Graphs and Combinatorics (2011) 27:865–869DOI 10.1007/s00373-010-1008-8

ORIGINAL PAPER

On the Perfect Matchings of Near Regular Graphs

Xinmin Hou

Received: 11 January 2010 / Revised: 7 December 2010 / Published online: 5 January 2011© Springer 2011

Abstract Let k, h be positive integers with k ≤ h. A graph G is called a [k, h]-graphif k ≤ d(v) ≤ h for any v ∈ V (G). Let G be a [k, h]-graph of order 2n such that k ≥ n.Hilton (J. Graph Theory 9:193–196, 1985) proved that G contains at least �k/3� dis-joint perfect matchings if h = k. Hilton’s result had been improved by Zhang and Zhu(J. Combin. Theory, Series B, 56:74–89, 1992), they proved that G contains at least�k/2� disjoint perfect matchings if k = h. In this paper, we improve Hilton’s resultfrom another direction, we prove that Hilton’s result is true for [k, k +1]-graphs. Spe-cifically, we prove that G contains at least � n

3 �+1+(k −n) disjoint perfect matchingsif h = k + 1.

Keywords Perfect matching · Regular graph · Near regular graph

1 Introduction

The reader can refer to [1] for standard graph theoretic terms not defined in this paper.A matching M of G is a 1-regular subgraph of G. A matching M is called a perfectmatching of G if V (M) = V (G). Clearly, if a graph G has a perfect matching, then|V (G)| ≡ 0 (mod 2). A vertex v is saturated by a matching M if v ∈ V (M).

To determine whether a graph G has a perfect matching is a fundamental problemin graph theory. A component of a graph is odd or even according to whether it has anodd or even order. For X ⊆ V (G), let qG(X) denote the number of odd components

The work was supported by NNSF of China (No.10701068 and 11071233).

X. Hou (B)Department of Mathematics, University of Science and Technology of China,Hefei, 230026 Anhui, Chinae-mail: xmhou@ustc.edu.cn

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of G − X . Tutte [6] gave a necessary and sufficient condition for a graph G to have aperfect matching. Tutte’s Theorem is as follows.

Theorem 1.1 (Tutte [6]) A graph G has a perfect matching if and only if qG(X) ≤ |X |for any X ⊆ V (G).

For regular graphs, there are many significant results about the perfect matchings.The first sufficient condition for a regular graph to have a perfect matching was givenby Petersen [5], which states that: Every 3-regular 2-edge-connected graph has aperfect matching. Hilton [3] proved the following theorem.

Theorem 1.2 (Hilton [3]) Let G be a k-regular graph of order 2n. If k ≥ n, then Gcontains at least �k/3� disjoint perfect matchings.

Hilton’s result had been improved by Zhang and Zhu [7], the result is as follows.

Theorem 1.3 (Zhang and Zhu [7]) Let G be a k-regular graph of order 2n. If k ≥ n,the G contains at least �k/2� disjoint perfect matchings.

Let k, h be positive integers with k ≤ h. A graph G is called a [k, h]-graph ifk ≤ dG(v) ≤ h for any v ∈ V (G). If k = h, then a [k, h]-graph is a k-regular graph.In this paper, we improve Hilton’s Theorem from another direction, we prove thatHilton’s result is true for [k, k + 1]-graphs. The main theorem is as follows.

Theorem 1.4 Let G be a [k, k + 1]-graph of order 2n and k ≥ n. Then G containsat least � n

3 � + 1 + (k − n) disjoint perfect matchings.

2 Proof of Theorem 1.4

The minimum degree and the maximum degree of a graph G is denoted by δ(G) and�(G), respectively. For disjoint vertex subsets X, Y ⊆ V (G), let EG[X, Y ] denote theset of edges in G with one end in X and the other one in Y , and eG(X, Y ) = |EG[X, Y ]|,let NG(X) = {v | v ∈ V (G) − X and v is adjacent with some vertex in X} and if thegraph G is clear from the context, we omit the subscript for convenience. Let G[X ]denote the subgraph of G induced by X ⊆ V (G). The following lemmas will be usedin our proofs.

Lemma 2.1 (Berge [1985]) Let G be a graph and M be a maximum matching. Then

|V (G)| − maxX⊆V (G)

{qG(X) − |X |} = 2|M |. (1)

Equality (1) is called Berge–Tutte formula, an extension of Tutte’s Theorem.

Lemma 2.2 (Dirac [2]) Let G be a graph. If δ(G) ≥ |V (G)|2 , then G contains a Ham-

ilton cycle.

The following lemma is fundamental in this paper, the idea of the proof comes fromKaterinis (the proof of Theorem 5 in [4]).

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Lemma 2.3 Let G be a connected [k, k + 1]-graph and |V (G)| ≡ d (mod 2). If

k ≥ |V (G)|3 , then G contains a matching of order |V (G)|−d

2 .

Proof Argue by contradiction. Suppose G has no such a matching. Let M be a max-imum matching of G. By Berge-Tutte formula (1), there exists a subset X ⊆ V (G)

such that |V (G)| − (qG(X) − |X |) = 2|M | < |V (G)| − d. So, qG(X) − |X | > d.Since qG(X) − |X | ≡ |V (G)|(mod 2), qG(X) − |X | ≥ d + 2.

Let Q be the subgraph composed by all the odd components of G − X and l =maxC∈Q |V (C)|. For each integer 1 ≤ i ≤ l, let Qi = {C ∈ Q | |V (C)| = i} andqi = |Qi |. For each C ∈ Qi , e(V (C), X) ≥ (k − i + 1)i . Since G is connected,

e(V (Q), X) ≥k−1∑

i=1

(k − i + 1)iqi +l∑

i=k

qi .

On the other hand, e(X, V (Q)) ≤ (k + 1)|X | as �(G) = k + 1. So, we have

(k + 1)|X | ≥k−1∑

i=1

(k − i + 1)iqi +l∑

i=k

qi

≥ kq1 + (k + 1)

k−1∑

i=2

qi +l∑

i=k

qi , (2)

the second inequality holds since (k − i + 1)i ≥ k + 1 when 2 ≤ i ≤ k − 1. By (2)and |X | ≤ qG(X) − d − 2 = ∑l

i=1 qi − d − 2, we have

q1 + kl∑

i=k

qi ≥ (k + 1)(d + 2). (3)

Case 1 q1 = 0.

By (3),

l∑

i=k

qi ≥ k + 1

k(d + 2) > d + 2.

This means that Q has at least d + 3 components of order at least k. So, |V (G)| ≥|V (Q)| + |X | ≥ (d + 3)k + 1 ≥ |V (G)| + 1, a contradiction.

Case 2 q1 = 0.

Then Q1 = ∅. Since N (V (C)) ⊆ X for any C ∈ Q1, |X | ≥ |N (V (C))| ≥ k. Weclaim that Q has at most one component of order at least k. Suppose to the contrarythat Q has two components of order at least k. Then |V (G)| ≥ |V (Q)| + |X | ≥q1 + 2k + k = 3k + q1 > |V (G)|, a contradiction.

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If Q contains no component of order at least k, then, by (3), q1 ≥ (k + 1)(d + 2)

and so |V (G)| ≥ q1 +|X | ≥ (k + 1)(d + 2)+ k ≥ 3k + 2 > |V (G)|, a contradiction.If Q contains exactly one component of order ≥ k, then, by (3), q1 + k ≥ (k +

1)(d+2) and so |V (G)| ≥ q1+k+|X | ≥ (k+1)(d+2)+k > |V (G)|, a contradictionagain. This completes the proof of the lemma. ��Proof of Theorem 1.4 If k > n, we can easily find k − n disjoint perfect matchingsby Lemma 2.2. So it is sufficient to show that an [n, n + 1]-graph G has at least� n

3 � + 1 disjoint perfect matchings. Let M = {M1, M2, . . . , Mt } be a maximum setof disjoint perfect matchings in G. Argue by contradiction. Suppose t < � n

3 � + 1. LetH = G −∪t

i=1 Mi . Then H is an [h, h + 1]-graph with h = n − t > n − (� n3 � + 1

) =� 2n

3 � − 1.If H is connected, since h ≥ 2n

3 , by Lemma 2.3, H contains a perfect matchingMt+1, contradicts the maximality of M.

So H must be disconnected. Since δ(H) = h ≥ 2n3 , each component of H is of

order at least 2n3 + 1. This implies that H has exactly two components, each of which

is of odd order (otherwise, by Lemma 2.3, each component has a perfect matching,the union of the two perfect matchings in the two components is a perfect matchingof H , a contradiction with the maximality of M). Let H1 and H2 be the two com-ponents of H and V1 = V (H1) and V2 = V (H2). Assume that |V1| ≤ |V2|. Then2n3 + 1 ≤ |V1| ≤ |V2| ≤ 4n

3 − 1.We claim that at least one Mi (1 ≤ i ≤ t) such that eMi (V1, V2) ≥ 3. Note that

G = H ∪ M1 ∪ · · · ∪ Mt . Then eG(V1, V2) = ∑ti=1 eMi (V1, V2). Since δ(G) = n,

eG(V1, V2) ≥ |V1|(n − |V1| + 1). Hence

eG(V1, V2)

t≥ |V1|(n − |V1| + 1)

t≥ 3,

the second inequality holds since |V1|(n−|V1|+1) ≥ n for 1 ≤ |V1| ≤ n and t ≤ � n3 �.

So there must exist at least one 1 ≤ i ≤ t such that eMi (V1, V2) ≥ 3.Suppose eM1(V1, V2) ≥ 3. Let {x1x2, y1 y2, z1z2} ⊆ EM1(V1, V2) with x j , y j , z j ∈

Vj for j = 1, 2. Since δ(Hj ) = h ≥ 2n3 ≥ |V (Hj )|

2 , by Lemma 2.2, Hj contains aHamilton cycle C j for j = 1, 2. Choose a matching Fj from C j such that x j is notsaturated by Fj , this can be done since C j is a Hamilton cycle of odd length. LetM0 = F1 ∪ F2 ∪ {x1x2}. Then M0 is a perfect matching of H ∪ M1. Let H ′ =(H ∪ M1) − M0. Then �(H ′) = δ(H ′) + 1 = δ(H) + 1 = h + 1 ≥ 2n

3 + 1 andH ′ is connected since H ′[V1], H ′[V2] are Hamiltonian and H ′[V1] and H ′[V2] areconnected by y1 y2 and z1z2. By Lemma 2.3, H ′ contains a perfect matching Mt+1.Then {M0, M2, . . . , Mt , Mt+1} is a bigger set of disjoint perfect matchings in G thanM, a contradiction. This completes the proof of the Theorem. ��

References

1. Bondy, J.A., Murty, U.S.R.: Graph theory with applications. North-Holland, New York (1976)2. Dirac, G.A.: Some theorems on abstract graphs. Proc. London Math. Soc. 2, 69–81 (1952)3. Hilton, A.J.W.: Factorizations of regular graphs of high degree. J. Graph Theory 9, 193–196 (1985)

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4. Katerinis, P.: Maximum matchings in a regular graph of specified connectivity and bounded order.J. Graph Theory 11, 53–58 (1987)

5. Petersen, J.: Die theorie der regularen graphen. Acta Math. 15, 163–220 (1891)6. Tutte, W.T.: The factorization of linear graphs. J. London Math. Soc. 22, 107–111 (1947)7. Zhang, C.Q., Zhu, Y.J.: Factorizations of regular graphs. J. Combin. Theory, Series B 56, 74–89 (1992)

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