Facets from Gadgets. Adam N. Letchford Anh Vu. December 2017

Transcription

1 Facets from Gadgets Adam N. Letchford Anh Vu December 2017 Abstract We present a new tool for generating cutting planes for N P-hard combinatorial optimisation problems. It is based on the concept of gadgets small subproblems that are glued together to form hard problems which we borrow from the literature on computational complexity. Using gadgets, we are able to derive huge (exponentially large) new families of strong (and sometimes facet-defining) cutting planes, accompanied by efficient separation algorithms. We illustrate the power of this approach on the stable set and clique partitioning problems. Keywords: branch-and-cut, gadgets, stable set problem, clique partitioning problem. 1 Introduction A popular approach to solving N P-hard combinatorial optimisation problems exactly is branch-and-cut [30], in which valid linear inequalities (also called cutting planes or simply cuts) are used to strengthen linear programming relaxations of the problem. Algorithms for generating cuts are called separation algorithms, and the strongest possible cuts are ones that define facets of the associated polyhedron (see, e.g., [9, 22]). The purpose of this paper is to present a new tool for generating cutting planes. It is based on the concept of gadgets small subproblems that are glued together to form hard problems which we borrow from the literature on computational complexity theory (see, e.g., [13, 33]). Using gadgets, we are able to derive huge (exponentially large) new families of strong cutting planes. We also obtain efficient separation algorithms as a by-product of the analysis. We illustrate the power of the gadget approach on two well-known and much-studied N P-hard problems: the stable set problem [16] and the clique Corresponding author. Department of Management Science, Lancaster University, Lancaster LA1 4YX, United Kingdom. a.n.letchford@lancaster.ac.uk Also at Lancaster. a.vu@lancaster.ac.uk 1

2 partitioning problem [18, 25]. Interestingly, for both problems, some of the new cutting planes that we derive are not only facet-defining, but also have rank greater than one, in the sense of Chvátal [8]. The structure of the paper is as follows. In Section 2, we review the relevant literature. In Section 3, we define gadgets and show how they can be used both to derive cuts and solve the associated separation problem. The applications to the stable set and clique partitioning problems are in Sections 4 and 5, respectively. Finally, in Section 6, we make some remarks about the possibility of applying our approach to other combinatorial optimisation problems. Throughout the paper, we use the following (standard) graph-theoretic notation and terminology. G = (V, E) will denote a generic (simple, loopless) undirected graph. The complement of G is the graph with node set V and edge set { {i, j} V : {i, j} / E }. Given a node i V, we let n(i) denote the neighbours of i in G, i.e., the set of nodes adjacent to i. Given a set S V, we let G[S] denote the subgraph of G induced by S. A clique (respectively, stable set) is a set of pairwise adjacent (respectively, non-adjacent) nodes. A cycle is a connected subgraph in which all nodes have degree two. Given a cycle C, we let V (C) denote its vertex set and E(C) its edge set. Given a positive integer n, we let K n = (V n, E n ) denote the complete undirected graph on n nodes, where V n = {1,..., n} and E n = {e V n : e = 2}. A directed graph, cycle or path will be called a digraph, dicycle or dipath, respectively. In a digraph, nodes are linked by arcs rather than edges. 2 Literature Review We now review the relevant literature. For reasons which will become apparent, we start by describing the satisfiability problem in Subsection 2.1. We then cover gadgets in Subsection 2.2. The remaining two subsections concern the stable set and clique partitioning problems. 2.1 Satisfiability The satisfiability problem (SAT) was the first ever problem to be proved N P- complete [10]. In this problem, we have a collection of Boolean variables, say x 1,..., x n, and a collection of logical disjunctions, called clauses, that they should satisfy. An example of a clause is x 1 x 2 x 3, which should be read as x 1 is true or x 2 is false or x 3 is false. In this context, or is always intended to be inclusive. The task is to check whether there exists an assignment of truth values to the variables that satisfies all of the clauses. The special case of SAT in which every clause contains exactly k terms is called k-sat. It is well known that 2-SAT can be solved in linear time. One way to do it as as follows [1]. Replace each clause with two implications. For example, the clause x 1 x 2 is replaced with the statements x 1 implies 2

3 x 2 and x 2 implies x 1. Construct a digraph with 2n nodes, where the first n nodes represent x 1,..., x n and the others represent x 1,..., x n. Represent every implication by a directed arc. For example, the first of the above implications is represented by an arc from node x 1 to node x 2. Compute the strongly connected components of the digraph. The instance is satisfiable if and only there does not exist an index i such that both node x i and node x i lie in the same component. 2.2 Gadgets Gadgets (also sometimes called units or components) are used heavily in computational complexity theory, in order prove hardness results. For example, suppose we wish to prove that 3-SAT is N P-complete. Since we know that SAT is N P-complete, it suffices to show that any clause with more than 3 terms is equivalent to a small (polynomially bounded) number of clauses with only 3 terms (possibly with the help of additional variables). This is indeed the case. For example, suppose we have a clause of the form p i=1 x i, with p 4. Let q = p/2 and replace the clause with two shorter clauses of the form y q i=1 x i and ȳ p i=q+1 x i, where y is an additional Boolean variable. Using this gadget repeatedly, if necessary, we eventually obtain O(k) clauses with only three terms each, using only O(k) additional variables. Instead of giving more details, we refer the reader to [13, 33]. 2.3 The stable set problem Let G be an undirected graph. The maximum cardinality of a stable set in G is called the stability number (or independence number) of G and denoted by α(g). The stable set problem (SSP) is the problem of computing α(g). A generalisation is the maximum weight stable set problem, in which are also given a weight vector w Q n +, and seek a stable set of maximum total weight. The SSP is equivalent to the maximum clique problem on the complement of G. The SSP is therefore not only N P-hard in the strong sense, but hard even to approximate [21]. The simplest formulation of the SSP is the following. For each i v, let x i be a binary variable, taking the value 1 if and only if node i is selected. Then: max i V x i (1) s.t. x i + x j 1 ({i, j} E) (2) x i {0, 1} (i V ). (3) The constraints (2) are called edge inequalities. The convex hull of feasible solutions to (1) (3) is called the stable set polytope and denoted by STAB(G). A wide range of valid inequalities are 3

4 3 2 Figure 1: Odd hole inequality (left) and odd antihole inequality (right). known for this polytope (see the survey [16]). For example, Padberg [29] introduced the following inequalities: clique inequalities i C x i 1 for each maximal clique C V ; odd hole inequalities i S x i S /2 for each set of nodes S inducing an odd hole (chordless cycle of odd length) in G; odd antihole inequalities i S x i 2 for each set of nodes S inducing the complement of an odd hole in G. See Fig. 1 for a representation of odd hole and odd antihole inequalities with S = 7. Small circles represent nodes in S and lines represent edges in G[S]. Padberg showed that the clique inequalities define facets of STAB(G). He also showed that any odd hole or odd antihole inequality either defines a facet, or can be lifted (strengthened) to obtain a facet-defining inequality. The separation problem for the clique inequalities is N P-hard, but some effective separation heuristics exist [27, 32]. The separation problem for the odd hole inequalities can be solved in polynomial time, by reduction to a series of V shortest (s, t)-path problems in a graph that is twice the size of G [14]. If we use the Fibonacci heap variant of Dijkstra s algorithm [12] to solve those ( shortest-path problems, ) the total time taken for odd hole separation is O V E + V 2 log V. Separation algorithms for other inequalities can be found in, e.g., [2, 4, 5, 15, 16, 24, 32]. We omit details for brevity. 2.4 The clique partitioning problem The clique partitioning problem (CPP) was first defined by Marcotorchino [25]. We are given an integer n 2 and a rational weight w e for each edge e E n. The task is to partition V n into subsets (cliques), such that the sum of the weights of the edges that have both end-nodes in the same clique is maximised. The CPP was proved to be N P-hard in the strong sense in [34]. 4

5 3 2 Figure 2: A 2-chorded odd cycle inequality (left) and an odd wheel inequality (right). Marcotorchino formulated the CPP as the following 0-1 LP: max e E n w e x e (4) s.t. x ik + x jk x ij 1 ({i, j} E n, k V n \ {i, j}) (5) x e {0, 1} (e E n ). (6) Here, the binary variable x ij takes the value 1 if and only if nodes i and j are in the same clique. The inequalities (5) are called transitivity inequalities. The convex hull of feasible solutions to (4) (6) is called the clique partitioning polytope and denoted by CPP n. Grötschel and Wakabayashi [19] showed that the transitivity inequalities (5), and the non-negativity inequalities x e 0 for all e E n, define facets of CPP n. They also introduced several other families of facet-defining inequalities. Of those, we will be especially interested in the 2-chorded odd cycle (2-COC) inequalities. These take the form: x e C /2, (7) e E(C) x e e H where C is a cycle in K n with C 5 and odd, and H is the set of 2-chords of C, i.e., the set of edges in E n which connect pairs of nodes in V (C) that have a distance 2 in C. Chopra and Rao [6] found two additional families of facet-defining inequalities for CPP n. Of those, we will be interested in the odd wheel (OW) inequalities. These take the form: x vh x e C /2, (8) v V (C) e E(C) where C is a cycle in K n with C 3 and odd, and h is any arbitrary node in V n \ V (C). Fig. 2 gives a representation of a 2-COC inequality with C = 7 and an OW inequality with C = 5. Solid and dotted lines represent edges whose variables have coefficient 1 and 1, respectively. 5

6 e(7) f(6) f(7) e(1) e(2) f(1) f(5) f(4) e(6) e(4) f(2) e(3) f(3) 3 e(5) Figure 3: An OCW inequality with c = 7. Müller [26] and Caprara & Fischetti [3] independently discovered a family of valid inequalities that includes the 2-COC and OW inequalities as special cases. We will follow Müller in calling them odd closed walk (OCW) inequalities. Suppose we are given a sequence e(1),..., e(c) of edges in E n such that c 3 and odd and such that, for i = 1,..., c, edge e(i) and edge e(i + 1) have exactly one end-node in common (indices taken modulo c). Also, for i = 1,..., c, let f(i) be the unique edge that shares one end-node with e(i) and the other end-node with e(i + 1). The OCW inequality takes the form: c c x e(i) x f(i) c/2. (9) i=1 i=1 Fig. 3 gives a representation of an OCW inequality that is not a 2-COC or OW inequality. Deza et al. [11] showed that the separation problem for the OW inequalities can be solved in O ( n 4) time, by reduction to n 2 shortest-path computations in K n 1. The complexity of separation for the 2-COC inequalities is unknown, but there exists an efficient separation algorithm for the more general OCW inequalities [3, 26]. The algorithm involves the solution of O ( n 2) shortest-path problems in a graph with O ( n 2) nodes and O ( n 3) edges. Using the algorithm in [12] to solve these shortest-path problems, one obtains a total running time of O ( n 5). Separation algorithms for other inequalities can be found in, e.g., [2, 11, 18, 23, 28]. Again, we omit details, for brevity. 3 Using Gadgets to Generate Cutting Planes In this section, we present our gadget approach in its general form. In Subsection 3.1, we define our gadgets formally. In Subsection 3.2, we show how they can be used to generate cuts. Subsections 3.3 and 3.4 are devoted to exact separation and heuristic separation, respectively. Finally, in Subsec- 6

7 tion 3.5, we relate our approach to two other general schemes for generating cuts. 3.1 Gadgets Consider any combinatorial optimisation problem that can be formulated as an integer linear program (ILP). Let S Z n + be the set of feasible solutions and let P = conv(s) be the associated polytope. We will need the following two definitions. Definition 1 Let α T x β be any valid linear inequality that defines a proper face of P. A point x S satisfying α T x = β will be called a root of the inequality. Definition 2 Let Z 0 and Z 1 denote the non-negative even and odd integers, respectively. A valid inequality α T x β will be called a gadget if it satisfies the following properties: it has integral coefficients; it defines a proper face of P ; there exists a pair {i, j} {1,..., n} and a pair p, q {0, 1} such that all roots of the inequality with x i Z p also have x j Z q. We will say that the pair {i, j} is associated with the given gadget. The following example shows that it is possible for several different pairs of variables to be associated with the same gadget. Example 1 Suppose the inequality x 1 + x 2 + 3x 3 3 is valid for P. Then all roots of the inequality with x 1 odd have x 2 and x 3 even, all roots with x 2 odd have x 1 and x 3 even, and all roots with x 3 odd have both x 1 and 2 even. Thus, the given inequality is a gadget, and it is associated with the pairs {1, 2}, {1, 3} and {2, 3}. Two special kinds of gadget will prove to be useful later on: Definition 3 A gadget will be called an exclusive or (XOR) gadget if there exists a pair {i, j} {1,..., n} such that, in all roots, exactly one of the variables x i and x j is odd. Definition 4 A gadget will be called an equality (EQ) gadget if there exists a pair {i, j} {1,..., n} such that, in all roots, the variables x i and x j have the same parity. Example 2 Suppose the inequality x 1 +2x 2 +3x 3 5 is valid for P. Then, in all roots, exactly one of the variables x 1 and x 3 must be odd. Thus, the given inequality is an XOR gadget, and it is associated with the pair {1, 3}. 7

8 3.2 Cutting planes Now we show how to use gadgets to derive cutting planes. Proposition 1 (Gadget Dicycle Inequalities) Consider an arbitrary collection of gadgets, of the form α r x β r for r R. Construct a directed multigraph, say G +, as follows. For i = 1,..., n and p {0, 1}, we have a node v p i. This node represents the statement x i Z p. There are directed arcs for each gadget in R and each pair {i, j} associated with that gadget (e.g., if all roots of the gadget with x i odd also have x j even, then we have an arc from vi 1 to vj 0, and another arc from vj 1 to v0 i ). Suppose there is an index i {1,..., n} such that there is a dicycle in G + passing through both vi 0 and vi 1. Let r(1),..., r(t) be the gadgets that correspond to the arcs in the dicycle. Then the gadget dicycle (GD) inequality t α r(s) x s=1 t β r(s) 1 s=1 is valid for P. Proof. Suppose that a solution x S were a root of all t gadgets simultaneously. The presence of a dipath in G + from vi 0 to vi 1 implies that, if x i is even, then it must be odd. Similarly, the dipath from vi 1 to v0 i implies that, if x i is odd, then it must be even. Since this yields a contradiction, at least one of the t gadgets must have a positive slack. We will see in the following two sections that, for the SSP and CPP, the number of facet-defining GD inequalities can be exponential in n. On the other hand, GD inequalities can sometimes be strengthened. In particular, it is possible that two or more arcs in the dicycle represent the same gadget. In this case, one can strengthen the GD inequality by including just one copy of each distinct gadget in the summation. 3.3 Exact separation We now show that the separation problem for the GD inequalities can be solved exactly in polynomial time. Proposition 2 (Separation of GD Inequalities) If R is a collection of gadgets whose size is bounded by a polynomial in n, then the separation problem for the corresponding GD inequalities can be solved in polynomial time. 8

9 Proof. Let x Q n + be the point to be separated. If any of the gadgets themselves are violated, then we immediately have a cutting plane. So assume that x satisfies all of the given gadgets. Construct the digraph G +, and set the weight of each arc to the slack of the associated gadget. Note that all of the arcs have non-negative weight. By construction, there is a oneto-one correspondence between violated GD inequalities and dicycles that have weight less than 1 and pass through some node vi 0 and its complement vi 1. The problem of finding a minimum-weight dicycle passing through a given pair can be reduced to two shortest (s, t)-dipath problems. So the separation problem can be reduced to 2n shortest-dipath calls in G +. Note that, if the digraph G + has parallel arcs between a pair of nodes, we can omit all apart from the one with smallest weight. So we can assume that G + contains O ( n 2) arcs. We can therefore solve each shortest-dipath subproblem in O ( n 2) time. Thus, the time taken to compute the minimumweight dicycle is O ( n 3). For a given combinatorial optimisation problem and a given family of gadgets, however, the total separation time could be larger or smaller, depending on the precise number of gadgets, the number of pairs associated with each gadget, and the time taken to compute the slacks of the gadgets. Note that any dicycle with weight less than 1 represents a violated inequality. As a result, the separation algorithm can potentially find up to n violated inequalities in a single call. Moreover, whenever a violated (or even near-violated) inequality has been found, one should check whether two or more arcs in the dicycle represent the same gadget. If so, then the inequality can be strengthened as mentioned in the previous subsection. 3.4 Heuristic separation for a special case We now consider the special case in which all of the gadgets in R are XOR and/or EQ gadgets. In this case, we can use an undirected multigraph G + instead of a directed one. For example, if an XOR gadget is associated with the pair {i, j}, all we need is an edge connecting vi 0 and vj 1 and an edge connecting vi 1 and vj 0. Then, every path in G+ linking some node vi 0 with its complement node vi 1 represents a GD inequality. We call these special GD inequalities gadget path (GP) inequalities. The use of an undirected graph in place of a directed one does not, in itself, lead to a significantly faster separation algorithm. On the other hand, it leads to a simple and fast separation heuristic. This heuristic is described in Algorithm 1. The following proposition shows that the heuristic can be implemented so that it runs efficiently. Proposition 3 Every step in Algorithm 1 from the computation of T onward can be implemented to take only O ( F 2) time. 9

10 Algorithm 1: Separation heuristic for GD inequalities. Input: An integer n 3, a collection R of XOR and EQ gadgets, and a fractional point x Q n + that satisfies all of the gadgets in R. Let F = { i {1,..., n} : 0 < x i < 1} denote the set of fractional variables; Construct the undirected version of G +, but include the nodes v 0 i and v 1 i only if i F ; for each remaining collection of parallel edges in G + do delete all apart from one of smallest weight (breaking ties arbitrarily); end Compute a minimum-weight spanning forest T in G + ; for each i F do if there is a path from v 0 i to v 1 i in T then if the weight of the path is less than 1 then output the corresponding violated GD inequality; end end end Proof. If we use a simple variant of Prim s algorithm [31], we can compute T in O ( F 2) time. There are then F candidates for the node vi 0. Consider a fixed candidate vi 0. Traverse T in a breadth-first manner, starting at v0 i. Along the way, record the total weight of the path in T from vi 0 to each node visited. This traversal of T takes O( F ) time. Once it is done, we can check in constant time if there is path from vi 0 to vi 1 with weight less than 1. So the time taken per candidate node is O( F ). In our preliminary computational experiments, this heuristic has proven to be remarkably effective. It is not clear to us whether the heuristic can be adapted to the (more general) GD inequalities. 3.5 Relationship with other approaches We now briefly relate our approach to two other general schemes for generating cuts, due to Caprara & Fischetti [3] and Borndoerfer & Weismantel [2]. Caprara & Fischetti [3] define a family of cuts, called {0, 1 2 }-cuts, for general ILPs. In Section 4 of their paper, they show that, if each constraint in the ILP has exactly two odd left-hand side coefficients, then the separation problem for {0, 1 2 }-cuts can be solved in polynomial time. Now, a constraint with two odd left-hand side coefficients is either an XOR gadget or an EQ gadget, depending on whether its right-hand side is odd or even (see 10

11 Example 2.) From this it can be shown that their result is a special case of ours. Moreover, our approach is more general, since (a) we do not impose restrictions on the parities of the left-hand side coefficients in the gadgets, and (b) we do not restrict ourselves to XOR and EQ gadgets only. Borndoerfer & Weismantel [2] present a related approach based on odd cycles in so-called conflict graphs. As far as we can tell, their approach is incomparable to ours. It is however shown in Section 4 of [23] that, if we are allowed to define new variables that are integer linear combinations of the original variables, then any cutting plane that can be derived using the conflict graph approach can also be derived using the above-mentioned approach of Caprara and Fischetti. Thus, if we allow such new variables, then our approach also generalises the conflict graph approach. 4 Application to the Stable Set Problem We now apply the gadget approach to the SSP. In Subsection 4.1, we define several kinds of gadget. In Subsection 4.2, we use our gadgets to obtain a new and exponentially large family of facet-defining inequalities. Then, in Subsection 4.3, we present separation algorithms and analyse their running time. 4.1 Gadgets for the SSP We know of many gadgets for the SSP. The ones that we have found most useful, in terms of deriving new valid inequalities for STAB(G), are all XOR and EQ gadgets. The following lemma introduces a trivial XOR gadget. Lemma 1 Every edge inequality (2) is an XOR gadget for STAB(G). Proof. In any root, exactly one of x i and x j must be equal to 1. The following theorem presents some more complex XOR gadgets. Theorem 1 Let i and j be a pair of non-adjacent nodes in G, let S be a subset of V \ {i, j}, and suppose that the inequality k S α kx k β defines a facet of STAB(G[S]), where the α k and β are all positive integers. Define { } β (i) = max α k x k : x STAB(G), x i = 1 k S { } β (j) = max α k x k : x STAB(G), x j = 1 k S { } β (i, j) = max α k x k : x STAB(G), x i = x j = 1, k S 11

12 and suppose that β (i, j) < β. Let = β (i) + β (j) β β (i, j). If = 1, then the inequality ( 2β 2β (i) + 1 ) x i + ( 2β 2β (j) + 1 ) x j + 2 α k x k 2β + 1 is an XOR gadget. If 2, then the inequality ( β β (i) + 1 ) x i + ( β β (j) + 1 ) x j + α k x k β + 1 is an XOR gadget. If 3, then the inequality ( β (j) β (i, j) 1 ) x i + ( β (i) β (i, j) 1 ) x j + α k x k β + 1 is also an XOR gadget. Proof. One can check that, if x i = x j = 0, all three inequalities reduce to k S α kx k β + ɛ for some positive ɛ. Thus, there are no roots with x i = x j = 0. One can also check that, if x i = x j = 1, the inequalities reduce to k S α kx k β (i, j) + ɛ for some positive ɛ. Thus, there are no roots with x i = x j = 1 either. If x i = 1 and x j = 0, the inequalities reduce to k S α kx k β (i). Therefore there exist roots with x i = 1 and x j = 0. A similar argument shows that there exist roots with x i = 0 and x j = 1. Here are two examples of XOR gadgets that can be derived in this way. Example 3 Let i and j be a pair of non-adjacent nodes in G, and let C V \ {i, j} be a clique such that (a) each node in C is adjacent to exactly one of i and j, (b) C n(i) 2 and (d) C n(j) 2. We have β = β (i) = β (j) = 1 and β (i, j) = 0. Thus, = 1 and the inequality k S k S k S x i + x j + 2 k C x k 3 (10) is an XOR gadget. Example 4 Let i and j be a pair of non-adjacent nodes in G, and let S V \ {i, j} be a set of 7 nodes such that G[S] is an odd hole. Suppose the nodes in S are labelled v(1),..., v(7), where nodes v(k) and v(k + 1) are adjacent for k = 1,..., 6. Suppose that node i is adjacent to nodes v(k) for k {1, 3, 5, 7}, and node j is adjacent to nodes v(k) for k {1, 2, 4, 6, 7}. We have β = β (i) = 3, β (j) = 2 and β (i, j) = 0. Thus, = 2 and the inequality x i + 2x j + k S x k 4 (11) is an XOR gadget. 12

13 Note that the gadget (11) does not have two odd left-hand side coefficients, as required in the method of Caprara and Fischetti. The following theorem presents some analogous EQ gadgets. Theorem 2 Let i, j, S, α, β, β (i), β (j), β (i, j) and be as in Theorem 1. If = 1, then the inequality ( 2β 2β (i) 1 ) x i + ( 2β 2β (j) 1 ) x j + 2 α k x k 2β is an EQ gadget. If = 2, then the inequality ( β β (i) 1 ) x i + ( β β (j) 1 ) x j + α k x k β is an EQ gadget. If 3, then the inequalities ( β β (i) 1 ) x i + ( β (i) β (i, j) + 1 ) x j + α k x k β ( β (j) β (i, j) + 1 ) x i + ( β β (j) 1 ) x j + α k x k β are EQ gadgets. Proof. If x i = x j = 0, all four inequalities reduce to k S α kx k β. Therefore there exist roots with x i = x j = 0. If x i = x j = 1, they all reduce to k S α kx k β (i, j). Therefore there also exist roots with x i = x j = 1. If x i = 1 and x j = 0, the inequalities reduce to k S α kx k β (i) + ɛ for some positive ɛ. Thus, there are no roots with x i = 1 and x j = 0. A similar argument shows that there are no roots with x i = 0 and x j = 1. Here are two examples of EQ gadgets that can be derived in this way. Example 5 Let i and j be a pair of non-adjacent nodes in G, and let C be a maximal subset of n(i) n(j) such that (a) C is a clique and (b) C 2. We have β = 1 and β (i) = β (j) = β (i, j) = 0. Thus, = 1 and the inequality x i + x j + 2 k C x k 2 (12) is an EQ gadget. Example 6 Let i and j be a pair of non-adjacent nodes in G, and let A be a subset of n(i) n(j) such that G[A] is an odd antihole. We have β = 2 and β (i) = β (j) = β (i, j) = 0. Thus, = 2 and the inequality k S k S k S k S x i + x j + k A x k 2 (13) is an EQ gadget. Again, the gadget (13) does not have two odd left-hand side coefficients. 13

14 Figure 4: Graphs for Examples 7 and Cutting planes for the SSP One can check that, if we use only the edge inequalities (2) as gadgets, the GP inequalities are precisely the odd hole inequalities. Thus, even with such a simple gadget, we obtain an exponentially large number of non-redundant GP inequalities. It turns out that, if we are allowed to use more complicated gadgets, we can obtain various lifted odd hole inequalities. Here are two examples. Example 7 Let G be the graph on the left of Fig. 4. Suppose we use the XOR gadget x 1 + x 7 1 and the following two EQ gadgets of type (12): x 1 + x ( x 2 + x 3 + x 8 ) 2 x 4 + x ( x 5 + x 6 + x 8 ) 2. One can check that, in G +, we have a path starting at v1 1, passing through v4 1 and v1 7, and ending at v0 1. This yields the GP inequality 2 7 x i + 4x 8 4. i=1 Dividing this by two, we obtain 7 i=1 x i + 2x 8 2. This is a lifting of the odd hole inequality x 1 + x 3 + x 4 + x 5 + x 7 2, and it can be shown to define a facet of STAB(G) (either by hand, or with the help of a software package such as PORTA [7]). Example 8 Let G be the graph on the right of Fig. 4. Suppose we use the following EQ gadget of type (12): x 1 + x ( x 2 + x 3 ) 2; and the following XOR gadget of type (10): x 1 + x x k 3. k=5

15 One can check that, in G +, the first gadget links v1 1 and v1 2, and the second links v2 1 with v0 1. This yields the GP inequality 2 8 x i 4. i=1 Dividing this by two, we obtain 8 i=1 x i 2. This is a lifting of the odd hole inequality x 1 + x 2 + x 4 + x 6 + x 7 2, and it can be shown to define a facet of STAB(G). One can check that the number of lifted odd hole inequalities obtainable in this way grows at least exponentially with n. It turns out that, with the aid of other gadgets, such as (11) and (13), we can obtain facet-defining GP inequalities that are not lifted odd hole inequalities. Example 9 Let G be the (unique) graph with 13 nodes and 31 edges such that: {1,..., 5} and {6,..., 10} induce 5-antiholes (which are also 5-holes); node 11 is adjacent to nodes 1,..., 10; node 12 is adjacent to nodes 1,..., 5; node 13 is adjacent to nodes 6,..., 10; nodes 12 and 13 are adjacent. (See Fig. 5.) Suppose we use the XOR gadget x 12 + x 13 1 and the following two EQ gadgets of type (13): x 11 + x 12 + x 11 + x x i 2 i=1 10 i=6 x i 2. One can check that, in G +, we have a path starting at v11 1, passing through v12 1 and v0 13, and ending at v0 11. This yields the GP inequality 2 ( ) 10 x 11 + x 12 + x 13 + x i 4. (14) This can be shown to define a facet of STAB(G). One can check that G does not contain any holes of cardinality 9, which means that the GP inequality is not a lifted odd hole inequality. i=1 15

16 11 {1,..., 5} {6,..., 10} Figure 5: Graph for Example 9. Readers who are familiar with the concept of Chvátal rank (see [8]) may find the following result of interest. Proposition 4 The GP inequality (14) has Chvátal rank larger than 1 with respect to the linear system defined by the clique and non-negativity inequalities. Proof. Since STAB(G) is full-dimensional and the inequality in question defines a facet, we can use the method in [20]. If we maximise the lefthand side of the inequality (14) subject to the clique and non-negativity inequalities, we obtain a fractional LP solution with x i equal to 0.25 for nodes 1,..., 10 and to 0.5 for nodes 11, 12 and 13. The profit of this solution is ( ) + (0.5 6) = 5.5. Since the right-hand side of (14) is less than 5.5, the inequality must have rank larger than Separation for the SSP Now we make some remarks on the separation problem for the GP inequalities for STAB(G). A first observation is that, if we include only the edge inequalities (2) as our gadgets, we effectively duplicate the exact odd hole separation routine of Gerards & Schrijver [14]. As mentioned in Subsection 2.3, this routine takes O ( V E + V 2 log V ) time. If we wish to add other gadgets in our collection, the running time increases. For example: If we also include our gadgets with S = 2, the time taken to compute the slacks of the gadgets increases to O ( E 2), and the total running time increases to O ( V 3 + E 2). If we also include our gadgets with S = 3, both the time to compute the slacks and the total running time increase to O ( V E 2). If we also include our gadgets with S = 4, both the time to compute the slacks and the total running time increase to O ( E 3). 16

17 In practice, rather than imposing restrictions on the cardinalities of S, it would be easier and faster to use greedy heuristics to find a good XOR gadget and EQ gadget for each pair i, j of non-adjacent nodes (where a good gadget means one with small slack). 5 Application to the Clique Partitioning Problem We now apply the gadget approach to the CPP. The structure of this section is identical to that of the previous one. 5.1 Gadgets for the CPP The following lemmas describe four simple, yet remarkably powerful, gadgets for the CPP. Lemma 2 For any {i, j} E n, the inequality x ij 0 (15) is a gadget for CPP n. This gadget is associated with n 2 distinct pairs of variables. Proof. Validity is trivial. If x is a root, then nodes i and j cannot lie in the same clique. This implies that, for any node k V \ {i, j}, the variables x ik and x jk cannot both be 1. There are n 2 candidates for k. Lemma 3 For any {i, j} E n, the inequality x ij 1 (16) is an EQ gadget for CPP n. This gadget too is associated with n 2 distinct pairs of variables. Proof. Validity is again trivial. If x is a root, then nodes i and j must lie in the same clique. This implies that, for any node k V \ {i, j}, the variables x ik and x jk must take the same value. Lemma 4 For any {i, j} E n and any k V n \ {i, j}, the transitivity inequality (5) is itself a gadget for the CPP. In particular, all roots satisfy the following implications: If x ij = 1 then x ik = 1 and x jk = 1; If x ik = 0 then x ij = 0 and x jk = 1; If x jk = 0 then x ij = 0 and x ik = 1. 17

18 Proof. One can check that there are two kinds of roots: either (a) x ij is 0 and exactly one of x ik and x jk are 1, or (b) all three variables are 1. In either case, the stated implications hold. Lemma 5 For any {i, j} E n and any k V n \ {i, j}, the inequality is an XOR gadget for CPP n. x ik + x jk 2x ij 1 (17) Proof. Validity follows from the fact that (17) is the sum of the transitivity inequality (5) and the inequality (15). One can check that in all roots, exactly one of the variables x ik and x jk is equal to one. Observe that the XOR gadget (17) is the sum of two other gadgets, namely, (5) and (15). This means that any valid inequality that can be derived using the four kinds of gadgets presented in this subsection can also be derived using only the first three kinds. Nevertheless, we will see that the XOR gadget is useful for the purposes of exposition, and also when devising separation algorithms. 5.2 Cutting planes for the CPP The following proposition shows that the XOR gadgets (17) alone are rather powerful. Proposition 5 Every GP inequality derived from the XOR gadgets (17) is equivalent to an OCW inequality (9) and vice-versa. Proof. Let G + be the graph obtained when R consists of the XOR gadgets. Note that there are two nodes in G + for each edge e E n, namely ve 0 and ve. 1 Moreover, G + is bipartite. Consider a path in G + that goes from one node to its complement node. Let c denote the number of edges in the path. Since G + is bipartite, c must be odd. Let e(1),..., e(c + 1) be the edges in E n that correspond to the first c nodes in the path, and note that the last node in the path corresponds to e(1). The ith edge in the path then corresponds to the XOR gadget x e(i) + x e(i+1) 2x f(i) 1, where f(i) is the edge having one end-node in common with e(i) and e(i + 1). The GP inequality can therefore be written as c ( ) xe(i) + x e(i+1) 2x f(i) c 1. i=1 Dividing this by two, we obtain the OCW inequality (9). Similarly, given any OCW inequality, one can construct a path in G + that yields a GP inequality that is twice the given OCW inequality. 18

19 Figure 6: Two facet-defining GP inequalities. It turns out that, if we use the EQ gadgets (16) and XOR gadgets (17) in combination, we can obtain facet-defining GP inequalities that are not OCW inequalities. Here are two examples. Example 10 Suppose we use the XOR gadgets x 12 + x 13 2x 23 1, x 13 + x 14 2x 34 1, x 14 + x 15 2x 45 1; and the EQ gadget x One can check that G + contains a path starting at the node v12 1, passing through v0 13, v1 14, v0 15, and ending at v0 12. This yields the GP inequality. x x x 14 + x 15 2x 23 + x 25 2x 34 2x (18) This can be shown to define a facet of CPP 5. Example 11 Suppose we use the XOR gadgets x 17 + x 27 2x 12 1, x 27 + x 37 2x 23 1, x 37 + x 47 2x 34 1, x 47 + x 57 2x 45 1, x 57 + x 67 2x 56 1, and the EQ gadget x One can check that G + contains a path starting at v17 1, passing through nodes v0 27,..., v0 67, and ending at v0 17. This yields the GP inequality x 17 +2x 27 +2x 37 +2x 47 +2x 57 +x 67 2x 12 2x 23 2x 34 2x 45 2x 56 5 (19) This can be shown to define a facet of CPP 7. The inequalities (18) and (19) are represented in Figure 6. Caprara & Fischetti [3] showed out that every OCW inequality has Chvátal rank 1. On the other hand, we have the following result. Proposition 6 The GP inequalities (18) and (19) have Chvátal rank larger than 1. 19

20 Figure 7: Basic fractional solutions to LP relaxations. Proof. Since CPP n is full-dimensional and the inequalities in question define facets of CPP 5 and CPP 7, respectively, we can again use the method in [20]. When n = 5, if we maximise the left-hand side of the inequality (18) subject to the transitivity and non-negativity inequalities, we obtain the fractional LP solution displayed on the left of Figure 7. (Here, the dotted lines represent variables with value 1/2.) The profit of this solution is 4. Since the right-hand side of (18) is less than 4, the inequality must have rank larger than 1. Similarly, when n = 7, if we maximise the left-hand side of the inequality (19) subject to the transitivity and non-negativity inequalities, we obtain the solution shown on the right of Figure 7. The profit of this solution is 6, whereas the right-hand side of (19) is less than 6. We leave it as an open question whether one can derive still more facetdefining inequalities by using a combination of the first three kinds of gadgets mentioned in the previous subsection. 5.3 Separation for the CPP Now we make some remarks on the separation problem for the GD and GP inequalities for the CPP. A major difference between the CPP and the SSP is that there are ( ) n 2 variables rather than V. Moreover, the number of gadgets described in Subsection 5.1 is O ( n 3). As a result, even if we use only those gadgets, the multigraph G + has O ( n 2) nodes and O ( n 3) edges. If we use the algorithm in [12], each shortest-path computation can be performed in O ( n 3) time. This leads to a total running time of O ( n 5). Since a running time of O ( n 5) is rather high, it may be preferable to use the GP separation heuristic described in Subsection 3.4. One can check that, for the CPP, the heuristic takes O ( f 2) time, where f is the number of fractional variables. We have performed some preliminary experiments with various cutting planes for the CPP. Our results indicate that, for typical CPP instances, the OW, OCW and GP inequalities tend to close around 83%, 88% and 92% 20

21 of the gap, respectively, between the LP upper bound and the optimum. However, we were using our heuristic to separate the GP inequalities, so the true gap closed by them may be higher. 6 Conclusions Up to now, gadgets have been used solely to prove hardness results. In this paper, we have shown that they can also be useful for deriving new cutting planes, along with accompanying separation algorithms. The results that we obtained with the stable set and clique partitioning problems suggest that this new approach has potential. A natural topic for future research is whether the gadget approach can be usefully applied to other N P-hard combinatorial optimisation problems. Problems worth consideration include, e.g., the TSP, the max-cut problem, the linear ordering problem [17] and the transitive acyclic subdigraph problem [26]. Another interesting topic is the possibility of defining gadgets that impose more complex structural properties on the roots, rather than merely imposing relationships between the parities of individual pairs of variables. References [1] B. Aspvall, M.F. Plass & R.E. Tarjan (1979) A linear-time algorithm for testing the truth of certain quantified boolean formulas. Inform. Proc. Lett., 8, [2] R. Borndörfer & R. Weismantel (2000) Set packing relaxations of integer programs. Math. Program., 88, [3] A. Caprara & M. Fischetti (1996) {0, 1 2 }-Chvátal-Gomory cuts. Math. Program., 74, [4] E. Cheng & W.H. Cunningham (1997) Wheel inequalities for stable set polytopes. Math. Program., 77, [5] E. Cheng & S. de Vries (2002). Antiweb-wheel inequalities and their separation problems over the stable set polytopes. Math. Program., 92, [6] S. Chopra & M.R. Rao (1993) The partition problem. Math. Program., 59, [7] T. Christof & A. Loebl, PORTA (polyhedron representation transformation algorithm). Software package, available for download at 21

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