1 The Minimum Spanning Tree (MST) Problem


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1 IEOR269: Lecture 3 1 Integer Programming and Combinatorial Optimization Lecture #3 IEOR 269 Feb 5, 2010 Lecturer: Dorit Hochbaum Scribe: Dogan Bilguvar Handouts: Game of Fiver and Fire Station Problems 1 The Minimum Spanning Tree (MST) Problem Definition: Let there be a graph G = (V, E) with weight c ij assigned to each edge e = (i, j) E, and let n = V. The MST problem is finding a tree that connects all the vertices of V and is of minimum total edge cost. The MST has the following linear programming (LP) formulation: Decision Variables: { 1 if edge e=(i,j) is selected X ij = 0 o/w min s.t i,j S X ij 0 c ij.x ij X ij = n 1 X ij S 1 ( S V ) The optimum solution to the (LP) formulation is an integer solution, so it is enough to have nonnegativity constraints instead of binary variable constraints (Proof will come later). One significant property of this (LP) formulation is that it contains exponential number of constraints with respect to the number of vertices. With Ellipsoid Method, however, this model can be solved optimally in polynomial time. As we shall prove later, given a polynomial time separation oracle, the ellipsoid method finds the optimal solution to an LP in polynomial time. A separation oracle is an algorithm that given a vector, it finds a violated constraint
2 IEOR269: Lecture 3 2 or asserts that the vector is feasible. There are other problem formulations with the same nature, meaning that nonnegativity constraints are enough instead of the binary constraints. Before proving that the MST formulation is in fact correct, we will take a look at these other formulations: 2 General Matching Problem Definition: Let there be a graph G = (V, E). A set M E is called a matching if v V there exists at most one e M adjacent to v. General Matching Problem is finding a feasible matching M so that M is maximized. This Maximum Cardinality Matching is also referred to as the Edge Packing Problem. 2.1 Maximum Matching Problem in Bipartite Graphs Figure 1: Bipartite Graph Definition: In a bipartite graph, G = (V, E B ) the set of vertices V can be partitioned into two disjoint sets V 1 and V 2 such that every edge connects a vertex in V 1 to another one in V 2. That is, no two vertices in V 1 have an edge between them, and likewise for V 2 (Please see Figure 1). The formulation of Maximum Matching Problem in Bipartite Graph is as follows: { 1 if edge e=(i,j) is selected X ij = 0 o/w
3 IEOR269: Lecture 3 3 max s.t B X ij X ij 1 B j V X ij 0 Remark: Assignment Problem is a Matching Problem on a complete Bipartite Graph with edge weights not necessarily Maximum Matching Problem in NonBipartite Graphs When the graph G = (V, E) is nonbipartite, the formulation above, replacing E B by E does not have integer extreme points and therefore does not solve the integer problem unless we add the integrality requirement. In a nonbipartite graph there are odd cycles. In that case, the optimum solution could be noninteger. As an example, consider a cycle with 3 vertices i, j, k. The best integer solution for this 3node cycle would be 1, as we can select at most one edge. However, in case of Linear Model, the optimum would be 1.5, where each edge is assigned 1/2 (Please see Figure 2). Figure 2: Odd Cycle Problem In order to exclude fractional solutions involving odd cycles, we need to add the following (exponentially many) constraints to the above formulation. j S X ij S 1 2 S V (where S is odd) Jack Edmonds showed that adding this set of constraints is sufficient to guarantee the integrality of the solutions to the LP formulation of the general matching problem.
4 IEOR269: Lecture Constraint Matrix Analysis for Matching Problems In the Bipartite Matching Problem, the constraint matrix is totally unimodular. Each column has exactly two 1 s, moreover the set rows can be partitioned into two sets such that on each set there is only one 1 on each column (Figure 3). In the Nonbipartite Matching Problem, however, although each column has again two 1 s, it is impossible to partition the rows of the constraint matrix as in the bipartite case (see Figure 4). Figure 3: Bipartite Graph Constraint Matrix Figure 4: NonBipartite Graph Constraint Matrix Indeed, the nonbipartite matching constraint matrix (without the odd sets constraints) is not totally unimodular. For example for a graph that is a triangle, the determinant of the corresponding constraint matrix is 2, hence it is not unimodular: Traveling Salesperson Problem (TSP) Given a graph G = (V, E) with each edge (i, j) E having a weight c ij. Our aim is to find a tour which visits each node exactly once with minimum total weighted cost. Definition: Degree of a node is the number of adjacent edges to the node. In the Figure 5, degree of the nodes are as follows:
5 IEOR269: Lecture 3 5 deg 1 = 2, deg 2 = 3, deg 3 = 2, deg 4 = 3, deg 5 = 3, deg 6 = 1 Figure 5: Example Graph Remark: deg i = 2 E i V Definition: A directed Graph is a graph where each edge is directed from one vertex to another, and is called an arc. A directed Graph is denoted by G = (V, A). To convert an undirected graph into a directed one, we replace each edge (i, j) E by two arcs (i, j)&(j, i) in A. Definition: The indegree of a node is the number of the incoming arcs to the node; the outdegree of a node is the number of the outgoing arcs from the node. For each node in a cycle, the outdegree and indegree are both equal to IP Formulation for TSP It is easier to give a formulation for the TSP if we represent each edge with two arcs. Decision Variable: { 1 if arc a=(i,j) is traversed X ij = 0 o/w The formulation is:
6 IEOR269: Lecture 3 6 min (i,j) A s.t (1) (i,j) A j V (2) (i,j) A i V (3) i S j S c ij.x ij X ij = 1 X ij = 1 X ij 1 (4)X ij {0, 1} i V j V S V, S = V/S (i, j) A Constraint sets (1) and (2) state that the outdegree and the indegree, respectively, for each node will be 1. Constraint set (3) is the Subtour Elimination constraints which prevents a solution with several (not connected) subtours instead of a single tour. This formulation has an exponential number of constraints, and the binary requirement cannot be omitted without it there are optimal fractional solutions (try to find an example). There are other alternative formulations to the TSP problem some of which are compact. For instance, the Tucker formulation (given in the book of Papadimitriou and Stiglitz) is compact has a polynomial number of constraints. However, that formulation is inferior in terms of the quality of the respective LP relaxation and is not used in practice. We will see later in the course, the 1tree formulation of the TSP, also with an exponential number of constraints. 4 Discussion of LPFormulation for MST Theorem 4.1 The formulation is correct, i.e. X ij is a (0 1) vector and the edges on which X ij = 1 form a minimum spanning tree. Proof: Let E be the set of edges for which the solution to (LP) is positive:. E = {e E X e > 0} Proposition 4.1 X e 1, e E. Proof of Proposition 4.1: For edge e = (i, j) set S = {i, j}. Then, the corresponding constraint yields X ij 2 1 = 1. Proposition 4.2 (V, E ) contains no cycles.
7 IEOR269: Lecture 3 7 Proof of Proposition 4.2: Step 1: There cannot be a cycle with only integer values (1 s). Proof of Step 1: Contradiction! Let the set of vertices the cycle passes through be denoted by R, then: R e=(i,j) i,j R X e R 1 Before Step 2, let s prove a lemma which we will use later: Definition: We call a set of vertices R tight if X e = R 1 e=(i,j) i,j R Lemma 4.1 No fractional tight cycles share a fractional edge. Proof of Lemma 4.1: Suppose there were two such cycles C 1 and C 2 with k and l vertices, respectively, sharing p edges as shown in Figure 6. Figure 6: 2 tight fractional cycles Let P denote the ppath consisting of the common edges, i.e. contains (p + 1) vertices. The total value of X e summed on these edges of the union of cycles is: X e = (l 1) + (k 1) X e e C 1 C 2 e P For all subset S V, we had the following constraint in our formulation: X ij S 1 i,j S
8 IEOR269: Lecture 3 8 Let define S = C 1 C 2, then: (l 1) + (k 1) e P X e k + l (p + 1) 1 p e P X e X e = 1 e P Done! Step 2: There is no fractional tight cycles. Proof of Step 2: Suppose there were fractional tight cycles in (V, E ). Consider the one containing the least number of fractional edges. Each such cycle contains at least two fractional edges e 1 &e 2 (else, the corresponding constraint is not tight). Let c e1 c e2 and θ = min {1 X e1, X e2 }. The solution {X e} with: (X e 1 = X e1 + θ), (X e 2 = X e2 θ) and (X e = X e, e e 1, e 2 ) is feasible and at least as good as the optimal solution {X e }. The feasibility comes from the Lemma 4.1. We are sure that we do not violate any other nontight fractional cycles which share the edge e 1. Otherwise, we should have selected θ as the minimum slack on that nontight fractional cycle and update X e accordingly. But, this would incur a solution where two fractional tight cycle share the fractional edge e 1 which cannot be the case due to Lemma 4.1. If there were nontight fractional cycles, then we could repeat the same modification for the fractional edges without violating feasibility. Therefore, there are no fractional cycles in (V, E ). The last step to prove that (LP) is the correct form will be to show that the resulting graph is connected. (Assignment 2)
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