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Published on January 5, 2008

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Cryptography and Privacy Preserving Operations Lecture 1:  Cryptography and Privacy Preserving Operations Lecture 1 Lecturer: Moni Naor Weizmann Institute of Science What is Cryptography? :  What is Cryptography? Traditionally: how to maintain secrecy in communication Alice and Bob talk while Eve tries to listen Alice Bob Eve History of Cryptography:  History of Cryptography Very ancient occupation Many interesting books and sources, especially about the Enigma David Kahn, The Codebreakers, 1967 Gaj and Orlowski, Facts and Myths of Enigma: Breaking Stereotypes, Eurocrypt 2003 Not the subject of this course Modern Times:  Modern Times Up to the mid 70’s - mostly classified military work Exception: Shannon, Turing* Since then - explosive growth Commercial applications Scientific work: tight relationship with Computational Complexity Theory Major works: Diffie-Hellman, Rivest, Shamir and Adleman (RSA) Recently - more involved models for more diverse tasks. How to maintain the secrecy, integrity and functionality in computer and communication system. Cryptography and Complexity:  Complexity Theory - Study the resources needed to solve computational problems computer time, memory Identify problems that are infeasible to compute. Cryptography - Find ways to specify security requirements of systems Use the computational infeasibility of problems in order to obtain security. The development of these two areas is tightly connected! The interplay between these areas is the subject of the course Cryptography and Complexity Key Idea of Cryptography:  Key Idea of Cryptography Use the intractability of some problems for the advantage of constructing secure system Short Course Outline:  Short Course Outline First part of this short course: Alice and Bob want to cooperate Eve wants to interfere One-way functions, pseudo-random generators pseudo-random functions Encryption Second part Alice and bob do not quite trust each other Zero-knowledge protocols Secure function evaluation Three Basic Issues in Cryptography:  Three Basic Issues in Cryptography Identification Authentication Encryption Example: Identification:  Example: Identification When the time is right, Alice wants to send an `approve’ message to Bob. They want to prevent Eve from interfering Bob should be sure that Alice indeed approves Alice Bob Eve Rigorous Specification of Security:  Rigorous Specification of Security To define security of a system must specify: What constitute a failure of the system The power of the adversary computational access to the system what it means to break the system. Specification of the Problem:  Specification of the Problem Alice and Bob communicate through a channel Bob has two external states {N,Y} Eve completely controls the channel Requirements: If Alice wants to approve and Eve does not interfere – Bob moves to state Y If Alice does not approve, then for any behavior from Eve, Bob stays in N If Alice wants to approve and Eve does interfere - no requirements from the external state Can we guarantee the requirements?:  Can we guarantee the requirements? No – when Alice wants to approve she sends (and receives) a finite set of bits on the channel. Eve can guess them. To the rescue - probability. Want that Eve will succeed with low probability. How low? Related to the string length that Alice sends… Example: Identification:  Example: Identification Alice Bob Eve X X ?? Suppose there is a setup period:  Suppose there is a setup period There is a setup where Alice and Bob can agree on a common secret Eve only controls the channel, does not see the internal state of Alice and Bob (only external state of Bob) Simple solution: Alice and Bob choose a random string X R {0,1}n When Alice wants to approve – she sends X If Bob gets any symbols on channel – compares to X If equal moves to Y If not equal moves permanently to N Eve’s probability of success:  Eve’s probability of success If Alice did not send X and Eve put some string X’ on the channel, then Bob moves to Y only if X= X’ Prob[X=X’] ≤ 2-n Good news: can make it a small as we wish What to do if Alice and Bob cannot agree on a uniformly generated string X? Less than perfect random variables:  Less than perfect random variables Suppose X is chosen according to some distribution Px over some set of symbols Γ What is Eve’s best strategy? What is her probability of success (Shannon) Entropy:  (Shannon) Entropy Let X be random variable over alphabet Γ with distribution Px The (Shannon) entropy of X is H(X) = - ∑ x Γ Px (x) log Px (x) Where we take 0 log 0 to be 0. Represents how much we can compress X Examples:  Examples If X=0 (constant) then H(x) = 0 Only case where H(x) = 0 is when x is constant All other cases H(x) >0 If X  {0,1} and Prob[X=0] = p and Prob[X=1]=1-p, then H(X) = -p log p + (1-p) log (1-p) ≡ H(p) If X  {0,1}n and is uniformly distributed, then H(X) = - ∑ x  {0,1}n 1/2n log 1/2n = 2n/2n n = n Properties of Entropy:  Properties of Entropy Entropy is bounded H(X) ≤ log | Γ | with equality only if X is uniform over Γ Does High Entropy Suffice for Identification?:  Does High Entropy Suffice for Identification? If Alice and bob agree on X  {0,1}n where X has high entropy (say H(X) ≥ n/2 ), what are Eve’s chances of cheating? Can be high: say Prob[X=0n ] = 1/2 For any x 1{0,1} n-1 Prob[X=x ] = 1/2n Then H(X) = n/2+1/2 But Eve can cheat with probability at least ½ by guessing that X=0n Another Notion: Min Entropy:  Another Notion: Min Entropy Let X be random variable over alphabet Γ with distribution Px The min entropy of X is Hmin(X) = - log max x Γ Px (x) The min entropy represents the most likely value of X Property: Hmin(X) ≤ H(X) Why? High Min Entropy and Passwords:  High Min Entropy and Passwords Claim: if Alice and Bob agree on such that Hmin(X) ≥ m, then the probability that Eve succeeds in cheating is at most 2-m Proof: Make Eve deterministic, by picking her best choice, X’ = x’. Prob[X=x’] = Px (x’) ≤ max x Γ Px (x) = 2 –Hmin(X) ≤ 2-m Conclusion: passwords should be chosen to have high min-entropy! Slide23:  Good source on Information Theory: T. Cover and J. A. Thomas, Elements of Information Theory One-time vs. many times:  One-time vs. many times This was good for a single identification. What about many identification? Later… A different scenario – now Charlie is involved:  A different scenario – now Charlie is involved Bob has no proof that Alice indeed identified If there are two possible verifiers, Bob and Charlie, they can each pretend to each other to be Alice Can each have there own string But, assume that they share the setup phase Whatever Bob knows Charlie know Relevant when they are many of possible verifiers! The new requirement:  The new requirement If Alice wants to approve and Eve does not interfere – Bob moves to state Y If Alice does not approve, then for any behavior from Eve and Charlie, Bob stays in N Similarly if Bob and Charlie are switched Alice Bob Eve Charlie Can we achieve the requirements?:  Can we achieve the requirements? Observation: what Bob and Charlie received in the setup phase might as well be public Therefore can reduce to the previous scenario (with no setup)… To the rescue - complexity Alice should be able to perform something that neither Bob nor Charlie (nor Eve) can do Must assume that the parties are not computationally all powerful! Function and inversions:  Function and inversions We say that a function f is hard to invert if given y=f(x) it is hard to find x’ such that y=f(x’) x’ need not be equal to x We will use f-1(y) to denote the set of preimages of y To discuss hard must specify a computational model Use two flavors: Concrete Asymptotic One-way functions - asymptotic:  One-way functions - asymptotic A function f: {0,1}* → {0,1}* is called a one-way function, if f is a polynomial-time computable function Also polynomial relationship between input and output length for every probabilistic polynomial-time algorithm A, every positive polynomial p(.), and all sufficiently large n’s Prob[A[f(x)]  f-1(f(x)) ] ≤ 1/p(n) Where x is chosen uniformly in {0,1}n and the probability is also over the internal coin flips of A One-way functions – concrete version:  One-way functions – concrete version A function f:{0,1}n → {0,1}n is called a (t,ε) one-way function, if f is a polynomial-time computable function (independent of t) for every t-time algorithm A, Prob[A[f(x)]  f-1(f(x)) ] ≤ ε Where x is chosen uniformly in {0,1}n and the probability is also over the internal coin flips of A Can either think of t and ε as being fixed or as t(n), ε(n) Complexity Theory and One-way Functions:  Complexity Theory and One-way Functions Claim: if P=NP then there are no one-way functions Proof: for any one-way function f: {0,1}n → {0,1}n consider the language Lf : Consisting of strings of the form {y, b1, b2…bk} There is an x  {0,1}n such that y=f(x) and The first k bits of x are b1, b2…bk Lf is NP – guess x and check If Lf is P then f is invertible in polynomial time: Self reducibility A few properties and questions concerning one-way functions:  A few properties and questions concerning one-way functions Major open problem: connect the existence of one-way functions and the P=NP? question If f is one-to-one it is a called a one-way permutation. In what complexity class does the problem of inverting one-way permutations reside? good exercise! If f’ is a one-way function, is f’ where f’(x) is f(x) with the last bit chopped a one-way function? If f is a one-way function, is fL where fL(x) consists of the first half of the bits of f(x) a one-way function? good exercise! If f is a one way function is g(x) = f(f(x)) necessarily a one-way function? good exercise! Solution to the password problem:  Solution to the password problem Assume that f: {0,1}n → {0,1}n is a (t,ε) one-way function Adversary’s run times is bounded by t Setup phase: Alice chooses xR {0,1}n computes y=f(x) Gives y to Bob and Charlie When Alice wants to approve – she sends x If Bob gets any symbols on channel – call them z; compute f(z) and compares to y If equal moves to state Y If not equal moves permanently to state N Eve’s and Charlie’s probability of success:  Eve’s and Charlie’s probability of success If Alice did not send x and Eve (Charlie) put some string x’ on the channel to Bob, then: Bob moves to state Y only if f(x’)=y=f(x) But we know that Prob[A[f(x)]  f-1(f(x)) ] ≤ ε or else we can use Eve to break the one-way function Good news: if ε can be made as small as we wish, then we have a good scheme. Can be used for monitoring Similar to the Unix password scheme f(x) stored in login file DES used as the one-way function. y y x’ y x’ Eve A’ Reductions:  Reductions This is a simple example of a reduction Simulate Eve’s algorithm in order to break the one-way function Most reductions are much more involved Cryptographic Reductions:  Cryptographic Reductions Show how to use an adversary for breaking primitive 1 in order to break primitive 2 Important Run time: how does T1 relate to T2 Probability of success: how does 1 relate to 2 Access to the system 1 vs. 2 Are one-way functions essential to the two guards password problem? :  Are one-way functions essential to the two guards password problem? Precise definition: for every probabilistic polynomial-time algorithm A controlling Eve and Charlie every polynomial p(.), and all sufficiently large n’s Prob[Bob moves Y | Alice does not approve] ≤ 1/p(n) Recall observation: what Bob and Charlie received in the setup phase might as well be public Claim: can get rid of interaction: given an interactive identification protocol possible to construct a noninteractive one. In new protocol: Alice’ sends Bob’ the random bits Alice used to generate the setup information Bob’ simulates the conversation between Alice and Bob in original protocol and accepts only if simulated Bob accepts. Probability of cheating is the same One-way functions are essential to the two guards password problem:  One-way functions are essential to the two guards password problem Are we done? Given a noninteracive identification protocol want to define a one-way function Define function f(r) as the mapping that Alice does in the setup phase between her random bits r and the information y given to Bob and Charlie Problem: the function f(r) is not necessarily one-way… Can be unlikely ways to generate it. Can be exploited to invert. Example: Alice chooses x, x’ {0,1}n if x’= 0n set y=x o.w. set y=f(x) The protocol is still secure, but with probability 1/2n not complete The resulting function f(x,x’) is easy to invert: given y {0,1}n set inverse as (y, 0n ) One-way functions are essential to the two guards password problem…:  One-way functions are essential to the two guards password problem… However: possible to estimate the probability that Bob accepts on a given string from Alice Second attempt: define function f(r) as the mapping that Alice does in the setup phase between her random bits r and the information given to Bob and Charlie, plus a bit indicating that probability of Bob accepts given r is greater than 2/3 Theorem: the two guards password problem has a solution if and only if one-way functions exist Examples of One-way functions:  Examples of One-way functions Examples of hard problems: Subset sum Discrete log Factoring (numbers, polynomials) into prime components How do we get a one-way function out of them? Easy problem Subset Sum:  Subset Sum Subset sum problem: given n numbers 0 ≤ a1, a2 ,…, an ≤ 2m Target sum T Find subset S⊆ {1,...,n} ∑ i S ai,=T (n,m)-subset sum assumption: for uniformly chosen a1, a2 ,…, an R{0,…2m -1} and S⊆ {1,...,n} For any probabilistic polynomial time algorithm, the probability of finding S’⊆ {1,...,n} such that ∑ i S ai= ∑ i S’ ai is negligible, where the probability is over the random choice of the ai‘s, S and the inner coin flips of the algorithm Subset sum one-way function f:{0,1}mn+n → {0,1}m f(a1, a2 ,…, an , b1, b2 ,…, bn ) = (a1, a2 ,…, an , ∑ i=1n bi ai mod 2m ) Exercise:  Exercise Show a function f such that if f is polynomial time invertible on all inputs, then P=NP f is not one-way Discrete Log Problem:  Discrete Log Problem Let G be a group and g an element in G. Let y=gz and x the minimal non negative integer satisfying the equation. x is called the discrete log of y to base g. Example: y=gx mod p in the multiplicative group of Zp In general: easy to exponentiate via repeated squaring Consider binary representation What about discrete log? If difficult, f(g,x) = (g, gx ) is a one-way function Integer Factoring:  Integer Factoring Consider f(x,y) = x • y Easy to compute Is it one-way? No: if f(x,y) is even can set inverse as (f(x,y)/2,2) If factoring a number into prime factors is hard: Specifically given N= P • Q , the product of two random large (n-bit) primes, it is hard to factor Then somewhat hard – there are a non-negligible fraction of such numbers ~ 1/n2 from the density of primes Hence a weak one-way function Alternatively: let g(r) be a function mapping random bits into random primes. The function f(r1,r2) = g(r1) • g(r2) is one-way Weak One-way function :  Weak One-way function A function f: {0,1}n → {0,1}n is called a weak one-way function, if f is a polynomial-time computable function There exists a polynomial p(¢), for every probabilistic polynomial-time algorithm A, and all sufficiently large n’s Prob[A[f(x)]  f-1(f(x)) ] ≤ 1-1/p(n) Where x is chosen uniformly in {0,1}n and the probability is also over the internal coin flips of A Exercise: weak exist if strong exists:  Exercise: weak exist if strong exists Show that if strong one-way functions exist, then there exists a a function which is a weak one-way function but not a strong one What about the other direction?:  What about the other direction? Given a function f that is guaranteed to be a weak one-way Let p(n) be such that Prob[A[f(x)]  f-1(f(x)) ] ≤ 1-1/p(n) can we construct a function g that is (strong) one-way? An instance of a hardness amplification problem Simple idea: repetition. For some polynomial q(n) define g(x1, x2 ,…, xq(n) )=f(x1), f(x2), …, f(xq(n)) To invert g need to succeed in inverting f in all q(n) places If q(n) = p2(n) seems unlikely (1-1/p(n))p2(n) ≈ e-p(n) But how to we show? Sequential repetition intuition – not a proof. Want: Inverting g with low probability implies inverting f with high probability:  Want: Inverting g with low probability implies inverting f with high probability Given a machine A that inverts g want a machine A’ operating in similar time bounds inverts f with high probability Idea: given y=f(x) plug it in some place in g and generate the rest of the locations at random z=(y, f(x2), …, f(xq(n))) Ask machine A to invert g at point z Probability of success should be at least (exactly) A’s Probability of inverting g at a random point Once is not enough How to amplify? Repeat while keeping y fixed Put y at random position (or sort the inputs to g ) Proof of Amplification for Repetition of Two:  Proof of Amplification for Repetition of Two Concentrate on repetition of two g(x1, x2 )=f(x1), f(x2) Goal: show that the probability of inverting g is roughly squared the probability of inverting f just as would be sequentially Claim: Let (n) be a function that for some p(n) satisfies 1/p(n) ≤ (n) ≤ 1-1/p(n) Let ε(n) be any inverse polynomial function suppose that for every polynomial time A and sufficiently large n Prob[A[f(x)]  f-1(f(x)) ] ≤ (n) Then for every polynomial time B and sufficiently large n Prob[B[g(x1, x2 )]  g-1(g(x1, x2 )) ] ≤ 2(n) + ε(n) Proof of Amplification for Two Repetition:  Proof of Amplification for Two Repetition Suppose not, then given a better than 2+ε algorithm B for inverting g construct the following: B’(y) Inversion algorithm for f Repeat t times Choose x’ at random and compute y’=f(x’) Run B(y,y’). Check the results If correct: Halt with success Output failure Inner loop Helpful for constructive algorithm Probability of Success:  Probability of Success Define S={y=f(x) | Prob[Inner loop successful| y ] > β} Since the choices of the x’ are independent Prob[B’ succeeds| xS] > 1-(1- β)t Taking t= n/β means that when yS almost surely A will invert it Hence want to show that Prob[ yS] > (n) The success of B:  The success of B Fix the random bits of B. Define P={(y1, y2)| B succeeds on (y1,y2)} y1 y2 P P= P ⋂ {(y1,y2 )| y1,y2 S} ⋃ P ⋂ {(y1,y2 )| y1 S} ⋃ P ⋂ {(y1,y2 )| y2 S} Well behaved part want to bound P by a square S is the only success..:  S is the only success.. But Prob[B[y1, y2]  g-1(y1, y2) | y1 S] ≤ β and similarly Prob[B[y1, y2]  g-1(y1, y2) | y2 S] ≤ β so Prob[(y1, y2) P and y1,y2 S] ≥ Prob[(y1, y2) P ] - 2β ≥ 2+ ε - 2β Setting β =ε/3 we have Prob[(y1, y2) P and y1,y2 S] ≥ 2+ ε/3 Contradiction:  Contradiction But Prob[(y1, y2) P and y1,y2 S] ≤ Prob[y1 S] Prob[y2 S] = Prob2[y S] So Prob[y S] ≥ √(α2+ ε/3) > α The Encryption problem::  The Encryption problem: Alice would want to send a message m  {0,1}n to Bob Set-up phase is secret They want to prevent Eve from learning anything about the message Alice Bob Eve m The encryption problem:  The encryption problem Relevant both in the shared key and in the public key setting Want to use many times Also add authentication… Other disruptions by Eve What does `learn’ mean?:  What does `learn’ mean? If Eve has some knowledge of m should remain the same Probability of guessing m Min entropy of m Probability of guess whether m is m0 or m1 Probability of computing some function f of m Ideally: the message sent is a independent of the message m Implies all the above Shannon: achievable only if the entropy of the shared secret is at least as large as the message m entropy If no special knowledge about m then |m| Achievable: one-time pad. Let rR {0,1}n Think of r and m as elements in a group To encrypt m send r+m To decrypt z send m=z-r Pseudo-random generators:  Pseudo-random generators Would like to stretch a short secret (seed) into a long one The resulting long string should be usable in any case where a long string is needed In particular: as a one-time pad Important notion: Indistinguishability Two probability distributions that cannot be distinguished Statistical indistinguishability: distances between probability distributions New notion: computational indistinguishability ...References:  ...References Books: O. Goldreich, Foundations of Cryptography - a book in three volumes. Vol 1, Basic Tools, Cambridge, 2001 Pseudo-randomness, zero-knowledge Vol 2, about to come out (Encryption, Secure Function Evaluation) Other volumes in www.wisdom.weizmann.ac.il/~oded/books.html M. Luby, Pseudorandomness and Cryptographic Applications, Princeton University Pres , References:  References Web material/courses S. Goldwasser and M. Bellare, Lecture Notes on Cryptography, http://www-cse.ucsd.edu/~mihir/papers/gb.html Wagner/Trevisan, Berkeley www.cs.berkeley.edu/~daw/cs276 Ivan Damgard and Ronald Cramer, Cryptologic Protocol Theory http://www.daimi.au.dk/~ivan/CPT.html Salil Vadhan, Pseudorandomness http://www.courses.fas.harvard.edu/~cs225/Lectures-2002/ Naor, Foundations of Cryptography and Estonian Course www.wisdom.weizmann.ac.il/~naor Recap of Lecture 1:  Recap of Lecture 1 Key idea of cryptography: use computational intractability for your advantage One-way functions are necessary and sufficient to solve the two guard identification problem Notion of Reduction between cryptographic primitives Amplification of weak one-way functions Things are a bit more complex in the computational world (than in the information theoretic one) Is there an ultimate one-way function?:  Is there an ultimate one-way function? If f1:{0,1}* → {0,1}* and f2:{0,1}* → {0,1}* are guaranteed to: Be polynomial time computable At least one of them is one-way. then can construct a function g:{0,1}* → {0,1}* which is one-way: g(x1, x2 )= (f1(x1),f2 (x2 )) If an 5n2 time one-way function is guaranteed to exist, can construct an O(n2 log n) one-way function g: Idea: enumerate Turing Machine and make sure they run 5n2 steps g(x1, x2 ,…, xlog (n) )=M1(x1), M2(x2), …, Mlog n(xlog (n)) If a one-way function is guaranteed to exist, then there exists a 5n2 time one-way: Idea: concentrate on the prefix 1/p(n) Conclusions:  Conclusions Be careful what you wish for Problem with resulting one-way function: Cannot learn about behavior on large inputs from small inputs Whole rational of considering asymptotic results is eroded Construction does not work for non-uniform one-way functions Encryption: easy when you share very long strings Started with the notion of pseudo-randomness The Encryption problem::  The Encryption problem: Alice would want to send a message m  {0,1}n to Bob Set-up phase is secret They want to prevent Eve from learning anything about the message Alice Bob Eve m The encryption problem:  The encryption problem Relevant both in the shared key and in the public key setting Want to use many times Also add authentication… Other disruptions by Eve What does `learn’ mean?:  What does `learn’ mean? If Eve has some knowledge of m should remain the same Probability of guessing m Min entropy of m Probability of guess whether m is m0 or m1 Probability of computing some function f of m Ideally: the message sent is a independent of the message m Implies all the above Shannon: achievable only if the entropy of the shared secret is at least as large as the message m entropy If no special knowledge about m then |m| Achievable: one-time pad. Let rR {0,1}n Think of r and m as elements in a group To encrypt m send r+m To decrypt z send m=z-r Pseudo-random generators:  Pseudo-random generators Would like to stretch a short secret (seed) into a long one The resulting long string should be usable in any case where a long string is needed In particular: as a one-time pad Important notion: Indistinguishability Two probability distributions that cannot be distinguished Statistical indistinguishability: distances between probability distributions New notion: computational indistinguishability Computational Indistinguishability :  Computational Indistinguishability Definition: two sequences of distributions {Dn} and {D’n} on {0,1}n are computationally indistinguishable if for every polynomial p(n) and sufficiently large n, for every probabilistic polynomial time adversary A that receives input y  {0,1}n and tries to decide whether y was generated by Dn or D’n |Prob[A=‘0’ | Dn ] - Prob[A=‘0’ | D’n ] | < 1/p(n) Without restriction on probabilistic polynomial tests: equivalent to variation distance being negligible ∑β  {0,1}n |Prob[ Dn = β] - Prob[ D’n = β]| < 1/p(n) Pseudo-random generators:  Pseudo-random generators Definition: a function g:{0,1}* → {0,1}* is said to be a (cryptographic) pseudo-random generator 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 is random the output is indistinguishable from random For any probabilistic polynomial time adversary A that receives input y of length ℓ(n) and tries to decide whether y= g(x) or is a random string from {0,1}ℓ(n) for any polynomial p(n) and sufficiently large n |Prob[A=`rand’| y=g(x)] - Prob[A=`rand’| yR {0,1}ℓ(n)] | < 1/p(n) Important issues: Why is the adversary bounded by polynomial time? Why is the indistinguishability not perfect? Pseudo-random generators:  Pseudo-random generators Definition: a function g:{0,1}* → {0,1}* is said to be a (cryptographic) pseudo-random generator 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 is indistinguishable from random For any probabilistic polynomial time adversary A that receives input y of length ℓ(n) and tries to decide whether y= g(x) or is a random string from {0,1}ℓ(n) for any polynomial p(n) and sufficiently large n |Prob[A=`rand’| y=g(x)] - Prob[A=`rand’| yR {0,1}ℓ(n)] | < 1/p(n) Want to use the output a pseudo-random generator whenever long random strings are used Especially encryption have not defined the desired properties yet. Anyone who considers arithmetical methods of producing random numbers is, of course, in a state of sin. J. von Neumann Important issues:  Important issues Why is the adversary bounded by polynomial time? Why is the indistinguishability not perfect? Construction of pseudo-random generators:  Construction of pseudo-random generators Idea: given a one-way function there is a hard decision problem hidden there If balanced enough: looks random Such a problem is a hardcore predicate Possibilities: Last bit First bit Inner product Hardcore Predicate:  Hardcore Predicate Definition: let f:{0,1}* → {0,1}* be a function. We say that h:{0,1}* → {0,1} is a hardcore predicate for f if It is polynomial time computable For any probabilistic polynomial time adversary A that receives input y=f(x) and tries to compute h(x) for any polynomial p(n) and sufficiently large n |Prob[A(y)=h(x)] -1/2| < 1/p(n) where the probability is over the choice y and the random coins of A Sources of hardcoreness: not enough information about x not of interest for generating pseudo-randomness enough information about x but hard to compute it Exercises:  Exercises Assume one-way functions exist Show that the last bit/first bit are not necessarily hardcore predicates Generalization: show that for any fixed function h:{0,1}* → {0,1} there is a one-way function f:{0,1}* → {0,1}* such that h is not a hardcore predicate of f Show a one-way function f such that given y=f(x) each input bit of x can be guessed with probability at least 3/4 Single bit expansion:  Single bit expansion Let f:{0,1}n → {0,1}n be a one-way permutation Let h:{0,1}n → {0,1} be a hardcore predicate for f Consider g:{0,1}n → {0,1}n+1 where g(x)=(f(x), h(x)) Claim: g is a pseudo-random generator Proof: can use a distinguisher for g to guess h(x) f(x), h(x)) f(x), 1-h(x)) Hardcore Predicate With Public Information:  Hardcore Predicate With Public Information Definition: let f:{0,1}* → {0,1}* be a function. We say that h:{0,1}* x {0,1}* → {0,1} is a hardcore predicate for f if h(x,r) is polynomial time computable For any probabilistic polynomial time adversary A that receives input y=f(x) and public randomness r and tries to compute h(x,r) for any polynomial p(n) and sufficiently large n |Prob[A(y,r)=h(x,r)] -1/2| < 1/p(n) where the probability is over the choice y of r and the random coins of A Alternative view: can think of the public randomness as modifying the one-way function f: f’(x,r)=f(x),r. Example: weak hardcore predicate:  Example: weak hardcore predicate Let h(x,i)= xi I.e. h selects the ith bit of x For any one-way function f, no polynomial time algorithm A(y,i) can have probability of success better than 1-1/2n of computing h(x,i) Exercise: let c:{0,1}* → {0,1}* be a good error correcting code |c(x)| is O(|x|) distance between any two codewords c(x) and c(x’) is a constant fraction of |c(x)| It is possible to correct in polynomial time errors in a constant fraction of |c(x)| Show that for h(x,i)= c(x)i and any one-way function f, no polynomial time algorithm A(y,i) can have probability of success better than a constant of computing h(x,i) Inner Product Hardcore bit:  Inner Product Hardcore bit The inner product bit: choose r R {0,1}n let h(x,r) = r ∙x = ∑ xi ri mod 2 Theorem [Goldreich-Levin]: for any one-way function the inner product is a hardcore predicate Proof structure: There are many x’s for which A returns a correct answer on ½+ε of the r ’s take an algorithm A that guesses h(x,r) correctly with probability ½+ε over the r‘s and output a list of candidates for x No use of the y info Choose from the list the/an x such that f(x)=y The main step! Why list?:  Why list? Cannot have a unique answer! Suppose A has two candidates x and x’ On query r it returns at `random’ either r ∙x or r ∙x’ Prob[A(y,r) = r ∙x ] =½ +½Prob[r∙x = r∙x’] = ¾ Warm-up (1):  Warm-up (1) If A returns a correct answer on 1-1/2n of the r ’s Choose r1, r2, … rn R {0,1}n Run A(y,r1), A(y,r2), … A(y,rn) Denote the response z1, z2, … zn If r1, r2, … rn are linearly independent then: there is a unique x satisfying ri∙x = zi for all 1 ≤i ≤n Prob[zi = A(y,ri)= ri∙x]≥ 1-1/2n Therefore probability that all the zi‘s are correct is at least ½ Do we need complete independence of the ri ‘s? `one-wise’ independence is sufficient Can choose r R {0,1}n and set ri∙ = r+ei ei =0i-110n-i All the ri `s are linearly independent Each one is uniform in {0,1}n Warm-up (2):  Warm-up (2) If A returns a correct answer on 3/4+ε of the r ’s Can amplify the probability of success! Given any r  {0,1}n Procedure A’(y,r): Repeat for j=1, 2, … Choose r’ R {0,1}n Run A(y,r+r’) and A(y,r’), denote the sum of responses by zj Output the majority of the zj’s Analysis Pr[zj = r∙x]≥ Pr[A(y,r’)=r∙x ^ A(y,r+r’)=(r+r’)∙x]≥½+2ε Does not work for ½+ε since success on r’ and r+r’ is not independent Each one of the events ‘zj = r∙x’ is independent of the others Therefore by taking sufficiently many j’s can amplify to a value as close to 1 as we wish Need roughly 1/ε2 examples Idea for improvement: fix a few of the r’ The real thing:  The real thing Choose r1, r2, … rk R {0,1}n Guess for j=1, 2, … k the value zj =rj∙x Go over all 2k possibilities For all nonempty subsets S {1,…,k} Let rS = ∑ j S rj The implied guess for zS = ∑ j S zj For each position xi for each S {1,…,k} run A(y,ei-rS) output the majority value of {zs +A(y,ei-rS) } Analysis: Each one of the vectors ei-rS is uniformly distributed A(y,ei-rS) is correct with probability at least ½+ε Claim: For every pair of nonempty subset S ≠T {1,…,k}: the two vectors rS and rT are pair-wise independent Therefore variance is as in completely independent trials I is the number of correct A(y,ei-rS), VAR(I) ≤ 2k(½+ε) Use Chebyshev’s Inequality Pr[|I-E(I)|≥ λ√VAR(I)]≤1/λ2 Need 2k = n/ε2 to get the probability of error to 1/n So process is successful simultaneously for all positions xi,i{1,…,n} S T Analysis:  Analysis Number of invocations of A 2k ∙ n ∙ (2k-1) = poly(n, 1/ε) ≈ n3/ε4 Size of resulting list of candidates for x for each guess of z1, z2, … zk unique x 2k =poly(n, 1/ε) ) ≈ n/ε2 Conclusion: single bit expansion of a one-way permutation is a pseudo-random generator guesses positions subsets Reducing the size of the list of candidates:  Reducing the size of the list of candidates Idea: bootstrap Given any r  {0,1}n Procedure A’(y,r): Choose r1, r2, … rk R {0,1}n Guess for j=1, 2, … k the value zj =rj∙x Go over all 2k possibilities For all nonempty subsets S {1,…,k} Let rS = ∑ j S rj The implied guess for zS = ∑ j S zj for each S {1,…,k} run A(y,r-rS) output the majority value of {zs +A(y,r-rS) For 2k = 1/ε2 the probability of error is, say, 1/8 Fix the same r1, r2, … rk for subsequent executions They are good for 7/8 of the r’s Run warm-up (2) Size of resulting list of candidates for x is ≈ 1/ε2


Application: Diffie-Hellman : 

Application: Diffie-Hellman The Diffie-Hellman assumption Let G be a group and g an element in G. Given g, a=gx and b=gy it is hard to find c=gxy for random x and y is probability of poly-time machine outputting gxy is negligible More accurately: a sequence of groups Don’t know how to verify given c’ whether it is equal to gxy Exercise: show that under the DH Assumption Given a=gx , b=gy and r  {0,1}n no polynomial time machine can guess r ∙gxy with advantage 1/poly for random x,y and r Application: if subset is one-way, then it is a pseudo-random generator:  Application: if subset is one-way, then it is a pseudo-random generator Subset sum problem: given n numbers 0 ≤ a1, a2 ,…, an ≤ 2m Target sum y Find subset S⊆ {1,...,n} ∑ i S ai,=y Subset sum one-way function f:{0,1}mn+n → {0,1}m f(a1, a2 ,…, an , x1, x2 ,…, xn ) = (a1, a2 ,…, an , ∑ i=1n xi ai mod 2m ) If m<n then we get out less bits then we put in. If m>n then we get out more bits then we put in. Theorem: if for m>n subset sum is a one-way function, then it is also a pseudo-random generator Subset Sum Generator:  Subset Sum Generator Idea of proof: use the distinguisher A to compute r ∙x For simplicity: do the computation mod P for large prime P Given r  {0,1}n and (a1, a2 ,…, an ,y) Generate new problem(a’1, a’2 ,…, a’n ,y’) : Choose c R ZP Let a’i = ai if ri=0 and ai =ai+c mod P if ri=1 Guess k R {o,…,n} - the value of ∑ xi ri the number of locations where x and r are 1 Let y’ = y+c k mod P Run the distinguisher A on (a’1, a’2 ,…, a’n ,y’) output what A says Xored with parity(k) Claim: if k is correct, then (a’1, a’2 ,…, a’n ,y’) is R pseudo-random Claim: for any incorrect k, (a’1, a’2 ,…, a’n ,y’) is R random y’= z + (k-h)c mod P where z = ∑ i=1n xi a’i mod P and h=∑ xi ri Therefore: probability to guess correctly r ∙x is 1/n∙(½+ε) + (n-1)/n (½)= ½+ε/n random pseudo-random Prob[A=‘0’|pseudo]= ½+ε Prob[A=‘0’|random]= ½ correct k incorrect k Interpretations of the Goldreich-Levin Theorem:  Interpretations of the Goldreich-Levin Theorem A tool for constructing pseudo-random generators The main part of the proof: A mechanism for translating `general confusion’ into randomness Diffie-Hellman example List decoding of Hadamard Codes works in the other direction as well (for any code with good list decoding) List decoding, as opposed to unique decoding, allows getting much closer to distance `Explains’ unique decoding when prediction was 3/4+ε Finding all linear functions agreeing with a function given in a black-box Learning all Fourier coefficients larger than ε If the Fourier coefficients are concentrated on a small set – can find them True for AC0 circuits Decision Trees Composing PRGs:  Composing PRGs Composition Let g1 be a (ℓ1, ℓ2 )-pseudo-random generator g2 be a (ℓ2, ℓ3)-pseudo-random generator Consider g(x) = g2(g1(x)) Claim: g is a (ℓ1, ℓ3 )-pseudo-random generator Proof: consider three distributions on {0,1}ℓ3 D1: y uniform in {0,1}ℓ3 D2: y=g(x) for x uniform in {0,1}ℓ1 D3: y=g2(z) for z uniform in {0,1}ℓ2 By assumption there is a distinguisher A between D1 and D2 A must either distinguish between D1 and D3 - can use A use to distinguish g2 or distinguish between D2 and D3 - can use A use to distinguish g1 ℓ2 ℓ1 ℓ3 triangle inequality Composing PRGs:  Composing PRGs When composing a generator secure against advantage ε1 and a a generator secure against advantage ε2 we get security against advantage ε1+ε2 When composing the single bit expansion generator n times Loss in security is at most ε/n Hybrid argument: to prove that two distributions D and D’ are indistinguishable: suggest a collection of distributions D= D0, D1,… Dk =D’ such that If D and D’ can be distinguished, there is a pair Di and Di+1 that can be distinguished. Difference ε between D and D’ means ε/k between some Di and Di+1 Use such a distinguisher to derive a contradiction Exercise:  Exercise Let {Dn} and {D’n} be two distributions that are Computationally indistinguishable Polynomial time samplable Suppose that {y1,… ym} are all sampled according to {Dn} or all are sampled according to {D’n} Prove: no probabilistic polynomial time machine can tell, given {y1,… ym}, whether they were sampled from {Dn} or {D’n} Existence of PRGs:  Existence of PRGs What we have proved: Theorem: if pseudo-random generators stretching by a single bit exist, then pseudo-random generators stretching by any polynomial factor exist Theorem: if one-way permutations exist, then pseudo-random generators exist A harder theorem to prove Theorem [HILL]: if one-way functions exist, then pseudo-random generators exist Exercise: show that if pseudo-random generators exist, then one-way functions exist Next-bit Test:  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(yi,y2,…, yi)= yi+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 Next-block Undpredictable:  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 yi+1, for any polynomial p(n) and sufficiently large n |Prob[A(y1,y2,…, yi)= yi+1] | < 1/p(n) Exercise: show how to convert a next-block unpredictable generator into a pseudo-random generator. y1 y2, … , Pseudo-Random Generators concrete version:  Pseudo-Random Generators concrete version Gn: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  Three Basic issues in cryptography:  Three Basic issues in cryptography Identification Authentication Encryption Solve in a shared key environment S S B A Identification - Remote login using pseudo-random sequence:  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 Gn(S) For each identification session - send next block of Gn(S) Gn(S) Problems...:  Problems... More than two parties Malicious adversaries - add noise Coordinating the location block number Better approach: Challenge-Response Challenge-Response Protocol:  Challenge-Response Protocol B selects a random location and sends to A A sends value at random location A B What’s this? Desired Properties:  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 Authenticating Messages:  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[2n] Problems and Solutions:  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 A B Use this! Encryption of Messages:  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 Encryption of Messages:  Encryption of Messages If a very long random looking string available - can use as in one-time encryption A B Use this! Pseudo-random Functions:  Pseudo-random Functions Concrete Treatment: F: 0,1k  0,1n  0,1m key Domain Range Denote Y= FS (X) A family of functions Φk ={FS | S0,1k  is (t, , q)-pseudo-random if it is Efficiently computable - random access and... (t,,q)-pseudo-random:  (t,,q)-pseudo-random The tester A that can choose adaptively X1 and get Y1= FS (X1) X2 and get Y2 = FS (X2 ) … Xq and get Yq= FS (Xq) Then A has to decide whether FS R Φk or FS R R n  m =  F | F :0,1n  0,1m  (t,,q)-pseudo-random:  (t,,q)-pseudo-random For a function F chosen at random from (1) Φk ={FS | 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 Fk  - ProbA ‘1’  FR R n  m     Equivalent/Non-Equivalent Definitions:  Equivalent/Non-Equivalent Definitions Instead of next bit test: for XX1,X2 ,, Xq chosen by A, decide whether given Y is Y= FS (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 Application to the basic issues in cryptography:  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= FS (X) A verifies Authentication: A to B: Y= FS (M) replay attack Encryption: A chooses XR 0,1n A to B: <X , Y= FS (X)  M > Goal:  Goal Construct an ensemble {Φk | kL  such that for any {tk, 1/k, qk | kL  polynomial in k, for all but finitely many k’s Φk is a (tk, k, qk )-pseudo-random family Construction:  Construction Construction via Expansion Expand n or m Direct constructions Effects of Concatenation:  Effects of Concatenation Given ℓ Functions F1 , F2 ,, Fℓ decide whether they are ℓ random and independent functions OR FS1 , FS2 ,, FSℓ for S1,S2 ,, Sℓ R0,1k Claim: If Φk ={FS | 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 Proof: Hybrid Argument:  Proof: Hybrid Argument i=0 FS1 , FS2 ,, FSℓ p0 … i R1, R2 ,  , Ri-1,FSi , FSi+1 ,, FSℓ pi … i=ℓ R1, R2 ,  , Rℓ pℓ  pℓ - p0     i pi+1 - pi  /ℓ ...Hybrid Argument:  ...Hybrid Argument Can use this i to distinguish whether FS R Φk or FS R R n  m Generate FSi+1 ,, FSℓ Answer queries to first i-1 functions at random (consistently) Answer query to FSi , using (black box) input Answer queries to functions i+1 through ℓ with FSi+1 ,, FSℓ Running time of test - t’  ℓq Doubling the domain:  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)  G0(S) G1(S) Let FS (n+1)(bx)  FGb(s) (n)(x) Claim:  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) FS0 (n) , FS1 (n) for independent S0, S1 (3) Random D 1 2 2 ...Proof:  ...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 Getting from G to F(n) :  Getting from G to F(n) Idea: Use recursive construction FS (n)(bnbn-1 b1)  FGb1(s) (n-1)(bn-1bn-2 b1)  Gbn(Gbn-1 (  Gb1(S)) ) Each evaluation of FS (n)(x): n invocations of G Tree Description:  Tree Description G0(S) G1(S) S G0(G0(S)) G1(G0(G0(S))) Each leaf corresponds to an X. Label on leaf – value of pseudo-random function Security claim:  Security claim If G is (t qn ,) p.r, then F(n) is (t, ’  nq,q) p.r Proof: Hybrid argument by levels Di : truly random labels for nodes at level i. Pseudo-random from i down Each Di - a collection of q functions  i pi+1 - pi  ’/n q Hybrid:  Hybrid S0 S1 ?S G0(S0) G1(G0(S0)) n-i i Di …Proof of Security:  …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 Disadvantages:  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. Applications of Pseudo-random Functions:  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? References:  References Blum-Micali : SIAM J. Computing 1984 Yao: Blum, Blum, Shub: SIAM J. Computing, 1988 Goldreich, Goldwasser and Micali: J. of the ACM, 1986 ...References:  ...References Books: O. Goldreich, Foundations of Cryptography - a book in three volumes. Vol 1, Basic Tools, Cambridge, 2001 Pseudo-randomness, zero-knowledge Vol 2, about to come out (Encryption, Secure Function Evaluation) Other volumes in www.wisdom.weizmann.ac.il/~oded/books.html M. Luby, Pseudorandomness and Cryptographic Applications, Princeton University Pres , References:  References Web material/courses S. Goldwasser and M. Bellare, Lecture Notes on Cryptography, http://www-cse.ucsd.edu/~mihir/papers/gb.html Wagner/Trevisan, Berkeley www.cs.berkeley.edu/~daw/cs276 Ivan Damgard and Ronald Cramer, Cryptologic Protocol Theory http://www.daimi.au.dk/~ivan/CPT.html Salil Vadhan, Pseudorandomness http://www.courses.fas.harvard.edu/~cs225/Lectures-2002/ Naor, Foundations of Cryptography and Estonian Course www.wisdom.weizmann.ac.il/~naor

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