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Some observations on maximum weight stable sets in certain P5-free graphs
European Journal of Operational Research 184 (2008) 849–859 www.elsevier.com/locate/ejor
Discrete Optimization
Some observations on maximum weight stable sets in certain P5-free graphs Raffaele Mosca
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Dipartimento di Scienze, Universita´ degli Studi ‘‘G. d’Annunzio’’, Pescara 65127, Italy Received 6 June 2006; accepted 1 December 2006 Available online 20 December 2006
Abstract The maximum weight stable set problem (MWS) is the weighted version of the maximum stable set problem (MS), which is NP-hard. The class of P5-free graphs – i.e., graphs with no induced path of five vertices – is the unique minimal class, defined by forbidding a single connected subgraph, for which the computational complexity of MS is an open question. At the same time, it is known that MS can be efficiently solved for ðP 5 ; F Þ-free graphs, where F is any graph of five vertices different to a C5. In this paper we introduce some observations on P5-free graphs, and apply them to introduce certain subclasses of such graphs for which one can efficiently solve MWS. That extends or improves some known results, and implies – together with other known results – that MWS can be efficiently solved for ðP 5 ; F Þ-free graphs where F is any graph of five vertices different to a C5. Ó 2006 Elsevier B.V. All rights reserved. Keywords: Maximum weight stable set problem; P5-free graphs; Polynomial time algorithm
1. Introduction A stable set of a graph G is a subset of pairwise nonadjacent vertices of G. The maximum weight stable set (MWS) problem is the following: Given a graph G ¼ ðV ; EÞ and a weight function w on V, determine a stable set of G having maximum weight (where the weight of a stable set S is given by the sum of the weights of each s 2 S). Let aw ðGÞ denote the value of a maximum weight stable set of G. The MWS problem is called MS problem if all vertices v have the same weight wðvÞ ¼ 1.
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It is well known that MS is NP-hard [26]. In particular, it remains difficult for triangle-free, cubic or planar graphs (see e.g. [36,21,22]). Then, the results in [1,2] imply that the class of P5-free graphs is the unique minimal class, defined by forbidding a single connected subgraph, for which the computational complexity of MS is an open question (see also [4]). On the other hand, MS (or MWS) can be efficiently solved for several graph classes. In particular, polynomial algorithms are known to solve MWS for P4-free graphs [14,15], 2K2-free graphs [18,19], and claw-free graphs [35,31,37]. Let us focus on the class of P5-free graphs. General structure properties of such graphs can be found in [5,6,17]. In particular, it is known that MS can be efficiently solved for ðP 5 ; F Þ-free graphs,
0377-2217/$ - see front matter Ó 2006 Elsevier B.V. All rights reserved. doi:10.1016/j.ejor.2006.12.011
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where F is any graph of five vertices different to a C5. In this paper we introduce some observations on P5-free graphs, and apply them to introduce certain subclasses of such graphs for which one can efficiently solve MWS. Two of such subclasses are: ðP 5 ; S k Þ-free graphs, where a S k consists of a stable set of k þ 1 vertices, a P 1 þ P 2 , and a join between them; ðP 5 ; P 3 Þ-free graphs, where a P 3 consists of a P 1 þ P 3 , a P 1 þ P 2 , and a join between them. That extends some known results, according to the fact that S k and P 3 contains induced subgraphs of five vertices (in particular, since S k contains a K 2;3 , one obtains that one can efficiently solve MWS for ðP 5 ; F Þ-free graphs, where F is any graph of five vertices different to a C5). Also, we point out that by a similar approach one obtains a Oðn4 mÞ time algorithm to solve MWS for ðP 5 ; F 1 Þ-free graphs where F1 is a graph of six vertices a; b; c; d; e; f and edges ab; ac; ad; bc; bd; ce; de; df : that improves a recent result (see among other results in [11]) stating that one can solve MWS for ðP 5 ; F 1 Þ-free graphs in Oðn7 mÞ time.
2. Preliminaries For any missing notation or reference, let us refer to [10]. Throughout this paper, let G ¼ ðV ; EÞ be a finite undirected graph without self-loops and multiple edges and let jV j ¼ n; jEj ¼ m. The complement of G is the graph, denoted as co-G, having the same vertices as G and such that two vertices are adjacent in co-G if and only if they are nonadjacent in G. Let U ; W be two subsets of V. Let N U ðW Þ ¼ fu 2 U n W juv 2 Eg be the set of neighbors of W in U. If U ¼ V , then we write N ðW Þ instead of N V ðW Þ. If W ¼ fvg, then we write N U ðvÞ instead of N U ðfvgÞ. Also, we write N U ½v ¼ N U ðvÞ [ fvg. Let us say that U has a join (a co-join, respectively) to W, if each vertex in U is adjacent (is nonadjacent) to each vertex in W. If U ¼ fug and fug has a join to W, then let us also say that u dominates W. Then let us say that a vertex v 2 V n U contacts U if v is adjacent to some vertex of U. Throughout this paper, all subgraphs are understood as induced subgraphs. Let G½U denote the subgraph of G induced by the vertex subset U. For any graph F, G is F-free if G contains no induced subgraph isomorphic to F. A component of G is a nonempty vertex-set W V such that
G½W induces a maximal connected subgraph of G. A component is trivial if it is a singleton. A P k has vertices v1 ; v2 ; . . . ; vk and edges vj vjþ1 for 1 6 j < k. A Ck has vertices v1 ; v2 ; . . . ; vk and edges vj vjþ1 for 1 6 j 6 k (index arithmetic modulo k). A 2K2 is a graph of four vertices a; b; c; d and edges ab; cd. A claw is a graph of four vertices and edges ab; ac; ad. A K p;q is a complete bipartite graph whose sides have respectively p and q vertices. A Sk is a graph of k vertices and no edges. Given two graphs G1 and G2, let G1 þ G2 denote the graph consisting of G1, G2, and a co-join between them. A clique of G is a set of pairwise adjacent vertices of G. One can verify that G is P3-free if and only if each component of G is a clique. A vertex z 2 V distinguishes vertices x; y 2 V if zx 2 E and zy 62 E or zy 2 E and zx 62 E. Let us also say that a vertex z distinguishes a vertex-set U V , z 62 U , if z distinguishes two vertices in U; in particular, z well distinguishes U, if z distinguishes two nonadjacent vertices in U. A vertex-set M V is a module if no vertex from V nM distinguishes M. Graph G is prime if it contains only trivial modules, i.e. ;, V and one-elementary vertex sets. A nontrivial module is called a homogeneous set. It is well known that in a connected graph G with connected complement co-G, the maximal homogeneous sets are pairwise disjoint which means that every vertex is contained in at most one homogeneous set. The existence and uniqueness of the modular decomposition tree is based on this property, and linear time algorithms were designed to determine this tree – see e.g. [32,30]. The tree contains the vertices of the graph as its leaves, and the internal nodes are of three types: they represent a join or co-join operation, or a prime subgraph. Observation 1. Let G ¼ ðV ; EÞ be a graph, and let W V induce a nontrivial connected subgraph of G. If a vertex v 2 V n W distinguishes W, then v distinguishes two adjacent elements of W. Observation 2. Let G ¼ ðV ; EÞ be a prime graph, and let W V induce a nontrivial connected subgraph of G. If W is not a clique, then there exists a vertex v 2 V n W which well distinguishes W. That comes by considering the components of co-G. The anti-neighborhood (or non-neighborhood) of U V is AðU Þ ¼ fv 2 V n U : v is nonadjacent to any element of U g. If U ¼ fu1 ; . . . ; uh g, then we write Aðu1 ; . . . ; uh Þ instead of Aðfu1 ; . . . ; uh gÞ. Obviously, the MWS problem on a graph G with vertex
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weight function w can be reduced to the same problem on anti-neighborhoods of vertices in the following way: aw ðGÞ ¼ maxfwðvÞ þ aw ðG½AðvÞÞjv 2 V g: For different graph classes including that of P5-free (see e.g. [29]), that holds by considering the antineighborhoods of edges as well. Theorem 1 [29]. Let G be a connected P5-free graph. If one can solve MWS for every subgraph induced by the anti-neighborhood of each edge of G in Oðnh Þ time, then one can solve MWS for G in maxfOðn3þh Þ; Oðn5 Þg time. Let us conclude by shortly listing some results concerning MWS for ðP 5 ; F Þ-free graphs, where F is a graph of five vertices different to a C5 (a drawn list of five-vertex graphs can be found e.g. in [25]). If F contains an isolated vertex or an isolated edge, then MWS can be efficiently solved for ðP 5 ; F Þ-free graphs by the observation and the result above (let us recall that MWS is efficiently solvable for ðP 5 ; C 3 Þ-free graphs and for ðP 5 ; C 4 Þ-free graphs – see e.g. [13]). If F admits a dominating vertex, then MWS can be efficiently solved for ðP 5 ; F Þ-free graphs by combining an augmenting approach with the previous facts (see e.g. [34]); in particular, concerning (P5, gem)-free graphs, where a gem has edges ab; ac; ad; ae; bc; cd; de, see e.g. [7] for a very efficient algorithm. If F is a banner (a banner has edges ab; bc; bd; ce; de), then MWS can be efficiently solved for ðP 5 ; F Þ-free graphs by combining different approaches [9] (see also [12,27] for other approaches concerning MS). If F is a bull (a bull has edges ab; ac; bc; bd; ce), then MWS can be efficiently solved for ðP 5 ; F Þ-free graphs as a subcase of a general decomposition result introduced in [16]. If F is either a co-chair (a co-chair has edges ab; bc; bd; cd; ce; de) or a co-banner (a co-banner has edges ab; bc; cd; ce; de), then MWS can be efficiently solved for ðP 5 ; F Þ-free graphs by applying some of the structure results introduced in [13]. If F is a fork (a fork – or chair – has edges ab; bc; cd; ce), then MWS can be (very) efficiently solved for ðP 5 ; F Þ-free graphs by combining the results in [8,28]. If F is a house (a house – or co-P5 – has edges ab; ac; bc; bd; ce; de), then MWS can be efficiently
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solved for ðP 5 ; F Þ-free graphs by a decomposition approach introduced in [20,24]. If F is a co-ðP 2 þ P 3 Þ, i.e., F has edges ab; ac; ad; bc; be; ce; de, then MWS can be efficiently solved for ðP 5 ; F Þ-free graphs by a clique separator decomposition approach introduced in [11]. It remains the case in which F is a K 2;3 . In this case MS can be efficiently solved for ðP 5 ; F Þ-free graphs by an augmenting approach (see e.g. [23,33,34]). 3. Some observations on P5-free graphs Throughout this section assume that G ¼ ðV ; EÞ is a prime P5-free graph. For any vertex-set X V , let us write J ðX Þ ¼ fv 2 V n X : fvg has a join to X g. Lemma 1. Let X V with G½X connected. Let Q ¼ fQ1 ; . . . ; Qm g be a family of nontrivial components of G½AðX Þ. Let D ¼ fv 2 V : v distinguishes some element of Q g. Then: (i) D J ðX Þ; (ii) D* is partitioned into fD1 ; . . . ; Dm g, where each d i 2 Di distinguishes Qi and does not distinguish Qj for every j 6¼ i; (iii) Di has a join to Dj, for 1 6 i < j 6 m; (iv) for any ðd 1 ; . . . ; d m Þ 2 D1 Dm , there exists di dominating Qj for every j 6¼ i. Proof. About (i). Let d 2 D and let Q be a nontrivial component of G½AðX Þ distinguished by d. Then by Observation 1 d 62 N ðX Þ n J ðX Þ, otherwise two adjacent vertices of X, d and two adjacent vertices of Q induce a P5. Then, since G is connected, d 2 J ðX Þ. About (ii). Let d 2 D . By Observation 1, d cannot distinguish two elements of Q otherwise a P5 arises. About (iii). Assume to the contrary there exist d i 2 Di and d j 2 Dj which are nonadjacent. Let x 2 X . By Observation 1: let xi ; y i 2 Qi such that di is adjacent to xi and nonadjacent to yi; let xj ; y j 2 Qj such that dj is adjacent to xj and nonadjacent to yj. To avoid that y j ; xj ; d j ; x; d i induce a P5, di is adjacent either to xj or to yj, that is, by statement (ii) di dominates Qj. By symmetry, dj dominates Qi . Then y j ; d i ; x; d j ; y i induce a P5 (contradiction). About (iv). The proof is by induction on m. For m ¼ 2, the assertion follows by Observation 1, statement (iii) and the fact that G is P5-free. Assume
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the assertion holds for m 1, and let us prove that it holds for m. Let us consider an element ðd 2 ; . . . ; d m Þ 2 D2 Dm . By the inductive assumption one has without loss of generality that d2 dominates Qj for every j > 2. Let d 1 2 D1 . If d2 dominates D1, then d2 is the required element. Then, by statement (ii), assume that fd 2 g has a co-join to Q1. Notice that then d1 dominates Q2 (otherwise a P5 arises). If d1 dominates Qj for every j > 2, then d1 is the required element. Otherwise, by the inductive assumption on ðd 1 ; d 3 ; . . . ; d m Þ, one has without loss of generality that d3 dominates Q1 and Qj for every j > 3; also, d3 dominates Q2 , otherwise two adjacent vertices in Q2 ; d 2 ; d 3 and a vertex in Q1 induce a P5. Then d3 is the required element. h Throughout the next lemmas of this section assume that: X V with G½X connected; Q is a non-clique component of G½AðX Þ; D is the set of vertices of G which distinguish Q. Lemma 2. Let d 2 D well distinguish Q. Then d dominates ðN ðX Þ n J ðX ÞÞ [ fv 2 J ðX Þ : fvg has a co-join to Qg. Proof. Let x 2 X . Let z 2 N ðX Þ n J ðX Þ. Assume to the contrary that d is nonadjacent to z. By (i) of Lemma 1, z does not distinguish Q. Then: if fzg has a co-join to Q, then by Observation 1 d; x; z and two adjacent vertices of Q induce a P5 (contradiction); if fzg has a join to Q, then, since d well distinguishes Q, d; x; z and two nonadjacent vertices of Q induce a P5 (contradiction). Let z 2 fv 2 J ðX Þ : fvg has a co-join to Qg. Then d is adjacent to z, otherwise by Observation 1 d; x; z and two adjacent vertices of Q induce a P5. h Lemma 3. Let d 2 D well distinguish Q. Then N Q ðdÞ \ N Q ðcÞ 6¼ ; for every c 2 D n N ½d. Proof. Let c 2 D n N ½d. Assume to the contrary N Q ðdÞ \ N Q ðcÞ ¼ ;. Then, since G is P5-free and G½Q is connected: fN Q ðdÞ; N Q ðcÞg is a partition of Q, and N Q ðdÞ has a join to N Q ðcÞ. But that contradicts the fact that d well distinguishes Q. h Lemma 4. Let d 2 D well distinguish Q, such that N Q ðdÞ is maximal under inclusion in fN Q ðcÞ : c well distinguish Qg. Then:
(i) no vertex of D n N ½d distinguishes any component of G½Q n N ½d; (ii) G½Q n N ½d is P3-free. Proof. Let x 2 X . About (i). Assume to the contrary a vertex c 2 D n N ½d distinguishes a component K of G½Q n N ½d. Then, by Observation 1, d; x; c and two adjacent vertices of K induce a P5 (contradiction). About (ii). Let S ¼ Q n N ðdÞ and T ¼ Q \ N ðdÞ. Notice that since G is P5-free all the elements from a component K of G½S have the same neighborhood in T, say T ðKÞ. Then by Observation 2 to prove the assertion it is sufficient to show that no vertex in D well distinguishes Q n N ðdÞ. Assume to the contrary there exists c 2 D and two nonadjacent vertices a; b in a nontrivial component K of G½S such that c is adjacent to a and is nonadjacent to b (then c 6¼ d). Then c dominates T ðKÞ otherwise, say t 2 T ðKÞ, vertices x; c; a; t; b induce a P5. If T ðKÞ ¼ T , then one gets a contradiction to the definition of d (in fact one would have that c well distinguishes Q, with N Q ðdÞ N Q ðcÞ). Then assume that T ðKÞ T . By Observation 1 there exist two adjacent vertices k 1 ; k 2 2 K such that c is adjacent to k1 and is nonadjacent to k2. By statement (i), c is adjacent to d. Then c dominates T n T ðKÞ otherwise, say t 2 T n T ðKÞ, vertices k 2 ; k 1 ; c; d; t induce a P5. But then again one gets a contradiction to the definition of d. h 4. An approach for MWS in P5-free graphs Let us first introduce two definitions. Definition 1. Let F be any graph. Let us say that a P5-free graph G ¼ ðV ; EÞ is F-depending if there exists a partition of V, say fV 1 ; V 2 ; V 3 g, such that: (i) G½V 1 and G½V 2 are respectively F-free; (ii) G½V 3 is P3-free, and no element of V 1 [ V 2 distinguishes any component of G½V 3 . Definition 2. Let F be any graph. Let us denote as F* the graph consisting of P 1 þ F , P 1 þ P 2 , and a join between them. Let us denote as F+ the graph consisting of P 1 þ F , P 1 þ P 1 , and a join between them. Let F be any graph. Throughout this section assume that G is a prime ðP 5 ; F Þ-free graph.
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Lemma 5. Let ðx; yÞ be an edge of G, Q be a nonclique component of G½Aðx; yÞ, D be the set of vertices of G which distinguish Q. Let d 2 D well distinguish Q. Then D n N ½d admits a partition, say fX ; Y g, such that there exist qX 2 N Q ðdÞ dominating X, and qY 2 N Q ðdÞ dominating Y. Proof. Let S ¼ Q n N ðdÞ and T ¼ Q \ N ðdÞ. Let s 2 S such that there exists t 2 T nonadjacent to s* (according to the fact that d well distinguishes Q). Let X ¼ fc 2 D n N ½d : c is adjacent to s g. Then t* is adjacent to each vertex a of X, otherwise s ; a; y; d; t induce a P5. Let Y ¼ fc 2 D n N ½d : c is nonadjacent to s g. Let t 2 T \ N ðs Þ (such a vertex does exist since G is P5-free and G½Q is connected). Then t is adjacent to each vertex a of Y, otherwise a; y; d; t; s induce a P5. Since fX ; Y g is a partition of D n N ½d, the lemma follows. h Theorem 2. One of the following cases occurs: (i) for every edge ðx; yÞ of G, G½Aðx; yÞ is P3-free; (ii) there exists a vertex v of G such that each component of G½AðvÞ induces either a P3-free or a F-depending graph. Proof. Assume that statement (i) does not occur, and let us verify that then statement (ii) occurs. Then there exists an edge ðx; yÞ of G such that G½Aðx; yÞ is not P3-free. Let Q ¼ fQ1 ; . . . ; Qm g be the family of non-clique components of G½Aðx; yÞ. Let D ¼ fv 2 V : v distinguishes some element of Q g. According to Lemma 1, D is partitioned into fD1 ; . . . ; Dm g, where each vertex d i 2 Di distinguishes Qi and does not distinguish Qj for every j 6¼ i. Let ðd 1 ; . . . ; d m Þ 2 D1 ; . . . ; Dm . By Observation 2 one can assume without loss of generality that for i ¼ 1; . . . ; m: di well distinguishes Qi ; N Q ðd i Þ is maximal under inclusion in fN Qi ðcÞ : c well distinguishes Qi g. By Lemma 1, one can assume without loss of generality that d1 dominates Qj for every j > 1. Furthermore, by Lemma 2, one has that V n N ½d 1 ¼ ðD1 [ Q1 [ H [ KÞ n N ½d 1 , where H is the vertex-set of the clique-components of G½Aðx; yÞ, and K is the set of vertices in N ðxÞ\ N ðyÞ dominating Q1. By Lemma 5, D1 n N ½d 1 admits a partition, say fX ; Y g, such that there exist qX 2 N Q ðdÞ dominating X, and qY 2 N Q ðdÞ dominating Y. By definition of K and by (i) of Lemma 1, one obtains that
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G½ðK [ X Þ n N ½d 1 and G½ðK [ Y Þ n N ½d 1 are respectively F-free, otherwise a F arises involving x; y; d 1 and respectively either qX or qY . By (ii) of Lemma 4 and by construction, ðQ1 [ H Þ n N ½d 1 induces a P3-free graph. Furthermore, by (ii) of Lemma 1 and (i) of Lemma 4, no element of D1 n N ½d 1 distinguishes any component of G½ðQ1 [ H Þn N ½d 1 . By definition of K, no element of K n N ½d 1 distinguishes any component of G½Q1 n N ½d 1 . Also, no element k of K n N ½d 1 distinguishes any component of G½H n N ½d 1 : in fact otherwise, d 1 ; x; k and two elements of H n N ½d 1 would induce a P5. Then V n N ½d 1 is a F-depending graph with partition fðK [ X Þ n N ½d 1 ; Y n N ½d 1 ; ðQ1 [ H Þ n N ½d 1 g. Then statement (ii) occurs with v ¼ d 1 . h In what follows, let us prove that if one can efficiently solve MWS for F-depending graphs, then one can efficiently solve MWS for ðP 5 ; F Þfree graphs. To this end, let us assume that one can solve MWS for every F-depending graph in Oðnt Þ time for some natural t. Then let us introduce a recursive algorithm to solve MWS for ðP 5 ; F Þ-free graphs. The following notation will be used. Let G ¼ ðV ; EÞ be a graph. Then, let G0 ¼ ðV 0 ; E0 Þ denote the characteristic graph of G, that is: each vertex of G0 corresponds to a maximal homogeneous set of G, in order that two vertices of G0 are adjacent (nonadjacent) if the corresponding maximal homogeneous sets of G have a join in G (a co-join in G). In particular for each element h 2 V 0 , let Gh denote the maximal homogeneous set of G corresponding to h. Finally, let us say that G is gray if each of its components induces either a P3-free or a F-depending graph. Then by the above assumption one can solve MWS for every gray graph in maxfOðmÞ; Oðnt Þg time. Algorithm Green(G,w) Input: a ðP 5 ; F Þ-free graph G ¼ ðV ; EÞ with a weight function w on V. Output: the value a of a maximum weight stable set of G. begin a :¼ 0 while G contains a vertex v such that G½AðvÞ is gray do begin a :¼ maxfa; aw ðG½fvg [ AðvÞÞg (G½fvg [ AðvÞ is gray); G :¼ G½V n v; end;
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if G is prime then a :¼ maxfa; aw ðGÞg (G enjoys (i) of Theorem 2; then cf. Theorem 1); if G is not prime then begin compute the characteristic graph of G, say G0 ¼ ðV 0 ; E0 Þ; for each element h 2 V 0 do begin w0 ðhÞ :¼ GreenðGh ; wÞ; end; a :¼ maxfa; GreenðG0 ; w0 Þg; end; end. The correctness of the above algorithm is based on the results from modular decomposition theory mentioned in Section 2. Concerning the computational complexity of the above algorithm, one has that Theorems 1 and 2, and the mentioned results from modular decomposition theory imply that the most expansive step is either that of computing a maximum weight stable set of a P5-free graph G such that the anti-neighborhoods of each edge of G induce P3-free graphs (cf. Theorem 1), or that of computing a maximum stable set in a gray graph. Then, the computational complexity may be estimated as maxfOðn5 Þ; Oðntþ1 Þg time. Theorem 3. Let F be any graph. If one can solve MWS for every F-depending graph in Oðnt Þ time for some natural t, then one can solve MWS for ðP 5 ; F Þfree graphs in maxfOðn5 Þ; Oðntþ1 Þg time. Let us point out that the facts of this section (which come by applying the results of Section 2 to the anti-neighborhood of an edge) can be directly referred to ðP 5 ; F þ Þ-free graphs once one considers the anti-neighborhood of a single vertex. In particular, the step of computing a maximum weight stable set of a P5-free graph G such that the anti-neighborhoods of each edge of G induce P3-free graphs is reduced to that of computing a maximum weight stable set of a P5-free graph G such that the anti-neighborhoods of each vertex of G induce P3-free graphs. That can be formalized as follows. Theorem 4. Let F be any graph. If one can solve MWS for every F-depending graph in Oðnt Þ for some natural t, then one can solve MWS for ðP 5 ; F þ Þ-free graphs in maxfOðnmÞ; Oðntþ1 Þg time.
5. Application to certain P5-free graphs In this section let us apply the results of the previous sections to introduce some subclasses of P5-free graphs for which one can solve MWS in polynomial time. In particular, by the previous section, it is sufficient to look for graphs F such that one can solve MWS for every F-depending graph in polynomial time. Observation 3. Let F be any graph. Let G ¼ ðV ; EÞ be a F-depending graph, and let fV 1 ; V 2 ; V 3 g be a partition of V according to Definition 1. Then by (ii) of Definition 1, to solve MWS for G one can assume without loss of generality that V3 is a stable set. 5.1. ðP 5 ; S k Þ-Free graphs In this subsection let us prove that one can efficiently solve MWS for ðP 5 ; S k Þ-free graphs. That extends some of the results mentioned in Section 2, according to the fact that S k contains induced subgraphs of five vertices. Also, that extends (to the weighted case as well) the fact that one can efficiently solve MS for ðP 5 ; K 2;n Þ-free graphs (actually, in general one can efficiently solve MS for ðP 5 ; K n;n Þfree graphs, see e.g. [23,34]). Theorem 5. For any natural k, one can solve MWS for every Sk-depending graphs in Oðn2k2 mÞ time. Proof. Let G ¼ ðV ; EÞ be an Sk-depending graph, and let fV 1 ; V 2 ; V 3 g be a partition of V according to Definition 1. Then G½V 1 [ V 2 is S 2k1 -free. Let T be the family of stable sets of G½V 1 [ V 2 . Then jTj is Oðn2k2 Þ. One can solve MWS in G by solving MWS in G½V 3 and in G½V 3 n N ðT Þ for every T 2 T. By the above, that can be done in Oðn2k2 mÞ time. h Corollary 1. For any natural k, one can solve MWS for ðP 5 ; S k Þ-free graphs in maxfOðn5 Þ; Oðn2k1 mÞg time. 5.2. A note on ðP 5 ; K 2;3 Þ-free graphs A K 2;3 is a S þ 2 . By Theorems 4 and 5, one can solve MWS for ðP 5 ; K 2;3 )-free graphs in Oðn3 mÞ time. However, one can obtain a slight improvement as follows. Given any graph F, let us say that a F-depending graph G ¼ ðV ; EÞ, with a partition fV 1 ; V 2 ; V 3 g of V
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according to Definition 1, is basic if V1 and V2 are respectively cliques. Lemma 6. Let G be a prime ðP 5 ; K 2;3 Þ-free graph. Then, for every vertex v of G, G½AðvÞ is either P3-free or a basic S2-depending graph. Proof. That comes by applying the argument of Lemma 2, with the assumption that G is (P5,K 2;3 )free, to the anti-neighborhood of a single vertex x instead of that of an edge ðx; yÞ. By referring to the mentioned argument, in detail: G½ðX [ KÞn N ½d 1 is a clique; G½Y n N ½d 1 is a clique. Let us only show that D1 n N ½d 1 has a join to K n N ½d 1 (the other facts are quite immediate). Assume to the contrary that there exists d 2 D1 n N ½d 1 nonadjacent to some vertex k 2 K n N ½d 1 ; let q 2 Q1 be adjacent to both d and k, according to Lemma 3; then x; d 1 ; d; k; q induce a K 2;3 – contradiction. h One can easily verify that one can solve MWS for every basic S2-depending graph in OðnmÞ. Then by results from modular decomposition theory one obtains: Corollary 2. One can solve MWS for ðP 5 ; K 2;3 Þ-free graphs in Oðn2 mÞ time. 5.3. ðP 5 ; P 3 Þ-Free graphs In this subsection let us prove that one can efficiently solve MWS for ðP 5 ; P 3 Þ-free graphs. That extends some of the results mentioned in Section 2, according to the fact that P 3 contains induced subgraphs of five vertices. Lemma 7. Let G ¼ ðV ; EÞ be a connected P5-free graph, and let fX ; Y g be a partition of V such that G½Y is P3-free. Let CX be the family of component of G½X which are distinguished by some vertex of Y, and let W ¼ fx 2 X : x belongs to no element of C X g. Then there exists y 2 Y such that: (i) y contacts each element of CðX Þ; (ii) at most one element of CðX Þ is not dominated by y; (iii) W n N ½y has a co-join to Y n N ½y. Proof. About (i). The proof is by induction on k ¼ jCðX Þj. If k ¼ 1, then the assertion follows since G is connected. Then let us assume the assertion is true for k 1, and prove it is true for k. Assume to the contrary that there exists no y 2 Y contacting each element of CðX Þ. Let CðX Þ ¼ fX 1 ; . . . ; X k g.
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Then by the inductive assumption, there exists y 2 Y contacting each element of fX 1 ; . . . ; X k1 g. Let y k 2 Y be partial to Xk, and let ak ; bk be vertices of Xk, with ak adjacent to bk, such that yk is adjacent to ak and nonadjacent to bk. Let us show that fy k g has a co-join to fyg [ X 1 [ [ X k1 . In fact, otherwise: if yk is adjacent to y, then to avoid a P5 involving ak and bk one has that yk contacts each element of fX 1 ; . . . ; X k1 g (contradiction); if yk is nonadjacent to y, then yk is adjacent to some vertex x of X 1 [ [ X k1 , but then y; x; y k ; ak ; bk induce a P5 (contradiction). Then let P be the set of interior vertices of a shortest path from fy k g [ X k to fyg [ X 1 [ [ X k1 . Since G is P5-free, jP j 6 2. Assume jP j ¼ 2. Let P ¼ fu; vg, such that u contacts fy k g [ X k . If u 2 X , then u does not contact Xk; then u is adjacent to yk; then ak ; y k ; u; v and a vertex of fyg [ X 1 [ [ X k1 induce a P5. If u 2 Y , then v 2 X , otherwise y k ; u; v induce a P3 of G½Y ; then v is adjacent to y and does not contact any element of fX 1 ; . . . ; X k1 g; then y k ; u; v; y and a vertex of X 1 [ [ X k1 induce a P5. Assume jP j ¼ 1. Let P ¼ fug. If u 2 X , then u does not contact X 1 [ [ X k ; then u is adjacent to both yk and y; then a P5 arises. Then assume u 2 Y . If u is adjacent to yk, then u is nonadjacent to y, otherwise a P3 arises in G½Y ; then u contacts some element of fX 1 ; . . . ; X k1 g; in particular, u contacts each element of fX 1 ; . . . ; X k1 g, otherwise a P5 arises involving y; on the other hand, u contacts Xk, otherwise bk ; ak ; y k ; u and a vertex of X 1 [ [ X k1 induce a P5; but this is a contradiction. If u is nonadjacent to yk, then u contacts Xk; if u is adjacent to y, then u contacts each element of fX 1 ; . . . ; X k1 g (otherwise a P5 arises involving yk) – a contradiction; if u is nonadjacent to y, then u contacts some element of fX 1 ; . . . ; X k1 g, but then a P5 arises involving yk and y. About (ii). It follows since G is P5-free. About (iii). Let Y ¼ fy 2 Y : y enjoys statement ðiÞg. Let y 2 Y such that N W ðy Þ is maximal under inclusion in fN W ðyÞ : y 2 Y g. Let us prove that W n N ½y has a co-join to Y n N ½y . Assume to the contrary there exists y 2 Y n N ½y adjacent to a vertex x 2 W n N ½y . By construction, one can assume without loss of generality that fxg is a trivial component of G½X . Note that y 62 Y , otherwise: by definition of y* there exists x 2 W adjacent to y* and nonadjacent to y; then by definition of Y* a P5 arises.
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Then y 2 ðY n N ½y Þ n Y . Then fy; xg has a cojoin to fy g [ ðX n W Þ. Let P be the set of interior vertices of a shortest path from fy; xg to fy g [ ðX n W Þ. Since G is P5-free, jP j 6 2. Assume jP j ¼ 2. This case leads to a contradiction, similarly to the corresponding case of the argument concerning statement (i). Assume jP j ¼ 1. Let P ¼ fug. If u 2 X , then u does not contact either x or X n W ; then u is adjacent to both y and y*; then a P5 arises. Then assume u 2 Y . Assume that u is adjacent to y. Then u is nonadjacent to y*, otherwise a P3 arises in G½Y ; then u contacts X n W ; it follows that u 2 Y , otherwise a P5 arises involving y; but then (by the above, recalling that u is nonadjacent to y*) u is nonadjacent to x; then a P5 arises. Assume that u is nonadjacent to y. Then u is adjacent to x. If u is nonadjacent to y*, then a P5 arises involving y and x. If u is adjacent to y*, then u 2 Y , otherwise a P5 arises involving y and x; then, by definition of y* there exists x 2 W adjacent to y* and nonadjacent to u; then either y; x; u; y ; x (if x* is nonadjacent to y) or x ; y; x; u and an element of X n W (if x* is adjacent to y) induce a P5. h Lemma 8. Let G ¼ ðV ; EÞ be a P5-free graph, such that V admits a partition fX ; Y g, where X is a stable set, and G½Y is P3-free. Then one can solve MWS for G in OðnmÞ time. Proof. Let us show that there exists x 2 X such that one can solve MWS for G½V n N ðxÞ in OðmÞ time. Assume without loss of generality that G is connected. Then, since G is P5-free, there exists x 2 X contacting each component of G½Y ; in particular, at most one component of G½Y is not dominated by x. If x dominates each component of G½Y , then G½V n N ðxÞ is a stable set. If there exists a component of G½Y , say Q, not dominated by x, then one can solve MWS in G½V n N ðxÞ by solving MWS in G½V n ðN ðxÞ [ QÞ and in G½V n ðN ðxÞ [ N ðqÞÞ for every q 2 Q n N ðxÞ: since G½Y is P3-free, each of such graph has no edge. Then the assertion follows. Now, to prove the lemma, one can iterate the above argument to each component of G½V n fxg. h Lemma 9. Let G ¼ ðV ; EÞ be a P5-free graph, such that V admits a partition fX ; Y g, where G½X and G½Y are respectively P3-free. Then one can solve MWS for G in Oðn2 mÞ time.
Proof. Let us show that there exists y 2 Y such that one can solve MWS for G½V n N ðyÞ in OðnmÞ time. Assume without loss of generality that G is connected. Let CX be the family of components of G½V 1 which are distinguished by some vertex of Y, and W ¼ fx 2 X : x belongs to no element of CX}. Then, by Lemma 7, there exists y 2 Y such that: (i) y contacts each element of CX; (ii) at most one element of CX is not dominated by y; (iii) W n N ½y has a co-join to V 2 n N ½y. If y dominates each element of CX, then G½V n N ½y is a P3-free graph. Then, one can solve MWS for G½V n N ðyÞ in OðmÞ time. If there exists an element of CX, say Q, not dominated by y, then one can solve MWS in G½V n N ðyÞ by solving MWS in G½V n ðN ðyÞ [ QÞ and in G½V n ðN ðyÞ [ N ðqÞÞ for every q 2 Q n N ðyÞ. Since G½X is P3-free, each of such graphs is P3-free. Then, one can solve MWS for G½V n N ðyÞ in OðnmÞ time. Now, to prove the lemma, one can iterate the above argument to each component of G½V n fyg. h Theorem 6. One can solve MWS P3-depending graph in Oðn3 mÞ time.
for
every
Proof. Let G ¼ ðV ; EÞ be a P3-depending graph, and let fV 1 ; V 2 ; V 3 g be a partition of V according to Definition 1. Assume without loss of generality that G is connected. To prove the assertion, it is sufficient to show that there exists a vertex v 2 V (which can be easily detected) such that one can solve MWS for G½V n N ðvÞ in Oðn2 mÞ time. Then the assertion will follow by iteratively applying such an argument to each component of G½V n fvg. Let us consider the following two exhaustive cases. Case 1. G½V 1 [ V 2 is connected. Let C V 1 be the family of component of G½V 1 which are distinguished by some vertex of V 2 , and W ¼ fx 2 V 1 : x belongs to no element of C V 1 g. Then, by Lemma 7, there exists y 2 V 2 such that: (i) y contacts each element of C V 1 ; (ii) at most one element of C V 1 is not dominated by y; (iii) W n N ½y has a co-join to V 2 n N ½y. If y dominates each element of C V 1 , then V n N ½y admits a partition into a stable set (i.e., V 3 ) and a vertex-set inducing a P3-free graph (i.e., ðV 1 [ V 2 Þ n N ½y). Then, by Lemma 8, one can solve MWS for G½V n N ðyÞ in OðnmÞ time.
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If there exists an element of C V 1 , say Q, not dominated by y, then one can solve MWS in G½V n N ðyÞ by solving MWS in G½V n ðN ðyÞ [ QÞ and in G½V n ðN ðyÞ [ N ðqÞÞ for every q 2 Q n N ðyÞ. Since G½V 1 is P3-free, the vertex-set of each of such graphs admits a partition into a stable set and a vertex-set inducing a P3-free graph. Then, by Lemma 8, one can solve MWS for G½V n N ðyÞ in Oðn2 mÞ time. Case 2. G½V 1 [ V 2 is not connected. Let C V 1 [V 2 be the family of components of G½V 1 [ V 2 which are distinguished by some vertex of V 3 . Assume without loss of generality that each component Q of G½V 1 [ V 2 not belonging to C V 1 [V 2 is a trivial component of G½V 1 [ V 2 : in fact, otherwise, one can contract Q into one vertex q, with wðqÞ ¼ aw ðG½QÞ; that can be done in Oðn2 mÞ time, by Lemma 9. Since G is connected P5-free, there exists a vertex z 2 Z contacting each component of G½V 1 [ V 2 ; in particular, at most one nontrivial component of G½V 1 [ V 2 is not dominated by z. If z dominates each component of G½V 1 [ V 2 , then G½V n N ðzÞ is a stable set. Then assume there exists a component Q of G½V 1 [ V 2 not dominated by z. By the above, Q belongs to C V 1 [V 2 . If C V 1 [V 2 ¼ fQg, then one can apply an argument similar to that of Case 1 (by recalling that ðV 1 [ V 2 Þ n Q is a stable set) to show that there exists a vertex q 2 Q such that one can solve MWS for G½V n N ðqÞ in Oðn2 mÞ time. If C V 1 [V 2 6¼ fQg, then let Q 0 be an element of C V 1 [V 2 n fQg. Let us prove that no element of Z n fzg distinguishes Q n N ðzÞ. Assume to the contrary there exists z 2 Z n fzg distinguishing Q. Then let a; b be two vertices of Q n N ðzÞ, with a adjacent to b, such that z is adjacent to a and nonadjacent to b. Since z dominates Q 0 : z does not contact Q 0 (otherwise a P5 arises involving b; a and z); there exists z0 2 Z n N ðzÞ distinguishing Q 0 . Let a0 ; b0 be two vertices of Q 0 , with a 0 adjacent to b 0 , such that z 0 is adjacent to a 0 and nonadjacent to b 0 . Then z 0 contacts (and thus dominates, since G is P5-free) Q, otherwise z0 ; a0 ; z and two vertices of Q induce a P5. But then b0 ; a0 ; z0 ; a; z induce a P5. Then the assertion is proved. Then, one can contract Q n N ðzÞ into a single vertex q with wðqÞ ¼ aw ðG½Q n N ðzÞÞ. That can be done in Oðn2 mÞ time, by referring to Lemma 9 for each component of G½Q n N ðzÞ. Then G½V n N ½z
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can be reduced to a bipartite graph (for which one can solve MWS in OðmÞ). Then the lemma follows. h Corollary 3. One can solve MWS for ðP 5 ; P 3 Þ-free graphs in Oðn4 mÞ time. 5.4. A note on possible extensions Let us point out the following general fact. Lemma 10. Let F be any graph. If one can solve MWS for every F-depending graph in Oðnt Þ time (for some natural t), then one can solve MWS for every ðP 1 þ F Þ-depending graph in Oðntþ2 Þ time. Proof. Let G ¼ ðV ; EÞ be an ðP 1 þ F Þ-depending graph. Let fV 1 ; V 2 ; V 3 g be a partition of V according to Definition 1. One can solve MWS in G by solving MWS: (i) in G½V n ðN ðxÞ [ N ðyÞÞ for each pair of nonadjacent vertices x 2 V 1 and y 2 V 2 ; (ii) in G½V n V 1 ; (iii) in G½V n V 2 . Concerning step (i): such graphs are Fdepending. Concerning step (ii): note that for each vertex v 2 V 2 , one has that G½ðV 2 [ V 3 Þ n N ðvÞ is an F-depending graph. Concerning step (iii): this step is similar to step (ii), by symmetry. Then, by the assumption, the lemma follows. h Then one obtains the following fact which admits a repeated application. Theorem 7. Let F be any graph. If one can solve MWS for ðP 5 ; F Þ-free graphs in polynomial time, then one can solve MWS for ðP 5 ; ðP 1 þ F Þ Þ-free graphs in polynomial time. 5.5. A note on ðP 5 ; F 1 Þ-free graphs In this subsection let us prove that, by an approach similar to that introduced in Section 3, one can solve MWS for ðP 5 ; F 1 Þ-free graphs in Oðn4 mÞ time. That improves a recent result (see among other results in [11]) stating that one can solve MWS for ðP 5 ; F 1 Þ-free graphs in Oðn7 mÞ time. Throughout this subsection assume that G is a prime ðP 5 ; F 1 Þ-free graph. Throughout the next two lemmas of this subsection assume that: ðx; yÞ is an edge of G; Q is a non-clique component of G½Aðx; yÞ; D is the set of vertices of G which distinguish Q. Lemma 11. Let d 2 D well distinguish Q. Then d dominates fv 2 N ðxÞ \ N ðyÞ : fvg has a join to Qg.
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Proof. Let z 2 N ðxÞ \ N ðyÞ such that fzg has a join to Q. Then d is adjacent to z otherwise, since d well distinguishes Q, one has that x; y; z; d and two nonadjacent vertices of Q induce a F1. h Lemma 12. Let d 2 D well distinguish Q. Then D n N ½d is a stable set. Proof. Let S ¼ Q n N ðdÞ and T ¼ Q \ N ðdÞ. Let s 2 S such that there exists t 2 T nonadjacent to s* (according to the fact that d well distinguishes Q). Let c 2 D n N ½d and let us prove that: c is nonadjacent to s* and dominates T \ N ðs Þ. Assume to the contrary that c is adjacent to s*. Then, to avoid that s ; c; y; d; t induce a P5, c is adjacent to t*, but then x; y; d; c; t ; s induce a F 1 (contradiction). Then c is nonadjacent to s*. That implies that each element t 2 T \ N ðs Þ is adjacent to c, otherwise c; y; d; t; s induce a P5. Then c is nonadjacent to s* and dominates T \ N ðs Þ. Let c1 ; c2 2 D n N ½d. Assume to the contrary that c1 is adjacent to c2. Let t 2 T \ N ðs Þ (such a vertex does exist, since G is P5-free and G½Q is connected). Then by the above paragraph one has that c1 ; c2 ; y; d; t; s induce a F1 (contradiction). h A graph G ¼ ðV ; EÞ is almost bipartite if there exists a partition of V, say fV 1 ; V 2 g such that: V1 is a stable set; G½V 2 is P3-free; no element of V1 distinguishes any component (clique) of G½V 2 . Theorem 8. One of the following cases occurs: (i) for every edge ðx; yÞ of G, G½Aðx; yÞ is P3-free; (ii) there exists a vertex v of G such that each component of G½AðvÞ induces either a P3-free or a almost bipartite graph. Proof. The proof is very similar to that of Theorem 2. Assume that statement (i) does not occur, and let us verify that then statement (ii) occurs. Then there exists an edge ðx; yÞ of G such that G½Aðx; yÞ is not P3-free. Let Q ¼ fQ1 ; . . . ; Qm g be the family of non-clique components of G½Aðx; yÞ. Let D ¼ fv 2 V : v distinguishes some element of Q g. According to Lemma 1, D* is partitioned into fD1 ; . . . ; Dm g, where each vertex d i 2 Di distinguishes Qi and does not distinguish Qj for every j 6¼ i. Let ðd 1 ; . . . ; d m Þ 2 D1 ; . . . ; Dm . By Observation 2 one can assume without loss of generality that for i ¼ 1; . . . ; m: di well distinguishes Qi ; N Q ðd i Þ is maximal under inclusion in fN Qi ðcÞ : c well
distinguishes Qi g. By Lemma 1, one can assume without loss of generality that d1 dominates Qj for every j > 1. Furthermore, by Lemmas 2 and 11, one has that V n N ½d 1 ¼ ðD1 [ Q1 [ H Þ n N ½d 1 where H is the vertex-set of the clique-components of G½Aðx; yÞ. By Lemma 12, D1 n N ½d 1 is a stable set. By (ii) of Lemma 4 and by construction, ðQ1 [ H Þ n N ½d 1 induces a P3-free graph (in particular, Q1 has a cojoin to H). By (i) of Lemma 4, no element of D1 n N ½d 1 distinguishes any component of Q1 n N ½d 1 . Furthermore, D1 n N ½d 1 has a co-join to H n N ½d 1 : in fact assume to the contrary that there exists d 2 D1 n N ½d 1 adjacent to some vertex h 2 H n N ½d 1 ; let q 2 Q1 be adjacent to both d and d1, according to Lemma 3; then x; y; d 1 ; d; h; q induce a F1 (contradiction). Then statement (ii) occurs with v ¼ d 1 . h Observation 4. One can solve MWS for almost bipartite graphs in OðmÞ time. By Observation 4, by considering Theorem 8 instead of Theorem 2, and by re-denoting a graph G as gray if each of its components induces either a P3-free or a almost bipartite graph, one can directly verify that Algorithm Green(G, w) – with input a ðP 5 ; F 1 Þ-free graph – correctly solves MWS for ðP 5 ; F 1 Þ-free graphs in Oðn4 mÞ time. Theorem 9. One can solve MWS for ðP 5 ; F 1 Þ-free graphs in Oðn4 mÞ time. Acknowledgement I am indebted and grateful to the referees for their remarks, which were helpful to improve both the contents and the presentation of the manuscript. References [1] V.E. Alekseev, On the local restriction effect on the complexity of finding the graph independence number, in: Combinatorial-algebraic Methods in Applied Mathematics, Gorkiy University Press, Gorkiy, 1983, pp. 3–13 (in Russian). [2] V.E. Alekseev, A polynomial algorithm for finding largest independent sets in fork-free graphs, Discrete Analysis of Operations Research Series 1 (6) (1999) 3–19 (in Russian) (see also [3] for the English version). [3] V.E. Alekseev, A polynomial algorithm for finding largest independent sets in fork-free graphs, Discrete Applied Mathematics 135 (2004) 3–16. [4] V.E. Alekseev, On easy and hard hereditary classes of graphs with respect to the independent set problem, Discrete Applied Mathematics 132 (2004) 17–26.
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Discrete Optimization
Some observations on maximum weight stable sets in certain P5-free graphs Raffaele Mosca
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Dipartimento di Scienze, Universita´ degli Studi ‘‘G. d’Annunzio’’, Pescara 65127, Italy Received 6 June 2006; accepted 1 December 2006 Available online 20 December 2006
Abstract The maximum weight stable set problem (MWS) is the weighted version of the maximum stable set problem (MS), which is NP-hard. The class of P5-free graphs – i.e., graphs with no induced path of five vertices – is the unique minimal class, defined by forbidding a single connected subgraph, for which the computational complexity of MS is an open question. At the same time, it is known that MS can be efficiently solved for ðP 5 ; F Þ-free graphs, where F is any graph of five vertices different to a C5. In this paper we introduce some observations on P5-free graphs, and apply them to introduce certain subclasses of such graphs for which one can efficiently solve MWS. That extends or improves some known results, and implies – together with other known results – that MWS can be efficiently solved for ðP 5 ; F Þ-free graphs where F is any graph of five vertices different to a C5. Ó 2006 Elsevier B.V. All rights reserved. Keywords: Maximum weight stable set problem; P5-free graphs; Polynomial time algorithm
1. Introduction A stable set of a graph G is a subset of pairwise nonadjacent vertices of G. The maximum weight stable set (MWS) problem is the following: Given a graph G ¼ ðV ; EÞ and a weight function w on V, determine a stable set of G having maximum weight (where the weight of a stable set S is given by the sum of the weights of each s 2 S). Let aw ðGÞ denote the value of a maximum weight stable set of G. The MWS problem is called MS problem if all vertices v have the same weight wðvÞ ¼ 1.
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Tel.: +39 854211628. E-mail address: [email protected]
It is well known that MS is NP-hard [26]. In particular, it remains difficult for triangle-free, cubic or planar graphs (see e.g. [36,21,22]). Then, the results in [1,2] imply that the class of P5-free graphs is the unique minimal class, defined by forbidding a single connected subgraph, for which the computational complexity of MS is an open question (see also [4]). On the other hand, MS (or MWS) can be efficiently solved for several graph classes. In particular, polynomial algorithms are known to solve MWS for P4-free graphs [14,15], 2K2-free graphs [18,19], and claw-free graphs [35,31,37]. Let us focus on the class of P5-free graphs. General structure properties of such graphs can be found in [5,6,17]. In particular, it is known that MS can be efficiently solved for ðP 5 ; F Þ-free graphs,
0377-2217/$ - see front matter Ó 2006 Elsevier B.V. All rights reserved. doi:10.1016/j.ejor.2006.12.011
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where F is any graph of five vertices different to a C5. In this paper we introduce some observations on P5-free graphs, and apply them to introduce certain subclasses of such graphs for which one can efficiently solve MWS. Two of such subclasses are: ðP 5 ; S k Þ-free graphs, where a S k consists of a stable set of k þ 1 vertices, a P 1 þ P 2 , and a join between them; ðP 5 ; P 3 Þ-free graphs, where a P 3 consists of a P 1 þ P 3 , a P 1 þ P 2 , and a join between them. That extends some known results, according to the fact that S k and P 3 contains induced subgraphs of five vertices (in particular, since S k contains a K 2;3 , one obtains that one can efficiently solve MWS for ðP 5 ; F Þ-free graphs, where F is any graph of five vertices different to a C5). Also, we point out that by a similar approach one obtains a Oðn4 mÞ time algorithm to solve MWS for ðP 5 ; F 1 Þ-free graphs where F1 is a graph of six vertices a; b; c; d; e; f and edges ab; ac; ad; bc; bd; ce; de; df : that improves a recent result (see among other results in [11]) stating that one can solve MWS for ðP 5 ; F 1 Þ-free graphs in Oðn7 mÞ time.
2. Preliminaries For any missing notation or reference, let us refer to [10]. Throughout this paper, let G ¼ ðV ; EÞ be a finite undirected graph without self-loops and multiple edges and let jV j ¼ n; jEj ¼ m. The complement of G is the graph, denoted as co-G, having the same vertices as G and such that two vertices are adjacent in co-G if and only if they are nonadjacent in G. Let U ; W be two subsets of V. Let N U ðW Þ ¼ fu 2 U n W juv 2 Eg be the set of neighbors of W in U. If U ¼ V , then we write N ðW Þ instead of N V ðW Þ. If W ¼ fvg, then we write N U ðvÞ instead of N U ðfvgÞ. Also, we write N U ½v ¼ N U ðvÞ [ fvg. Let us say that U has a join (a co-join, respectively) to W, if each vertex in U is adjacent (is nonadjacent) to each vertex in W. If U ¼ fug and fug has a join to W, then let us also say that u dominates W. Then let us say that a vertex v 2 V n U contacts U if v is adjacent to some vertex of U. Throughout this paper, all subgraphs are understood as induced subgraphs. Let G½U denote the subgraph of G induced by the vertex subset U. For any graph F, G is F-free if G contains no induced subgraph isomorphic to F. A component of G is a nonempty vertex-set W V such that
G½W induces a maximal connected subgraph of G. A component is trivial if it is a singleton. A P k has vertices v1 ; v2 ; . . . ; vk and edges vj vjþ1 for 1 6 j < k. A Ck has vertices v1 ; v2 ; . . . ; vk and edges vj vjþ1 for 1 6 j 6 k (index arithmetic modulo k). A 2K2 is a graph of four vertices a; b; c; d and edges ab; cd. A claw is a graph of four vertices and edges ab; ac; ad. A K p;q is a complete bipartite graph whose sides have respectively p and q vertices. A Sk is a graph of k vertices and no edges. Given two graphs G1 and G2, let G1 þ G2 denote the graph consisting of G1, G2, and a co-join between them. A clique of G is a set of pairwise adjacent vertices of G. One can verify that G is P3-free if and only if each component of G is a clique. A vertex z 2 V distinguishes vertices x; y 2 V if zx 2 E and zy 62 E or zy 2 E and zx 62 E. Let us also say that a vertex z distinguishes a vertex-set U V , z 62 U , if z distinguishes two vertices in U; in particular, z well distinguishes U, if z distinguishes two nonadjacent vertices in U. A vertex-set M V is a module if no vertex from V nM distinguishes M. Graph G is prime if it contains only trivial modules, i.e. ;, V and one-elementary vertex sets. A nontrivial module is called a homogeneous set. It is well known that in a connected graph G with connected complement co-G, the maximal homogeneous sets are pairwise disjoint which means that every vertex is contained in at most one homogeneous set. The existence and uniqueness of the modular decomposition tree is based on this property, and linear time algorithms were designed to determine this tree – see e.g. [32,30]. The tree contains the vertices of the graph as its leaves, and the internal nodes are of three types: they represent a join or co-join operation, or a prime subgraph. Observation 1. Let G ¼ ðV ; EÞ be a graph, and let W V induce a nontrivial connected subgraph of G. If a vertex v 2 V n W distinguishes W, then v distinguishes two adjacent elements of W. Observation 2. Let G ¼ ðV ; EÞ be a prime graph, and let W V induce a nontrivial connected subgraph of G. If W is not a clique, then there exists a vertex v 2 V n W which well distinguishes W. That comes by considering the components of co-G. The anti-neighborhood (or non-neighborhood) of U V is AðU Þ ¼ fv 2 V n U : v is nonadjacent to any element of U g. If U ¼ fu1 ; . . . ; uh g, then we write Aðu1 ; . . . ; uh Þ instead of Aðfu1 ; . . . ; uh gÞ. Obviously, the MWS problem on a graph G with vertex
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weight function w can be reduced to the same problem on anti-neighborhoods of vertices in the following way: aw ðGÞ ¼ maxfwðvÞ þ aw ðG½AðvÞÞjv 2 V g: For different graph classes including that of P5-free (see e.g. [29]), that holds by considering the antineighborhoods of edges as well. Theorem 1 [29]. Let G be a connected P5-free graph. If one can solve MWS for every subgraph induced by the anti-neighborhood of each edge of G in Oðnh Þ time, then one can solve MWS for G in maxfOðn3þh Þ; Oðn5 Þg time. Let us conclude by shortly listing some results concerning MWS for ðP 5 ; F Þ-free graphs, where F is a graph of five vertices different to a C5 (a drawn list of five-vertex graphs can be found e.g. in [25]). If F contains an isolated vertex or an isolated edge, then MWS can be efficiently solved for ðP 5 ; F Þ-free graphs by the observation and the result above (let us recall that MWS is efficiently solvable for ðP 5 ; C 3 Þ-free graphs and for ðP 5 ; C 4 Þ-free graphs – see e.g. [13]). If F admits a dominating vertex, then MWS can be efficiently solved for ðP 5 ; F Þ-free graphs by combining an augmenting approach with the previous facts (see e.g. [34]); in particular, concerning (P5, gem)-free graphs, where a gem has edges ab; ac; ad; ae; bc; cd; de, see e.g. [7] for a very efficient algorithm. If F is a banner (a banner has edges ab; bc; bd; ce; de), then MWS can be efficiently solved for ðP 5 ; F Þ-free graphs by combining different approaches [9] (see also [12,27] for other approaches concerning MS). If F is a bull (a bull has edges ab; ac; bc; bd; ce), then MWS can be efficiently solved for ðP 5 ; F Þ-free graphs as a subcase of a general decomposition result introduced in [16]. If F is either a co-chair (a co-chair has edges ab; bc; bd; cd; ce; de) or a co-banner (a co-banner has edges ab; bc; cd; ce; de), then MWS can be efficiently solved for ðP 5 ; F Þ-free graphs by applying some of the structure results introduced in [13]. If F is a fork (a fork – or chair – has edges ab; bc; cd; ce), then MWS can be (very) efficiently solved for ðP 5 ; F Þ-free graphs by combining the results in [8,28]. If F is a house (a house – or co-P5 – has edges ab; ac; bc; bd; ce; de), then MWS can be efficiently
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solved for ðP 5 ; F Þ-free graphs by a decomposition approach introduced in [20,24]. If F is a co-ðP 2 þ P 3 Þ, i.e., F has edges ab; ac; ad; bc; be; ce; de, then MWS can be efficiently solved for ðP 5 ; F Þ-free graphs by a clique separator decomposition approach introduced in [11]. It remains the case in which F is a K 2;3 . In this case MS can be efficiently solved for ðP 5 ; F Þ-free graphs by an augmenting approach (see e.g. [23,33,34]). 3. Some observations on P5-free graphs Throughout this section assume that G ¼ ðV ; EÞ is a prime P5-free graph. For any vertex-set X V , let us write J ðX Þ ¼ fv 2 V n X : fvg has a join to X g. Lemma 1. Let X V with G½X connected. Let Q ¼ fQ1 ; . . . ; Qm g be a family of nontrivial components of G½AðX Þ. Let D ¼ fv 2 V : v distinguishes some element of Q g. Then: (i) D J ðX Þ; (ii) D* is partitioned into fD1 ; . . . ; Dm g, where each d i 2 Di distinguishes Qi and does not distinguish Qj for every j 6¼ i; (iii) Di has a join to Dj, for 1 6 i < j 6 m; (iv) for any ðd 1 ; . . . ; d m Þ 2 D1 Dm , there exists di dominating Qj for every j 6¼ i. Proof. About (i). Let d 2 D and let Q be a nontrivial component of G½AðX Þ distinguished by d. Then by Observation 1 d 62 N ðX Þ n J ðX Þ, otherwise two adjacent vertices of X, d and two adjacent vertices of Q induce a P5. Then, since G is connected, d 2 J ðX Þ. About (ii). Let d 2 D . By Observation 1, d cannot distinguish two elements of Q otherwise a P5 arises. About (iii). Assume to the contrary there exist d i 2 Di and d j 2 Dj which are nonadjacent. Let x 2 X . By Observation 1: let xi ; y i 2 Qi such that di is adjacent to xi and nonadjacent to yi; let xj ; y j 2 Qj such that dj is adjacent to xj and nonadjacent to yj. To avoid that y j ; xj ; d j ; x; d i induce a P5, di is adjacent either to xj or to yj, that is, by statement (ii) di dominates Qj. By symmetry, dj dominates Qi . Then y j ; d i ; x; d j ; y i induce a P5 (contradiction). About (iv). The proof is by induction on m. For m ¼ 2, the assertion follows by Observation 1, statement (iii) and the fact that G is P5-free. Assume
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the assertion holds for m 1, and let us prove that it holds for m. Let us consider an element ðd 2 ; . . . ; d m Þ 2 D2 Dm . By the inductive assumption one has without loss of generality that d2 dominates Qj for every j > 2. Let d 1 2 D1 . If d2 dominates D1, then d2 is the required element. Then, by statement (ii), assume that fd 2 g has a co-join to Q1. Notice that then d1 dominates Q2 (otherwise a P5 arises). If d1 dominates Qj for every j > 2, then d1 is the required element. Otherwise, by the inductive assumption on ðd 1 ; d 3 ; . . . ; d m Þ, one has without loss of generality that d3 dominates Q1 and Qj for every j > 3; also, d3 dominates Q2 , otherwise two adjacent vertices in Q2 ; d 2 ; d 3 and a vertex in Q1 induce a P5. Then d3 is the required element. h Throughout the next lemmas of this section assume that: X V with G½X connected; Q is a non-clique component of G½AðX Þ; D is the set of vertices of G which distinguish Q. Lemma 2. Let d 2 D well distinguish Q. Then d dominates ðN ðX Þ n J ðX ÞÞ [ fv 2 J ðX Þ : fvg has a co-join to Qg. Proof. Let x 2 X . Let z 2 N ðX Þ n J ðX Þ. Assume to the contrary that d is nonadjacent to z. By (i) of Lemma 1, z does not distinguish Q. Then: if fzg has a co-join to Q, then by Observation 1 d; x; z and two adjacent vertices of Q induce a P5 (contradiction); if fzg has a join to Q, then, since d well distinguishes Q, d; x; z and two nonadjacent vertices of Q induce a P5 (contradiction). Let z 2 fv 2 J ðX Þ : fvg has a co-join to Qg. Then d is adjacent to z, otherwise by Observation 1 d; x; z and two adjacent vertices of Q induce a P5. h Lemma 3. Let d 2 D well distinguish Q. Then N Q ðdÞ \ N Q ðcÞ 6¼ ; for every c 2 D n N ½d. Proof. Let c 2 D n N ½d. Assume to the contrary N Q ðdÞ \ N Q ðcÞ ¼ ;. Then, since G is P5-free and G½Q is connected: fN Q ðdÞ; N Q ðcÞg is a partition of Q, and N Q ðdÞ has a join to N Q ðcÞ. But that contradicts the fact that d well distinguishes Q. h Lemma 4. Let d 2 D well distinguish Q, such that N Q ðdÞ is maximal under inclusion in fN Q ðcÞ : c well distinguish Qg. Then:
(i) no vertex of D n N ½d distinguishes any component of G½Q n N ½d; (ii) G½Q n N ½d is P3-free. Proof. Let x 2 X . About (i). Assume to the contrary a vertex c 2 D n N ½d distinguishes a component K of G½Q n N ½d. Then, by Observation 1, d; x; c and two adjacent vertices of K induce a P5 (contradiction). About (ii). Let S ¼ Q n N ðdÞ and T ¼ Q \ N ðdÞ. Notice that since G is P5-free all the elements from a component K of G½S have the same neighborhood in T, say T ðKÞ. Then by Observation 2 to prove the assertion it is sufficient to show that no vertex in D well distinguishes Q n N ðdÞ. Assume to the contrary there exists c 2 D and two nonadjacent vertices a; b in a nontrivial component K of G½S such that c is adjacent to a and is nonadjacent to b (then c 6¼ d). Then c dominates T ðKÞ otherwise, say t 2 T ðKÞ, vertices x; c; a; t; b induce a P5. If T ðKÞ ¼ T , then one gets a contradiction to the definition of d (in fact one would have that c well distinguishes Q, with N Q ðdÞ N Q ðcÞ). Then assume that T ðKÞ T . By Observation 1 there exist two adjacent vertices k 1 ; k 2 2 K such that c is adjacent to k1 and is nonadjacent to k2. By statement (i), c is adjacent to d. Then c dominates T n T ðKÞ otherwise, say t 2 T n T ðKÞ, vertices k 2 ; k 1 ; c; d; t induce a P5. But then again one gets a contradiction to the definition of d. h 4. An approach for MWS in P5-free graphs Let us first introduce two definitions. Definition 1. Let F be any graph. Let us say that a P5-free graph G ¼ ðV ; EÞ is F-depending if there exists a partition of V, say fV 1 ; V 2 ; V 3 g, such that: (i) G½V 1 and G½V 2 are respectively F-free; (ii) G½V 3 is P3-free, and no element of V 1 [ V 2 distinguishes any component of G½V 3 . Definition 2. Let F be any graph. Let us denote as F* the graph consisting of P 1 þ F , P 1 þ P 2 , and a join between them. Let us denote as F+ the graph consisting of P 1 þ F , P 1 þ P 1 , and a join between them. Let F be any graph. Throughout this section assume that G is a prime ðP 5 ; F Þ-free graph.
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Lemma 5. Let ðx; yÞ be an edge of G, Q be a nonclique component of G½Aðx; yÞ, D be the set of vertices of G which distinguish Q. Let d 2 D well distinguish Q. Then D n N ½d admits a partition, say fX ; Y g, such that there exist qX 2 N Q ðdÞ dominating X, and qY 2 N Q ðdÞ dominating Y. Proof. Let S ¼ Q n N ðdÞ and T ¼ Q \ N ðdÞ. Let s 2 S such that there exists t 2 T nonadjacent to s* (according to the fact that d well distinguishes Q). Let X ¼ fc 2 D n N ½d : c is adjacent to s g. Then t* is adjacent to each vertex a of X, otherwise s ; a; y; d; t induce a P5. Let Y ¼ fc 2 D n N ½d : c is nonadjacent to s g. Let t 2 T \ N ðs Þ (such a vertex does exist since G is P5-free and G½Q is connected). Then t is adjacent to each vertex a of Y, otherwise a; y; d; t; s induce a P5. Since fX ; Y g is a partition of D n N ½d, the lemma follows. h Theorem 2. One of the following cases occurs: (i) for every edge ðx; yÞ of G, G½Aðx; yÞ is P3-free; (ii) there exists a vertex v of G such that each component of G½AðvÞ induces either a P3-free or a F-depending graph. Proof. Assume that statement (i) does not occur, and let us verify that then statement (ii) occurs. Then there exists an edge ðx; yÞ of G such that G½Aðx; yÞ is not P3-free. Let Q ¼ fQ1 ; . . . ; Qm g be the family of non-clique components of G½Aðx; yÞ. Let D ¼ fv 2 V : v distinguishes some element of Q g. According to Lemma 1, D is partitioned into fD1 ; . . . ; Dm g, where each vertex d i 2 Di distinguishes Qi and does not distinguish Qj for every j 6¼ i. Let ðd 1 ; . . . ; d m Þ 2 D1 ; . . . ; Dm . By Observation 2 one can assume without loss of generality that for i ¼ 1; . . . ; m: di well distinguishes Qi ; N Q ðd i Þ is maximal under inclusion in fN Qi ðcÞ : c well distinguishes Qi g. By Lemma 1, one can assume without loss of generality that d1 dominates Qj for every j > 1. Furthermore, by Lemma 2, one has that V n N ½d 1 ¼ ðD1 [ Q1 [ H [ KÞ n N ½d 1 , where H is the vertex-set of the clique-components of G½Aðx; yÞ, and K is the set of vertices in N ðxÞ\ N ðyÞ dominating Q1. By Lemma 5, D1 n N ½d 1 admits a partition, say fX ; Y g, such that there exist qX 2 N Q ðdÞ dominating X, and qY 2 N Q ðdÞ dominating Y. By definition of K and by (i) of Lemma 1, one obtains that
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G½ðK [ X Þ n N ½d 1 and G½ðK [ Y Þ n N ½d 1 are respectively F-free, otherwise a F arises involving x; y; d 1 and respectively either qX or qY . By (ii) of Lemma 4 and by construction, ðQ1 [ H Þ n N ½d 1 induces a P3-free graph. Furthermore, by (ii) of Lemma 1 and (i) of Lemma 4, no element of D1 n N ½d 1 distinguishes any component of G½ðQ1 [ H Þn N ½d 1 . By definition of K, no element of K n N ½d 1 distinguishes any component of G½Q1 n N ½d 1 . Also, no element k of K n N ½d 1 distinguishes any component of G½H n N ½d 1 : in fact otherwise, d 1 ; x; k and two elements of H n N ½d 1 would induce a P5. Then V n N ½d 1 is a F-depending graph with partition fðK [ X Þ n N ½d 1 ; Y n N ½d 1 ; ðQ1 [ H Þ n N ½d 1 g. Then statement (ii) occurs with v ¼ d 1 . h In what follows, let us prove that if one can efficiently solve MWS for F-depending graphs, then one can efficiently solve MWS for ðP 5 ; F Þfree graphs. To this end, let us assume that one can solve MWS for every F-depending graph in Oðnt Þ time for some natural t. Then let us introduce a recursive algorithm to solve MWS for ðP 5 ; F Þ-free graphs. The following notation will be used. Let G ¼ ðV ; EÞ be a graph. Then, let G0 ¼ ðV 0 ; E0 Þ denote the characteristic graph of G, that is: each vertex of G0 corresponds to a maximal homogeneous set of G, in order that two vertices of G0 are adjacent (nonadjacent) if the corresponding maximal homogeneous sets of G have a join in G (a co-join in G). In particular for each element h 2 V 0 , let Gh denote the maximal homogeneous set of G corresponding to h. Finally, let us say that G is gray if each of its components induces either a P3-free or a F-depending graph. Then by the above assumption one can solve MWS for every gray graph in maxfOðmÞ; Oðnt Þg time. Algorithm Green(G,w) Input: a ðP 5 ; F Þ-free graph G ¼ ðV ; EÞ with a weight function w on V. Output: the value a of a maximum weight stable set of G. begin a :¼ 0 while G contains a vertex v such that G½AðvÞ is gray do begin a :¼ maxfa; aw ðG½fvg [ AðvÞÞg (G½fvg [ AðvÞ is gray); G :¼ G½V n v; end;
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if G is prime then a :¼ maxfa; aw ðGÞg (G enjoys (i) of Theorem 2; then cf. Theorem 1); if G is not prime then begin compute the characteristic graph of G, say G0 ¼ ðV 0 ; E0 Þ; for each element h 2 V 0 do begin w0 ðhÞ :¼ GreenðGh ; wÞ; end; a :¼ maxfa; GreenðG0 ; w0 Þg; end; end. The correctness of the above algorithm is based on the results from modular decomposition theory mentioned in Section 2. Concerning the computational complexity of the above algorithm, one has that Theorems 1 and 2, and the mentioned results from modular decomposition theory imply that the most expansive step is either that of computing a maximum weight stable set of a P5-free graph G such that the anti-neighborhoods of each edge of G induce P3-free graphs (cf. Theorem 1), or that of computing a maximum stable set in a gray graph. Then, the computational complexity may be estimated as maxfOðn5 Þ; Oðntþ1 Þg time. Theorem 3. Let F be any graph. If one can solve MWS for every F-depending graph in Oðnt Þ time for some natural t, then one can solve MWS for ðP 5 ; F Þfree graphs in maxfOðn5 Þ; Oðntþ1 Þg time. Let us point out that the facts of this section (which come by applying the results of Section 2 to the anti-neighborhood of an edge) can be directly referred to ðP 5 ; F þ Þ-free graphs once one considers the anti-neighborhood of a single vertex. In particular, the step of computing a maximum weight stable set of a P5-free graph G such that the anti-neighborhoods of each edge of G induce P3-free graphs is reduced to that of computing a maximum weight stable set of a P5-free graph G such that the anti-neighborhoods of each vertex of G induce P3-free graphs. That can be formalized as follows. Theorem 4. Let F be any graph. If one can solve MWS for every F-depending graph in Oðnt Þ for some natural t, then one can solve MWS for ðP 5 ; F þ Þ-free graphs in maxfOðnmÞ; Oðntþ1 Þg time.
5. Application to certain P5-free graphs In this section let us apply the results of the previous sections to introduce some subclasses of P5-free graphs for which one can solve MWS in polynomial time. In particular, by the previous section, it is sufficient to look for graphs F such that one can solve MWS for every F-depending graph in polynomial time. Observation 3. Let F be any graph. Let G ¼ ðV ; EÞ be a F-depending graph, and let fV 1 ; V 2 ; V 3 g be a partition of V according to Definition 1. Then by (ii) of Definition 1, to solve MWS for G one can assume without loss of generality that V3 is a stable set. 5.1. ðP 5 ; S k Þ-Free graphs In this subsection let us prove that one can efficiently solve MWS for ðP 5 ; S k Þ-free graphs. That extends some of the results mentioned in Section 2, according to the fact that S k contains induced subgraphs of five vertices. Also, that extends (to the weighted case as well) the fact that one can efficiently solve MS for ðP 5 ; K 2;n Þ-free graphs (actually, in general one can efficiently solve MS for ðP 5 ; K n;n Þfree graphs, see e.g. [23,34]). Theorem 5. For any natural k, one can solve MWS for every Sk-depending graphs in Oðn2k2 mÞ time. Proof. Let G ¼ ðV ; EÞ be an Sk-depending graph, and let fV 1 ; V 2 ; V 3 g be a partition of V according to Definition 1. Then G½V 1 [ V 2 is S 2k1 -free. Let T be the family of stable sets of G½V 1 [ V 2 . Then jTj is Oðn2k2 Þ. One can solve MWS in G by solving MWS in G½V 3 and in G½V 3 n N ðT Þ for every T 2 T. By the above, that can be done in Oðn2k2 mÞ time. h Corollary 1. For any natural k, one can solve MWS for ðP 5 ; S k Þ-free graphs in maxfOðn5 Þ; Oðn2k1 mÞg time. 5.2. A note on ðP 5 ; K 2;3 Þ-free graphs A K 2;3 is a S þ 2 . By Theorems 4 and 5, one can solve MWS for ðP 5 ; K 2;3 )-free graphs in Oðn3 mÞ time. However, one can obtain a slight improvement as follows. Given any graph F, let us say that a F-depending graph G ¼ ðV ; EÞ, with a partition fV 1 ; V 2 ; V 3 g of V
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according to Definition 1, is basic if V1 and V2 are respectively cliques. Lemma 6. Let G be a prime ðP 5 ; K 2;3 Þ-free graph. Then, for every vertex v of G, G½AðvÞ is either P3-free or a basic S2-depending graph. Proof. That comes by applying the argument of Lemma 2, with the assumption that G is (P5,K 2;3 )free, to the anti-neighborhood of a single vertex x instead of that of an edge ðx; yÞ. By referring to the mentioned argument, in detail: G½ðX [ KÞn N ½d 1 is a clique; G½Y n N ½d 1 is a clique. Let us only show that D1 n N ½d 1 has a join to K n N ½d 1 (the other facts are quite immediate). Assume to the contrary that there exists d 2 D1 n N ½d 1 nonadjacent to some vertex k 2 K n N ½d 1 ; let q 2 Q1 be adjacent to both d and k, according to Lemma 3; then x; d 1 ; d; k; q induce a K 2;3 – contradiction. h One can easily verify that one can solve MWS for every basic S2-depending graph in OðnmÞ. Then by results from modular decomposition theory one obtains: Corollary 2. One can solve MWS for ðP 5 ; K 2;3 Þ-free graphs in Oðn2 mÞ time. 5.3. ðP 5 ; P 3 Þ-Free graphs In this subsection let us prove that one can efficiently solve MWS for ðP 5 ; P 3 Þ-free graphs. That extends some of the results mentioned in Section 2, according to the fact that P 3 contains induced subgraphs of five vertices. Lemma 7. Let G ¼ ðV ; EÞ be a connected P5-free graph, and let fX ; Y g be a partition of V such that G½Y is P3-free. Let CX be the family of component of G½X which are distinguished by some vertex of Y, and let W ¼ fx 2 X : x belongs to no element of C X g. Then there exists y 2 Y such that: (i) y contacts each element of CðX Þ; (ii) at most one element of CðX Þ is not dominated by y; (iii) W n N ½y has a co-join to Y n N ½y. Proof. About (i). The proof is by induction on k ¼ jCðX Þj. If k ¼ 1, then the assertion follows since G is connected. Then let us assume the assertion is true for k 1, and prove it is true for k. Assume to the contrary that there exists no y 2 Y contacting each element of CðX Þ. Let CðX Þ ¼ fX 1 ; . . . ; X k g.
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Then by the inductive assumption, there exists y 2 Y contacting each element of fX 1 ; . . . ; X k1 g. Let y k 2 Y be partial to Xk, and let ak ; bk be vertices of Xk, with ak adjacent to bk, such that yk is adjacent to ak and nonadjacent to bk. Let us show that fy k g has a co-join to fyg [ X 1 [ [ X k1 . In fact, otherwise: if yk is adjacent to y, then to avoid a P5 involving ak and bk one has that yk contacts each element of fX 1 ; . . . ; X k1 g (contradiction); if yk is nonadjacent to y, then yk is adjacent to some vertex x of X 1 [ [ X k1 , but then y; x; y k ; ak ; bk induce a P5 (contradiction). Then let P be the set of interior vertices of a shortest path from fy k g [ X k to fyg [ X 1 [ [ X k1 . Since G is P5-free, jP j 6 2. Assume jP j ¼ 2. Let P ¼ fu; vg, such that u contacts fy k g [ X k . If u 2 X , then u does not contact Xk; then u is adjacent to yk; then ak ; y k ; u; v and a vertex of fyg [ X 1 [ [ X k1 induce a P5. If u 2 Y , then v 2 X , otherwise y k ; u; v induce a P3 of G½Y ; then v is adjacent to y and does not contact any element of fX 1 ; . . . ; X k1 g; then y k ; u; v; y and a vertex of X 1 [ [ X k1 induce a P5. Assume jP j ¼ 1. Let P ¼ fug. If u 2 X , then u does not contact X 1 [ [ X k ; then u is adjacent to both yk and y; then a P5 arises. Then assume u 2 Y . If u is adjacent to yk, then u is nonadjacent to y, otherwise a P3 arises in G½Y ; then u contacts some element of fX 1 ; . . . ; X k1 g; in particular, u contacts each element of fX 1 ; . . . ; X k1 g, otherwise a P5 arises involving y; on the other hand, u contacts Xk, otherwise bk ; ak ; y k ; u and a vertex of X 1 [ [ X k1 induce a P5; but this is a contradiction. If u is nonadjacent to yk, then u contacts Xk; if u is adjacent to y, then u contacts each element of fX 1 ; . . . ; X k1 g (otherwise a P5 arises involving yk) – a contradiction; if u is nonadjacent to y, then u contacts some element of fX 1 ; . . . ; X k1 g, but then a P5 arises involving yk and y. About (ii). It follows since G is P5-free. About (iii). Let Y ¼ fy 2 Y : y enjoys statement ðiÞg. Let y 2 Y such that N W ðy Þ is maximal under inclusion in fN W ðyÞ : y 2 Y g. Let us prove that W n N ½y has a co-join to Y n N ½y . Assume to the contrary there exists y 2 Y n N ½y adjacent to a vertex x 2 W n N ½y . By construction, one can assume without loss of generality that fxg is a trivial component of G½X . Note that y 62 Y , otherwise: by definition of y* there exists x 2 W adjacent to y* and nonadjacent to y; then by definition of Y* a P5 arises.
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Then y 2 ðY n N ½y Þ n Y . Then fy; xg has a cojoin to fy g [ ðX n W Þ. Let P be the set of interior vertices of a shortest path from fy; xg to fy g [ ðX n W Þ. Since G is P5-free, jP j 6 2. Assume jP j ¼ 2. This case leads to a contradiction, similarly to the corresponding case of the argument concerning statement (i). Assume jP j ¼ 1. Let P ¼ fug. If u 2 X , then u does not contact either x or X n W ; then u is adjacent to both y and y*; then a P5 arises. Then assume u 2 Y . Assume that u is adjacent to y. Then u is nonadjacent to y*, otherwise a P3 arises in G½Y ; then u contacts X n W ; it follows that u 2 Y , otherwise a P5 arises involving y; but then (by the above, recalling that u is nonadjacent to y*) u is nonadjacent to x; then a P5 arises. Assume that u is nonadjacent to y. Then u is adjacent to x. If u is nonadjacent to y*, then a P5 arises involving y and x. If u is adjacent to y*, then u 2 Y , otherwise a P5 arises involving y and x; then, by definition of y* there exists x 2 W adjacent to y* and nonadjacent to u; then either y; x; u; y ; x (if x* is nonadjacent to y) or x ; y; x; u and an element of X n W (if x* is adjacent to y) induce a P5. h Lemma 8. Let G ¼ ðV ; EÞ be a P5-free graph, such that V admits a partition fX ; Y g, where X is a stable set, and G½Y is P3-free. Then one can solve MWS for G in OðnmÞ time. Proof. Let us show that there exists x 2 X such that one can solve MWS for G½V n N ðxÞ in OðmÞ time. Assume without loss of generality that G is connected. Then, since G is P5-free, there exists x 2 X contacting each component of G½Y ; in particular, at most one component of G½Y is not dominated by x. If x dominates each component of G½Y , then G½V n N ðxÞ is a stable set. If there exists a component of G½Y , say Q, not dominated by x, then one can solve MWS in G½V n N ðxÞ by solving MWS in G½V n ðN ðxÞ [ QÞ and in G½V n ðN ðxÞ [ N ðqÞÞ for every q 2 Q n N ðxÞ: since G½Y is P3-free, each of such graph has no edge. Then the assertion follows. Now, to prove the lemma, one can iterate the above argument to each component of G½V n fxg. h Lemma 9. Let G ¼ ðV ; EÞ be a P5-free graph, such that V admits a partition fX ; Y g, where G½X and G½Y are respectively P3-free. Then one can solve MWS for G in Oðn2 mÞ time.
Proof. Let us show that there exists y 2 Y such that one can solve MWS for G½V n N ðyÞ in OðnmÞ time. Assume without loss of generality that G is connected. Let CX be the family of components of G½V 1 which are distinguished by some vertex of Y, and W ¼ fx 2 X : x belongs to no element of CX}. Then, by Lemma 7, there exists y 2 Y such that: (i) y contacts each element of CX; (ii) at most one element of CX is not dominated by y; (iii) W n N ½y has a co-join to V 2 n N ½y. If y dominates each element of CX, then G½V n N ½y is a P3-free graph. Then, one can solve MWS for G½V n N ðyÞ in OðmÞ time. If there exists an element of CX, say Q, not dominated by y, then one can solve MWS in G½V n N ðyÞ by solving MWS in G½V n ðN ðyÞ [ QÞ and in G½V n ðN ðyÞ [ N ðqÞÞ for every q 2 Q n N ðyÞ. Since G½X is P3-free, each of such graphs is P3-free. Then, one can solve MWS for G½V n N ðyÞ in OðnmÞ time. Now, to prove the lemma, one can iterate the above argument to each component of G½V n fyg. h Theorem 6. One can solve MWS P3-depending graph in Oðn3 mÞ time.
for
every
Proof. Let G ¼ ðV ; EÞ be a P3-depending graph, and let fV 1 ; V 2 ; V 3 g be a partition of V according to Definition 1. Assume without loss of generality that G is connected. To prove the assertion, it is sufficient to show that there exists a vertex v 2 V (which can be easily detected) such that one can solve MWS for G½V n N ðvÞ in Oðn2 mÞ time. Then the assertion will follow by iteratively applying such an argument to each component of G½V n fvg. Let us consider the following two exhaustive cases. Case 1. G½V 1 [ V 2 is connected. Let C V 1 be the family of component of G½V 1 which are distinguished by some vertex of V 2 , and W ¼ fx 2 V 1 : x belongs to no element of C V 1 g. Then, by Lemma 7, there exists y 2 V 2 such that: (i) y contacts each element of C V 1 ; (ii) at most one element of C V 1 is not dominated by y; (iii) W n N ½y has a co-join to V 2 n N ½y. If y dominates each element of C V 1 , then V n N ½y admits a partition into a stable set (i.e., V 3 ) and a vertex-set inducing a P3-free graph (i.e., ðV 1 [ V 2 Þ n N ½y). Then, by Lemma 8, one can solve MWS for G½V n N ðyÞ in OðnmÞ time.
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If there exists an element of C V 1 , say Q, not dominated by y, then one can solve MWS in G½V n N ðyÞ by solving MWS in G½V n ðN ðyÞ [ QÞ and in G½V n ðN ðyÞ [ N ðqÞÞ for every q 2 Q n N ðyÞ. Since G½V 1 is P3-free, the vertex-set of each of such graphs admits a partition into a stable set and a vertex-set inducing a P3-free graph. Then, by Lemma 8, one can solve MWS for G½V n N ðyÞ in Oðn2 mÞ time. Case 2. G½V 1 [ V 2 is not connected. Let C V 1 [V 2 be the family of components of G½V 1 [ V 2 which are distinguished by some vertex of V 3 . Assume without loss of generality that each component Q of G½V 1 [ V 2 not belonging to C V 1 [V 2 is a trivial component of G½V 1 [ V 2 : in fact, otherwise, one can contract Q into one vertex q, with wðqÞ ¼ aw ðG½QÞ; that can be done in Oðn2 mÞ time, by Lemma 9. Since G is connected P5-free, there exists a vertex z 2 Z contacting each component of G½V 1 [ V 2 ; in particular, at most one nontrivial component of G½V 1 [ V 2 is not dominated by z. If z dominates each component of G½V 1 [ V 2 , then G½V n N ðzÞ is a stable set. Then assume there exists a component Q of G½V 1 [ V 2 not dominated by z. By the above, Q belongs to C V 1 [V 2 . If C V 1 [V 2 ¼ fQg, then one can apply an argument similar to that of Case 1 (by recalling that ðV 1 [ V 2 Þ n Q is a stable set) to show that there exists a vertex q 2 Q such that one can solve MWS for G½V n N ðqÞ in Oðn2 mÞ time. If C V 1 [V 2 6¼ fQg, then let Q 0 be an element of C V 1 [V 2 n fQg. Let us prove that no element of Z n fzg distinguishes Q n N ðzÞ. Assume to the contrary there exists z 2 Z n fzg distinguishing Q. Then let a; b be two vertices of Q n N ðzÞ, with a adjacent to b, such that z is adjacent to a and nonadjacent to b. Since z dominates Q 0 : z does not contact Q 0 (otherwise a P5 arises involving b; a and z); there exists z0 2 Z n N ðzÞ distinguishing Q 0 . Let a0 ; b0 be two vertices of Q 0 , with a 0 adjacent to b 0 , such that z 0 is adjacent to a 0 and nonadjacent to b 0 . Then z 0 contacts (and thus dominates, since G is P5-free) Q, otherwise z0 ; a0 ; z and two vertices of Q induce a P5. But then b0 ; a0 ; z0 ; a; z induce a P5. Then the assertion is proved. Then, one can contract Q n N ðzÞ into a single vertex q with wðqÞ ¼ aw ðG½Q n N ðzÞÞ. That can be done in Oðn2 mÞ time, by referring to Lemma 9 for each component of G½Q n N ðzÞ. Then G½V n N ½z
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can be reduced to a bipartite graph (for which one can solve MWS in OðmÞ). Then the lemma follows. h Corollary 3. One can solve MWS for ðP 5 ; P 3 Þ-free graphs in Oðn4 mÞ time. 5.4. A note on possible extensions Let us point out the following general fact. Lemma 10. Let F be any graph. If one can solve MWS for every F-depending graph in Oðnt Þ time (for some natural t), then one can solve MWS for every ðP 1 þ F Þ-depending graph in Oðntþ2 Þ time. Proof. Let G ¼ ðV ; EÞ be an ðP 1 þ F Þ-depending graph. Let fV 1 ; V 2 ; V 3 g be a partition of V according to Definition 1. One can solve MWS in G by solving MWS: (i) in G½V n ðN ðxÞ [ N ðyÞÞ for each pair of nonadjacent vertices x 2 V 1 and y 2 V 2 ; (ii) in G½V n V 1 ; (iii) in G½V n V 2 . Concerning step (i): such graphs are Fdepending. Concerning step (ii): note that for each vertex v 2 V 2 , one has that G½ðV 2 [ V 3 Þ n N ðvÞ is an F-depending graph. Concerning step (iii): this step is similar to step (ii), by symmetry. Then, by the assumption, the lemma follows. h Then one obtains the following fact which admits a repeated application. Theorem 7. Let F be any graph. If one can solve MWS for ðP 5 ; F Þ-free graphs in polynomial time, then one can solve MWS for ðP 5 ; ðP 1 þ F Þ Þ-free graphs in polynomial time. 5.5. A note on ðP 5 ; F 1 Þ-free graphs In this subsection let us prove that, by an approach similar to that introduced in Section 3, one can solve MWS for ðP 5 ; F 1 Þ-free graphs in Oðn4 mÞ time. That improves a recent result (see among other results in [11]) stating that one can solve MWS for ðP 5 ; F 1 Þ-free graphs in Oðn7 mÞ time. Throughout this subsection assume that G is a prime ðP 5 ; F 1 Þ-free graph. Throughout the next two lemmas of this subsection assume that: ðx; yÞ is an edge of G; Q is a non-clique component of G½Aðx; yÞ; D is the set of vertices of G which distinguish Q. Lemma 11. Let d 2 D well distinguish Q. Then d dominates fv 2 N ðxÞ \ N ðyÞ : fvg has a join to Qg.
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Proof. Let z 2 N ðxÞ \ N ðyÞ such that fzg has a join to Q. Then d is adjacent to z otherwise, since d well distinguishes Q, one has that x; y; z; d and two nonadjacent vertices of Q induce a F1. h Lemma 12. Let d 2 D well distinguish Q. Then D n N ½d is a stable set. Proof. Let S ¼ Q n N ðdÞ and T ¼ Q \ N ðdÞ. Let s 2 S such that there exists t 2 T nonadjacent to s* (according to the fact that d well distinguishes Q). Let c 2 D n N ½d and let us prove that: c is nonadjacent to s* and dominates T \ N ðs Þ. Assume to the contrary that c is adjacent to s*. Then, to avoid that s ; c; y; d; t induce a P5, c is adjacent to t*, but then x; y; d; c; t ; s induce a F 1 (contradiction). Then c is nonadjacent to s*. That implies that each element t 2 T \ N ðs Þ is adjacent to c, otherwise c; y; d; t; s induce a P5. Then c is nonadjacent to s* and dominates T \ N ðs Þ. Let c1 ; c2 2 D n N ½d. Assume to the contrary that c1 is adjacent to c2. Let t 2 T \ N ðs Þ (such a vertex does exist, since G is P5-free and G½Q is connected). Then by the above paragraph one has that c1 ; c2 ; y; d; t; s induce a F1 (contradiction). h A graph G ¼ ðV ; EÞ is almost bipartite if there exists a partition of V, say fV 1 ; V 2 g such that: V1 is a stable set; G½V 2 is P3-free; no element of V1 distinguishes any component (clique) of G½V 2 . Theorem 8. One of the following cases occurs: (i) for every edge ðx; yÞ of G, G½Aðx; yÞ is P3-free; (ii) there exists a vertex v of G such that each component of G½AðvÞ induces either a P3-free or a almost bipartite graph. Proof. The proof is very similar to that of Theorem 2. Assume that statement (i) does not occur, and let us verify that then statement (ii) occurs. Then there exists an edge ðx; yÞ of G such that G½Aðx; yÞ is not P3-free. Let Q ¼ fQ1 ; . . . ; Qm g be the family of non-clique components of G½Aðx; yÞ. Let D ¼ fv 2 V : v distinguishes some element of Q g. According to Lemma 1, D* is partitioned into fD1 ; . . . ; Dm g, where each vertex d i 2 Di distinguishes Qi and does not distinguish Qj for every j 6¼ i. Let ðd 1 ; . . . ; d m Þ 2 D1 ; . . . ; Dm . By Observation 2 one can assume without loss of generality that for i ¼ 1; . . . ; m: di well distinguishes Qi ; N Q ðd i Þ is maximal under inclusion in fN Qi ðcÞ : c well
distinguishes Qi g. By Lemma 1, one can assume without loss of generality that d1 dominates Qj for every j > 1. Furthermore, by Lemmas 2 and 11, one has that V n N ½d 1 ¼ ðD1 [ Q1 [ H Þ n N ½d 1 where H is the vertex-set of the clique-components of G½Aðx; yÞ. By Lemma 12, D1 n N ½d 1 is a stable set. By (ii) of Lemma 4 and by construction, ðQ1 [ H Þ n N ½d 1 induces a P3-free graph (in particular, Q1 has a cojoin to H). By (i) of Lemma 4, no element of D1 n N ½d 1 distinguishes any component of Q1 n N ½d 1 . Furthermore, D1 n N ½d 1 has a co-join to H n N ½d 1 : in fact assume to the contrary that there exists d 2 D1 n N ½d 1 adjacent to some vertex h 2 H n N ½d 1 ; let q 2 Q1 be adjacent to both d and d1, according to Lemma 3; then x; y; d 1 ; d; h; q induce a F1 (contradiction). Then statement (ii) occurs with v ¼ d 1 . h Observation 4. One can solve MWS for almost bipartite graphs in OðmÞ time. By Observation 4, by considering Theorem 8 instead of Theorem 2, and by re-denoting a graph G as gray if each of its components induces either a P3-free or a almost bipartite graph, one can directly verify that Algorithm Green(G, w) – with input a ðP 5 ; F 1 Þ-free graph – correctly solves MWS for ðP 5 ; F 1 Þ-free graphs in Oðn4 mÞ time. Theorem 9. One can solve MWS for ðP 5 ; F 1 Þ-free graphs in Oðn4 mÞ time. Acknowledgement I am indebted and grateful to the referees for their remarks, which were helpful to improve both the contents and the presentation of the manuscript. References [1] V.E. Alekseev, On the local restriction effect on the complexity of finding the graph independence number, in: Combinatorial-algebraic Methods in Applied Mathematics, Gorkiy University Press, Gorkiy, 1983, pp. 3–13 (in Russian). [2] V.E. Alekseev, A polynomial algorithm for finding largest independent sets in fork-free graphs, Discrete Analysis of Operations Research Series 1 (6) (1999) 3–19 (in Russian) (see also [3] for the English version). [3] V.E. Alekseev, A polynomial algorithm for finding largest independent sets in fork-free graphs, Discrete Applied Mathematics 135 (2004) 3–16. [4] V.E. Alekseev, On easy and hard hereditary classes of graphs with respect to the independent set problem, Discrete Applied Mathematics 132 (2004) 17–26.
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