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Some results on maximum stable sets in certain P5-free graphs
Discrete Applied Mathematics 132 (2004) 175 – 183
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Some results on maximum stable sets in certain P5-free graphs Ra)aele Mosca Dipartimento di Scienze, Universita degli Studi “G. D’Annunzio”, Viale Pindaro 42, Pescara 65127, Italy Received 19 March 1998; received in revised form 26 March 2002; accepted 15 April 2002
Abstract We introduce some large classes of P5 -free graphs for which the maximum stable set problem can be e.ciently solved. The solution algorithms are based on the search of augmenting vertices. Then a method for extending some of them to the weighted case is provided. ? 2003 Elsevier B.V. All rights reserved. Keywords: Stable set; Computational complexity
1. Introduction Let G = (V; E) be a graph. If G does not admit any induced subgraph isomorphic to a graph H , then G is H -free. A stable set in G is a subset of pairwise non-adjacent vertices of G. The maximum stable set problem (MS) is that of computing a stable set of G of maximum cardinality. If one associates a positive real number w(v) with each vertex v of G, and for any U ⊆ V denotes as weight w(U ) the sum of w(u) for u ∈ U , then the weighted maximum stable set problem (WMS) is that of computing a stable set S of G of maximum weight. MS is a well known NP-hard problem and remains di.cult for cubic and planar graphs. However, polynomial-time exact algorithms are known for several classes of graphs, which are usually called good classes. For example, MS can be e.ciently solved for perfect graphs and for partial k-trees with ?xed k. Then some good classes can be characterized in terms of forbidden induced subgraphs, such as claw-free graphs [16,19], 2K2 -free graphs [12] which
This work has been partially supported by the German Research Community, DFG, n. BR1446/4-1. E-mail address: [email protected] (R. Mosca).
0166-218X/$ - see front matter ? 2003 Elsevier B.V. All rights reserved. doi:10.1016/S0166-218X(03)00399-8
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are a special class of those P5 -free, and chair-free graphs [2] (a claw is formed by four vertices a; b; c; d and edges ab; ac; ad; a 2K2 is formed by four vertices a; b; c; d and edges ab; cd; a chair is formed by ?ve vertices a; b; c; d; e and edges ab; ac; ad; de). Actually, the results in [1,2] imply that the class of P5 -free is the unique minimal class de?ned by forbidding a single connected subgraph for which the computational complexity of MS is an open question. On the other hand, several results on P5 -free graphs are known in the literature. In [4,5] the authors give structural properties of such graphs in terms of dominating subgraphs, and provide a characterization of graphs with no induced paths of a given length. In [18] the author shows that P5 -free graphs are irredundance perfect. Then strong structural properties are known for special classes containing (P5 , claw)-free graphs [8], and for (P5 , diamond)-free graphs [3,7]. Structural properties based on graph decomposition are introduced in [13] for (P5 ; PJ 5 )-free graphs, in order to generalize a characterization of (2K2 ; C4 )-free graphs. In [6] the authors prove that the Strong Perfect Graph Conjecture holds true for any (P5 ; F)-free graph, where F is any connected ?ve-vertex graph containing no induced 2K2 . Then e.cient algorithms to solve MS have been designed for classes of P5 -free graphs, such as (P5 , house)-free [13,14], (P5 , banner)-free [10,15], (P5 , co-chair)-free and (P5 , co-banner)-free [11], (P5 , gem)-free graphs [9]. The contribution of this paper is mainly given by the introduction of large good classes of P5 -free graphs and of a general method to approach WMS for P5 -free graphs. 2. Notation and preliminary Let G = (V; E) be a graph and U; W be subsets of V . Let us write NU (W ) = {v ∈ U \ W : (v; w) ∈ E for some w ∈ W }. If W is a singleton, i.e., W = {w}, then we write NU (w) instead of NU ({w}); if U = V , then we write N (W ) instead of NV (W ). Furthermore, let us denote as G[W ] the subgraph of G induced by W . Let us say that W dominates U if U ⊆ W ∪ N (W ); if W = {w}, then we will say that w dominates U . Let S be a stable set of G, and v ∈ V \ S: let us write H (v; S) = {w ∈ V \ (S ∪ {v} ∪ N (v)): NS (w) ⊆ NS (v)}. A graph G weighted by a positive real function w will be denoted as G = (V; E; w). Then let us denote as Pk a path with k vertices, as Ck a cycle with k vertices, and as Km; n a bipartite graph G = (V1 ; V2 ; E) with |V1 | = m; |V2 | = n, and E = V1 × V2 . Let us report two results on P5 -free graphs which will be useful in the next sections. Lemma 2.1. Let G = (B1 ; B2 ; F) be a connected bipartite graph with |B1 | = k. Then G is P5 -free if and only if there exists a bijective function f : {1; : : : ; k} → B1 such that N (f(i)) ⊇ N (f(i + 1)), for i = 1; : : : ; k − 1; moreover N (f(1)) ≡ B2 . For the above lemma one can refer to di)erent known results (see e.g. [4,21]).
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Denition 2.2. Let G = (V; E) be a P5 -free graph with n vertices, and S be a maximal stable set of G. A vertex v of V \ S is augmenting for S if H (v; S) contains a stable set T with |T | ¿ |NS (v)|. Theorem 2.3 (Mosca [17]). Let G = (V; E) be a P5 -free graph with n vertices, and S be a maximal stable set of G. Then: (i) S is maximum if and only if there exists no augmenting vertex for S; (ii) if one can compute in time O(nh ) (where h is a natural number) a maximum stable set of G[H (v; Z)] for every maximal stable set Z of G and for every vertex v ∈ V \ Z, then one can compute a maximum stable set of G in time max{O(nh+2 ); O(n4 )}. By the proof of the above result, one realizes that, while the ?rst statement can be easily generalized for WMS, the second cannot be directly extended to the weighted case (apart from pseudo-polynomial considerations). 3. An augmenting approach for WMS in P5 -free graphs Let G = (V; E) be a graph, and S be a maximal stable set of G. Consider the following equivalence relation R(S) on V \ S: for any u; v ∈ V \ S, uR(S)v
if NS (u) = NS (v):
Let us denote as MR(S) the set of the equivalence classes obtained with respect to R(S). Then let us de?ne a binary relation 6 on MR(S) as follows: for any Mi ; Mj ∈ MR(S) Mi 6 Mj
if NS (u) ⊆ NS (v)
for any u ∈ Mi ; v ∈ Mj :
It is immediate to see that (MR(S) ; 6 ) is a poset. Let us denote as V (Mi ) the set of vertices of G belonging to any Mi ∈ MR(S) . Two elements M1 ; M2 of MR(S) are incomparable if neither M1 6 M2 nor M2 6 M1 . As usual, a chain of poset (MR(S) ; 6 ) is a subset C of MR(S) such that each pair of elements of C is in 6 . Two chains C1 ; C2 of (MR(S) ; 6 ) are incomparable if there exists no element of S adjacent to both an element of V (C1 ) and V (C2 ) (in particular each element of C1 is incomparable with each element of C2 ). To focus the properties of the above poset with respect to P5 -free graphs, let us point out the proposition below.
Proposition 3.1. Let G = (V; E) be a connected P5 -free graph, S be a maximal stable set of G, and T = ∅ be any stable set of G. Then there exist h ¿ 1 pairwise incom parable chains C1 ; : : : ; Ch of (MR(S) ; 6 ), such that T ⊆ S ∪ V (C1 ) ∪ · · · ∪ V (Ch ). Proof. If |T | = 1, then the assertion immediately follows. Suppose |T | ¿ 1. Then let M1 ; M2 be any two elements of MR(S) such that M1 and M2 are incomparable and V (M1 ) ∩ T = ∅; V (M2 ) ∩ T = ∅: let t1 ∈ V (M1 ) ∩ T , and t2 ∈ V (M2 ) ∩ T . Note that there exists no vertex s ∈ S adjacent to both vertices in V (M1 ) and in V (M2 ),
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otherwise, since by construction there exist two distinct vertices s1 ; s2 ∈ S such that (s1 ; t1 ); (s2 ; t2 ) ∈ E, and (s1 ; t2 ); (s2 ; t1 ) ∈ E, then s1 ; t1 ; s; t2 ; s2 would induce a P5 of G. In particular, there is no M3 ∈ MR(S) such that M3 6 M1 and M3 6 M2 . Furthermore, there M3 : in fact, let is no M3 ∈ MR(S) such that V (M3 )∩T = ∅, and both M1 6 M3 and M2 6 t3 ∈ V (M3 ) ∩ T ; then by construction there would exist two distinct vertices s1 ; s2 ∈ S such that (s1 ; t1 ); (s2 ; t2 ); (s1 ; t3 ); (s2 ; t3 ) ∈ E, that is, G would contain an induced P5 . Then the proposition follows. Now let us prove a theorem which extends Theorem 2.3 to the weighted case. That is possible by de?ning a method for checking just once if a vertex of G is augmenting or not. Also, one obtains an improvement of the computational complexity, with respect to that exposed in Theorem 2.3. Theorem 3.2. Let G = (V; E; w) be a P5 -free graph with n vertices, and d be a natural number. If one can solve WMS in H (v; S) in time O(nd ) for every maximal stable set S of G and for every v ∈ V \ S, then one can solve WMS in G in time max{O(nd+1 ); O(n3 )}. Proof. Let S be a maximal stable set of G. It is well known that, based on poset (MR(S) ; 6 ), one can de?ne a (topological) total order ¡ on MR(S) , such that, for any M1 ; M2 ∈ MR(S) , if M1 6 M2 then M1 ¡ M2 . Then, let MR(S) = {M1 ; M2 ; : : : ; Mr }, with Mi ¡ Mi+1 for any index i, and let us write M (k) = M1 ∪ M2 ∪ · · · ∪ Mk . The method is based on the idea of iteratively solving WMS in G[S ∪ V (M (k) )]. Let us denote by S0∗ the set S; then, for any index k, denote by Sk∗ a maximum weighted stable of G[S ∪ V (M (k) )]. Let us prove the following fact. ∗ Fact 1. For any index k, if w(Sk∗ ) ¿ w(Sk−1 ) then there exists v ∈ Mk such that ∗ ∗ ). G[{v} ∪ H (v; Sk−1 )] contains a stable set of weight w(Sk∗ \ Sk−1
Let k = 1. One clearly has w(S1∗ ) ¿ w(S0∗ ) = w(S). If w(S1∗ ) ¿ w(S0∗ ), then, since NS (x) = NS (y) for any x; y ∈ M1 , one has that G[S1∗ ∪ S0∗ ] has exactly one non-empty component H , and then: H ∩ S1∗ ⊆ M1 and each vertex v ∈ M1 dominates H ∩ S0∗ . Then Fact 1 follows for k = 1. ∗ ∗ Let k ¿ 1. Suppose w(Sk∗ ) ¿ w(Sk−1 ): then G[Sk∗ ∪ Sk−1 ] has non-empty components H1 ; H2 ; : : : ; Hp . Recalling that Mi ¡ Mk for i ¡ k, let Z be the set of vertices of S which are adjacent to all those in V (Mk ) and non-adjacent to all those in V (Mi ) ∗ for i ¡ k. One clearly has Z ⊆ Sk−1 . Furthermore, since Sk∗ ∩ V (Mk ) = ∅, there ∗ ∗ ], say H1 , such that Z ⊆ H1 ∩ Sk−1 exists one non-empty component of G[Sk∗ ∪ Sk−1 ∗ ∗ ∗ (k) and (V (Mk ) ∩ Sk ) ⊆ H1 ∩ Sk . Hence, since Sk ⊆ S ∪ V (M ), one has that Hi ⊆ ∗ S ∪ V (M (k−1) ), for i ¿ 2. But, by de?nition of Sk−1 , that implies that one can suppose ∗ w.l.o.g. that Hi is empty for any i ¿ 2. Thus G[Sk∗ ∪ Sk−1 ] has exactly one non-empty component, that is H1 ; then, since in H1 the vertices of Sk∗ ∩V (Mk ) are the only vertices adjacent to Z, Lemma 2.1 implies that there exists a vertex v ∈ V (Mk ) dominating
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∗ ∗ ∗ H1 ∩ Sk−1 , and hence G[{v} ∪ H (v; Sk−1 )] contains a stable set of weight w(Sk∗ \ Sk−1 ). Then Fact 1 is proved. Fact 1 implies that one can compute Sk∗ , whenever for every v ∈ Mk one can compute ∗ )]. a maximum weighted stable set S(v) of G[{v} ∪ H (v; Sk−1 ∗ ∗ (v)) + w(S ∗ Denote as Hk = max{w(S(v)) − w(NSk−1 k−1 \ NSk−1 (v)): v ∈ Mk }. ∗ ∗ ∗ Then one has w(Sk ) = max{w(Sk−1 ); Hk }. In particular: if w(Sk∗ ) ¿ w(Sk−1 ), then ∗ ∗ ∗ Sk = S(u) ∪ Sk−1 \ NSk−1 (u), where u ∈ Mk is the vertex individuated by Hk ; otherwise, ∗ one can let Sk∗ = Sk−1 . Therefore, to compute a maximum weighted stable set of G one can apply the following algorithm, whose output Sr∗ is an optimum solution.
Algorithm 1 Step 1: Compute a maximal stable set S of G, and set S0∗ := S; Step 2: for k := 1; : : : ; r do begin compute Sk∗ ; end. ) and (MR(S) ; ¡ ) About the computational complexity of Algorithm 1: (MR(S) ; 6 3 can be detected in time O(n ); thus Step 1 can be executed in time O(n2 ), while Step 2 in time O(nd+1 ). Then the theorem follows.
4. Large good classes of P5 -free graphs for WMS and MS In this section we show that the primitive approach used in Theorem 2.3 can be extended in order to detect several good classes of P5 -free graphs. In particular, by Theorem 3.2 for some of them one can de?ne an e.cient algorithm for the weighted case as well. Notation 4.1. Let F be a graph. Let us denote as F (1) the graph obtained from F by adding two new vertices v; s, such that s dominates F, while v is adjacent only to s. In general, let F (h) be the graph obtained from F by adding h + 1 new vertices v; s1 ; : : : ; sh such that {s1 ; : : : ; sh } induce a stable set, si dominates F for i = 1; : : : ; h, while v is adjacent only to {s1 ; : : : ; sh } (see for example the graphs F (1) = K3(1) and F (2) = K3(2) in Fig. 1). s2
v
s
v
(a)
(b) (1)
s1 (2)
Fig. 1. (a) The graph K3 ; (b) the graph K3 .
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Theorem 4.2. Let F be a graph. If one can solve WMS for (P5 ; F)-free graphs in time O(nd ), then one can solve WMS for (P5 ; F (1) )-free graphs in time O(nd+2 ). Proof. Let G be a (P5 ; F (1) )-free graph with n vertices. Consider the problem of ?nding a maximum weighted stable, say S , of G[H (v; S)]. By de?nition of H (v; S) one has that NS (v) is a maximal stable set of G[NS (v) ∪ H (v; S)]; in particular, NS (v) and S induce a connected bipartite subgraph of G. Thus, by Lemma 2.1 there exists s ∈ NS (v) such that S ⊆ N (s). It follows that S can be found by solving WMS in G[N (s) ∩ H (v; S)] for all s ∈ NS (v). But G[N (s) ∩ H (v; S)] is F-free, otherwise vertices v; s and subgraph F would induce a F (1) of G (contradiction). Thus one can compute S in time O(nd+1 ), and therefore, by Theorem 3.2, ?nd a maximum weighted stable set of G in time O(nd+2 ). As WMS is easily solvable in classes of graphs with very small forbidden induced subgraphs, Theorem 4.2 can be used as a tool to de?ne other good classes of P5 -free graphs with respect to the weighted case. In particular, in view of a repeated application, the computational complexity of the method for classes of graphs with large forbidden subgraphs is a)ected by the complexity of WMS for classes with smaller forbidden subgraphs. For example, as one can compute a maximum weighed stable set in (P5 , diamond)-free graphs in time O(n+m) [7], one has that, given a natural number p ¿ 4, then one can solve WMS for (P5 ; Kp )-free graphs and for (P5 ; Kp -e)-free graphs in time O(n2p−4 ) and O(n2p−6 ), respectively. Now let us show that a result similar to Theorem 4.2 can be stated for (P5 ; F (h) )-free graphs, with h ¿ 1, as well. In particular from one hand, we extend the method introduced in Theorem 4.2; from the other hand, to this aim we can only consider MS, leaving the weighted case to pseudo-polynomial considerations. First, let us report [17] below a method based on Theorem 2.3 to solve MS for P5 -free graphs. Algorithm Alpha (input: a P5 -free graph G = (V; E); output: a maximum stable set of G) Step 1: Set S = ∅; Step 2: check whether there exists a vertex v ∈ V \ S either non-adjacent to S or augmenting for S; Step 3: if the answer is no then S is maximal and, by Theorem 2.3, S is maximum; STOP; otherwise if v is non-adjacent to S, then S := S ∪ {v}; if v is augmenting for S, then S := (S \ NS (v)) ∪ {v} ∪ Tv {Tv denotes a stable set of H (v; S) such that |Tv | ¿ |NS (v)|}; go to Step 2. Let G = (V; E) be a (P5 ; F (h) )-free graph with n vertices and S be a maximal stable set of G. Suppose that one can solve MS for (P5 ; F)-free graphs in time O(nd ). The goal is to show that one can carry out Step 2 of the above algorithm in polynomial time. Let us say that a vertex v ∈ V is a trivial augmenting vertex
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for S if: • v is augmenting for S; • |NS (v)| ¡ h. Then one can check if an element v ∈ V is a trivial augmenting vertex for S in time O(nh ), by verifying if G[H (v; S)] contains a stable set of |NS (v)| elements. Suppose that G does not admit trivial augmenting vertices for S, and that there exists v ∈ V \ S augmenting for S (then, in particular, h 6 |NS (v)|). Thus G[H (v; S)] contains a stable set T with |NS (v)| 6 |T |. Since G is P5 -free, by Lemma 2.1 let us write T = {t1 ; : : : ; tr }, with NS (ti ) ⊆ NS (ti+1 ) for any index i. Since G does not admit trivial augmenting vertices for S, one has |NS (tk )| ¿ k for k = 1; : : : ; h. For any t ∈ H (v; S), let us write M (t) = {w ∈ H (v; S): NS (w) ⊇ NS (t), and |NS (w)| ¿ h}. Then T ⊆ {t1 ; : : : ; th−1 } ∪ (M (th−1 ) \ N ({t1 ; : : : ; th−1 })). Note that M (th−1 ) is F (h) -free, otherwise G would contain an induced F (h) . Now, since Step 2 of Algorithm Alpha considers all the vertices in V \ S, to check if S admits an augmenting vertex one has not to solve MS in H (w; S) for every w ∈ V \S. In fact, for every w ∈ V \ S it is su.cient to verify: (i) if w is a trivial augmenting vertex for S, and then (ii) if w is augmenting, by assuming that S does not admit trivial augmenting vertices. That can be formalized by the following procedure (whose input is any vertex v of V \ S) which can be executed in time O(nh+d ). Procedure Green (v) Input: a vertex v of V \ S. Output: a possible proof that v is augmenting. begin set S ∗ := ∅; if |NS (v)| ¡ h, then if H (v; S) contains a stable set Q of |NS (v)| elements then set S ∗ := Q, and stop (v is augmenting for S); if |NS (v)| ¿ h, then for every stable set U of h−1 elements of G[H (v; S)], i.e. U ={t1 ; : : : ; th−1 }, with NS (ti ) ⊆ NS (ti+1 ), and |NS (ti+1 )| ¿ h − 1, do begin solve MS in G[M (th−1 ) \ N ({t1 ; : : : ; th−1 })]; let S be a maximum stable set of G[M (th−1 )\N ({t1 ; : : : ; th−1 })]; if |S ∪ {t1 ; : : : ; th−1 }| ¿ |S ∗ |, then S ∗ := S; end; if |S ∗ | ¿ |NS (v)|, then v is augmenting for S; end. Note that, given an augmenting vertex v (for S), Procedure Green(v) could not recognize it as an augmenting vertex: that can happen whenever H (v; S) contains a trivial augmenting vertex.
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To summarize, in order to de?ne an e.cient method to solve MS in (P5 ; F (h) )-free graphs, one can rewrite Step 2 of Algorithm Alpha as follows: Step 2: for every v ∈ V \ S do if v is non-adjacent to any vertex of S then go to Step 3; else begin Procedure Green(v); if v is augmenting for S, then go to Step 3; end; By the structure of the new Algorithm Alpha, then one can formally state the following: Theorem 4.3. Let F be a graph. If one can solve MS for (P5 ; F)-free graphs in time O(nd ), then one can solve MS for (P5 ; F (h) )-free graphs in time O(nd+h+2 ). By the above theorem one can iteratively de?ne several good classes of P5 -free graphs. In particular, MS can be solved in time O(nr+s ) for (P5 ; Kr; s )-free graphs. 5. Concluding remark In this paper we introduce a general method (Theorem 4.3) for de?ning classes of P5 -free graphs for which the maximum stable set problem can be solved in polynomial time. In certain cases, such a result holds for the weighted case as well (Theorem 4.2). Actually, by the nature of the method exposed, large good classes of P5 -free graphs can be de?ned by considering at the same time: (i) Instances for which MS or WMS can be e.ciently solved, such as for example chair-free graphs, (P5 , diamond)-free graphs, mK2 -free graphs (a mK2 is a graph formed by m disjoint edges; for such graphs WMS can be solved in time O(n2m+1 ) by results in [1,20]), etc. (ii) The possibility of “composing” good instances by iterating the application of the method. For example, one obtains that WMS is solvable in polynomial time for (P5 , gem)free, (P5 , dart)-free and (P5 ; Dm; 1 )-free graphs, and MS is solvable in polynomial time for (P5 ; Dm; l )-free graphs, where a gem is obtained from a P4 by adding a dominating vertex, a dart is obtained from a diamond by adding a pendent edge at a degree 3 vertex in the diamond, and Dm; l is obtained from a mK2 ∪ lK2 by adding all edges between the two parts. Acknowledgements I would like to thank Prof. Andreas BrandstNadt and Prof. Van Bang Le for their valuable help and suggestions, Prof. Claudio Arbib for his encouragement and
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www.elsevier.com/locate/dam
Some results on maximum stable sets in certain P5-free graphs Ra)aele Mosca Dipartimento di Scienze, Universita degli Studi “G. D’Annunzio”, Viale Pindaro 42, Pescara 65127, Italy Received 19 March 1998; received in revised form 26 March 2002; accepted 15 April 2002
Abstract We introduce some large classes of P5 -free graphs for which the maximum stable set problem can be e.ciently solved. The solution algorithms are based on the search of augmenting vertices. Then a method for extending some of them to the weighted case is provided. ? 2003 Elsevier B.V. All rights reserved. Keywords: Stable set; Computational complexity
1. Introduction Let G = (V; E) be a graph. If G does not admit any induced subgraph isomorphic to a graph H , then G is H -free. A stable set in G is a subset of pairwise non-adjacent vertices of G. The maximum stable set problem (MS) is that of computing a stable set of G of maximum cardinality. If one associates a positive real number w(v) with each vertex v of G, and for any U ⊆ V denotes as weight w(U ) the sum of w(u) for u ∈ U , then the weighted maximum stable set problem (WMS) is that of computing a stable set S of G of maximum weight. MS is a well known NP-hard problem and remains di.cult for cubic and planar graphs. However, polynomial-time exact algorithms are known for several classes of graphs, which are usually called good classes. For example, MS can be e.ciently solved for perfect graphs and for partial k-trees with ?xed k. Then some good classes can be characterized in terms of forbidden induced subgraphs, such as claw-free graphs [16,19], 2K2 -free graphs [12] which
This work has been partially supported by the German Research Community, DFG, n. BR1446/4-1. E-mail address: [email protected] (R. Mosca).
0166-218X/$ - see front matter ? 2003 Elsevier B.V. All rights reserved. doi:10.1016/S0166-218X(03)00399-8
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are a special class of those P5 -free, and chair-free graphs [2] (a claw is formed by four vertices a; b; c; d and edges ab; ac; ad; a 2K2 is formed by four vertices a; b; c; d and edges ab; cd; a chair is formed by ?ve vertices a; b; c; d; e and edges ab; ac; ad; de). Actually, the results in [1,2] imply that the class of P5 -free is the unique minimal class de?ned by forbidding a single connected subgraph for which the computational complexity of MS is an open question. On the other hand, several results on P5 -free graphs are known in the literature. In [4,5] the authors give structural properties of such graphs in terms of dominating subgraphs, and provide a characterization of graphs with no induced paths of a given length. In [18] the author shows that P5 -free graphs are irredundance perfect. Then strong structural properties are known for special classes containing (P5 , claw)-free graphs [8], and for (P5 , diamond)-free graphs [3,7]. Structural properties based on graph decomposition are introduced in [13] for (P5 ; PJ 5 )-free graphs, in order to generalize a characterization of (2K2 ; C4 )-free graphs. In [6] the authors prove that the Strong Perfect Graph Conjecture holds true for any (P5 ; F)-free graph, where F is any connected ?ve-vertex graph containing no induced 2K2 . Then e.cient algorithms to solve MS have been designed for classes of P5 -free graphs, such as (P5 , house)-free [13,14], (P5 , banner)-free [10,15], (P5 , co-chair)-free and (P5 , co-banner)-free [11], (P5 , gem)-free graphs [9]. The contribution of this paper is mainly given by the introduction of large good classes of P5 -free graphs and of a general method to approach WMS for P5 -free graphs. 2. Notation and preliminary Let G = (V; E) be a graph and U; W be subsets of V . Let us write NU (W ) = {v ∈ U \ W : (v; w) ∈ E for some w ∈ W }. If W is a singleton, i.e., W = {w}, then we write NU (w) instead of NU ({w}); if U = V , then we write N (W ) instead of NV (W ). Furthermore, let us denote as G[W ] the subgraph of G induced by W . Let us say that W dominates U if U ⊆ W ∪ N (W ); if W = {w}, then we will say that w dominates U . Let S be a stable set of G, and v ∈ V \ S: let us write H (v; S) = {w ∈ V \ (S ∪ {v} ∪ N (v)): NS (w) ⊆ NS (v)}. A graph G weighted by a positive real function w will be denoted as G = (V; E; w). Then let us denote as Pk a path with k vertices, as Ck a cycle with k vertices, and as Km; n a bipartite graph G = (V1 ; V2 ; E) with |V1 | = m; |V2 | = n, and E = V1 × V2 . Let us report two results on P5 -free graphs which will be useful in the next sections. Lemma 2.1. Let G = (B1 ; B2 ; F) be a connected bipartite graph with |B1 | = k. Then G is P5 -free if and only if there exists a bijective function f : {1; : : : ; k} → B1 such that N (f(i)) ⊇ N (f(i + 1)), for i = 1; : : : ; k − 1; moreover N (f(1)) ≡ B2 . For the above lemma one can refer to di)erent known results (see e.g. [4,21]).
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Denition 2.2. Let G = (V; E) be a P5 -free graph with n vertices, and S be a maximal stable set of G. A vertex v of V \ S is augmenting for S if H (v; S) contains a stable set T with |T | ¿ |NS (v)|. Theorem 2.3 (Mosca [17]). Let G = (V; E) be a P5 -free graph with n vertices, and S be a maximal stable set of G. Then: (i) S is maximum if and only if there exists no augmenting vertex for S; (ii) if one can compute in time O(nh ) (where h is a natural number) a maximum stable set of G[H (v; Z)] for every maximal stable set Z of G and for every vertex v ∈ V \ Z, then one can compute a maximum stable set of G in time max{O(nh+2 ); O(n4 )}. By the proof of the above result, one realizes that, while the ?rst statement can be easily generalized for WMS, the second cannot be directly extended to the weighted case (apart from pseudo-polynomial considerations). 3. An augmenting approach for WMS in P5 -free graphs Let G = (V; E) be a graph, and S be a maximal stable set of G. Consider the following equivalence relation R(S) on V \ S: for any u; v ∈ V \ S, uR(S)v
if NS (u) = NS (v):
Let us denote as MR(S) the set of the equivalence classes obtained with respect to R(S). Then let us de?ne a binary relation 6 on MR(S) as follows: for any Mi ; Mj ∈ MR(S) Mi 6 Mj
if NS (u) ⊆ NS (v)
for any u ∈ Mi ; v ∈ Mj :
It is immediate to see that (MR(S) ; 6 ) is a poset. Let us denote as V (Mi ) the set of vertices of G belonging to any Mi ∈ MR(S) . Two elements M1 ; M2 of MR(S) are incomparable if neither M1 6 M2 nor M2 6 M1 . As usual, a chain of poset (MR(S) ; 6 ) is a subset C of MR(S) such that each pair of elements of C is in 6 . Two chains C1 ; C2 of (MR(S) ; 6 ) are incomparable if there exists no element of S adjacent to both an element of V (C1 ) and V (C2 ) (in particular each element of C1 is incomparable with each element of C2 ). To focus the properties of the above poset with respect to P5 -free graphs, let us point out the proposition below.
Proposition 3.1. Let G = (V; E) be a connected P5 -free graph, S be a maximal stable set of G, and T = ∅ be any stable set of G. Then there exist h ¿ 1 pairwise incom parable chains C1 ; : : : ; Ch of (MR(S) ; 6 ), such that T ⊆ S ∪ V (C1 ) ∪ · · · ∪ V (Ch ). Proof. If |T | = 1, then the assertion immediately follows. Suppose |T | ¿ 1. Then let M1 ; M2 be any two elements of MR(S) such that M1 and M2 are incomparable and V (M1 ) ∩ T = ∅; V (M2 ) ∩ T = ∅: let t1 ∈ V (M1 ) ∩ T , and t2 ∈ V (M2 ) ∩ T . Note that there exists no vertex s ∈ S adjacent to both vertices in V (M1 ) and in V (M2 ),
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otherwise, since by construction there exist two distinct vertices s1 ; s2 ∈ S such that (s1 ; t1 ); (s2 ; t2 ) ∈ E, and (s1 ; t2 ); (s2 ; t1 ) ∈ E, then s1 ; t1 ; s; t2 ; s2 would induce a P5 of G. In particular, there is no M3 ∈ MR(S) such that M3 6 M1 and M3 6 M2 . Furthermore, there M3 : in fact, let is no M3 ∈ MR(S) such that V (M3 )∩T = ∅, and both M1 6 M3 and M2 6 t3 ∈ V (M3 ) ∩ T ; then by construction there would exist two distinct vertices s1 ; s2 ∈ S such that (s1 ; t1 ); (s2 ; t2 ); (s1 ; t3 ); (s2 ; t3 ) ∈ E, that is, G would contain an induced P5 . Then the proposition follows. Now let us prove a theorem which extends Theorem 2.3 to the weighted case. That is possible by de?ning a method for checking just once if a vertex of G is augmenting or not. Also, one obtains an improvement of the computational complexity, with respect to that exposed in Theorem 2.3. Theorem 3.2. Let G = (V; E; w) be a P5 -free graph with n vertices, and d be a natural number. If one can solve WMS in H (v; S) in time O(nd ) for every maximal stable set S of G and for every v ∈ V \ S, then one can solve WMS in G in time max{O(nd+1 ); O(n3 )}. Proof. Let S be a maximal stable set of G. It is well known that, based on poset (MR(S) ; 6 ), one can de?ne a (topological) total order ¡ on MR(S) , such that, for any M1 ; M2 ∈ MR(S) , if M1 6 M2 then M1 ¡ M2 . Then, let MR(S) = {M1 ; M2 ; : : : ; Mr }, with Mi ¡ Mi+1 for any index i, and let us write M (k) = M1 ∪ M2 ∪ · · · ∪ Mk . The method is based on the idea of iteratively solving WMS in G[S ∪ V (M (k) )]. Let us denote by S0∗ the set S; then, for any index k, denote by Sk∗ a maximum weighted stable of G[S ∪ V (M (k) )]. Let us prove the following fact. ∗ Fact 1. For any index k, if w(Sk∗ ) ¿ w(Sk−1 ) then there exists v ∈ Mk such that ∗ ∗ ). G[{v} ∪ H (v; Sk−1 )] contains a stable set of weight w(Sk∗ \ Sk−1
Let k = 1. One clearly has w(S1∗ ) ¿ w(S0∗ ) = w(S). If w(S1∗ ) ¿ w(S0∗ ), then, since NS (x) = NS (y) for any x; y ∈ M1 , one has that G[S1∗ ∪ S0∗ ] has exactly one non-empty component H , and then: H ∩ S1∗ ⊆ M1 and each vertex v ∈ M1 dominates H ∩ S0∗ . Then Fact 1 follows for k = 1. ∗ ∗ Let k ¿ 1. Suppose w(Sk∗ ) ¿ w(Sk−1 ): then G[Sk∗ ∪ Sk−1 ] has non-empty components H1 ; H2 ; : : : ; Hp . Recalling that Mi ¡ Mk for i ¡ k, let Z be the set of vertices of S which are adjacent to all those in V (Mk ) and non-adjacent to all those in V (Mi ) ∗ for i ¡ k. One clearly has Z ⊆ Sk−1 . Furthermore, since Sk∗ ∩ V (Mk ) = ∅, there ∗ ∗ ], say H1 , such that Z ⊆ H1 ∩ Sk−1 exists one non-empty component of G[Sk∗ ∪ Sk−1 ∗ ∗ ∗ (k) and (V (Mk ) ∩ Sk ) ⊆ H1 ∩ Sk . Hence, since Sk ⊆ S ∪ V (M ), one has that Hi ⊆ ∗ S ∪ V (M (k−1) ), for i ¿ 2. But, by de?nition of Sk−1 , that implies that one can suppose ∗ w.l.o.g. that Hi is empty for any i ¿ 2. Thus G[Sk∗ ∪ Sk−1 ] has exactly one non-empty component, that is H1 ; then, since in H1 the vertices of Sk∗ ∩V (Mk ) are the only vertices adjacent to Z, Lemma 2.1 implies that there exists a vertex v ∈ V (Mk ) dominating
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∗ ∗ ∗ H1 ∩ Sk−1 , and hence G[{v} ∪ H (v; Sk−1 )] contains a stable set of weight w(Sk∗ \ Sk−1 ). Then Fact 1 is proved. Fact 1 implies that one can compute Sk∗ , whenever for every v ∈ Mk one can compute ∗ )]. a maximum weighted stable set S(v) of G[{v} ∪ H (v; Sk−1 ∗ ∗ (v)) + w(S ∗ Denote as Hk = max{w(S(v)) − w(NSk−1 k−1 \ NSk−1 (v)): v ∈ Mk }. ∗ ∗ ∗ Then one has w(Sk ) = max{w(Sk−1 ); Hk }. In particular: if w(Sk∗ ) ¿ w(Sk−1 ), then ∗ ∗ ∗ Sk = S(u) ∪ Sk−1 \ NSk−1 (u), where u ∈ Mk is the vertex individuated by Hk ; otherwise, ∗ one can let Sk∗ = Sk−1 . Therefore, to compute a maximum weighted stable set of G one can apply the following algorithm, whose output Sr∗ is an optimum solution.
Algorithm 1 Step 1: Compute a maximal stable set S of G, and set S0∗ := S; Step 2: for k := 1; : : : ; r do begin compute Sk∗ ; end. ) and (MR(S) ; ¡ ) About the computational complexity of Algorithm 1: (MR(S) ; 6 3 can be detected in time O(n ); thus Step 1 can be executed in time O(n2 ), while Step 2 in time O(nd+1 ). Then the theorem follows.
4. Large good classes of P5 -free graphs for WMS and MS In this section we show that the primitive approach used in Theorem 2.3 can be extended in order to detect several good classes of P5 -free graphs. In particular, by Theorem 3.2 for some of them one can de?ne an e.cient algorithm for the weighted case as well. Notation 4.1. Let F be a graph. Let us denote as F (1) the graph obtained from F by adding two new vertices v; s, such that s dominates F, while v is adjacent only to s. In general, let F (h) be the graph obtained from F by adding h + 1 new vertices v; s1 ; : : : ; sh such that {s1 ; : : : ; sh } induce a stable set, si dominates F for i = 1; : : : ; h, while v is adjacent only to {s1 ; : : : ; sh } (see for example the graphs F (1) = K3(1) and F (2) = K3(2) in Fig. 1). s2
v
s
v
(a)
(b) (1)
s1 (2)
Fig. 1. (a) The graph K3 ; (b) the graph K3 .
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Theorem 4.2. Let F be a graph. If one can solve WMS for (P5 ; F)-free graphs in time O(nd ), then one can solve WMS for (P5 ; F (1) )-free graphs in time O(nd+2 ). Proof. Let G be a (P5 ; F (1) )-free graph with n vertices. Consider the problem of ?nding a maximum weighted stable, say S , of G[H (v; S)]. By de?nition of H (v; S) one has that NS (v) is a maximal stable set of G[NS (v) ∪ H (v; S)]; in particular, NS (v) and S induce a connected bipartite subgraph of G. Thus, by Lemma 2.1 there exists s ∈ NS (v) such that S ⊆ N (s). It follows that S can be found by solving WMS in G[N (s) ∩ H (v; S)] for all s ∈ NS (v). But G[N (s) ∩ H (v; S)] is F-free, otherwise vertices v; s and subgraph F would induce a F (1) of G (contradiction). Thus one can compute S in time O(nd+1 ), and therefore, by Theorem 3.2, ?nd a maximum weighted stable set of G in time O(nd+2 ). As WMS is easily solvable in classes of graphs with very small forbidden induced subgraphs, Theorem 4.2 can be used as a tool to de?ne other good classes of P5 -free graphs with respect to the weighted case. In particular, in view of a repeated application, the computational complexity of the method for classes of graphs with large forbidden subgraphs is a)ected by the complexity of WMS for classes with smaller forbidden subgraphs. For example, as one can compute a maximum weighed stable set in (P5 , diamond)-free graphs in time O(n+m) [7], one has that, given a natural number p ¿ 4, then one can solve WMS for (P5 ; Kp )-free graphs and for (P5 ; Kp -e)-free graphs in time O(n2p−4 ) and O(n2p−6 ), respectively. Now let us show that a result similar to Theorem 4.2 can be stated for (P5 ; F (h) )-free graphs, with h ¿ 1, as well. In particular from one hand, we extend the method introduced in Theorem 4.2; from the other hand, to this aim we can only consider MS, leaving the weighted case to pseudo-polynomial considerations. First, let us report [17] below a method based on Theorem 2.3 to solve MS for P5 -free graphs. Algorithm Alpha (input: a P5 -free graph G = (V; E); output: a maximum stable set of G) Step 1: Set S = ∅; Step 2: check whether there exists a vertex v ∈ V \ S either non-adjacent to S or augmenting for S; Step 3: if the answer is no then S is maximal and, by Theorem 2.3, S is maximum; STOP; otherwise if v is non-adjacent to S, then S := S ∪ {v}; if v is augmenting for S, then S := (S \ NS (v)) ∪ {v} ∪ Tv {Tv denotes a stable set of H (v; S) such that |Tv | ¿ |NS (v)|}; go to Step 2. Let G = (V; E) be a (P5 ; F (h) )-free graph with n vertices and S be a maximal stable set of G. Suppose that one can solve MS for (P5 ; F)-free graphs in time O(nd ). The goal is to show that one can carry out Step 2 of the above algorithm in polynomial time. Let us say that a vertex v ∈ V is a trivial augmenting vertex
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for S if: • v is augmenting for S; • |NS (v)| ¡ h. Then one can check if an element v ∈ V is a trivial augmenting vertex for S in time O(nh ), by verifying if G[H (v; S)] contains a stable set of |NS (v)| elements. Suppose that G does not admit trivial augmenting vertices for S, and that there exists v ∈ V \ S augmenting for S (then, in particular, h 6 |NS (v)|). Thus G[H (v; S)] contains a stable set T with |NS (v)| 6 |T |. Since G is P5 -free, by Lemma 2.1 let us write T = {t1 ; : : : ; tr }, with NS (ti ) ⊆ NS (ti+1 ) for any index i. Since G does not admit trivial augmenting vertices for S, one has |NS (tk )| ¿ k for k = 1; : : : ; h. For any t ∈ H (v; S), let us write M (t) = {w ∈ H (v; S): NS (w) ⊇ NS (t), and |NS (w)| ¿ h}. Then T ⊆ {t1 ; : : : ; th−1 } ∪ (M (th−1 ) \ N ({t1 ; : : : ; th−1 })). Note that M (th−1 ) is F (h) -free, otherwise G would contain an induced F (h) . Now, since Step 2 of Algorithm Alpha considers all the vertices in V \ S, to check if S admits an augmenting vertex one has not to solve MS in H (w; S) for every w ∈ V \S. In fact, for every w ∈ V \ S it is su.cient to verify: (i) if w is a trivial augmenting vertex for S, and then (ii) if w is augmenting, by assuming that S does not admit trivial augmenting vertices. That can be formalized by the following procedure (whose input is any vertex v of V \ S) which can be executed in time O(nh+d ). Procedure Green (v) Input: a vertex v of V \ S. Output: a possible proof that v is augmenting. begin set S ∗ := ∅; if |NS (v)| ¡ h, then if H (v; S) contains a stable set Q of |NS (v)| elements then set S ∗ := Q, and stop (v is augmenting for S); if |NS (v)| ¿ h, then for every stable set U of h−1 elements of G[H (v; S)], i.e. U ={t1 ; : : : ; th−1 }, with NS (ti ) ⊆ NS (ti+1 ), and |NS (ti+1 )| ¿ h − 1, do begin solve MS in G[M (th−1 ) \ N ({t1 ; : : : ; th−1 })]; let S be a maximum stable set of G[M (th−1 )\N ({t1 ; : : : ; th−1 })]; if |S ∪ {t1 ; : : : ; th−1 }| ¿ |S ∗ |, then S ∗ := S; end; if |S ∗ | ¿ |NS (v)|, then v is augmenting for S; end. Note that, given an augmenting vertex v (for S), Procedure Green(v) could not recognize it as an augmenting vertex: that can happen whenever H (v; S) contains a trivial augmenting vertex.
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To summarize, in order to de?ne an e.cient method to solve MS in (P5 ; F (h) )-free graphs, one can rewrite Step 2 of Algorithm Alpha as follows: Step 2: for every v ∈ V \ S do if v is non-adjacent to any vertex of S then go to Step 3; else begin Procedure Green(v); if v is augmenting for S, then go to Step 3; end; By the structure of the new Algorithm Alpha, then one can formally state the following: Theorem 4.3. Let F be a graph. If one can solve MS for (P5 ; F)-free graphs in time O(nd ), then one can solve MS for (P5 ; F (h) )-free graphs in time O(nd+h+2 ). By the above theorem one can iteratively de?ne several good classes of P5 -free graphs. In particular, MS can be solved in time O(nr+s ) for (P5 ; Kr; s )-free graphs. 5. Concluding remark In this paper we introduce a general method (Theorem 4.3) for de?ning classes of P5 -free graphs for which the maximum stable set problem can be solved in polynomial time. In certain cases, such a result holds for the weighted case as well (Theorem 4.2). Actually, by the nature of the method exposed, large good classes of P5 -free graphs can be de?ned by considering at the same time: (i) Instances for which MS or WMS can be e.ciently solved, such as for example chair-free graphs, (P5 , diamond)-free graphs, mK2 -free graphs (a mK2 is a graph formed by m disjoint edges; for such graphs WMS can be solved in time O(n2m+1 ) by results in [1,20]), etc. (ii) The possibility of “composing” good instances by iterating the application of the method. For example, one obtains that WMS is solvable in polynomial time for (P5 , gem)free, (P5 , dart)-free and (P5 ; Dm; 1 )-free graphs, and MS is solvable in polynomial time for (P5 ; Dm; l )-free graphs, where a gem is obtained from a P4 by adding a dominating vertex, a dart is obtained from a diamond by adding a pendent edge at a degree 3 vertex in the diamond, and Dm; l is obtained from a mK2 ∪ lK2 by adding all edges between the two parts. Acknowledgements I would like to thank Prof. Andreas BrandstNadt and Prof. Van Bang Le for their valuable help and suggestions, Prof. Claudio Arbib for his encouragement and
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