Mathematical Problems in Engineering

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Bilevel Programming, Equilibrium, and Combinatorial Problems with Applications to Engineering 2016

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Volume 2016 |Article ID 5384190 | https://doi.org/10.1155/2016/5384190

Zhihong He, Xiaoying Wang, Caiming Zhang, "Complementary Cycles in Irregular Multipartite Tournaments", Mathematical Problems in Engineering, vol. 2016, Article ID 5384190, 7 pages, 2016. https://doi.org/10.1155/2016/5384190

Complementary Cycles in Irregular Multipartite Tournaments

Academic Editor: Sergii V. Kavun
Received29 Mar 2016
Accepted17 Jul 2016
Published15 Aug 2016

Abstract

A tournament is a directed graph obtained by assigning a direction for each edge in an undirected complete graph. A digraph is cycle complementary if there exist two vertex disjoint cycles and such that . Let be a locally almost regular -partite tournament with and such that all partite sets have the same cardinality , and let be a -cycle of . In this paper, we prove that if has no cycle factor, then contains a pair of disjoint cycles of length and , unless is isomorphic to , , , or .

1. Introduction

Denote the vertex set and arc set of a digraph by and , respectively. For a subset of , we define as the subdigraph induced by . If is an arc of a digraph , then we write and say dominates . If and are two disjoint subsets of or subdigraphs of such that every vertex of dominates every vertex of , then dominates , denoted by . We write if there is no arc from to .

The out-neighborhood of a vertex is the set of vertices dominated by , and the in-neighborhood is the set of vertices dominating . The numbers and are the outdegree and indegree of , respectively. The minimum outdegree and the minimum indegree of are denoted by and and . The local irregularity is defined by over all vertices of and the global irregularity is defined by . Clearly, . If , then is regular and if , then is almost regular. If , then is locally almost regular.

By a path or cycle we mean a directed path or cycle. A cycle of length is an -cycle. The independence number is the maximum size among all independent sets of vertices of . A digraph is strongly connected or strong if, for each pair of vertices and , there is a path from to in . A digraph with at least vertices is -connected if for any set of at most vertices, the subdigraph obtained by deleting is strong. The connectivity of , denoted by , is defined to be the largest value of such that is -connected.

A cycle factor is a spanning subdigraph consisting of disjoint cycles. A cycle factor with the minimum number of cycles is called a minimal cycle factor. If we replace every arc of by , then we call the resulting digraph, denoted by , the converse digraph of .

A -partite graph is a graph whose vertices are or can be partitioned into different independent sets. Equivalently, it is a graph that can be colored with colors, so that no two endpoints of an edge have the same color. A complete -partite graph is a graph that can be partitioned into independent sets, as above, so that every pair of vertices from two different independent sets have different colors.

A -partite or multipartite tournament is an orientation of a complete -partite graph. If is a -partite tournament with the partite sets such that , then we define . If the vertex of belongs to the partite set , then we define .

A digraph is cycle complementary if there exist two vertex disjoint cycles and such that . The problem of complementary cycles in tournaments was almost completely solved by Reid [1] in 1985 and by Song [2] in 1993. The authors proved that every -connected tournament on at least vertices has complementary cycles of length and for all .

There exists only the following conjecture.

Conjecture 1 (see Yeo [3] 1999). A regular -partite tournament with and has a pair of vertex disjoint cycles of length and for all .

In 2005, Volkmann [4] confirmed the first conjecture for , unless is a regular -partite tournament with two vertices in each partite set. In 2004, Volkmann [5] confirmed conjecture for . In 2009, He et al. [6] confirmed conjecture for . In 2013, Volkmann [7] confirmed every almost regular c-partite tournament with such that all partite sets have the same cardinality . If or and , then D contains a pair of vertex disjoint directed cycles of lengths and . In 2014, Bai et al. [8] confirmed every regular bipartite tournament contains two complementary cycles. In 2014, He et al. [9] characterized all locally almost regular multipartite tournaments whose partite sets have the same cardinality with that contain two complementary cycles of length 3 and . Let be a locally almost regular -partite tournament with and such that all partite sets have the same cardinality, and let be a -cycle of . We prove if has no cycle factor, then contains a pair of disjoint cycles of length and , unless is isomorphic to , , , or .

2. Preliminary Results

The following results play important roles in the proof of our main theorem (Theorem 12), where the first one is immediate.

Lemma 2 (see [7]). If is a regular -partite tournament with the partite sets , then .

Lemma 3 (see Yeo [10] 1998). If is a multipartite tournament, then

Lemma 4 (see Reid [1] 1985 and Song [2] 1993). If is a -connected tournament with at least vertices, then contains two complementary cycles of length and for all , unless it is isomorphic to the tournament .

Lemma 5 (see Bondy [11] 1976). Each strong -partite tournament with contains an -cycle for each .

Lemma 6 (see Yeo [12] 1999). Let be a -partite tournament with the partite sets such that for some . If then is Hamiltonian.

Lemma 7 (see Volkmann [5] 2004). Let be a regular -partite tournament with . Then contains two complementary cycles of length and , unless is isomorphic to .

Lemma 8 (see Volkmann [4] 2005). Let be a regular c-partite tournament with and . Then contains two complementary cycles of length and , unless is isomorphic to , , or .

Lemma 9 (see Tewes [13] 2002). If is a multipartite tournament, then

Lemma 10 (see Yeo [12] 1999 and Gutin and Yeo [14] 2000). A digraph has no cycle factor if and only if its vertex set can be partitioned into four subsets , and such thatwhere is an independent set.

Lemma 11 (see Volkmann and Winzen [15] 2004). Let be the partite sets of a -partite tournament with no cycle factor such that . According to Lemma 10, the vertex set can be partitioned into subsets and satisfying (4) such that with integers . Let be the partite set with the property that . If , , and , thenif or case part (1) of lemma and if and case part (2) of lemma.

3. Main Result

Theorem 12. Let be a locally almost regular -partite tournament with and such that all partite sets have the same cardinality, and let be a -cycle of . We prove that if has no cycle factor, then according to Lemma 10, the vertex set can be partitioned into subsets and satisfying (4) such that and , where is an independent set. If , then contains a pair of disjoint cycles of length and , unless is isomorphic to , , , or .

Proof. First of all notice that the condition that all partite sets have the same cardinality shows that . Let be the partite sets of a locally almost regular -partite tournament such that . If is even, then it is easy to say that is regular. Thus, if , then is a regular -partite tournament, and by Lemma 7, contains two complementary cycles of length and , unless is isomorphic to .
According to Lemma 3, we have If and , which means that is a tournament, then (7) yields . In addition, if and , then it can be shown that too. Hence Lemma 4 implies that contains two complementary cycles of length and , unless is isomorphic to .
Therefore we only need to consider the case and . In view of Lemma 3, and thus Lemma 5 ensures the existence of a -cycle in . If we define in the case the -partite tournament by , then , and . If are the partite sets of such that , then and . With exception of the cases and and , and , and and , this leads to Applying Lemma 6, we conclude that has a Hamiltonian cycle , and we obtain the desired result that . Since is a regular -partite tournament for and , and , and , and and , by Lemma 8, contains two complementary cycles of length and , unless is isomorphic to , or . Therefore, there only remain the cases and .
Let , , , be the partite sets of , , and . Then and for each . Then , and . If are the partite sets of such that , then and . Then by Lemma 10, the vertex set can be partitioned into subsets , such that , and is an independent set. Let, without loss of generality, (if , then we consider ). By Lemma 9, .
Case 1. and . Then . Otherwise and for which contradicts . Consider , for . So .
Assume that . We claim that , or else since , induces a -cycle, and thus there is a vertex such that . It follows that and , a contradiction to . Assume that . If is bipartite, then since , each partite set of consists of two vertices. Hence, induces a -cycle and thus there is a vertex such that . It follows that and , a contradiction to . If is -partite, then there exists a vertex such that and or and . This leads to and , a contradiction to . If is -partite, then such that and ; then and , a contradiction to .
If , then . Note that , induces a -cycle. So there is a vertex such that and , a contradiction to .
If , then replacing by and by in Lemma 11 leads us to , a contradiction to .
Case 2. One has and . We discuss this case by the value of , where
Case 2.1. Assume that ; that is, .
Note that since . If , for some , then . Then and for , a contradiction. Therefore, . Assume, without loss of generality, that , , and . Since , , and there exists one vertex in , say, such that .
Case 2.1.1. Assume that has a Hamiltonian path.
() Assume that there is a Hamiltonian path which does not end at or begin at , say, with in . If does not dominate , then has at least one positive neighbor in , and there are the complementary cycles and . If , then there exist at least two arcs from to . Let, without loss of generality, with and . This leads to the complementary cycles and , all indices are taken modulo 5.
() If there is a Hamiltonian path ending at in , say, with , then we need only to consider . The other cases are easily reduced to (). Thus has at least one positive neighbor in , say, . This leads to the complementary cycles and .
() If there is a Hamiltonian path beginning at in , say, with , then we need only to consider . The other cases are easily reduced to () or ().
Assume that . Then , and we obtain a -cycle . Next we are to find its complementary cycle. Since , dominates at least one vertex of , and there exists at least one vertex of , say, , such that with , and . This leads to the complementary cycles and if or if .
Now assume that . If there is a vertex of that dominates , then let take the place of , the above case being reduced to () of Case . Therefore, we let . Then we have .
If and , then there exists an arc between and . Let, without loss of generality, . This leads to the contradiction .
If and , then and and . If , then , , , and . This leads to the complementary cycles and . If , then , , , and . This leads to the complementary cycles and .
If , let, without loss of generality, ; then we have , , , , , and . This leads to the complementary cycles and .
Case 2.1.2. Now we assume that has no Hamiltonian path. Then let, without loss of generality, .
()Assume that dominates one vertex , say, .
(a) If , then since has no Hamiltonian path, , , and .
If there is one vertex of that dominates , let, without loss of generality, ; then let take the place of , the case being reduced to () of Case . Therefore, let . Then we have . Since , or . In the former case there are complementary cycles and ; in the latter case complementary cycles are and .
(b) Assume that .
If , then since has no Hamiltonian path, . This leads to the Hamiltonian path , a contradiction.
If and , then since has no Hamiltonian path, . This leads to the Hamiltonian path , a contradiction.
If and , then since has no Hamiltonian path, , , and . If or and , then let take the place of , the case being reduced to () of Case . If and , then let take the place of , the case being reduced to () of Case .
() Assume that . Then we have .
(a) If dominates one vertex of , say, , then since , has no Hamiltonian path, and .
(i) If and dominates one vertex of , then let take the place of , the case being reduced to () of Case .
(ii) If and , then let take the place of , the case being reduced to () of Case .
(iii) Assume that ; then we have .
If dominates one vertex of , say, , then we obtain the complementary cycles and .
If , then we have . Then let take the place of , the case being reduced to () of Case .
(b) Assume that . Since has no Hamiltonian path, .
(i) Assume that . Then .
If , then dominates at least one vertex of ; then let take the place of , the case being reduced to () of Case .
If and there exists one vertex of that dominates at least one vertex of , say, , then we obtain the complementary cycles and .
If and , then . This leads to the complementary cycles and .
(ii) Assume that .
If and dominates one vertex of , then let take the place of , the case being reduced to () of Case .
Assume that . Then we have and . If there exists one vertex of that dominates , say, , then let take the place of , the case being reduced to () of Case . Therefore, let . This leads to the complementary cycles and .
Now assume that . Then we have .
Let dominate one vertex of , say, .
If , then we have the complementary cycles and .
If , then since , dominates at least one vertex of . This leads to the complementary cycles and if or if .
Now let .
If dominates one vertex of , say, , then we obtain the complementary cycles and .
If , then we have and . This leads to the complementary cycles and .
Case 2.2. Assume that . If , then for . Therefore, we have and for , a contradiction. Hence, let, without loss of generality, . If , then the case is equivalent to the case of “, , and ,” which has been proved in () of Case and Case . If , then , a contradiction. So, let, without loss of generality, and . Then and .
() Assume that there is an arc from to , say, . Then we have .
(a) Assume that has a Hamiltonian path.
If there is a Hamiltonian path which does not end at or begin at , let, without loss of generality, with be a Hamiltonian path in ; then has at least one positive neighbor in , say, , and we have the complementary cycle .
If there is a Hamiltonian path ending at , let, without loss of generality, with be a Hamiltonian path in ; then we need only to consider . The other cases are easily reduced to the above case which does not end at . Thus and there is the complementary cycle .
If there is a Hamiltonian path beginning at , let, without loss of generality, with be a Hamiltonian path in ; then we need only to consider . The other cases are easily reduced to the above case which does not begin at . Thus and we have the complementary cycle .
(b) Now we assume that has no Hamiltonian path. Let, without loss of generality, .
If , then since has no Hamiltonian path, , , and . Then we have and dominates at least one vertex of , say, . This leads to the complementary cycle .
If and , then and . This leads to and the complementary cycle .
If and , then and . This leads to the complementary cycle .
() Assume that . Then we have , , and .
(a) Assume that has a Hamiltonian path. Then, since , there exists a Hamiltonian ending at or ; let, without loss of generality, with be a Hamiltonian path . This leads to the complementary cycle .
(b) Assume that has no Hamiltonian path.
If , then or , say, . Since has no Hamiltonian path, and . This leads to and we obtain the complementary cycle .
If , then or , say, . Since has no Hamiltonian path, and . This leads to and we obtain the complementary cycle .
Case 2.3. Assume that . If , then there exists a vertex such that or . Each case leads to and , a contradiction to . Therefore, . Let, without loss of generality, .
() Assume that . Let, without loss of generality, and . Then . Let, without loss of generality, . Since , , , or , say, . Then we have
Assume that there is an arc from to , say, .
If , then we have and dominates at least two vertices of . This leads to the complementary cycles and if , or if .
If , then , and dominates at least two vertices of . This leads to the complementary cycles and if , or if .
Now assume that . Then we have , , and .
If , then we have and . This leads to the complementary cycle .
If , then and we obtain the complementary cycle .
() Assume that . Let, without loss of generality, . Then , , and . Since , . Let, without loss of generality, .
(a) Assume that . Let, without loss of generality, . Then .
(i) Assume that there is an arc from to , say, . Then we have .
If has a Hamiltonian path, then it ends at (or ). Let, without loss of generality, with be a Hamiltonian path in . Then dominates at least two vertices of . This leads to the complementary cycle if or if .
Now assume that has no Hamiltonian path.
If , then . Since has no Hamiltonian path, . Then we have . If , then we have the complementary cycle . If , then and we have the complementary cycle .
If , then dominates one vertex of , say, . Since has no Hamiltonian path, and . Then we have . If , then we have the complementary cycle . If , then and we have the complementary cycle .
(ii) Assume that . Then there is an arc from to , say, . This leads to the -cycle .
If , then and dominates at least two vertices of . This leads to the complementary cycle if or