Cryptography and Privacy Preserving Operations

Lecture 2: Pseudo-randomness

Lecturer: Moni Naor

Weizmann Institute of Science

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)• Encryption: easy when you share very long strings• Started with the notion of pseudo-randomness

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

• 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

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

• 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?• 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 rR {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

• 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: IndistinguishabilityTwo probability distributions that cannot be distinguished– Statistical indistinguishability: distances between probability

distributions– New notion: computational indistinguishability

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 generatorsDefinition: 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’| yR {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

• Why is the adversary bounded by polynomial time?• Why is the indistinguishability not perfect?

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 PredicateDefinition: 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

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

• 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 whereg(x)=(f(x), h(x))

Claim: g is a pseudo-random generatorProof: can use a distinguisher for g to guess h(x)

f(x), h(x)) f(x), 1-h(x))

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• 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• 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: Algorithm A’ for inverting f• There are many x’s for which A returns a correct answer (r ∙x ) on ½+ε of the

r ’s • Reconstruction algorithm R: 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 by R (except feeding to A)

• Choose from the list the/an x such that f(x)=y

The main step!

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

A: algorithm for guessing r¢xR: Reconstruction algorithm that outputs a list of candidates for xA’: algorithm for inverting f on a given y

y,r1

z1 =r1 ¢ x

y

x1 ,x2 xk

A

R

?

z2 =r2 ¢ xA ?

zk =rk ¢ xA ?

y,r2

y,rk

z1, z2, zk

y

Check whether f(xi)=y xi=x

A’

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)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• 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

AnalysisNumber 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

x f(x) h(x,r)

n n+1

Reducing the size of the list of candidatesIdea: bootstrapGiven 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 executionsThey 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 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

• 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+mn 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 GeneratorIdea 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-randomClaim: 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

• A tool for constructing pseudo-random generatorsThe 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

CompositionLet • 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

ordistinguish between D2 and D3 - can use A use to distinguish g1

ℓ2

ℓ1

ℓ3

triangle inequality

Composing PRGsWhen 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 timesLoss 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

From single bit expansion to many bit expansion

• Can make r and f(m)(x) public – But not any other internal state

• Can make m as large as needed

x f(x) h(x,r)

OutputInternal Configuration

r

f(2)(x)

f(3)(x)

Input

h(f(x),r)

h(f (2)(x),r)

h(f (m-1)(x),r)f(m)(x)

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

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 proveTheorem [HILL]: if one-way functions exist, then pseudo-

random generators existExercise: show that if pseudo-random generators exist, then

one-way functions exist

Next-bit TestDefinition: 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 ifand only if it is a pseudo-random generator

Next-block UndpredictableSuppose 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 testFor 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.

G:

S y1 y2, … ,

Pseudo-Random Generatorsconcrete version

Gn:0,1m 0,1n

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

• Identification• Authentication• EncryptionSolve in a shared key environment

S S

Identification - Remote login using pseudo-random sequence

A and B share key S0,1k

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)

G:

S

Problems...

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

Challenge-Response Protocol

• B selects a random location and sends to A • A sends value at random location

What’s this ?

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

• A wants to send message M0,1n to B• B should be confident that A is indeed the sender of M

One-time application:S =(a,b) -

where a,bR 0,1n

To authenticate M: supply aM bComputation is done in GF[2n]

Problems and Solutions

• Problems - same as for identification• If a very long random string available -

– can use for one-time authentication– Works even if only random looking

a,b

Use this !

Encryption of Messages

• A wants to send message M0,1n to B• only B should be able to learn M

One-time application:S = a

where aR 0,1n

To encrypt M: send a M

Encryption of Messages

• If a very long random looking string available - – can use as in one-time encryption

Use this !

Pseudo-random Functions

Concrete Treatment:F: 0,1k 0,1n 0,1m

key Domain Range

Denote Y= FS (X)

A family of functions Φk ={FS | S0,1k is (t, , q)-pseudo-random if it is

• Efficiently computable - random access and...

(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 Φkor

– FS R R n m = F | F :0,1n 0,1m

(t,,q)-pseudo-random

For a function F chosen at random from(1) Φk ={FS | S0,1k

(2) R n m = F | F :0,1n 0,1m

For all t-time machines A that choose q locations and try to distinguish (1) from (2)

ProbA ‘1’ FR Fk

- ProbA ‘1’ FR R n m

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• A pseudo-random sequence generator g:0,1m

0,1n – a pseudo-random function on small domain 0,1log n0,1 with

key in 0,1m

Application to the basic issues in cryptography

Solution using a shared key SIdentification:

B to A: X R 0,1n

A to B: Y= FS (X) A verifies

Authentication:A to B: Y= FS (M)

replay attack

Encryption:A chooses XR 0,1n A to B: <X , Y= FS (X) M >

Goal

• Construct an ensemble {Φk | kL such that

• for any {tk, 1/k, qk | kL polynomial in k, for all but finitely many k’s

Φk is a (tk, k, qk )-pseudo-random family

Construction

• Construction via Expansion– Expand n or m

• Direct constructions

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 R0,1k

Claim: If Φk ={FS | S0,1k is (t,,q)-pseudo-random: cannot distinguish two cases

– using q queries – in time t’=t - ℓq – with advantage better than ℓ

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

Can use this i to distinguish whether – FS R Φkor 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’ ℓq

Doubling the domain

• Suppose F(n): 0,1k 0,1n 0,1m

which is (t,,q)-p.r.• Want F(n+1): 0,1k 0,1n+1

0,1m which is (t’,’,q’)-p.r.Use G: 0,1k 0,12k which is (t ,) p.r

G(S) G0(S) G1(S)

Let FS (n+1)(bx) FGb(s)

(n)(x)

Claim

If G is (tq,1)-p.r and F(n)is (t2q,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

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)

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

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 claimIf G is (t qn ,) p.r, then F(n)is (t, ’ nq,q) p.rProof: 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

S0 S1

?S

G0(S0)

G1(G0(S0))

n-i

i

Di

…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

• 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 pseudo-

random functions exist.

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

• Blum-Micali : SIAM J. Computing 1984 • Yao:• Blum, Blum, Shub: SIAM J. Computing, 1988• Goldreich, Goldwasser and Micali: J. of the

ACM, 1986

...ReferencesBooks:• 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

• ,

ReferencesWeb material/courses• S. Goldwasser and M. Bellare, Lecture Notes on Cryptography,

http://www-cse.ucsd.edu/~mihir/papers/gb.html

• Wagner/Trevisan, Berkeleywww.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