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Dense Graphs and Arbitrary Bipartite Graphs University of Chicago Autumn 2003 CS37101-1 Markov Chain Monte Carlo Methods Lecture 7: November 11, 2003 Estimating the permanent Eric Vigoda We refer the reader to Jerrum's book [1] for the analysis of a Markov chain for generating a random matching of an arbitrary graph. Here we'll look at how to extend the argument to sample perfect matchings in dense graphs and arbitrary bipartite graphs. Both of these results are due to Jerrum and Sinclair [2]. Dense graphs We'll need some de¯nitions and notations ¯rst: De¯nition 7.1 A graph G = (V; E) is said to be dense if for every v 2 V , degree(v) > n=2, where n = jV j. De¯nition 7.2 Let G be a graph and let MG be the set of all perfect matchings of G. Similarly, for two distinct vertices u; v, let NG(u; v) be the set of all perfect matchings of the graph Gnfu; vg. We refer to matchings in NG(u; v) as near-perfect matchings, or matchings with holes u and v. Whenever G is clearly understood from the context, we shall use simply M and N (u; v). Jerrum's book presents the analysis of a Markov chain for sampling matchings (not neces- sarily perfect). The Markov chain for sampling perfect matchings in dense graphs is only a slight modi¯cation of the original chain. The state space ­ is the set of perfect and near- perfect matchings, i.e., ­ = M [ [u;v2V N (u; v). The near-perfect matchings are needed to navigate between perfect matchings. ² For Xt 2 ­, 1. Choose e = (u; v) uniformly at random from E. 0 2. If u; v are both unmatched, let X = Xt [ e. 0 If e 2 Xt, then let X = Xt n e. 0 If u is unmatched and (v; w) 2 Xt, let X = Xt [ e n (v; w). 7-1 Lecture 7: November 11, 2003 7-2 0 0 3. If X 2 ­, with probability 1=2 set Xt+1 = X , else Xt+1 = Xt. The chain is symmetric, thus it's stationary distribution ¼ is uniform over ­. For this Markov chain to e±ciently (in polynomial time) sample perfect matchings, we have to have a polynomial bound on the ratio of perfect matchings to near-perfect matchings. j­j 2 2 Lemma 7.3 For a dense graph, jMj · n . Thus, ¼(M) ¸ 1=n . Proof: We shall prove that jMj ¸ jN (u; v)j for every u; v. We do this by de¯ning an injective map fu;v : N (u; v) ! M. Case 1: If u and v are adjacent, then fu;v(M) = M [ f(u; v)g for M 2 N (u; v). Clearly, this map is injective. Case 2: If u and v are not adjacent, then for M 2 N (u; v) we claim there exists e = fw; xg 2 M such that w 2 N(u) and x 2 N(v). (The set N(u) denotes the neighbors of vertex u.) This will then give an augmenting path of length three to transform M into a perfect matching. To prove this claim, let S(v) = fw j fw; xg 2 M; x 2 N(v)g. Since v is unmatched, every vertex in S(v) is matched and therefore jS(v)j = jN(v)j > n=2. Since we also know jN(u)j > n=2, we have S(v) \ N(u) =6 ;. Let z 2 S(v) \ ¡(u) and fw; zg 2 M. Finally, let fu;v(M) = M [ fv; wg [ fz; ug n fw; zg. Again, fu;v is injective. Theorem 7.4 For dense graphs, ¿(²) · poly(n; log 1=²). Proof:(sketch): Our aim is to use the canonical paths and associated encoding ´t which were we used for the chain on all matchings. We will take it for granted that the reader remembers the paths and encoding that were used earlier. However, there are two problems we must address. By taking a brief look at the various cases, we will see that there are two scenarios where the proof from last class uses matchings with 4 holes. This is the issue we must resolve since 4-hole matchings are not contained in ­. Consider a pair of perfect matchings I; F . The canonical path used previously as well as the associated encoding are legitimate in the sense that both always contain at most two holes. Suppose I 2 N (u; v) and F 2 M. Looking at I © F , we have an augmenting path from u to v and a set of alternating cycles. We de¯ne the canonical path γIF to unwind, i.e., augment, the path, and then unwind the cycles in some canonical order. Notice that after we unwind the path, we are at a perfect matching. It is easy to verify that this path never has more than two holes (the holes arise in the midst of unwinding an alternating cycle). However, the associated encoding will have 4 holes. After we unwind the path, the encoding will have holes at u and v. During the unwinding of any cycle, we will get a further two holes in the encoding. Despite this fact, we keep the same encoding as before, but we now Lecture 7: November 11, 2003 7-3 0 0 have that ´t : cp(t) ! ­ , where ­ is the set of matchings with at most 4 holes and cp(t) is the set of canonical paths that traverse t. The encoding is simply an ingredient in the proof, used to bound jcp(t)j. Therefore, there is no inherent reason why the encoding must be contained in ­. It will su±ce to simply prove that j­0j · n2j­j. Finally, consider a pair of near-perfect matchings I; F . The canonical path will visit 4-hole matchings. To avoid this, we construct di®erent canonical paths. We ¯rst choose a random M 2 M and the new canonical path is the concatenation of γIM and γMF . Now we want to bound the congestion. We ¯rst analyze the congestion created by paths between I 2 ­ and F 2 M. For every t = M ! M 0, we can de¯ne an injective map 0 0 2 ´t : cp(t) ! ­ . Thus, we can bound jcp(t)j · j­ j · n j­j. (The second inequality follows from Lemma 7.3.) In order to account for the congestion added by flows between pairs of 2 near-perfect matchings we use the fact that jN = [u;vN (u; v)j · n jMj. Fix a near-perfect matching I and a perfect matching M. For each near-perfect matching F , we use γIM with 2 probability 1=jMj. Summing over F , the expected extra load on γIM is jN j=jMj · n . Therefore, paths between pairs of near-perfect matchings increase the congestion by a factor of n2. If ¼ is uniform on ­, we can bound the congestion through t = M ! M 0: 1 ½ = ¼(I)¼(F )jγ j ¼(M)P (M; M 0) IF (I;F )2cp(t) n X · j­jm j­j2 (I;FX)2cp(t) jcp(t)j j­0j = mn · mn · mn3 j­j j­j For dense graphs we can generate a random perfect matching in polynomial time, by running the chain for the mixing time. If the ¯nal state is a perfect matching (with probability at least 1=n2), then it's a random perfect matching. Otherwise, we run the chain again. General Bipartite Graphs Consider a bipartite graph G = (V1; V2; E). We'll present the algorithm of Jerrum, Sinclair and Vigoda [3] for generating a random perfect of an arbitrary bipartite graph (and thus approximate the permanent of any non-negative matrix). Once we remove the min-degree condition, the ratio of perfect matchings to near-perfect matchings can be exponentially small. To ensure perfect matchings are likely in the station- ary distribution we introduce suitable weights on matchings and modify the Markov chain. Lecture 7: November 11, 2003 7-4 The weight of a matching is a function of the location of its holes (if any). Let 1 for M 2 M w¤(M) = w¤(u; v) = jMj for M 2 N (u; v) ( jN (u;v)j We then modify the Markov chain introduced for dense graphs so that the stationary dis- tribution is proportional to the weights. To do this we simply add what is known as a Metropolis ¯lter. More precisely we simply modify the last step of the Markov chain to only move to the proposed new matching with an appropriate¤ 0 probability: 0 w (X ) 0 ¤ 3. If X 2 ­, with probability 1=2 minf1; w (Xt) g set Xt+1 = X , else Xt+1 = Xt. 0 It is straightforward to verify that, for M 2 ­, ¼(M) = w(M)=Z where Z = M 02­ w(M ): There is an obvious concern with our proposed weights. In order to computePthe weights, we need to be able to compute the number of perfect matchings and near-perfect matchings { this was our original task. We will tackle this problem later. Let us ¯rst show that if we can obtain the weights, then we can generate a random perfect matching e±ciently. To motivate the above weights, let's look at the e®ect on the stationary distribution. We have 1 1 jMj ¼(N (u; v)) = w¤(M) == = jMj=Z: Z Z jN (u; v)j M2NX(u;v) M2NX(u;v) Similarly, ¼(M) = jMj=Z: Therefore, each hole pattern is equally likely, i.e., for all u 2 V1; v 2 V2; ¼(M) = ¼(N (u; v)): Hence, 1 ¼(M) = : n2 + 1 Thus, in the stationary distribution we output a random perfect matching with probability 1=(n2 +1).
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