Communcaton Complexty 16:198:671 09 February 2010 Lecture 4 Lecturer: Troy Lee Scrbe: Rajat Mttal 1 Homework problem : Trbes We wll solve the thrd queston n the homework. The goal s to show that the nondetermnstc complexty (N 1 (f), N 0 (f)) of TRIES n functon s O( n) and determnstc complexty s Ω(n). Hence the gap between N 1 (f), N 0 (f) and D(f) s essentally tght. The trbes functon s gven by TRIES n (x, y) = n n =1 j=1 x j y j Let s calculate the nondetermnstc complexty N 1 (f) frst. It s enough to specfy postons n all the n clauses where both x j, y j take the value one. It takes (log n) bts to specfy a poston n n long clause. Snce there are n such clauses N 1 (f) n log n = O( n log n). In the artcle [1] an upper bound of O( n) s clamed. The log factor, however, s actually requred. One can create a foolng set F = {(x, x) : x has exactly one one n each block}. There are n n many such nputs and they all satsfy TRIES n (x, x) = 1. For (x, x), (x, x ) F where x x, however, we have TRIES n (x, x ) = 0. One can show that N 0 (f) = O( n). In case the functon evaluates to zero the prover can gve the ndex of an nput to of the AND gate whch s not satsfed wth log( n) bts, and then the players can verfy wth O( n) bts of communcaton that ther nputs do not ntersect on ths block. For the determnstc complexty lower bound, we wll use tensor product trck. Frst notce that every clause s solvng a set ntersecton problem..e., gven two strngs, fnd f ther s a common ndex (poston) where both are one. Set ntersecton problem s the negaton of the dsjontness problem. The dsjontness problem s DISJ n = n =1 (x y ) The matrx for DISJ 1, the dsjontness functon on one bt, s gven by 1 1 1 0 We clam that DISJ n = n =1DISJ 1. Gven two matrces A,, ther tensor product s (A )[(x 1, x 2 ), (y 1, y 2 )] = A(x 1, y 1 ) (x 2, y 2 ) 1
So when the matrces have entres from {0, 1}, the tensor product A expresses the AND of all pars from A and. Two n-bt strngs x, y are dsjont f and only f x 1,..., x n/2, y 1,..., y n/2 are dsjont AND x n/2 +1,..., x n, y n/2 +1,..., y n are dsjont, we see that by expandng ths expresson recursvely that DISJ) n = n =1DISJ 1. Also for the tensor product we know rk(a ) = rk(a) rk(). Notce that the matrx for DISJ 1 s full rank, so the rank of DISJ n s 2 n. Snce set ntersecton s the negaton of dsjontness problem, DISJ n = J SI n, where J s the all 1 s matrx. y the subaddtvty of rank rk(disj n ) rk(j) + rk(si n ) rk(si n ) 2 n 1 The TRIES n functon s the AND of n many set ntersecton problems. Thus the communcaton matrx for TRIES n s the tensor product of n copes of the communcaton matrx for SI n. Agan by multplcatvty of rank under tensor product we have. rk(tries n (x, y)) (2 n 1) n, and so D(TRIES n ) log(rk(tries n )) = Ω(n). 2 Randomzed Communcaton Complexty In the last lecture, we defned randomzed communcaton complexty. There are two versons, publc con and prvate con. We wll focus on publc con randomzed complexty (recall that by Newman s theorem the two models are equvalent, up to an addtve logarthmc factor). A publc con protocol s a probablty dstrbuton over determnstc protocols {P r }. A protocol has error at most ɛ f Pr r [P r (x, y) = f(x, y)] 1 ɛ for all (x, y). The cost of a randomzed protocol s max r D(P r ). As usual the randomzed complexty R ɛ (f) s the mnmum cost of a protocol whch computes f wth error at most ɛ. 2.1 Lower bounds on randomzed communcaton complexty Consder a randomzed protocol P whch wth probablty p r runs the determnstc protocol P r. We dentfy P r wth the oolean matrx whose (x, y) entry s the output of P r on (x, y). Let M P be the matrx whose (x, y) entry represents the probablty that the protocol outputs 1. In other words, M P = r p rp r. Consder a functon wth communcaton matrx F. If the protocol P computes F wth error at most ɛ, then we have that F M P ɛ. The fact that a randomzed protocol s a convex combnaton of determnstc protocols allows us to qute generally transform a lower bound technque for determnstc complexty wth nce propertes nto a lower bound for randomzed complexty. Theorem 1. Suppose there exst a functon Φ : X Y R, whch satsfes the followng propertes 1. Φ(F ) C P (F ) 2
2. Φ(x + y) Φ(x) + Φ(y) 3. Φ(cx) = c Φ(x) Let Φ ɛ (F ) = mn G {Φ(G) : G F ɛ} (remember ths notaton, we wll use t for dfferent functons also). Then R ɛ (F ) log Φ ɛ (F ) Proof : The matrx for protocol P s ɛ away from the matrx for F. So we get Φ ɛ (F ) Φ(M P ). Now we compute Φ(M P ). Φ(M P ) = Φ( r p r P r ) p r Φ(P r ) p r C P (P r ) p r 2 D(Pr) p r 2 D(Pr ) (where D(P r ) s maxmum for r*) 2 D(Pr ) 2 Rɛ(F ) So we get Φ ɛ (F ) 2 Rɛ(F ). In other words R ɛ (F ) log Φ ɛ (F ). 2.2 Examples for Theorem 1 Lets take example of rk(f ) and see f t fts n the framework of Theorem 1. We know rk(f ) C P (F ) and also that rk(x+y) rk(x)+rk(y). ut the thrd property doesn t work because rk(cx) = rk(x). Lets look at rank as an optmzaton program and try to relax t to some other combnatoral quantty whch wll ft nto theorem s framework. Notce that C D (F ) = mn α such that F = α x y t where x, y {0, 1} 2n α {0, 1} 3
Rank can be wrtten as a relaxaton of ths program rk(f ) = mn α such that F = α x y t where x, y R 2n α {0, 1} We see that rank relaxes the constrant that the x, y are oolean vectors and allows them to be real vectors, but keeps the constrant that α {0, 1}, a quadratc constrant. Let us relax C D (F ) n a dfferent way. µ(f ) = mn α such that F = α x y t where x, y {0, 1} 2n α R Here we keep x, y to be oolean vectors, but relax the constrant on α. Ths quantty can now be wrtten as a lnear program, albet a huge one as there are 2 2n+1 many varables, one for each rectangle. It s easy to verfy that µ satsfes the propertes of theorem 1 and hence R ɛ (F ) log µ ɛ (F ). Whle not obvous from the defnton, t actually turns out µ(f ) 2 = O(rk(F )) for a oolean matrx F. We can even consder a tghter relaxaton of C D. Call t µ +. µ + (F ) = mn α such that F = α x y t where x, y {0, 1} 2n α 0. Notce that µ + s only defned for nonnegatve matrces (otherwse we can thnk of t as takng the value + ). ecause of ths t does not strctly satsfy condton (3) from the Theorem 1. However, t does satsfy ths for nonnegatve constants c and one can easly see that ths s all that s needed for the theorem to go through, thus we also have R ɛ (F ) log µ ɛ +(F ). 2.3 Dual of a norm Any functon Φ : X Y R whch satsfes these two condtons s called a semnorm. 1. Φ(A + ) Φ(A) + Φ() 2. Φ(cA) = c Φ(A) 4
Notce the smlarty wth condtons n Theorem 1. The dual of a semnorm Φ s defned as Φ (A) = max A, Φ() = max A,. Φ()=1 Here A, s the nner product of A, thought of as long vectors. We can use these norms to gve lower bound on randomzed communcaton complexty. Example of norms and dual norms : l 1 (v) = v. The dual norm s l 1(v) = max P u =1 v u = l (v) (the best u s the one whch puts all ts weght on the coordnate of v of largest magntude). l 2 (v) = v2 l 2(v) = l 2 (v). In general l p s dual to l q f 1 p + 1 q = 1. Now consder the dual norm of µ. Ths norm s defned as µ (A) = max A,. :µ()=1 If µ() = 1, then we can express = α R where α = 1 and each R s a rank one oolean matrx. Then we see that A, α R α A, R max A, R. Ths last value s achevable by settng α = 1 where R realzes ths maxmum. Thus we see that µ (A) = max A, R. Expressng Φ n terms of ts dual norm can be useful for showng lower bounds as t s phrased as a maxmzaton problem. Ths to lower bound Φ(A) we just have to exhbt a vector whch has non-neglgble nner product wth A and small dual norm. For ths framework we wll also need the fact that the dual of a dual norm s agan the orgnal norm. Theorem 2. Φ (v) = Φ(v) for any norm Φ. For example, ths theorem gves that µ(a) = max A, µ (). We wll use ths formulaton next to show a lower bound on the randomzed complexty of nner product. 5
3 Approxmate norms From Theorem 1 the dea of an approxmate norm naturally arose. Recall that Φ ɛ (A) = mn{φ() : A ɛ}. Note that Φ ɛ s not tself a norm n general. Let us see how we can lower bound Φ ɛ. We have Φ ɛ (A) = mn max A +, : ɛ Φ () max A, ɛl 1 () Φ () In fact t turns out that Φ ɛ s equal to ths expresson, as we wll dscuss n the next lecture. Ths means that we do not lose any power by showng a lower bound n ths fashon. Next we wll use ths expresson wth the norm µ to see a lower bound on the nner product functon. 3.1 Lower bound on Inner Product The nner product functon s defned as The matrx for IP 1 s gven by IP n (x, y) = ( 1) x,y 1 1 1-1 Ths s the famlar 2-by-2 Hadamard matrx, an orthogonal matrx (.e. has orthogonal rows) wth entres from { 1, +1}. We can agan use the tensor product to express the communcaton matrx for nner product. As we are workng over { 1, +1} the tensor product now takes the xor of the respectve entres. Ths means that IP n = IP n 1 and s also orthogonal. Now we want to lower bound µ 2ɛ (IP n ) max IP n, 2ɛl 1 (). µ () (The transformaton from {0, 1} valued matrces to { 1, +1} means we have to work wth 2ɛ approxmatons rather than ɛ.) We choose = IP n. Then the numerator s (1 2ɛ)2 2n. It remans to upper bound µ (IP n ). µ (IP n ) = max x,y {0,1} 2n xt IP n y max 2n/2 IP n y y {0,1} 2n 6
by the Cauchy-Schwarz nequalty. Let us now evaluate IP n y 2 = y t IP t nip n y = 2 n y t y 2 2n. Puttng everythng together we have µ 2ɛ (IP n ) (1 2ɛ)2 n/2. Ths shows that R ɛ (IP n ) n/2 + log(1 2ɛ). References [1] T. S. Jayram, R. Kumar, D. Svakumar: Two applcatons of nformaton complexty. n Proceedngs of the 35th annual ACM symposum on Theory of computng, 2003, 673-682. [2] E. Kushlevtz, N. Nssan: Communcaton complexty. Cambrdge Unversty Press, 1997. [3] T. Lee, A. Shrabman: Lower bounds n communcaton complexty. Foundatons and Trends n Theoretcal Computer Scence 3(4): 263-398 (2009). 7