12.1 Formulation of General Perfect Matching
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1 CSC5160: Combinatorial Optimization and Approximation Algorithms Topic: Perfect Matching Polytope Date: 22/02/2008 Lecturer: Lap Chi Lau Scribe: Yuk Hei Chan, Ling Ding and Xiaobing Wu In this lecture, the focus is on general perfect matching problem where the goal is to prove that it can be solved in polynomial time by linear programming. Based on the LP formulation for bipartite matching studied in Lecture 10, we add some valid inequalities to establish a new formulation. Then we prove that for general perfect matching, all vertex solutions of the linear program are integral. This shows that the new formulation defines the matching polytope, and the general perfect matching problem can be solved by linear programming. Finally, we will show that this linear program can be solved in polynomial time, by providing a polynomial time separation oracle Formulation of General Perfect Matching Given a weighted general graph G = (V, E), for any edge e, w e represents the weight of the edge and x e is the indicator variable of e. If an edge e is in the matching, we set x e = 1, otherwise x e = 0. Base on these definitions, the weighted perfect matching problem can be formulated as follows: e δ(v) max e E w e x e x e = 1 v V (12.1.1) x e = {0, 1} e E (12.1.2) Note that this formulation is the same as the weighted perfect matching on bipartite graphs. Since integer linear programming is NP-hard in general, so in the following we try to relax the above integer linear program Relaxing the integrality constraint Similar to the case of bipartite matching, we replace the constraint x e {0, 1} with 0 x e 1. Since the degree constraint already implies x e 1, we formulate the linear programming as follows: e δ(v) max e E w e x e x e = 1 v V x e 0 e E (12.1.3) Unlike the case for bipartite matching, this formulation does not guarantee an integral solution. A simple example is shown in Figure , in which (0.5, 0.5, 0.5) is a solution to the linear program. Since the three degree constraints are tight and independent, this is a basic solution, which means 1
2 not all vertex solutions are integral (Recall basic solution is equivalent to vertex solution see Lecture 9). To have a better formulation, we add valid inequalities that kill the fractional vertex solutions while preserving all integral solutions. x1 x x Odd-set Inequalities Figure : K 3 with solution (0.5, 0.5, 0.5) The problem in the above example is that, for a set of odd number of vertices S, there can only be at most edges, that is S 2 e E(S) x e S 2 where E(S) denotes the set of edges with both endpoints in S. S V : S odd (12.1.4) Since we are focussing on perfect matching, we can assume V is even and simplify (12.1.4). Let S = V S. Since V is even, S is odd whenever S is odd. We have x e S 1 2 so e E(S) e E(S) e E(S) E(S) In a perfect matching we have x e S 1 2 x e S + S 2 e E S V : S odd 1 S V : S odd (12.1.5) x e = V 2. (12.1.6) Subtracting (12.1.5) from (12.1.6) we obtain the odd-set inequalities x e 1 S V : S odd (12.1.7) e δ(s) where δ(s) denotes the set of edges with one end in S and one end in V S. 2
3 x1 x1 x2 x3 x2 x3 Figure : Fractional solution (0.5, 0.5, 0.5) in the polytope (left) and how it is removed by the inequality x 1 + x 2 + x 3 1 (right) In 1965, Edmonds proved that the inclusion of the odd-set inequalities is enough to kill all fractional solutions in the perfect matching polytope, that is, with the odd-set inequalities, the polytope is the same as the convex hull of all perfect matchings. He also devised a polynomial time algorithm to find a perfect matching in his famous paper Paths, trees, and flowers [1]. This was well before the Ellipsoid method, invented in 1972, which solves exponential sized LP in polynomial time LP Formulation of General Perfect Matching After relaxing the integral constraint and adding odd-set inequalities, we get the LP formulation of general perfect matching problem as described below: max w e x e e δ(v) e δ(s) x e = 1 x e 1 x e 0 e E(G) v V S : S odd e E In the next section, we will prove that all the vertex solutions of this linear program are integral. That means that every vertex solution corresponds to a matching. Note that in this LP formulation, there are exponentially many inequalities. We need a polynomial time separation oracle in order to solve the LP in polynomial time via the Ellipsoid method. This will be discussed in Section
4 12.2 Matching Polytope In this section, we will prove that each basic solution of the linear program in the previous section corresponds to a matching. Before doing that, we first introduce the concept of convex combination. Definition A point y in R n is a convex combination of x 1, x 2,..., x k if there exist λ 1, λ 2,..., λ k > 0 such that y = λ 1 x 1 + λ 2 x λ k x k and λ 1 + λ λ k = 1. Based on this definition, we have the following useful fact about basic solutions that will be used later. Fact A feasible solution is a vertex solution if and only if it is not a convex combination of other feasible solutions. Geometrically, a point p is a convex combination of other points p 1, p 2,..., p k if p is in the convex hull of p 1,..., p k. So, it is intuitively clear that a vertex solution is not a convex combination of other feasible solutions. After realizing this fact, the proof of integrality can be reduced to proving that every feasible solution can be written as a convex combination of matchings (0-1 solutions of this linear program). Because if we can prove that, we can conclude that there is no vertex solution which is not a matching (a 0-1 solution), and hence every vertex solution is a matching. Indeed, we only need to prove that every basic feasible solution is a convex combination of matchings, because Lemma Every feasible solution is a convex combination of basic feasible solutions. Proof: We just outline an informal proof here. Suppose y is a fractional solution but not a basic solution (Figure 12.2). Then there must be at least one direction ɛ i such that both y + ɛ i and y ɛ i are feasible. Extend y along this dimension as much as you can without leaving the polytope and stop at two points y i + and yi. It is clear that y is a convex combination of these two points. Also, both y i + and yi will have at least one more constraint that goes tight, compared to y. As basic solutions are unique solution of n linearly independent tight constraints (for n variables) and each time at least one more inequality goes tight, by induction (on number of tight inequalities) y i + and yi can be written as a convex combination of basic feasible solutions and thus y can be written as a convex combination of basic feasible solutions Reduction Now we construct a minimal example of such graph. As in the case of bipartite graph, we leave out edges that are 0 for they do not participate in the matching and do not affect the existance of a matching in the smaller graph. In case any edge is 1 we reduce the graph by removing the two vertices and edges incident to these vertices. This is because we can add them back easily. Consider Figure On the top we have a fractional solution with an edge of one. We leave that out and in the smaller graph, if we can find a convex combination of matchings for the reduced graph (middle), then by adding the edge of 1 back to matching we know the original graph can also be written as a convex combination of matchings (bottom). After these operations we are left with a graph such that for each edge e, we have 0 < x e < 1. 4
5 y 4 y 1 y i y y i + y 3 y 2 Figure : y is a convex combination of y + i and y i, and y+ i (y i ) is a convex combination of basic solutions y 1 and y 2 (y 3 and y 4 ), so y is a convex combination of y 1, y 2, y 3 and y Contraction As the odd-set constraints e δ(s) x e 1 when S is a singleton, is implied by the degree constraint x e = 1, e δ(v) in the following we will focus on odd-set with at least three vertices. We claim that if there is a tight odd-set inequality, we can break the graph into 2 smaller graphs, and project the feasible fractional solution of perfect matching in the original graph into feasible fractional solutions of perfect matching in the smaller graphs. By induction, each projected solution can be written as convex combination of matchings in the smaller graphs. Then we show how to use these matchings to write the original feasible fractional solution as a convex combination of matchings in the original graphs. This type of divide and conquer argument is quite common in combinatorial optimization. Now we describe the details when we see a tight odd-set constraint. Edges involved in a tight oddset inequality separates the graph into 2 components, each with an odd number of vertices. We construct the two smaller graphs by contracting one component into a single vertex (Figure ). Also, we define the fractional value of each edge in the smaller graphs to be the same as the fractional value of the corresponding edge in the original graph (this is what we meant by projecting the original solution into the two smaller graphs). As both odd-sets contains at least three vertices, the resulting graphs are smaller than the original one, and there are even number of vertices in both graphs. So, to apply induction to show that the two projected solutions can be written as convex combination of matchings in the smaller graphs, we 5
6 Figure : Recovering the matching from the reduced graph only need to check that the projected solutions are feasible fractional solutions of perfect matching in the smaller graphs. There are two types of contstraints to check: degree constraints and odd-set constraints. For degree constraints, note that degree of vertices in individual components remain unchanged, while the degree of the new vertices (the contracted components) are exactly one, by the assumption that the odd-set inequality is tight. So all the degree constraints are satisfied in the smaller graphs. For odd-set inequalities, if the odd-set does not include the contracted vertex, then the inequality is clearly not violated. If the odd-set does include the contracted vertex, this corresponds to the set of vertices in the contracted component and there are odd number of them. Thus an odd-set in the smaller graph corresponds to an odd-set in the original graph, which means the odd-set inequality is still satisfied. 0.3 A B A B Combining matchings Figure : Contracting into smaller subgraphs We have written the projected solutions as convex combinations of matchings in the smaller graphs. Now we want to combine the matchings in order to write the original feasible fractional solution as a convex combination of matchings in the original graph. The idea is simple: just combine the 6
7 matching. The easier case is that the weights of the corresponding edges are equal, then we can simply join them and this shows the original fractional solution is a convex combination of matchings (Figure ) Figure : Combine when the weighing of the matchings are equal The slightly more difficult case is when weighing of matchings do not match. In this case we can further breakdown the weighing of the matching to their common denominator, then we can join the both sides (Figure ). (Here we use the fact that the solution from LP is rational so we can ask for the common denominator) x 0.3 x 0.3 x 0.3 x = x 0.6 x 0.2 x 0.4 x 0.5 x 0.1 x 0.1 Figure : Combine when the weighing of the matchings are not equal To summarize, if there are tight odd-set inequalities, then we can apply this divide and conquer procedure to write the fractional solution as a convex combination of matchings. 7
8 Basic Solution We keep contracting the graph if there are any tight odd-set inequalities. fractional solutions when there are no tight odd-set inequalities: max w e x e S : S odd e δ(v) e δ(s) e E(G) x e = 1 x e 1 x e 0 An edge of 1, reduce it. Tight odd-set, contract it. An edge of 0, delete it. Let us examine the After removal of edges of 0 and contraction of tight odd-set, the only constraint that can be tight is the degree constraint. If there are 2n vertices after contraction, there can be at most 2n tight constraints. Since every edge is fractional, each vertex has degree at least 2, which means there are at least 2n edges. By Lemma , we can focus on basic solutions, which means we only need to consider the case when there are exactly 2n edges. (with > 2n edges (variables) and 2n tight inequalities the solution cannot be basic.) This implies that each vertex has degree 2, so the graph is a disjoint union of cycles. With odd-set inequalities that rule out the possibility of odd cycles, the graph is actually a disjoint union of even cycles (so it is bipartite). In the previous lecture, it has been shown that even cycles can be decomposed into matchings, therefore each fractional solution is a convex combination of matchings. To summarize, through a few steps of reduction, most importantly the contraction step, the problem is reduced to a very simple problem when there are only disjoint union of cycles. In this simple case, we can write it as a convex combination of matchings, and hence the general case as well. This completes the proof of integrality Separation Oracle In the previous section, we proved that the general matching polytope is integral. In this section, we are going to discuss how to solve this problem by linear programming. Based on the Ellipsoid method, the crucial step is to decide, given x R n, whether x P. If not, we also need to find a violated inequality. In particular, given a fractional solution, we want to check whether it satisfies all the following constraints. If not, return one constraint violated by this fractional 8
9 solution. Remember, we should do this in polynomial time. max w e x e e δ(v) e δ(s) x e = 1 x e 1 x e 0 e E(G) v V S : S odd e E In the above constraints, note that constraints of the first and third kind can be checked one by one. However, we cannot simply follow the same way for the second constraints, for there are exponentially many of them, implying that if we simply check them one by one, then the whole task cannot be completed in polynomial time. In the remaining part of this section, we focus on the odd-set constraints. Now consider a graph G(V, E), V = 2n with a capacity function c : E R +, and for each edge e E, c(e) = x e. Consider a u-v cut which splits the nodes of G into two sets U and V, such that u U and v V, then this cut is said to be odd(even) if both U and V are odd(even). If the second constraint is satisfied, then each odd cut in the graph has total capacity at least 1. In other words, the minimum odd cut of the graph is at least 1. Next we will show how to use Gomory-Hu Tree introduced in Lecture 7 to achieve this goal. The remaining part of this section is organized as follows: first, a brief review of Gomory-Hu tree is given in Section , and then we give the separation oracle and finally show the correctness of this separation oracle in Section Gomory-Hu Tree Figure : A graph (left) and its Gomory-Hu tree representation (right) A Gomory-Hu tree is a compact representation of all minimum s-t cuts in undirected graphs. For any two vertices s and t in the graph, there is a unique s-t path P in the Gomory-Hu tree. The capacity of a s-t min-cut is equal to the bottleneck capacity of P, that is the capacity of the edge having the smallest capacity among all edges in P. Furthermore, the cut in the Gomory-Hu tree is the cut of the original graph. Suppose we want to find the min a-e cut in Figure Just look at the unique path from a to e in the Gomory-Hu tree, and the edge with smallest capacity 9
10 is cd (with capacity 13). By removing the edge cd, we get two components: U = {e, f, d} and V = {a, c, b}. So the min a-e cut is 13, and C = {(u, v) u U, v V } is one of the min a-e cuts (minimum cuts are not necessarily unique) Separation Oracle for General Matching To check if a fractional solution satisfies all odd-set constraints, we want to check whether the minimum odd cut has capacity at least 1. In order to use the Gomory-Hu tree to achieve this goal, we first give the following definition. Definition odd (even) edge: Given a Gomory-Hu tree T, an edge e is odd (even) if T e consists of two components, both of which has an odd (even) number of vertices. Figure : A Gomory-Hu tree. All the odd edges are in red. Based on Definition , if any odd edge in the Gomory-Hu tree of a graph has capacity less than 1, then we find a violating odd cut. Given a fractional solution, a polynomial time algorithm to check whether it satisfies all odd set constraints consists of the following three steps: 1. Compute a Gomory-Hu tree based on this fractional solution. 2. For each edge in the Gomory-Hu tree, check whether it is odd or even. 3. For each odd edge in the Gomory-Hu tree, check whether its capacity is at least 1. If there is an odd edge with capacity less than 1, we return a violating odd cut Correctness In this section, we will show the correctness of this separation oracle. Specifically, given a graph G with even number of vertices, a Gomory-Hu tree T and an odd cut C, we consider the following two cases: Case 1 There is an odd edge in the Gomory-Hu tree T crossing C. As Figure shows, suppose the odd edge crossing C be uv with value x. If we have checked that all the odd edges in Gomory-Hu tree T has capacity at least 1, then any u v cut has capacity at least 1. Since C is an u v cut, this implies that C has capacity at least 1. 10
11 u x v C Figure : Case 1 U None of such single isolated vertex will appear in U cut C Figure : Case 2 Case 2 There is no odd edge in the Gomory-Hu tree T crossing C. Acutally, this case never exists. Suppose the odd cut C splits the graph into two components U and G U. Now we consider all the nodes U in the Gomory-Hu tree T, as Figure shows. Note that there is no single isolated nodes in U. This is because T is a tree, so each node is connected to some other node. If a node is connected to some node that is also in U, then this node is part of a tree component with more than one node; if it is connected to some node in T U by an edge e, then e is an odd edge, contradicting the assumption. Since there are only even edges crossing C, we argue that each tree component inside U has an even number of vertices. Consider any even edge across the cut. By removing this edge, we have an even number of vertices isolated. Assume the isolated component does not completely lies in U (otherwise the component is even). Since every edge crossing the cut is an even edge, disconnecting them does not change the number of vertices from even to odd. This implies that U has an even number of vertices, contradicting the assumption that U is one of the components created by the odd cut C. 11
12 12.4 Summary Problem Solution LP-formulation Polynomial time Vertex solution LP-solver integral Separation oracle Figure : The Big Picture Figure illustrates the procedure of solving a problem by linear programming. First of all, we have to find a good LP formulation. In this general matching case, we add all the odd-set inequalities. Next, we should prove that the LP can be solved in polynomial time. Since sometimes we could write exponential number of constraints in linear program, we have to provide a polynomial time separation oracle. The black box in Figure is ellipsoid algorithm. After LP formulation is done and separation oracle is given, Ellipsoid method will give us a vertex solution. As long as we can prove that the vertex solution is integral, we solve our original problem, in other words, we get the optimal solution of the original problem. References [1] Jack Edmonds, Paths, Trees, and Flowers, Canadian Journal of Mathematics 17, p , Available at Google Scholar (through keyword: Paths, trees and flowers ). [2] Alexander Schrijver, Ch. 25 The matching polytope, Combinatorial Optimization, Springer,
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