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ISRN Discrete Mathematics
Volume 2013 (2013), Article ID 164535, 4 pages
http://dx.doi.org/10.1155/2013/164535
Research Article

Removable Cycles Avoiding Two Connected Subgraphs

Department of Mathematics, University of Pune, Pune 411007, India

Received 28 December 2012; Accepted 23 January 2013

Academic Editors: M. Chlebík and P. Lam

Copyright © 2013 Y. M. Borse and B. N. Waphare. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We provide a sufficient condition for the existence of a cycle in a connected graph which is edge-disjoint from two connected subgraphs and of such that is connected.

1. Introduction

Prompted by Hobbs' conjecture, Jackson [1] proved that if is a 2-connected simple graph of minimum degree at least and , then contains a cycle such that is 2-connected. Lemos and Oxley [2] generalized this result as follows.

Theorem 1. Let be a 2-connected simple graph and let be a subgraph of such that is either 2-connected or . Suppose that for all in and has a cycle that is edge-disjoint from . Then there exists a cycle in that is edge-disjoint from such that is 2-connected.

Borse and Waphare [3] obtained the following result for the class of connected graphs which is analogous to the above theorem and is an improvement of a result due to Sinclair [4].

Theorem 2. Let be a connected simple graph and let be a connected subgraph of such that there is a cycle in that is edge-disjoint from . Suppose that for all . Then there exists a cycle in that is edge-disjoint from such that is connected.

The problem of the existence of cycles in graphs deletion of whose edges preserve connectedness is well studied in the literature (see [110]). In this paper, we improve Theorem 2 as follows.

Theorem 3. Let be a connected simple graph and let and be connected subgraphs of . Suppose that there are at least two cycles in and further, no two edges of form an edge-cut of . Then there exists a cycle in that is edge-disjoint from both and such that is connected.

Let be a connected graph of minimum degree at least three and let and be connected subgraphs of . If has only one cycle, say , then is not connected if a subset of forms an edge-cut of . Therefore, in Theorem 3, we must assume that contains at least two cycles. The following example shows that the condition in Theorem 3 regarding an edge-cut of is necessary. Let and be two disjoint copies of with . Suppose that is an edge of for . Let for . Let be the graph obtained from and by adding two new vertices , and seven new edges , , , , , , and ; see Figure 1. Then is simple, connected and the minimum degree of is 3. Let ,   , and . Then , and are the only cycles of . For , it is easy to see that is not connected as the set is a 2-edge-cut of for each .

164535.fig.001
Figure 1

As a consequence of Theorem 3, we get the following result.

Corollary 4. Let be an Eulerian 3-edge-connected simple graph containing at least four mutually edge-disjoint cycles. Then there exist three mutually edge-disjoint cycles , and such that is connected for .

We refer to [11] for the terms that are not defined here.

2. Proof

In this section, we prove Theorem 3 and its corollary. First we prove the following lemma.

Lemma 5. Let , and be as in the hypothesis of Theorem 3. If there is an edge of joining and , then there exists a cycle in that is edge-disjoint from both and such that is connected.

Proof. Let be an edge of , having one end vertex in and the other end vertex in . Let . Then is a connected subgraph of . Let and be cycles in . If or does not contain , then it is a cycle in . If belongs to both and , then contains a cycle. Thus, in any case, there is a cycle in . Since is connected for every subset of edges of the graph with ,   for all . Hence, by Theorem 2, there exists a cycle in such that is connected. Obviously, is edge-disjoint from both and .

The number of edges of a cycle or a path is denoted by . If and are distinct vertices of a cycle of a graph , then and denote the edge-disjoint paths along which join to .

Proof of Theorem 3. Suppose that the result is false and let be a counterexample. Then there exist connected subgraphs and of satisfying the hypothesis of the theorem such that for every cycle in that is edge-disjoint from , the graph is not connected. Choose and such that is maximum. By a hypothesis, for all . Therefore, by Theorem 2, is not connected. Thus is vertex disjoint from .
Claim 1. If is a cycle in , then every component of contains or .
Let be a cycle in . Since is connected, it is contained in a component of for . If ; then, by Theorem 2, there exists a cycle in such that is connected, which is a contradiction. Therefore . We prove that and are the only components of . Let . It is sufficient to prove that is an empty set. Assume that is nonempty. Since is connected, each component of contains a vertex of . If a component of contains exactly one vertex of , then the edges incident with that vertex form a -edge-cut of , which is a contradiction. Hence each component of contains at least two vertices of . Hence . Therefore there is a component of different from and such that . Let . Let be a path in between two vertices of . Then contains at least two cycles which are edge-disjoint with . By the maximality of , we have and . By Lemma 5, there is no edge between and . Suppose that . Then . It follows that has exactly three components and hence we get two edges of , with each having one end in and the other in such that is an edge-cut of , a contradiction. Thus .
Let . Then . By traversing from along on both sides, let be the first vertex of on one side and let be the first vertex of on the other side. By Lemma 5, . We may assume that . Then . Let . Then is connected and . By the maximality of ,   contains at most one cycle. Let be the component of containing . Then and . If , then contains two cycles, which is a contradiction. Hence . This implies that there is a component of such that . Traverse the path beginning at , let be the second vertex of on this path. Let be a path in . Let . Then is a cycle. If does not meet , then intersects in at least two vertices giving a cycle which is disjoint from , a contradiction. Hence contains a vertex of . If this path contains another vertex of , then contains two cycles, where is an path in . This is a contradiction. Hence . This implies that the path meets in a vertex . Let be an path in . Then contains two cycles, which is a contradiction. This proves the claim.
Claim 2. If and are cycles in , then .
Assume that the claim is false. Let and be cycles in such that . Then is edge-disjoint from . By Claim 1, each component of contains or . Thus has exactly two components. We may assume that is contained in a component of for . Obviously, is contained in or . We may assume that is contained in . By the maximality of , it follows that . By Claim 1, has only two components and , with each containing or . We may assume that is contained in for . Since is connected, it is contained in . Let be the set of all edges in , having one end vertex in and the other end vertex in . Then . By a hypothesis, . Let be the set of all edges in , having one end vertex in and the other end vertex in . Then and .
If is connected, then is connected, a contradiction. Therefore we may assume that is disconnected. Obviously, is a component of . Let . Note that ,   ,   , and . Hence . Let . Then is vertex disjoint from both and . If contains a cycle then contains two cycles and hence we get two cycles in . By the maximality of , we get a cycle that is edge-disjoint from both and such that is connected, which is a contradiction. Thus is a forest. Let . Then is a forest. Every edge of with one end vertex in and the other end vertex in is either in or in and hence or . As for all and , at most one vertex of is isolated. Hence contains an edge. Let be a nontrivial tree in . Let and be two pendant vertices of . Since , for . We may assume that . Suppose that . Let and let be the path in . Then contains at least two cycles. These cycles are edge-disjoint from and . Now, a contradiction follows from the maximality of . Hence . Similarly, . This implies that for . If has a third pendant vertex , then is in or , which is a contradiction. Hence has exactly two pendant vertices and therefore it is a path. As for all ,   . This implies that has no internal vertex. Hence is isomorphic to . Thus every component of is isomorphic to either or . Let be an edge of with and . Suppose that has an another edge . Then for each . We may assume that and . Since , we may assume by suitable labeling that at least one edge of is not in . It follows that the subgraph is edge-disjoint from the subgraph and further, it contains at least two cycles. Obviously, is connected and contains more edges than . A contradiction follows from the maximality of . Hence contains exactly one edge. Since ,   . As at most one vertex of is isolated, . Let . We may assume that ,   ,   , and . Obviously, or . We may assume that . The subgraph contains two cycles and further, it is edge-disjoint from the subgraphs and . A contradiction follows from the maximality of . The proof of Claim 2 is completed.
By Claim 2, any two cycles in has at least two vertices in common. This implies that there exist two vertices and and three mutually internally disjoint , paths , , and in . We may assume that . Let . Then is a cycle in . By Claim 1, has exactly two components each containing or . Let be the component of containing for . We may assume that the path is contained in the component . Hence the vertices and are in . It follows from the maximality of that . Let be the set of all edges in that have one end vertex in and the other end vertex in . Then . By a hypothesis, . In fact, . By Lemma 5, no edge of has an end vertex in as . Obviously, or . Let if ; otherwise . Thus is a cycle in that is edge-disjoint from both and and further, and for some . By Claim 1, has exactly two components and . We may assume that is contained in and is contained in . As ,   is connected and hence contains . So the vertices and do not belong to . Thus has at least two components, one of which is .
By a hypothesis, . Further, every component of contains at least one vertex of and hence contains at least one vertex of . Suppose that there is a component of containing two vertices and of with . Let be an path in . Obviously, contains at least two cycles. Thus there are at least two cycles in . Further, . Since and are connected and are vertex disjoint subgraphs, by the maximality of , there exists a cycle in that is edge-disjoint from and hence edge-disjoint from such that is connected, which is a contradiction. Thus has exactly one vertex in common with every component of that is different from . Since and are common vertices of and , there are two components and of , both different from such that and .
Suppose that , , and are the only components of . Let . Then . Suppose that does not meet . Then, as meets each of and only once, the corresponding two edges of form a 2-edge-cut of , which is a contradiction. Therefore meets . Since is a path and for . If meets , then is connected, which is a contradiction as is a cycle in . Thus does not meet . Hence . This shows that . Traverse the path beginning at and suppose that meets at the first time in the vertex . Let be the last vertex of on the subpath of . Then the subpath of contains at most two edges of . Extend the path to a path by adding a path in and an path in . Thus . As contains at least two cycles, there are at least two cycles that are edge-disjoint from the connected subgraphs and of . Since is contained in ,   is contained in , and is connected, by the maximality of , there exists a cycle that is edge-disjoint from such that is connected, which is a contradiction. Thus has at least four components.
Since and are the only components of , the path meets every component of except . Also, . It follows that we can choose two components and of and two vertices ,   such that for the subpath of we have . Let and let and be paths in and , respectively, from to and to . Then contains at least two cycles and further, it is edge-disjoint from , and . Due to the maximality of , there exists a cycle that is edge-disjoint from both and such that is connected, which is a contradiction. This completes the proof of the theorem.

Proof of Corollary 4. Let , , and be mutually edge-disjoint cycles in . By Theorem 3, there exists a cycle in such that is connected. Since is eulerian, every component of is eulerian. Hence has a cycle decomposition containing the cycle . Since and are edge-disjoint from and contains at least one cycle other than . As is edge-disjoint from , by Theorem 3, there exists a cycle in such that is connected. Obviously, each component of is eulerian. Further, is a cycle in . There exists a cycle decomposition of containing . It is easy to see that contains at least one edge. Hence contains at least one edge which is not in . This implies that contains a cycle with . Hence, by Theorem 3, there exists a cycle in such that is connected. Obviously, ,   , and are mutually edge-disjoint cycles.

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