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

Lecture 2 : Authentication and Communication

Complexity

Lecturer: Moni Naor

Weizmann Institute of ScienceWeb 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 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:– Amplification of weak one-way functions– Distributionally one-way functions

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 in cryptography– Many variants

• Relevant both in communication and in storage

The authentication problemone-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

Alice Bob

Eve

m

Specification of the ProblemAlice and Bob communicate through a channelBob has an external register R N N (no message) ⋃ {0,1}n Eve completely controls the channelRequirements:• Completeness: If Alice wants to send m {0,1}n and Eve

does not interfere – Bob has value m in RR • Soundness: If Alice wants to send m and Eve does interfere

– RR is either NN or m (but not m’ ≠m )– If Alice does not want to send a message RR is NN

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, the

probability that Bob is in state m’ ≠ m or NN is at most ε

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 RR to m’• If not equal, register R R stays in NN

• 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|– Easy to compute h(m) given h and m

Universal hash functions• Given that for hH we have h: {0,1}n → {0,1}k we know that

ε≥2-k • A family where this is an equality is called universal2

Definition: a family of functions H= {h| h: {0,1}n → {0,1}k } is called Strongly Universal2 or pair-wise independent if: – for all m1, m2 {0,1}n and y1, y2 {0,1}k we have

Prob[h(m1) = y1 and h(m2) = y2 ] = 2-2k Where the probability is over a randomly chosen h H

In particular Prob[h(m2) = y2 | h(m1) = y1 ] = 2-k

when a strongly universal2 family is used in the protocol, Eve’s probability of cheating is at most 2-k

Constructing universal hash functionsThe linear polynomial construction: • fix a finite field F of size at least the message space 2n

– Could be either GF[2n] or GF[P] for some prime P ≥ 2n • 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 universal2 Proof: for every m1, m2 , y1, y2 F there are unique a, b

F such thata∙m1+b = y1

a∙m2+b = y2

Size: each hH represented by 2n bits

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(2n) More precisely, to obtain a d-wise independence family

|H| should be Ω(2n└d/2┘)Theorem: see

N. Alon and J. Spencer, The Probabilistic MethodChapter 15 on derandomization, proposition 2.3

An almost perfect solutionBy 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 δ-

Universal2 if for all m1, m2 {0,1}n where m1 ≠ m2

we have

Prob[h(m1) = h(m2) ] ≤ δ

An almost perfect solutionIdea: combine • a family of δ-Universal2 functions H1= {h| {0,1}n → {0,1}k }with • a Strongly Universal2 family H2= {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 h1

H1 and h2 H2

h(x) = h2(h1(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 h1 Or in h2

Or– y’ ≠ h(m) which means that Eve succeeds with probability at most 2-k

• Collision in h2

Size: each hH represented by log |H1 |+ log |H2|

Constructing almost universal hash functionsThe polynomial evaluation construction {0,1}n → {0,1}k : • fix a finite field F of size at least the target space 2k

– Could be either GF[2k] or GF[P] for some prime P ≥ 2k • Let n = ℓ ∙ k• Treat each (non-zero) message m{0,1}n as a degree (ℓ-1)-

polynomial over F. Denote by Pm

• 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-Universal2 for δ= (ℓ-1)/2k Proof: the maximum number of points where two different degree (ℓ-1)

polynomials agree is ℓ-1 Size: each hH represented by k bits

m

Parameters for authentication

• To authenticate an n bits message with probability of error ε

Need:• Secret key length: (log n + log 1 / ε )

• Added tag length (log 1 / ε)

log n Lower bound does not hold for interactive protocols

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 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?

public encodingpx

AuthenticatorsHow to authenticate a large and unreliable memory with a small and secret memory

• Encoding Algorithm E:– Receives a vector x 2 {0,1}n, encodes it into:

• a public encoding px

• a small secret encoding sx. Space complexity: s(n)• Decoding Algorithm D:

– Receives a public encoding p and decodes it into a vector x 2 {0,1}n

• Consistency Verifier V:– Receives public px’ and secret sx encodings, verifies whether decoder output = encoder input– Makes few queries to public encoding: query complexity: t(n)

• An adversary sees (only) the public encoding and can change it

E

secret encoding sx

V

Dpublic encodingpy

acceptreject

x {0,1}n xy

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

– 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 authenticatorSuppose 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 computationally indistinguishable iffor 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 FunctionsLet {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 | k0,1}s } is pseudo-random if it is

• Efficiently computable - random access and...

Pseudo-random

Any polynomial time 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– Fk R Φnor– Fk R Rm ℓ = { F | F :{0,1}m {0,1}ℓ }

Not important for us

Pseudo-random

For a function F chosen at random from(1) Φn = {Fk | k0,1s

(2) Rm ℓ = { F | F :{0,1}m {0,1}ℓ }

For all polynomial time machines A that choose q locations and try to distinguish (1) from (2) for all polynomial 1/p(n)

ProbA ‘1’ FR Fk

- ProbA ‘1’ FR R n m 1/p(n)

Equivalent/Non-Equivalent Definitions

• Instead of next bit test: for XX1,X2 ,,

Xqchosen by A, decide whether given Y is – Y= FS (X) or

– YR0,1m

• Adaptive vs. Non-adaptive• Unpredictability vs. pseudo-randomness

Really what we need

Existence of Pseudo-Random functions and Authenticators

• If one-way functions exist so pseudo-random generators

• If pseudo-random generators exist, so do pseudo-random functions

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/

Probability of error

So are we done

Two problems:• Need

– computational bounded adversary and – one-way functions

• Efficiency: the evaluation of a pseudo-random function might be a burden to add to every memory fetch operation

Communication Complexity

Alice

x2X

Boby2Y

Let f:X x Y Z

Input is split between two participantsWant to compute outcome: z=f(x,y)

while exchanging as few bits as possible

A protocol is defined by the communication tree

z0 z1 z2 z3 z4 z5 z6 z7 ...

Alice: 0

Bob: 1

Alice: 0

Bob: 1

z5

A ProtocolA 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 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 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 of the multiset x [ y If the size is even then element ranked |x[ y|/2

Claim: D(MED) is O(log2 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 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 (x1,y1) 2 R and

(x2,y2) 2 R implies that (x1,y2) 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 determine z=f(x,y)

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 (x1,y1), (x2,y2) 2 S either

• f(x1,y2)≠z or• f(x2,y1)≠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) ≥ log2 t

z

zx1

x2

y1 y2

ApplicationsEquality: 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

ALICE

BOBy

x

Random coins

Probabilistic Communication Complexity• Alice an Bob have each, in addition to their inputs, access to random strings of

arbitrary length rA and rB (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, rA ) or bv(y, rB ) – Each leaf is labeled with an element z 2 Z

Take all probabilities over the choice of rA and rB • 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)=0Pr[P(x,y)=0]=1

and for all (x,y) s.t. f(x,y)=1

Pr[P(x,y)=1]¸ 1-

Measuring Probabilistic Communication ComplexityFor input (x,y) can consider as the cost of protocol P on input (x,y) either • worst-case depth• average depth over rA and rB

Cost of a protocol: maximum cost over all inputs (x,y)

The appropriate measure of probabilistic communication complexity:

• R0(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< <½R(f) = R1/3(f):

• R1(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. R1(f) = R½

1(f):

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 2RH

Pr[(h(x)=h(y)]· Protocol: Alice: pick random h 2RH 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 n2 · q · 2n2 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 source of bits.Possible view: distribution over deterministic protocols

Let Rpub(f): be the minimum cost of a public coins protocol computing f

correctly with probability at least 1- for any input (x,y)

Example: Rpub(Equality) = (-log )

Theorem: for any Boolean f: R(f) is R

pub(f)+O(log n + log 1/

Proof: choose t = 8n/2 assignments to the public string…

Simulating large sample spaces• 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

Pr[Ax,y] · e-22t < 2-2n

Pr[[x,y A x,y] · x,y Pr[A x,y] <22n 2-2n=1

Good 1-

Bad

Collection that should resemble probability of success on ALL inputs

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?

mA

mB

f(x,y)

ALICE

BOB

CAROL

y

x

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 probability of error

Provided the random bits are chosen independently than the inputs

Simultaneous Messages Protocols

• Suggested by Yao 1979

mA

mB

f(x,y)

x {0,1}n

y {0,1}n

ALICE

BOB

CAROL

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

– |mA| x |mB| = O(n) – Let C be a good error correcting code– Alice and bob arrange C(x) and

C(y) in an |mA| x |mB| rectangle• Alice sends a random row• Bob send a random columns• Carol compares the intersection

Simultaneous Messages Protocols• Lower bounds for the equality function:

– |mA| + |mB| = (√n) [Newman Szegedy 1996]

– |mA| x |mB| = (n) [Babai Kimmel 1997]

• Idea: for each x 2 {0,1}n find a `typical’ multiset of messages Tx = {w1, w2,…,wt} where

t 2 O(|mB|)

Each wi is a message in the original protocol, |mA| bits

Property: for each message mB the behavior on Tx approximates the real behavior

Average behavior of Carol on w1, w2,…,wt is close to its average response during protocol

Over random i, and randomness of Carol

Over randomness of Alice and Carol

Simultaneous Messages ProtocolsHow to find for each x 2 {0,1}n such a `typical’ Tx of size t

• Claim: a random choice of wi’s is goodProof by Chernoff

– Need to `take care’ of every mB (2|mB| possibilities )

• Claim: for x x’ we have Tx Tx’– Otherwise behaves the same when y = x for x and x’

• Let Sx be the mB’s for which protocol mostly says ’1’ • Let Wx be the mB’s for which protocol mostly says ’0’• Then for y=x the distribution should be mostly on Sx

• Conclusion: t ¢ |mA| ¸ n and we get|mA| x |mB| = (n) [Babai Kimmel 1997]

General issue

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

• What happens to the pigeon-hole principal when one-way functions (one-way 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

Faculty members in Cryptography and Complexity

• Prof. Uri Feige

• Prof. Oded Goldreich

• Prof. Shafi Goldwasser

• Prof. Moni Naor

• Dr. Omer Reingold

• Prof. Ran Raz

• Prof. Adi Shamir

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שפי גולדווסר

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