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Second kind maximum matching graph
Discrete Mathematics 323 (2014) 27–34
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Discrete Mathematics journal homepage: www.elsevier.com/locate/disc
Second kind maximum matching graph✩ Yan Liu ∗ , Zhengbiao Liu Department of Mathematics, South China Normal University, Guangzhou, 510631, PR China
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Article history: Received 5 January 2012 Received in revised form 18 January 2014 Accepted 21 January 2014 Available online 6 February 2014 Keywords: Maximum matching Second kind maximum matching graph Forbidden induced subgraph
abstract The second kind maximum matching graph M2 (G) of a graph G is the graph whose vertices are the maximum matchings of G such that two vertices M1 and M2 of M2 (G) are adjacent if and only if the symmetric difference of M1 and M2 induces either a cycle or a path of length 2. In this paper, we prove that the class of second kind maximum matching graphs has no forbidden induced subgraphs, and we characterize the graphs whose second kind maximum matching graphs are trees, or cycles, or complete graphs. © 2014 Published by Elsevier B.V.
1. Introduction and preliminaries The reader is referred to [1] for undefined terms and concepts. Only finite, undirected, simple graphs are considered. A graph is trivial if it has no edges and all other graphs are nontrivial. Let G be a graph. An edge subset M ⊆ E (G) is a matching of G if no two edges in M are incident with a common vertex. A matching M of G is a perfect matching if every vertex of G is incident with an edge in M. A matching M of G is a maximum matching if |M ′ | ≤ |M | for any matching M ′ of G. The symmetric difference of two subsets M1 and M2 of a set is denoted by M1 ⊕ M2 , that is, M1 ⊕ M2 = M1 ∪ M2 − M1 ∩ M2 . The maximum matching problem plays an important role in graph theory and combinatorial optimization due to its wide range of applications. For further understanding of maximum matchings, several graphs have been defined whose vertex set is the set of maximum matchings of a graph G. In the maximum matching graph, denoted by M (G), two maximum matchings of G are adjacent if their symmetric difference forms a path of length 2 (see Fig. 1). For a graph with no perfect matchings, the maximum matching graph has many good properties, as discussed in [3,5–8,12]. However, when G has a perfect matching, M(G) is trivial. In the perfect matching graph, denoted by P M(G), two perfect matchings of G are adjacent if their symmetric difference forms a single cycle; see [10,13–15] for results on this topic. In this paper, we combine the definitions of P M (G) and M (G), and introduce a new graph on the maximum matching, the second kind maximum matching graph, denoted by M2 (G). The vertex set of M2 (G) is the set of maximum matchings of G and two maximum matchings are adjacent in M2 (G) if their symmetric difference forms either a single cycle or a path of length 2. Thus M (G) is a spanning subgraph of M2 (G), and P M (G) = M2 (G) when G has a perfect matching. The definition of second kind maximum matching graph is thus useful for all graphs (see Fig. 1). The Gallai–Edmonds Structure Theorem is useful in this paper; stating it requires additional terminology. Denote by D(G) the set of all vertices in G that are missed by at least one maximum matching of G, denote by A(G) the set of vertices in V (G) − D(G) adjacent to at least one vertex in D(G), and let C (G) = V (G) − A(G) − D(G) (see Fig. 2). For
✩ This work is supported by the National Natural Science Foundation of China (No. 11061027).
∗
Corresponding author. E-mail address: [email protected] (Y. Liu).
0012-365X/$ – see front matter © 2014 Published by Elsevier B.V. http://dx.doi.org/10.1016/j.disc.2014.01.011
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Fig. 1. A graph G with M (G) and M2 (G).
Fig. 2. Gallai–Edmonds decomposition of G : C (G) = {v1 , v2 , v3 , v4 , v5 , v6 }, A(G) = {u1 , u2 }, D(G) = {x, y1 , y2 , y3 , z1 , z2 , z3 , z4 , z5 }.
Fig. 3. A graph G with P M (GC ), M2 (GAD ), and M2 (G).
convenience, we denote by GC , GAD and GD the subgraphs of G induced by C (G), A(G) ∪ D(G) and D(G), respectively. Thus G has a perfect matching if and only if D(G) = ∅ (it follows that A(G) = ∅), that is, G = GC . A connected graph G is factor-critical if G − v has a perfect matching for each vertex v ∈ V (G). Factor-critical graphs have an odd number of vertices and are not bipartite. Furthermore, if G is a factor-critical graph, then every maximum matching of G covers all vertices but one, D(G) = V (G), and δ(G) ≥ 2. A bipartite graph G with bipartition (A, B) has positive surplus (as viewed from A) if |NG (X )| > |X | for all non-empty subset X of A. So by the well-known Hall’s Theorem [1], if a bipartite graph G with bipartition (A, B) has positive surplus (as viewed from A), then every maximum matching of G covers all vertices in A and there exists a maximum matching missing v for each v ∈ B; it follows that A = A(G) and B = D(G). Let M be a maximum matching of G and V (M ) the set of vertices covered by M. The number of vertices missed by M is denoted by def(G). That is, def(G) = |V (G) − V (M )|. Clearly, if a bipartite graph G with bipartition (A, B) has positive surplus (as viewed from A), then def(G) = |B| − |A|. Proposition 1.1 (Gallai–Edmonds Structure Theorem [2,4]). If G is a graph, then (1) every component of GD is factor-critical; (2) GC has a perfect matching;
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Fig. 4. H is obtained from G by the method mentioned above.
Fig. 5. The structural relation of Gm−3 and Gm for some graph G.
Fig. 6. A graph G satisfying M2 (G) = Kn+1 .
(3) a matching M of G is a maximum matching if and only if M contains a maximum matching of each component of GD , contains a perfect matching of GC , and matches all vertices in A(G) with vertices in distinct components of GD ; (4) def(G) = c (GD ) − |A(G)|, where c (GD ) is the number of components of GD ; (5) for any non-empty subset X of A(G), there exist at least |X | + 1 components G′ of GD such that NG (X ) ∩ V (G′ ) ̸= ∅. By Proposition 1.1(3), a maximum matching of G uses no edges induced by A(G). Hence in this paper we delete all edges induced by A(G); this does not change M2 (G). By Proposition 1.1(1), if G is bipartite, then GD is trivial; so GAD is a bipartite graph with bipartition (A(G), D(G)) and has positive surplus as viewed from A(G) by Proposition 1.1(5). Proposition 1.2 ([12]). If G is a graph, then M (GAD ) is connected, each component of M (G) is isomorphic to M (GAD ), and the number of components of M (G) equals the number of perfect matchings of GC unless C (G) = ∅, in which case the number of components is 1. Proposition 1.3 ([13]). For every graph G, if G has at least two perfect matchings, then P M (G) is connected. Proposition 1.4 ([13]). For every graph G, if G has at least three perfect matchings, then there exists a Hamilton cycle C of P M(G) containing e for any e ∈ E (P M(G)).
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2. Structure of second kind maximum matching graphs The Cartesian product of graphs G1 and G2 , denoted by G1 G2 , is the graph with vertex set V (G1 ) × V (G2 ) such that
(x, u)(y, v) is an edge if x = y and uv ∈ E (G2 ) or if xy ∈ E (G1 ) and u = v . Proposition 2.1. If G is a graph, then M2 (G) ∼ = P M(GC )M2 (GAD ).
Proof. By Proposition 1.1, every maximum matching of G can be expressed as an ordered pair consisting of a perfect matching of GC and a maximum matching of GAD . To change a maximum matching of G via an alternating path of length 2 or an alternating cycle, we must change one component of the ordered pair in this way, since no maximum matching has an edge joining a vertex in C (G) to a vertex outside C (G) by Proposition 1.1. Hence by the definition of Cartesian product, we have M2 (G) ∼ = M2 (GC )M2 (GAD ). Since GC has a perfect matching, we have P M(G) = M2 (G). Hence M2 (G) ∼ = P M(GC )M2 (GAD ) (see Fig. 3). The connectivity κ(G) of G is the fewest number of vertices whose removal from G results in a disconnected or trivial graph. The following lemma is useful for the proofs of Corollary 2.3. Lemma 2.2 ([11]). If G1 and G2 are two nontrivial graphs, then
κ(G1 G2 ) = min{κ(G1 )|V (G2 )|, κ(G2 )|V (G1 )|, δ(G1 ) + δ(G2 )}. Corollary 2.3. If G is a graph with no isolated vertices, then M2 (G) is connected. Proof. If C (G) = ∅ or GC has only one perfect matching, then M (G) = M (GAD ) and M (GAD ) is connected by Proposition 1.2, so M2 (G) is connected. If D(G) = ∅, then G = GC and M2 (G) = P M (G); by Proposition 1.3, M2 (G) is connected. If GC has at least two perfect matchings and D(G) ̸= ∅, then P M (GC ) is connected by Proposition 1.3 and GAD has at least two maximum matchings, since G has no isolated vertices; by Proposition 2.1 and Lemma 2.2, M2 (G) is connected. 3. Forbidden induced subgraphs of second kind maximum matching graphs A graph is a forbidden induced subgraph for a class of graphs if it does not appear as an induced subgraph of any graph in the class. Eroh [3] and Liu [9] discovered many forbidden induced subgraphs for the class of maximum matching graphs. In contrast, we prove next that the class of second kind maximum matching graphs has no forbidden induced subgraphs. Theorem 3.1. For every graph G, there exists a graph H such that M2 (H ) contains G as an induced subgraph. Proof. For each vertex v ∈ V (G), we construct a cycle Cv of length lv , where lv = 4 if dG (v) ≤ 1, and lv = 2dG (v) otherwise. Note that Cv has a perfect matching. If NG (v) = {u1 , . . . , us }, where s ≥ 2, then we choose a perfect matching Mv of Cv and name its edges ev u1 , . . . , ev us (see w in Fig. 4). If NG (v) = {u}, then we choose a perfect matching Mv of Cv and name its edges ev u and ev ′ v (see y in Fig. 4). If dG (v) = 0, then we choose a perfect matching Mv of Cv and name its edges ev ′ v and ev ′′ v (see x in Fig. 4). Now we construct a new graph H. Let H be the union of the cycles Cv for v ∈ V (G), arranged so that (1) if uv ̸∈ E (G), then V (Cu ) ∩ V (Cv ) = ∅; and (2) if uv ∈ E (G), then E (Cu ) ∩ E (Cv ) = {ev u } and |V (Cu ) ∩ V (Cv )| = 2 (note that Cu and Cv have only one common edge, which is euv , and exactly two common vertices which are endvertices of euv ). Fig. 4 shows an example of the graph H obtained from a graph G as just defined. Let M = v∈V (G) Mv . Then M is a perfect matching of H. By the construction of H, we have E (Cu ) ∩ E (Cw ) ̸= ∅ (in this case, E (Cu ) ∩ E (Cw ) = {euw }) if and only if uw ∈ E (G). Let Mv′ = M ⊕ E (Cv ) for each v ∈ V (G); also Mv′ is a perfect matching = {Mv′ |v ∈ V (G)}. of H. Let M such that α(v) = Mv′ for each vertex v of G. By the definition of Mv′ , α is a bijection. If Let α be a mapping from V (G) to M ′ ′ ′ ′ = (M ⊕ E (Cu ))⊕(M ⊕ E (Cw )) = E (Cu )⊕ E (Cw ), Mu and Mw are adjacent in M2 (H ), then Mu′ ⊕ Mw forms a cycle. Since Mu′ ⊕ Mw also E (Cu ) ⊕ E (Cw ) forms a cycle. Hence E (Cu ) ∩ E (Cw ) ̸= ∅. By the construction of H, we have E (Cu ) ∩ E (Cw ) = {euw }, so ′ uw ∈ E (G). If uw ∈ E (G), then E (Cu ) ∩ E (Cw ) = {euw } and E (Cu ) ⊕ E (Cw ) forms a cycle. Since Mu′ ⊕ Mw = E (Cu ) ⊕ E (Cw ), ′ ′ . we have Mu and Mw are adjacent in M2 (H ). Therefore, G is isomorphic to the subgraph of M2 (H ) induced by M 4. Some characterizations First, we characterize the graphs whose second kind maximum matching graphs are trees. We need two lemmas. An edge of G is a pendant edge if it has an endvertex of degree 1. Lemma 4.1 ([9]). If G is a bipartite graph with bipartition (A, B), then G has positive surplus (as viewed from A) if and only if G has a spanning forest F such that dF (u) = 2 for all u ∈ A.
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In fact, in Lemma 4.1, also F has positive surplus (as viewed from A) and def(F ′ ) = 1 for each component F ′ of F , so def(G) = def(F ) = c (F ), where c (F ) is the number of components of F . Thus we have the following corollary. Corollary 4.1. If a bipartite graph G with bipartition (A, B) has positive surplus (as viewed from A), then G is connected and dG (u) = 2 for all u ∈ A if and only if G is a tree and def(G) = 1. Lemma 4.2 ([6]). If G is a graph, then M (G) is a tree if and only if either C (G) = ∅ or GC has exactly one perfect matching, and GAD is a bipartite graph with bipartition (A(G), D(G)) such that def(GAD ) = 1 and the last graph Gm in the graph sequence (G0 = GAD , G1 , . . . , Gm ) is K1 , where Gi+1 is obtained from Gi by deleting all vertex pairs that can induce a pendant edge in Gi for 0 ≤ i ≤ m − 1. If a bipartite graph G with bipartition (A, B) has positive surplus (as viewed from A) and def(G) = 1, then dG (u) ≥ 2 for any u ∈ A and |B| = |A| + 1. Thus G is connected and there is at most one vertex of degree 1 in NG (u) for any vertex u ∈ A (otherwise, if dG (v1 ) = dG (v2 ) = 1 and v1 , v2 ∈ NG (u), then |NG (A − {u})| ≤ |B − {v1 , v2 }| = |A| − 1, a contradiction). So the set of all pendant edges of G, denoted by M0 , is a matching of G. Further, let G1 = G − V (M0 ), A1 = A − V (M0 ) and B1 = B − V (M0 ). It is easy to check that G1 has the bipartition (A1 , B1 ) and has the same properties with G (i.e., also G1 has positive surplus (as viewed from A1 ) and def(G1 ) = 1). If G1 also has a vertex of degree 1, then we continue to construct a graph G2 by the same method as above until some graph Gm satisfies that δ(Gm ) ≥ 2 or Gm is K1 . Thus we obtain a graph sequence (G0 = G, G1 , . . . , Gm ) and a matching sequence (M0 , M1 , . . . , Mm−1 ) such that Gi has the bipartition (Ai , Bi ), Ai = Ai−1 − V (Mi−1 ) and Bi = Bi−1 − V (Mi−1 ) for 2 ≤ i ≤ m (see Fig. 5). Theorem 4.3. If G is a graph with no isolated vertices, then M2 (G) is a tree if and only if G satisfies one of the following two conditions: (1) G has at most two perfect matchings; (2) def(GAD ) = 1, GAD is a tree and either C (G) = ∅ or GC has exactly one perfect matching. Proof. ⇐ If G has at most two perfect matchings, then M2 (G) = K1 or K2 by Proposition 1.3; so M2 (G) is a tree. If def(GAD ) = 1, GAD is a tree and either C (G) = ∅ or GC has exactly one perfect matching, then M2 (G) ∼ = M2 (GAD )by Proposition 2.1 and M2 (GAD ) = M (GAD ); so GAD is a bipartite graph with bipartition (A(G), D(G)) and has positive surplus (as viewed from A(G)) by Proposition 1.1(5). Therefore, the last graph Gm in the graph sequence (G0 = GAD , G1 , . . . , Gm ) is K1 , where Gi+1 is obtained from Gi by deleting all vertex pairs that induce a pendant edge in Gi for 0 ≤ i ≤ m − 1. By Lemma 4.2, M (GAD ) is a tree. Thus M2 (G) is a tree. ⇒ If M2 (G) is a tree, then GC has at most two perfect matchings by Proposition 1.4 and Proposition 2.1. (Otherwise, M2 (GC ) has a cycle. Then M2 (G) has a cycle, a contradiction.) Case 1: GC has exactly two perfect matchings. Thus M2 (GC ) = K2 . If D(G) ̸= ∅, then GAD has at least two maximum matchings, since G has no isolated vertices. By Corollary 2.3, M2 (GAD ) is connected. By Proposition 2.1, M2 (G) has a cycle, a contradiction. Thus D(G) = ∅. Hence A(G) = ∅. It follows that G satisfies condition (1). Case 2: C (G) = ∅ or GC has a unique perfect matching. By Proposition 2.1, we have M2 (G) ∼ = M2 (GAD ). By Proposition 1.2, M(G) is connected. So M(G) is a connected spanning subgraph of M2 (G). Since M2 (G) is a tree, we have M (G) = M2 (G). Hence M (G) is a tree. By Lemma 4.2, GAD is a bipartite graph with def(GAD ) = 1, has positive surplus (as viewed from A(G)) and the last graph Gm in the graph sequence (G0 = GAD , G1 , . . . , Gm ) is K1 , where Gi+1 = Gi − V (Mi ), Gi+1 has the bipartition (Ai+1 , Bi+1 ), Mi is the set of all pendant edges of Gi , Ai+1 = Ai − V (Mi ), and Bi+1 = Bi − V (Mi ) for 0 ≤ i ≤ m − 1. Therefore, we only need to show that GAD has no cycles. We prove it by contradiction. Suppose that Gt −1 has a cycle but Gt has no cycles without loss of generality. Since Gt is connected and has no cycles, Gt is a tree. By Corollary 4.1, dGt (v) = 2 for any v ∈ At . Since Gt −1 has a cycle, there exists a vertex w ∈ At −1 such that dGt −1 (w) ≥ 3 by Corollary 4.1. Thus w ̸∈ V (Gt ). Since dGt −1 (w) ≥ 3, we can assume that w1 , w2 ∈ NGt −1 (w)∩Bt and ww3 ∈ Mt −1 . Since Gt has positive surplus (as viewed from At ), we have D(Gt ) = Bt . So we can assume that Mw1 and Mw2 are two maximum matchings of Gt missing w1 and w2 , respectively. Since def(Gt ) = 1, Mw1 ⊕ Mw2 forms a path P of even length joining w1 and w2 . Let C = P + ww1 + ww2 . Thus C is an even cycle. Let M = Mw1 ∪ (Mt −1 − ww3 + ww1 ) ∪ Mt −2 ∪ · · · ∪ M0 and M ′ = Mw2 ∪ (Mt −1 − ww3 + ww2 ) ∪ Mt −2 ∪ · · · ∪ M0 . Thus M and M ′ are two maximum matchings of GAD , and M ⊕ M ′ forms C . Hence MM ′ ∈ E (M2 (G)). Since MM ′ ̸∈ E (M (G)), we have M2 (G) ̸= M (G), a contradiction. Second, we characterize the graphs whose second kind maximum matching graphs are complete graphs. Lemma 4.4. If M , M1 , and M2 are three maximum matchings of G such that M1 ⊕ M forms a cycle C and M2 = M − uv + uw , then M1 M2 ̸∈ E (M2 (G)). Proof. If uv ∈ E (C ), then M1 ⊕ M2 = C − uv + uw ; so M1 ⊕ M2 induces a path of length more than 2. If uv ̸∈ E (C ), then C is an (M1 , M2 ) alternating cycle; so M1 ⊕ M2 consists of a cycle and a path of length 2. Hence M1 M2 ̸∈ E (M2 (G)). Lemma 4.5. If G is a graph with no isolated vertices, then M ⊕ M ′ forms a cycle for any two maximum matchings M and M ′ of G that are adjacent in M2 (G) if and only if G has a perfect matching.
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Proof. If G has a perfect matching, then the statement holds clearly. Conversely, it suffices to prove that D(G) = ∅. We prove it by contradiction. Suppose that D(G) ̸= ∅. Let u ∈ D(G) and uv ∈ E (G). Thus there exists a maximum matching M of G missing u. By the maximality of M, there exists an edge vw ∈ M. Let M ′ = M − vw + uv , also M ′ is a maximum matching of G. Hence MM ′ ∈ E (M2 (G)). But M ⊕ M ′ is not a cycle, a contradiction. So D(G) = ∅. Thus G has a perfect matching. Lemma 4.6 ([6]). If G is a graph with no isolated vertices, then M (G) is a complete graph if and only if either C (G) = ∅ or GC has a unique perfect matching and GAD = K3 or GAD = K1,n . Theorem 4.7. If G is a graph with no isolated vertices, then M2 (G) is a complete graph if and only if G satisfies one of the following two conditions: (1) C (G) = ∅ or GC has a unique perfect matching and GAD = K3 or GAD = K1,n ; (2) G = GC and E (C1 ) ⊕ E (C2 ) forms a cycle for any perfect matching M of G and any two M-alternating cycles C1 and C2 . Proof. ⇒ If M2 (G) is a complete graph, either M1 ⊕ M2 forms a cycle for any two maximum matchings M1 and M2 of G or M1 ⊕ M2 forms a path of length 2 for any two maximum matchings M1 and M2 of G by Lemma 4.4. Case 1: M1 ⊕ M2 forms a path of length 2 for any two maximum matchings M1 and M2 of G. Thus M2 (G) = M (G). By Lemma 4.6, either C (G) = ∅ or GC has a unique perfect matching and GAD = K3 or GAD = K1,n . Case 2: M1 ⊕ M2 forms a cycle for any two maximum matchings M1 and M2 of G. By Lemma 4.5, we have G = GC , i.e., G has a perfect matching. Set V (M2 (G)) = {M1 , . . . , Mm }. Since M2 (G) is a complete graph, we can assume that M1 ⊕ Mi forms an M1 -alternating cycle Ci for 2 ≤ i ≤ m. Thus all M1 -alternating cycles are C2 , . . . , Cm . Since E (Ci ) ⊕ E (Cj ) = Mi ⊕ Mj for 2 ≤ i < j ≤ m, E (Ci ) ⊕ E (Cj ) forms a cycle. ⇐ If C (G) = ∅ or GC has a unique perfect matching and GAD = K3 or GAD = K1,n , then M2 (G) = M2 (GAD ) and M2 (GAD ) = M (GAD ); so M2 (G) is a complete graph by Lemma 4.6. If G = GC , then G has a perfect matching; so the symmetric difference of any two adjacent maximum matchings of G forms a cycle. We choose a perfect matching M of G and find all M-alternating cycles C1 , . . . , Ct . Let Mi = M ⊕ E (Ci ) for 1 ≤ i ≤ t, also Mi is a perfect matching of G and NM2 (G) (M ) = {M1 , . . . , Mt }. If the symmetric difference of any two Malternating cycles is a cycle, then Mi ⊕ Mj = (M ⊕ E (Ci )) ⊕ (M ⊕ E (Cj )) = E (Ci ) ⊕ E (Cj ) forms a cycle, it follows that Mi and Mj are adjacent in M2 (G) for any 1 ≤ i < j ≤ t. Claim. V (M2 (G)) = {M , M1 , . . . , Mt }. We prove it by contradiction. Let A = V (M2 (G)) − {M , M1 , M2 , . . . , Mt } ̸= ∅. By Corollary 2.3, M2 (G) is connected. Thus there exist a perfect matching M ′ ∈ A and some l (1 ≤ l ≤ t) such that M ′ Ml ∈ E (M2 (G)). Note that M ′ M ̸∈ E (M2 (G)). Denote by C the cycle formed by Ml ⊕ M ′ . Since both C and Cl are Ml -alternating cycles, E (C ) ⊕ E (Cl ) forms a cycle. So M ⊕ M ′ = (Ml ⊕ E (Cl )) ⊕ (Ml ⊕ E (C )) = E (Cl ) ⊕ E (C ), it follows that M ′ M ∈ E (M2 (G)), a contradiction. Hence A = ∅. Thus M2 (G) is a complete graph (see Fig. 6). Finally, we characterize the graphs whose second kind maximum matching graphs are cycles. Lemma 4.8 ([8]). If G is a graph and M is a maximum matching of G, then the degree of M in M (G) is dM(G) (M ) =
dG (v).
v∈D(G)−V (M )
Lemma 4.9 ([6]). If G is a graph with no isolated vertices, then M (G) is a cycle if and only if either C (G) = ∅ or GC has a unique perfect matching and GAD = 2K1,2 , C2n+1 or K1,3 . Theorem 4.10. If G is a graph with no isolated vertices, then (1) when D(G) = ∅, M2 (G) is a cycle if and only if G satisfies one of the following two conditions: (a) G has exactly three perfect matchings; (b) G has exactly four perfect matchings and there exists a perfect matching M such that there are exactly two M-alternating cycles C1 and C2 that are disjoint. (2) when D(G) ̸= ∅ and GC has at most one perfect matching, M2 (G) is a cycle if and only if GAD ∈ {B∗ , C2n+1 , 2K1,2 , K1,3 }, where B∗ is shown in Fig. 7. (3) when D(G) ̸= ∅ and GC has at least two perfect matchings, M2 (G) is a cycle if and only if GAD = K1,2 and GC has exactly two perfect matchings. Proof. ⇒ If M2 (G) is a cycle, then dM2 (G) (M ) = 2 for any maximum matching M of G. Case 1: D(G) = ∅. Let M be a perfect matching of G. Since dM2 (G) (M ) = 2, there exist exactly two M-alternating cycles C1 and C2 . Let Mi = M ⊕ E (Ci ) for 1 ≤ i ≤ 2. Note that M1 and M2 are two perfect matchings of G and MM1 , MM2 ∈ E (M2 (G)).
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C2n u1
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v0
Fig. 7. Graph B∗ .
If E (C1 ) ∩ E (C2 ) = ∅, we have M1 ⊕ E (C2 ) = M2 ⊕ E (C1 ). Let M ′ = M1 ⊕ E (C2 ). So M ′ M1 , M ′ M2 ∈ E (M2 (G)). It follows that MM1 M ′ M2 M is a 4-cycle of M2 (G). Hence G satisfies the condition (b) of (1) in Theorem 4.10. We can assume that E (C1 ) ∩ E (C2 ) ̸= ∅. Since M1 ⊕ M2 = E (C1 ) ⊕ E (C2 ), E (C1 ) ⊕ E (C2 ) must form the union of some disjoint (M1 , M2 )-alternating cycles. If E (C1 ) ⊕ E (C2 ) consists of at least two (M1 , M2 )-alternating cycles, then there exist at least three M1 -alternating cycles, including C1 . Thus dM2 (G) (M1 ) ≥ 3, a contradiction. Therefore, E (C1 ) ⊕ E (C2 ) forms a cycle. Thus MM1 M2 M is a cycle. It follows that M2 (G) = C3 . Hence G satisfies the condition (a) of (1) in Theorem 4.10. Case 2: D(G) ̸= ∅ and GC has at most one perfect matching. By Proposition 2.1, we have M2 (G) ∼ = M2 (GAD ). Thus M2 (GAD ) is a cycle. By Proposition 1.2, M(G) = M(GAD ) and M(GAD ) is connected. Since M (GAD ) is a spanning subgraph of M2 (GAD ), M (GAD ) is a cycle or a path. Subcase 1: M (GAD ) is a cycle. By Lemma 4.9, we have GAD = 2K1,2 , C2n+1 or K1,3 . Subcase 2: M (GAD ) is a path. Let M (GAD ) = M1 M2 . . . Mm , where M1 , . . . , Mm are all maximum matchings of GAD and M1 Mm ̸∈ E (M (GAD )). Thus M2 (GAD ) = M (GAD ) + M1 Mm = M1 . . . Mm M1 . It follows that M1 ⊕ Mm forms an even cycle C2n . Since M(GAD ) is a path, GAD is a bipartite graph with bipartition (A(G), D(G)), has positive surplus as viewed from A(G), and def(GAD ) = 1 by Lemma 4.2. Thus |D(G)| = |A(G)| + 1. Since M1 ⊕ Mm is a cycle and def(GAD ) = 1, there exists exactly one vertex v0 in D(G) missed by both M1 and Mm , and all other vertices are covered by both M1 and Mm . Since dM(GAD ) (M1 ) = dM(GAD ) (Mm ) = 1, we have dGAD (v0 ) = 1 by Lemma 4.8. Hence we can assume that v0 u1 ∈ E (G). It follows that u1 is covered by both M1 and Mm . Furthermore, since dM(GAD ) (Mi ) = 2 for any 2 ≤ i ≤ m − 1, we have dGAD (v) = 2 for any v ∈ D(G) − v0 by Lemma 4.8. Claim 1. u1 ∈ V (C2n ). To the contrary, if u1 ̸∈ V (C2n ), then there exists a vertex v1 ∈ D(G) − v0 such that u1 v1 ∈ M1 ∩ Mm . It follows that no cycles on 2n vertices that contains v1 . Let M = M1 − u1 v1 + u1 v0 and M ′ = Mm − u1 v1 + u1 v0 ; so M ⊕ M ′ = M1 ⊕ Mm , it follows that MM ′ ̸∈ E (M (GAD )), but MM ′ ∈ E (M2 (GAD )), a contradiction. Claim 2. GAD = B∗ (see Fig. 7). By Claim 1, we have dGAD (u1 ) ≥ 3. Since dG (v0 ) = 1 and dG (v) = 2 for any v ∈ D(G) − v0 , we have
|E (GAD )| =
dG (v) = 2|A(G)| + 1.
v∈D(G)
Note that dGAD (u) ≥ 2 for any u ∈ A(G). It follows that dGAD (u1 ) = 3 and dGAD (u) = 2 for any u ∈ A(G) − u1 . Thus C2n + u1 v0 (that is, B∗ ) is a component of GAD . Since GAD is connected, we have GAD = B∗ . Let A1 = A(G) − V (C2n ) and D1 = D(G) − V (C2n ) − v0 . Case 3: D(G) ̸= ∅ and GC has at least two perfect matchings. Since D(G) ̸= ∅, GAD has at least two maximum matchings. By Proposition 1.2, M2 (GAD ) has an edge. If GC has at least three perfect matchings, then P M (GC ) has a cycle by Proposition 1.4; so M2 (G) is not a cycle by Proposition 2.1, a contradiction. Similarly, if GAD has at least three maximum matchings, then we can show that also M2 (G) is not a cycle. Therefore, GAD = K1,2 and GC have exactly two maximum matchings. ⇐ Conversely, we consider the following cases. Case 1: D(G) = ∅. Thus A(G) = ∅. It follows that G = GC . If G has exactly three perfect matchings, then by Proposition 1.4, we have M2 (G) = C3 . If G has exactly four perfect matchings and there exists a perfect matching M such that there are exactly two M-alternating cycles C1 and C2 that are disjoint, then M , M1 = M ⊕ E (C1 ), M2 = M ⊕ E (C2 ), and M ′ = M1 ⊕ E (C2 ) are the four perfect matchings of G such that MM1 , MM2 , M ′ M1 , M ′ M2 ∈ E (M2 (G)) and M1 M2 , MM ′ ̸∈ E (M2 (G)). Hence M2 (G) = MM1 M ′ M2 M (i.e., M2 (G) = C4 ). Case 2: D(G) ̸= ∅. If GC has at most one perfect matching and GAD ∈ {B∗ , C2n+1 , 2K1,2 , K1,3 }, then M2 (G) ∼ = M2 (GAD ) by Proposition 2.1; so it is easy to show that M2 (B∗ ) = Cn+2 , M2 (C2n+1 ) = C2n+1 , M2 (K1,3 ) = C3 , and M2 (2K1,2 ) = C4 . Therefore, M2 (G) is a cycle. If GAD = K1,2 and GC has exactly two perfect matchings, then it is easy to show that M2 (G) = C4 .
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Discrete Mathematics journal homepage: www.elsevier.com/locate/disc
Second kind maximum matching graph✩ Yan Liu ∗ , Zhengbiao Liu Department of Mathematics, South China Normal University, Guangzhou, 510631, PR China
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Article history: Received 5 January 2012 Received in revised form 18 January 2014 Accepted 21 January 2014 Available online 6 February 2014 Keywords: Maximum matching Second kind maximum matching graph Forbidden induced subgraph
abstract The second kind maximum matching graph M2 (G) of a graph G is the graph whose vertices are the maximum matchings of G such that two vertices M1 and M2 of M2 (G) are adjacent if and only if the symmetric difference of M1 and M2 induces either a cycle or a path of length 2. In this paper, we prove that the class of second kind maximum matching graphs has no forbidden induced subgraphs, and we characterize the graphs whose second kind maximum matching graphs are trees, or cycles, or complete graphs. © 2014 Published by Elsevier B.V.
1. Introduction and preliminaries The reader is referred to [1] for undefined terms and concepts. Only finite, undirected, simple graphs are considered. A graph is trivial if it has no edges and all other graphs are nontrivial. Let G be a graph. An edge subset M ⊆ E (G) is a matching of G if no two edges in M are incident with a common vertex. A matching M of G is a perfect matching if every vertex of G is incident with an edge in M. A matching M of G is a maximum matching if |M ′ | ≤ |M | for any matching M ′ of G. The symmetric difference of two subsets M1 and M2 of a set is denoted by M1 ⊕ M2 , that is, M1 ⊕ M2 = M1 ∪ M2 − M1 ∩ M2 . The maximum matching problem plays an important role in graph theory and combinatorial optimization due to its wide range of applications. For further understanding of maximum matchings, several graphs have been defined whose vertex set is the set of maximum matchings of a graph G. In the maximum matching graph, denoted by M (G), two maximum matchings of G are adjacent if their symmetric difference forms a path of length 2 (see Fig. 1). For a graph with no perfect matchings, the maximum matching graph has many good properties, as discussed in [3,5–8,12]. However, when G has a perfect matching, M(G) is trivial. In the perfect matching graph, denoted by P M(G), two perfect matchings of G are adjacent if their symmetric difference forms a single cycle; see [10,13–15] for results on this topic. In this paper, we combine the definitions of P M (G) and M (G), and introduce a new graph on the maximum matching, the second kind maximum matching graph, denoted by M2 (G). The vertex set of M2 (G) is the set of maximum matchings of G and two maximum matchings are adjacent in M2 (G) if their symmetric difference forms either a single cycle or a path of length 2. Thus M (G) is a spanning subgraph of M2 (G), and P M (G) = M2 (G) when G has a perfect matching. The definition of second kind maximum matching graph is thus useful for all graphs (see Fig. 1). The Gallai–Edmonds Structure Theorem is useful in this paper; stating it requires additional terminology. Denote by D(G) the set of all vertices in G that are missed by at least one maximum matching of G, denote by A(G) the set of vertices in V (G) − D(G) adjacent to at least one vertex in D(G), and let C (G) = V (G) − A(G) − D(G) (see Fig. 2). For
✩ This work is supported by the National Natural Science Foundation of China (No. 11061027).
∗
Corresponding author. E-mail address: [email protected] (Y. Liu).
0012-365X/$ – see front matter © 2014 Published by Elsevier B.V. http://dx.doi.org/10.1016/j.disc.2014.01.011
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Fig. 1. A graph G with M (G) and M2 (G).
Fig. 2. Gallai–Edmonds decomposition of G : C (G) = {v1 , v2 , v3 , v4 , v5 , v6 }, A(G) = {u1 , u2 }, D(G) = {x, y1 , y2 , y3 , z1 , z2 , z3 , z4 , z5 }.
Fig. 3. A graph G with P M (GC ), M2 (GAD ), and M2 (G).
convenience, we denote by GC , GAD and GD the subgraphs of G induced by C (G), A(G) ∪ D(G) and D(G), respectively. Thus G has a perfect matching if and only if D(G) = ∅ (it follows that A(G) = ∅), that is, G = GC . A connected graph G is factor-critical if G − v has a perfect matching for each vertex v ∈ V (G). Factor-critical graphs have an odd number of vertices and are not bipartite. Furthermore, if G is a factor-critical graph, then every maximum matching of G covers all vertices but one, D(G) = V (G), and δ(G) ≥ 2. A bipartite graph G with bipartition (A, B) has positive surplus (as viewed from A) if |NG (X )| > |X | for all non-empty subset X of A. So by the well-known Hall’s Theorem [1], if a bipartite graph G with bipartition (A, B) has positive surplus (as viewed from A), then every maximum matching of G covers all vertices in A and there exists a maximum matching missing v for each v ∈ B; it follows that A = A(G) and B = D(G). Let M be a maximum matching of G and V (M ) the set of vertices covered by M. The number of vertices missed by M is denoted by def(G). That is, def(G) = |V (G) − V (M )|. Clearly, if a bipartite graph G with bipartition (A, B) has positive surplus (as viewed from A), then def(G) = |B| − |A|. Proposition 1.1 (Gallai–Edmonds Structure Theorem [2,4]). If G is a graph, then (1) every component of GD is factor-critical; (2) GC has a perfect matching;
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Fig. 4. H is obtained from G by the method mentioned above.
Fig. 5. The structural relation of Gm−3 and Gm for some graph G.
Fig. 6. A graph G satisfying M2 (G) = Kn+1 .
(3) a matching M of G is a maximum matching if and only if M contains a maximum matching of each component of GD , contains a perfect matching of GC , and matches all vertices in A(G) with vertices in distinct components of GD ; (4) def(G) = c (GD ) − |A(G)|, where c (GD ) is the number of components of GD ; (5) for any non-empty subset X of A(G), there exist at least |X | + 1 components G′ of GD such that NG (X ) ∩ V (G′ ) ̸= ∅. By Proposition 1.1(3), a maximum matching of G uses no edges induced by A(G). Hence in this paper we delete all edges induced by A(G); this does not change M2 (G). By Proposition 1.1(1), if G is bipartite, then GD is trivial; so GAD is a bipartite graph with bipartition (A(G), D(G)) and has positive surplus as viewed from A(G) by Proposition 1.1(5). Proposition 1.2 ([12]). If G is a graph, then M (GAD ) is connected, each component of M (G) is isomorphic to M (GAD ), and the number of components of M (G) equals the number of perfect matchings of GC unless C (G) = ∅, in which case the number of components is 1. Proposition 1.3 ([13]). For every graph G, if G has at least two perfect matchings, then P M (G) is connected. Proposition 1.4 ([13]). For every graph G, if G has at least three perfect matchings, then there exists a Hamilton cycle C of P M(G) containing e for any e ∈ E (P M(G)).
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2. Structure of second kind maximum matching graphs The Cartesian product of graphs G1 and G2 , denoted by G1 G2 , is the graph with vertex set V (G1 ) × V (G2 ) such that
(x, u)(y, v) is an edge if x = y and uv ∈ E (G2 ) or if xy ∈ E (G1 ) and u = v . Proposition 2.1. If G is a graph, then M2 (G) ∼ = P M(GC )M2 (GAD ).
Proof. By Proposition 1.1, every maximum matching of G can be expressed as an ordered pair consisting of a perfect matching of GC and a maximum matching of GAD . To change a maximum matching of G via an alternating path of length 2 or an alternating cycle, we must change one component of the ordered pair in this way, since no maximum matching has an edge joining a vertex in C (G) to a vertex outside C (G) by Proposition 1.1. Hence by the definition of Cartesian product, we have M2 (G) ∼ = M2 (GC )M2 (GAD ). Since GC has a perfect matching, we have P M(G) = M2 (G). Hence M2 (G) ∼ = P M(GC )M2 (GAD ) (see Fig. 3). The connectivity κ(G) of G is the fewest number of vertices whose removal from G results in a disconnected or trivial graph. The following lemma is useful for the proofs of Corollary 2.3. Lemma 2.2 ([11]). If G1 and G2 are two nontrivial graphs, then
κ(G1 G2 ) = min{κ(G1 )|V (G2 )|, κ(G2 )|V (G1 )|, δ(G1 ) + δ(G2 )}. Corollary 2.3. If G is a graph with no isolated vertices, then M2 (G) is connected. Proof. If C (G) = ∅ or GC has only one perfect matching, then M (G) = M (GAD ) and M (GAD ) is connected by Proposition 1.2, so M2 (G) is connected. If D(G) = ∅, then G = GC and M2 (G) = P M (G); by Proposition 1.3, M2 (G) is connected. If GC has at least two perfect matchings and D(G) ̸= ∅, then P M (GC ) is connected by Proposition 1.3 and GAD has at least two maximum matchings, since G has no isolated vertices; by Proposition 2.1 and Lemma 2.2, M2 (G) is connected. 3. Forbidden induced subgraphs of second kind maximum matching graphs A graph is a forbidden induced subgraph for a class of graphs if it does not appear as an induced subgraph of any graph in the class. Eroh [3] and Liu [9] discovered many forbidden induced subgraphs for the class of maximum matching graphs. In contrast, we prove next that the class of second kind maximum matching graphs has no forbidden induced subgraphs. Theorem 3.1. For every graph G, there exists a graph H such that M2 (H ) contains G as an induced subgraph. Proof. For each vertex v ∈ V (G), we construct a cycle Cv of length lv , where lv = 4 if dG (v) ≤ 1, and lv = 2dG (v) otherwise. Note that Cv has a perfect matching. If NG (v) = {u1 , . . . , us }, where s ≥ 2, then we choose a perfect matching Mv of Cv and name its edges ev u1 , . . . , ev us (see w in Fig. 4). If NG (v) = {u}, then we choose a perfect matching Mv of Cv and name its edges ev u and ev ′ v (see y in Fig. 4). If dG (v) = 0, then we choose a perfect matching Mv of Cv and name its edges ev ′ v and ev ′′ v (see x in Fig. 4). Now we construct a new graph H. Let H be the union of the cycles Cv for v ∈ V (G), arranged so that (1) if uv ̸∈ E (G), then V (Cu ) ∩ V (Cv ) = ∅; and (2) if uv ∈ E (G), then E (Cu ) ∩ E (Cv ) = {ev u } and |V (Cu ) ∩ V (Cv )| = 2 (note that Cu and Cv have only one common edge, which is euv , and exactly two common vertices which are endvertices of euv ). Fig. 4 shows an example of the graph H obtained from a graph G as just defined. Let M = v∈V (G) Mv . Then M is a perfect matching of H. By the construction of H, we have E (Cu ) ∩ E (Cw ) ̸= ∅ (in this case, E (Cu ) ∩ E (Cw ) = {euw }) if and only if uw ∈ E (G). Let Mv′ = M ⊕ E (Cv ) for each v ∈ V (G); also Mv′ is a perfect matching = {Mv′ |v ∈ V (G)}. of H. Let M such that α(v) = Mv′ for each vertex v of G. By the definition of Mv′ , α is a bijection. If Let α be a mapping from V (G) to M ′ ′ ′ ′ = (M ⊕ E (Cu ))⊕(M ⊕ E (Cw )) = E (Cu )⊕ E (Cw ), Mu and Mw are adjacent in M2 (H ), then Mu′ ⊕ Mw forms a cycle. Since Mu′ ⊕ Mw also E (Cu ) ⊕ E (Cw ) forms a cycle. Hence E (Cu ) ∩ E (Cw ) ̸= ∅. By the construction of H, we have E (Cu ) ∩ E (Cw ) = {euw }, so ′ uw ∈ E (G). If uw ∈ E (G), then E (Cu ) ∩ E (Cw ) = {euw } and E (Cu ) ⊕ E (Cw ) forms a cycle. Since Mu′ ⊕ Mw = E (Cu ) ⊕ E (Cw ), ′ ′ . we have Mu and Mw are adjacent in M2 (H ). Therefore, G is isomorphic to the subgraph of M2 (H ) induced by M 4. Some characterizations First, we characterize the graphs whose second kind maximum matching graphs are trees. We need two lemmas. An edge of G is a pendant edge if it has an endvertex of degree 1. Lemma 4.1 ([9]). If G is a bipartite graph with bipartition (A, B), then G has positive surplus (as viewed from A) if and only if G has a spanning forest F such that dF (u) = 2 for all u ∈ A.
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In fact, in Lemma 4.1, also F has positive surplus (as viewed from A) and def(F ′ ) = 1 for each component F ′ of F , so def(G) = def(F ) = c (F ), where c (F ) is the number of components of F . Thus we have the following corollary. Corollary 4.1. If a bipartite graph G with bipartition (A, B) has positive surplus (as viewed from A), then G is connected and dG (u) = 2 for all u ∈ A if and only if G is a tree and def(G) = 1. Lemma 4.2 ([6]). If G is a graph, then M (G) is a tree if and only if either C (G) = ∅ or GC has exactly one perfect matching, and GAD is a bipartite graph with bipartition (A(G), D(G)) such that def(GAD ) = 1 and the last graph Gm in the graph sequence (G0 = GAD , G1 , . . . , Gm ) is K1 , where Gi+1 is obtained from Gi by deleting all vertex pairs that can induce a pendant edge in Gi for 0 ≤ i ≤ m − 1. If a bipartite graph G with bipartition (A, B) has positive surplus (as viewed from A) and def(G) = 1, then dG (u) ≥ 2 for any u ∈ A and |B| = |A| + 1. Thus G is connected and there is at most one vertex of degree 1 in NG (u) for any vertex u ∈ A (otherwise, if dG (v1 ) = dG (v2 ) = 1 and v1 , v2 ∈ NG (u), then |NG (A − {u})| ≤ |B − {v1 , v2 }| = |A| − 1, a contradiction). So the set of all pendant edges of G, denoted by M0 , is a matching of G. Further, let G1 = G − V (M0 ), A1 = A − V (M0 ) and B1 = B − V (M0 ). It is easy to check that G1 has the bipartition (A1 , B1 ) and has the same properties with G (i.e., also G1 has positive surplus (as viewed from A1 ) and def(G1 ) = 1). If G1 also has a vertex of degree 1, then we continue to construct a graph G2 by the same method as above until some graph Gm satisfies that δ(Gm ) ≥ 2 or Gm is K1 . Thus we obtain a graph sequence (G0 = G, G1 , . . . , Gm ) and a matching sequence (M0 , M1 , . . . , Mm−1 ) such that Gi has the bipartition (Ai , Bi ), Ai = Ai−1 − V (Mi−1 ) and Bi = Bi−1 − V (Mi−1 ) for 2 ≤ i ≤ m (see Fig. 5). Theorem 4.3. If G is a graph with no isolated vertices, then M2 (G) is a tree if and only if G satisfies one of the following two conditions: (1) G has at most two perfect matchings; (2) def(GAD ) = 1, GAD is a tree and either C (G) = ∅ or GC has exactly one perfect matching. Proof. ⇐ If G has at most two perfect matchings, then M2 (G) = K1 or K2 by Proposition 1.3; so M2 (G) is a tree. If def(GAD ) = 1, GAD is a tree and either C (G) = ∅ or GC has exactly one perfect matching, then M2 (G) ∼ = M2 (GAD )by Proposition 2.1 and M2 (GAD ) = M (GAD ); so GAD is a bipartite graph with bipartition (A(G), D(G)) and has positive surplus (as viewed from A(G)) by Proposition 1.1(5). Therefore, the last graph Gm in the graph sequence (G0 = GAD , G1 , . . . , Gm ) is K1 , where Gi+1 is obtained from Gi by deleting all vertex pairs that induce a pendant edge in Gi for 0 ≤ i ≤ m − 1. By Lemma 4.2, M (GAD ) is a tree. Thus M2 (G) is a tree. ⇒ If M2 (G) is a tree, then GC has at most two perfect matchings by Proposition 1.4 and Proposition 2.1. (Otherwise, M2 (GC ) has a cycle. Then M2 (G) has a cycle, a contradiction.) Case 1: GC has exactly two perfect matchings. Thus M2 (GC ) = K2 . If D(G) ̸= ∅, then GAD has at least two maximum matchings, since G has no isolated vertices. By Corollary 2.3, M2 (GAD ) is connected. By Proposition 2.1, M2 (G) has a cycle, a contradiction. Thus D(G) = ∅. Hence A(G) = ∅. It follows that G satisfies condition (1). Case 2: C (G) = ∅ or GC has a unique perfect matching. By Proposition 2.1, we have M2 (G) ∼ = M2 (GAD ). By Proposition 1.2, M(G) is connected. So M(G) is a connected spanning subgraph of M2 (G). Since M2 (G) is a tree, we have M (G) = M2 (G). Hence M (G) is a tree. By Lemma 4.2, GAD is a bipartite graph with def(GAD ) = 1, has positive surplus (as viewed from A(G)) and the last graph Gm in the graph sequence (G0 = GAD , G1 , . . . , Gm ) is K1 , where Gi+1 = Gi − V (Mi ), Gi+1 has the bipartition (Ai+1 , Bi+1 ), Mi is the set of all pendant edges of Gi , Ai+1 = Ai − V (Mi ), and Bi+1 = Bi − V (Mi ) for 0 ≤ i ≤ m − 1. Therefore, we only need to show that GAD has no cycles. We prove it by contradiction. Suppose that Gt −1 has a cycle but Gt has no cycles without loss of generality. Since Gt is connected and has no cycles, Gt is a tree. By Corollary 4.1, dGt (v) = 2 for any v ∈ At . Since Gt −1 has a cycle, there exists a vertex w ∈ At −1 such that dGt −1 (w) ≥ 3 by Corollary 4.1. Thus w ̸∈ V (Gt ). Since dGt −1 (w) ≥ 3, we can assume that w1 , w2 ∈ NGt −1 (w)∩Bt and ww3 ∈ Mt −1 . Since Gt has positive surplus (as viewed from At ), we have D(Gt ) = Bt . So we can assume that Mw1 and Mw2 are two maximum matchings of Gt missing w1 and w2 , respectively. Since def(Gt ) = 1, Mw1 ⊕ Mw2 forms a path P of even length joining w1 and w2 . Let C = P + ww1 + ww2 . Thus C is an even cycle. Let M = Mw1 ∪ (Mt −1 − ww3 + ww1 ) ∪ Mt −2 ∪ · · · ∪ M0 and M ′ = Mw2 ∪ (Mt −1 − ww3 + ww2 ) ∪ Mt −2 ∪ · · · ∪ M0 . Thus M and M ′ are two maximum matchings of GAD , and M ⊕ M ′ forms C . Hence MM ′ ∈ E (M2 (G)). Since MM ′ ̸∈ E (M (G)), we have M2 (G) ̸= M (G), a contradiction. Second, we characterize the graphs whose second kind maximum matching graphs are complete graphs. Lemma 4.4. If M , M1 , and M2 are three maximum matchings of G such that M1 ⊕ M forms a cycle C and M2 = M − uv + uw , then M1 M2 ̸∈ E (M2 (G)). Proof. If uv ∈ E (C ), then M1 ⊕ M2 = C − uv + uw ; so M1 ⊕ M2 induces a path of length more than 2. If uv ̸∈ E (C ), then C is an (M1 , M2 ) alternating cycle; so M1 ⊕ M2 consists of a cycle and a path of length 2. Hence M1 M2 ̸∈ E (M2 (G)). Lemma 4.5. If G is a graph with no isolated vertices, then M ⊕ M ′ forms a cycle for any two maximum matchings M and M ′ of G that are adjacent in M2 (G) if and only if G has a perfect matching.
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Proof. If G has a perfect matching, then the statement holds clearly. Conversely, it suffices to prove that D(G) = ∅. We prove it by contradiction. Suppose that D(G) ̸= ∅. Let u ∈ D(G) and uv ∈ E (G). Thus there exists a maximum matching M of G missing u. By the maximality of M, there exists an edge vw ∈ M. Let M ′ = M − vw + uv , also M ′ is a maximum matching of G. Hence MM ′ ∈ E (M2 (G)). But M ⊕ M ′ is not a cycle, a contradiction. So D(G) = ∅. Thus G has a perfect matching. Lemma 4.6 ([6]). If G is a graph with no isolated vertices, then M (G) is a complete graph if and only if either C (G) = ∅ or GC has a unique perfect matching and GAD = K3 or GAD = K1,n . Theorem 4.7. If G is a graph with no isolated vertices, then M2 (G) is a complete graph if and only if G satisfies one of the following two conditions: (1) C (G) = ∅ or GC has a unique perfect matching and GAD = K3 or GAD = K1,n ; (2) G = GC and E (C1 ) ⊕ E (C2 ) forms a cycle for any perfect matching M of G and any two M-alternating cycles C1 and C2 . Proof. ⇒ If M2 (G) is a complete graph, either M1 ⊕ M2 forms a cycle for any two maximum matchings M1 and M2 of G or M1 ⊕ M2 forms a path of length 2 for any two maximum matchings M1 and M2 of G by Lemma 4.4. Case 1: M1 ⊕ M2 forms a path of length 2 for any two maximum matchings M1 and M2 of G. Thus M2 (G) = M (G). By Lemma 4.6, either C (G) = ∅ or GC has a unique perfect matching and GAD = K3 or GAD = K1,n . Case 2: M1 ⊕ M2 forms a cycle for any two maximum matchings M1 and M2 of G. By Lemma 4.5, we have G = GC , i.e., G has a perfect matching. Set V (M2 (G)) = {M1 , . . . , Mm }. Since M2 (G) is a complete graph, we can assume that M1 ⊕ Mi forms an M1 -alternating cycle Ci for 2 ≤ i ≤ m. Thus all M1 -alternating cycles are C2 , . . . , Cm . Since E (Ci ) ⊕ E (Cj ) = Mi ⊕ Mj for 2 ≤ i < j ≤ m, E (Ci ) ⊕ E (Cj ) forms a cycle. ⇐ If C (G) = ∅ or GC has a unique perfect matching and GAD = K3 or GAD = K1,n , then M2 (G) = M2 (GAD ) and M2 (GAD ) = M (GAD ); so M2 (G) is a complete graph by Lemma 4.6. If G = GC , then G has a perfect matching; so the symmetric difference of any two adjacent maximum matchings of G forms a cycle. We choose a perfect matching M of G and find all M-alternating cycles C1 , . . . , Ct . Let Mi = M ⊕ E (Ci ) for 1 ≤ i ≤ t, also Mi is a perfect matching of G and NM2 (G) (M ) = {M1 , . . . , Mt }. If the symmetric difference of any two Malternating cycles is a cycle, then Mi ⊕ Mj = (M ⊕ E (Ci )) ⊕ (M ⊕ E (Cj )) = E (Ci ) ⊕ E (Cj ) forms a cycle, it follows that Mi and Mj are adjacent in M2 (G) for any 1 ≤ i < j ≤ t. Claim. V (M2 (G)) = {M , M1 , . . . , Mt }. We prove it by contradiction. Let A = V (M2 (G)) − {M , M1 , M2 , . . . , Mt } ̸= ∅. By Corollary 2.3, M2 (G) is connected. Thus there exist a perfect matching M ′ ∈ A and some l (1 ≤ l ≤ t) such that M ′ Ml ∈ E (M2 (G)). Note that M ′ M ̸∈ E (M2 (G)). Denote by C the cycle formed by Ml ⊕ M ′ . Since both C and Cl are Ml -alternating cycles, E (C ) ⊕ E (Cl ) forms a cycle. So M ⊕ M ′ = (Ml ⊕ E (Cl )) ⊕ (Ml ⊕ E (C )) = E (Cl ) ⊕ E (C ), it follows that M ′ M ∈ E (M2 (G)), a contradiction. Hence A = ∅. Thus M2 (G) is a complete graph (see Fig. 6). Finally, we characterize the graphs whose second kind maximum matching graphs are cycles. Lemma 4.8 ([8]). If G is a graph and M is a maximum matching of G, then the degree of M in M (G) is dM(G) (M ) =
dG (v).
v∈D(G)−V (M )
Lemma 4.9 ([6]). If G is a graph with no isolated vertices, then M (G) is a cycle if and only if either C (G) = ∅ or GC has a unique perfect matching and GAD = 2K1,2 , C2n+1 or K1,3 . Theorem 4.10. If G is a graph with no isolated vertices, then (1) when D(G) = ∅, M2 (G) is a cycle if and only if G satisfies one of the following two conditions: (a) G has exactly three perfect matchings; (b) G has exactly four perfect matchings and there exists a perfect matching M such that there are exactly two M-alternating cycles C1 and C2 that are disjoint. (2) when D(G) ̸= ∅ and GC has at most one perfect matching, M2 (G) is a cycle if and only if GAD ∈ {B∗ , C2n+1 , 2K1,2 , K1,3 }, where B∗ is shown in Fig. 7. (3) when D(G) ̸= ∅ and GC has at least two perfect matchings, M2 (G) is a cycle if and only if GAD = K1,2 and GC has exactly two perfect matchings. Proof. ⇒ If M2 (G) is a cycle, then dM2 (G) (M ) = 2 for any maximum matching M of G. Case 1: D(G) = ∅. Let M be a perfect matching of G. Since dM2 (G) (M ) = 2, there exist exactly two M-alternating cycles C1 and C2 . Let Mi = M ⊕ E (Ci ) for 1 ≤ i ≤ 2. Note that M1 and M2 are two perfect matchings of G and MM1 , MM2 ∈ E (M2 (G)).
Y. Liu, Z. Liu / Discrete Mathematics 323 (2014) 27–34
C2n u1
33
v0
Fig. 7. Graph B∗ .
If E (C1 ) ∩ E (C2 ) = ∅, we have M1 ⊕ E (C2 ) = M2 ⊕ E (C1 ). Let M ′ = M1 ⊕ E (C2 ). So M ′ M1 , M ′ M2 ∈ E (M2 (G)). It follows that MM1 M ′ M2 M is a 4-cycle of M2 (G). Hence G satisfies the condition (b) of (1) in Theorem 4.10. We can assume that E (C1 ) ∩ E (C2 ) ̸= ∅. Since M1 ⊕ M2 = E (C1 ) ⊕ E (C2 ), E (C1 ) ⊕ E (C2 ) must form the union of some disjoint (M1 , M2 )-alternating cycles. If E (C1 ) ⊕ E (C2 ) consists of at least two (M1 , M2 )-alternating cycles, then there exist at least three M1 -alternating cycles, including C1 . Thus dM2 (G) (M1 ) ≥ 3, a contradiction. Therefore, E (C1 ) ⊕ E (C2 ) forms a cycle. Thus MM1 M2 M is a cycle. It follows that M2 (G) = C3 . Hence G satisfies the condition (a) of (1) in Theorem 4.10. Case 2: D(G) ̸= ∅ and GC has at most one perfect matching. By Proposition 2.1, we have M2 (G) ∼ = M2 (GAD ). Thus M2 (GAD ) is a cycle. By Proposition 1.2, M(G) = M(GAD ) and M(GAD ) is connected. Since M (GAD ) is a spanning subgraph of M2 (GAD ), M (GAD ) is a cycle or a path. Subcase 1: M (GAD ) is a cycle. By Lemma 4.9, we have GAD = 2K1,2 , C2n+1 or K1,3 . Subcase 2: M (GAD ) is a path. Let M (GAD ) = M1 M2 . . . Mm , where M1 , . . . , Mm are all maximum matchings of GAD and M1 Mm ̸∈ E (M (GAD )). Thus M2 (GAD ) = M (GAD ) + M1 Mm = M1 . . . Mm M1 . It follows that M1 ⊕ Mm forms an even cycle C2n . Since M(GAD ) is a path, GAD is a bipartite graph with bipartition (A(G), D(G)), has positive surplus as viewed from A(G), and def(GAD ) = 1 by Lemma 4.2. Thus |D(G)| = |A(G)| + 1. Since M1 ⊕ Mm is a cycle and def(GAD ) = 1, there exists exactly one vertex v0 in D(G) missed by both M1 and Mm , and all other vertices are covered by both M1 and Mm . Since dM(GAD ) (M1 ) = dM(GAD ) (Mm ) = 1, we have dGAD (v0 ) = 1 by Lemma 4.8. Hence we can assume that v0 u1 ∈ E (G). It follows that u1 is covered by both M1 and Mm . Furthermore, since dM(GAD ) (Mi ) = 2 for any 2 ≤ i ≤ m − 1, we have dGAD (v) = 2 for any v ∈ D(G) − v0 by Lemma 4.8. Claim 1. u1 ∈ V (C2n ). To the contrary, if u1 ̸∈ V (C2n ), then there exists a vertex v1 ∈ D(G) − v0 such that u1 v1 ∈ M1 ∩ Mm . It follows that no cycles on 2n vertices that contains v1 . Let M = M1 − u1 v1 + u1 v0 and M ′ = Mm − u1 v1 + u1 v0 ; so M ⊕ M ′ = M1 ⊕ Mm , it follows that MM ′ ̸∈ E (M (GAD )), but MM ′ ∈ E (M2 (GAD )), a contradiction. Claim 2. GAD = B∗ (see Fig. 7). By Claim 1, we have dGAD (u1 ) ≥ 3. Since dG (v0 ) = 1 and dG (v) = 2 for any v ∈ D(G) − v0 , we have
|E (GAD )| =
dG (v) = 2|A(G)| + 1.
v∈D(G)
Note that dGAD (u) ≥ 2 for any u ∈ A(G). It follows that dGAD (u1 ) = 3 and dGAD (u) = 2 for any u ∈ A(G) − u1 . Thus C2n + u1 v0 (that is, B∗ ) is a component of GAD . Since GAD is connected, we have GAD = B∗ . Let A1 = A(G) − V (C2n ) and D1 = D(G) − V (C2n ) − v0 . Case 3: D(G) ̸= ∅ and GC has at least two perfect matchings. Since D(G) ̸= ∅, GAD has at least two maximum matchings. By Proposition 1.2, M2 (GAD ) has an edge. If GC has at least three perfect matchings, then P M (GC ) has a cycle by Proposition 1.4; so M2 (G) is not a cycle by Proposition 2.1, a contradiction. Similarly, if GAD has at least three maximum matchings, then we can show that also M2 (G) is not a cycle. Therefore, GAD = K1,2 and GC have exactly two maximum matchings. ⇐ Conversely, we consider the following cases. Case 1: D(G) = ∅. Thus A(G) = ∅. It follows that G = GC . If G has exactly three perfect matchings, then by Proposition 1.4, we have M2 (G) = C3 . If G has exactly four perfect matchings and there exists a perfect matching M such that there are exactly two M-alternating cycles C1 and C2 that are disjoint, then M , M1 = M ⊕ E (C1 ), M2 = M ⊕ E (C2 ), and M ′ = M1 ⊕ E (C2 ) are the four perfect matchings of G such that MM1 , MM2 , M ′ M1 , M ′ M2 ∈ E (M2 (G)) and M1 M2 , MM ′ ̸∈ E (M2 (G)). Hence M2 (G) = MM1 M ′ M2 M (i.e., M2 (G) = C4 ). Case 2: D(G) ̸= ∅. If GC has at most one perfect matching and GAD ∈ {B∗ , C2n+1 , 2K1,2 , K1,3 }, then M2 (G) ∼ = M2 (GAD ) by Proposition 2.1; so it is easy to show that M2 (B∗ ) = Cn+2 , M2 (C2n+1 ) = C2n+1 , M2 (K1,3 ) = C3 , and M2 (2K1,2 ) = C4 . Therefore, M2 (G) is a cycle. If GAD = K1,2 and GC has exactly two perfect matchings, then it is easy to show that M2 (G) = C4 .
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