Lecture 9: Pseudo-random generators against space bounded computation,

Size: px

Start display at page:

Download "Lecture 9: Pseudo-random generators against space bounded computation,"

Nickolas Henderson
5 years ago
Views:

1 Lecture 9: Pseudo-radom geerators agaist space bouded computatio, Primality Testig Topics i Pseudoradomess ad Complexity (Sprig 2018) Rutgers Uiversity Swastik Kopparty Scribes: Harsha Tirumala, Jiyu Zhag 1 Pseudo Radom Geerators agaist small space brachig programs I the previous lectures, we have see costructios of efficiet radomess extractors which help extract radom bits from a weakly radom source. I this lecture, we will use these radomess extractors to show that read-oce brachig programs operatig o low space ca be simulated (with very little error) usig at most O(log 2 ) radom bits. 1.1 defiitios Defiitio 1. A radom variable X o {0, 1} has miimum etropy H (X) k if x {0, 1} P r[x = x] 2 k Defiitio 2. A fuctio E : {0, 1} {0, 1} d {0, 1} m is a (k, ɛ) extractor if : radom variable X with H (X) k, (E(X, U d ), U m ) ɛ The radomess extractor E helps extract the radomess hidde i a weak radom source X by ivestig d bits of radomess (which are recovered i the process). We have already see the existece of expader-based extractors with the followig guaratees : E : {0, 1} {0, 1} d {0, 1} m which is a (k, ɛ) extractor for d = O( k + log( 1 ɛ )) We will use the above expader-based extractor to costruct a pseudo-radom geerator that fools LOGSP ACE read-oce brachig programs. Fact 3. If X 1 ad X 2 are idepedet radom variables with H (X 1 ) k 1 ad H (X 2 ) k 2, the f(x 1, X 2 ) has H k 1 + k 2 Theorem 4. For ay δ > 0, there is a pseudoradom geerator takig a uiformly radom seed of legth O(log 2 ) ad producig bits that δ-fool ay s = O(log ) space read-oce brachig program. Proof. The costructio is as follows. Let t, d be costats to be fixed later. Let G 0 : {0, 1} t {0, 1} be a fuctio that returs the first bit of the iput; i.e. G 0 (x) = x 1 1

2 Let G i {0, 1} t+id {0, 1} 2i be a fuctio such that G i (x, y) = G i 1 (x)g i 1 (E i 1 (x, y)), where x is the first t + (i 1)d bits of iput, ad E i : {0, 1} t+(i 1)d {0, 1} d {0, 1} t+(i 1)d is a (t + (i 1)d 2s, ɛ )-extractor, with ɛ to be chose later. We ca take E i to be the adjacecy map of a good absolute eigevalue expader. Let k = t + (i 1)d 2s, ad let = t + (i 1)d. If E i : {0, 1} {0, 1} d {0, 1} t+(i 1)d, the we get d = O( k + log( 1 ɛ )) = O(2s log( 1 ɛ )). To fool a read-oce brachig program B : {0, 1} {0, 1}, we will eed to use G log, which will use t + d log radom bits. It will tur out that we ca take ɛ = 1 poly(), so d = O(log ). So, we will eed a O(log 2 ) legth radom seed. Claim 5. G i (ɛ + 2 s )(2 i+1 1)-fools read-oce brachig programs of space s. We will prove the claim by iductio o i. For v i layer 2 i 1 of the brachig program, let w G i 1 (x), ad let p v = P r[b(w) = v]. Let X v = X B(w)=v. Claim 6. H (X v ) t + (i 1)d log( 1 p v ). Call v uimportat if p v 2 2s. The the probability of edig up i ay uimportat state at all is give by Σ v p v 2 s 2 2s 2 s. For importat states v, we have H (X v ) t + (i 1)d 2s. Pick v accordig to p v. By iductio, this is (ɛ + 2 s )(2 i+1 1)close to the v chose from B(z), for uiformly radom z. If v is good, the E(x, v) is ɛ -close to U t+(i 1)d, so G i 1 (E(x v, y)) is ɛ -close o G i 1 (U t+(i 1)d ). Let B v be the legth 2 i 1 brachig program startig at v. We kow B v (G i 1 (U t+(i 1)d )) is ɛ -close to B v (G i 1 (U 2 i 1)). By iductio, B v (G i 1 (U t+(i 1)d )) is (ɛ + 2 s )(2 i 1)-close to B v (U 2 i 1). So, for v chose from the distributio of G i 1 (U t+(i 1)d ) ad v chose from the distributio of B(U 2 i 1), we have that (v, B v (G i 1 (E(x, y)))) is (ɛ +2 s )(2 i 1)+2 s +ɛ +(ɛ +2 s )(2 i 1)-close to (v, B v (U 2 i 1)). Sice i log, provided that s is a sufficietly large multiple of log ad ɛ 2 log is sufficietly small relative to ɛ, it follows that G log ɛ-fools ay brachig program of size s, usig O(log 2 ) radom bits. 2 Primality Testig Problem: Give a iteger, decide if is prime. Moreover, we wat to decide this i time poly(log ). (Sice the iput size is log ) 2

3 History of Algorithmic Approach 1976 Miller-Rabi (Radomized) 1999 Agrawal, Biswas (Radomized) 2002 Agrawal, Biswas (Determiistic) Radomized Algorithm[AB99] We will ow discuss the radomized algorithm due to Agrawal ad Biswas. Cosider the polyomial (x + 1), which expaded to be x + ( ) 1 x ( 1) x + 1. We have the followig facts: 1. If is prime, the for all 0 < i <, ( ) i. (By we mea divides ( i) ) The we have 2. If is prime, (x + 1) x + 1 (mod ). Note that the cogruece relatio above is a cogruece of polyomials with iteger coefficiets: A(x) B(x) (mod ) if A(x) B(x) = C(x) for some C(x) Z[x]. Lemma 7. is prime iff (x + 1) x + 1 (mod ) Proof: Give the fact 2 above, we wat to prove the other directio by cotrapositio. To be specific, we wat to show that if is composite, the i, 0 < i < such that ( i). Here is a quick observatio: if p, the ( ) p = ( 1) ( p+1) p 1 where the above is divided by p below, so o loger divides ( p). A Naive Algorithm A aive algorithm is to thik that we ca utilize methods i polyomial idetity testig. To be specific, the algorithm goes as below: 1. Pick radom x {0,..., 1} 2. Check if (x + 1) x 1 0 (mod ) But this approach does t work. Why? Because i polyomial idetity testig we eed to be prime as prerequisite to esure that we choose x from a field.(this may ot seem obvious but we have aother reaso as follows). I additio, we require that the degree of the polyomial is low i the sese that it must be less tha the field size(or the umber of possible values) of x. While i the above aive algorithm, the size ad degree are both, ad i fact ca be extremely large. Istead, we have the followig modified algorithm due to Agrawal ad Biswas. 3

4 Agrawal ad Biswas Algorithm 1. Pick radom polyomial Q(x) of degree d where d is of size O(log ). 2. Check if (x + 1) x 1 0 (mod, Q(x)). 3. If yes, the output Prime, else output Composite. Proof of Correctess: It is easy to see that there s o false egative give the lemma above. That is to say, whe is prime it always outputs the correct aswer. We d like to show that if is ideed composite, the with high probability over the choice of Q(x), (x + 1) x 1 0 (mod, Q(x)) holds. Now cosider a prime p where p (so is composite). Let p i be the largest power of p that divides, so = p i s for some s. Cosider ( ) p = ( 1) ( p i +1), it s easy to see that ( ) i p i (p i 1) 1 p 0 (mod p). i This idicates that give composite iteger, the both (x + 1) x 1 0 (mod ) ad (x + 1) x 1 0 (mod p) hold. Now give is composite, we wat to show that if we pick Q(x) at radom, the with high probability (x + 1) x 1 0 (mod Q(x), p). By above discussio we have (x + 1) x 1 0 (mod Q(x), ) with high probability. Notice that pickig a Q(x) of degree d = O(log ) ad do operatios modulo p is the same as pickig a Q(x) of degree d with coefficiets i Z p. (that is to say, pick Q(x) from F p [x]). Assume that the algorithm outputs Prime with probability at least α, that is, which is equivalaet to P r [(x + Q(x) Z[x],deg(Q) d 1) x 1 0 (mod Q(x), )] α the P r [Q(x) (x + Q(x) F 1) x 1] α p[x],deg(q) d P r[q(x) is ot irreducible] + P r [Q(x) is irreducible but Q(x) (x + Q Q 1) x 1] 1 α (Note that (x + 1) x 1 has uique factorizatio of irreducible polyomials. Also i above we write rage Q below P r just for coveiece, the rage is the same as above.) Fact 8. Therefore, P r Q(x) F p[x],deg(q)=d [Q(x) is a irreducible polyomial ] 1 d P r [Q(x) is irreducible ad Q(x) (x + Q 1) x 1] 1 d (1 α) Now we prove by cotradictio. If α 1 1 have at least pd 1 1,the the above probability is at least. So we such Q(x) that divides (x + 1) x 1. Give (x + 1) x 1, there are at most /d irreducible factors of degree at most d. Therefore, we should have pd 1 d. But sice 4

5 we have d = poly(log ), we reach a cotradictio. Therefore we have P r [(x + Q(x) Z[x],deg(Q) d 1) x 1 0 (mod Q(x), )] 1 The remaiig key issue i aalysis of this algorithm is that how to compute (x + 1) x 1 mod Q(x) mod p i a reasoable time. But we shall see that this ca be doe by repeatedly squarig, ad it is i time O(log ). 5

Lecture 9: Expanders Part 2, Extractors

Lecture 9: Expanders Part 2, Extractors Lecture 9: Expaders Part, Extractors Topics i Complexity Theory ad Pseudoradomess Sprig 013 Rutgers Uiversity Swastik Kopparty Scribes: Jaso Perry, Joh Kim I this lecture, we will discuss further the pseudoradomess