Minimum d-Transversals of Maximum-Weight Stable Sets in Trees

Given an integer d and a weighted tree T , we show how to ﬁnd in polynomial time a minimum d -transversal of all maximum-weight stable sets in T , i.e., a set of vertices of minimum size having at least d vertices in common with every maximum-weight stable set. Our proof relies on new structural results for maximum-weight stable sets on trees.


Definitions and previous results
Given an undirected simple loopless graph G =( V, E) and a positive weight function w on its vertices, a stable set S of G is a set of pairwise non adjacent vertices. The weight of a stable set is the sum of the weights of its vertices. A stable set S is a maximum-weight stable set if no other stable set has a strictly larger weight. We will denote by α w (G) the weighted stability number, i.e., the weight of a maximum-weight stable set in G.
A subset T ′ ⊆ V is a d-transversal of G if for every maximum-weight stable set S we have |S ∩ T ′ |≥d. Thus, a d-transversal is a subset of vertices which intersects every maximum-weight stable set in at least d vertices.
We denote by τ d (G) the minimum cardinality of a d-transversal in G.A d-transversal will be minimum if it is of minimum size.
Some properties and results concerning d-transversals of maximum matchings (i.e., subsets of edges intersecting every maximum matching in at least d edges) were given in [2,5] where the following is proved (here ν(G) denotes the size of a maximum matching): Theorem 1.1 For every fixed d ∈{ 1, 2,...,ν(G)}, finding a minimum dtransversal of maximum matchings is NP-hard even if G is bipartite.
This result implies that finding minimum d-transversals of maximum stable sets (i.e., stable sets with maximum cardinality)i sNP-hard (matchings in G are stable sets in the line-graph L(G)). In [1] it is shown how to find such minimum d-transversals of maximum-cardinality stable sets in polynomial time when G is bipartite. NP-hard results concerning maximum-weight stable sets in bipartite graphs can be found in [3,4].
Here we are interested in maximum-weight stable sets: we show how to find minimum d-transversals of maximum-weight stable sets in polynomial 1 The research work of these two authors was supported by the French ANR project DOPAGE (ANR-09-JCJC-0068). 2 Email: bentz@lri.fr 3 Email: marie-christine.costa@ensta.fr 4 Email: dominique.dewerra@epfl.ch 5 Email: chp@cnam.fr 6 Email: bernard.ries@dauphine.fr 2 time in trees. Our proof mainly relies on the definition of a partition of the given tree having useful properties, and on new structural results concerning maximum-weight stable sets in trees, which allow us to reduce the minimum d-transversal problem in weighted trees to a matching problem.
2 Structural results concerning maximum-weight stable sets in trees Let T =(V, E) be a weighted tree. We first give some preliminary properties.
Given an arbitrary vertex r of T , we denote by T r the tree rooted at vertex r and by T r (v) the subtree of T r rooted at vertex v, for every v ∈ T . We also define the following function: for every vertex v ∈ T , p r (v, δ) is the weight of a maximum-weight stable set in T r that contains v if δ = 1, and that does not contain v if δ = 0. Note that we have α w (T r (v)) = max{p r (v, 0),p r (v, 1)}, and that all values p r (v, δ) can be easily computed by dynamic programming formulas, starting from the leaves of T r . This also allows us to find forced, excluded, and free vertices.

Definition 2.1 A vertex of a tree T is:
• forced if it belongs to all the maximum-weight stable sets of T , • excluded if it does not belong to any maximum-weight stable set of T , • free otherwise.
The sets of forced, excluded and free vertices can be easily computed using the following property: Note that excluded vertices belong neither to maximum-weight stable sets nor to minimal inclusionwise transversals: hence, they can be removed from the tree. Moreover, when looking for a minimum d-transversal T ′ , we include min{d, f } forced vertices in T ′ (and possibly other vertices if needed), where f is the total number of forced vertices. So, when d ≤ f , we can trivially solve the problem by using only forced vertices. As a consequence, we can assume without loss of generality that d>f, and that we look for a d ′ -transversal in a forest having only free vertices, where d ′ = d − f . Since vertices of weight 0 do not belong to minimal inclusionwise transversals, they can be removed as well. Finally, any maximum-weight stable set in a forest is the disjoint union of maximum-weight stable sets in each one of the trees of the forest, so it is sufficient to focus on maximum-weight stable sets in trees having only free vertices and positive weights. We define the following partition of the tree, that will allow us to establish the structure of such stable sets.
We give some useful results.

Proposition 2.4 For any vertex
This property implies that, in a tree containing only free vertices, we have p r (v, 0) ≤ p r (v, 1) for any vertex v and any root r.
In other words, the partition of V is unique, and does not depend on the choice of the root. From now on we will denote by V = {V 1 ,...,V q } this unique partition.
Remark 2.7 Lemma 2.6 and the uniqueness of the partition V imply that, given any two adjacent vertices u and v of V with u ∈ V i and v ∈ V j for some i and j Let (B, W) denote the bipartition of V .
Lemma 2.8 Let T be a weighted tree containing only free vertices, and let u, v be two adjacent vertices in T with u ∈ V i and v ∈ V j for some i and j = i. Then, a stable set S of T has maximum weight in T if and only if w(S ∩ T 1 )=α w (T 1 ) and w(S ∩ T 2 )=α w (T 2 ), where T 1 = T v (u) and Proof (Sketch.) Actually, it is sufficient to show that α w (T 1 )+α w (T 2 )= α w (T ). The inequality α w (T ) ≤ α w (T 1 )+α w (T 2 ) simply comes from the fact that any stable set in T is the union of a stable set in T 1 and of a stable set in T 2 . The inequality α w (T ) ≥ α w (T 1 )+α w (T 2 ) comes from the fact that p v (u, 0) = p v (u, 1) implies that there exist two maximum-weight stable sets S 1 and S 2 of T 1 and T 2 respectively that are compatible (i.e., S 1 ∪ S 2 is a stable set for T ). ✷ Lemma 2.9 Let T be a weighted tree containing only free vertices. Then, a vertex set S is a maximum-weight stable set if and only if S is stable and for any j ∈{1,...,q} either Proof (Sketch.) To show that any maximum-weight stable set S satisfies ..,q}, we only need to use the uniqueness of the partition V and the fact that for every j ∈{ 1,...,q} and v ∈ V j , we have by definition p v (u, 0) <p v (u, 1) for every u ∈ V j \{v}.
We use induction to show that any stable set S satisfying S ∩ V j = B ∩ V j or S ∩ V j = W ∩ V j for each j ∈{1,...,q} has maximum weight. When q =1, this trivially follows from the above argument. Assume now that q>1. We consider two vertices u ∈ V i and v ∈ V j for i = j, and apply induction on T v (u) and T u (v) (this is possible since a straightforward consequence of Lemma 2.8 is that any vertex in T v (u)o rT u (v) is free): finally, we conclude the proof by using Lemma 2.8. ✷

Finding minimal d-transversals in weighted trees
From the partition V, we define the following auxiliary digraph T * =(V * ,A * ): (Notice that T * is a directed tree.) From T * we define the following partial order (V, ): for u ∈ W and v ∈ B, u v if and only if either u, v ∈ V j for some j ∈{ 1,...,q} or u ∈ V i , v ∈ V j , i = j, and there exists a (directed) path from V i to V j in T * . Let d max be the minimum size of a maximum-weight stable set in T . Then, our main result is the following: Lemma 3.1 Let T be a weighted tree containing only free vertices. Then for any 0 ≤ d ≤ d max , a set T ′ is a minimum d-transversal in T iff it consists of d disjoint pairs of vertices (u, v) such that u ∈ W , v ∈ B and u v.
Proof (Sketch.) First we show that a set T ′ consisting of d disjoint pairs of vertices (u, v) with u ∈ W , v ∈ B and u v is a d-transversal. Each of these pairs (u, v) with u ∈ V i and v ∈ V j has at least one vertex in common with any maximum-weight stable set S, because of Lemma 2.9: either S ∩ V i = W ∩ V i 5 and so u ∈ S,o rS ∩ V i = B ∩ V i and then we have S ∩ V j = B ∩ V j (because of the definition of T * ), and so v ∈ S. Then we have to show that at least 2d vertices are needed in any dtransversal T ′ (which is obvious, since from Lemma 2.9 we must have |B∩T ′ |≥ d and |W ∩ T ′ |≥d), and that there always exist d max such pairs. We consider a maximum matching M in the bipartite graphG =(W, B,Ẽ) with edge set E = {uv : u ∈ W, v ∈ B, u v}, and show how to construct, using the structure of M , a maximum-weight stable set with the same size. Finally, we assume by contradiction that it exists a d-transversal T ′ not containing d such pairs, and show how to construct, using the structure of T ′ , a maximum-weight stable set having at most d − 1 vertices in common with T ′ . ✷ We observe that, from Lemma 3.1, there is a bijection between matchings of size d inG and d-transversals in T . This yields: Theorem 3.2 Let T =( V, E) be a weighted tree. Then, finding a minimum d-transversal in T can be done in polynomial time, by solving a maximum matching problem inG.

Open problems
It would be interesting to study this problem in chordal graphs, or in other classes of bipartite graphs: for instance, in [2], minimum d-transversals of maximum matchings were studied in grids.