Lecturer: Moni Naor Weizmann Institute of Science On necessary and sufficient cryptographic assumptions: the case of memory checking Lecture 1: One-way functions Lecturer: Moni Naor Weizmann Institute of Science Give the handouts (course web page)

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 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 these talks Kahn Book was important to the development of the science of crypto, since this is what people (like Diffie) knew The two historical reference are an early one and a late one.

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. Key idea in cryptography: Use the computational infeasibility of problems in order to obtain security The development of these two areas is tightly connected!

Key Idea of Cryptography Use the intractability of some problems for the advantage of constructing secure system Almost any cryptographic task requires using this idea. Our goal is to investigate the tasks of memory checking and sublinear authentication and see whether they require it

Lectures Outline One-way functions and their essential role in cryptography The authentication problem Communication Complexity Memory Checking Lower bound for memory checking and authentication

Three Basic Issues in Cryptography Identification Authentication Encryption Won’t discuss much

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 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 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? 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 only with low probability. How low? Related to the string length that Alice sends…

Identification X X Alice Bob ?? Eve

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 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 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

H(X) = - ∑ x Γ Px (x) log Px (x) (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 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 Entropy is bounded H(X) ≤ log | Γ | with equality only if X is uniform over Γ

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 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 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!

Good source on Information Theory: T. Cover and J. A. Thomas, Elements of Information Theory

One-time vs. many times This was good for a single identification. What about many sessions of identification? Later…

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 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 Charlie Alice Bob Eve

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 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 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 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 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 Major open problem: connect the existence of one-way functions and the P=NP? 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? If f is a one way function is g(x) = f(f(x)) necessarily a one-way function?

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 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. A’ Eve y y y x’ x’

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 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? 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 the 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 Are we done? Given a noninteracive identification protocol want to define a one-way function Define f(r): the setup phase mapping between the random bits r of Alice 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… 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 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 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 Show a function f such that if f is polynomial time invertible on all inputs, then P=NP f is not one-way

x is called the discrete log of y to base g. 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 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

Prob[A[f(x)]  f-1(f(x)) ] ≤ 1-1/p(n) 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 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? 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 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 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 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 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)

P={(y1, y2)| B succeeds on (y1,y2)} The success of B Fix the random bits of B. Define P={(y1, y2)| B succeeds on (y1,y2)} P= P ⋂ {(y1,y2 )| y1,y2 S} ⋃ P ⋂ {(y1,y2 )| y1 S} ⋃ P ⋂ {(y1,y2 )| y2 S} y1 Well behaved part y2 want to bound P by a square P

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

Prob[y S] ≥ √(α2+ ε/3) > α 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

Is there an ultimate one-way function? Aka `universal’ 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 Robust Combiner 1/p(n)

Ultimate one-way function conclusions Original proof due to L. Levin 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 Notion of robust combiner seems fundamental See “Robust Combiners for Oblivious Transfer and Other Primitives” by Harnik, Kilian, Naor , Reingold and Rosen, Eurocrypt 2005

On One-Way Functions A function f is one-way if: it is computable in poly-time the probability of successfully finding an inverse in poly-time is negligible (on a random input) A function f is distributionally one-way if: No poly-time algorithm can successfully find a random inverse (on a random input) Theorem [Impagliazzo Luby 89]: distributionally one-way functions exist iff one-way functions exists

Distributionally one-way functions Definition: function f:{0,1}*  {0,1}* is distributionally one-way if for some constant c>0 for every pptm M the distributions: x◦f(x) M(f(x))◦f(x) Are statistically distinguishable by a least n-c

Existence of distributionally one-way functions Theorem: distributionally one-way functions exist iff one-way functions exists Proof: Suppose f be a distributionally one-way function with constant c Construct a weak one-way function g Need H - a family of pairwise independent functions Let g(x,h,i) = hf(x), h, i, prefi(h(x))i Claim, if f be a distributionally one-way function, then g is a weak one-way function Guess of (log) the number of pre-images

Application: bit commitment Sender - Input b{0,1} Receiver - Two Phases Commit Reveal At the end of protocol: Receiver obtains b decides valid or not

Receiver can verify b was the value in the box Commitment Protocol Commit Phase b Sender Receiver Sender is bound to b Reveal Phase Sender b Receiver Receiver can verify b was the value in the box

Following Commit Phase Receiver should not have gained any information about b Information theoretic? Computationally? Sender should be bound to b No two different and valid openings exist It is computationally infeasible to find two different valid openings

Both worlds? Cannot have best of both worlds: Information theoretic secrecy following commit Distribution of conversation independent of b Perfect binding No two different and valid openings exist whp

One-way functions are essential for bit-commitment Protocol may be interactive Let f(b, rA, rB) be the transcript of the commit phase with sender having input b and random string rA receiver having input b and random string rB Claim: f is a distributionally one-way function Suppose not, then given a typical transcript c=f(b, rA, rB) Can find random inverses in f-1(c) Run an find a small collection of invreses in f-1(c) Two possibilities: Most of the inverses have the same value b Can break the secrecy Different values of b occur Can break the binding Make sure that the receiver accepts the value when inputs are string are b, rA and rB

Existence of one-way functions is equivalent: The existence of one-way functions is equivalent to Pseudo-random generators [HILL] Pseudo-random functions and permutations Block ciphers Bit commitment ~ zero-knowledge Signature Schemes (Non trivial) shared-key encryption Goal of these talk: add two other items to the list: Sub-linear authentication Memory Checking

References Books: O. Goldreich, Foundations of Cryptography - a book in three volumes. Vol 1, Basic Tools, Cambridge, 2001 One-way Functions, Pseudo-randomness, zero-knowledge See other volumes in www.wisdom.weizmann.ac.il/~oded/books.html R. Impagliazzo and M. Luby, One-way Functions are Essential for Complexity Based Cryptography, FOCS 1989: