3 5 8 5 09 6 43 1 10 5 64 0 73 5 0 1 508187510 67 6 59 4 62 9 20 5 56 3 68 5 5 29 4 75 2 13 5 00 8 52 8 79 4 16 3 7 7 3 2 85 3 39 0 61 0 97 5 05 4 4 3 34999811150 05 69 7 72 36 8 90 92 75 6 3

RSA

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RSA Factoring Challenge. . .

Name : RSA−704 Digits : 212 740375634 79 5 61 7 12 8 28 0 46 7 96 0 97 4 29 5 7 31 4 25 9 31 8 88 8 92 3 12 8 90 8 49 3 62 3 2 63 8 97 276503402 82 6 62 7 68 9 19 9 64 1 96 2 51 1 78 4 3 99 5 89 4 33 0 50 2 12 7 58 5 37 0 11 8 96 8 0 98 2 86 733173273 10 8 93 0 90 0 5 52 5 05 1 16 8 77 0 6 32 9 90 7 23 9 63 8 07 8 6 71 0 08 6 09 6 96 2 5 37 9 34 6 50 5 63 7 96 3 5 9

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The effort took approximately 30 2.2GHz-Opteron-CPU years according to the submitters, over five months of calendar time. 29/83

✞

☎

1634733645 80 9 25 3 84 8 44 3 13 3 88 3 86 50 9 08 5 98 4 17 8 36 7 00 3 30 9231218111 08 5 23 8 93 3 31 00 1 04 5 08 1 51 2 12 11 8 16 7 51 1 57 9

39807508642 4 06 4 93 7 39 7 12 55 0 05 5 03 8 64 91 1 99 0 64 3 62 34252670840 6 38 5 18 95 7 59 4 63 88 9 57 2 61 7 68 58 3 31 7

✝

✆

✝ ✆

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☎

Name : RSA−640 Digits : 193 310741824 04 9 00 4 37 2 13 5 07 5 00 3 58 8 85 6 79 3 0 03 7 34 6 02 2 84 2 72 7 54 5 72 0 16 1 94 8 82 320644051 80 8 15 0 45 5 63 4 68 2 96 7 17 2 32 8 67 8 2 43 7 91 6 27 2 83 8 03 3 41 5 47 1 07 3 10 8 50 1919548529 0 07 3 37 7 2 48 2 27 8 35 2 57 4 23 8 64 5 40 1 46 9 17 3 66 0 24 7 76 5 23 4 66 0 9

http://www.rsa.com/rsalabs/node.asp?id=2093

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✆ 31/83

RSA

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RSA Security: How to use RSA

Outline 1 2

Two plaintexts M1 and M2 are encrypted into ciphertexts C1 and C2 .

3

But, RSA is deterministic!

4

If C1 = C2 then we know that M1 = M2 ! Also, side-channel attacks are possible against RSA, for example by measuring the time taken to encrypt. 5

6 RSA

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Software – GPG


GPG

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Key generation: Bob > gpg --gen-key

gpg is a public domain implementation of pgp.

Please select what kind of key you want: (1) RSA and RSA (default) (2) DSA and Elgamal (3) DSA (sign only) (4) RSA (sign only) Your selection? 1 What keysize do you want? (2048) Key is valid for? (0) Key does not expire at all Real name: Bobby Email address: [email protected] Comment: recipient You need a Passphrase to protect your secret key. Enter passphrase: Bob rocks Repeat passphrase: Bob rocks

Supported algorithms: Pubkey: RSA, RSA-E, RSA-S, ELG-E, DSA Cipher: 3DES, CAST5, BLOWFISH, AES, AES192, AES256, TWOFISH, CAMELLIA128, CAMELLIA192, CAMELLIA256 Hash: MD5, SHA1, RIPEMD160, SHA256, SHA384, SHA512, SHA224 Compression: Uncompressed, ZIP, ZLIB, BZIP2 http://www.gnupg.org

GPG

.

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GPG

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Key generation: Alice

Exporting the Key

> gpg --gen-key Please select what kind of key you want: (1) RSA and RSA (default) (2) DSA and Elgamal (3) DSA (sign only) (4) RSA (sign only) Your selection? 1 What keysize do you want? (2048) Key is valid for? (0) Key does not expire at all Real name: Alice Email address: [email protected] Comment: sender You need a Passphrase to protect your secret key. Enter passphrase: Alice is cute Repeat passphrase: Alice is cute GPG

> gpg --armor --export Bobby -----BEGIN GPG PUBLIC KEY BLOCK----Version: GnuPG v1.4.11 (Darwin) mQENBE83U28BCADTVOkHpNjWzk7yEzMhiNJcmOtmUYfn4hzgYTDsP2otI0UhfJ4q EZCuPoxECIZ479k3YpBvZM2JC48Ht9j1kVnDPLCrongyRdSko0AwG7OYAyHWa7/U SeGwjZ+0MUuM3SwqHdo1/0XS3P8LABTQNXtrQf9kF8UNLIaHr1IvBcae1K44MPL6 ................................................................ EBHmAM7iiWgWI6/6qEmN46ZQEmoR86vWhQL3LQ6p/FUaBA== =FZ78 -----END GPG PUBLIC KEY BLOCK-----

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Encryption

GPG

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Decryption

We can encrypt a message using Bobby’s key: > cat message Attack at dawn > gpg --recipient bobby --armor --encrypt message > cat message.asc -----BEGIN PGP MESSAGE----Version: GnuPG v1.4.11 (Darwin)

Bobby can now decrypt the message using his private key: > gpg --decrypt message.asc You need a passphrase to unlock the secret key for user: "Bobby (recipient) " 2048-bit RSA key, ID D95291EF, created 2012-02-12 (main key ID 9974031B)

hQEMA97v9lbZUpHvAQf/a9QklXMiMzBWy5yyZBtNrg7FcrIqx+gXVVUXNN86tZtE RF42elwU6QwamDzfcOHqp+3zeor4Y5xN+/pL91xti6uwFOhgGrCGJq//AfUKgQyk MH2e4gR8Y1BuPm9b1c7uzXxRMMOUBBt75KquYGOBLybsP29ttD9iL/ZJl1zSPjSj El7O0Gp7PqEBotStVOtuknYW/fX0zXndU8XNllKnsnZn21Xm0rMQcFMu8Do/tF5I lRfTEcL4S9tV4vshgXhNSpTg9sZs1UZynvU2cJqyYkCtgT7TdtrK3fTa8UN+CYQv U2QRnaNtFhYwBMonFqhefNzDqeZb+P0RqOuoDllYuNJRAViJ3CLjT7kwgBgRtNfY RkGArQQmgrknW2jq/Y2GZTE8CC7pNXY8U3KYMl9hRA6U5fMp08ndFp8vowBbB2sw zjxjSY7ZeIR2uwxdLYydtW4m =B+JA -----END PGP MESSAGE----GPG

Enter passphrase: Bob rocks gpg: encrypted with 2048-bit RSA key, ID D95291EF, created 2012-02-12 "Bobby (recipient) " Attack at dawn

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GPG

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The keyring

The keyring. . .

> gpg --list-keys /Users/collberg/.gnupg/pubring.gpg ---------------------------------pub 2048R/9974031B 2012-02-12 uid Bobby (recipient) sub 2048R/D95291EF 2012-02-12

> gpg --list-secret-keys /Users/collberg/.gnupg/secring.gpg ---------------------------------sec 2048R/9974031B 2012-02-12 uid Bobby (recipient) ssb 2048R/D95291EF 2012-02-12

pub uid sub

sec uid ssb

2048R/4EC8A0CB 2012-02-12 Alice (sender) 2048R/B901E082 2012-02-12

GPG

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Sign and Encrypt

2048R/4EC8A0CB 2012-02-12 Alice (sender) 2048R/B901E082 2012-02-12

GPG

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Check Signature and Decrypt

Bob can sign his message before sending it to Alice:

Alice can now decrypt the message and check the signature:

> gpg -se --recipient alice --armor message > gpg --decrypt message.asc You need a passphrase to unlock the secret key for user: "Bobby (recipient) " 2048-bit RSA key, ID 9974031B, created 2012-02-12

You need a passphrase to unlock the secret key for user: "Alice (sender) " 2048-bit RSA key, ID B901E082, created 2012-02-12 (main key ID 4EC8A0CB)

Enter passphrase: Bob rocks

Enter passphrase: Alice is cute

> cat message.asc -----BEGIN PGP MESSAGE----Version: GnuPG v1.4.11 (Darwin)

GPG

hQEMA7osp1S5AeCCAQgAsSqSs+Urf0f3KHTtP7cqTwugpcJ9oUAGkw/KQ0DHIE0v ................................................................ 8XEAaCwZ8aZK1lXhqBSd/9hCm9Mup2NECihO8crVyff7NTWFyaTBeGAm10q3y46o QpIgPbcdYZqIt8e/8wPU6xlMZUStzxBKLB+Rj/Zg35ZVioYL =oiv8 -----END PGP MESSAGE-----

gpg: encrypted with 2048-bit RSA key, ID B901E082, created 2012-02-12 "Alice (sender) " Attack at dawn gpg: Signature made Sat Feb 11 23:10:59 2012 MST using RSA key ID 9974031B gpg: Good signature from "Bobby (recipient) " 43/83

GPG

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Symmetric Encryption Only

Deleting Keys

> gpg --cipher-algo=AES --armor --symmetric message Enter passphrase: sultana Repeat passphrase: sultana > cat message.asc -----BEGIN PGP MESSAGE----Version: GnuPG v1.4.11 (Darwin)

> gpg --delete-secret-keys bobby sec 2048R/9974031B 2012-02-12 Bobby (recipient) Delete this key from the keyring? (y/N) y This is a secret key! - really delete? (y/N) y

jA0EBwMCgZ3PBfSZxJlg0ksBBooTMLEVQ2q9HkTR5y9FIoX9nbsyohrOXeQLFlcf wtWcg+dZvlMS6D7OE3wZCeW2LX50kYcU17MUc8wnJLDAzAdRqPAgDma+sP4= =UtI4 -----END PGP MESSAGE-----

> gpg --delete-keys bobby pub 2048R/9974031B 2012-02-12 Bobby (recipient)

> gpg message.asc gpg: AES encrypted data Enter passphrase: sultana gpg: encrypted with 1 passphrase

Delete this key from the keyring? (y/N) y

> cat message Attack at dawn GPG

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Generating Primes

GPG

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Generating Random Numbers

Generate a prime number of the given number of bits: Generate 100 (base64 encoded) random bytes:

> gpg --gen-prime 1 16 C4B7 > gpg --gen-prime 1 1024 D34D4347ED013242EE06811BC561C6587D75ADE33D1BEC954D648E22 9D88B5E0AF1394459FB48B135B99C8BA8C50E5331C6226CBF6D70031 4A8CC84C7B363BE7DD7BBBB29E545D199339263F5FB2E9F1B84BA9D5 05B5B79858FC6149CF09E6C56D9730C3BD1E62B378C8DFAF4233B8DC BA999A21EC9C4BF8C60AACDCBC607AC5

GPG

> gpg --armour --gen-random 0 100 e0zAVl6jbe/Dma9VF20lMgZxE1RA4S8TwNwu6KP8+o1kjdtBm2 AjKFSVsj/d3zG/9KqmNj7j6symEUZ3e0fWZaWqLBxzJuSur5sK C8omfPus2QtYJJNOgVbpJ7X9L4t1iNJtnw==

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GPG

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Print Message Digests

Goal: Read a message encrypted with gpg

> gpg --print-mds message MD5 = 36 D1 A5 12 17 CD 34 FC 04 F5 6C C4 91 39 C7 59 SHA1 = 6DA4 473A 00CE 7AB6 7B6F 884D 1E75 6633 C21A 56DB RMD160 = D1DE 4194 C0CD 3AED 30F3 38CD 68F3 800F CCF0 3B87 SHA224 = B4E94780 1AA1A9C3 418F72D8 651BA995 83284003 EBEE183A 589702EE SHA256 = B83EF405 07696578 9D4BBDA7 D7932700 5F2AE6CB A2696FDE 69694D12 AFE70E4A SHA384 = 7AC39A0C 945844F1 1316BB46 C9FC7EEA E892A178 2D20E4CA E7BE686C 1A091C8C F1BBDFD1 3D42BEA2 88AF5A4F E3705474 SHA512 = 9CA1EB88 F064CB0D 536254B2 5755919F 45564276 96CA27A0 389E4817 53F81DC2 3222488D 7D11F3DD C066B9E8 027F3870 395A2561 157DDC38 BD679D37 C2E361CC

GPG

Decrypt the message itself (OR)

2

Determine symmetric key used to encrypt the message by other means (OR)

3

Get recipient to help decrypt message (OR)

4

Obtain private key of recipient.

http://www.schneier.com/paper-attacktrees-fig7.html

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Goal: Read a message encrypted with gpg. . .

GPG

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Goal: Read a message encrypted with gpg. . . Determine symmetric key by other means: 1 Fool sender into encrypting message using public key whose private key is known (OR)

Decrypt the message itself: 1 Break asymmetric encryption (OR) 1 2

1

Convince sender that fake key (with known private key) is the key of the intended recipient 2 Convince sender to encrypt with more than one key—the real key of the recipient and a key whose private key is known. 3 Have the message encrypted with a different public key in the background, unbeknownst to the sender. 1

Brute force break asymmetric encryption (OR) Mathematically break asymmetric encryption (OR) 1 Break RSA (OR) 2 Factor RSA modulus/calculate Elgamal discrete log

3

Cryptanalyze asymmetric encryption (OR) 1 General cryptanalysis of RSA/Elgamal (OR) 2 Exploit weakness in RSA/Elgamal (OR) 3 Timing attack on RSA/Elgamal

2

2 3 4

Break symmetric-key encryption

5

Brute force break symmetric-key encryption 2 Cryptanalysis of symmetric-key encryption 1

Determine state of randseed during encryption (OR) Implant virus that alters the state of randseed. (OR) 3 Implant software that affects the choice of symmetric key. 1 2 6

GPG

Have the recipient sign the encrypted publc key (OR) Monitor the sender’s computer memory (OR) Monitor the receiver’s computer memory (OR) Determine key from pseudo-random number generator (OR)

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GPG

Implant virus that that exposes public key. 52/83



Get recipient to help decrypt message:

GPG

Obtain private key of recipient:

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Goal: Read a message encrypted with PGP

GPG

Goal: Read a message encrypted with PGP. . .

What immediately becomes apparent from the attack tree is that breaking the RSA or IDEA encryption algorithms are not the most profitable attacks against PGP. There are many ways to read someone’s PGP-encrypted messages without breaking the cryptography. You can capture their screen when they decrypt and read the messages (using a Trojan horse like Back Orifice, a TEMPEST receiver, or a secret camera), grab their private key after they enter a passphrase (Back Orifice again, or a dedicated computer virus), recover their passphrase (a keyboard sniffer, TEMPEST receiver, or Back Orifice), or simply try to brute force their passphrase (I can assure you that it will have much less entropy than the 128-bit IDEA keys that it generates). GPG

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In the scheme of things, the choice of algorithm and the key length is probably the least important thing that affects PGP’s overall security. PGP not only has to be secure, but it has to be used in an environment that leverages that security without creating any new insecurities. http://www.schneier.com/paper-attacktrees-fig7.html

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GPG

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Outline 1 2

3 4

5

6

Elgamal


The Elgamal cryptosystem relies on the inherent difficulty of calculating discrete logarithms. It is a probabilistic scheme: a particular plaintext can be encrypted into multiple different ciphertexts; ⇒ ciphertexts become twice the length of the plaintext.

Elgamal

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Elgamal: Algorithm

Elgamal

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Elgamal: Algorithm Notes

Bob (Key generation): Pick a prime p. Find a generator g for Zp . 3 Pick a random number x between 1 and p − 2. 4 Compute y = g x mod p. PB = (p, g , y ) is Bob’s RSA public key. SB = x is Bob’ RSA private key. 1 2

Alice must choose a different random number k for every message, or she’ll leak information. Bob doesn’t need to know the random value k to decrypt. Each message has p − 1 possible different encryptions.

Alice (encrypt and send a message M to Bob):

The division in the decryption can be avoided by use of Lagrange’s theorem :

Get Bob’s public key PB = (p, g , y ). 2 Pick a random number k between 1 and p − 2. 3 Compute the ciphertext C = (a, b): 1

a b

= =

M = b · (ax )−1 mod p

g k mod p My k mod p

= b · ap−1−x mod p

Bob (decrypt a message C = (a, b) received from Alice): 1 Elgamal

Compute M = b(ax )−1 mod p. 59/83

Elgamal

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Elgamal: Finding the generator

Elgamal Example: Key generation

Computing the generator is, in general, hard.

1

Pick a prime p = 13.

We can make it easier by choosing a prime number with the property that we can factor p − 1.

2

Find a generator g = 2 for Z13 (see next slide).

3

Pick a random number x = 7.

Then we can test that, for each prime factor pi of p − 1:

4

Compute y = g x mod p = 27 mod 13 = 11.

g (p−1)/pi mod p 6= 1 If g is not a generator, then one of these powers will 6= 1.

Elgamal

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Powers of Integers, Modulo 13

Elgamal

a2 1 4 9 3 12 10 10 12 3 9 4 1

a3 1 8 1 12 8 8 5 5 1 12 5 12

a4 1 3 3 9 1 9 9 1 9 3 3 1

a5 1 6 9 10 5 2 11 8 3 4 7 12

a6 1 12 1 1 12 12 12 12 1 1 12 1

a7 1 11 3 4 8 7 6 5 9 10 2 12

PB = (p, g , y ) = (13, 2, 11) is Bob’s public key.

6

SB = x = 7 is Bob’ private key.

Elgamal

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Elgamal Example: Encryption

2 is a primitive root modulo 13 because for each integer i ∈ Z13 = {1, 2, 3, . . . , 12} there’s an integer k, such that i = 2k mod 13: a1 1 2 3 4 5 6 7 8 9 10 11 12

5

a8 1 9 9 3 1 3 3 1 3 9 9 1

a9 1 5 1 12 5 5 8 8 1 12 8 12

a10 1 10 3 9 12 4 4 12 9 3 10 1

a11 1 7 9 10 8 11 2 5 3 4 6 12

Encrypt the plaintext message M = 3. Alice gets Bob’s public key PB = (p, g , y ) = (13, 2, 11). To encrypt:

a12 1 1 1 1 1 1 1 1 1 1 1 1

1 2

Pick a random number k = 5: Compute: a b

= =

g k mod p = 25 mod 13 = 6 My k mod p = 3 · 115 mod 13 = 8

The ciphertext C = (a, b) = (6, 8).

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Elgamal

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Elgamal Example: Decryption

In-Class Exercise

Bob’s private key is SB = x = 7.

Pick the prime p = 13.

Bob receives the ciphertext C = (a, b) = (6, 8) from Alice.

Find the generator g = 2 for Z13 .

Bob computes the plaintext M:

Pick a random number x = 9. Compute

M = b · (ax )−1 mod p

y = g x mod p = 29 mod 13 = 5

= b · ap−1−x mod p

PB = (p, g , y ) = (13, 2, 5) is Bob’s public key.

= 8 · 613−1−7 mod 13

SB = x = 9 is Bob’ private key.

= 8 · 65 mod 13 = 3

Elgamal

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Elgamal Correctness

1

Encrypt the message M = 11 using the random number k = 10.

2

Decrypt the ciphertext from 1.

Elgamal

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Elgamal Security

Show that M = b · (ax )−1 mod p decrypts. We have that a = g k mod p b = My k mod p y

= g x mod p

The security of the scheme depends on the hardness of solving the discrete logarithm problem.

We get

Generally believed to be hard.

b · (ax )−1 mod p = (My k ) · ((g k )x )−1 mod p = (My k ) · (g kx )−1 mod p = (M((g x )k ) · (g kx )−1 mod p = Mg kx · (g kx )−1 mod p = Mg kx · g −kx mod p = M mod p Elgamal

= M

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Elgamal

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Outline 1 2

3 4

5

6

Key Exchange


Diffie-Hellman Key Exchange

A key exchange protocol (or key agreement protocol ) is a way for parties to share a secret (such as a symmetric key) over an insecure channel. With an active adversary (who can modify messages) we can’t reliably share a secret. With a passive adversary (who can only eavesdrop on messages) we can share a secret. A passive adversary is said to be honest but curious .

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Diffie-Hellman: Algorithm 1

All parties (set-up): 1 2

2

2

Pick a random x ∈ Zp , x > 0. Compute X = g x mod p.

3

Send X to Bob.

Based on modular exponentiation . The secret K1 = K2 shared by Alice and Bob at the end of the protocol would typically be a shared symmetric key.

3

4 5 71/83

Pick p, a prime number. Pick g , a generator for Zp .

Alice : 1

A classic key exchange protocol.


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Bob : 1 2

Pick a random y ∈ Zp , x > 0. Compute Y = g y mod p.

3

Send Y to Alice

Alice computes the secret: K1 = Y x mod p. Bob computes the secret: K2 = X y mod p.


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Example

In-Class Exercise

1

Pick p = 13, a prime number.

2

Pick g = 2, a generator for Z13 . Alice :

3

Let p = 19. Let g = 10.

Pick a random x = 3. 2 Compute X = g x mod p = 23 mod 13 = 8. 1 4

Let Bob’s secret y = 15.

Bob : 1 2

5

Let Alice’s secret x = 7.

Pick a random y = 7. Compute Y = g y mod p = 27 mod 13 = 11.

Alice computes: K1 =

6

Bob computes: K2 =

7

⇒ K1 = K2 = 5.

Yx

Xy

mod p = mod p =

113

87

mod 13 = 5.

1

Compute K1 .

2

Compute K2 .

mod 13 = 5.


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Diffie-Hellman Correctness


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Diffie-Hellman Correctness. . . Alice has

Alice has computed

K1 = Y x mod p X

= (g y )x mod p

= g x mod p

= (g x )y mod p

K1 = Y x mod p.

= X y mod p

Bob has computed Y

Bob has

= g y mod p

K2 = X y mod p

K2 = X y mod p.

= (g x )y mod p = X y mod p ⇒ K1 = K2 .


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Diffie-Hellman Security

Diffie-Hellman: Man-In-The-Middle attack

1

The security of the scheme depends on the hardness of solving the discrete logarithm problem.

Alice : 1

2

Eve : Intercept X = g x mod p from Alice. Pick a number t in Zp . 3 Send T = g t mod p to Bob.

Generally believed to be hard. Diffie-Hellman Property :

1 2

Given p, X = g x , Y = g y

3

Bob : 1

computing

4

K = g xy mod p

Send Y = g y mod p to Alice

Eve : Intercept Y = g y mod p from Bob. Pick a number s in Zp . 3 Send S = g s mod p to Alice. 1 2

is thought to be hard.


Send X = g X mod p to Bob.

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Diffie-Hellman: Man-In-The-Middle attack. . .


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Diffie-Hellman: Man-In-The-Middle attack. . .

7 8

Alice : Send C = EK1 (M) to Bob Eve : Intercept C . Decrypt:M = DK1 (C ) Re-encrypt:C ′ = EK2 (M) 4 Send C ′ to Bob 1

5

Alice and Eve : 1

6

2 3

Compute K1 = g xS mod p

Bob and Eve : 1

Compute K2 = g yT mod p

9 10

Bob : Send C = EK2 (M) to Alice Eve : Intercept C . Decrypt:M = DK2 (C ) 3 Re-encrypt:C ′ = EK1 (M) 4 Send C ′ to Alice. 1 2


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Outline 1 2

3 4

5

6

Readings and References


Summary

Chapter 8.1.1-8.1.5 in Introduction to Computer Security, by Goodrich and Tamassia.

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Acknowledgments Additional material and exercises have also been collected from these sources:

Summary

1

Igor Crk and Scott Baker, 620—Fall 2003—Basic Cryptography.

2

Bruce Schneier, Applied Cryptography.

3

Pfleeger and Pfleeger, Security in Computing.

4

William Stallings, Cryptography and Network Security.

5

Bruce Schneier, Attack Trees, Dr. Dobb’s Journal December 1999, http://www.schneier.com/paper-attacktrees-ddj-ft.html .

6

Barthe, Gr´egoire, Beguelin, Hedin, Heraud, Olmedo, Verifiable Security of Cryptographic Schemes, http://www.irisa.fr/celtique/blazy/seminar/20110204.pdf .

7

http://homes.cerias.purdue.edu/~crisn/courses/cs355_Fall_2008/lect18.pdf 83/83

Summary

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Encrypting and Decrypting

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