Jump to content

Pseudorandom generator theorem

From Wikipedia, the free encyclopedia
This is an old revision of this page, as edited by 68.43.229.106 (talk) at 21:47, 27 November 2005. The present address (URL) is a permanent link to this revision, which may differ significantly from the current revision.

Pseudorandomness

In computational complexity a distribution is considered pseudorandom is no efficient computation can distinguish it from the true uniform distribution by a non-negligible advantage. Formally, a family of distributions Dn is pseudorandom if for any polynomial size circuit C, and any ε polynomial in n

|  ProbxU [C(x)=1] − ProbxD [C(x)=1] | ≤ ε

Pseudorandom generators

A function Gl: {0,1}l → {0,1}m, where l < m is a pseudorandom generator if:

  • Gl can be computed in time polynomial in l
  • Gl(x) is pseudorandom

Pseudorandom generator theorem

It can be shown that if there is a pseudorandom generator Gl: {0,1}l → {0,1}l+1, i.e. a generator that adds only one pseudorandom bit, then for any m = poly(l), there is a pseudorandom generator G'l: {0,1}l → {0,1}m.

The idea of the proof is as follows: first s bits from uniform distributin Ul are picked and used as the seed to the first instance of Gl, which is known to be a pseudorandom generator. Next, the output of the first instance of Gl is divided into two parts: first l bits are fed into the second instance of Gl as a seed, while the last bit becomes the first bit of the output. Repeating this process for m times yields an output of m pseudorandom bits.

It can be shown that such G'l, that consists of m instances of Gl, is indeed a pseudorandom generator by using hybrid approach and proof by contradiction as follows:

Let's consider m+1 intermediate distributions Hi: 0  ≤  i  ≤  m, where first i bits are chosen from the uniform distribution, and last m − i bits are chosen from the output of G'l. Thus, H0 is the full output of G'l and Hm is a true uniform distribution Um. Therefore, distributions Hi and Hi+1 will be different in only one bit (bit number i+1).

Now, assume that G'l is not a pseudorandom distribution, that is, there exists some circuit C that can distinguish between G'l and Um with an advantage ε  =   1/poly(l). In other words this circuit can distinguish between H0 and Hm. Therefore, exists such i that the circuit C can distinguish between Hi and Hi+1 by at least ε / m. Note, that since m are polynomial in l, then ε / m is also polynomial in l and is still a non-negligible advantage.

Now, assume we are given l+1 bits that are either output of Gl or a drawn from uniform distribution. Let's reuse the approach of building large pseudorandom generators out of instances of Gl and construct a string of pseudorandom bits of length m−i−1 in the same way the G'l was constructed above using the first l given bits as the seed. Then, let's create a string consisting of i bits drawn from uniform, last one of the given bits, and created m−i−1 bits. The resulting output is either Hi or Hi+1, since the i+1 bit is either drawn from uniform or from Gl. Since by assumption we can distinguish between Hi and Hi+1 with non-negligible advantage, then we can distinguish betwenn U and Gl, which implies that Gl is not a pseudorandom generator, which is a contradiciton to the hypothesis. Q.E.D.

Existence of pseudorandom generators

In complexity theory existence of pseudorandom generators is related to existence of one-way functions and hard-core predicates. Formally, pseudorandom generators exist if and only if one-way functions exist, or

PRG ↔ OWF

Definitions

One-way functions

Intuitively one-way functions are functions that are easy to compute and hard to invert. In other words the complexity (or circuit size) of the function is much smaller than one of its inverse. Formally: A function ƒ:  {0,1}n → {0,1}n is (S,ε) one-way if for any circuit C of size ≤ S,

Prob[ƒ(C(ƒ(x))) = ƒ(x)] ≤ ε .

Moreover, ƒ is a one-way function if

  • ƒ can be computed in polynomial time
  • ƒ is (poly(n), 1/poly(n)) one-way

Hard-core predicate

Function B: {0,1}n → {0,1} is a hard-core predicate for function ƒ if

  • B can be computed
  • for any polynomial size circuit C and any non-negligible ε = 1/poly(n), Probx~U [C(ƒ(x))  = B(x)] ≤ 1/2+ε

In other words it is hard to predict B(x) from function ƒ(x).

Proof

Here an outline of the proof is given. Please see references for detailed proofs.

PRG → OWF

Consider a pseudorandom generator Gl: {0,1}l → {0,1}2l. Let's create the following one-way function ƒ:  {0,1}n → {0,1}n that uses the first half of the output of Gl as its output. Formally,

ƒ(x,y) → Gl(x)

A key observation that justifies such selection, is that the image of the function is of size 2n and is a negligible fraction of the pre-image universe of size 22n.

To prove that ƒ is indeed a one-way function let's construct an argument by contradiction. Assume there exists a circuit C that inverts ƒ with advantage ε:

Prob[ƒ(C(ƒ(x,y)))  = ƒ(x,y)] > ε

Then we can create the following algorithm that will distinguish Gl from uniform, which contradicts the hypothesis. The algorithm would take an input of 2n bits z and compute (x,y) = C(z). If Gl(z) = z the algorithm would accept, otherwise it rejects.

Now, if z is drawn from uniform distribution probability of the algorithm above accepting is ≤ 1/2l, since the size of image is 1/2l of the size of the pre-image. However, if z was drawn from the output of Gl then probability of acceptance is > ε by assumption of existence of circuit C. Therefore, the advantage that circuit C has in distinguishing between the uniform U and output of Gl is > ε − 1/2, which is non-negligible and thus contradicts our assumption of Gl being a pseudorandom generator. Q.E.D.

OWF → PRG

For this case we prove a weaker version of the theorem:

One-way permutation → pseudorandom generator

One-way permutation is a one-way function that is also a permutation of the input bits. A pseudorandom generator can be constructed from one-way permutation ƒ as follows:

Gl: {0,1}l→{0,1}l+1  =  ƒ(x).B(x), where B is hard-core predicate of ƒ and "." is a concatination operator. Note, that by theorem proven above, it is only needed to show existence of a generator that adds just one pseudorandom bit.

Hard-core predicate → PRG

First, let's show that if B is a hard-core predicate for ƒ then Gl is indeed pseudorandom. Again, we'll use an argument by contradiction.

Assume that Gl is not a pseudorandom generator, that is there exists circuit C of polynomial size that distinguishes Gl(x) =ƒ(x).B(x) from Ul+1 with advantage ≥ε, where ε is non-negligible. Note, that since ƒ(x) is a permutation, then if x is drawn from uniform distribution, then so if ƒ(x). Therefore, Ul+1 is equivalent to ƒ(x).b, where b is a bit drawn independently from a uniform distribution. Formally,

Probx~U [C(G(x))=1] − Probx~U,b~U [C(x.b)=1]  ≥ ε

Let's construct the following algorithm C':

1. Given z=f(x) guess bit b 
2. Run C on z.b
3. IF C(z.b)=1
4.     output b
5. ELSE
6.     output 1-b

Given the output of ƒ the algorithm first guesses bit b by tossing a random coin, i.e. Prob[b=0] = Prob[b=1] = 0.5. Then, algorithm (circuit) C is run on f(x).b and if the result is 1 then b is outputted, otherwise the inverse of b is returned.

Then probability of C' guessing B(x) correctly is:

Probx~U [C'(z)=B(x)] =

Prob[b=B(x) ∧ C(z.b)=1] + Prob[bB(x) ∧ C(z.b)=0] =

Prob[b=B(x)]⋅Prob[C(z.b)=1 | b=B(x)] + Prob[bB(x)]⋅Prob[C(z.b)=0 | bB(x)] =

1/2⋅Prob[C(z.b)=1 | b=B(x)] + 1/2⋅Prob[C(z.b)=0 | bB(x)] =

(1−1/2)⋅Prob[C(z.b)=1 | b=B(x)] + 1/2⋅(1−Prob[C(z.b)=1 | bB(x)]) =

1/2+Probz.b~G(x) [C(z.b)=1] − 1/2⋅(Prob[C(z.b)=1 | b=B(x)]+Prob[C(z.b)=1 | bB(x)]) =

1/2+Probz.b~G(x) [C(z.b)=1] − Probz.b~U [C(z.b)=1] ≥ 1/2+ε

This implies that circuit C' can predict B(x) with probability more than 1/2 + ε, which means that B cannot be a hard-core predicate for ƒ and the hypothesis is contradicted. Q.E.D.

OWP → hard-core predicate

The outline of the proof is as follows:

If ƒ{0,1}n→{0,1}n is a one-way permutation, then so is ƒ'{0,1}2n→{0,1}2n. Then B(x,y)=xy is a hard-core predicate for ƒ', where is a vector dot product. To prove that it is indeed hard-core let's assume otherwise, and show a contradiction with the hypothesis of ƒ being one-way. If B is not a hard-core predicate, then there exists a circuit C that predicts it, so

Probx,y[C(ƒ(x),y)=xy] ≥  1/2+ε. That fact can be used to recover x by cleverly constructing permutations y that isolate bits in x. In fact, there exists a polynomial time algorithm that lists O(1/&epsilon2) candidates that include all valid x. Thus, an algorithm can reverse ƒ(x) in polynomial time, which contradicts the hypothesis.

References

  • W. Diffe, M.E. Hellman. "New Directions in Cryptography." IEEE Transactions on Information Theory, IT-22, pp.644-654, 1976.
  • A.C. Yao. "Theory and Application of Trapdoor Functions." 23rd IEEE Symposium on Foundations of Computer Science, pp.80-91, 1982.
  • M. Blum and S. Micali "How to Generate Cryptographically Strong Sequences of Pseudo-Random Bits." SIAM Journal on Computing, v13, pp.850-864, 1984.
  • J. Hastad, R. Impagliazzo, L.A. Levin and M. Luby. "A Pseudorandom Generator from any One-way Function." SIAM Journal on Computing, v28 n4, pp.-1364-1396, 1999.