CS 598CSC: Combinatorial Optimization Lecture date: 2/16/2009 Instructor: Chandra Chekuri Scribe: Vivek Srikumar 1 Perfect Matching and Matching Polytopes 0 E0 0 jEj Let G = (V; E) be a graph. For a set E ⊆ E, let χ denote the characteristic vector of E in R . We define two polytopes: M Pperfect matching(G) = convexhull(fχ jM is a perfect matching in Gg) M Pmatching(G) = convexhull(fχ jM is a perfect in Gg) Edmonds gave a description of these polytopes. Recall that for bipartite graphs, Pperfect matching(G) is given by x (δ(v)) = 1 8v 2 V x(e) ≥ 0 8e 2 E and Pmatching(G) is given by x (δ(v)) ≤ 1 8v 2 V x(e) ≥ 0 8e 2 E 1 1 1 We saw an example of a non-bipartite graph, for which ( 2 ; 2 ; 2 ) is an optimum solution. 1 1 2 2 1 2 Hence, (perfect) matching polytope for non-bipartite graphs are not captured by the simple con- straints that work for bipartite graphs. Theorem 1 (Edmonds) Pperfect matching(G) is determined by the following set of inequalities. x(e) ≥ 0; e 2 E x (δ(v)) = 1; v 2 V x (δ(U)) ≥ 1; U ⊆ V; jUj ≥ 3; jUj odd Edmonds gave a proof via an algorithmic method. In particular, he gave a primal-dual algorithm for the minimum cost perfect matching problem, which, as a by product showed that for any cost vector c on the edges, there is a minimum cost perfect matching whose cost is equal the minimum value of cx subjecto to the above set of inequalities. This implies that the polytope is integral. We describe a short non-algorithmic proof that was given later [1] (Chapter 25). Proof: Let Q(G) denote the polytope described by the inequalities in the theorem statement. It is easy to verify that for each graph G, Pperfect matching(G) ⊆ Q(G). Suppose there is a graph G such that Q(G) 6⊆ Pperfect matching(G). Among all such graphs, choose the one that minimizes jV j + jEj. Let G be this graph. In particular, there is a basic feasible solution (vertex) x of Q(G) such that x is not in Pperfect matching(G). We claim that x(e) 2 (0; 1); 8e 2 E. If x(e) = 0 for some e, then deleting e from G gives a smaller counter example. If x(e) = 1 for some e, then deleting e and its end points from G gives a smaller counter example. We can assume that jV j is even, for otherwise Q(G) = ; (why?) and Pperfect matching(G) = ; as well. Since 0 < x(e) < 1 for each e and x(δ(v) = 1 for all v, we can assume that deg(v) ≥ 2; 8v 2 V . Suppose jEj = 2jV j. Then deg(v) = 2; 8v 2 V and therefore, G is a collection of vertex disjoint cycles. Then, either G has an odd cycle in its collection of cycles, in which case, Q(G) = ; = Pperfect matching(G), or G is a collection of even cycles and, hence bipartite and Q(G) = Pperfect matching(G). Thus jEj > 2jV j. Since x is a vertex of Q(G), there are jEj inequalities in the system that are tight and determine x. Therefore, there is some odd set U ⊆ V such that x (δ(U)) = 1. Let G0 = G=U, where U is shrunk to a node, say u0. Define G00 = G=U¯, where U¯ = V −U is shrunk to a node u00; see Figure 1. The vector x when restricted to G0 induces x0 2 Q(G0) and similarly x induces U U¯ U¯ U e1 e e u0 1 1 u00 eh eh eh (a) G (b) G0 (c) G00 Figure 1: 00 00 0 00 0 0 x 2 Q(G ). Since G and G are smaller than G, we have that Q(G ) = Pperfect matching(G ) 00 00 0 and Q(G ) = Pperfect matching(G ). Hence, x can be written as a convex combination of perfect matchings in G0 and x00 can be written as a convex combination of perfect matchings in G00. The vector x is rational since we chose it as a vertex of Q(G), therefore, x0; x00 are also rational; hence, k 0 0 1 P Mi 0 0 0 0 9 integer k such that x = k i=1 χ , where M1;M2; ··· ;Mk are perfect matchings in G and k 00 00 1 P Mi 00 00 00 00 x = k i=1 χ , where M1 ;M2 ; ··· ;Mk are perfect matchings in G . (Note that k is the same in both expressions.) 0 0 0 Let e1; e2; ··· ; eh be edges in δ(U). Since x (δ(u ) = 1 and u is in every perfect matching, 0 0 0 we have that ej is in exactly kx (ej) = kx(ej) matchings of M1; ··· ;Mk. Similarly, ej is in ex- 00 00 Pn actly kx(ej) matchings of M1 ; ··· ;Mk . Note that j=1 kx(ej) = k and moreover, exactly one of 0 00 0 e1; ··· ; eh can be in Mi and Mi . We can, therefore, assume (by renumbering if necessary) that Mi 00 0 00 and Mi share exactly one edge from e1; ··· ; eh. Then, Mi = Mi [ Mi is a perfect matching in G. 1 Pk Mi Hence, x = k i=1 χ , which implies that x 2 Pperfect matching(G), contradicting our assumption. 2 Now, we use the above theorem to derive the following: Theorem 2 Pmatching(G) is determined by x(e) ≥ 0; e 2 E x (δ(v)) ≤ 1; v 2 V jUj − 1 x (E[U]) ≤ ; U ⊆ V; jUj odd 2 Here E[U] is the set of edges with both end points in U. Proof: We can use a reduction of weighted matching to weighted perfect matching that is obtained as follows: Given G = (V; E), create a copy G0 = (V 0;E0) of G. And let G~ be the graph (V;~ E~) defined as V~ = V [ V 0, E~ = E [ E0 [ f(v; v0) j v 2 V g. G G0, copy of G v v0 The following claim is easy to prove. Claim 3 There is an injective mapping from the matchings in G to the perfect matchings in G~. Corollary 4 The maximum weight matching problem is poly-time equivalent to maximum weight perfect matching problem. We can use the above idea to establish the theorem. Let x be feasible for the system of inequalities in the theorem. We show that x can be written a convex combination of matchings in G. It is clear that χM satisfies the inequalities for every matching M. From G, create G~ as above + 0 0 + and define a fractional solutionx ~ : E~ ! R as follows: first, we define x : E ! R as the copy of x on E. That is, x0(e0) = x(e), where e0 is the copy of e. Then, 8 x(e); if e 2 E <> x~ = x0(e); if e 2 E0 :>1 − x (δ(v)) ; if e = vv0 G G0 0:3 0:3 0:2 0:4 0:2 0:1 0:1 Claim 5 x~ belongs to Pperfect matching(G~). Assuming the claim, we see thatx ~ can be written as a convex combination of perfect matchings in G~. Each perfect matching in G~ induces a matching in G and it is easy to verify that x can therefore be written as a convex combination of matchings in G. It only remains to verify the claim. From the previous theorem, it suffices to show that x~ δ~(U) ≥ 1; 8U ⊆ V;~ jUjodd G G0 U W X0 Let U ⊆ V~ and jUj odd. Let W = U \ V and X0 = U \ V 0, where X0 is the copy of X ⊆ V . First we consider the case that X0 = ; and jW j is odd. Then x~ δ~(U) =x ~ δ~(W ) X = x~ δ~(v) − 2~x (E[W ]) v2W = jW j − 2x (E[W ]) jW j − 1 ≥ jW j − 2 2 ≥ 1 For the general case, we claim thatx ~ δ~(U) ≥ x~ δ~(W n X) +x ~ δ~(X0 n W 0) . Without loss of generality, W n X is odd. Thenx ~ δ~(U) ≥ x~ δ~(W n X) ≥ 1 from above. The claim can be verified as follows: X0 n W 0 W \ X X0 W W 0 \ X0 W n X Notice that only edges between W and X0 are between W \X and X0\W 0. Let A = W \X; A0 = W 0 \ X0. Then x~ δ~(U) =x ~ δ~(W [ X0) =x ~ δ~(W n X) +x ~ δ~(X0 n W 0) + x (δ(A)) − 2x((δ (E[A; W n X])) + x δ0(A0) − 2x δ0 E[A0;X0 n W 0] The claim follows from the observation that x(δ(A)) ≥ x (E[A; W n A]) + x (δ(E[A; X n W ]). 2 Corollary 6 Pperfect matching(G) is also determined by x(e) ≥ 0; e 2 E x (δ(v)) = 1; v 2 V jUj − 1 x (E[U]) ≤ ; U ⊆ V; jUj odd 2 We note that although the system in the above corollary and the earlier theorem both determine Pperfect matching(G), they are not identical. 2 Separation Oracle for Matching Polytope The inequality systems that we saw for Pperfect matching(G) and Pmatching(G) have an exponential number of inequalities. Therefore, we cannot use them directly to solve the optimization problems of interest, namely, the maximum weight matching problem or the minimum weight perfect match- ing problem.