On necessary and sufficient cryptographic assumptions the case

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On necessary and sufficient cryptographic assumptions: the case of memory checking Lecture 2 :

On necessary and sufficient cryptographic assumptions: the case of memory checking Lecture 2 : Authentication and Communication Complexity Lecturer: Moni Naor Weizmann Institute of Science Web site of lectures: www. wisdom. weizmann. ac. il/~naor/COURSE/ens. html

Recap of Lecture 1 • Key idea of cryptography: use computational intractability for your

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 • Equivalence of the existence of one-way functions:

Existence of one-way functions is equivalent: The existence of one-way functions is equivalent to

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 – Implies 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

Authentication • Verifying that a string has not been modified – A central problem

Authentication • Verifying that a string has not been modified – A central problem in cryptography – Many variants • Relevant both in communication and in storage

The authentication problem one-time version • Alice would want to send a message m

The authentication problem one-time version • Alice would want to send a message m {0, 1}n to Bob • They want to prevent Eve from interfering – Bob should be sure that the message m’ he receives is equal to the message m Alice sent m Alice Eve Bob

Specification of the Problem Alice and Bob communicate through a channel Bob has an

Specification of the Problem Alice and Bob communicate through a channel Bob has an external register R N (no message) ⋃ {0, 1}n Eve completely controls the channel Requirements: • Completeness: If Alice wants to send m {0, 1}n and Eve does not interfere – Bob has value m in R • Soundness: If Alice wants to send m and Eve does interfere – R is either N or m (but not m’ ≠m ) – If Alice does not want to send a message R is N Since this is a generalization of the identification problem – must use shared secrets and probability or complexity Probabilistic version: • for any behavior from Eve, for any message m {0, 1}n,

Authentication using hash functions • Suppose that – H= {h| h: {0, 1}n →

Authentication using hash functions • Suppose that – H= {h| h: {0, 1}n → {0, 1}k } is a family of functions – Alice and Bob share a random function h H – To authenticate message m {0, 1}n Alice sends (m, h(m)) – When receiving (m’, z) Bob computes h(m’) and compares to z • If equal, moves register R to m’ • If not equal, register R stays in N • What properties do we require from H – hard to guess h(m’) - at most ε • But clearly not sufficient: one-time pad. – hard to guess h(m’) even after seeing h(m) - at most ε • Should be true for any m’ – Short representation for h - must have small log|H|

Universal hash functions • Given that for h H we have h: {0, 1}n

Universal hash functions • Given that for h H we have h: {0, 1}n → {0, 1}k we know that ε≥ 2 -k • A family where this is an equality is called universal 2 Definition: a family of functions H= {h| h: {0, 1}n → {0, 1}k } is called Strongly Universal 2 or pair-wise independent if: – for all m 1, m 2 {0, 1}n and y 1, y 2 {0, 1}k we have Prob[h(m 1) = y 1 and h(m 2) = y 2 ] = 2 -2 k Where the probability is over a randomly chosen h H In particular Prob[h(m 2) = y 2 | h(m 1) = y 1 ] = 2 -k when a strongly universal 2 family is used in the protocol, Eve’s probability of cheating is at most 2 -k

Constructing universal hash functions The linear polynomial construction: • fix a finite field F

Constructing universal hash functions The linear polynomial construction: • fix a finite field F of size at least the message space 2 n – Could be either GF[2 n] or GF[P] for some prime P ≥ 2 n • The family H of functions h: F→ F is defined as H= {ha, b(m) = a∙m + b | a, b F} Claim: the family above is strongly universal 2 Proof: for every m 1, m 2 , y 1, y 2 F there are unique a, b F such that a∙m 1+b = y 1 a∙m 2+b = y 2 Size: each h H represented by 2 n bits

Lower bound on size of strongly universal hash functions Theorem: let H= {h| h:

Lower bound on size of strongly universal hash functions Theorem: let H= {h| h: {0, 1}n → {0, 1} } be a family of pair-wise independent functions. Then |H| is Ω(2 n) More precisely, to obtain a d-wise independence family |H| should be Ω(2 n└d/2┘) Theorem: see N. Alon and J. Spencer, The Probabilistic Method

An almost perfect solution By allowing ε to be slightly larger than 2 -k

An almost perfect solution By allowing ε to be slightly larger than 2 -k we can get much smaller families Definition: a family of functions H= {h| h: {0, 1}n → {0, 1}k } is called δUniversal 2 if for all m 1, m 2 {0, 1}n where m 1 ≠ m 2 we have Prob[h(m 1) = h(m 2) ] ≤ δ

An almost perfect solution Idea: combine • a family of δ-Universal 2 functions H

An almost perfect solution Idea: combine • a family of δ-Universal 2 functions H 1= {h| {0, 1}n → {0, 1}k } with • a Strongly Universal 2 family H 2= {h| {0, 1}k → {0, 1}k } Consider the family H where each h H is {0, 1}n → {0, 1}k is defined by h 1 H 1 and h 2 H 2 h(x) = h 2(h 1(x)). As before Alice sends m, h(m) Claim : probability of cheating is at most δ + 2 -k Proof: when Eve sends m’, y’ we must have m ≠ m‘ but either – y’ =h(m), which means that Eve succeeds with probability at most δ + 2 -k • Collision in h 1 Or in h 2 Or – y’ ≠ h(m) which means that Eve succeeds with probability at most 2 -k • Collision in h 2 Size: each h H represented by log |H 1 |+ log |H 2|

Constructing almost universal hash functions The polynomial evaluation construction {0, 1}n → {0, 1}k

Constructing almost universal hash functions The polynomial evaluation construction {0, 1}n → {0, 1}k : • fix a finite field F of size at least the target space 2 k – Could be either GF[2 k] or GF[P] for some prime P ≥ 2 k • Let n = ℓ ∙ k • Treat each (non-zero) message m {0, 1}n as a degree (ℓ 1)- polynomial over F. Denote by Pm m • The family H of functions h: Fℓ → F is defined by all elements in F: H= {hx (m)= Pm (x)| x F} Claim: the family above is δ-Universal 2 for δ= (ℓ-1)/2 k Proof: the maximum number of points where two different degree (ℓ-1) polynomials agree is ℓ-1 Size: each h H represented by k bits

Parameters for authentication • To authenticate an n bits message with probability of error

Parameters for authentication • To authenticate an n bits message with probability of error ε Need: • Secret key length: (log n + log 1 / ε ) log n Lower bound does not hold for interactive protocols • Added tag length (log 1 / ε)

Authentication for Storage • Large file residing on a remote server • Verifier stores

Authentication for Storage • Large file residing on a remote server • Verifier stores a small secret `representative’ of file – Fingerprint – When retrieving the file should identify corruption • The size of the fingerprint – A well understood problem

Sub-linear Authentication What about sub-linear authentication: – Do you have to read the whole

Sub-linear Authentication What about sub-linear authentication: – Do you have to read the whole file to figure out whether it has been corrupted? – Encode the information you store (Authenticators). – How large a fingerprint do you need? How much of the encoding do you need to read?

Authenticators How to authenticate a large and unreliable memory with a small and secret

Authenticators How to authenticate a large and unreliable memory with a small and secret memory secret encoding sx x {0, 1}n E V accept reject public encoding px py • Encoding Algorithm E: • Decoding Algorithm D: • Consistency Verifier V: • An adversary sees (only) the public encoding and can change it D x y – Receives a vector x 2 {0, 1}n, encodes it into: • a public encoding px • a small secret encoding sx. Space complexity: s(n) – Receives a public encoding p and decodes it into a vector x 2 {0, 1}n – Receives public px’ and secret sx encodings, verifies whether decoder output = encoder input – Makes few queries to public encoding: query complexity: t(n)

Power of the Adversary • We have seen the access the Adversary has to

Power of the Adversary • We have seen the access the Adversary has to the system • Distinguish between computationally – all powerful and – Bounded Adversaries Dr. Evil

Pretty Good Authenticator • Idea: encode X using a good error correcting code C

Pretty Good Authenticator • Idea: encode X using a good error correcting code C – Actually erasures are more relevant – As long as a certain fraction of the symbols of C(X) is available, can decode X – Good example: Reed Solomon code • Add to each symbol a tag Fk(a, i), a function of • secret information k 2 {0, 1}s, • symbol a 2 • location i • Verifiers picks random location i reads symbol ’a’ and tag t – Check whether t=Fk(a, i) and rejects if not • Decoding process removes all inappropriate tags and uses the decoding procedure of C

How good is the authenticator Suppose it is impossible to forge tags • If

How good is the authenticator Suppose it is impossible to forge tags • If adversary changes fraction of symbols – Probability of being caught per test is • If the code C can recover X from 1 - of the symbols – then the probability of false `accept’ is – Can make it smaller by repetition • How to make the tags unforgeable? – Easy: Need the range of Fk(a, i) to be large enough to make guessing unlikely – Need that for random k 2 {0, 1}s any adversary • given many tags {(aj, ij) Fk(aj, ij)}j • Hard to guess Fk(a, i) for any (a, i) not in the list

Computational Indistinguishability Definition: two sequences of distributions {Dn} and {D’n} on {0, 1}n are

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

Pseudo-random Functions Let {s(n), m(n), ℓ(n)} be a sequence of parameters: F: {0, 1}s

Pseudo-random Functions Let {s(n), m(n), ℓ(n)} be a sequence of parameters: F: {0, 1}s {0, 1}m {0, 1}ℓ key Domain Range Denote Y= Fk (X) A family of functions Φn ={Fk | k 0, 1}s } is pseudo-random if it is • Efficiently computable - random access and. . .

Pseudo-random Any polynomial time tester A that can choose adaptively – X 1 and

Pseudo-random Any polynomial time tester A that can choose adaptively – X 1 and get Y 1= FS (X 1) – X 2 and get Y 2 = FS (X 2 ) … – Xq and get Yq= FS (Xq) • Then A has to decide whether Not important for us – Fk R Φn or – Fk R Rm ℓ = { F | F : {0, 1}m {0, 1}ℓ }

Pseudo-random For a function F chosen at random from (1) Φn = {Fk |

Pseudo-random For a function F chosen at random from (1) Φn = {Fk | k 0, 1 s ℓ m (2) R = { F | F : {0, 1} } For all polynomial time machines A that choose q locations and try to distinguish (1) from (2) for all polynomial 1/p(n) Prob A ‘ 1’ F R Fk - Prob A ‘ 1’ F R R n m 1/p(n)

Equivalent/Non-Equivalent Definitions • Instead of next bit test: for X X 1, X 2

Equivalent/Non-Equivalent Definitions • Instead of next bit test: for X X 1, X 2 , , Xq chosen by A, decide whether given Y is – Y= FS (X) or – Y R 0, 1 m Really what we need • Adaptive vs. Non-adaptive • Unpredictability vs. pseudo-randomness

Existence of Pseudo-Random functions and Authenticators • If one-way functions exist so pseudorandom generators

Existence of Pseudo-Random functions and Authenticators • If one-way functions exist so pseudorandom generators • If pseudo-random generators exist, so do Probability of pseudo-random functions error Authenticators • Conclusion: If one-way functions exist, – so do sublinear authenticators with • Secret memory: sufficient to store a key • Query complexity log n or log n log 1/

So are we done Two problems: • Need – computational bounded adversary and –

So are we done Two problems: • Need – computational bounded adversary and – one-way functions • Efficiency: the evaluation of a pseudorandom function might be a burden to add to every memory fetch operation

Communication Complexity x 2 X Alice Bob y 2 Y Let f: X x

Communication Complexity x 2 X Alice Bob y 2 Y Let f: X x Y Z Input is split between two participants Want to compute outcome: z=f(x, y) while exchanging as few bits as possible

A protocol is defined by the communication tree Alice: 0 Bob: 1 z 0

A protocol is defined by the communication tree Alice: 0 Bob: 1 z 0 z 1 z 2 z 3 z 4 zz 55 z 6 z 7. . .

A Protocol A protocol P over domain X x Y with range Z is

A Protocol A protocol P over domain X x Y with range Z is a binary tree where – Each internal node v is labeled with either • av: X {0, 1} or • bv: Y {0, 1} – Each leaf is labeled with an element z 2 Z • The value of protocol P on input (x, y) is the label of the leaf reached by starting from the root and walking down the tree. • At each internal node labeled av walk – left if av(x)=0 – right if av(x)=1 • At each internal node labeled bv walk – left if bv(y)=0 – right if bv(y)=1 – The cost of protocol P on input (x, y) is the length of the path taken on input (x, y) – The cost of protocol P is the maximum path length

Motivation for studying communication complexity • • • Originally for studying VLSI questions Connection

Motivation for studying communication complexity • • • Originally for studying VLSI questions Connection with Turing Machines Data structures and the cell probe model Boolean circuit depth … New application: lower bound for the authentication problem

Communication Complexity of a function • For a function f: X x Y Z

Communication Complexity of a function • For a function f: X x Y Z the (deterministic) communication complexity of f (D(f)) is the minimum cost of protocol P over all protocols that compute f Observation: For any function f: X x Y Z D(f) ≤ log |X| + log |Z| Example: let x, y µ {1, …, n} and let f(x, y)=max{x [ y} Then D(f) · 2 log n

Median let x, y µ {1, …, n} and let MED(x, y) be the

Median let x, y µ {1, …, n} and let MED(x, y) be the median of the multiset x [ y If the size is even then element ranked |x[ y|/2 Claim: D(MED) is O(log 2 n) protocol idea: do a binary search on the value, each party reporting how many are above the current guess Exercise: D(MED) is O(log n) protocol idea: each party proposes a candidate See which one is larger - no need to repeat bits

Combinatorial Rectangles • A combinatorial rectangle in X x Y is a subset R

Combinatorial Rectangles • A combinatorial rectangle in X x Y is a subset R µ X x Y such that R= A x B for some A µ X and B µ Y Proposition: R µ X x Y is a combinatorial rectangle iff (x 1, y 1) 2 R and (x 2, y 2) 2 R implies that (x 1, y 2) 2 R For Protocol P and node v let Rv be the set of inputs (x, y) reaching v Claim: For any protocol P and node v the set Rv is a combinatorial rectangle Claim: For any given the transcript of an exchange between Alice an Bob possible (but not x and y) possible to

Fooling Sets • For f: X x Y Z a subset R µ X

Fooling Sets • For f: X x Y Z a subset R µ X x Y is f-monochromatic if f is fixed on R • Observation: any protocol P induces a partition of X x Y into f-monochromatic rectangles. The number of rectangles is the number of leaves in P • A set Sµ X x Y is a fooling set for f if there exists a z 2 Z where – For every (x, y) 2 S, f(x, y)=z – For every distinct (x 1, y 1), (x 2, y 2) 2 S either • f(x 1, y 2)≠z or • f(x 2, y 1)≠z y 1 x 2 y 2 z z Property: no two elements of a fooling set S can be in the same monochromatic rectangle Lemma: if f has a fooling set of size t, then D(f) ≥ log 2 t

Applications Equality: Alice and Bob each hold x, y 2 {0, 1}n – want

Applications Equality: Alice and Bob each hold x, y 2 {0, 1}n – want to decide whether x=y or not. • Fooling set for Equality S={(w, w)|w 2 {0, 1}n } Conclusion: D(Equality) ¸ n Disjointness: let x, y µ {1, …, n} and let – DISJ(x, y)=1 if |x y|¸ 1 and – DISJ(x, y)=0 otherwise • Fooling set for Disjointness S={(A, comp(A))|A µ {1, …, n} } Conclusion: D(DISJ) ¸ n

Probabilistic Protocols Random coins x y ALICE BOB

Probabilistic Protocols Random coins x y ALICE BOB

 • Probabilistic Communication Complexity Alice an Bob have each, in addition to their

• Probabilistic Communication Complexity Alice an Bob have each, in addition to their inputs, access to random strings of arbitrary length r. A and r. B (respectively) A probabilistic protocol P over domain X x Y with range Z is a binary tree where – Each internal node v is labeled with either av(x, r. A ) or bv(y, r. B ) – Each leaf is labeled with an element z 2 Z Take all probabilities over the choice of r. A and r. B • P computes f with zero error if for all (x, y) Pr[P(x, y)=f(x, y)]=1 • P computes f with error if for all (x, y) Pr[P(x, y)=f(x, y)]¸ 1 - • For Boolean f, P computes f with one-sided error if for all (x, y) s. t. f(x, y)=0 Pr[P(x, y)=0]=1 and for all (x, y) s. t. f(x, y)=1

Measuring Probabilistic Communication Complexity For input (x, y) can consider as the cost of

Measuring Probabilistic Communication Complexity For input (x, y) can consider as the cost of protocol P on input (x, y) either • worst-case depth • average depth over r. A and r. B Cost of a protocol: maximum cost over all inputs (x, y) The appropriate measure of probabilistic communication complexity: • R 0(f): minimum (over all protocols) of the average cost of a randomized protocol that computes f with zero error. • R (f): minimum (over all protocols) of the worst-case cost of a randomized protocol that computes f with error. – Makes sense: if 0< <½. R(f) = R 1/3(f): • R 1(f): minimum (over all protocols) of the worst-case cost of a randomized protocol that computes f with one-sided error. – Makes sense: if 0< <1. R 1(f) = R½ 1(f):

Equality • Idea: pick a family of hash functions H={h|h: {0, 1}n {1…m}} such

Equality • Idea: pick a family of hash functions H={h|h: {0, 1}n {1…m}} such that for all x≠y, for random h 2 RH Pr[(h(x)=h(y)]· Protocol: Alice: pick random h 2 RH and send <h, h(x)> Bob: compare h(x) to h(y) and announce the result This is a one-sided error protocol with cost log|H|+ log m Constructing H: Fact: over any two polynomials of degree d agree on at most d points Fix prime q such that n 2 · q · 2 n 2 map x to a polynomial Wx of degree d=n/log q over GF[q] H={hz|z 2 GF[q]} and hz(x)=Wx(z) = d/q= n/q log q · 1/n log n

Public coins model • What if Alice and Bob have access to a joint

Public coins model • What if Alice and Bob have access to a joint source of bits. Possible view: distribution over deterministic protocols Let R pub(f): be the minimum cost of a public coins protocol computing f correctly with probability at least 1 - for any input (x, y) Example: R pub(Equality) = (-log ) Theorem: for any Boolean f: R + (f) is R pub(f)+O(log n + log 1/ ) Proof: choose t = 8 n/ 2 assignments to the public string…

Simulating large sample spaces Collection that should resemble probability of success on ALL inputs

Simulating large sample spaces Collection that should resemble probability of success on ALL inputs Bad Good 1 - • Want to find among all possible public random strings a small collection of strings on which the protocol behave similarly on all inputs • Choose m random strings • For input (x, y) event Ax, y is more than ( + ) of the m strings fail the protocol 2 t -2 Pr[Ax, y] · e < 2 -2 n Pr[[x, y A x, y] · x, y Pr[A 2 n 2 -2 n=1 ] <2 x, y

Number of rounds • So far have not been concerned with the number of

Number of rounds • So far have not been concerned with the number of rounds • It is known that there functions with a large gap in the communication complexity between protocols with few and many rounds • What is the smallest number of rounds?

What happens if we flatten the tree? x ALICE m. A CAROL y BOB

What happens if we flatten the tree? x ALICE m. A CAROL y BOB m. B f(x, y)

The simultaneous messages model: • Alice receives x and Bob who receives inputs y

The simultaneous messages model: • Alice receives x and Bob who receives inputs y • They simultaneously send a message to a referee Carol who initially get no input • Carol should compute f(x, y) Several possible models: • Deterministic: all lower bounds for deterministic protocols for f(x, y) are applicable here • Shared (Public) random coins: – Equality has a good protocol • Consider public string as hash functions h, • Alice sends h(x) Bob sends h(y) and Charlie compares the outcome • The complexity can be as little as O(1) is after constant Provided the random bits are chosen independently probability of error than the inputs

Simultaneous Messages Protocols x {0, 1}n ALICE m. A CAROL y {0, 1}n BOB

Simultaneous Messages Protocols x {0, 1}n ALICE m. A CAROL y {0, 1}n BOB m. B • Suggested by Yao 1979 f(x, y)

Simultaneous Messages Protocols • For the equality function: • There exists a protocol where

Simultaneous Messages Protocols • For the equality function: • There exists a protocol where – |m. A| x |m. B| = O(n) – Let C be a good error correcting code – Alice and bob arrange C(x) and C(y) in an |m. A| x |m. B| rectangle • Alice sends a random row • Bob send a random columns • Carol compares the intersection

Simultaneous Messages Protocols • Lower bounds for the equality function: – |m. A| +

Simultaneous Messages Protocols • Lower bounds for the equality function: – |m. A| + |m. B| = (√n) [Newman Szegedy 1996] – |m. A| x |m. B| = (n) [Babai Kimmel 1997] • Idea: for each x 2 {0, 1}n find a `typical’ multiset of messages Tx = {w 1, w 2, …, wt} where t 2 O(|m. B|) Each wi is a message in the original protocol, |m. A| bits Over random i, and randomne ss of Carol Property: for each message m. B the behavior on Tx approximates the real behavior Average behavior of Carol on w 1, w 2, …, wt is close to its average Over randomness of response during protocol Alice and Carol

Simultaneous Messages Protocols How to find for each x 2 {0, 1} such a

Simultaneous Messages Protocols How to find for each x 2 {0, 1} such a `typical’ T n size t • Claim: a random choice of wi’s is good Proof by Chernoff – Need to `take care’ of every m. B (2|m. B| possibilities ) • Claim: for x x’ we have Tx Tx’ – Otherwise behaves the same when y = x for x and x’ • Let Sx be the m. B’s for which protocol mostly says ’ 1’ • Let Wx be the m. B’s for which protocol mostly says ’ 0’ • Then for y=x the distribution should be mostly on Sx • Conclusion: t ¢ |m. A| ¸ n and we get |m. A| x |m. B| = (n) [Babai Kimmel 1997] x of

General issue • What do combinatorial lower bounds men when complexity issues are involved?

General issue • What do combinatorial lower bounds men when complexity issues are involved? • What happens to the pigeon-hole principal when one-way functions (oneway hashing) are involved? • Does the simultaneous message lower bound hold when one-way functions exist – Issue is complicated by the model – Can define Consecutive Message Protocol model with iff results

And now for something completely different

And now for something completely different

Faculty members in Cryptography and Complexity • Prof. Uri Feige אורי פייגה • Prof.

Faculty members in Cryptography and Complexity • Prof. Uri Feige אורי פייגה • Prof. Oded Goldreich עודד גולדרייך • Prof. Shafi Goldwasser שפי גולדווסר • Prof. Moni Naor • Dr. Omer Reingold • Prof. Ran Raz מוני נאור עומר ריינגולד רן רז עדי שמיר • Prof. Adi Shamir One of the most active groups in the world!