Table of Contents Author Guidelines Submit a Manuscript
Mathematical Problems in Engineering
Volume 2016, Article ID 5384190, 7 pages
http://dx.doi.org/10.1155/2016/5384190
Research Article

Complementary Cycles in Irregular Multipartite Tournaments

1School of Mathematics and Information Science, Yantai University, Yantai 264005, China
2Chifeng University, Chifeng 024000, China
3College of Computer Science and Technology, Shandong University, Jinan 250101, China

Received 29 March 2016; Accepted 17 July 2016

Academic Editor: Sergii V. Kavun

Copyright © 2016 Zhihong He et al. 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

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 if .
If and , then dominates at least two vertices of . This leads to the complementary cycle if or if .
If and , then . This leads to the complementary cycle if or if .
(b) Assume that . Then let, without loss of generality, . Then we have .
(i) Assume that there is an arc from to , say, . Then there is an arc from to .
If there is an arc from to , say, , then we have . Let, without loss of generality, . Then and dominates at least one vertex of . This leads to the complementary cycle if or if .
Otherwise, . Then we have -cycle . Let, without loss of generality, . Then .
If , then we have the complementary cycle .
If , then and we have the complementary cycle .
(ii) Assume that . Then and we have -cycle .
If , then , , and . This leads to the complementary cycle .
Now assume that . Then .
If , then and we have the complementary cycle .
If , then and we have the complementary cycle .
Case 2.4. Assume that .
Let . If , then there exists one vertex such that , a contradiction to . Therefore, . Let, without loss of generality, , . Since , let, without loss of generality, the set induce the -cycle , and the set induces the -cycle such that . In addition, implies . It is evident that and are complementary cycles in .
If , then there exist two vertices such that . Since there is an arc between and , or , a contradiction.
Therefore, and . Thus or .
() Assume that first. Let, without loss of generality, .
(a) If , then and .
(i) If , then there exists one vertex such that , a contradiction.
(ii) If, without loss of generality, and , then . Then we have , , and . This leads to the complementary cycles and .
(b) If, without loss of generality, and we let , then , , and .
(i) Let .
Assume that . Then and . In addition, implies .
If or , say, , then and . Since , or . In the former case there exist two complementary cycles and , and in the latter case two complementary cycles are and .
If , then . In addition, if , we have the complementary cycles and ; if and there exists an arc from to , say, , then we have the complementary cycles and ; if , then , and we have the complementary cycles and .
Now assume that or . Then and . Otherwise, there exists one vertex such that , a contradiction. Therefore, let, without loss of generality, . Then we have , , and . Let, without loss of generality, . This leads to the contradiction .
(ii) Let . Then and . Let, without loss of generality, . If , let, without loss of generality, , ; then , , and . It is evident that and are complementary cycles in . If , let, without loss of generality, ; then . This leads to the contradiction .
() Assume that . Let, without loss of generality, and .
(a) If , then and . This leads to the contradiction .
(b) If , then , , , and . This leads to the complementary cycles and .
(c) If, without loss of generality, , then , , and . This leads to and .
If or , say, , then and . This leads to the complementary cycles and .
If , then . This leads to the complementary cycles and .
Thus we finish proof of case
Case 3. One has and . Then .
The case is equivalent to the case of “, , and ,” which has been proved in Case .

4. Concluding Remarks

In this paper, we proved the existence of complementary cycles in a kind of irregular multipartite tournaments. Problems on complementary cycles in irregular multidigraphs have important significance.

In 2002, Volkmann gave a new type of connectivity condition for a multipartite tournament to be cycle complementary. They proposed the following conjecture.

Conjecture 13 (see Volkmann [16] 2002). Let be a multipartite tournament. If