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Foundations of Cryptography Lecture 11 Lecturer: Moni Naor

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Recap of Lecture 10 Pseudo-randomness of subset sum Composing pseudo-random generators Hybrid arguments The next-bit test Pseudo-random functions

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Next-bit Test Definition : a function g:{0,1} * → {0,1}* is said to pass the next bit test if It is polynomial time computable It stretches the input |g(x)|>|x| – denote by ℓ(n) the length of the output on inputs of length n If the input (seed) is random, then the output passes the next-bit test For any prefix 0≤ i< ℓ(n), for any probabilistic polynomial time adversary A that receives the first i bits of y= g(x) and tries to guess the next bit, or any polynomial p(n) and sufficiently large n |Prob[A(y i,y 2,…, y i )= y i+1 ] – 1/2 | < 1/p(n) Theorem : a function g:{0,1} * → {0,1}* passes the next bit test if and only if it is a pseudo-random generator

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Next- block Undpredictable Suppose that the function G maps a given a seed into a sequence of blocks let ℓ(n) be the length of the number of blocks given a seed of length n If the input (seed) is random, then the output passes the next-block unpredicatability test For any prefix 0≤ i< ℓ(n), for any probabilistic polynomial time adversary A that receives the first i blocks of y= g(x) and tries to guess the next block y i+1, for any polynomial p(n) and sufficiently large n |Prob[A(y 1,y 2,…, y i )= y i+1 ] | < 1/p(n) Homework : show how to convert a next-block unpredictable generator into a pseudo-random generator. G: S y 1 y 2, …,

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Pseudo-Random Generators concrete version G n : 0,1 m 0,1 n A cryptographically strong pseudo-random sequence generator - if passes all polynomial time statistical tests (t, )- pseudo-random - no test A running in time t can distinguish with advantage

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Three Basic issues in cryptography Identification Authentication Encryption Solve in a shared key environment S S

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Identification - Remote login using pseudo-random sequence A and B share key S 0,1 k In order for A to identify itself to B Generate sequence G n (S) For each identification session - send next block of G n (S) G n (S) G: S

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Problems... More than two parties Malicious adversaries - add noise Coordinating the location block number Better approach: Challenge-Response

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Challenge-Response Protocol B selects a random location and sends to A A sends value at random location What’s this?

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Desired Properties Very long string - prevent repetitions Random access to the sequence Unpredictability - cannot guess the value at a random location –even after seeing values at many parts of the string to the adversary’s choice. –Pseudo-randomness implies unpredictability Not the other way around for blocks

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Authenticating Messages A wants to send message M 0,1 n to B B should be confident that A is indeed the sender of M One-time application: S a,b) - where a,b R 0,1 n To authenticate M: supply aM b Computation is done in GF[2 n ]

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Problems and Solutions Problems - same as for identification If a very long random string available - –can use for one-time authentication –Works even if only random looking a,b Use this!

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Encryption of Messages A wants to send message M 0,1 n to B only B should be able to learn M One-time application: S a - where a R 0,1 n To encrypt M send a M

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Encryption of Messages If a very long random looking string available - –can use as in one-time encryption Use this!

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Pseudo-random Functions Concrete Treatment: F: 0,1 k 0,1 n 0,1 m key Domain Range Denote Y= F S (X) A family of functions Φ k ={F S | S 0,1 k is (t, , q)- pseudo-random if it is Efficiently computable - random access and...

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(t, ,q)- pseudo-random The tester A that can choose adaptively –X 1 and get Y 1 = F S (X 1 ) –X 2 and get Y 2 = F S (X 2 ) … –X q and get Y q = F S (X q ) Then A has to decide whether – F S R Φ k or – F S R R n m = F | F : 0,1 n 0,1 m

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(t, ,q)- pseudo-random For a function F chosen at random from (1) Φ k ={F S | S 0,1 k (2) R n m = F | F : 0,1 n 0,1 m For all t -time machines A that choose q locations and try to distinguish (1) from (2) Prob A ‘1’ F R F k - Prob A ‘1’ F R R n m

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Equivalent/Non-Equivalent Definitions Instead of next bit test: for X X 1,X 2, , X q chosen by A, decide whether given Y is –Y= F S (X) or –Y R 0,1 m Adaptive vs. Non-adaptive Unpredictability vs. pseudo-randomness A pseudo-random sequence generator g: 0,1 m 0,1 n –a pseudo-random function on small domain 0,1 log n 0,1 with key in 0,1 m

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Application to the basic issues in cryptography Solution using a shared key S Identification: B to A: X R 0,1 n A to B: Y= F S (X) A verifies Authentication: A to B: Y= F S (M) replay attack Encryption: A chooses X R 0,1 n A to B:

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Goal Construct an ensemble {Φ k | k L such that for any {t k, 1/ k, q k | k L polynomial in k, for all but finitely many k’s Φ k is a (t k, k, q k )- pseudo-random family

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Construction Construction via Expansion –Expand n or m Direct constructions

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Effects of Concatenation Given ℓ Functions F 1, F 2, , F ℓ decide whether they are –ℓ random and independent functions OR –F S 1, F S 2, , F S ℓ for S 1,S 2, , S ℓ R 0,1 k Claim: If Φ k ={F S | S 0,1 k is (t, ,q)- pseudo-random: cannot distinguish two cases –using q queries –in time t’=t - ℓ q –with advantage better than ℓ

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Proof: Hybrid Argument i=0 F S 1, F S 2, , F S ℓ p 0 … i R 1, R 2, , R i-1,F S i, F S i+1, , F S ℓ p i … i=ℓ R 1, R 2, , R ℓ p ℓ p ℓ - p 0 i p i+1 - p i /ℓ

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...Hybrid Argument Can use this i to distinguish whether – F S R Φ k or F S R R n m Generate F S i+1, , F S ℓ Answer queries to first i-1 functions at random (consistently) Answer query to F S i, using (black box) input Answer queries to functions i+1 through ℓ with F S i+1, , F S ℓ Running time of test - t’ ℓ q

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Doubling the domain Suppose F (n) : 0,1 k 0,1 n 0,1 m which is (t, ,q)- p.r. Want F (n+1) : 0,1 k 0,1 n+1 0,1 m which is (t’, ’,q’)- p.r. Use G: 0,1 k 0,1 2k which is (t, ) p.r G(S) G 0 (S) G 1 (S) Let F S (n+1) (bx) F G b (s) (n) (x)

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Claim If G is (t q, 1 ) -p.r and F (n) is (t 2q, 2,q) -p.r, then F (n+1) is (t, 1 2 2,q) -p.r Proof: three distributions (1) F (n+1) (2) F S 0 (n), F S 1 (n) for independent S 0, S 1 (3) Random D 1 2 2

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...Proof Given that (1) and (3) can be distinguished with advantage 1 2 2, then either (1) and (2) with advantage 1 –G can be distinguished with advantage 1 or (2) and (3) with advantage 2 2 –F (n) can be distinguished with advantage 2 Running time of test - t’ q

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Getting from G to F (n) Idea: Use recursive construction F S (n) (b n b n-1 b 1 ) F G b 1 (s) (n-1) (b n-1 b n-2 b 1 ) G b n (G b n-1 ( G b 1 (S)) ) Each evaluation of F S (n) (x) : n invocations of G

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Tree Description G 0 (S) G 1 (S) S G 0 (G 0 (S)) G 1 (G 0 (G 0 (S))) Each leaf corresponds to an X. Label on leaf – value of pseudo- random function

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Security claim If G is (t qn, ) p.r, then F (n) is (t, ’ n q ,q) p.r Proof: Hybrid argument by levels D i : – truly random labels for nodes at level i. – Pseudo-random from i down Each D i - a collection of q functions i p i+1 - p i ’/n q

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Hybrid S0S0 S1S1 ?S?S G 0 (S 0 ) G 1 (G 0 (S 0 )) n-i i Di

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…Proof of Security Can use this i to distinguish concatenation of q sequence generators G from random. The concatenation is (t,q ) p.r Therefore the construction is (t, ,q) p.r

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Disadvantages Expensive - n invocations of G Sequential Deterioration of But does the job! From any pseudo-random sequence generator construct a pseudo-random function. Theorem: one-way functions exist if and only if pseud- random functions exist.

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Applications of Pseudo-random Functions Learning Theory - lower bounds –Cannot PAC learn any class containing pseudo-random function Complexity Theory - impossibility of natural proofs for separating classes. Any setting where huge shared random string is useful Caveat: what happens when the seed is made public?

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Application to Signatures Shared secret seed - can get authentication What about public-key? Can we use the techniques? Yes!? – Private key is S – Public key is commitment to F S –To sign M - provide F S (M) and a proof of consistency with the commitment

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Pseudo-Random Permutations Block-Ciphers : Shared-key encryption schemes where: the encryption of every plaintext block is a ciphertext block of the same length. ey CC Plaintext Ciphertext

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Block Ciphers Advantages –Saves up on memory and communication bandwidth –Easy to incorporate within existing systems. Main Disadvantage –Every block is always encrypted in the same way. Important Examples: DES, AES

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Modeling Block Ciphers Pseudo-random Permutations F : 0,1 k 0,1 n 0,1 n Key Domain Range F -1 : 0,1 k 0,1 n 0,1 n Key Range Domain Want: –X= F S -1 (F S (X)) Correct inverse –Efficiently computable

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The Test The tester A that can choose adaptively –X 1 and get Y 1 = F S (X 1 ) –Y 2 and get X 2 = F S -1 (Y 2 ) … –X q and get Y q = F S (X q ) Then A has to decide whether – F S R Φ k or – F S R P (n) = F | 1-1 F : 0,1 n 0,1 n Can choose to evaluate or invert any point!

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(t, ,q)- pseudo-random For a function F chosen at random from (1) Φ k ={F S | S 0,1 k (2) P (n) = F | 1-1 F : 0,1 n 0,1 n For all t-time machines A that choose q locations and try to distinguish (1) from (2) Pr A= ‘1’ F R F k - Pr A= ‘1’ F R P (n)

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Construction of Pseudo-Random Permutations Possible to construct p.r. permutation from p.r. functions (and vice versa..) Based on 4 Feistal Permutations

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Feistal Permutation Any f : 0,1 n 0,1 n defines a Feistal Permutation D f (L,R)=(R, L f(R)) Feistal permutations are as easy to invert as to compute: D f -1 (L,R)=(R f(L),L) Many Block Cipher based on such permutations where the function f is derived from secret key

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Feistal Permutation f L1L1 R1R1 L2L2 R2R2

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Composing Feistal Permutations Make the function f: 0,1 n 0,1 n a pseudo-random function F S R Φ k = {F S | S 0,1 k This defines a keyed family of permutations 0,1 2n 0,1 2n Clearly it is not pseudo-random –Right block goes unchanged to left block What about composing two such keyed permutations With independent keys Not pseudo-random: D S 2 (D S 1 (L,R)= (F S 1 (L) R, F S 2 (F S 1 (L) R) R) -For two inputs sharing the same left block Looks pretty good for random attacks!

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Main Construction Let F 1, F 2,F 3,F 4 R PRF, then the composition of D F 1, D F 2, D F 3, D F 4 is a pseudo-random permutation. Each F i : 0,1 n 0,1 n Resulting Permutation 0,1 2n 0,1 2n. F 1 and F 4 can be ``combinatorial”: –pair-wise independent. –low probability of collision on first block Error probability is ~ q 2 /2 n

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References Blum-Micali : SIAM J. Computing 1984 Yao: Blum, Blum, Shub: SIAM J. Computing, 1988 Goldreich, Goldwasser and Micali: J. of the ACM, 1986 Luby-Rackoff: SIAM J. Computing, 1988 Naor-Reingold: Journal of Cryptology, 1999

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...References O. Goldreich, The Foundations of Cryptography - a book in preparation, www.wisdom.weizmann.ac.il/~oded/foc-book.html M. Luby, Pseudorandomness and Cryptographic Applications, Princeton University Press. S. Goldwasser and M. Bellare Lecture Notes on Cryptography, www-cse.ucsd.edu/~mihir/papers/gb.html

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Foundations of Cryptography Lecture 9 Lecturer: Moni Naor.

Foundations of Cryptography Lecture 9 Lecturer: Moni Naor.

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