On the order of almost regular bipartite graphs without perfect matchings

A graph G is almost regular or more precisely is a (d, d+ 1)-graph, if the degree of each vertex of G is either d or d + 1. Let d ≥ 2 be an integer, and let G be a connected bipartite (d, d + 1)-graph with partite sets X and Y such that |X| = |Y |. If the order of G is at most 4d + 4, then we show in this paper that G contains a perfect matching. Examples will demonstrate that the given bound on the order of G is best possible. We shall assume that the reader is familiar with standard terminology on graphs (see, e.g., Chartrand and Lesniak [2]). In this paper, all graphs are finite and simple. The vertex set of a graph G is denoted by V (G), and n = n(G) = |V (G)| is called the order of G. The neighborhood NG(x) of a vertex x is the set of vertices adjacent with x, and the number dG(x) = |NG(x)| is the degree of x in the graph G. If d ≤ dG(x) ≤ d+1 for each vertex x in a graph G, then we speak of an almost regular graph or more precisely of a (d, d + 1)-graph. If M is a matching in a graph G with the property that every vertex is incident with an edge of M , then M is a perfect matching. We denote by Kr,s the complete bipartite graph with partite sets X and Y , where |X| = r and |Y | = s. If G is a graph and A ⊆ V (G), then we denote by G[A] the subgraph induced by A and by q(G − A) the number of odd components in the subgraph G−A. As an extension of a theorem of Wallis [10] on regular graphs, Zhao [11] in 1991 proved the following result. Theorem 1 (Zhao [11] 1991) Let d ≥ 2 be an integer, and let G be a (d, d+1)-graph without an odd component. If |V (G)| ≤ 3d + 3, then G has a perfect matching. For supplements, extensions or generalizations of Theorem 1, see the articles by Caccetta and Mardiyono [1], Volkmann [9] and Klinkenberg and Volkmann [3, 4, 5]. 166 LUTZ VOLKMANN AND AXEL ZINGSEM In this paper, we will prove an analogue to Zhao’s theorem for bipartite graphs. The proof of our main theorem is based on Tutte’s famous 1-factor theorem [7] (for a proof see e.g., [8]). Theorem 2 (Tutte [7] 1947) A nontrivial graph G has a perfect matching (or a 1-factor) if and only if q(G− S) ≤ |S| for every proper subset S of V (G). Theorem 3 Let d ≥ 2 be an integer, and let G be a connected bipartite (d, d + 1)graph of order n with partite sets X and Y such that |X| = |Y |. If n ≤ 4d+ 4, then G contains a perfect matching. Proof. Suppose to the contrary that G does not contain a perfect matching. Then, Theorem 2 implies that there exists a non-empty set A ⊂ V (G) such that q(G−A) ≥ |A| + 1. Since n is even, the numbers q(G− A) and |A| of the same parity, and we deduce that q(G−A) ≥ |A| + 2. (1) We call an odd component of G−A large if it has at least 2d+ 1 vertices, and small otherwise. If we denote by α and β the number of large and small components, respectively, then we deduce from (1) that α + β = q(G− A) ≥ |A| + 2. (2) If U is a small component of G− A of minimum order, then we observe that n ≥ |A| + α(2d+ 1) + β|V (U)|. (3) Since G is a bipartite (d, d + 1)-graph, it is easy to verify that there are at least d edges of G joining each small component of G − A with A. Using the hypothesis that G is connected, we deduce that α + dβ ≤ |A|(d + 1). (4) Next we distinguish four cases. Case 1: Assume that α ≥ 3. The hypothesis n ≤ 4d + 4 and (3) lead to the contradiction 4d + 4 ≥ n ≥ 3(2d + 1). Case 2: Assume that α = 2. Inequality (2) yields β ≥ |A| ≥ 1, and thus we obtain by (3) 4d + 4 ≥ n ≥ |A| + 2(2d + 1) + β|V (U)| ≥ 4d + 2 + |A|(1 + |V (U)|) and therefore |A| = |V (U)| = 1. However, now the only vertex of the small component U has only one neighbor, a contradiction to d ≥ 2. ALMOST REGULAR BIPARTITE GRAPHS 167 Case 3: Assume that α = 1. Inequality (2) yields β ≥ |A| + 1, and thus we obtain by (4) |A| ≥ d + 1. (5) Applying (3), we arrive at 4d + 4 ≥ n ≥ |A| + α(2d + 1) + β|V (U)| ≥ d + 1 + 2d + 1 + β = 3d + 2 + β and thus β ≤ d + 2. (6) Using the hypothesis n ≤ 4d+4, we altogether observe that β = d+2 = |A|+1, each small component consists of a single vertex, the large component is of order exactly 2d + 1 and n = 4d + 4. Since G is a connected (d, d+ 1)-graph, there are at least d + 2d edges in G joining the small components of G− A with A and at least one edge in G joining the large component of G− A with A. In addition, there are at most d + 2d + 1 edges in G joining A with the odd components of G−A. Consequently, all vertices in A are of degree d+ 1, and the subgraph G[A] induced by A is empty. Since there is only one edge, say uv, connecting the large component W of order 2d + 1 with A, the large component W has a bipartition V ′, V ′′ such that |V ′′| = |V ′| + 1 = d + 1. Without loss of generality, let u in W . Suppose that u ∈ V ′. This implies that every vertex of V ′′ is connected with every vertex of V ′ in W , and we arrive at the contradiction dG(u) = d + 2. Thus u ∈ V ′′, and now X = V ′ ∪ A and Y = V (G) − (V ′ ∪ A) is a bipartition of G with |X| = 2d+ 1 and |Y | = 2d+ 3. Since G is connected, this is a contradiction to the hypothesis that |X| = |Y |. Case 4: Assume that α = 0. Inequality (2) yields β ≥ |A| + 2, and thus (4) leads to |A| ≥ 2d. (7) Applying the bound β ≥ |A| + 2, we obtain β ≥ |A| + 2 ≥ 2d + 2. (8) According to (3) and (7), we arrive at 4d + 4 ≥ n ≥ |A| + α(2d+ 1) + β|V (U)| ≥ 2d + β (9) and thus 2d + 4 ≥ β. (10) The inequalities (8) and (10) show that 2d + 2 ≤ β ≤ 2d + 4. Subcase 4.1: Assume that β = 2d + 4. In view of (9), it follows that |A| = 2d, and hence (4) yields the contradiction d(2d + 4) = dβ ≤ |A|(d + 1) = 2d(d + 1). 168 LUTZ VOLKMANN AND AXEL ZINGSEM Subcase 4.2: Assume that β = 2d + 3. In view of (9), it follows that |A| ≤ 2d + 1. Because of |A| ≥ 2d and the fact that n is even, we deduce that |A| = 2d + 1. As n ≤ 4d + 4, we conclude that all small components of G − A are isolated vertices. Consequently, there are at least 2d + 3d edges in G joining the small components of G− A with A. In addition, there are at most 2d + 3d + 1 edges in G joining A with the odd components of G − A. Therefore, the subgraph G[A] is empty. Thus X = A and Y = V (G)−A is a bipartition of G with |X| = 2d+ 1 and |Y | = 2d+ 3. Since G is connected, this is a contradiction to the hypothesis that |X| = |Y |. Subcase 4.3: Assume that β = 2d + 2. By (2) and (7), it follows that 2d + 2 = β ≥ |A| + 2 ≥ 2d + 2 and thus |A| = 2d. Hence there are at least 2d + 2d edges in G joining the small components of G− A with A, and there are at most 2d + 2d edges in G joining A with the odd components of G− A. Therefore the subgraph G[A] is empty. If the small components of G−A are isolated vertices, then we arrive a contradiction as above. Otherwise, the hypothesis n ≤ 4d+4 shows that there is exactly one small component of order three and that the remaining 2d+1 small components are of order one. Hence there are at least 3d − 4 + d(2d + 1) = 2d + 4d − 4 edges in G joining the small components of G − A with A, and there are at most 2d + 2d edges in G joining A with the odd components of G − A. This leads to a contradiction when d ≥ 3. In the remaining case that d = 2, we obtain |A| = 4, β = 6 and n = 12. A straightforward calculation leads to the contradiction that G has a bipartition X, Y with |X| = |Y | = 6, and the proof of Theorem 3 is complete. The following family of examples will show that the bound presented in Theorem 3 is best possible. Example 4 Let d ≥ 2 be an integer, and let Kd+1,d+2 be the complete bipartite graph with the partite sets {x1, x2, . . . , xd+2} and {y1, y2, . . . , yd+1}. If we delete in the graph Kd+1,d+2 the edges x1y1, x2y2, . . . , xd+1yd+1 and xd+2yd+1, then we denote the resulting graph by H1. In addition, let Kd+1,d+2 be the complete bipartite graph with the partite sets {u1, u2, . . . , ud+2} and {v1, v2, . . . , vd+1}. If we delete the edges u1v1, u2v2, . . . , ud+1vd+1 and ud+2vd+1, then we denote the resulting graph by H2. Now let H be the disjoint union of H1 and H2 together with the edge yd+1vd+1. It is straightforward to verify that H is a connected bipartite (d, d + 1)-graph of order |V (H)| = 4d + 6 with a partition X, Y such that |X| = |Y | = 2d + 3 without a perfect matching. Corollary 5 Let d ≥ 2 be an integer, and let G be a bipartite (d, d + 1)-graph of order n with partite sets X and Y such that |X| = |Y |. If n ≤ 4d + 4 and if G has no odd component, then G contains a perfect matching. Proof. Since G is a bipartite (d, d + 1)-graph, each component of G has order at ALMOST REGULAR BIPARTITE GRAPHS 169 least 2d. Hence G consists of at most two components when d ≥ 3 and at most three components when d = 2. In the case that G is connected, the desired result follows from Theorem 3. If d = 2 and G has three components, then all components are isomorphic to K2,2, and G contains a perfect matching. Assume next that G consists of exactly two components G1 and G2 such that, without loss of generality, 2d ≤ n(G1) ≤ n(G2) ≤ 2d + 4. Case 1: Assume that n(G1) = 2d. It follows that G1 is isomorphic to Kd,d and thus G1 has a perfect matching. The hypothesis |X| = |Y | implies that G2 has a bipartition X2, Y2 with |X2| = |Y2|. Therefore, according to Theorem 3, the component G2 has also a perfect matching, and we are done. Case 2: Assume that n(G1) = n(G2) = 2d + 2. If G1 and G2 have partite sets X1, Y1 and X2, Y2 such that |X1| = |Y1| = |X2| = |Y2| = d + 1, then it follows from Theorem 3 that G1 and G2 have perfect matchings and so also G contains a perfect matching. In the remaining case, the components G1 and G2 have partite sets X1, Y1 and X2, Y2 such that, without loss of generality, |X1| = |X2| = d and |Y1| = |Y2| = d + 2. However, since G is (d, d + 1)-graph, this is impossible, and the proof of Corollary 5 is complete. Note that the case d = 1 in Theorem 3 is trivial, since each (1, 2)-graph without an odd component has a perfect matching. Finally notice that by a classical and well-known theorem of König [6], each d-regular bipartite graph contains a perfect matching for d ≥ 1.