Fair Cryptography. Cryptography CS 507 Erkay Savas Sabanci University
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1 Fair Cryptography Cryptography CS 507 Erkay Savas Sabanci University
2 Conflicting Interests Private Industry wants to secure communications Citizens want privacy Law enforcement wants wiretapping ability Law enforcement sees encryption as a threat. 1/13/2003 Erkay Savas 2
3 Capstone Project U.S Government cryptography standards Computer Security Act of 1987 Contains Digital signature: DSS Hash function: SHS Data encryption: DES, Skipjack and Clipper Key escrowing procedure Key exchange: not specified 1/13/2003 Erkay Savas 3
4 Players NIST National Institute of Standards and Technology Selects and propose official standards NSA National Security Agency Designs cryptographic algorithms VLSI Technologies, Inc. Manufactures the chip Company no longer exists Mykotronx, Inc. Programs the chip. 1/13/2003 Erkay Savas 4
5 Escrowing Merriam-Webster gives the definition: es crow es-"kro, es- Etymology: Middle French escroue scroll Date: 1594 A deed, a bond, money, or a piece of property held in trust by a third party to be turned over to the grantee only upon fulfillment of a condition 1/13/2003 Erkay Savas 5
6 Key Escrowing Decryption key is stored with two escrow agency Identities of the escrow agencies are not yet decided. Escrow agencies give the key to a law enforcement agency which has the proper authorization Each escrow agency stores 80-bit strings. XORing the two pieces of 80-bit to get the decryption key. 1/13/2003 Erkay Savas 6
7 Overview of Clipper Proposal Key is escrowed at time of manufacture Clipper is a tamper proof hardware Use is voluntary in private sector 1/13/2003 Erkay Savas 7
8 Keys Session Key K Encrypts the message Unique to each session Unit Key U Encrypts the session key. Embedded on chip, escrowed Family Key F Encrypts law enforcement block Common to all users 1/13/2003 Erkay Savas 8
9 Secure Communication Agree on a session key K Protocol unspecified A public-key method, e.g. Diffie-Hellman key exchange algorithm can be used Message M is encrypted with K Message block: E(M, K) K is not escrowed. 1/13/2003 Erkay Savas 9
10 Law enforcement block: LEAF LEAF Law Enforcement Access Field. Session key is encrypted with unit key: E(K, U) Serial number of sender: N Authentication string: A All encrypted with family key: F E({E(K, U), N, A}, F) LEAF is sent at least once during session 1/13/2003 Erkay Savas 10
11 Receiver Decrypts LEAF to get authentication string Checks authentication string Decrypts message using session key K Authentication string prevents users removing LEAF Details are not specified. 1/13/2003 Erkay Savas 11
12 Court-authorized Wiretapping 1. Uses family key to decrypt the law enforcement block 2. Now has serial number and encrypted session key 3. Obtains wiretap authorization from court 4. Presents authorization to escrow agencies 5. Obtains unit key to decrypt session key K 6. Uses session key to decrypt message 1/13/2003 Erkay Savas 12
13 Skipjack Algorithm Block cipher 64-bit input, 64-bit output 80-bit key 32 rounds of a nonlinear function Same modes as DES DES: 64-bit IO, 56-bit key, 16 rounds 1/13/2003 Erkay Savas 13
14 Skipjack is classified Not implemented in software Only authorized manufacturers can implement it in hardware Not open to public scrutiny One-month review by a team of five independent cryptographers: Secure E. F. Brickell Sandia Labs D. E. Denning Georgetown University S. T. Kent BBN Corporation D. P. Maher AT & T W. Tuchman Amperif Corporation 1/13/2003 Erkay Savas 14
15 Unit Key Generation Details are not specified Programmed inside a vault at chip factory Two agents are involved Each agent generates random 80-bit string: S 1 and S 2. Serial number N padded to get three 64-bit strings: N 1, N 2, and N 3. 1/13/2003 Erkay Savas 15
16 Unit Key Generation Triple encryption to get three 64-bit blocks R 1 = E(D(E(N 1, S 1 ), S 2 ), S 1 ) R 2 = E(D(E(N 2, S 1 ), S 2 ), S 1 ) R 3 = E(D(E(N 3, S 1 ), S 2 ), S 1 ) Concatenate: R = R 1 R 2 R 3 Two 80-bit strings U 1 and U 2 are taken from R. Unit key U = U 1 U 2 Escrow agencies get U 1 and U 2. 1/13/2003 Erkay Savas 16
17 Security of Clipper System Points of attack (external or internal) Escrow agencies Chip factory Courts Other abuses Wiretap beyond authorized period Intentional trapdoor 1/13/2003 Erkay Savas 17
18 Advantages vs. Disadvantages Advantages of key escrowing Police can wiretap suspected criminals Citizens obtain secure communication Disadvantages of key escrowing Government invasion of privacy Criminals will use unescrowed cryptography Hinders economic competitiveness in security and communications 1/13/2003 Erkay Savas 18
19 Clipper Politics Large opposition Distrust of government Industry sees restriction Is it really effective against criminals? 1/13/2003 Erkay Savas 19
20 Two Levels of Debate 1. Policy Debate about key escrowing. Good or bad? How widespread? 2. Implementation Debate about a given system e.g. Clipper 1/13/2003 Erkay Savas 20
21 Fair Public-Key Cryptography Public-key cryptosystems with escrowed keys Proposed by Silvio Micali of MIT Software implementation Users generate their own keys Users choose their cryptosystems Fair cryptography does not require Tamper-proof hardware Secret algorithm 1/13/2003 Erkay Savas 21
22 Desired Properties Secure communication for ordinary user Access to message by authorized agency Audit trail: No undetectable or unauthorized wiretaps. Law enforcement knows it can access message with authorization Time-limited access 1/13/2003 Erkay Savas 22
23 Overview of Fair Cryptography Choose public-key cryptosystem Distribute pieces of private key to escrow agencies Escrow agencies can verify that they have pieces of private key corresponding to user s public key without disclosing pieces of key. 1/13/2003 Erkay Savas 23
24 Diffie-Hellman Prime p and element g common to all users Alice: Private key x, public key A = g x (mod p) Bob: Private key y, public key B = g y (mod p) Alice and Bob both calculate the shared key: K = g xy (mod p) 1/13/2003 Erkay Savas 24
25 Fair Diffie-Hellman Assume 5 trustees (escrow agencies) Alice chooses 5 secret numbers x 1, x 2, x 3, x 4, x 5, [1, p-1] Let x = Σ x i (mod p-1) Private key: x Public key: A = g x (mod p) x i s are private pieces of key Public pieces of key t i = g x i (mod p) 1/13/2003 Erkay Savas 25
26 Fair Diffie-Hellman KDC gets t Trustee i gets x i and t i It signs t i if t i = g x i (mod p) Sends t i to KDC After receiving all five public pieces, the KDC verifies that t = (t 1 t 2 t 3 t 4 t 5 ) (mod p) If it does, it approves the public key Alice and Bob can communicate as usual. 1/13/2003 Erkay Savas 26
27 RSA Cryptosystem Pick two primes: p and q. The modulus is the product: n = pq Euler s function Φ (n) = (p-1)(q-1) Select e (1, Φ (n)) with gcd(e, Φ (n)) = 1 Compute d e -1 mod Φ (n) Public key: e Private key: d Encryption: C M e mod n Decryption: M C d mod n 1/13/2003 Erkay Savas 27
28 Toward Fair RSA Let p be a prime and t x 2 mod p The equation x 2 t mod p has two solutions ±x. If p 3 mod 4, then x can be found easily x t (p+1)/4 mod p If n = pq, then the equation x 2 t mod n has four solutions ±x and ±y. Solving x 2 t mod n is hard; If you can find both roots x and y, then you can factor n. 1/13/2003 Erkay Savas 28
29 Towards Fair RSA Sincex 2 t mod n and y 2 t mod n, we have x 2 - y 2 0 mod n (x - y)(x + y) 0 mod n Thus, either gcd(n, x y) or gcd(n, x + y) is a nontrivial factor of n. Jacobi symbol: J(a, n) = +1 or 1 Jacobi symbol is multiplicative: J(ab, n)=j(a, n) J(b, n) Of the four square roots modulo n, two are Jacobi +1, and two are Jacobi 1. If you know a Jacobi +1 root and Jacobi 1 root of t (1, n), then you can can factor n and find the private key d. 1/13/2003 Erkay Savas 29
30 Fair RSA Alice chooses primes p and q n = pq and p, q 3 mod 4 Alice chooses five Jacobi +1 integers x 1, x 2, x 3, x 4, x 5, Z * n where Z * n is set of integers modulo n which is relatively prime to n. Let x x 1 x 2 x 3 x 4 x 5 mod n and s x 12 x 22 x 32 x 42 x 52 mod n 1/13/2003 Erkay Savas 30
31 Fair RSA Alice computes y, a Jacobi 1 root of s y 2 s Πx i2 mod n (Π x i ) 2 mod n x 2 x and y are Jacobi +1 and Jacobi 1 roots of s, respectively. Anyone who knows x and y can factor n. Alice sends x i and y to trustee i. Trustee checks that x i is Jacobi +1 and that y is Jacobi -1 1/13/2003 Erkay Savas 31
32 Fair RSA By sharing the x i2 s, trustees can verify that (Π x i ) 2 y mod n and thus that Π x i is a Jacobi +1 root of s y 2 mod n. Trustees together can produce a Jacobi +1 and a Jacobi 1 root of s. Therefore, trustees can factor n, and recover private key. Cooperation of all 5 trustees is necessary to factor n. 1/13/2003 Erkay Savas 32
33 Clipper vs. Fair PKC Clipper: Requires users to trust key generation (distribution) center Requires special hardware No software implementation Automatic key registration Process is invisible to users Possible abuse by authorities 1/13/2003 Erkay Savas 33
34 Clipper vs. Fair PKC Fair PKC: Users generate own keys No secure hardware is needed Software implementations are possible Key registration is more complicated Users have more responsibility Abuse by users is more of a concern 1/13/2003 Erkay Savas 34
35 Additional Features in Fair PKC A subset of trustees can construct private key in case a trustee loses data or is not available Threshold schemes Time-limited access Keys with expiration periods 1/13/2003 Erkay Savas 35
36 Open Questions How can cryptography best meet the needs of society? Will the ordinary cryptography become illegal? Key escrowing is ineffective against criminals unless ordinary cryptography is illegal. Privacy will be the first roadkill along the information superhighway (Brock N Meeks, April 1994) Ordinary cryptography would be legal but unusual and unsupported Criminals may use unescrowed keys even if it is illegal to do so. 1/13/2003 Erkay Savas 36
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