Lecture 8: Pesudorandom Generators (II) 1 Pseudorandom Generators for Bounded Computation

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1 Expander Graphs in Computer Science WS 2010/2011 Lecture 8: Pesudorandom Generators (II) Lecturer: He Sun 1 Pseudorandom Generators for Bounded Computation Definition 8.1 Let M be a randomized TM that on input w requires l( w ) random bits. The family { n } n=1 of functions ( n : {0, 1} s(n) {0, 1} l(n) ) is an ε enerator for M if for any w {0, 1}, it holds that ( Pr M(w, r) = 1] Pr M w, w (z) ) = 1 ] < ε r {0,1} l( w ) z {0,1} s( w ) Instead of a family of functions, it is more convenient to view, the ε enerator for M as a TM that on input a strin of lenth s(n) outputs a strin of lenth l(n). We call w the seed of the PG. 2 andomized Lospace TM In this lecture, we present a PG that fools all randomized lospace TMs. Before proceedin any further, we have to clarify a point reardin how the random bits are accessed in a randomized space bounded computation (lospace in our case). There are two varyin definitions of randomized space bounded TMs based on the manner the random bits are accessed: The TM obtains the random bits as and when required by it. The strin of random bits is fed as an auxiliary off-line input (on a separate tape) in addition to the reular input to the TM. In this case, the head accessin the random bits on this tape can move back and forth on the tape. It is to be noted that in the case of randomized time bounded computation it is immaterial which convention we observe. For randomized space bounded computation, we consider only TMs of the first kind or equivalently consider TMs of the second kind in which the head on the random tape is restricted in the sense that it can only move riht alon the tape (i. e., the random tape is one-way read-only tape). We also assume that all randomized space S TMs halt in time less than 2 S. 3 Nisan s Generators The main task of this lecture is to construct the followin PG. 1

2 2 Lecture 8: Pesudorandom Generators (II) 2 Lecture 8: Pesudorandom Generators (II) 3 Nisan s Generators Theorem 8.2 There exists an n-space-bounded TM G : {0, 1} O(lo2 m) {0, 1} m such that for Theorem all randomized 8.2 There S-space-bounded exists a SPACE(n) TM M, TM G is G, ag 2 S : {0, -enerator 1} O(lo2 m) for M, {0, 1} where m such m that = 2 S forand n = all O(lo randomized 2 m). lospace TM M, G is a 1 n enerator for M.1 Before discussin the space-bounded PGs, let us look at the structure of computation tableaubefore of a randomized discussin the TM. space-bounded For an S space-bounded PGs, we studytm them, structure the of computation tableau of M consists of a randomized of at most TM. 2 S rows For aand space each S-bounded row corresponds TM M, the tocomputation a confiuration tableau of M, of Masconsists shown in Fiure of at 1. most 2 S rows and each row corresponds to a confiuration of M, as shown in 1. S 2 S andom bits Fiure 1: The computation tableau of a space S-bounded TM Fiure 1: The computation tableau of a space S-bounded TM By the definition of randomized TM, each step of M s processin requires one random By the definition of randomized TMs, each step of M s processin bit and the number of random bits required for is at most 2 S requires one random. For simplicity we assume bit and that the M requires number 2 S of random random bits. bits Consider required the for tableau M is at divided most into 2 S. two For simplicity with each we assume half thatrequirin the runnin random time strins of Mr 1 isand 2 S. r 2 Consider respectively theeach tableau of lenth divided r = into 2 S /2. two If rhalves 1 and r 2 with are chosen each half requirin independently, random then strins by rdefinition 1 and r 2 the respectively randomized each TMofcan lenth output r = the 2 S correct /2. If ranswer. 1 and r 2 However, are chosen independently, we would like then to choose by definition r 1 and Mr 2 in cansuch output a manner the correct that their answer. behavior However, is not sinificantly we would like to choose different r 1 from and the r 2 incase such when a manner they arethat chosen their independently. behavior is not sinificantly different from the case when To put theythins are chosen more formally, independently. we have 2 alorithms A 1 and such that Formally, we assume that A 1 and are the upper and lower halves respectively of the computation tableau of M and x 1 We shall infact prove a more eneral 1 and x result 2 are independently chosen random strins, see Fiure 2. that for all randomized space S TM M, G is a 1 enerator for 2 Moreover, S M. alorithms A 1 and are of the followin form: Alorithm A 1 takes an input x 1 of lenth r and outputs a strin b 1 of lenth c. Alorithm takes as input the output b 1 of A 1 and another strin x 2 of lenth r and outputs a strin b 2 of lenth c.

3 Lecture 8: Pesudorandom Generators (II) 3 A 1 x 1 b 1 x 2 b 2 Fiure 2: A 1, with random inputs What we are in search of is a enerator that supplies strins x 1 and x 2 in a fashion better than choosin them independently. For notational brevity, iven a function, we use l (z) and r (z) to express the left half and the riht half of the strin (z), i. e., (z) = l (z) r (z) and l (z) = r (z). See Fiure 3. A 1 l (z) b 1 r (z) b 2 Fiure 3: A 1, with inputs from enerator Definition 8.3 A function : {0, 1} t {0, 1} r {0, 1} r is defined to be a ε-enerator for communication c if for all functions A 1 : {0, 1} r {0, 1} c and : {0, 1} c {0, 1} r {0, 1} c, we have that b Pr (A 1 (x 1 ), x 2 ) = b] Pr (A 1 ( l (z)), r (z)) = b] < ε. x 1,x 2 {0,1} r z {0,1} t For simplicity, we call a 2 c -enerator for communication c a c-enerator. For the present we shall assume the followin lemma hold and use expanders to prove this lemma in the end of the lecture. Lemma 8.4 There exists a constant k > 0 such that for all r, c, there exists a polynomial time computable c-enerator of the form : {0, 1} r+kc {0, 1} r {0, 1} r. Now we break the tableau into several components, called A 1,,, S, and let the output of be the random strins for every pair of consecutive components (i 1 and i ) such that their behavior is not sinificantly different from usin pure random bits, see Fiure 4. Moreover, we hope that every application of causes error at most 1/2 2S. Because we invoke at most 2 S 1 times, so the total error is at most 2 S 1 1/2 2S. Notice that in the above framework there are 2 S 1 components each of which require +ks random bits. We use the same approach to reduce the number of random bits required by every

4 4 Lecture 8: Pesudorandom Generators (II) A 1 + ks A 3 + ks A 4 S 1 + ks S Fiure 4: The first level of the recursive construction A ks 2 A ks A 4 S + S 2 + ks 2 S ks S Fiure 5: ecursive construction of G S pair of components from + ks each to + 2kS total. We perform this operation recursively till there is only one component left. See Fiure 5. Formally let i : {0, 1} +iks {0, 1} +(i 1)kS {0, 1} +(i 1)kS be an S-enerator for i = 1,, S. Define G i : {0, 1} +iks {0, 1} 2i inductively as follows: G i (z) = { z i = 0 G i 1 ( l i (z) ) G i 1 ( r i (z)) i > 0 Lemma 8.5 The PG G S runs in space O ( + ks 2). Lemma 8.6 For all space S-bounded TM M, G S is a 2 S -enerator for M. Proof: Since each application of i incurs an error of at most 1/2 2S, and there are 2 S S = 2 S 1 applications of i, therefore by the union bound the error probability is at most (2 S 1)/2 2S 1/2 S.

5 Lecture 8: Pesudorandom Generators (II) 5 4 Proof of Lemma 8.4 Lemma 8.7 Let G = (V, E) be a d-reular raph with spectral expansion λ. subsets S, T V we have E(S, T ) S E V T V λ S V T V. Then for any Proof: Apply the Expander Mixin Lemma on the double coverin of G. Now we prove Lemma 8.4. Proof: Let G = (V, E) be a d = 2 6c -reular amanujan expander on V = 2 r vertices. The enerator : {0, 1} r+6c {0, 1} r {0, 1} r works as follows: On input z = (x, i) {0, 1} r {0, 1} d, output ( l (z), r (z) ) = (x, y) where y is the vertex reached by takin the i-th ede out of x. Let b be any output of the alorithms (A 1, ). Fix b. For any b {0, 1} c, define S b = { x {0, 1} r A 1 (x) = b }, T b = { x {0, 1} r (b, x) = b }. Thus for truly random strins x 1 and x 2, the probability that alorithms (A 1, ) outputs b is expressed by Pr (A 1 (x 1 ), x 2 ) = b] = Prx 1 S b x 2 T b ] x 1,x 2 = S b V T b V. On the other hand, when usin the output of to instead truly random bits, the probability that (A 1, ) outputs b becomes ] Pr (A 1 ( l (z)), r (z)) = b = Pr x S b z {0,1} r+d x,i ith ede out of x leads to T b ] Therefore = E(S b, T b ). E Pr (A 1 (x 1 ), x 2 ) = b] Pr (A 1 ( l (z)), r (z)) = b] x 1,x 2 z {0,1} r+d ( = Sb V T b V E(S b, T b ) ) E b {0,1} c S b V T b V E(S b, T b ) E. Because S b and T b are subsets of V, by Lemma 8.7 we know that S b V T b V E(S b, T b ) E λ S b V T b λ, V

6 6 Lecture 8: Pesudorandom Generators (II) where λ is the spectral expansion of G and satisfies λ 2 d 1/d. Therefore Pr (A 1 (x 1 ), x 2 ) = b] Pr (A 1 ( l (z)), r (z)) = b] x 1,x 2 z {0,1} r+d λ 2 d 1 d <2 c 1 2 c c