6.842 Randomness and Computation February 18, Lecture 4

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1 6.842 Randomness and Computaton February 18, 2014 Lecture 4 Lecturer: Rontt Rubnfeld Scrbe: Amartya Shankha Bswas Topcs 2-Pont Samplng Interactve Proofs Publc cons vs Prvate cons 1 Two Pont Samplng 1.1 Error Reducton Let s say we are gven a language L and an algorthm A n RP whch uses random bts r R {0, 1} R x L = P r[a(x, r) = 1] 1 2 x L = P r[a(x, r) = 1] = 0 How do we reduce error? Repeat A wth k dfferent values of r {r 1...r k }. Let a = A(x, r ) {1...k} and r = {r 1,..., r k }. Defne A (x, r ) = k a. Clam 1 Gven r R {0, 1} kr, error probablty s reduced to 1 2 k.e. Proof x L = P r[a (x, r) = 1] k x L = P r[a (x, r) = 1] = 0 If x L, A (x, r) = k A(x, r ) = k 0 = 0 If x L, A (x, r) = k A(x, r ) = P r[a (x, r) = 0] 1 2 A uses k R random bts. Can we do better? k = P r[a (x, r) = 1] k 1.2 Usng Parwse Independence to Reduce Randomness Defnton 2 A famly of hash functons H = {h : A B} s parwse ndependent f a 1 a 2 A and b 1 b 2 B and gven h R H P r[h(a 1 ) = b 1 h(a 2 ) = b 2 ] = 1 B 2 (1) Consder the famly of parwse ndependent hash functons H : {0, 1} k+2 {0, 1} R. Let h R H samplng h requres O(k + R) random bts. 1

2 Algorthm Pck h R H for = k+2 r = h() f A(x, r ) = 1 Output 1 (Accept) Output 0 (Reject) If x L A(x, r ) = 0 for all random strngs r. So, the algorthm outputs Reject. If x L, Defne 0, f A(x, r ) = 0. c(r ) = 1, otherwse. (2) E[c(r )] = P r[c(r ) = 1] > 1 2 Let Y = q=2 k+2 c(r ) = E[ Y q ] = E[Y ] q > 1 2 Chebyshev s Inequalty If X s a random varable and E[X] = µ then P r[ X µ ɛ] var[x] Lemma 3 If X 1, X 2,..., X n are parwse ndependent random varables, V ar[ n X ] = Proof n V ar[x ]. n n n V ar[ X ] = E[( X ) 2 ] E[( X )] 2 (3) = E[( n X X j )] ( E[X ]) 2 (4),j = E[X X j ] E[X ]E[X j ] (5),j,j = (E[X 2 ] E[X ] 2 ) (E[X X j ] E[X ]E[X j ]) (6) j = V ar[x ] 0 = V ar[x ] (7) Snce parwse ndependence = E[X X j ] = E[X ]E[X j ] j. So, f X = X and µ = E[X], then P r[ X µ > ɛ] = V ar[ n X] n ɛ = 2 V ar[x] ɛ = V ar[x] 2 2

3 Parwse Independent Tal Inequalty If X s a random varable and E[X] = µ, P r[ X µ ɛ] var[x] So, P r[ Y q = 0] P r[ Y/q E[Y/q] E[Y/q]] 1 q E[ Y q ]2 < 4 q = 1 2 k. Remark tme of the algorthm. Usng ths algorthm reduces the randomness complexty but greatly ncreases the runnng The runnng tme s now O(2 k+2 T A (n)) rather than O(k T A (n)). 2 Interactve Proofs Generalzaton of NP 2.1 NP vs IP Defnton 4 NP s the class of all languages L for whch an yes (x L) answer s verfable n polynomal tme by a determnstc Turng Machne. Defnton 5 Consder a model wth a Prover P and a Verfer V. V s bounded n polynomal tme and can toss cons (non-determnstc). P has unbounded tme and s determnstc. (No pont beng randomzed snce tme s unbounded) V and P can send nformaton to each other through conversaton tapes. V s random bts are prvate P doesn t know what they are. An Interactve Proof System for a language L s a protocol such that gven nput x, P tres to convnce V that x L and at the end V ether accepts or rejects the proof. It must satsfy the followng condtons 1. If x L and V and P follow the protocol, P r consv [V accepts] If x L and V follows the protocol, no matter what P does, P r consv [V rejects] 2 3 Defnton 6 IP s the class of languages L such that there exsts an Interactve Proof System for L. Known NP IP and IP = P SP ACE 3

4 2.2 Graph Isomorphsm and Graph Non-somorphsm Graph Isomorphsm Input Graphs G and H. G = H ( ψ S VG s.t. (u, v) E G (ψ(u), ψ(v)) E G )) Output 1 f G = H, 0 else. Graph Isomorphsm s n NP snce G = H can be proven by provdng ψ. Can be verfed n polynomal tme.x So, Graph Isomorphsm s n IP Graph Nonsomorphsm Input Graphs G and H. Output 1 f G = H, 0 else. Protocol Repeat k tmes 1. V computes G and H whch are random permutatons of G and H. 2. V flps a con and wth equal probablty Accept. Heads : Sends (G, G ) to P Tals : Sends (G, H ) to P P reples ndcatng whether the par of graphs t receved were somorphc or not. If (V sends (G, G ) and P sends =) or f (V sends (G, H ) and P sends =) Contnue. If (V sends (G, G ) and P sends =) or f (V sends (G, H ) and P sends =) Reject. If x L = G = H, then P wll follow protocol and always answer correctly and V wll contnue tll the loop ends and then Accept. If x L = G = H, then (G, G ) and (G, H ) are ndstngushable by P. So, P wll return a value that causes Reject wth probablty 1 2 at every teraton. Hence, P r[v accepts] = 1 = P r[v rejects] = k 2 k 1, f x L So, P r[v accepts] = 2 k, f x L. Hence, Graph Nonsomorphsm n n IP. (8) Remark Ths protocol only works f V has prvate cons. If P can see V s random bts, V can be made to accept for all nputs. 4

5 2.3 Arthur-Merln Protocol The Arthutr-Merln protocol s an nteractve proof system where the Verfer s cons are publc. (Goldwasser, Spser 1986) Arthur-Merln protocol IP wth prvate cons. 5