Graphs and CombinatoricsDOI 10.1007/s00373-013-1358-0

ORIGINAL PAPER

Contractible Edges in k-Connected Graphswith Some Forbidden Subgraphs

Yingqiu Yang · Liang Sun

Received: 25 September 2011 / Revised: 18 November 2012© Springer Japan 2013

Abstract In 2001, Kawarabayashi proved that for any odd integer k ≥ 3, if a k-connected graph G is K −

4 -free, then G has a k-contractible edge. He pointed out, bya counterexample, that this result does not hold when k is even. In this paper, we haveproved the following two results on the subject: (1) For any even integer k ≥ 4, if ak-connected graph G is K −

4 -free and dG(x) + dG(y) ≥ 2k + 1 hold for every twoadjacent vertices x and y of V (G), then G has a k-contractible edge. (2) Let t ≥ 3,k ≥ 2t − 1 be integers. If a k-connected graph G is (K1 + (K2 ∪ K1,t ))-free anddG(x) + dG(y) ≥ 2k + 1 hold for every two adjacent vertices x and y of V (G), thenG has a k-contractible edge.

Keywords Fragment · Minimum fragment · Contractible edge · k-connected graph

Mathematics Subject Classification (2000) 05C40

1 Introduction

In this paper, we only consider finite undirected simple graphs. For the notations andterminologies not defined here we refer the reader to [6]. Let G = (V (G), E(G))

be a simple graph with vertex set V (G) and edge set E(G). For a vertex x of V (G),we write dG(x) for the degree of x . We denote the neighborhood of x by NG(x)

and let NG[x] = NG(x) ∪ {x}. Let S be a subset of V (G). We write NG(S) =

The project is supported by National Natural Science Foundation of China (Grant Number: 11071016).

Y. Yang (B) · L. SunSchool of Mathematics, Beijing Institute of Technology, Beijing 100081, Chinae-mail: yyq0227@yahoo.com.cn

L. Sune-mail: lsunn@sina.com

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(⋃

x∈S NG(x))− V (S) for the neighborhood of S in G. Let G[S] denote the subgraphof G induced by S. By E(x) we denote the set of edges incident with x . We denotethe minimum degree of G by δ(G). For any two disjoint nonempty subsets A and Bof V (G), we write EG(A, B) for the set of edges between A and B. For two graphs Gand H with V (G) ∩ V (H) = ∅, the union of G and H is G ∪ H , and the join G + Hof G and H is the graph obtained from G ∪ H by joining each vertex of G to eachvertex of H . We write Kn, K1,n−1 and Pn for the complete graphs, stars and paths oforder n, respectively.

Let G be a k-connected graph. For a subset T of V (G), if G − T has at least twocomponents, then T is a cutset of G. If |T | = k, then T is said to be a smallest cutsetor a k-cutset of G. Let G be a noncomplete k-connected graph, and T a smallest cutsetof G. If F is an union of at least one but not all components of G − T , then F is calleda fragment or a T -fragment of G. Set F = G − T − F . Then F is also a T -fragmentof G. An inclusion-minimal fragment is called an end.

An edge of a k-connected graph G is said to be a k-contractible edge or simply acontractible edge if its contraction yields again a k-connected graph. Otherwise, wecall it noncontractible edge. It is obvious that two endvertices of a noncontractibleedge of G are contained in a smallest cutset of G, if G is not complete.

Let K −n denote the graph obtained from Kn by removing just one edge.

If a graph H is not isomorphic to any subgraph of G, then H is a forbidden subgraphof G, or G is H -free.

The following are several results about the contractible edges in k-connected graphswith forbidden subgraphs.

Theorem A [17] If G is a k-connected triangle-free graph, then G contains acontractible edge.

Theorem B [9] Let k ≥ 3 be an odd integer. If a k-connected graph G is K −4 -free,

then G has a contractible edge.

Theorem C [10] Let k ≥ 5 be an odd integer. If a k-connected graph G is (K1 +(K2 ∪ P3))-free, then G has a contractible edge.

Clearly, Theorems B and C are extensions of Theorem A when k is odd. Some otherrelated results can be found in [1–3,18].

By the results of [4,7,8,11,14,16,19,20], Su et al. [15] proved the following resultabout the contractible edges in k-connected graphs with a degree sum condition.

Theorem D [15] Let G be a noncomplete k-connected graph, and k �= 7. If dG(x) +dG(y) ≥ 2� 5k

4 − 1 hold for every two adjacent vertices x and y of V (G), then G hasa contractible edge.

In [9], Kawarabayashi gave an example to show that the result in Theorem B is nottrue when k is even. In this paper, we explore the existence of k-contractible edges inK −

4 -free k-connected graphs when k is even, and the following result is obtained.

Theorem 1 Let k ≥ 4 be an even integer, and G a K −4 -free graph with connectivity

k. If dG(x) + dG(y) ≥ 2k + 1 hold for every two adjacent vertices x and y of V (G),then G contains a contractible edge.

Moreover, we generalize Theorem C and obtain the following result.

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Theorem 2 Let t ≥ 3, k ≥ 2t − 1 be integers. Let G be a (K1 + (K2 ∪ K1,t ))-freegraph with connectivity k. If dG(x) + dG(y) ≥ 2k + 1 hold for every two adjacentvertices x and y of V (G), then G contains a contractible edge.

Clearly, the forbidden subgraph conditions of Theorems 1 and 2 relax the bound ofdegree sum in Theorem D to 2k + 1.

2 Some Lemmas

Let G be a k-connected graph. For e = xy ∈ E(G), suppose that S is a smallest cutsetof G such that {x, y} ⊆ S. If F is a fragment of G − S, then we call it a fragment withrespect to e. For any X ⊆ E(G), F is called a fragment with respect to X , if F is afragment with respect to some edge e ∈ X . Among all fragments with respect to e, aninclusion-minimal fragment is called an end (or a minimal fragment) with respect to e.A fragment with respect to e with minimum cardinality is called a minimum fragmentwith respect to e. Similarly, we have the concept of an end (resp. minimum fragment)with respect to X . We denote the cardinality of a minimum fragment with respect toe by ϕ(e). Denote J (G) = {e ∈ E(G) | e is noncontractible and ϕ(e) ≥ � k+1

2 �},Q1(G) = {e ∈ E(G) | e is noncontractible and ϕ(e) ≥ k + 1}, Q2(G) = {e ∈ E(G)

| e is noncontractible and ϕ(e) ≥ k}, M = {e ∈ E(G) | e is not contained in anytriangle of G }, N = {e = xy ∈ E(G) | e is contained in some triangles of G, andevery triangle xyzx containing e = xy satisfies that dG(z) ≥ k + 1}. It is obvious thatQ1(G) ∪ Q2(G) ⊆ J (G).

Lemma 1 [3,4,12,13] Let G be a k-connected graph. Suppose that F is a nonemptysubset of J (G). If B is an end with respect to F, then E(x)∩ F = ∅ for all x ∈ V (B).

Lemma 2 [4,5] Let G be a k-connected graph. If there is a vertex x ∈ V (G) suchthat E(x) ⊆ J (G), then G has a contractible edge.

Lemma 3 [3] Let G be a graph, and W a subset of V (G). Then

∑

x∈V (G)−W

|NG(x) ∩ W | =∑

y∈W

dG(y) − 2|E(G[W ])|.

Lemma 4 Let k ≥ 4 be an integer. Let G be a K −4 -free k-connected graph, and S a

k-cutset of G with E(G[S]) �= ∅. Suppose that A is a fragment of G − S and |A| ≥ 2.

1. If δ(G) ≥ k + 1, then |A| ≥ k + 1.2. If dG(x)+ dG(y) ≥ 2k + 1 hold for every two adjacent vertices x and y of G, then

|A| ≥ k.

The proof of Lemma 4 is easy and is left to reader. �Lemma 5 Let G be a K −

4 -free k-connected graph such that dG(x)+dG(y) ≥ 2k +1hold for every two adjacent vertices x and y of G. If e is a noncontractible edge of Gsuch that e ∈ M ∪ N, then e ∈ Q2(G).

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Proof Assume that e = uv is a noncontractible edge of G. Then there is a k-cutset Sof G such that {u, v} ⊆ S. Let A be a fragment of G − S. We choose S and A suchthat A is a minimum fragment with respect to e = uv. By the minimality of A, A isconnected.

First suppose that e ∈ M . If V (A) = {x, y}, then xy ∈ E(G) and dG(x)+dG(y) ≥2k + 1. This contradicts that e is not contained in any triangles. Thus |A| ≥ 3. Sincee ∈ E(G[S]) �= ∅, we have |A| ≥ k by Lemma 4. Thus e ∈ Q2(G).

Now, suppose that e = uv ∈ N . Because G is a K −4 -free k-connected graph, uv is

contained in an unique triangle uvzu with dG(z) ≥ k + 1. If |A| = 1 and A = {z},then e is contained in a triangle uvzu and dG(z) = k, a contradiction. So |A| ≥ 2.Since e ∈ E(G[S]) �= ∅, we have |A| ≥ k by Lemma 4. Thus e ∈ Q2(G). �Lemma 6 Let k ≥ 4 be an integer and G be a K −

4 -free k-connected graph.

(a) If δ(G) ≥ k + 1, then every vertex of G is incident with a contractible edge.(b) If dG(x) + dG(y) ≥ 2k + 1 for all adjacent vertices x and y, then every vertex

of G with degree k is incident with a contractible edge.

Proof (a) On the contrary, suppose x ∈ V (G) with the property that every edgeof E(x) is noncontractible. We first prove that E(x) ⊆ Q1(G). Let xy be anarbitrary edge of E(x). Since xy is a noncontractible edge in G, x and y arecontained in a k-cutset of G. Let A be a minimum fragment with respect to xy.Then T = NG(A) is a k-cutset of G such that {x, y} ⊆ T . Clearly, |A| ≥ 2 sinceδ(G) ≥ k + 1. Thus |A| ≥ k + 1 by Lemma 4. Therefore, we have xy ∈ Q1(G).Since the choice of xy is arbitrary from E(x), we have E(x) ⊆ Q1(G) ⊆ J (G).Assume that B is an end with respect to E(x). Then S = NG(B) is a k-cutset ofG such that some edge xy of E(x) is contained in E(G[S]). By Lemma 1, wehave that E(u) ∩ E(x) = ∅ for all u ∈ V (B). This contradicts that x ∈ NG(B).

(b) By contradiction. Assume that x is a vertex of G with degree k such that anyedge of E(x) is noncontractible. By the degree sum condition, for any vertexy ∈ NG(x), we have dG(y) ≥ k + 1. Now, we claim that E(x) ⊆ Q2(G).Let xy be an arbitrary edge of E(x). If xy is not contained in any triangle, thenxy ∈ M . If xy is contained in a triangle xyzx , then there is only one such trianglesince G is K −

4 -free, and dG(z) ≥ k + 1 since xz ∈ E(G), dG(x) = k anddG(x) + dG(z) ≥ 2k + 1. Thus xy ∈ N . Then xy ∈ Q2(G) by Lemma 5.Therefore E(x) ⊆ Q2(G) ⊆ J (G). Similar to the proof of (a), we can get acontradiction and prove (b).

�

3 Proof of Theorem 1

Let k ≥ 4 be an even integer. Assume that G is a K −4 -free graph with connectivity k

and dG(x)+ dG(y) ≥ 2k + 1 hold for every two adjacent vertices x and y of V (G). Ifδ(G) ≥ k + 1, then G contains a contractible edge by Lemma 6(a). If δ(G) = k and xis a vertex of G with degree k, then E(x) contains a contractible edge by Lemma 6(b).Thus, G contains a contractible edge. This completes the proof of Theorem 1. �

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4 Proof of Theorem 2

By contradiction. Let t ≥ 3, k ≥ 2t − 1 be integers. Suppose that G is a k-connectedgraph without K1 + (K2 ∪ K1,t ) and dG(x) + dG(y) ≥ k + 1 hold for every twoadjacent vertices x and y of V (G), but G has no contractible edge. Especially, G isnot complete.

Claim 1 Let S be a k-cutset of G, and A a minimum fragment of G − S. Then |A| ∈{1, 2} or |A| ≥ � k+1

2 �.

Proof Assume that |A| ≥ 3. Now, we prove that |A| ≥ � k+12 �. Denote H = G[S ∪

V (A)]. It is obvious that NG(u) = NH (u) for every u ∈ V (A). Thus dG(u) = dH (u).By the minimality of A, A is connected. Since |A| ≥ 3, A contains a path P3 = x1x2x3with three vertices. Denote W = {x1, x2, x3}, and Q = V (A) ∪ S − W .

Define

ξ = |{u|u ∈ NG(x1) ∩ NG(x2) ∩ NG(x3)}|,γ1 = |{u|u ∈ NG(x2) ∩ NG(x3) − NG [x1]}|,γ2 = |{u|u ∈ NG(x1) ∩ NG(x3) − NG[x2]}|,γ3 = |{u|u ∈ NG(x1) ∩ NG(x2) − NG [x3]}|.

Now we discuss two cases.

Case 1 ξ ≥ 1.Since G is (K1 + (K2 ∪ K1,t ))-free, we have (1) 1 ≤ ξ ≤ t ; (2) if x1x3 ∈ E(G),

then 1 ≤ ξ + γi ≤ t, i = 1, 2, 3; (3) if x1x3 �∈ E(G), then 1 ≤ ξ + γi ≤ t, i = 1, 3.

Subcase 1.1 x1x3 ∈ E(G).Because the sum of degrees of every two adjacent vertices is greater than or equal

to 2k + 1, we have dG(x1)+ dG(x2)+ dG(x3) ≥ 3k + 2. Thus, by Lemma 3, we have

3k + 2 ≤∑

x∈W

dG(x) = 2|E(G[W ])| + |E(W, Q)|

≤ 3 × 2 + 3ξ + 2(γ1 + γ2 + γ3) + (|A| + k − 3 − ξ − (γ1 + γ2 + γ3))

= 3 + 2ξ + γ1 + γ2 + γ3 + |A| + k. (∗)

Suppose that ξ + γ1 = t . If γ2 �= 0 or γ3 �= 0, then H contains K1 + (K2 ∪ K1,t ),a contradiction. Thus γ2 = γ3 = 0. Hence, by (∗), we have

3k + 2 ≤ 3 + 2ξ + γ1 + |A| + k= 3 + ξ + t + |A| + k≤ 3 + 2t + |A| + k.

Thus, |A| ≥ 2k − 2t − 1. This implies that |A| ≥ � k+12 �.

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Now, suppose that 1 ≤ ξ + γ1 ≤ t − 1, then 1 ≤ ξ + γi ≤ t − 1, i = 2, 3. By (∗),we get

3k + ξ − 1 ≤ (ξ + γ1) + (ξ + γ2) + (ξ + γ3) + |A| + k≤ 3(t − 1) + |A| + k.

Thus, |A| ≥ 2k − 3t + 2 + ξ ≥ 2k − 3t + 3. Therefore, |A| ≥ � k+12 �.

Subcase 1.2 x1x3 �∈ E(G).We may assume that 1 ≤ ξ + γ1 ≤ t . By the degree sum condition, we have

k + |A| ≥ |NG(x2)| + |NG(x3)| − |NG(x2) ∩ NG(x3)|≥ 2k + 1 − ξ − γ1≥ 2k + 1 − t.

Hence, |A| ≥ k + 1 − t . So we have |A| ≥ � k+12 �.

Case 2 ξ = 0.If γ1 ≥ t , then γ3 = 0 obviously. By the degree sum condition, we have

k + |A| ≥ |NG(x1)| + |NG(x2)| − |NG(x1) ∩ NG(x2)|≥ 2k + 1 − 1= 2k.

Thus, we have |A| ≥ k ≥ � k+12 �.

Now suppose that γ1 ≤ t − 1. Then, by the degree sum condition, we have

k + |A| ≥ |NG(x2)| + |NG(x3)| − |NG(x2) ∩ NG(x3)|≥ 2k + 1 − γ1 − 1≥ 2k − t + 1.

Thus, |A| ≥ k + 1 − t . Hence, |A| ≥ � k+12 �. �

Claim 2 If x is a vertex of G with degree k, then E(x) ⊆ J (G).

Proof Let xy be an edge in E(x). If xy is not contained in any triangle, then we havexy ∈ M . If xy is contained in some triangles, then any triangle xyzx containing xysatisfies that dG(z) ≥ k +1 by the degree sum condition. It follows that xy ∈ N . Thusxy ∈ M ∪ N . Let A be a minimum fragment with respect to xy. Then S = NG(A) isa k-cutset in G such that {x, y} ⊆ S.

First suppose that xy ∈ M . Then |A| ≥ 2. If |A| = 2 and A = {a, b}, then a andb are adjacent in G. Thus dG(a) + dG(b) ≥ 2k + 1, and axya or bxyb is a triangle.This contradicts that xy ∈ M . Hence |A| ≥ 3. Therefore, xy ∈ J (G) by Claim 1.

Now, suppose that xy ∈ N . If |A| = 1 and A = {a}, then axya is a triangleand dG(a) = k, contradicting that xy ∈ N and dG(x) + dG(a) ≥ 2k + 1. Hence,|A| ≥ 2. Assume that |A| = 2 and A = {a, b}. Then a and b are adjacent in G, and

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dG(a)+dG(b) ≥ 2k +1. If dG(a) = dG(b) = k +1, then S ⊆ NG(a)∩ NG(b). ThusG[A ∪ S] contains K1 + (K2 ∪ K1,t ), a contradiction. Hence, we have dG(a) = k + 1and dG(b) = k, or dG(a) = k and dG(b) = k + 1. Since dG(x) = k, we haveG[{x, y, a, b}] ∼= K −

4 . Now, without loss of generality, we assume that dG(a) =k + 1, dG(b) = k. Then, NG(a) = S ∪ {b}, NG(b) = (S − {x}) ∪ {a}. So we have|(NG(a)∩NG(b))∩S−{x, y}| ≥ k−2 ≥ t . Thus G[A∪S] contains K1+(K2∪K1,t ),a contradiction. Hence |A| ≥ 3. Thus xy ∈ J (G) by Claim 1.

Hence xy ∈ J (G) for any edge xy ∈ E(x). Therefore E(x) ⊆ J (G). �Claim 3 If δ(G) ≥ k + 1, then E(G) = J (G).

Proof Let e = xy be an arbitrary edge in G. Suppose that A is a minimum fragmentwith respect to e = xy. Let S = NG(A) be a k-cutset of G such that {x, y} ⊆ S. Sinceδ(G) ≥ k + 1, we have |A| ≥ 2. If A = {a, b}, then we have S ∪ {b} ⊆ NG(a) andS ∪ {a} ⊆ NG(b). Thus G[A ∪ S] contains K1 + (K2 ∪ K1,t ), a contradiction. So,|A| ≥ 3, and thus xy ∈ J (G) by Claim 1. Therefore, E(G) = J (G). �

Now, we continue the proof of Theorem 2. If δ(G) ≥ k + 1, then by Claim 3, wehave E(G) = J (G). Let B be an end with respect to J (G). Then E(x) ∩ J (G) = ∅for all x ∈ V (B) by Lemma 1. This contradicts that E(G) = J (G). Now we assumethat δ(G) = k, and x is a vertex of G with degree k. Then we have E(x) ⊆ J (G)

by Claim 2. By Lemma 2, G contains a k-contractible edge, a contradiction. Thiscompletes the proof of Theorem 2. �Acknowledgments The authors would like to thank the referees for their valuable suggestions.

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