Complementary cycles of any length in regular bipartite tournaments
CComplementary cycles of any length in regular bipartitetournaments
St´ephane Bessy ∗ LIRMM, Univ Montpellier, CNRS, France [email protected]
Jocelyn ThiebautUniv. Orl´eans, INSA Centre Val de Loire,LIFO EA 4022, F-45067 Orl´eans, France [email protected]
February 10, 2021
Abstract
Let D be a k -regular bipartite tournament on n vertices. We show that, for every p with2 ≤ p ≤ n/ − D has a cycle C of length 2 p such that D \ C is hamiltonian unless D isisomorphic to the special digraph F k . This statement was conjectured by Manoussakis, Songand Zhang [K. Zhang, Y. Manoussakis, and Z. Song. Complementary cycles containing a fixedarc in diregular bipartite tournaments. Discrete Mathematics , 133(1-3):325–328,1994]. In thesame paper, the conjecture was proved for p = 2 and more recently Bai, Li and He gave a prooffor p = 3 [Y. Bai, H. Li, and W. He. Complementary cycles in regular bipartite tournaments. Discrete Mathematics , 333:14–27, 2014].
Keywords:
Cycle factor, complementary cycles, regular bipartite tournaments
Throughout all the paper, we are dealing with directed graphs or digraphs. Notations not explicitlystated follows [2].A cycle-factor of a digraph D is a spanning subdigraph of D whose components are vertex-disjoint (directed) cycles. For some positive integer k , a k -cycle-factor of D is a cycle-factor of D with k vertex-disjoint cycles; it can also be considered as a partition of D into k hamiltoniansubdigraphs. In particular, a 1-cycle-factor is a hamiltonian cycle of D . The cycles of a 2-cycle-factor are often called complementary cycles .A tournament is an orientation of a complete graph. A lot of work has been done on cycle-factors in tournament. For instance, the classical result of Camion [5] states that a tournament isstrong if and only if it admits an hamiltonian cycle (i.e. a 1-cycle-factor). Reid [12] proved thatevery 2-connected tournament with at least 6 vertices and not isomorphic to T has a 2-cycle-factor, where T is the Paley tournament on 7 vertices: it has vertex set { , , , , , , } and isthe union of the three directed cycles: the cycle 1 , , , , , ,
7, the cycle 1 , , , , , , ∗ This work was supported by ANR under contract DIGRAPH ANR-19-CE48-0013-02 a r X i v : . [ m a t h . C O ] F e b ycle 1 , , , , , ,
4. This was then extended by Chen, Gould and Li [6] who proved that every k -connected tournament with at least 8 k vertices contains a k -cycle-factor.On the other hand, finding cycles of many lengths in different digraphs is a natural problem inGraph Theory [3]. For example, Moon proved in [11] that every vertex of a strong tournament is ina cycle of every length. Concerning cycle-factor with prescribed lengths in tournaments, Song [13],extended the results of Reid [12], and proved that every 2-connected tournament with at least 6vertices and not isomorphic to T has a 2-cycle-factor containing cycles of lengths p and | V ( T ) | − p for all p such that 3 ≤ p ≤ | V ( T ) | −
3. Li and Shu [9] finally refined the previous result byproving that any strong tournament with at least 6 vertices, a minimum out-degree or a minimumin-degree at least 3, and not isomorphic to T , has 2-cycle-factor containing cycles of lengths p and | V ( T ) | − p for all p such that 3 ≤ p ≤ | V ( T ) | −
3. Recently, K¨uhn, Osthus and Townsend [8] haveextended these results by showing that every O ( k )-connected tournament admits a k -cycle-factorwith prescribed lengths.In this paper, we focus on cycle-factors in k -regular bipartite tournaments. A k -regular bipartitetournament is an orientation of a complete bipartite graph K k, k where every vertex has out-degree k exactly. The existing results concerning this class of digraphs try to extend what isknown about cycle-factors in tournaments. Thus, Zhang and Song [16] proved that any k -regularbipartite tournament with k ≥ Theorem 1.
For k ≥ let D be a k -regular bipartite tournament not isomorphic to F k . Thenfor every p with ≤ p ≤ k , D has a 2-cycle-factor containing cycles of length p and | V ( D ) | − p . The digraph F k corresponds to the k -regular bipartite tournament consisting of four indepen-dent sets K, L , M and N each of cardinality k with all possible arcs from K to L , from L to M ,from M to N and from N to K . In fact, every cycle of F k has length 0 (mod 4). Thus, forinstance, F k has no 2-cycle-factor of length 6 and 4 k −
6. Zhang, Manoussakis and Song provedtheir conjecture when p = 2 in their original paper [15]. In 2014, Bai, Li and He proved theconjecture for p = 3 [1].Our proof of Theorem 1 runs by induction on p and so we will use as basis cases the resultsof Theorem 1 for p = 2 [15] and p = 3 [1]. To perform induction step, we will need also a weakerform of Theorem 1 given by the following lemma. Lemma 1.
For k ≥ let D be a k -regular bipartite tournament. If D contains a cycle-factorwith a cycle C of length p with ≤ p ≤ k , then D contains a (2 p, | V ( D ) | − p ) -cycle-factor ( C (cid:48) , C (cid:48)(cid:48) ) . Moreover, if p is at least 3 and even and D [ C ] is not isomorphic to F p , then D [ C (cid:48) ] isnot isomorphic to F p neither Theorem 1 and Lemma 1 will both need the following result due to H¨aggkvist and Manoussakisto be proven.
Theorem 2 (H¨aggkvist and Manoussakis [7] and Manoussakis [10]) . A bipartite tournamentcontaining a cycle-factor has either a hamiltonian cycle or a cycle-factor consisting of cycles C , . . . , C m such that for any ≤ i < j ≤ m , there is no arc from C j to C i . Section 2 contains introducing tools and definitions we use for the proofs. Lemma 1 andTheorem 1 are proven in Section 3 and 4 respectively. Finally, in Section 5 we give some concludingremarks concerning cycle-factors in bipartite tournaments.2
Definitions and Notations
Generic definitions
Throughout the paper, all digraphs are simple and loopless. Notations notgiven here are consistent with [4]. The vertex set of a digraph D is denoted by V ( D ) and its arcsset by A ( D ). Given a digraph D and a set X of vertices such that X ⊆ V ( D ), we denote by D [ X ]the subdigraph with vertex set X , and arc set { uv ∈ A ( D ) : u ∈ X, v ∈ X } . If H is a subdigraphof D we abusively write D [ H ] for D [ V ( H )]. In the following, we say that two digraphs D and D are isomorphic if there exists a bijection ϕ : V ( D ) → V ( D ) such that, for every ordered pair x, y of vertices in D , xy is an arc of D if and only if ϕ ( x ) ϕ ( y ) is an arc of D .The complement digraph of a digraph D , denoted by D , corresponds to the digraph with vertexset V ( D ) and arc set { uv : uv / ∈ A ( D ) } .For any vertices u and v such that uv is an arc of D , we say that v is an out-neighbor of u and u is an in-neighbor of v . The out-neighborhood (resp. in-neighborhood ) of u in D , denoted N + D ( u )(resp. N − D ( u )), corresponds to the set of vertices which are out-neighbor (resp. in-neighbor ) of u . The out-degree (resp. in-degree ) of a vertex u , denoted d + D ( u ) (resp. d − D ( u )), is the size of its out-neighborhood (resp. in-neighborhood ). We say that D is regular if, for any vertices u and v of D , we have d + ( u ) = d − ( u ) = d + ( v ) = d − ( v ). If, in addition, we have d + ( u ) = d − ( u ) = k , we saythat D is k -regular . For two vertex-disjoint sets A and B , if there are all the possible arcs goingfrom A to B , then we say that A dominates B . The number of arcs from A to B is denoted by e ( A, B ). We simply write e ( A ) instead of e ( A, A ) to denote the number of arcs linking two verticesof A .Similarly, if there is no arc from u to v , we say that there is an anti-arc from u to v . Moreover, wesay that v is an anti-out-neighbor of u and u is an anti-in-neighbor of v . The anti-out-neighborhood (resp. anti-in-neighborhood ) of u in D , denoted N + D ( u ) (resp. N − D ( u )), corresponds to the setof vertices which are anti-out-neighbor (resp. anti-in-neighbor ) of u . The anti-out-degree (resp. anti-in-degree ) of a vertex u , denoted d + D ( u ) (resp. d − D ( u )), is the size of its anti-out-neighborhood(resp. anti-in-neighborhood ). For two vertex-disjoint sets A and B if there are all the possibleanti-arcs going from A to B (that is there are no arcs from A to B ), we say that A anti-dominates B . If there is no ambiguity, we omit the reference to the considered digraph in the previous nota-tions ( N + ( u ) instead of N + D ( u ), etc ...).Given a digraph D and a set { u , . . . , u t } of t disjoint vertices of D , we say that P = u , . . . , u t is a directed path of length t − D if u i u i +1 ∈ A ( D ) for 1 ≤ i ≤ t −
1. The vertices u , . . . , u t − are called the internal vertices of P . In addition, if we also have u t u ∈ A ( D ), then u , . . . , u t is a directed cycle of length t . A cycle of length 2 is also called a digon . In the paper, path and cycle always means directed path and directed cycle, respectively. Symmetrically, given a digraph D anda set of t disjoint vertices { u , . . . , u t } of D , we say that u , . . . , u t is an anti-path if u i u i +1 / ∈ A ( D )for any 1 ≤ i ≤ t −
1. In addition, if u t u / ∈ A ( D ), then we obtain an anti-cycle . A digraph D is strongly connected (or strong for short) if we have a path from u to v for any vertices u and v of D . If D is not strong, a strongly connected component (or strong component for short) of D is aset X of vertices of D such that D [ X ] is strong and X is maximal by inclusion for that. A strongcomponent X is an initial strong component (resp. a terminal strong component ) of D if there isno arc from V \ X to X (resp. from X to V \ X ) in D . It is well-known that every non-strongdigraph contains at least one initial and one terminal strong component. Given a set X of verticesand a cycle C , we denote by C ( X ) the set of the successors of X along C . If X is a singleton { x } ,we simply write C ( x ) instead of C ( { x } ).Finally for a digraph D and integers n , . . . , n k such that n + · · · + n k = | V ( D ) | , a ( n , . . . , n k ) -cycle -factor is a k -cycle-factor ( C , . . . , C k ) of D such that for each i = 1 , . . . , k the cycle C i haslength n i . The cycle C will be called the first cycle of the cycle-factor.3 D D
MST a b c dM ( a ) M ( c ) M ( d ) M ( b ) abc d Figure 1: A 2-regular bipartite tournament D and the contracted digraph according to the redmatching M . In D M , we only keep the vertices of S and the green arcs from T to S . Note that thecycle a, c, b in D M (with bold arcs) corresponds to the cycle a, M ( a ) , c, M ( c ) , b, M ( b ) in D (alsodepicted with bold arcs). Bipartite tournaments and contracted digraphs A bipartite tournament is an orientationof a complete bipartite graph. Let D be a k -regular bipartite tournament with bipartition ( S, T ).We have | S | = | T | = 2 k , and for any vertex u of D we have d + ( u ) = d − ( u ) = k . Moreover, the(unoriented) graph on S ∪ T containing an edge for every arcs from S to T is a bipartite graphwhere every vertex has degree k . Hence, by Hall’s Theorem [4], it admits a perfect matching. Let M be a set of arcs of D corresponding to such a perfect matching. For each vertex u of S , thevertex M ( u ) denotes the only vertex of T such that the arc uM ( u ) is an arc of M .We extend this notation to sets that is, given a subset X of S , we define M ( X ) by M ( X ) = (cid:83) x ∈ X M ( x ).Now, given a perfect matching M of D made of arcs from S to T , we define the contracteddigraph according to M , denoted D M and obtained by contracting the arcs of M and only keepingthe arcs of D from T to S . More formally, the new digraph D M has vertex set S and arc set { uv : u ∈ S, v ∈ S and M ( u ) v ∈ A ( D ) } . As the vertex set of D M is S , we also consider verticesof D M as vertices of D . Notice that D M has 2 k vertices and that for every vertex u of D M wehave N + D M ( u ) = N + D ( M ( u )) and so, u has out-degree k exactly. Similarly, u has in-neighborhood { v ∈ S : M ( v ) ∈ N − D ( u ) } and so has in-degree k exactly. Notice also that D M does not containany parallel arc but may contains cycles on 2 vertices. See Figure 1 which depicts an example ofcontracted digraph according a matching.Let D (cid:48) be a sub-digraph of D and denote by M (cid:48) the arcs of M with both extremities in D (cid:48) .If M (cid:48) is also a perfect matching of D (cid:48) , then we abusively denote by D (cid:48) M the contracted digraph D (cid:48) M (cid:48) .If now M is a perfect matching of D made of arcs from T to S , then we can symmetrically define D M by exchanging S and T in the previous definitions.Structurally, u , . . . , u t is a cycle in D M if and only if u , M ( u ) , . . . , u t , M ( u t ) is a cycle of D .Thus, to prove Theorem 1, if D is not isomorphic to F k , then for every p with 2 ≤ p ≤ k − p, k − p )-cycle-factor in D M . Finally, note that the graph D M contains thesame information than D but, most of the time, it will be easier to identify particular structuresin the former. 4 From cycle-factor to 2-cycle-factor
The aim of this section is to prove Lemma 1 which states that, given a cycle-factor with a cycleof length 2 p , we can “merge” the other cycles in order to obtain a (2 p, | V ( D ) | − p )-cycle-factor.Moreover, in the case where p is even, we could ask that the new cycle of length 2 p is not isomorphicto F p is the former was not. This condition will be useful in the induction step to prove Theorem 1. Lemma 1.
For k ≥ let D be a k -regular bipartite tournament. If D contains a cycle-factorwith a cycle C of length p with ≤ p ≤ k , then D contains a (2 p, | V ( D ) | − p ) -cycle-factor ( C (cid:48) , C (cid:48)(cid:48) ) . Moreover, if p is at least 3 and even and D [ C ] is not isomorphic to F p , then D [ C (cid:48) ] isnot isomorphic to F p neitherProof. As the cases where p = 2 and p = 3 of Theorem 1 are already proven in [15] and [1], weassume that p ≥ C (cid:48) of D containing a cycle C of length 2 p , such that D [ C ] is not isomorphicto F p if p is even, and such that C (cid:48) has a minimum total number of cycles. We denote by C theset of cycles of C (cid:48) different from C . Thus, we want to show that |C| = 1. By Theorem 2, if |C| (cid:54) = 1then we can assume that C = { C , . . . , C (cid:96) } with (cid:96) ≥ C i dominates C j whenever i < j .Let ( S, T ) denotes the bipartition of D and for every i , we denote by c i the number of vertices of V ( C i ) ∩ S , that is C i is of length 2 c i . Claim 1.1.
We have e ( C, C ) = c (2 k − c ) and e ( C (cid:96) , C ) = c (cid:96) (2 k − c (cid:96) ) .Proof. A vertex x in C is an in-neighbor of every vertex in C , . . . , C (cid:96) . Hence, we have (cid:80) x ∈ C d − D ( x ) = e ( C ) + e ( C, C ). Thus we get 2 kc = c + e ( C, C ) and the first result holds. The other equalityis obtained similarly by reasoning on the out-neighborhood of C (cid:96) .Now, using Claim 1.1 we have the following. (cid:16) c (cid:96) (cid:88) x ∈ T ∩ C (cid:96) d + C ( x ) + 1 c (cid:88) x ∈ T ∩ C d + C ( x ) (cid:17) + (cid:16) c (cid:96) (cid:88) x ∈ S ∩ C (cid:96) d + C ( x ) + 1 c (cid:88) x ∈ S ∩ C d + C ( x ) (cid:17) = 1 c (cid:96) e ( C (cid:96) , C ) + 1 c e ( C, C ) = 4 k − ( c + c (cid:96) )Hence, either we have (cid:16) c (cid:96) (cid:88) x ∈ T ∩ C (cid:96) d + C ( x ) + 1 c (cid:88) x ∈ T ∩ C d + C ( x ) (cid:17) ≥ k − ( c + c (cid:96) )2or we have (cid:16) c (cid:96) (cid:88) x ∈ S ∩ C (cid:96) d + C ( x ) + 1 c (cid:88) x ∈ S ∩ C d + C ( x ) (cid:17) ≥ k − ( c + c (cid:96) )2Without loss of generality, we can assume that the former holds (otherwise we exchange in thatfollows the role of S and T ).Denote by M the set of arcs of the digraph induced by the cycle-factor C, C , . . . , C (cid:96) and goingfrom S to T in D . It is clear that M forms a perfect matching of D and that C M ∪ C M is acycle-factor of D M , where C M = { C M , . . . , C M(cid:96) } . Moreover, notice that the length of C M is p andfor i with 1 ≤ i ≤ (cid:96) the length of C Mi is c i . By the previous assumption, in D M we have thefollowing e ( C M(cid:96) , C M ) c (cid:96) + e ( C M , C M ) c ≥ k − ( c + c (cid:96) )2 (1)5ow, we will find suitable vertices in C M to design the desired 2-cycle-factor. To do so, let W (resp. R ) be the set of pairs { x, y } of distinct vertices of C M which are “well connected” to C M (resp. from C M(cid:96) ), that is such that d + C M ( x ) + d + C M ( y ) > c (resp. d − C M(cid:96) ( x ) + d − C M(cid:96) ( y ) > c (cid:96) ). Wedenote by w (resp. r ) the cardinal of W (resp. R ). Claim 1.2.
We have w + r ≥ p ( p − .Proof. For every pair { x, y } of distinct vertices of V ( C M ), we have d + C M ( x ) + d + C M ( y ) ≤ c and,if { x, y } is not a pair of W , we have more precisely d + C M ( x ) + d + C M ( y ) ≤ c . Thus, in total, (cid:88) { x,y } pairof V ( C M ) (cid:16) d + C M ( x ) + d + C M ( y ) (cid:17) = (cid:88) { x,y }∈ W (cid:16) d + C M ( x ) + d + C M ( y ) (cid:17) + (cid:88) { x,y } / ∈ W (cid:16) d + C M ( x ) + d + C M ( y ) (cid:17) ≤ wc + (cid:16) p ( p − − w (cid:17) c (2)and (cid:88) { x,y } pairof V ( C M ) (cid:16) d + C M ( x ) + d + C M ( y ) (cid:17) = ( p − e ( C M , C M )Thus, we get ( p − e ( C M , C M ) c ≤ w + p ( p − C M(cid:96) to C M and R , we obtain( p − e ( C M(cid:96) , C M ) c (cid:96) ≤ r + p ( p − p − k − c + c (cid:96) ≤ w + r + p ( p − C M ∪ C M ∪ C M(cid:96) is a subgraph of D M , we have p + c + c (cid:96) ≤ k and so 2 k − ( c + c (cid:96) ) / ≥ k + p/
2. With the previous inequality we obtain w + r ≥ ( p − k − p/ k ≥ p ,we get the result, that is w + r ≥ p ( p − / { x, x (cid:48) } of distinct vertices of C M , we color { x, x (cid:48) } in white if it is a pair of W , and we color { x, x (cid:48) } in red if { y, y (cid:48) } ∈ R where y (resp. y (cid:48) ) is the out-neighbor of x (resp. x (cid:48) )along C M . Claim 1.3.
There exists a pair of vertices colored both in white and red.Proof. If w + r > p ( p − /
2, then we have colored more than p ( p − / C M . Thus, at least one pair have been colored both in white and red, yielding the result.Now, let suppose that w + r ≤ p ( p − /
2. By Claim 1.2, it means that we have w + r = p ( p − / p + c + c (cid:96) = 2 k and p = k . Notice that, as c and c (cid:96) are at least 2, we have k ≥ d + C M ( x ) + d + C M ( y ) = 2 c for every pair { x, y } of W and d + C M ( x ) + d + C M ( y ) = c for every pair { x, y } of vertices of C M which is not in W . In particular,6f { x, y } ∈ W we have exactly d + C M ( x ) = c and d + C M ( y ) = c . Similarly, if { x, y } ∈ R , then wehave d − C M(cid:96) ( x ) = c (cid:96) and d − C M(cid:96) ( y ) = c (cid:96) . In particular, we can prove that w (cid:54) = 0. Indeed, if it is notthe case, we have r = p ( p − /
2, that is, every pair of elements of C M is a pair of R . Then bythe previous remark, every vertex x of C M satisfies d − C M(cid:96) ( x ) = c (cid:96) , and C M(cid:96) dominates C M . So, theout-neighborhood of any vertex y of C M(cid:96) would contain the successor of y along C M(cid:96) and all thecycle C M , which is of length p = k , a contradiction to d + D M ( y ) = k . Similarly, we have r (cid:54) = 0.Now, let V W ( resp. V R ) be the collection of vertices which belong to at least one pair of W ( resp. R ). Thus, V W is not empty and for every vertex v ∈ V W , we have d + C M ( v ) = c . As everypair { x, y } of distinct vertices of C M satisfies d + C M ( x ) + d + C M ( y ) ∈ { c , c } , it is easy to see thatevery vertex w / ∈ V W satisfies d + C M ( w ) = 0, and that there is at most one vertex a which does notbelong to V W . With the same arguments we see that there is at most one vertex b which does notbelong to V R . So, as p = k ≥ C M containing neither a nor thepredecessor of b along C M . Thus, this pair will colored both in white and red.In the following, let { y , z } be a pair of vertices of V ( C M ) colored both in white and redand we will denote by y (cid:96) (resp. z (cid:96) ) the successor of y (resp. z ) along C M . Therefore we have { y , z } ∈ W and { y (cid:96) , z (cid:96) } ∈ R . Notice that { y , z } and { y (cid:96) , z (cid:96) } are two distinct pairs of verticesbut that y (cid:96) = z or z (cid:96) = y is possible. Claim 1.4.
For every i with ≤ i ≤ c − there exist y and z ∈ C M with y y, z z ∈ A ( D M ) andsuch that the sub-path of C M from y to z contains i internal vertices exactly. Similarly for every i (cid:48) with ≤ i (cid:48) ≤ c (cid:96) − there exist y (cid:48) and z (cid:48) ∈ C M(cid:96) with y (cid:48) y (cid:96) , z (cid:48) z (cid:96) ∈ A ( D M ) and such that the sub-pathof C M(cid:96) from z (cid:48) to y (cid:48) contains i (cid:48) internal vertices exactly.Proof. Suppose that for every vertex y of N + C M ( y ), the vertex z which is i + 1 vertices away from y along C M does not belong to N + C M ( z ). Thus we have | N + C M ( z ) | ≤ c − | N + C M ( y ) | , whichcontradicts d + C M ( y ) + d + C M ( z ) > c as { y , z } is a pair of W . The proof is similar for the pair { y (cid:96) , z (cid:96) } .Now, we can construct our 2-cycle-factor from the collection of cycles. To do so, we will builda cycle γ containing p vertices, such that D M [ V ( D M ) \ V ( γ )] contains a spanning cycle denotedby γ (cid:48) . Let s ( resp. s (cid:48) ) be the number of vertices in the path P (resp. P (cid:48) ) along C M from y (cid:96) to z ( resp. from z (cid:96) to y ). We have s + s (cid:48) = p , thus either we have s ≤ p/ s (cid:48) ≤ p/
2. We willsuppose that the former holds, since an analogous reasoning can be applied for the other case. Inthe following, we will denote by i the smallest index j such that s + (cid:80) ji =1 c i > p . Such indexexists, since s < p and s + (cid:80) (cid:96)i =1 c i > (cid:80) (cid:96)i =1 c i = 2 k − p ≥ p . The cycle γ (resp. γ (cid:48) ) will be obtainedas the union of the path P (resp. P (cid:48) ) and a path Q (resp. Q (cid:48) ) well chosen in C M . . . C M(cid:96) . Todesign Q and Q (cid:48) we consider several cases.First, assume that we have 1 < i < (cid:96) . According to Claim 1.4 applied with i = 0 and i (cid:48) = 0 there exists a pair of vertices { y, z } of C M such that y is the successor of z along C M and with y z, z y ∈ A ( D M ). Similarly, there is { y (cid:48) , z (cid:48) } in C M(cid:96) such y (cid:48) is the successor of z (cid:48) along C M(cid:96) and z (cid:48) z (cid:96) , y (cid:48) y (cid:96) ∈ A ( D M ). As C Mi dominates C Mj for any i < j , we consider in D M the path Q starting in y , containing every vertices of C M except z , every vertices of C Mj , forany 2 ≤ j ≤ i − p − s − (cid:80) i − i =1 c i consecutive vertices of C Mi and finally ending with thevertex y (cid:48) . Similarly we construct the path Q (cid:48) containing z , the remaining vertices of C Mi , everyvertices of C Mj , for any i < j < (cid:96) and every vertices of C M(cid:96) except y (cid:48) , that is Q (cid:48) ends in z (cid:48) . As z y , y (cid:48) y (cid:96) , y z and z (cid:48) z (cid:96) are arcs of D M γ = P ∪ Q and γ (cid:48) = P (cid:48) ∪ Q (cid:48) are cycles and they form a7 M C M‘ C M z y y ‘ z ‘ y z z y C Figure 2: An illustrative case of the proof of Claim 1.4. The red path contains i vertices and theorange one i (cid:48) vertices.2-cycle-factor of D M . To conclude this case, it remains to notice that the number of vertices in γ is s + ( c −
1) + ( (cid:80) i − i =2 c i ) + ( p − s − (cid:80) i − i =1 c i ) + 1 = p .In the case where i = 1, Claim 1.4 applied with i = p − s − i (cid:48) = 0 asserts that there exist { y, z } in C M such that there are p − s − y to z along C M with y z, z y ∈ A ( D M ).As we assume that p ≥ s ≤ p/
2, we know that p − s ≥
2. There also are { y (cid:48) , z (cid:48) } in C M(cid:96) such that y (cid:48) is the successor of z (cid:48) along C M(cid:96) and z (cid:48) z (cid:96) , y (cid:48) y (cid:96) ∈ A ( D M ). Thus we construct Q starting from y , containing p − s − C M and ending in y (cid:48) . The path Q (cid:48) starts in z ,contains the remaining vertices of C M , every vertices of C Mj , for any 1 < j < (cid:96) and every verticesof C M(cid:96) except y (cid:48) . That is, Q (cid:48) ends in z (cid:48) . As previously, we easily check that γ = P ∪ Q and γ (cid:48) = P (cid:48) ∪ Q (cid:48) form a 2-cycle-factor of D M of lengths p and 2 k − p .The case i = (cid:96) is symmetric to the previous one.To check the last part of the statement, we have to guarantee that D [ C (cid:48) ] is not isomorphic to F p , where C (cid:48) denote the cycle of D corresponding to γ (i.e. such that C (cid:48) M = γ ). To do so, noticethat, in all cases, we added y and y (cid:48) to P in order to close γ . In D M , as y ∈ C M and y (cid:48) ∈ C M(cid:96) ,we obtain that yy (cid:48) is an arc of D M and y (cid:48) y is not an arc of D M . Then C (cid:48) contains four vertices y , C (cid:48) ( y ) , y (cid:48) and C (cid:48) ( y (cid:48) ) such that yC ( y ), C (cid:48) ( y ) y (cid:48) , y (cid:48) C (cid:48) ( y (cid:48) ) and yy (cid:48) are arcs of D . However F p doesnot contain such a subdigraph. So D [ C (cid:48) ] is not isomorphic to F p . We prove a slightly stronger version of Theorem 1 where we ask for the first cycle of the cycle-factorto be different from F p if p is even (notice that F p is not defined for odd p ). Namely, we provethe following result. Theorem 3.
For k ≥ let D be a k -regular bipartite tournament not isomorphic to F k . Thenfor every p with ≤ p ≤ k , D has a 2-cycle-factor ( C , C ) where C has length p and if p iseven, C is not isomorphic to F p . We prove this statement by induction on p . By the result of Bai, Li and He [1] the statement istrue for p = 3 and the basis case for the induction holds. So for 3 ≤ p < k we consider D = ( V, A )8 k -regular bipartite tournament which admits a (2 p, k − p )-cycle-factor ( C , C ) where C isnot isomorphic to F p if p is even. In particular, notice that D is not isomorphic to F k . We wantto show that D admits a (2( p + 1) , k − p + 1))-cycle-factor whose first cycle is not isomorphicto F p +1) if p + 1 is even. In the following we call a good cycle-factor such a cycle-factor.We denote by ( S, T ) the bipartition of D and by ( C , C ) the (2 p, k − p )-cycle-factor of D , with D [ C ] not isomorphic to F p . We also denote by M u the arcs of C ∪ C going (up) from S to T and by M d the arcs of C ∪ C going (down) from T to S . It is clear that M u ans M d are perfectmatchings of D and that their union is C ∪ C . For M being either M u or M d , the digraph D M admits the 2-cycle-factor ( C M , C M ) with | C M | = p and | C M | = 2 k − p . Notice that, for even p , having C not isomorphic to F p is equivalent to having D M [ C M ] being not isomorphic to thebalanced complete bipartite digraph on p vertices.To form a good cycle-factor from ( C M , C M ), we will have a case-by-case study according to thestructure of the non-arc in the digraph D M . Prior to this study, we introduce some needed tools. In this subsection, we first consider a matching M of the k -regular bipartite tournament D madefrom arcs from S to T and we define an operation allowing some local change in M . Lemma 2. If D M contains an anti-cycle u , . . . , u t with t ≥ , then the set M (cid:48) of arcs definedin D by M (cid:48) = ( M \ (cid:83) ti =1 u i M ( u i )) ∪ ( (cid:83) t − i =1 u i +1 M ( u i ) ∪ u M ( u t )) is a perfect matching of D .Moreover, for every v / ∈ { u , . . . , u t } , we have N + D M (cid:48) ( v ) = N + D M ( v ) and, for every i with ≤ i ≤ t ,we have N + D M (cid:48) ( u i ) = N + D M ( u i − ) and N + D M (cid:48) ( u ) = N + D M ( u t ) .Proof. If u , . . . , u t is an anti-cycle of D M it follows by definition that u i +1 M ( u i ) is an arc of D for every i such that 1 ≤ i ≤ t − u M ( u t ). Thus, M (cid:48) is a perfect matching of D .For every i with 1 ≤ i ≤ t − M (cid:48) ( u i +1 ) = M ( u i ) in D (and M (cid:48) ( u ) = M ( u t )), then N + D M (cid:48) ( u i +1 ) = N + D M ( u i ) (and N + D M (cid:48) ( u ) = N + D M ( u t )). The out-neighborhood of the other verticesare unchanged.The “shifting” operation between matchings M and M (cid:48) described in the previous lemma iscalled a switch along the anti-cycle u , . . . , u t . The first easy observation we can make on the newcontracted digraph is the following. Corollary 3. If D M has a cycle-factor and contains an anti-cycle u , . . . , u t with t ≥ , then thedigraph obtained after the switch along the anti-cycle u , . . . , u t has a cycle-factor.Proof. Let C be the anti-cycle u , . . . , u t and let M (cid:48) be the perfect matching obtained after theswitch along C . Moreover, let ∆ be the subdigraph induced by the arcs of the cycle-factor in D M ,and let ∆ (cid:48) be the subdigraph induced by the switch of ∆ along C . By Lemma 2, we make a cyclicpermutation on the out-neighborhoods of the vertices of C . So for every vertex x in S we have d +∆ (cid:48) ( x ) = d +∆ ( x ) = 1 and d − ∆ (cid:48) ( x ) = d − ∆ ( x ) = 1. Therefore ∆ (cid:48) is a cycle-factor of D M (cid:48) .Claim 3.1 below is our main application of a switch along an anti-cycle in a contracted digraph.Before stating it, we need the following simple result. Lemma 4. If D M contains a cycle-factor { B , . . . , B l , B l +1 , . . . B l (cid:48) } such that | B | + · · · + | B l | = p + 1 and D M [ B ∪ · · · ∪ B l ] is strongly connected and not isomorphic to a balanced completebipartite digraph, then D admits a good cycle-factor.Proof. For i = 1 , . . . , l (cid:48) we denote by ˜ B i the cycle of D such that ˜ B Mi = B i . Since D M [ B ∪· · ·∪ B l ]is strongly connected, the digraph D [ ˜ B ∪ · · · ∪ ˜ B l ] is also strongly connected and admits a cycle-factor. So, by Theorem 2, it has a Hamiltonian cycle C which is of length 2 p + 2. Moreover, as9 M [ B ∪ · · · ∪ B l ] is not isomorphic to a balanced complete bipartite digraph, then D [ C ] is notisomorphic to F p +2 if p is odd. Using Lemma 1 on the cycle-factor { C, ˜ B l +1 , . . . , ˜ B l (cid:48) } , it provesthat D contains a good cycle-factor.Now, back to the proof of Theorem 3, we see a first case where it is possible to extend thecycle C to obtain a good cycle-factor. Recall that the digraph D M contains the 2-cycle-factor( C M , C M ), where here again M stands for the matching M u or the matching M d of D . Claim 3.1. If D M contains an anti-cycle H such that H \ V ( C M ) is an anti-path from x to y with x = C M ( y ) then D admits a good cycle-factor. Proof.
To shorten notations, we denote C M by C and C M by C (cid:48) . Let H = a . . . a t be an anti-cycle of D M such that a . . . a s is an anti-path of D M [ C (cid:48) ], a is the successor of a s along C (cid:48) and a s +1 . . . a t is an anti-path of D M [ C ]. Moreover, for every i with 1 ≤ i ≤ t we denote by b i theout-neighbor of a i along C or C (cid:48) . An illustrative case is depicted in Figure 3. a = b s = xa s = y a t a s +1 a t − a a s − CC b s − b b b s +1 b t − b t Figure 3: An illustrative case of the proof of Claim 3.1. The dashed arcs form the anti-cycle H and the blue arcs form the cycle-factor B of D M (cid:48) .We perform a switch exchange on H to obtain the digraph D M (cid:48) . By Corollary 3, the digraph D M (cid:48) contains a cycle-factor. More precisely, we pay attention to B the cycle-factor derived from C ∪ C (cid:48) . By Lemma 2, the out-neighbor in B of every vertex not in { a , . . . , a t } is its out-neighborin C ∪ C (cid:48) and the out-neighbor in B of every vertex a i in { a , . . . , a t } is b i − (where indices aregiven modulo t ). As there is only one arc of H from C to C (cid:48) and one arc of H from C (cid:48) to C ,the only arc of B from V ( C ) to V ( C (cid:48) ) of B is a s +1 b s and the only arc of B from V ( C (cid:48) ) to V ( C )is a b t . So B contains a subset B of cycles covering V ( C ) ∪ { a } and a subset B of cyclescovering V ( C (cid:48) ) \ { a } . Thus the cycles of B (resp. B ) form a cycle-factor of D M (cid:48) [ C ∪ { a } ] (resp. D M (cid:48) [ C (cid:48) \ { a } ]) and we denote by ˜ B (resp. ˜ B ) the corresponding cycle-factors of D . Moreover,for i = s + 1 , . . . t the arcs a i M ( a i ) belongs to D and as M ( a i ) = M (cid:48) ( a i +1 ) they link the cycles of˜ B in a strongly connected way. Thus D [ ˜ B ] is strongly connected and so by Theorem 2 it has anhamiltonian cycle B on 2( p + 1) vertices. In addition if p + 1 is even D [ B ] is not isomorphic to F p +1) as it contains C as a subdigraph, C being a cycle on 2 p vertices with p odd. Therefore10 B ∪ B forms a cycle-factor of D with a cycle, D [ B ], of length 2( p + 1) not isomorphic to F p +1) and we can conclude with Lemma 1.The next claim is an easy case where we can insert a vertex of C M into C M . Claim 3.2. If C M contains three consecutive vertices a , b and c (in this order along C M ) and C M contains two consecutive vertices x and y (in this order along C M ) such that ac , xb and by are arcs of D M then D admits a good cycle-factor. Proof.
It is clear that using the arcs ac , xb and by we can form a 2-cycle-factor of D M , with onecycle of length 2 k − ( p + 1) covering C M \ { b } and the other of length p + 1 covering C M ∪ { b } . Letus denote by ˜ C this latter one. If p +1 is even notice that the cycle of D corresponding to ˜ C cannotbe isomorphic to F p +1) . Indeed otherwise ˜ C would be isomorphic to a complete bipartite digraphbut ˜ C contains C has a subdigraph which is a cycle on p vertices with p odd, a contradiction.Now we can prove the following claim, that we will intensively use in the remaining of the proofof Theorem 3. Claim 4.1.
Assume that D M [ C M ] is not strongly connected and denote by S , . . . , S l its stronglyconnected components. If there exists an arc ab of C M such that there is an anti-path in D M [ C M ] from b to a and a ∈ S i , b ∈ S j for i (cid:54) = j , then D admits a good cycle-factor.Proof. In D M , we denote C M by C (cid:48) and C M by C . All the proof stands in D M . First assumethat there exists a vertex c ∈ C such that ac and cb are anti-arcs, then the anti-cycle formed bythe anti-path from b to a in C (cid:48) completed with the anti-arcs ac and cb satisfies the hypothesis ofClaim 3.1 and we can conclude.Hence, we assume that N + C ( a ) ∩ N − C ( b ) = ∅ . In particular, we have d + C ( a ) + d − C ( b ) ≤ p . Let usdenote by A the set of all the vertices of C (cid:48) for whom there is a anti-path from a to them, and by B the set C (cid:48) \ A . The set A dominates the set B , we have | A | + | B | = 2 k − p and also S i ⊆ A and S j ⊆ B (otherwise a and b would have been in the same connected component of D M [ C (cid:48) ]).Therefore, we have2 k − d + ( a ) + d − ( b ) ≤ ( d + C ( a ) + d − C ( b )) + d + C (cid:48) ( a ) + d − C (cid:48) ( b ) ≤ p + | A | − | B | − k − A, B ) is a partition of V ( C (cid:48) )with A \ { a } ⊆ N + ( a ) and B \ { b } ⊆ N − ( b ). As a consequence for every x A ∈ A and x B ∈ B there exists an anti-path from x B to x A . Another consequence is that d + C ( a ) + d − C ( b ) = p and that C admits a partition into N + C ( a ) and N − C ( b ).So let b (cid:48) be the successor of b along C (cid:48) , that is b (cid:48) = C (cid:48) ( b ), and assume first that b (cid:48) ∈ B . Hence, forevery x ∈ N + C ( a ) the arc xb exists in D M (as we assume that N + C ( a ) ∩ N − C ( b ) = ∅ ) and so bC ( x )is an anti-arc, otherwise we can insert b into C and shortcut the path abb (cid:48) using Claim 3.2. Thuswe have C ( N + C ( a )) ⊆ N + C ( b ) and In particular, we obtain d + C ( a ) ≤ d + C ( b ). Hence, we have2 k − d + ( a ) + d − ( b (cid:48) ) = d + C ( a ) + | A | − d − C ( b (cid:48) ) + d − B ( b (cid:48) ) ≤ d + C ( b ) + d − C ( b (cid:48) ) + | A | − | B | − d + C ( b ) + d − C ( b (cid:48) ) ≥ p + 1. In particular, there exists c ∈ C such that bc and cb (cid:48) are anti-arcsand we can apply Claim 3.1 to the anti-cycle bb (cid:48) c to obtain a good cycle-factor of D .Now, we assume that we have b (cid:48) ∈ A . And more generally we can assume that C (cid:48) has no twoconsecutive vertices lying in B . Indeed, otherwise considering a non empty path of C (cid:48) [ B ], vw itsfirst arc and u the predecessor of v along C (cid:48) we can proceed as before with a = u , b = v and11 (cid:48) = w . Symmetrically, C (cid:48) has no two consecutive vertices in A and C (cid:48) alternates between A and B . In particular, p is even and we write C (cid:48) = a b a b . . . a k − p/ b k − p/ with A = { a , . . . , a k − p/ } and B = { b , . . . , b k − p/ } (indices in C (cid:48) will be given modulo k − p/ A \ { a i } ⊆ N + ( a i ) and B \ { b i } ⊆ N − ( b i ) for every 1 ≤ i ≤ k − p/ A and B are two independent sets of D M .Notice that, p being even, we have p ≥ k − p/ > p/ ≥
2. So we obtain that k − p/ ≥ i with 1 ≤ i ≤ k − p/
2, we denote by B i the set N − C ( b i ) and by A i the set N + C ( a i ).As a i b i is an arc of C (cid:48) , as there exists an anti-path from b i to a i in D [ C (cid:48) ] and as a i and b i do notbelong to the same strongly connected component of D M [ C (cid:48) ] (because A dominates B ), we canargue as before with a i playing the role of a and b i the one of b . In particular, we obtain that( A i , B i ) is a partition of V ( C ) for every i ∈ { , . . . k − p/ } with | A i | = | B i | = p/
2. Moreover,assume that there exists x ∈ C and i ∈ { , . . . k − p/ } such that a i x and xb i +1 are anti-arcs.In this case, we modify C (cid:48) into the cycle C (cid:48)(cid:48) by replacing the subpath a i − b i − a i b i a i +1 b i +1 of C (cid:48) by a i − b i a i +1 b i − a i b i +1 . Now, a i b i +1 is an arc of C (cid:48)(cid:48) and there exists an anti-path P in D M [ C (cid:48)(cid:48) ]from b i +1 to a i . So we can conclude by applying Claim 3.1 to the anti-cycle P x . So it meansthat A i ∩ B i +1 = ∅ for every i ∈ { , . . . k − p/ } . We deduce then that all the A i coincide aswell as all the B i and in particular we have A anti-dominates A i and A i dominates B for every i ∈ { , . . . k − p/ } .To conclude, consider s, t ∈ [1 , k − p/
2] such that b s a t is an anti-arc (such an anti-arc existssince there exists an anti-path from b to A ). If there exists an anti-arc from x a ∈ A t to x b ∈ B t ,then we conclude with Claim 3.1 applied to the anti-cycle x a x b b t b s a t . Otherwise it means that A t dominates B t and so as A t dominates B , we have A t anti-dominates A . But then we obtain { b s } ∪ A t ∪ A \ { a t } ⊆ N − ( a t ) and d − ( a t ) ≥ k , a contradiction. ( C , C ) . Now, we define four different properties of the 2-cycle-factor ( C , C ) of D . For this, we first needthe next claim. Claim 4.2.
Let C be C or C . We have: • either D M u [ C M u ] or D M d [ C M d ] has exactly one terminal strong component • or | C | is congruent to 0 modulo 4 and D [ C ] is isomorphic to F | C | .Similarly, we have: • either D M u [ C M u ] or D M d [ C M d ] has exactly one initial strong component • or | C | is congruent to 0 modulo 4 and D [ C ] is isomorphic to F | C | Proof.
We prove the statement for terminal strong components. The proof for initial ones is similar.We denote by D (cid:48) the digraph D [ C ] with bipartition ( S (cid:48) , T (cid:48) ) where S (cid:48) = S ∩ V ( C ) and T (cid:48) = T ∩ V ( C ).So, assume that D (cid:48) M u has at least two terminal strong components and denote by A and B twosuch components. In D (cid:48) M u , it means that A dominates D (cid:48) M u \ A and that B dominates D (cid:48) M u \ B .In D (cid:48) , by considering that A and B are subsets of S (cid:48) , we then have C ( A ), the successors of thevertices of A along C , dominates S (cid:48) \ A and C ( B ) dominates S (cid:48) \ B . Notice that A and B aretwo disjoint sets of S (cid:48) and that C ( A ) and C ( B ) are two disjoint sets of T (cid:48) with | C ( A ) | = | A | and | C ( B ) | = | B | .Now, assume that there exists a vertex u of C ( A ) such that u (cid:48) = C ( u ) does not belong to B . Thenlet v be a vertex of C ( B ). As u (cid:48) / ∈ B , the arc vu (cid:48) belongs to D (cid:48) and then there exists an anti-arcfrom u to v in D (cid:48) M d . Moreover, for every vertex w of T (cid:48) , if C ( w ) does not belongs to A then there12s an anti-arc in D (cid:48) M d from w to u and if it belongs to A , then there is an anti-arc in D (cid:48) M d from w to v . Thus, D (cid:48) M d has exactly one initial strong component, containing v .Otherwise, in D (cid:48) every vertex u of C ( A ) satisfies C ( u ) ∈ B and similarly every vertex v of C ( B )satisfies C ( v ) ∈ A . In particular, we have | C ( A ) | ≤ | B | = | C ( B ) | ≤ | A | = | C ( A ) | and so | A | = | B | = | C ( A ) | = | C ( B ) | . Moreover, as C contains all the vertices of D (cid:48) , ( A, B ) is a partitionof S (cid:48) and ( C ( A ) , C ( B )) is a partition of T (cid:48) with C ( A ) dominates B and C ( B ) dominates A . Inparticular, | C | is congruent to 0 modulo 4. As all the arcs of C from T (cid:48) to S (cid:48) go from C ( A ) to B or from C ( B ) to A and that C ( A ) dominates B and C ( B ) dominates A in D (cid:48) , the sets C ( A )and C ( B ) respectively induce an independent set in D (cid:48) M d , that is a complete digraph in D (cid:48) M d . Asthey form a partition of the vertex set of D (cid:48) M d , it has exactly one initial strong component exceptif there is no arc between C ( A ) and C ( B ). In this later case, it means that all the arcs from A to C ( A ) and from B to C ( B ) are contained in D (cid:48) . But, then D (cid:48) [ C ] is isomorphic to F | C | .In the case where D M u [ C M u ] has exactly one initial strong component, we say that the 2-cycle-factor ( C , C ) has property Q up . And in the case where D M d [ C M d ] has exactly one initial strongcomponent, we say that ( C , C ) has property Q down . By Claim 4.2, we know that either ( C , C )has property Q up or Q down or that | C | is congruent to 0 modulo 4 and that D [ C ] is isomorphicto F | C | .Now, let us define another pair of properties for the 2-cycle-factor ( C , C ). As D is a bipartitetournament every vertex of C has an in-neighbor or an out-neighbor in C . Moreover, as D is k -regular and | C | = 2 p < k there exists at least one arc from C to C and one arc from C to C .So, it is easy to check that there exists an arc uv of C such that N − C ( u ) (cid:54) = ∅ and that N + C ( v ) (cid:54) = ∅ .If we have u ∈ S and v ∈ T we say that ( C , C ) has property P down . In this case, it means thatthe arc uv is an arc of M u , and in D M u the vertex u has an in-neighbor and an out-neighbor in C M u . Otherwise, that is when we have v ∈ S and u ∈ T , we say that ( C , C ) has property P up .In this case, let u (cid:48) be the predecessor of u along C , that is u = C ( u (cid:48) ). So, in D M u , u (cid:48) v is an arcof C M u , the vertex u (cid:48) has an anti-out-neighbor in C M u and v has an anti-in-neighbor in C M u . Claim 4.3. If ( C , C ) does not satisfies P down , then in D M u every vertex of C M u anti-dominates C M u or is anti-dominated by C M u .Proof. Let x be a vertex of C M u . Since D does not satisfies P down , then in D , either C ( x ) isdominated by C ∩ S or x dominates C ∩ T . In the first case, it means that there is no arc from x to C M u in D M u , while in the latter, it means that there is no arc from C M u to x .Notice that if ( C , C ) satisfies property Q up , then exchanging the role of S and T in D (andthen of M u and M d ) leads to ( C , C ) satisfies property Q down , and conversely. We also have thesimilar property with P up and P down .Thus, without loss of generality, we assume that ( C , C ) has the property P up . Then, we study,in this order, the three different cases: either ( C , C ) has property Q up , or D [ C ] is isomorphicto F | C | , or ( C , C ) satisfies property Q down . ( C , C ) has property Q up So, we know that D M u [ C M u ] has exactly one initial strong component. If it is not strong itself,there exists an arc of C M u entering into its unique initial strong component and we can directlyconclude with Claim 4.1. Then, we assume that D M u [ C M u ] is strongly connected. We considerseveral cases. 13 C AB ab a b Figure 4: Illustration of the anti-cycle when D M u [ C ] is not strongly connected but there exist twovertices a ∈ A and b ∈ B such that there is an anti-path P from b to a in D M u [ C ] (depicted inblue) and, there are two vertices a (cid:48) and b (cid:48) in C (cid:48) such that aa (cid:48) and bb (cid:48) are anti-arcs, and b (cid:48) is thesuccessor of a (cid:48) along C (cid:48) . We can use the red anti-path from a (cid:48) to b (cid:48) and P to form an anti-cycle. Case 1 : D M u [ C M u ] is strongly connected too . As ( C , C ) has the property P up , there exist u and v in D M u such that v is the successor of u along C M u , u has an anti-out-neighbor u (cid:48) in C M u and v has an anti-in-neighbor v (cid:48) in C M u . As both D M u [ C M u ] and D M u [ C M u ] are stronglyconnected, there exists an anti-path from v to u in D M u [ C M u ] and an anti-path from u (cid:48) to v (cid:48) in D M u [ C M u ]. So, we can form an anti-cycle in D M u which satisfies the hypothesis of Claim 3.1 andconclude that D contains a good cycle-factor. Case 2 : D M u [ C M u ] is not strongly connected. In what follows, to shorten the notation, wedenote C M u by C and C M u by C (cid:48) . So, as D M u [ C ] is not strong, there exists a partition ( A, B ) of V ( C ) such that there is no anti-arcs from A to B , thus we have A dominating B in D M u . Case 2.1 :
There exist two vertices a ∈ A and b ∈ B such that there is an anti-path P from b to a in D M u [ C ] . So, in D M u we have d + ( a ) + d − ( b ) = 2 k − d + ( a ) ≤ | A | − d + C (cid:48) ( a ) and d − ( b ) ≤ | B | − d − C (cid:48) ( b ) and | A | + | B | = p , we get d + C (cid:48) ( a ) + d − C (cid:48) ( b ) ≥ k − p If d + C (cid:48) ( a ) + d − C (cid:48) ( b ) > k − p = | C (cid:48) | then there exist two vertices a (cid:48) and b (cid:48) in C (cid:48) such that aa (cid:48) and bb (cid:48) are anti-arcs, and b (cid:48) is the successor of a (cid:48) along C (cid:48) . As D M u [ C (cid:48) ] is strongly connected,there exists an anti-path Q from a (cid:48) to b (cid:48) in D M u [ C (cid:48) ]. So the concatenation of the paths P , Q andthe arcs a (cid:48) a and bb (cid:48) forms an anti-cycle of D M u satisfying the conditions of Claim 3.1 and we canconclude that D contains a good cycle-factor. See Figure 4 which depicts this subcase.So, we assume that d + C (cid:48) ( a )+ d − C (cid:48) ( b ) = 2 k − p . It implies that d + A ( a ) = | A |− d − B ( b ) = | B |− a anti-dominates A and B anti-dominates b . We deduce that for every a (cid:48) ∈ A and b (cid:48) ∈ B there exists an anti-path from b (cid:48) to a (cid:48) and applying the same arguments to a (cid:48) and b (cid:48) we can assumethat A and B are both independent sets of D M u . As C has then to alternate between A and B ,we have | A | = | B | = p/
2, and in particular p is even.Now, let B (cid:48) be the set of vertices in V ( C (cid:48) ) which have an anti-out-neighbor in B . More for-mally, we define B (cid:48) = { c (cid:48) ∈ V ( C (cid:48) ) : ∃ b ∈ B such that c (cid:48) b is an anti-arc } . For any b ∈ B we have d − C (cid:48) ( b ) = k − p/ | B (cid:48) | ≥ k − p/
2. Similarly, we define A (cid:48) = { c (cid:48) ∈ C (cid:48) : ∃ a ∈ A, ac (cid:48) isan anti-arc } . We have the analogous result | A (cid:48) | ≥ k − p/ b (cid:48) z (cid:48) of C (cid:48) with b (cid:48) ∈ B (cid:48) and z (cid:48) having an anti-in-neighbor z in C .Then b (cid:48) has an anti-out-neighbor b in C and there exists an anti-path P from b to z in D M u [ C ].Moreover, as D M u [ C (cid:48) ] is strongly connected, there exists also an anti-path Q from z (cid:48) to b (cid:48) in D M u [ C ] and we can conclude with Claim 3.1 applied on the anti-cycle P ∪ Q .Now we can assume that every arc b (cid:48) z (cid:48) of C (cid:48) with b (cid:48) ∈ B (cid:48) satisfies N − ( z (cid:48) ) ∩ V ( C ) = ∅ . Sym-metrically, we can assume that every arc z (cid:48) a (cid:48) of C (cid:48) with a (cid:48) ∈ A (cid:48) satisfies N + ( z (cid:48) ) ∩ V ( C ) = ∅ .But now, we will obtain the contradiction that ( C , C ) cannot satisfy property P up . Indeed, inparticular, we have A (cid:48) ∩ C (cid:48) ( B (cid:48) ) = ∅ and as | A (cid:48) | ≥ k − p/ | C (cid:48) ( B (cid:48) ) | = | B (cid:48) | ≥ k − p/ A (cid:48) , C (cid:48) ( B (cid:48) )) is a partition of V ( C (cid:48) ). As ( C , C ) satisfies property P up , the cycle C (cid:48) shouldcontains two vertices u and v such that v is the successor of u along C (cid:48) , u has an anti-out-neighborin C and v has an anti-in-neighbor in C . By the previous arguments, we must have u / ∈ B (cid:48) and v / ∈ A (cid:48) , a contradiction to the fact that ( A (cid:48) , C (cid:48) ( B (cid:48) )) is a partition of V ( C (cid:48) ). Case 2.2 :
For every a ∈ A and b ∈ B , there are no anti-paths from b to a . Thus, the set A dominates the set B and B dominates A . Suppose without loss of generality that | B | ≤ | A | and let b be a vertex in B . If d + C (cid:48) ( b ) + d − C (cid:48) ( b ) > k − p , then we can find two vertices u and v in C (cid:48) such that ub and bv are anti-arcs, and v is the successor of u along C (cid:48) . In that case weconclude with Claim 3.1 considering the anti-cycle P ∪ b where P is an anti-path from v to u in D M u [ C (cid:48) ] (which exists as D M u [ C (cid:48) ] is strongly connected). Therefore, we can assume that for every b ∈ B, d + C (cid:48) ( b ) + d − C (cid:48) ( b ) ≤ k − p . Thus, we have d + B ( b ) + d − B ( b ) = ( d + ( b ) + d − ( b )) − ( d + C (cid:48) ( b ) + d − C (cid:48) ( b )) ≥ (2 k − − (2 k − p ) ≥ p − d + B ( b ) + d − B ( b ) ≤ | B | − ≤ | A | + | B | − p −
2, we have equalities everywhere in theprevious computation. In particular, we have | A | = | B | = p/ p is even. Moreover, for every b ∈ B we have d + B ( b ) = d − B ( b ) = p/ −
1, implying that B is an independent set. Symmetrically for A , as | A | = | B | either we can conclude as previously with Claim 3.1 or A is also an independentset. In this latter case, C would induce a complete bipartite graph in D M u and D [ C ] would beisomorphic to F p , a contradiction to our induction hypothesis.To conclude this section, notice that if D M u [ C M u ] has exactly one terminal strong component,then we can conclude similarly. Indeed, we have seen that if D M u [ C M u ] is strongly connected,then D admits a good cycle-factor, and if D M u [ C M u ] is not strong, then there exists an arc of C M u leaving its unique terminal component and we can directly conclude with Claim 4.1. D [ C ] is isomorphic to F k − p . As D [ C ] is not isomorphic to F p , by Claim 4.2 we can assume without loss of generality that D M u [ C M u ] has exactly one strong initial component. As usual, we denote C M u by C and C M u by C (cid:48) . As D [ C ] is isomorphic to F k − p , p is even and the digraph D M u [ C (cid:48) ] is the complete bipartitedigraph K ( k − p/ ,k − p/ . We denote by ( A, B ) its bipartition.
Claim 4.4.
If there exists an arc ab of C such that there is an anti-path in C from b to a , ananti-arc from a to A and an anti-arc from A to b , then D admits a good cycle-factor.Proof. First denote by a (cid:48) the end of an anti-arc from a to A and by b (cid:48) the beginning of an anti-arcfrom A to b . We can assume that there exists a vertex c of B such that b (cid:48) ca (cid:48) is a subpath of C (cid:48) .Indeed, as D M u [ C (cid:48) ] is isomorphic to a complete bipartite digraph, there exists a hamiltonian cycle15 C AB a b c ab Figure 5: Illustrative example of the Claim 4.4. C (cid:48)(cid:48) of D M u [ C (cid:48) ] starting with b (cid:48) ca (cid:48) . We can then consider C (cid:48)(cid:48) instead of C (cid:48) . So we assume that b (cid:48) , c and a (cid:48) are consecutive along C (cid:48) , and as a (cid:48) and b (cid:48) belong to A , a (cid:48) b (cid:48) is an anti-arc of D M u . Sowe perform a switch exchange along the anti-cycle formed by the anti-path from b to a in C , andthe anti-arcs aa (cid:48) , a (cid:48) b (cid:48) and b (cid:48) b . See Figure 4 which depicts this subcase. According to Lemma 2,we denote by M (cid:48) the new contracted matching in D . By Lemma 2 and Corollary 3, it is easyto see that we obtain a new cycle-factor C of D M (cid:48) containing the 3-cycle C = a (cid:48) bc , a cycle C s containing all the vertices of C (cid:48) \ { a (cid:48) , c } and other cycles included into C . Notice that D M (cid:48) [ C s ]is a bipartite complete digraph on 2 k − p − C (cid:48) s on p − C (cid:48)(cid:48) s on 2 k − p vertices (with 2 k − p ≥ C contains a vertexof A and C (cid:48) s a vertex of B , the union of these two cycles is strongly connected in D M (cid:48) . So usingTheorem 2 in D M u there exists a cycle of length p + 1 spanning C (cid:48) s ∪ C . Using this cycle and thecycles of ( C \ { C , C s } ) ∪ C (cid:48) s , by Lemma 1 we form a 2-cycle-factor of D M (cid:48) with one being of length p + 1. In particular, as p + 1 is odd, this cycle-factor of D M (cid:48) corresponds to a good cycle-factor of D . As A and B have a symmetric role, Claim 4.4 still holds by replacing A by B in its statement.Then, let us consider two cases, according to the structure of D M u [ C ]. Case 1: D M u [ C ] is strongly connected In this case, let v be a vertex of C and u its predecessoralong C . As | C | = p < k , the vertex u is the beginning of an anti-arc which ends in C (cid:48) , that is in A or B . So, if we cannot conclude with Claim 4.4, it means than v is dominated by A or B . As thisis true for every vertex v , by a direct counting argument, there exist a set X A of p/ C which are dominated by A and a set X B of p/ B . Moreover, novertex x of A dominates a vertex of X B (otherwise we would have d D Mu ( x ) > k ) and no vertex of B dominates a vertex of X B . So, for every vertex v of C we have d − C ( v ) = k − ( k − p/
2) = p/ v and its successor along C we obtain that every vertex v of C satisfies d + C ( v ) = p/
2. It means that in D , the bipartite tournament D [ C ] contains 2 p vertices, is not isomorphic to F p and satisfies d + ( u ) = d − ( u ) = p for each of its vertex u .Now, assume first that p ≥
5. By induction, provided that p ≥
5, the bipartite tournament D [ C ] has at least 10 vertices and admits a 2-cycle-factor ( C ind , C (cid:48) ind ) with C ind being of length 6.In D M u , we have a 2-cycle-factor ( F ind , F (cid:48) ind ) with F ind being of length 3. Let xy be an arc of F ind .As d + C (cid:48) ( x ) ≥ k − p > d − C (cid:48) ( y ) ≥ k − p >
0, there exist x (cid:48) and y (cid:48) in C (cid:48) such that xx (cid:48) and y (cid:48) y are arcs of D M u . Now, D M u [ C (cid:48) ] being a complete bipartite digraph on 2 k − p vertices, it admits a2-cycle-factor ( C s , C (cid:48) s ) such that C s contains x (cid:48) and y (cid:48) and is of length p −
2. So, using Lemma 4,as D M u [ C s ∪ F ind ] is strongly connected, D admits a good cycle-factor.To conclude, assume that p = 4. As every vertex x of C satisfies d + C ( x ) = d − C ( x ) = 2 and D M u [ C ] is strongly connected, it is easy to see that the anti-arcs of C form a cycle of length 4. If16 contains a vertex x such that x has an in-neighbor x (cid:48) in A and an out-neighbor y (cid:48) in B , thenwe form a good cycle-factor of D M u with : a cycle of length 3 in C \ x , a cycle of length p + 1 = 5containing x (cid:48) , x , y (cid:48) and another vertex x (cid:48)(cid:48) of A and another vertex y (cid:48)(cid:48) of B , and a cycle of length2 k − C (cid:48) \ { x (cid:48) , x (cid:48)(cid:48) , y (cid:48) , y (cid:48)(cid:48) } . Otherwise, it means that we can write V ( C ) = { a, a (cid:48) , b, b (cid:48) } such that the in- and out-neighborhood of a and a (cid:48) in C (cid:48) are exactly A and that the in- andout-neighborhood of b and b (cid:48) in C (cid:48) are exactly B . Moreover, as every pair of vertices are linkedby at least one arc in D M u [ C ], we can assume that aa (cid:48) and bb (cid:48) are arcs of D M u . Then, as k > p ,we have 2 k − p ≥ a , a , a in A and three vertices b , b , b in B . Then, we form a good cycle-factor by applying Lemma 1 to the cycles aa (cid:48) a b a and bb (cid:48) b a b of length p + 1 = 5 and a cycle covering C (cid:48) \ { a , a , a , b , b , b } . Case 2: D M u [ C ] is not strongly connected . However, as ( C , C ) satisfies P up , we knowthat D M u [ C ] contains only one strong initial component, denoted by Y . Let Y be V ( C ) \ Y . Inparticular, all the arcs from Y to Y exist in D M u . Moreover, there exists an arc xy of C with x ∈ Y and y ∈ Y . As Y is the only initial component of D M u [ C ] there exists an anti-path from y to x in D M u [ C ]. As | C | = p < k , there exists a vertex x (cid:48) in C (cid:48) such that xx (cid:48) is an anti-arc of D M u .Without loss of generality, we can assume that x (cid:48) ∈ A . By Claim 4.4, if y has an anti-in-neighborin A , then D admits a good cycle-factor. So we can assume that A dominates y . Similarly, y hasan anti-in-neighbor in C (cid:48) which must be in B and if we cannot conclude with Claim 4.4, it meansthat x dominates B . As x dominates also Y and y is dominated by Y , we have | Y | = | Y | = p/ x is exactly B ∪ S and the in-neighborhood of y is exactly A ∪ Y .Now, assume that z , the successor of y along C is in Y . Notice that zy is an anti-arc of D M u . As Y and y dominate z , it has at least one anti-in-neighbor in A and at least one anti-in-neighbor in B (otherwise, we would have d − ( z ) ≥ k − p/ p/ k + 1). As y has an anti-out-neighborin C (cid:48) , wherever it is, in A or B , we can conclude with Claim 4.4. Otherwise, it means that z is in Y . Symmetrically, the predecessor of x along C is in Y . Repeating the argument, we concludethat C alternates between Y and Y . Indeed C cannot induce a path of positive length in Y forinstance : the first vertex of such a path would be the end of an arc from Y to Y and the secondvertex of the path would lie in Y then. Also the conclusions we had for x and y respectively holdfor all vertices of Y and Y . So, we have Y is dominated by A and Y and is an independent setof D M u , and Y dominates B and Y and is also an independent set of D M u . We deduce that thein-neighborhood of B is exactly A ∪ Y and then the out-neighborhood of Y is A ∪ Y also. Inparticular, Y dominates Y and there is no anti-path from Y to Y , providing a contradiction, as Y is the only initial strong component of D M u [ C ]. ( C , C ) satisfies Q down As we assume that we are not in Case A, the digraph D M u [ C M u ] has at least two initial strongcomponents, and at least two terminal strong components (as noticed at the end of Case A).Besides, if ( C , C ) satisfies P down , then by exchanging the role of S and T , we are in the symmet-rical case of Case A. Then, we can assume that ( C , C ) does not satisfy P down . Once again, wedenote C M u by C and C M u by C (cid:48) . So, by Claim 4.3, for every vertex x of C (cid:48) either there is no arcfrom x to C or there is no arc from C to x .Before concluding the proof of Theorem 3, we need the two following claims. Claim 4.5. If D M u [ C (cid:48) ] contains a vertex x such that there is no arc between x and C , then D admits a good cycle-factor.Proof. Otherwise, let x be such a vertex and call x, y, z, t the subpath of C (cid:48) of length 3 startingfrom x . We know that either there is no arc from C to t or no arc from t to C . Assume that the17atter holds, the other case could be treated symmetrically. If there also exists an anti-arc from C to t , then we can find three vertices a , b and c in C such that a, b, c is a subpath of C of length2 and that { a, x } and { c, t } are independent sets of D M u . So, we will exchange some small pathsbetween C and C (cid:48) .First, notice that p >
3. Indeed, if we denote by A C (resp. B C ) the set of vertices of C (cid:48) whichanti-dominate C (resp. are anti-dominating by C ), we have V ( C (cid:48) ) = A C ∪ B C , x ∈ A C ∩ B C andthen | A C | + | B C | ≥ | A C ∪ B C | + 1 = | C (cid:48) | + 1 = 2 k − p + 1 ≥ k − p ≤
3. So, as | A C | ≤ k − | B C | ≤ k −
1, we have | A C | = | B C | = k − A C ∩ B C = { x } . Moreover, we have t ∈ A c \ { x } and so N + ( c ) contains B C ∪ { t } of size k , a contradiction.Now, to perform the path exchange, let us depict the situation in D : C contains the path aC ( a ) bC ( b ) cC ( c ) and C contains the path xC ( x ) yC ( y ) zC ( z ) tC ( t ). Moreover, in D , x dom-inates V ( C ) ∩ T , C ( x ) is dominated by V ( C ) ∩ S , there is an arc from c to C ( t ) and an arc from t to C ( c ). So, we replace in C the path aC ( a ) bC ( b ) cC ( c ) by aC ( x ) yC ( y ) zC ( z ) tC ( c ) to ob-tain the cycle ˜ C and we replace in C the path xC ( x ) y C ( y ) zC ( z ) tC ( t ) by xC ( a ) bC ( b ) cC ( t )to obtain the cycle ˜ C . The cycles ( ˜ C , ˜ C ) form a (2( p + 1) , k − p + 1))-cycle-factor of D . More-over, ˜ C is not isomorphic to F p +1) . Indeed, as p >
3, the cycle ˜ C contains the predecessor u of a along C (which is not C ( c ) then). Let call d the vertex of C with C ( d ) = u . As C ( x )is dominated by V ( C ) ∩ S in D , there is an arc from d to C ( x ). Thus, ˜ C is not isomorphic to F p +1) , as it contains the path dC ( d ) aC ( x ) and the arc dC ( x ) while F p +1) does not containsuch a sub-structure.So, we can assume that C dominates t . Call by Y the strong component of D M u [ C (cid:48) ] containing t . There exists Y term a terminal strong component of D M u [ C (cid:48) ] distinct from Y . Let Y (cid:48) be theunion of the strong components of D M u [ C (cid:48) ] different from Y term and Y . As Y term is a terminalstrong component of D M u [ C (cid:48) ], for any vertex u of Y term , we have k − − | C | ≤ d + C (cid:48) ( u ) ≤ | Y term | − | Y term | ≥ k − p (noticed that a symmetrical reasoning holds for initial strong componentsalso). Moreover, as C and Y term dominate t we must have exactly | Y term | = k − p and the in-neighborhood of t is exactly C ∪ Y term . In particular, z belongs to Y term and Y is the otherterminal strong component of D M u [ C (cid:48) ]. Moreover, as Y (cid:48) ∪ Y has size k and is dominated by Y term ,then Y (cid:48) ∪ Y is exactly the out-neighborhood of each vertex of Y term . Thus there is no arc from Y term to C . We look at two cases to conclude the proof.First assume that C dominates z . As z is dominated by Y (recall that Y is a terminal strongcomponent of D M u [ C (cid:48) ]) and Y has size at least k − p , then we have that Y ∪ C is exactly thein-neighborhood of z and that | Y | = k − p . Finally, Y (cid:48) is non empty (of size 2 k − p ) and thereis no arc from Y (cid:48) to { z, t } . To conclude, let uv be an arc of C (cid:48) with u ∈ Y term ∪ Y and v ∈ Y (cid:48) (such an arc exits as Y (cid:48) (cid:54) = ∅ ). By the previous arguments there exists an anti-path from v to u in D M u [ C (cid:48) ] and by Claim 4.1, we conclude that D admits a good cycle-factor.Now, if there is an anti-arc from C to z , then we will conclude as previously using a small pathexchange between C and C (cid:48) . Indeed, if such an anti-arc exists between a vertex b (cid:48) of C and z ,call a (cid:48) the predecessor of b (cid:48) along C . So, { a (cid:48) , x } and { b (cid:48) , z } are independent sets of D M u . Then,in D we exchange the path a (cid:48) C ( a (cid:48) ) b (cid:48) C ( b (cid:48) ) of C with the path a (cid:48) C (cid:48) ( x ) yC (cid:48) ( y ) zC (cid:48) ( z ) to obtain thecycle ˜ C . Similarly, we exchange the path xC (cid:48) ( x ) yC (cid:48) ( y ) zC (cid:48) ( z ) of C with the path xC ( a (cid:48) ) b (cid:48) C (cid:48) ( z )to obtain the cycle ˜ C . Then, ( ˜ C , ˜ C ) forms a good cycle-factor of D , as in particular, denotingby c (cid:48) the predecessor of a (cid:48) along C , ˜ C contains in D the path c (cid:48) C ( c (cid:48) ) a (cid:48) C (cid:48) ( x ) and the arc c (cid:48) C (cid:48) ( x )and so ˜ C is not isomorphic to F p +1) which does not contain such a sub-structure.The last claim will show that every arc of C (cid:48) is contained in a digon. Claim 4.6. If D M u [ C (cid:48) ] contains an arc xy such yx is not an arc of D M u , then D admits a goodcycle-factor. roof. Assume that xy is an arc of C (cid:48) such that yx is an anti-arc of D M u . If x and y are not inthe same strong component of D M u [ C (cid:48) ], then we conclude with Claim 4.1. Otherwise, we assumethat x and y lie in a same strong component Y of D M u [ C (cid:48) ]. As D M u [ C (cid:48) ] has at least two initialand two terminal strong components, there exist an initial strong component Y init and a terminalstrong component Y term of D M u [ C (cid:48) ] which are different from Y (with possibly Y init = Y term ). Aspreviously noticed, we have | Y init | ≥ k − p and | Y term | ≥ k − p . And as Y init and Y term are differentfrom Y then, Y dominates Y init and is dominated by Y term . In particular, as x and y lie in Y and xy is an arc of C (cid:48) , x has at least an anti-out-neighbor x (cid:48) in C and y has at least an anti-in-neighbor y (cid:48) in C (otherwise, considering Y init or Y term we would have d + D Mu ( x ) ≥ k + 1 or d − D Mu ( y ) ≥ k + 1).If x (cid:48) y or xy (cid:48) is not an arc of D M u then we conclude with Claim 3.1, using an anti-path from y to x in Y . So we assume that x (cid:48) y and xy (cid:48) are arcs of D M u . Similarly, if we have d + C ( x ) + d − C ( y ) > p then there exists a vertex u in C such that xu and uy are anti-arcs of D M u and we conclude withClaim 3.1. Thus we assume that we have d + C ( x )+ d − C ( y ) ≤ p and then that d + C (cid:48) ( x )+ d − C (cid:48) ( y ) ≥ k − p .If there is no vertex z in V ( C (cid:48) ) such that yxz is an anti-cycle, then we can partition V ( C (cid:48) ) \ { x, y } into two sets X and Y such that x anti-dominates X and Y anti-dominates y . Call Y X the setof strongly connected components of D M u [ C (cid:48) ] which are included into X . Notice that Y X is notempty as it contains all the terminal strong component of D M u [ C (cid:48) ], but does not contain anyinitial strong component of D M u [ C (cid:48) ] . Now, consider any arc uv of C (cid:48) going from a component Y of Y X to a component Y not belonging to Y X . As Y contains a vertex of { x, y } ∪ Y , there existsan anti-path from v to x . And as Y belongs to Y X , there is an anti-arc from x to u . So usingClaim 4.1, we can conclude that D admits a good cycle-factor.Thus we assume that there exists a vertex z in V ( C (cid:48) ) such that yxz is an anti-cycle of D M u .By Claim 4.5, we can assume that there exists an arc between z and C . Without loss of generalityassume that there is an arc zz (cid:48) from z to C . We will perform a switch exchange along theanti-cycle yxz and show that the 2-cycle-factor that we obtain will satisfy P down and Q down .Denote by M (cid:48) u the perfect matching of D obtain from M u by switch exchange along yxz (that is M (cid:48) u = ( M u \ { xM u ( x ) , yM u ( y ) , zM u ( z ) } ) ∪ { xM u ( y ) , yM u ( z ) , zM u ( x ) } ). So, when performing theswitch exchange along yxz , by Lemma 2, we obtain N + D M (cid:48) u ( x ) = N + D Mu ( y ), N + D M (cid:48) u ( y ) = N + D Mu ( z )and N + D M (cid:48) u ( z ) = N + D Mu ( x ). In D M u , call by P the sub-path of C (cid:48) going from the successor of y (along C (cid:48) ) to the predecessor of z and by P the sub-path of C (cid:48) going from the successor of z tothe predecessor of x . Then, after the switch exchange, the cycle C (cid:48) becomes in D M (cid:48) u the cycle C (cid:48)(cid:48) = xP zyP . We denote by C (cid:48) its corresponding cycle in D . Notice that the strong componentsof D M (cid:48) u [ C (cid:48)(cid:48) ] are the same than the ones of D M u [ C (cid:48) ]. Indeed, as the anti-cycle yxz becomes theanti-cycle xyz in D M (cid:48) u , the permutation of the anti-out-neighborhoods of x , y and z does notaffect the strong components of D M u [ C (cid:48) ] and their relationships. So, ( C , C (cid:48) ) still satisfies Q down .Finally, remind that in D M u , the vertex z has an out-neighbor z (cid:48) in C and y an in-neighbor y (cid:48) in C . As we have N + D M (cid:48) u ( y ) = N + D Mu ( z ), the arcs y (cid:48) y and yz (cid:48) belong to D M u . Thus, ( C , C (cid:48) ) nowsatisfies P down and we can conclude with the symmetrical case of Case A.Finally we can assume that every arc of C (cid:48) is in a digon. We write C (cid:48) = u , . . . , u l with l = 2 k − p . The indices of vertices of C (cid:48) will be given modulo l . Then we consider two cases: Case 1: p is odd. Then l = 2 k − p is also odd. As D M u [ C (cid:48) ] is not strongly connected,there exists i ∈ { , . . . , l } such that u i u i − is an arc of D M u . Without loss of generality, we canassume that u l u l − is an arc of D M u . We consider then the set X = { u , u , u , . . . , u l − p +1 } ,that is all the vertices u i with odd i between 1 and l − p + 1, except u . Notice that X has size( l − p +1 − / − k − p . If there is no arc between X and u , as u has no arc to C or no arc from C , we would have d + D Mu ( u ) ≥ k or d − D Mu ( u ) ≥ k , a contradiction. So, there exists an arc between19 and some u i ∈ X . If i = 1, then we consider the cycle-factor C (cid:48) on 2 k − p − C and the cycles with vertex sets { u , u , u } , { u , u } , . . . , { u l − p − , u l − p − } and the cycle-factor C on p + 1 vertices containing the cycles with vertex set { u l − p , u l − p +1 } , { u l − p +2 , u l − p +3 } , . . . , { u l − , u l } .Notice that D M u [ u l − p , . . . , u l ] is strongly connected and is not a complete bipartite graph, asit contains the cycle u l − , u l − u l . So, by Lemma 4, the digraph D admits a good cycle-factor.Now, if i = u l − p +1 , then we consider the cycle-factor C on p + 1 vertices containing the cycleswith vertex sets { u l − p +2 , u l − p +3 } , . . . , { u l − , u l } , { u , u } , and the cycle-factor C (cid:48) on 2 k − p − C and only the cycle with vertex set { u , u , . . . , u l − p , u l − p +1 } . We conclude aspreviously. Finally, if i ∈ { , , . . . , l − p − } , then we choose C to be the cycle-factor on p +1 verticescontaining the cycles with vertex set { u l − p , u l − p +1 } , { u l − p +2 , u l − p +3 } , . . . , { u l − , u l } and C (cid:48) the onecontaining the cycles with vertex set { u , u } , { u , u , . . . , u i } , { u i +1 , u i +2 } . . . , { u l − p − , u l − p − } .Once again, we conclude as previously. Case 2: p is even. If there exists an arc between two vertices at distance 2 along C (cid:48) , thenwe will proceed almost as in the case where p is odd. The difference here, is that C will con-tain cycles all of length 2 except one of length 3. Indeed, assume for instance that u l u l − isan arc of D M u . We consider once again the set X = { u , u , u , . . . , u l − p +1 } and show, aspreviously, that there exists an arc between a vertex u i of X and u . If i = 1, then we con-sider the cycle-factor C (cid:48) on 2 k − p − C and the cycles with vertex sets { u , u , u } , { u , u } , . . . , { u l − p − , u l − p − } and the cycle-factor C on p +1 vertices containing the cy-cles with vertex set { u l − p , u l − p +1 } , { u l − p +2 , u l − p +3 } , . . . , { u l − , u l − } , { u l − , u l − , u l } . Once again,we conclude with Lemma 4 that D admits a good cycle-factor. The cases where i = l − p + 1 andwhere i ∈ { , , . . . , l − p − } are similar to the corresponding cases where p is odd.Finally, assume that there is no arc between two vertices at distance 2 along C (cid:48) . Then, wedenote by A the vertices u i with odd indices and by B the vertices u i with even indices. Thesets A and B form two strong connected components of D M u [ C (cid:48) ] and so their are both initial andterminal strong components of D M u [ C (cid:48) ]. In particular, D M u contains all the arcs from A to B andall the arcs from B to A . The vertex u must have a neighbor in A . Indeed, otherwise, as there isno arc from u to C or from C to u , we will conclude that d + D Mu ( u ) ≥ k or d − D Mu ( u ) ≥ k , whichis not possible. So, assume that u is adjacent to a vertex u i of A . Then, instead C (cid:48) we considerthe cycle C (cid:48)(cid:48) = u , u , u i , u , . . . , u i − u u i +1 , . . . u l . As there exist all the possible arcs between A and B , then all the arcs of C (cid:48)(cid:48) are contained in a digon and there exists an arc between twovertices at distance 2 along C (cid:48)(cid:48) (the arc u u i ). Thus, we are in the previous case and we concludethat D M u admits a good anti-cycle. We finish this paper with some conjectures about the problem of cycle-factor in bipartite or mul-tipartite tournaments. First of all, we have to mention the two related conjectures appearing inthe original paper of Zhang and al. [15]. The first one adds a new hypothesis imposing an arc inthe 2-cycle-factor.
Conjecture 5 (Zhang and al. [15]) . Let D be a k -regular bipartite tournament, with k an integergreater than 2. Let uv be any specified arc of D . If D is isomorphic neither to F k nor to someother specified families of digraphs, then for every even p with 4 ≤ p ≤ | V ( D ) | − D has a cycle C of length p such that D \ C is hamiltonian and such that C goes through the arc uv .The second conjecture, conversely, imposes that the cycles contain specific vertices.20 onjecture 6 (Zhang and al. [15]) . Let D be a k -regular bipartite tournament, with k an integergreater than 2. Let u and v be two specified vertices of D . If D is isomorphic neither to F k nor tosome other specified families of digraphs, then for every even p with 4 ≤ p ≤ | V ( D ) | − D has acycle C of length p such that D \ C is hamiltonian and such that C contains u and D \ C contains v . Throughout the proof of Theorem 1, we intensively used the regularity of the bipartite tour-nament. It seems that we cannot get rid of this condition as we can easily find an infinite familyof bipartite tournament with | d + ( u ) − d − ( u ) | ≤ u and | d + ( u ) − d + ( v ) | ≤ { u, v } , which does not contain any cycle-factor. For instance, for any k ≥ F k , consisting of four independent sets K , L , M and N with | K | = | N | = k and | L | = | M | = k + 1 with all possible arcs from K to L , from L to M , from M to N and from N to K .Let D be a c -partite tournament, and denote by I , . . . , I c its independent sets. We say that D is k - fully regular if, for any distinct i and j with 1 ≤ i, j ≤ c , D [ I i ∪ I j ] is a k -regular bipartitetournament. In particular, all the sets I i have size 2 k .In [14], Yeo proved that if c ≥ c -partite regular tournament D , every vertex iscontained in a cycle of length l for l = 3 , . . . , | V ( D ) | . He also conjectured the following. Conjecture 7 (Yeo [14]) . Every regular c -partite tournaments D , with c ≥
5, contains a ( p, | V ( D ) |− p )-cycle-factor for all p ∈ { , . . . , | V ( D ) | − } .An extension of our results and a weaker form of Yeo’s Conjecture could be the following. Conjecture 8.
Let D be a k -fully regular c -partite tournament with c ≥
5. Then for every even p with 4 ≤ p ≤ | V ( D ) | − D has a ( p, | V ( D ) | − p )-cycle-factor.We can see that if c is even and there is at least one pair { I i , I j } such that D [ I i , I j ] is notisomorphic to F k , then our result implies the Conjecture 8, by properly partitioning the sets I i into two parts and applying Theorem 1 on the bipartite lying between the two parts. However,the case where c is odd seems more complicated to handle. References [1] Yandong Bai, Hao Li, and Weihua He. Complementary cycles in regular bipartite tournaments.
Discret. Math. , 333:14–27, 2014.[2] Jørgen Bang-Jensen and Gregory Z. Gutin.
Digraphs - Theory, Algorithms and Applications,Second Edition . Springer Monographs in Mathematics. Springer, 2009.[3] Jean-Claude Bermond and Carsten Thomassen. Cycles in digraphs- a survey.
J. Graph Theory ,5(1):1–43, 1981.[4] J. Adrian Bondy and Uppaluri S. R. Murty.
Graph Theory . Graduate Texts in Mathematics.Springer, 2008.[5] Paul Camion. Chemins et circuits hamiltoniens des graphes complets.
C. R. Acad. Sci. Paris ,249(21):2151–2152, 1959.[6] Guantao Chen, Ronald J. Gould, and Hao Li. Partitioning vertices of a tournament intoindependent cycles.
J. Comb. Theory, Ser. B , 83(2):213–220, 2001.217] Roland H¨aggkvist and Yannis Manoussakis. Cycles and paths in bipartite tournaments withspanning configurations.
Comb. , 9(1):33–38, 1989.[8] Daniela K¨uhn, Deryk Osthus, and Timothy Townsend. Proof of a tournament partitionconjecture and an application to 1-factors with prescribed cycle lengths.
Comb. , 36(4):451–469, 2016.[9] Hao Li and Jinlong Shu. The partition of a strong tournament.
Discret. Math. , 290(2/3):211–220, 2005.[10] Yannis Manoussakis.
Probl`emes extr´emaux dans les graphes orient´es . PhD thesis, 1987.1987PA112305.[11] J. W. Moon. On subtournaments of a tournament.
Can. Math. Bull. , 9(3):297–301, 1966.[12] K. Brooks Reid. Two complementary circuits in two-connected tournaments. In
Cycles ingraphs (Burnaby, B.C., 1982) , volume 115 of
North-Holland Math. Stud. , pages 321–334.North-Holland, 1985.[13] Zeng Min Song. Complementary cycles of all lengths in tournaments.
J. Comb. Theory, Ser.B , 57(1):18–25, 1993.[14] Anders Yeo. Diregular c -partite tournaments are vertex-pancyclic when c ≥ Journal ofGraph Theory , 32(2):137–152, 1999.[15] Ke Min Zhang, Yannis Manoussakis, and Zeng Min Song. Complementary cycles containinga fixed arc in diregular bipartite tournaments.
Discret. Math. , 133(1-3):325–328, 1994.[16] Ke Min Zhang and Zeng Min Song. Complementary cycles containing of fixed vertices inbipartite tournaments.