Seidel complementation on ($P_5$, $House$, $Bull$)-free graphs

arXiv:1003.5458v1 [cs.DM] 29 Mar 2010 SEIDEL COMPLEMENTATION ON (P5, House, Bull)-FREE GRAPHS J.L. FOUQUET AND J.M. VANHERPE Abstract. 1. Seidel complementation Notation 1.1. N(v) denotes the neighborhood of the vertex v, N[v] = N(v) ∪{v} and N(v) = V −N(v). Given a graph G = (V, E) and a vertex v of G, the seidel complement of G on v inverses all edges between N(v) and V −N[v]. More formaly Definition 1.2. Let = (V, E) be an undirected graph and let v be a vertex of G. The Seidel complement oat v on G, denoted G ∗v is defined as follows : G = v = (V, E1 ∪E2 ∪E3) where E1 = {xy|xy ∈E, x ∈N[v], y ∈N[v]}, E2 = E ∩N(v) 2 and E3 = {xy|xy /∈E, x ∈N(v), y ∈N(v)}. Remark 1.3. [3] • G ∗v ∗v = G. • If G is a cograph and v is a vertex of G then G ∗v is a cograph. • G is prime with respect to modular decomposition if and only if G ∗v is prime with respect to modular decomposition • G ∗v = G ∗v 2. seidel complementation on (P5, P5,Bull)-free graphs Theorem 2.1. A graph G is (P5, House, Bull)-free if and only if for all vertex v of G, G ∗v is (P5, House, Bull)-free. Proof Let v be a vertex of G. Assume that G ∗v contains an induced subgraph, say H,which is isomorphic to either a P5 or a House or a Bull. Claim 2.1.1. The vertex v does not belong to H. 1991 Mathematics Subject Classification. 035 C. P 5 Bull House Figure 1. Forbidden configurations for (P5, House, Bull)-free graphs. 1 2 J.L. FOUQUET AND J.M. VANHERPE v N(v) N(v) v N(v) N(v) v v N(v) N(v) N(v) N(v) v N(v) N(v) v N(v) N(v) v N(v) N(v) v N(v) N(v) v N(v) N(v) Figure 2. The possible cases when v is a vertex of H. Proof Assume not. Figure 2 describes all cases that can occur whenever v is a vertex of H. It is not difficult to check that, in all cases, (G ∗v) ∗v would contain a subgraph isomorphic to P5 or to House or to Bull, a contradiction since G ∗v ∗v = G and G is assumed to be (P5, House, Bull)-free. □ In the following we suppose that the vertices of H are in N(v) ∪N(v), moreover H has vertices in both sets N(v) and N(v), otherwise H would be an induced subgraph of G ∗v ∗v = G, a contradiction. Figure 3 (resp Figure 4, Figure 5) describes all cases that can occur whenever H is isomorphic to a P5 (resp. a Bull, a House) and has at most two vertices in N(v). In all cases we get a contradiction with Claim 2.1.1 or a forbidden configuration appears in G ∗v ∗v. When H has more than 2 vertices in N(v) we get a similar contradiction in considering G. □ 3. seidel complementation on the modular decomposition tree of (P5, P5,Bull)-free graphs 3.1. On (P5, P5)-free graphs. We shall say that an induced subgraph of a (P5,(P5))-free graph is a buoy[1] whenever we can find a partition of its vertex set into 5 subsets Ai, i = 1, . . . , 5 (subsript i is to be taken modulo 5, such that Ai and Ai+1 are joined by every possible edge, while no possible edge are allowed between Ai and Aj when j ̸= i + 1 mod[5], and such that the Ai’s are maximal for these properties. Theorem 3.1. [1, 2] Let G be a connected (P5,P5)-free graph. If G contains an induced C5 then every C5 of G is contained into a buoy, and this buoy is either equal to G or is an homogeneous set of G. Corollary 3.2. Every prime (P5,P5)-free graph is either a C5 or is C5-free 3 N(v) N(v) N(v) v v N(v) N(v) v N(v) N(v) v N(v) v N(v) N(v) v N(v) N(v) v N(v) N(v) N(v) N(v) v N(v) v N(v) Figure 3. Cases : H is a P5 and has at most 2 vertices in N(v). v N(v) N(v) v N(v) N(v) v v N(v) N(v) v N(v) N(v) N(v) v N(v) v N(v) N(v) N(v) N(v) v N(v) N(v) v N(v) N(v) Figure 4. Cases : H is a Bull and has at most 2 vertices in N(v). 3.2. On (P5, P5, Bull)-free graphs. Theorem 3.3. [1] Let G be a prime graph. G is a P5HB-free graph if and only if one of the following conditions is satisfied (i) G is isomorphic to a C5; (ii) G is bipartite and P5-free; (iii) G is bipartite and P5-free. In a prime P5-free biparte graph the neighborhoods of two distinct vertices can- not overlap properly, thus : Proposition 3.4. Let G = (V, E) be a prime graph of n vertices. G is bipartite and P5-free iffthe following conditions are verified: 4 J.L. FOUQUET AND J.M. VANHERPE v v N(v) N(v) N(v) N(v) v v N(v) N(v) N(v) v N(v) v N(v) N(v) N(v) N(v) v N(v) N(v) v N(v) N(v) Figure 5. Cases : H is a House and has at most 2 vertices in N(v). b1 w5 w4 w3 w2 b5 b4 b3 b2 w1 Figure 6. A Prime P5-free bipartite graph. (i) There exists a partition of V (G) into two stable sets B = {b1, b2, . . . b n 2 } and W = {w1, w2, . . . w n 2 }. (ii) The neighbors of bi (i = 1 . . . n 2 ) are precisely w1, . . . w n 2 −i+1. 3.3. Seidel complementation of a Prime P5-free bipartite graph. Proposition 3.5. Let G = (B ∪W, E) be a prime bipartite P5-free graph such that B = {b1, . . . b n 2 } and W = {w1, . . . , w n 2 }, then G ∗bi is a prime P5-free bipartite graph together with the bipartition : B′ = {bi−1, bi−2, . . . bi, w1, w2, . . . , w n 2 }, W ′ = {bi+1, . . . , b n 2 , w n 2 , w n 2 −1, . . . w n 2 −i+1} Proof We have N(bi) = {w n 2 , w n 2 −1, . . . w n 2 −i+1} and N(bi) = {b1,

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