13.2 Fully Polynomial Randomized Approximation Scheme for Permanent of Random 0-1 Matrices

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1 CS71 Randoness & Coputation Spring 018 Instructor: Alistair Sinclair Lecture 13: February 7 Disclaier: These notes have not been subjected to the usual scrutiny accorded to foral publications. They ay be distributed outside this class only with the perission of the Instructor Peranent of Rando Matrices In this lecture we focus on a siple algorith to approxiate the peranent of a rando atrix with 0, 1 entries. We start with the definition of the peranent of a atrix. Definition 13.1 Let A be an n n atrix such that each entry of A is either 0 or 1. The peranent of A, denoted by per(a), is defined as per(a) = a i,σ(i), σ where σ ranges over the perutations on {1,,..., n}. Note the siilarity between per(a) and det(a), which is defined as the sae su but with each ter weighted by sgn(σ). However, while we can copute the deterinant easily in polynoial tie (say, by Gaussian eliination), coputing the peranent is apparently very hard. Coputing the peranent of A is in fact equivalent to counting the nuber of perfect atchings in the bipartite graph G A, which has n vertices on each side, and an edge connecting i to j iff a ij = 1. It is an easy exercise to check that perfect atchings of G A are in 1-1 correspondence with non-zero ters of per(a). While the proble of checking the existence of a perfect atching in a bipartite graph is easily solved in polynoial tie by (e.g.) network flow techniques, counting the nuber of perfect atchings is #P - coplete [Val79]. Hence coputing per(a) for a atrix with 0, 1 entries is also #P -coplete, which eans that (under standard coplexity theoretic assuptions) it is not possible to obtain a polynoial tie algorith to copute the peranent of a atrix. The focus has therefore shifted to efficient approxiation algoriths with precise perforance guarantees. In this lecture we will present a fully polynoial randoized approxiation schee for the peranent of a randoly chosen atrix (i.e., the algorith works well with high probability over the choice of the input atrix, but ay behave arbitrarily badly on a vanishing fraction of inputs). 13. Fully Polynoial Randoized Approxiation Schee for Peranent of Rando 0-1 Matrices The goal of the lecture is to design an fpras for alost all 0-1 atrices A. In other words, we will devise an algorith that takes as input a atrix A and an accuracy paraeter ɛ and outputs a rando variable X A such that Pr[(1 ɛ)per(a) X A (1 + ɛ)per(a)] 3 4, 13-1

2 13- Lecture 13: February 7 for alost all A (the eaning of alost all will be ade precise later). Recall that fully polynoial requires that the run-tie of the algorith is polynoial in both 1 ɛ and the size of the input. Note that the constant 3 4 can be increased to 1 δ using only O(log δ 1 ) trials, by the edian technique. We reark that there is an fpras for the peranent of an arbitrary 0-1 atrix [JSV04], which we ay see later in the course. Here, however, we focus on a uch sipler algorith, due to Rasussen [R94], that works for alost all rando atrices. The algorith is as follows: Input: An n n atrix A with 0-1 entries. Output: A rando variable X A. 1. if n = 0, then X A = 1. else let W A = {j : a 1j = 1} be the set of 1 s in the first row.1. Pick j fro W A u.a.r... output W A X A1,j, where A 1,j is the (1, j)-inor of A (i.e., row 1 and colun j reoved fro A) Intuitively, the above procedure picks an eleent fro W A and assues all the sub-peranents have the sae value. Since per(a) = j a 1,j per(a 1,j ), the above algorith is an averaging schee, where it assues per(a 1,j ) is sae for all j such that a 1,j is 1. We first argue that X A is an unbiased estiator of per(a) i.e., per(a) = E[X A ]: We view the run of the algorith as a coputation tree, where the root of the tree is the atrix A, with branches at the root to each inor A 1,j with a 1j = 1 (thus the root has degree W A ); the coputation tree is continued recursively. Every path fro root to a leaf l (at depth n) of the coputation corresponds to a (generalized) diagonal of A such that every entry of the diagonal is 1. Hence the nuber of leaves of the coputation corresponds to per(a). Observe that the algorith reaches a particular leaf l 1 W Ak, where the W A k s are the sets along the path fro the root to l, and with probability Pr[l] = n k=1 for the leaf l it outputs the reciprocal of this probability. Hence E[X A ] = l leaves Pr[l] 1 = # of leaves = per(a). Pr[l] Reark 13. The above arguent is based on a technique of Knuth fro the 1970 s. The idea is that given an arbitrary tree, to produce an unbiased estiator for the nuber of leaves of the tree it suffices to navigate the tree top-down while keeping track of the probabilities along the path fro the root to the leaf, and output the reciprocal of the path probability. Proving that the algorith works for ost atrices aounts to showing that the rando variable X A is sufficiently concentrated about its ean; this is done by bounding the nd oent. Variance estiate. We first present an exaple that shows that the variance of the rando variable X A can be very bad in the worst case. Consider a atrix A with 1 s in the principal diagonal and the upper triangular atrix and 0 s in the lower triangular atrix, as shown below: A =

3 Lecture 13: February There is only one diagonal (the principal diagonal) such that all the entries of the diagonal are 1. Hence per(a) is 1, and there is only 1 leaf in the coputation tree of the algorith. The single coputation path to this leaf is chosen with probability 1 n!, which gives { n! with prob 1 X A = n! 0 otherwise Hence the estiator is alost surely zero, and we would need a huge nuber of trials (on the order of n!) to get a decent estiate of per(a). For the rest of this lecture we focus on rando atrices, where we show that the above algorith works with only polynoially any trials with high probability over the choice of atrix. Definition 13.3 Let A n denote the probability space of n n atrices such that every entry of the atrix is 0 or 1 with probability 1 independently. We will prove the following result. Theore 13.4 Let A A n and let ω(n) be any function such that ω(n). Then [ ] E[X Pr A ] An (E[X A ]) > n ω(n) 0, as n. Let us first interpret the result of Theore We can choose a function ω(n) that goes to infinity as slowly as we want. Since X A is an unbiased estiator of per(a), the nuber of trials of the algorith E[X A ] required to ɛ-approxiate per(a) is O( ɛ (E[X A ]) ), and the theore says with high probability it is bounded by O( n ω(n) ɛ ). Corollary 13.5 Given A A n, the above algorith repeated O( n ω(n) ɛ ) ties yields a fpras for per(a) with probability tending to 1 over the choice of A. Run-tie analysis. The loop in Step runs for n ties, and each iteration of the loop takes O(n) tie; hence each trial takes O(n ) tie. Cobining with the bound on the nuber of trials gives an fpras for per(a) that runs in tie O( n3 ω(n) ɛ ). The proof of Theore 13.4 will follow fro a sequence of leas. We begin with the following clai: Clai 13.6 The following assertions hold: 1. E An [E[X A ]] = E An [per(a)] = n! n.. E An [E[X A ]] = 1 4 n n (i + i). Proof: An n n atrix has n! diagonals, and for each diagonal the probability that all entries are 1 is 1 n. The first equation then follows by linearity of expectation. We now present an alternative arguent that proves both parts siultaneously. We ay write [ n ] E An [per(a)] = E W i,

4 13-4 Lecture 13: February 7 where the W i s are independent and each W i Bin(i, 1 ), i.e. is distributed as the # of heads in i tosses of a fair coin. Since E(W i ) = i and E(Wi ) = i + i, 4 we have E An [E[X A ]] = E[W i ] (by independence) = i = n! n. For part, a straightforward induction (Exercise!) shows that E An [E[XA ]] = n get E An [E[X A]] = = E[Wi ] i + i 4 = 1 4 n (i + i). E[W i ], fro which we Corollary 13.7 E An [E[XA ]] 1 n (E An [E[X A ]]) = 4 n (i + i) 1 n = 4 n i i + 1 i = n + 1. This Corollary shows that Theore 13.4 holds in expectation. To turn this into a high-probability stateent, we need to appeal to first and second oents (the first oent for the nuerator and the second oent for the denoinator). The second oent part is supplied by the following lea. To siplify notation we will write µ(n) = E An [per(a)]. Lea 13.8 [Main Lea] For any ω(n), we have Pr An [per(a) < µ(n) ω(n) ] 0 as n. We first prove Theore 13.4 assuing Lea Proof:(of Theore 13.4). We handle the nuerator and denoinator of the expression in Theore 13.4 as follows: Nuerator. Markov s inequality gives Pr An [E[XA] ω(n) E An [XA]] 1 0, as n. ω(n) Denoinator. Lea 13.8 gives [ 1 Pr An (E[X A ]) > ω(n) ] (E An [E[X A ]]) 0. The above line is a restateent of Lea 13.8 with per(a) = E[X A ] and taking reciprocals and squaring.

5 Lecture 13: February Putting these together we have Pr An [ E[X A ] (E[X A ]) > ω(n)3 Pr An [ E[X A ] (E[X A ]) > ω(n)3 (n + 1) E An [XA ] ] (E An [E[X A ]]) 0 ] 0 (by Corollary 13.7) Note that since ω(n) is an arbitrary function that goes to infinity, the sae is true of ω(n) 3. (Alternatively, we ay replace ω(n) in the above arguent by ω(n) 1/3.) Structure of proof of Main Lea We will consider generating a rando atrix A A n by first picking a nuber according to the binoial distribution Bin(n, 1 ), then distributing 1 s in the atrix uniforly at rando, setting all other entries to 0. Definition 13.9 We denote by A n, the probability space of rando n n, 0-1 atrices where the nuber of 1 s in the atrix is exactly and the 1 s are distributed uniforly at rando in the atrix. The reason we do this is that, for typical values of (note that will be sharply concentrated about its ean, n ), per(a) will be sharply concentrated about its ean in the odel A n,. This fact is expressed in the following lea. Lea Suppose = (n) satisfies n 3 { } 1. E An, [per(a)] = n! ( n ) n exp n + 1 n3 + O( ).. Then for A A n, we have. E An, [per(a) ] (E An, [per(a)]) = 1 + O( n3 ). Observe that fro part of Lea it follows that given n 3 n. Hence the peranent is tightly concentrated in A n,., we have Var An, [per(a)] (E An, [per(a)]) 0, as We now assue Lea and prove Lea 13.8; to coplete the entire analysis, we will then just need to go back and prove Lea Proof of Lea 13.8: We consider the following procedure to generate A A n : Pick M fro Bin(n, 1 ); pick A A n,m u.a.r. Let us denote by ω = ω (n) an arbitrary function of n that goes to with n. inequalities: We have the following Pr[M < n ω n] 0; this follows by Chebyshev s inequality or the Central Liit Theore because the standard deviation of M is Θ(n), so a deviation of ω n is ore than a constant ties the s.d. For any = (n) such that n 3, we have Pr An, [per(a) < 1 E A n, [per(a)]] < 4Var An, [per(a)] (E An, [per(a)]) (by Chebyshev s inequality) 0 (by part of Lea 13.10).

6 13-6 Lecture 13: February 7 Since both of the above events happen with probability tending to 0, we will assue henceforth that neither of the happens: i.e., we assue that M n ω n and per(a) 1 E A n, [per(a)]. Recall that µ(n) = E An [per(a)], and define µ(n, ) = E An, [per(a)]. We then have fro part 1 of Lea n per(a) µ(n, ( ω n) = 1 n µ(n) µ(n) n n! n! ) n { ω n n n exp n ω n O( 1 } n ) 1 ) n { ( )} (1 ω 1 exp 1 + O n n 1 exp{ ω 1}. Finally, given an arbitrary function ω(n) such that ω(n) as n, we choose ω (n) = 1 ω(n) log e. Observe that ω (n) as n, and fro the above analysis we have per(a) µ(n) 1 exp{ ω 1} = 1/ω(n), as required to prove Lea We now prove Lea Proof of Lea 13.10: The arguent to prove Lea is graph-theoretic. Naely, we work with the interpretation of per(a) (for an n n 0-1 atrix A) as the nuber of perfect atchings in the associated graph G A, as explained earlier. Given A A n,, the graph G A is a bipartite graph with n vertices on each side and exactly edges distributed uniforly. Let H be a fixed labeled sub-graph of G A with t n edges. Let q(t) = Pr[H is subgraph of G A ]. Then ) q(t) = ( n t t ( n To see this, note that ( n t) is the nuber of possible ways of choosing GA under the constraint that H is a subgraph of G A (i.e., t edges are fixed), and ( ) n is the nuber of possible ways of choosing out of n edges (i.e., choosing G A ). Hence we have ( 1)... ( t + 1) ( ) t q(t) = n (n 1)... (n t + 1) = exp { t n ( 1 1n ( )} n 3 ) + O for t n. (13.1) Exercise: Fill in the details of the above calculation. [Hint: Take logs, and use the approxiation ln(1 x) = x + O(x ).] Hence we have E An, [per(a)] = This copletes part 1 of Lea We now prove part of the Lea. We have ). {H : H a perfect atching } Pr[H is subgraph of G A ] = n! q(n) ( ) n n = n! exp{ n + 1 ( ) n 3 + O }. (13.) E An, [per(a) ] = H,H Pr[H, H are subgraphs of G A ], where H, H range over all pairs of perfect atchings in G A. We first calculate the nuber of perfect atchings H and H that overlap in exactly k edges. The expression for this is derived as follows: there are

7 Lecture 13: February n! perfect atchings H, and given a perfect atching H the nuber of perfect atchings H that overlap with H in exactly k edges is ( n k) D(n k), where D(n k) denotes the nuber of derangeents of (n k) ites (i.e., the nuber of perutations σ such that σ(x) x for all x). Hence the nuber of perfect atching pairs H, H that have exactly k overlapping edges is given by n! (n k) D(n k). It is well known that D(n) n! e (and the very sall error in this estiate can be absorbed into our other error ters). Observe also that if H and H overlap in exactly k edges, the union of H and H has exactly n k n edges. Hence we have ( ) n E An, [per(a) ] = n! D(n k) q(n k) k. = n! ( ) { ( )} n n 3 exp n αn + O ( ) n D(n k) k ( e α n ) k, where α = n( 1 1 n ). [Exercise: Perfor the algebraic anipulations to fill in the dots above, by plugging in the estiate (13.1) for q(n k).] We now copute the su in the above expression: ( ) ( n e α n ) k D(n k) k = n! e ( ) ( n (n k)! e α n k e ( 1 e α n ) k k! n! ( e 1 e α n ) k k! { e α n } = n! exp 1. Since e α = 1 + O( n n ), the above expression siplifies to n! exp{ expression for E An, [per(a) ] we get ( ) n E An, [per(a) ] = (n!) exp { n n O ) k 1 + O( n3 )}. Plugging this into the ( n 3 )}. (13.3) Finally, we divide expression (13.3) for E An, [per(a) ] by the square of expression (13.) for (E An, [per(a)]) and obtain the result claied in part of Lea Exercise: Here is a trivial approxiation algorith for the peranent of a rando atrix. On input A A n, let be the nuber of 1 s in A and siply output the approxiation to µ(n, ) (the expected value of per(a)) given in part 1 of Lea Why is this significantly weaker than a fpras for per(a)? References [JSV04] M. Jerru, A. Sinclair and E. Vigoda A polynoial-tie approxiation algorith for the peranent of a atrix with nonnegative entries, Journal of the ACM 51 (004), pp

8 13-8 Lecture 13: February 7 [R94] L. E. Rasussen, Approxiating the Peranent: a Siple Approach, Rando Structures & Algoriths 5 (1994), pp [Val79] L.G. Valiant, The coplexity of coputing the peranent, Theoretical Coputer Science 8 (1979), pp