On kernels by rainbow paths in arc-coloured digraphs

Ruijuan Li; Yanqin Cao; Xinhong Zhang

doi:10.1515/math-2021-0019

Article Open Access

On kernels by rainbow paths in arc-coloured digraphs

Ruijuan Li , Yanqin Cao and Xinhong Zhang

Published/Copyright: May 10, 2021

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$Open Mathematics$

From the journal Open Mathematics Volume 19 Issue 1

Abstract

In 2018, Bai, Fujita and Zhang [Discrete Math. 341 (2018), no. 6, 1523–1533] introduced the concept of a kernel by rainbow paths (for short, RP-kernel) of an arc-coloured digraph D , which is a subset S of vertices of D such that ( a ) there exists no rainbow path for any pair of distinct vertices of S , and ( b ) every vertex outside S can reach S by a rainbow path in D . They showed that it is NP-hard to recognize whether an arc-coloured digraph has an RP-kernel and it is NP-complete to decide whether an arc-coloured tournament has an RP-kernel. In this paper, we give the sufficient conditions for the existence of an RP-kernel in arc-coloured unicyclic digraphs, semicomplete digraphs, quasi-transitive digraphs and bipartite tournaments, and prove that these arc-coloured digraphs have RP-kernels if certain “short” cycles and certain “small” induced subdigraphs are rainbow.

Keywords: arc-coloured digraphs; kernels; kernels by rainbow paths

MSC 2010: 05C20; 05C12; 05C07

1 Introduction

For convenience of the reader, some necessary terminologies and notations not mentioned in this section can be found in Section 2. All digraphs considered in this paper are finite. In this paper, all paths, walks and cycles are always directed. For terminology and notation, we refer the reader to Bang-Jensen and Gutin [1].

Let D be a digraph. A kernel of D is a subset S ⊆ V ( D ) such that ( a ) for any pair of distinct vertices x , y ∈ S are non-adjacent, and ( b ) for each vertex v ∈ V ( D ) ⧹ S , there exists a vertex s ∈ S such that ( v , s ) ∈ A ( D ) . This notion was originally introduced in the game theory by von Neumann and Morgenstern [2] in 1944. Kernels have found many applications and several sufficient conditions for the existence of a kernel have been proved, see [3]. In this paper, we will need the following result.

Theorem 1.1

[2] Let D be a digraph. If D has no cycle, then D has a unique kernel.

Let D be a digraph and m a positive integer. An arc-colouring of D is a mapping C : A ( D ) → N , where N is the set of natural numbers. C ( D ) and C ( x , y ) denote the set of colours appearing on all the arcs of D and the colour appearing on the arc ( x , y ) ∈ A ( D ) . We call D an m-arc-coloured digraph if ∣ C ( D ) ∣ = m . An arc-coloured digraph is called monochromatic if all arcs are assigned the same colour. Define a kernel by monochromatic paths of an arc-coloured digraph D to be a subset S ⊆ V ( D ) such that ( a ) there exists no monochromatic path for any pair of vertices of S , and ( b ) for each vertex outside S can reach S by a monochromatic path.

The concept of a kernel by monochromatic paths of an arc-coloured digraph was introduced by Sands et al. [4] in 1982 as a generalization of the concept of a kernel. They showed that every 2-coloured digraph has a kernel by monochromatic paths. As a corollary, they showed that every 2-coloured tournament has a one-vertex kernel by monochromatic paths. They also proposed the problem asking whether a 3-coloured tournament with no rainbow triangle has a one-vertex kernel by monochromatic paths. In 1988, Shen [5] proved that for m ⩾ 3 every m -coloured tournament with no rainbow triangle and no rainbow transitive triangle has a one-vertex kernel by monochromatic paths, and also showed that the condition “with no rainbow triangle and no rainbow transitive triangle” cannot be improved for m ⩾ 5 . In 2004, Galeana-Sánchez and Rojas-Monroy [6] showed that the condition of Shen cannot be improved for m = 4 by constructing a family of counterexamples. For more results about kernels by monochromatic paths of an arc-coloured digraph can be found in [7,8,9, 10,11].

An arc-coloured digraph is called properly coloured if any two consecutive arcs have distinct colours. Define a kernel by properly coloured paths of an arc-coloured digraph D to be a subset S ⊆ V ( D ) such that ( a ) there exists no properly coloured path for any pair of vertices of S , and ( b ) for each vertex outside S can reach S by a properly coloured path.

The concept of a kernel by properly coloured paths of an arc-coloured digraph was introduced by Delgado-Escalante and Galeana-Sánchez [12] in 2009 as a generalization of the concept of a kernel. Bai et al. [13] showed in 2018 that it is NP-hard to recognize whether an arc-coloured digraph has a kernel by properly coloured paths. They conjecture that every arc-coloured digraph with all cycles properly coloured has a kernel by properly coloured paths and verified the conjecture for unicyclic digraphs, semicomplete digraphs and bipartite tournaments. In 2018, Delgado-Escalante et al. [14] gave some sufficient conditions for the existence of a kernel by properly coloured paths in arc-coloured tournaments, quasi-transitive digraphs and k -partite tournaments.

An arc-coloured digraph is called rainbow if all arcs have distinct colours. Define a kernel by rainbow paths (for short, RP-kernel) of an arc-coloured digraph D to be a subset S ⊆ V ( D ) such that ( a ) there exists no rainbow path for any pair of vertices of S , and ( b ) for each vertex outside S can reach S by a rainbow path.

The concept of an RP-kernel of an arc-coloured digraph was introduced by Bai et al. [13] in 2018 as a generalization of the concept of kernel. They showed that it is NP-hard to recognize whether an arc-coloured digraph has an RP-kernel. Recently, Bai et al. [15] proposed the following theorem.

Theorem 1.2

[15] It is NP-complete to decide whether an arc-coloured tournament has an RP-kernel.

Problem 1.3

[15] Is it true that every arc-coloured digraph with all cycles rainbow has an RP-kernel?

Just as other NP-complete problems, we give some sufficient conditions for the existence of an RP-kernel in arc-coloured unicyclic digraphs, semicomplete digraphs, quasi-transitive digraphs and bipartite tournaments and prove that these arc-coloured digraphs have RP-kernels if certain “short” cycles and certain “small” induced subdigraphs are rainbow.

2 Terminology and preliminaries

Let D be a digraph. V ( D ) and A ( D ) denote its vertex and arc sets. If ( x , y ) is an arc of D , sometimes we use the notation x → y to denote this arc and say that x dominates y . The out-neighbourhood (resp. in-neighbourhood) of a vertex x ∈ V ( D ) is the N D + ( x ) = { y ∣ ( x , y ) ∈ A ( D ) } (resp. N D − ( x ) = { y ∣ ( y , x ) ∈ A ( D ) } ). For a vertex x ∈ V ( D ) , the out-degree (resp. in-degree) of x is denoted by d D + ( x ) = ∣ N D + ( x ) ∣ (resp. d D − ( x ) = ∣ N D − ( x ) ∣ ). A vertex in D is called sink (resp. source) if d D + ( x ) = 0 (resp. d D − ( x ) = 0 ). An arc ( x , y ) ∈ A ( D ) is called asymmetrical (resp. symmetrical) if ( y , x ) ∉ A ( D ) (resp. ( y , x ) ∈ A ( D ) ). If S is a nonempty set of V ( D ) , then the subdigraph D [ S ] induced by S is the digraph having vertex set S , and whose arcs are all those arcs of D joining vertices of S .

For disjoint sets X and Y , X ⇒ Y means that every vertex of X dominates every vertex of Y and y ↛ x for any x ∈ X and y ∈ Y . If Y = { v } , we always denote X ⇒ v instead of X ⇒ { v } . X ⇏ Y means that there exists a vertex u ∈ Y such that X ⇏ u .

For two distinct vertices x , y ∈ V ( D ) , a path P from x to y is denoted by ( x , y ) -path, and ℓ ( P ) denotes the length of path P . Let S ⊆ V ( D ) . ( x , S ) -path in D denotes an ( x , s ) -path for some s ∈ S . ( S , x ) -path in D denotes an ( s , x ) -path for some s ∈ S . We always call a cycle C of length ℓ ( C ) an ℓ ( C ) -cycle.

A digraph D is called strong if there exists a path from x to y and a path from y to x in D for every choice of distinct vertices x , y of D . A strong component of a digraph D is a maximal induced subdigraph of D which is strong. For any non-strong digraph D , we can label its strong components D 1 , D 2 , … , D p , p ⩾ 2 , in such a way that there is no arc from D j to D i when j > i .

In the following proof, we use the definition below.

Definition 2.1

For an arc-coloured digraph D , the rainbow closure of D denoted by C r ( D ) , is a digraph such that:

V ( C r ( D ) ) = V ( D ) ;
A ( C r ( D ) ) = { ( u , v ) ∣ there exists a rainbow ( u , v ) -path in D } .

It is not hard to see the following simple and useful result.

Observation 2.2

An arc-coloured digraph D has an RP-kernel if and only if C r ( D ) has a kernel.

A digraph D is called a kernel-perfect digraph or KP-digraph when every induced subdigraph of D has a kernel. The following theorem gives a sufficient condition for a digraph to be a KP-digraph.

Theorem 2.3

[3] Let D be a digraph such that every cycle in D has at least one symmetrical arc. Then D is a KP-digraph.

3 Unicyclic digraphs

A digraph D is a unicyclic digraph if it contains only one cycle. In this section, we consider the sufficient conditions for the existence of an RP-kernel in an arc-coloured unicyclic digraph.

Theorem 3.1

Let D be an m -arc-coloured unicyclic digraph such that the unique cycle is rainbow. Then D has an RP-kernel.

Proof

Let D be an m -arc-coloured unicyclic digraph with the unique cycle C . We will show the result by constructing an RP-kernel S of D . If D is strong, then D = C . Since the cycle C is rainbow, each vertex of C forms an RP-kernel of C . The desired result follows directly.

Now assume D is not strong. Then D has strong components D 1 , D 2 , … , D k ( k ⩾ 2 ) such that there exists no arc from D i to D j for any i > j . Since D is unicyclic, then one of the strong components D 1 , D 2 , … , D k containing the unique cycle C and any other strong component is a single vertex. Let k = j 1 . If D k is a single vertex, say D k = D j 1 = { v j 1 } , we put v j 1 into S . If D k = D j 1 = C , we put an arbitrary vertex of C , say also v j 1 , into S . Since C is rainbow, V ( C ) ⧹ { v j 1 } can reach v j 1 by a rainbow path. In the i th step, say S = { v j 1 , v j 2 , … , v j i − 1 } . Let j i ∈ { 1 , 2 , … , j i − 1 − 1 } be the largest integer such that there exists no rainbow path from some vertex of D j i to S . If D j i is a single vertex, say D j i = { v j i } , we put v j i into S . If D j i = C , we put an arbitrary vertex of C , say also v j i , into S . The procedure can be completed after finite steps. The set S constructed by the aforementioned procedure is an RP-kernel of D .□

4 Semicomplete digraphs

A digraph D is semicomplete if for any pair of vertices there exists at least one arc between them. A tournament is a semicomplete digraph with no 2-cycle. In this section, we consider the sufficient conditions for the existence of an RP-kernel in an arc-coloured semicomplete digraph. Since each pair of vertices in a semicomplete digraph are adjacent, it follows that an RP-kernel of a semicomplete digraph consists of only one vertex.

Theorem 4.1

Let D be an m -arc-coloured semicomplete digraph with all 3-cycles are rainbow in D . Then D has a one-vertex RP-kernel.

Proof

Let v be a vertex of D with maximum in-degree. Since D is a semicomplete digraph, it follows that N D + ( v ) ∪ N D − ( v ) ∪ { v } = V ( D ) . Moreover, for any u ∈ N D + ( v ) , there exists a vertex w ∈ N D − ( v ) such that u → w . If not, then for any w ∈ N D − ( v ) , we have w → u . This implies that N D − ( v ) ∪ { v } ⊆ N D − ( u ) , which contradicts the choice of v . So ( u , w , v , u ) is a rainbow 3-cycle. It follows ( u , w , v ) is a rainbow ( u , v ) -path. Now for any u ∈ N D + ( v ) , there exists a rainbow ( u , v ) -path. Clearly, for any w ∈ N D − ( v ) , there exists a rainbow ( w , v ) -path. Combining with N D + ( v ) ∪ N D − ( v ) ∪ { v } = V ( D ) , we have for any x ∈ V ( D ) ⧹ { v } , there exists a rainbow ( x , v ) -path. Thus, { v } is a one-vertex RP-kernel of D .□

Corollary 4.2

[15] Let T be an m -arc-coloured tournament in which all 3-cycles are rainbow. Then T has a one-vertex RP-kernel.

5 Quasi-transitive digraphs

A digraph D is a quasi-transitive digraph if whenever { ( u , v ) , ( v , w ) } ⊆ A ( D ) , then either ( u , w ) ∈ A ( D ) or ( w , u ) ∈ A ( D ) . In this section, we consider the sufficient conditions for the existence of an RP-kernel in an arc-coloured quasi-transitive digraph.

Lemma 5.1

[1] Let D be a quasi-transitive digraph. If x and y are a pair of distinct vertices of D such that D has an ( x , y ) -path but x does not dominate y , then either y → x , or there exist vertices u , v ∈ V ( D ) ⧹ { x , y } such that x → u → v → y and y → u → v → x .

Let QT 4 be the quasi-transitive digraph, which has V ( QT 4 ) = { x , y , u , v } and A ( QT 4 ) = { ( x , u ) , ( u , v ) , ( v , y ) , ( y , u ) , ( v , x ) } (Figure 1).

$Figure 1 A quasi-transitive digraph QT 4 {{QT}}_{4} .$

Figure 1

A quasi-transitive digraph QT 4 .

Lemma 5.2

Let D be an m -arc-coloured quasi-transitive digraph with all 3-cycles and all induced subdigraphs QT 4 are rainbow in D . If x and y are a pair of distinct vertices of D such that D has a rainbow ( x , y ) -path but no rainbow ( y , x ) -path, then ( x , y ) ∈ A ( D ) .

Proof

Suppose to the contrary that ( x , y ) ∉ A ( D ) . Since there exists no rainbow ( y , x ) -path, then x , y are non-adjacent. Let P = ( x = x 0 , x 1 , x 2 , … , x n = y ) ⊆ D be a rainbow ( x , y ) -path. By Lemma 5.1, there exist vertices u , v ∈ V ( D ) ⧹ { x , y } such that x → u → v → y and y → u → v → x . This implies that D [ x , u , v , y ] is QT 4 which is rainbow. It follows ( y , u , v , x ) is a rainbow ( y , x ) -path, which is a contradiction. Thus, ( x , y ) ∈ A ( D ) .□

Theorem 5.3

Let D be an m -arc-coloured quasi-transitive digraph with all 3-cycles and all induced subdigraphs QT 4 are rainbow in D . Then C r ( D ) is a KP-digraph.

Proof

Suppose to the contrary that C r ( D ) is not a KP-digraph. By Theorem 2.3, there exists a cycle with no symmetrical arc. Let C = ( u 1 , u 2 , … , u ℓ , u 1 ) be a shortest cycle with no symmetrical arc in C r ( D ) . We will get a contradiction by showing that C has a symmetrical arc.

Claim 1

C ⊆ D and ℓ ⩾ 5 .

Proof

Since C has no symmetrical arc, for each i ∈ { 1 , 2 , … , ℓ } , there exists a rainbow ( u i , u i + 1 ) -path and no rainbow ( u i + 1 , u i ) -path in D . By Lemma 5.2, we have ( u i , u i + 1 ) ∈ A ( D ) . Then C ⊆ D .

Now we prove ℓ ⩾ 5 . Since C has no symmetrical arc, we have ℓ ⩾ 3 .

If ℓ = 3 , combining with C ⊆ D , we have C = ( u 1 , u 2 , u 3 , u 1 ) is a rainbow 3-cycle in D . This implies that ( u 2 , u 3 , u 1 ) is a rainbow ( u 2 , u 1 ) -path and hence ( u 2 , u 1 ) ∈ A ( C r ( D ) ) . Note that ( u 1 , u 2 ) ∈ A ( C ) , which contradicts that C has no symmetrical arc.

If ℓ = 4 , by the proof above, we have C = ( u 1 , u 2 , u 3 , u 4 , u 1 ) is a cycle in D . Since D is quasi-transitive, we have u 1 , u 3 are adjacent. If ( u 1 , u 3 ) ∈ A ( D ) , then ( u 1 , u 3 , u 4 , u 1 ) is a rainbow 3-cycle. This implies that ( u 4 , u 1 , u 3 ) is a rainbow ( u 4 , u 3 ) -path and hence ( u 4 , u 3 ) ∈ A ( C r ( D ) ) . Note that ( u 3 , u 4 ) ∈ A ( C ) , which contradicts that C has no symmetrical arc. If ( u 3 , u 1 ) ∈ A ( D ) , then ( u 1 , u 2 , u 3 , u 1 ) is a rainbow 3-cycle. This implies that ( u 2 , u 3 , u 1 ) is a rainbow ( u 2 , u 1 ) -path and hence ( u 2 , u 1 ) ∈ A ( C r ( D ) ) . Note that ( u 1 , u 2 ) ∈ A ( C ) , which contradicts that C has no symmetrical arc.

Thus, ℓ ⩾ 5 .□

By Claim 1, we have ℓ − 1 ⩾ 4 and C ⊆ D . Considering { ( u ℓ − 1 , u ℓ ) , ( u ℓ , u 1 ) } ⊆ A ( D ) , we have u 1 , u ℓ − 1 are adjacent.

If ( u 1 , u ℓ − 1 ) ∈ A ( D ) , then ( u ℓ − 1 , u ℓ , u 1 , u ℓ − 1 ) is a rainbow 3-cycle. This implies that ( u ℓ , u 1 , u ℓ − 1 ) is a rainbow ( u ℓ , u ℓ − 1 ) -path and hence ( u ℓ , u ℓ − 1 ) ∈ A ( C r ( D ) ) . Note that ( u ℓ − 1 , u ℓ ) ∈ A ( C ) , which contradicts that C has no symmetrical arc.

If ( u ℓ − 1 , u 1 ) ∈ A ( D ) , since ( u 1 , u 2 ) ∈ A ( D ) , there exists i ∈ { 3 , 4 , … , ℓ − 1 } such that ( u i , u 1 ) ∈ A ( D ) . Let

i 0 = min { i ∈ { 3 , 4 , … , ℓ − 1 } ∣ ( u i , u 1 ) ∈ A ( D ) } .

Considering { ( u i 0 − 1 , u i 0 ) , ( u i 0 , u 1 ) } ⊆ A ( D ) , we have u 1 , u i 0 − 1 are adjacent. By the choice of i 0 , we have ( u 1 , u i 0 − 1 ) ∈ A ( D ) . It follows ( u 1 , u i 0 − 1 , u i 0 , u 1 ) is a rainbow 3-cycle. This implies that ( u i 0 , u 1 , u i 0 − 1 ) is a rainbow ( u i 0 , u i 0 − 1 ) -path and hence ( u i 0 , u i 0 − 1 ) ∈ A ( C r ( D ) ) . Note that ( u i 0 − 1 , u i 0 ) ∈ A ( C ) , which contradicts that C has no symmetrical arc.

Thus, C r ( D ) is a KP-digraph.□

By Observation 2.2 and Theorem 5.3, the following corollary is direct.

Corollary 5.4

Let D be an m -arc-coloured quasi-transitive digraph with all 3-cycles and all induced subdigraphs QT 4 are rainbow in D . Then D has an RP-kernel.

6 Bipartite tournaments

A digraph D is a bipartite tournament if there exists a partition of V ( D ) into two sets { X , Y } such that there exists no arc between any two vertices in the same set and there exists an arc between any two vertices in different sets. In this section, we consider the sufficient conditions for the existence of an RP-kernel in an arc-coloured bipartite tournament. We begin with two simple observations.

Observation 6.1

[13] Let D be an m -arc-coloured digraph and v ∈ V ( D ) a source. Then D has an RP-kernel if and only if D − v has an RP-kernel.

Observation 6.2

1-arc-coloured bipartite tournament D = ( X , Y ) has an RP-kernel.

Proof

Obviously, a kernel of D is also an RP-kernel of D . We claim that either X or Y is a kernel of D and hence an RP-kernel of D . If X is not a kernel, then there exists v ∈ Y such that X ⇒ v . This implies that Y is a kernel of D .□

In the following, we may assume m ⩾ 2 .

Observation 6.3

Let D = ( X , Y ) be an m -arc-coloured bipartite tournament with min { ∣ X ∣ , ∣ Y ∣ } = 1 . Then D has an RP-kernel.

Proof

W.l.o.g., assume ∣ X ∣ = min { ∣ X ∣ , ∣ Y ∣ } = 1 . Obviously, D has no cycle. This implies that C r ( D ) also has no cycle. By Theorem 1.1, C r ( D ) has a unique kernel. By Observation 2.2, D has an RP-kernel.□

Now we consider the RP-kernel of an m -arc-coloured bipartite tournament D = ( X , Y ) with min { ∣ X ∣ , ∣ Y ∣ } = 2 . W.l.o.g., assume min { ∣ X ∣ , ∣ Y ∣ } = ∣ X ∣ = 2 and X = { x 1 , x 2 } . If X ⇒ Y , then Y is an RP-kernel of D . If Y ⇒ X , then X is an RP-kernel of D . So we assume X ⇏ Y and Y ⇏ X . Let

Y 0 = { y ∈ Y ∣ X ⇒ y } ,

Y 1 = { y ∈ Y ⧹ Y 0 ∣ there exists a rainbow ( y , Y 0 ) -path in D } ,

Y 2 = Y ⧹ ( Y 0 ∪ Y 1 ) .

By Observation 6.1, we also assume that D has no source in Y . By the definition of Y 0 , the following claim holds directly.

Each vertex in Y ⧹ Y 0 = Y 1 ∪ Y 2 has exactly one out-neighbour and one in-neighbour in X . ( ∗ )

We will show that D has an RP-kernel if every 4-cycle contained in D is coloured with at least three colours. For this purpose, we divide the proof into two lemmas.

Lemma 6.4

Let D = ( X , Y ) be an m -arc-coloured bipartite tournament with min { ∣ X ∣ , ∣ Y ∣ } = 2 satisfying that every 4-cycle contained in D is coloured with at least three colours. Y 0 , Y 1 and Y 2 are defined as above. If Y 0 ≠ ∅ , then D has an RP-kernel.

Proof

Note that there exists no rainbow ( y , Y 0 ) -path for any y ∈ Y 2 . If Y 2 = ∅ , then Y 0 is an RP-kernel of D . So we assume that Y 2 ≠ ∅ . Clearly, the following claim holds directly.

Claim 1

There exist rainbow paths from X ∪ Y 1 to Y 0 ; there exists no rainbow path from Y 0 to Y ⧹ Y 0 and there exists no rainbow path from Y 2 to Y 0 .

Claim 2

For some x i ∈ X , if x i has an in-neighbour in Y 2 , then all arcs from x i to Y 0 are assigned the common colour.

Proof

Suppose to the contrary that C ( x i , y 1 ) ≠ C ( x i , y 2 ) for some y 1 , y 2 ∈ Y 0 . Let y ∈ Y 2 with y → x i . Since C ( x i , y 1 ) ≠ C ( x i , y 2 ) , y can reach Y 0 by a rainbow path passing through ( y , x i ) as well as either ( x i , y 1 ) or ( x i , y 2 ) . This contradicts that there exists no rainbow ( y , Y 0 ) -path for any y ∈ Y 2 . Thus, all arcs from x i to Y 0 are assigned the common colour.□

For convenience, we will denote the common colour assigned the arcs from x i to Y 0 by C ( x i , Y 0 ) for x i ∈ X with an in-neighbour in Y 2 . By the definition of Y 2 , the following claim holds directly.

Claim 3

For any y ∈ Y 2 with y → x i for some x i ∈ X , C ( y , x i ) = C ( x i , Y 0 ) .

Let S ⊆ Y 2 be the maximal subset such that there exists no rainbow path for any pair of vertices of S in D . Let

R = { r ∈ Y 2 ⧹ S ∣ there exists no rainbow ( r , S ) -path in D } .

If R = ∅ , then Y 0 ∪ S is an RP-kernel of D . Assume that R ≠ ∅ and let r ∈ R ⊆ Y 2 be arbitrary. By ( ∗ ) , w.l.o.g., we assume

x 1 → r → x 2 .

By the choice of S , there exists a rainbow ( s , r ) -path P for some s ∈ S in D .

If ℓ ( P ) = 4 , w.l.o.g., assume P = ( s , x 2 , y , x 1 , r ) where y ∈ Y ⧹ Y 0 . It is clear that C ( x 1 , r ) ≠ C ( x 1 , Y 0 ) , since otherwise, in the rainbow path P , we replace the arc ( x 1 , r ) with ( x 1 , y 0 ) for any y 0 ∈ Y 0 and get a rainbow ( s , y 0 ) -path, which contradicts s ∈ Y 2 . Now we claim that Y 0 ∪ { r } is an RP-kernel of D . By Claim 1, it is sufficient to show that there exists a rainbow ( z , r ) -path for any z ∈ Y 2 ⧹ { s , r } . By ( ∗ ) , we have either z → x 1 or z → x 2 . If z → x 1 , by Claim 3, we have C ( z , x 1 ) = C ( x 1 , Y 0 ) . Combining with C ( x 1 , r ) ≠ C ( x 1 , Y 0 ) , we have ( z , x 1 , r ) is a rainbow ( z , r ) -path. If z → x 2 , by Claim 3, we have C ( z , x 2 ) = C ( s , x 2 ) = C ( x 2 , Y 0 ) . In the rainbow path P , we replace the arc ( s , x 2 ) with ( z , x 2 ) and get a rainbow ( z , r ) -path ( z , x 2 , y , x 1 , r ) . This implies that Y 0 ∪ { r } is an RP-kernel of D .

If ℓ ( P ) = 2 , now ( s , x 1 , r ) is the rainbow ( s , r ) -path. Note that s ∈ S ⊆ Y 2 and s → x 1 . Let y ∈ Y 2 with y → x 1 . By Claim 3, we have C ( y , x 1 ) = C ( s , x 1 ) = C ( x 1 , Y 0 ) . In the rainbow path ( s , x 1 , r ) , we replace the arc ( s , x 1 ) with ( y , x 1 ) and get a rainbow path ( y , x 1 , r ) . This means that all vertices dominating x 1 in Y 2 can reach r by a rainbow path. Let

Q 1 = { y ∈ Y 2 ⧹ { r } ∣ y → x 1 } , Q 2 = { y ∈ Y 2 ⧹ { r } ∣ y → x 2 } .

Clearly, each vertex of Q 1 can reach r by a rainbow path. By ( ∗ ) , we have Q 1 ∪ Q 2 = Y 2 ⧹ { r } and Q 1 ∩ Q 2 = ∅ . If Q 2 = ∅ , then Y 0 ∪ { r } is an RP-kernel of D . So assume that Q 2 ≠ ∅ . Also by ( ∗ ) ,

x 1 ⇒ Q 2 ⇒ x 2 .

If there exists a rainbow ( q 2 , r ) -path P ′ for some q 2 ∈ Q 2 , we claim that Y 0 ∪ { r } is an RP-kernel of D . Since x 1 → r → x 2 and x 1 → q 2 → x 2 , we have ℓ ( P ′ ) ≠ 2 and hence ℓ ( P ′ ) = 4 . W.l.o.g., assume P ′ = ( q 2 , x 2 , y , x 1 , r ) where y ∈ Y ⧹ Y 0 . It is sufficient to show that there exists a rainbow ( q 2 ′ , r ) -path for any q 2 ′ ∈ Q 2 ⧹ { q 2 , r } . By Claim 3, we have C ( q 2 , x 2 ) = C ( q 2 ′ , x 2 ) = C ( x 2 , Y 0 ) . In the rainbow path P ′ , we replace the arc ( q 2 , x 2 ) with ( q 2 ′ , x 2 ) and get a rainbow ( q 2 ′ , r ) -path ( q 2 ′ , x 2 , y , x 1 , r ) . This implies that Y 0 ∪ { r } is an RP-kernel of D .

If there exists a rainbow ( r , q 2 ) -path P ″ for some q 2 ∈ Q 2 , we claim that Y 0 ∪ { q 2 } is an RP-kernel of D . Since x 1 → r → x 2 and x 1 → q 2 → x 2 , we have ℓ ( P ″ ) ≠ 2 and hence ℓ ( P ″ ) = 4 . W.l.o.g., assume P ″ = ( r , x 2 , y , x 1 , q 2 ) where y ∈ Y ⧹ Y 0 . It is clear that C ( x 1 , q 2 ) ≠ C ( x 1 , Y 0 ) , since otherwise, in the rainbow path P ″ , we replace the arc ( x 1 , q 2 ) with ( x 1 , y 0 ) for any y 0 ∈ Y 0 and get a rainbow ( r , y 0 ) -path, which contradicts r ∈ Y 2 . Now it is sufficient to show that there exists a rainbow ( w , q 2 ) -path for any w ∈ Y 2 ⧹ { r , q 2 } . By ( ∗ ) , we have either w → x 1 or w → x 2 . If w → x 1 , by Claim 3, we have C ( w , x 1 ) = C ( x 1 , Y 0 ) . Combining with C ( x 1 , q 2 ) ≠ C ( x 1 , Y 0 ) , we have ( w , x 1 , q 2 ) is a rainbow ( w , q 2 ) -path. If w → x 2 , by Claim 3, we have C ( w , x 2 ) = C ( r , x 2 ) = C ( x 2 , Y 0 ) . In the rainbow path P ″ , we replace the arc ( r , x 2 ) with ( w , x 2 ) and get a rainbow ( w , q 2 ) -path ( w , x 2 , y , x 1 , q 2 ) . This implies that Y 0 ∪ { q 2 } is an RP-kernel of D .

If there exists no rainbow ( q 2 , r ) -path and no rainbow ( r , q 2 ) -path for any q 2 ∈ Q 2 , we claim that Y 0 ∪ Q 2 ∪ { r } is an RP-kernel of D .

Claim 4

If there exists no rainbow ( Q 2 , r ) -path and there exists no rainbow ( r , Q 2 ) -path, then there exists no rainbow path for any pair of vertices of Q 2 .

Proof

Suppose to the contrary that there exists a rainbow path for some q 2 , q 2 ′ ∈ Q 2 , say ( q 2 , x 2 , y , x 1 , q 2 ′ ) , where y ∈ Y ⧹ Y 0 . Note that y ≠ r since there exists no rainbow ( Q 2 , r ) -path. Since r , q 2 ∈ Y 2 , by Claim 3, we have C ( r , x 2 ) = C ( q 2 , x 2 ) = C ( x 2 , Y 0 ) . In the rainbow path ( q 2 , x 2 , y , x 1 , q 2 ′ ) , we replace the arc ( q 2 , x 2 ) with ( r , x 2 ) and get a rainbow ( r , q 2 ′ ) -path ( r , x 2 , y , x 1 , q 2 ′ ) , which contradicts that there exists no rainbow ( r , Q 2 ) -path.□

Recall that Q 1 ∪ Q 2 = Y 2 ⧹ { r } and each vertex of Q 1 can reach r by a rainbow path. By Claims 1 and 4, Y 0 ∪ Q 2 ∪ { r } is an RP-kernel of D .□

Lemma 6.5

Let D = ( X , Y ) be an m -arc-coloured bipartite tournament with min { ∣ X ∣ , ∣ Y ∣ } = 2 satisfying that every 4-cycle contained in D is coloured with at least three colours. Y 0 , Y 1 and Y 2 are defined as above. If Y 0 = ∅ , then D has an RP-kernel.

Proof

By ( ∗ ) , each vertex of Y has one out-neighbour and one in-neighbour in X . We give a partition of Y as follows

Y ′ = { y ∈ Y ∣ x 1 → y → x 2 } , Y ″ = { y ∈ Y ∣ x 2 → y → x 1 } .

If Y ′ = ∅ , then Y = Y ″ and x 2 ⇒ Y ⇒ x 1 . If there exists a rainbow ( x 2 , x 1 ) -path, then { x 1 } is an RP-kernel of D . If there exists no rainbow ( x 2 , x 1 ) -path, then { x 1 , x 2 } is an RP-kernel of D . If Y ″ = ∅ , we can prove that either { x 2 } or { x 1 , x 2 } is an RP-kernel of D . So we assume Y ′ ≠ ∅ and Y ″ ≠ ∅ .

In particular, we consider the following subsets of Y ′ and Y ″ , respectively

Y ∗ = { y ∈ Y ′ ∣ C ( x 1 , y ) ≠ C ( y , x 2 ) } , Y ∗ ∗ = { y ∈ Y ″ ∣ C ( x 2 , y ) ≠ C ( y , x 1 ) } .

Let y ′ ∈ Y ′ and y ″ ∈ Y ″ be arbitrary. If Y ∗ = ∅ , then C ( x 1 , y ′ ) = C ( y ′ , x 2 ) . Note that ( x 1 , y ′ , x 2 , y ″ , x 1 ) is a 4-cycle. Since every 4-cycle is coloured with at least three colours, we have ( y ′ , x 2 , y ″ , x 1 ) is a rainbow path. Clearly, ( y ″ , x 1 ) is a rainbow path. It follows that { x 1 } is an RP-kernel of D . Similarly, if Y ∗ ∗ = ∅ , we can prove that { x 2 } is an RP-kernel of D . So we assume Y ∗ ≠ ∅ and Y ∗ ∗ ≠ ∅ .

Case 1: Y ′ ⧹ Y ∗ ≠ ∅ or Y ″ ⧹ Y ∗ ∗ ≠ ∅ .

W.l.o.g., assume Y ′ ⧹ Y ∗ ≠ ∅ . Let y ′ ∈ Y ′ ⧹ Y ∗ be arbitrary. We assume that C ( x 1 , y ′ ) = C ( y ′ , x 2 ) = α . Let

Y α ′ = { y ′ ∈ Y ′ ∣ C ( x 1 , y ′ ) = C ( y ′ , x 2 ) = α } .

Clearly, Y α ′ ≠ ∅ . Let y ″ ∈ Y ″ be arbitrary. Note that x 1 ⇒ y ′ ⇒ x 2 ⇒ y ″ ⇒ x 1 . Since every 4-cycle in D is coloured with at least three colours, we have C ( x 2 , y ″ ) ≠ C ( y ″ , x 1 ) , C ( x 2 , y ″ ) ≠ α and C ( y ″ , x 1 ) ≠ α . Let C ( x 2 , y ″ ) = β and C ( y ″ , x 1 ) = γ . Then α , β , γ are pairwise distinct.

Define the following vertex subsets, which are shown in Figure 2, in which a box represents a set of vertices while dotted, dashed, thick dotted and solid arcs represent, respectively, the arcs coloured by α , β , γ and ω , where ω stands for an arbitrary colour not in { α , β , γ } . Let

Y c 1 c 2 ′ = { y ′ ∈ Y ′ ∣ C ( x 1 , y ′ ) = c 1 , C ( y ′ , x 2 ) = c 2 } ,

where ( c 1 , c 2 ) = { ( α , β ) , ( β , α ) , ( α , γ ) , ( γ , α ) , ( ω , α ) , ( ω , β ) , ( ω , γ ) } ;

Y ω ′ + = { y ′ ∈ Y ′ ∣ C ( x 1 , y ′ ) is arbitrary , C ( y ′ , x 2 ) ∉ { α , β , γ } } ; Y c 1 c 2 ″ = { y ″ ∈ Y ″ ∣ C ( x 2 , y ″ ) = c 1 , C ( y ″ , x 1 ) = c 2 } ,

where ( c 1 , c 2 ) = { ( β , γ ) , ( γ , β ) , ( γ , ω ) , ( ω , γ ) , ( β , ω ) , ( ω , β ) } ;

Y ω 1 ω 2 ″ = { y ″ ∈ Y ″ ∣ C ( x 2 , y ″ ) , C ( y ″ , x 1 ) ∉ { α , β , γ } and C ( x 2 , y ″ ) ≠ C ( y ″ , x 1 ) } .

Note that Y α ′ ≠ ∅ and Y β γ ″ ≠ ∅ . Since every 4-cycle is coloured with at least three colours, we have

Y ′ = Y α ′ ∪ Y α β ′ ∪ Y β α ′ ∪ Y α γ ′ ∪ Y γ α ′ ∪ Y ω α ′ ∪ Y ω β ′ ∪ Y ω γ ′ ∪ Y ω ′ + ,

Y ″ = Y β γ ″ ∪ Y γ β ″ ∪ Y β ω ″ ∪ Y γ ω ″ ∪ Y ω β ″ ∪ Y ω γ ″ ∪ Y ω 1 ω 2 ″ .

For convenience, a vertex in Y c 1 c 2 ′ (resp. Y c 1 c 2 ″ , Y α ′ , Y ω ′ + , Y ω 1 ω 2 ″ ) we denote by y c 1 c 2 ′ (resp. y c 1 c 2 ″ , y α ′ , y ω ′ + , y ω 1 ω 2 ″ ).

If Y ω 1 ω 2 ″ ≠ ∅ , then for any y ′ ∈ Y ′ ⧹ Y ω ′ + , ( y ′ , x 2 , y ω 1 ω 2 ″ , x 1 ) is a rainbow ( y ′ , x 1 ) -path; for any y ω ′ + ∈ Y ω ′ + , ( y ω ′ + , x 2 , y β γ ″ , x 1 ) is a rainbow ( y ω ′ + , x 1 ) -path. This implies that { x 1 } is an RP-kernel of D . So we assume Y ω 1 ω 2 ″ = ∅ .

Subcase 1.1: Y ω β ″ ∪ Y β ω ″ ≠ ∅ and Y ω γ ″ ∪ Y γ ω ″ ≠ ∅ .

Then for any y ′ ∈ Y α ′ ∪ Y β α ′ ∪ Y γ α ′ ∪ Y ω α ′ ∪ Y ω ′ + , ( y ′ , x 2 , y β γ ″ , x 1 ) is a rainbow ( y ′ , x 1 ) -path; for any y ′ ∈ Y α β ′ ∪ Y ω β ′ , ( y ′ , x 2 , y ″ , x 1 ) is a rainbow ( y ′ , x 1 ) -path, where y ″ ∈ Y ω γ ″ ∪ Y γ ω ″ ; for any y ′ ∈ Y α γ ′ ∪ Y ω γ ′ , ( y ′ , x 2 , y ″ , x 1 ) is a rainbow ( y ′ , x 1 ) -path, where y ″ ∈ Y ω β ″ ∪ Y β ω ″ . This implies that { x 1 } is an RP-kernel of D .

Subcase 1.2: Exactly one of the subsets Y ω β ″ ∪ Y β ω ″ and Y ω γ ″ ∪ Y γ ω ″ is not an empty set.

W.l.o.g., assume Y ω β ″ ∪ Y β ω ″ ≠ ∅ and Y ω γ ″ ∪ Y γ ω ″ = ∅ . Now Y ″ = Y β γ ″ ∪ Y γ β ″ ∪ Y β ω ″ ∪ Y ω β ″ .

Subcase 1.2.1: Y ω β ′ = ∅ and Y α β ′ = ∅ .

Now Y ′ = Y α ′ ∪ Y β α ′ ∪ Y α γ ′ ∪ Y γ α ′ ∪ Y ω α ′ ∪ Y ω γ ′ ∪ Y ω ′ + . For any y ′ ∈ Y α ′ ∪ Y β α ′ ∪ Y γ α ′ ∪ Y ω α ′ ∪ Y ω ′ + , ( y ′ , x 2 , y β γ ″ , x 1 ) is a rainbow ( y ′ , x 1 ) -path. For any y ′ ∈ Y α γ ′ ∪ Y ω γ ′ , ( y ′ , x 2 , y ″ , x 1 ) is a rainbow ( y ′ , x 1 ) -path, where y ″ ∈ Y ω β ″ ∪ Y β ω ″ . This implies that { x 1 } is an RP-kernel of D .

Subcase 1.2.2: Y ω β ′ ≠ ∅ and Y α β ′ ≠ ∅ .

Now Y ′ = Y α ′ ∪ Y α β ′ ∪ Y β α ′ ∪ Y α γ ′ ∪ Y γ α ′ ∪ Y ω α ′ ∪ Y ω β ′ ∪ Y ω γ ′ ∪ Y ω ′ + . Note that there exists no rainbow path for any pair of vertices of Y ω β ′ ∪ Y α β ′ since C ( Y ω β ′ ∪ Y α β ′ , x 2 ) = β ∈ C ( x 2 , Y ″ ) ∪ C ( Y ″ , x 1 ) . Since every 4-cycle is coloured with at least three colours, we have C ( x 1 , y ω β ′ ) ∉ C ( x 2 , y ω β ″ ) ∪ C ( y β ω ″ , x 1 ) for any y ω β ′ ∈ Y ω β ′ . For any y ′ ∈ Y α ′ ∪ Y β α ′ ∪ Y α γ ′ ∪ Y γ α ′ ∪ Y ω α ′ ∪ Y ω γ ′ , ( y ′ , x 2 , y ″ , x 1 , y ω β ′ ) is a rainbow ( y ′ , y ω β ′ ) -path, where y ″ ∈ Y ω β ″ ∪ Y β ω ″ . For any y ω ′ + ∈ Y ω ′ + , ( y ω ′ + , x 2 , y β γ ″ , x 1 , y α β ′ ) is a rainbow ( y ω ′ + , y α β ′ ) -path. This implies that Y ω β ′ ∪ Y α β ′ is an RP-kernel of D .

Subcase 1.2.3: Y ω β ′ ≠ ∅ and Y α β ′ = ∅ .

Now Y ′ = Y α ′ ∪ Y β α ′ ∪ Y α γ ′ ∪ Y γ α ′ ∪ Y ω α ′ ∪ Y ω β ′ ∪ Y ω γ ′ ∪ Y ω ′ + . By the proof above, there exists no rainbow path for any pair of vertices of Y ω β ′ and each vertex of Y α ′ ∪ Y β α ′ ∪ Y α γ ′ ∪ Y γ α ′ ∪ Y ω α ′ ∪ Y ω γ ′ can reach Y ω β ′ by a rainbow path passing through a vertex of Y ω β ″ ∪ Y β ω ″ . If ∣ C ( x 1 , Y ω β ′ ) ∣ ⩾ 2 , let y ω β 1 ′ , y ω β 2 ′ ∈ Y ω β ′ with C ( x 1 , y ω β 1 ′ ) ≠ C ( x 1 , y ω β 2 ′ ) . Let y ω ′ + ∈ Y ω ′ + be arbitrary. Note that either ( y ω ′ + , x 2 , y β γ ″ , x 1 , y ω β 1 ′ ) or ( y ω ′ + , x 2 , y β γ ″ , x 1 , y ω β 2 ′ ) is a rainbow ( y ω ′ + , Y ω β ′ ) -path. This implies that Y ω β ′ is an RP-kernel of D . If ∣ C ( x 1 , Y ω β ′ ) ∣ = 1 , let U = { y ω ′ + ∈ Y ω ′ + ∣ C ( y ω ′ + , x 2 ) = C ( x 1 , Y ω β ′ ) } . For any y ′ ∈ Y ω ′ + ⧹ U , C ( y ′ , x 2 ) ≠ C ( x 1 , Y ω β ′ ) and ( y ′ , x 2 , y β γ ″ , x 1 , y ω β ′ ) is a rainbow path from Y ω ′ + ⧹ U to Y ω β ′ . If there exists a vertex u ∈ U with C ( x 1 , u ) ≠ β , for any u ′ ∈ U ⧹ { u } , either ( u ′ , x 2 , y 1 ″ , x 1 , u ) or ( u ′ , x 2 , y 2 ″ , x 1 , u ) is a rainbow ( u ′ , u ) -path, where y 1 ″ ∈ Y ω β ″ ∪ Y β ω ″ and y 2 ″ ∈ Y β γ ″ ∪ Y γ β ″ . Note that there exists no rainbow path for any pair of vertices of Y ω β ′ ∪ { u } . This implies that Y ω β ′ ∪ { u } is an RP-kernel of D . If C ( x 1 , U ) = β , then there exists no rainbow path for any pair of vertices of Y ω β ′ ∪ U . This implies that Y ω β ′ ∪ U is an RP-kernel of D .

Subcase 1.2.4: Y ω β ′ = ∅ and Y α β ′ ≠ ∅ .

Now Y ′ = Y α ′ ∪ Y α β ′ ∪ Y β α ′ ∪ Y α γ ′ ∪ Y γ α ′ ∪ Y ω α ′ ∪ Y ω γ ′ ∪ Y ω ′ + . If Y ω α ′ ≠ ∅ or Y γ α ′ ≠ ∅ , then for any y ″ ∈ Y γ β ″ ∪ Y ω β ″ , ( y ″ , x 1 , y ′ , x 2 ) is a rainbow ( y ″ , x 2 ) -path, where y ′ ∈ Y γ α ′ ∪ Y ω α ′ ; for any y ″ ∈ Y β γ ″ ∪ Y β ω ″ , ( y ″ , x 1 , y α β ′ , x 2 ) is a rainbow ( y ″ , x 2 ) -path. This implies that { x 2 } is an RP-kernel of D . If Y ω α ′ = ∅ and Y γ α ′ = ∅ , now Y ′ = Y α ′ ∪ Y α β ′ ∪ Y β α ′ ∪ Y α γ ′ ∪ Y ω γ ′ ∪ Y ω ′ + . Note that there exists no rainbow path for any pair of vertices of Y α ′ ∪ Y α β ′ ∪ Y β α ′ . For any y ′ ∈ Y α γ ′ ∪ Y ω γ ′ , ( y ′ , x 2 , y ″ , x 1 , y α β ′ ) is a rainbow ( y ′ , Y α β ′ ) -path, where y ″ ∈ Y ω β ″ ∪ Y β ω ″ . For any y ω ′ + ∈ Y ω ′ + , ( y ω ′ + , x 2 , y β γ ″ , x 1 , y α β ′ ) is a rainbow ( y ω ′ + , Y α β ′ ) -path. This implies that Y α ′ ∪ Y α β ′ ∪ Y β α ′ is an RP-kernel of D .

Subcase 1.3: Y ω β ″ ∪ Y β ω ″ = ∅ and Y ω γ ″ ∪ Y γ ω ″ = ∅ .

Then Y ″ = Y β γ ″ ∪ Y γ β ″ . This implies that there exists no rainbow path for any pair of vertices of Y ω β ′ ∪ Y ω γ ′ ∪ Y α γ ′ ∪ Y α β ′ .

Subcase 1.3.1: Y α γ ′ ∪ Y α β ′ ≠ ∅ and Y ω β ′ ∪ Y ω γ ′ ≠ ∅ .

Now Y ′ = Y α ′ ∪ Y α β ′ ∪ Y β α ′ ∪ Y α γ ′ ∪ Y γ α ′ ∪ Y ω α ′ ∪ Y ω β ′ ∪ Y ω γ ′ ∪ Y ω ′ + . For any y ′ ∈ Y ′ ⧹ Y ω ′ + , ( y ′ , x 2 , y β γ ″ , x 1 , y ω β ′ ) and ( y ′ , x 2 , y β γ ″ , x 1 , y ω γ ′ ) are rainbow ( y ′ , Y ω β ′ ∪ Y ω γ ′ ) -paths; for any y ω ′ + ∈ Y ω ′ + , ( y ω ′ + , x 2 , y β γ ″ , x 1 , y α γ ′ ) and ( y ω ′ + , x 2 , y β γ ″ , x 1 , y α β ′ ) are rainbow ( y ′ , Y α γ ′ ∪ Y α β ′ ) -paths. This implies that Y ω β ′ ∪ Y ω γ ′ ∪ Y α γ ′ ∪ Y α β ′ is an RP-kernel of D .

Subcase 1.3.2: Y α γ ′ ∪ Y α β ′ = ∅ and Y ω β ′ ∪ Y ω γ ′ = ∅ .

Now Y ′ = Y α ′ ∪ Y β α ′ ∪ Y γ α ′ ∪ Y ω α ′ ∪ Y ω ′ + . For any y ′ ∈ Y ′ , ( y ′ , x 2 , y β γ ″ , x 1 ) is a rainbow ( y ′ , x 1 ) -path. This implies that { x 1 } is an RP-kernel of D .

Subcase 1.3.3: Y α γ ′ ∪ Y α β ′ ≠ ∅ and Y ω β ′ ∪ Y ω γ ′ = ∅ .

Now Y ′ = Y α ′ ∪ Y α β ′ ∪ Y β α ′ ∪ Y α γ ′ ∪ Y γ α ′ ∪ Y ω α ′ ∪ Y ω ′ + . If Y ω α ′ = ∅ , then Y ′ = Y α ′ ∪ Y α β ′ ∪ Y β α ′ ∪ Y α γ ′ ∪ Y γ α ′ ∪ Y ω ′ + . Note that there exists no rainbow path for any pair of vertices of Y α ′ ∪ Y α β ′ ∪ Y β α ′ ∪ Y α γ ′ ∪ Y γ α ′ . For any y ω ′ + ∈ Y ω ′ + , ( y ω ′ + , x 2 , y β γ ″ , x 1 , y α ′ ) is a rainbow ( y ω ′ + , y α ′ ) -path. This implies that Y α ′ ∪ Y α β ′ ∪ Y β α ′ ∪ Y α γ ′ ∪ Y γ α ′ is an RP-kernel of D . If Y ω α ′ ≠ ∅ , for any y ″ ∈ Y ″ , ( y ″ , x 1 , y ω α ′ , x 2 ) is a rainbow ( y ″ , x 2 ) -path. This implies that { x 2 } is an RP-kernel of D .

Subcase 1.3.4: Y α γ ′ ∪ Y α β ′ = ∅ and Y ω β ′ ∪ Y ω γ ′ ≠ ∅ .

Now Y ′ = Y α ′ ∪ Y β α ′ ∪ Y γ α ′ ∪ Y ω α ′ ∪ Y ω β ′ ∪ Y ω γ ′ ∪ Y ω ′ + . For any y ′ ∈ Y α ′ ∪ Y β α ′ ∪ Y γ α ′ ∪ Y ω α ′ , ( y ′ , x 2 , y β γ ″ , x 1 , y ω β ′ ) and ( y ′ , x 2 , y β γ ″ , x 1 , y ω γ ′ ) are rainbow ( y ′ , Y ω β ′ ∪ Y ω γ ′ ) -paths. If ∣ C ( x 1 , Y ω β ′ ∪ Y ω γ ′ ) ∣ ⩾ 2 , let y 1 ′ , y 2 ′ ∈ Y ω β ′ ∪ Y ω γ ′ with C ( x 1 , y 1 ′ ) ≠ C ( x 1 , y 2 ′ ) . Let y ω ′ + ∈ Y ω ′ + be arbitrary. Note that either ( y ω ′ + , x 2 , y β γ ″ , x 1 , y 1 ′ ) or ( y ω ′ + , x 2 , y β γ ″ , x 1 , y 2 ′ ) is a rainbow ( y ω ′ + , Y ω β ′ ∪ Y ω γ ′ ) -path. This implies that Y ω β ′ ∪ Y ω γ ′ is an RP-kernel of D . If ∣ C ( x 1 , Y ω β ′ ∪ Y ω γ ′ ) ∣ = 1 , let U = { y ω ′ + ∈ Y ω ′ + ∣ C ( y ω ′ + , x 2 ) = C ( x 1 , Y ω β ′ ∪ Y ω γ ′ ) } . Then for any y ′ ∈ Y ω ′ + ⧹ U , we have C ( y ′ , x 2 ) ≠ C ( x 1 , Y ω β ′ ∪ Y ω γ ′ ) . Note that ( y ′ , x 2 , y β γ ″ , x 1 , y ω β ′ ) and ( y ′ , x 2 , y β γ ″ , x 1 , y ω γ ′ ) are two rainbow ( y ′ , Y ω β ′ ∪ Y ω γ ′ ) -paths. If there exists a vertex u ∈ U with C ( x 1 , u ) ∉ { β , γ , C ( x 1 , Y ω β ′ ∪ Y ω γ ′ ) } , then for any u ′ ∈ U ⧹ { u } , ( u ′ , x 2 , y β γ ″ , x 1 , u ) is a rainbow ( u ′ , u ) -path. This implies that Y ω β ′ ∪ Y ω γ ′ ∪ { u } is an RP-kernel of D . If C ( x 1 , U ) ⊆ { β , γ , C ( x 1 , Y ω β ′ ∪ Y ω γ ′ ) } , there exists no rainbow path for any pair of vertices of Y ω β ′ ∪ Y ω γ ′ ∪ U . This implies that Y ω β ′ ∪ Y ω γ ′ ∪ U is an RP-kernel of D .

Case 2: Y ′ ⧹ Y ∗ = Y ″ ⧹ Y ∗ ∗ = ∅ .

Now Y ′ = Y ∗ , Y ″ = Y ∗ ∗ . This means C ( x 1 , y ′ ) ≠ C ( y ′ , x 2 ) and C ( x 2 , y ″ ) ≠ C ( y ″ , x 1 ) for any y ′ ∈ Y ′ and y ″ ∈ Y ″ . Let y ′ ∈ Y ′ be arbitrary. Assume C ( x 1 , y ′ ) = α and C ( y ′ , x 2 ) = β .

Define the following vertex subsets, which are shown in Figure 3 in which a box represents a set of vertices, while dotted, dashed, solid arcs represent, respectively, the arcs coloured by α , β and ω , where ω stands for an arbitrary colour not in { α , β } .

Y c 1 c 2 ′ = { y ′ ∈ Y ′ ∣ C ( x 1 , y ′ ) = c 1 , C ( y ′ , x 2 ) = c 2 } ,

where ( c 1 , c 2 ) ∈ { ( α , β ) , ( β , α ) , ( α , ω ) , ( β , ω ) , ( ω , α ) , ( ω , β ) } ;

Y ω 1 ω 2 ′ = { y ′ ∈ Y ′ ∣ C ( x 1 , y ′ ) , C ( y ′ , x 2 ) ∉ { α , β } and C ( x 1 , y ′ ) ≠ C ( y ′ , x 2 ) } .

Since every 4-cycle is coloured with at least three colours and Y α β ′ ≠ ∅ , Y ″ can be divided into the following vertex subsets.

Y c 1 c 2 ″ = { y ″ ∈ Y ″ ∣ C ( x 2 , y ″ ) = c 1 , C ( y ″ , x 1 ) = c 2 } ,

where ( c 1 , c 2 ) ∈ { ( α , ω ) , ( β , ω ) , ( ω , α ) , ( ω , β ) } and ω is an arbitrary colour not in { α , β } ;

Y ω 1 ω 2 ″ = { y ″ ∈ Y ″ ∣ C ( x 2 , y ″ ) , C ( y ″ , x 1 ) ∉ { α , β } and C ( x 2 , y ″ ) ≠ C ( y ″ , x 1 ) } .

Now,

Y ′ = Y α β ′ ∪ Y β α ′ ∪ Y α ω ′ ∪ Y β ω ′ ∪ Y ω α ′ ∪ Y ω β ′ ∪ Y ω 1 ω 2 ′ , Y ″ = Y α ω ″ ∪ Y β ω ″ ∪ Y ω α ″ ∪ Y ω β ″ ∪ Y ω 1 ω 2 ″ .

For convenience a vertex in Y c 1 c 2 ′ ( Y c 1 c 2 ″ , Y ω 1 ω 2 ′ , Y c 1 c 2 ″ ) we denote by y c 1 c 2 ′ ( y c 1 c 2 ″ , y ω 1 ω 2 ′ , y c 1 c 2 ″ ).

If Y ω 1 ω 2 ′ ≠ ∅ or Y ω 1 ω 2 ′ ∪ Y ω α ″ ∪ Y ω β ″ = ∅ , then for any y ″ ∈ Y ω α ″ ∪ Y ω β ″ , ( y ″ , x 1 , y ω 1 ω 2 ′ , x 2 ) is a rainbow ( y ″ , x 2 ) -path; for any y ″ ∈ Y α ω ″ ∪ Y β ω ″ ∪ Y ω 1 ω 2 ″ , ( y ″ , x 1 , y α β ′ , x 2 ) is a rainbow ( y ″ , x 2 ) -path. This implies that { x 2 } is an RP-kernel of D . So we assume Y ω 1 ω 2 ′ = ∅ and Y ω α ″ ∪ Y ω β ″ ≠ ∅ .

Subcase 2.1: Y ω α ″ ≠ ∅ and Y ω β ″ ≠ ∅ .

Then Y ″ = Y α ω ″ ∪ Y β ω ″ ∪ Y ω β ″ ∪ Y ω α ″ ∪ Y ω 1 ω 2 ″ . For any y ′ ∈ Y α β ′ ∪ Y ω β ′ , ( y ′ , x 2 , y ω α ″ , x 1 ) is a rainbow ( y ′ , x 1 ) -path. For any y ′ ∈ Y β α ′ ∪ Y ω α ′ , ( y ′ , x 2 , y ω β ″ , x 1 ) is a rainbow ( y ′ , x 1 ) -path. Since every 4-cycle is coloured with at least three colours, we have C ( y α ω ′ , x 2 ) ∉ C ( x 2 , Y ω α ″ ) for any y α ω ′ ∈ Y α ω ′ and C ( y β ω ′ , x 2 ) ∉ C ( x 2 , Y ω β ″ ) for any y β ω ′ ∈ Y β ω ′ . It follows that ( y α ω ′ , x 2 , y ω α ″ , x 1 ) is a rainbow ( y α ω ′ , x 1 ) -path and ( y β ω ′ , x 2 , y ω β ″ , x 1 ) is a rainbow ( y β ω ′ , x 1 ) -path. This implies that { x 1 } is an RP-kernel of D .

Subcase 2.2: Exactly one of the subsets Y ω α ″ and Y ω β ″ is not empty set.

W.l.o.g., assume Y ω α ″ ≠ ∅ and Y ω β ″ = ∅ . Now, Y ″ = Y α ω ″ ∪ Y β ω ″ ∪ Y ω α ″ ∪ Y ω 1 ω 2 ″ .

Subcase 2.2.1: Y β ω ′ ∪ Y ω β ′ ≠ ∅ .

We see that for any y ″ ∈ Y α ω ″ ∪ Y β ω ″ ∪ Y ω 1 ω 2 ″ , ( y ″ , x 1 , y α β ′ , x 2 ) is a rainbow ( y ″ , x 2 ) -path; for any y ω α ″ ∈ Y ω α ″ , ( y ω α ″ , x 1 , y ′ , x 2 ) is a rainbow ( y ω α ″ , x 2 ) -path, where y ′ ∈ Y β ω ′ ∪ Y ω β ′ . This implies that { x 2 } is an RP-kernel of D .

Subcase 2.2.2: Y β ω ″ ≠ ∅ .

We see that for any y ′ ∈ Y β α ′ ∪ Y ω α ′ ∪ Y β ω ′ , ( y ′ , x 2 , y β ω ″ , x 1 ) is a rainbow ( y ′ , x 1 ) -path; for any y ′ ∈ Y α β ′ ∪ Y α ω ′ ∪ Y ω β ′ , ( y ′ , x 2 , y ω α ″ , x 1 ) is a rainbow ( y ′ , x 1 ) -path. This implies that { x 1 } is an RP-kernel of D .

Subcase 2.2.3: Y β ω ′ ∪ Y ω β ′ = ∅ and Y β ω ″ = ∅ .

We see that Y ′ = Y α β ′ ∪ Y β α ′ ∪ Y α ω ′ ∪ Y ω α ′ and Y ″ = Y α ω ″ ∪ Y ω α ″ ∪ Y ω 1 ω 2 ″ . If Y ω 1 ω 2 ″ ≠ ∅ , then for any y ′ ∈ Y ′ ⧹ Y α ω ′ , ( y ′ , x 2 , y ω 1 ω 2 ″ , x 1 ) is a rainbow ( y ′ , x 1 ) -path; for any y α ω ′ ∈ Y α ω ′ , ( y α ω ′ , x 2 , y α ω ″ , x 1 ) is a rainbow ( y α ω ′ , x 1 ) -path. This implies that { x 1 } is an RP-kernel of D . If Y ω 1 ω 2 ″ = ∅ , then Y ″ = Y α ω ″ ∪ Y ω α ″ . For any y ′ ∈ Y α β ′ ∪ Y β α ′ ∪ Y ω α ′ , ( y ′ , x 2 , y ω α ″ ) is a rainbow ( y ′ , y ω α ″ ) -path. By the proof above, for any y α ω ′ ∈ Y α ω ′ , ( y α ω ′ , x 2 , y ω α ″ ) is a rainbow ( y α ω ′ , y ω α ″ ) -path. If ∣ C ( x 2 , Y ω α ″ ) ∣ ⩾ 2 , let y ω α 1 ″ , y ω α 2 ″ ∈ Y ω α ″ with C ( x 2 , y ω α 1 ″ ) ≠ C ( x 2 , y ω α 2 ″ ) . For any y α ω ″ ∈ Y α ω ″ , either ( y α ω ″ , x 1 , y α β ′ , x 2 , y ω α 1 ″ ) or ( y α ω ″ , x 1 , y α β ′ , x 2 , y ω α 2 ″ ) is a rainbow ( y α ω ″ , y ω α ″ ) -path. This implies that Y ω α ″ is an RP-kernel of D . If ∣ C ( x 2 , Y ω α ″ ) ∣ = 1 , let U = { y α ω ″ ∈ Y α ω ″ ∣ C ( y α ω ″ , x 1 ) = C ( x 2 , Y ω α ″ ) } . Note that there exists no rainbow path for any pair of vertices of Y ω α ″ ∪ U . For any y ″ ∈ Y α ω ″ ⧹ U , we have C ( y ″ , x 1 ) ≠ C ( x 2 , Y ω α ″ ) and ( y ″ , x 1 , y α β ′ , x 2 , y ω α ″ ) is a rainbow ( y ″ , y ω α ″ ) -path. This implies that Y ω α ″ ∪ U is an RP-kernel of D .

In any case, we can find an RP-kernel of D . This proof of Lemma 6.5 is complete.□

$Figure 2 An arc-coloured bipartite tournament for Case 1 of the proof of Lemma 6.5.$

Figure 2

An arc-coloured bipartite tournament for Case 1 of the proof of Lemma 6.5.

$Figure 3 An arc-coloured bipartite tournament for Case 2 of the proof of Lemma 6.5.$

Figure 3

An arc-coloured bipartite tournament for Case 2 of the proof of Lemma 6.5.

Theorem 6.6

Let D = ( X , Y ) be an m -arc-coloured bipartite tournament with min { ∣ X ∣ , ∣ Y ∣ } = 2 . If every 4-cycle contained in D is coloured with at least three colours, then D has an RP-kernel.

By Theorem 6.6, the following corollary is immediate.

Corollary 6.7

Let D = ( X , Y ) be an m -arc-coloured bipartite tournament with min { ∣ X ∣ , ∣ Y ∣ } = 2 . If every 4-cycle contained in D is rainbow, then D has an RP-kernel.

Remark 6.8

The condition “every 4-cycle is coloured with at least 3 colours” in Theorem 6.6 cannot be reduced. An arc-coloured bipartite tournament with ∣ X ∣ = 2 shown in Figure 4 satisfying “every 4-cycle is 2-arc-coloured” has no RP-kernel, in which solid and dotted arcs represent, respectively, the arcs coloured by two distinct colours. Large m -arc-coloured bipartite tournaments with no RP-kernel can be obtained by adding new vertices to Y and new colours such that these new vertices completely dominate X .

$Figure 4 An arc-coloured bipartite tournament with ∣ X ∣ = 2 | X| =2 satisfying “every 4-cycle is 2-arc-coloured” has no RP-kernel.$

Figure 4

An arc-coloured bipartite tournament with ∣ X ∣ = 2 satisfying “every 4-cycle is 2-arc-coloured” has no RP-kernel.

In the following proof, we consider min { ∣ X ∣ , ∣ Y ∣ } ⩾ 3 .

Lemma 6.9

[7] Let D = ( X , Y ) be a bipartite tournament. Then the following statements hold:

let C = ( u 0 , u 1 , u 2 , … , u n ) be a walk in D . For { i , j } ⊆ { 1 , 2 , … , n } , u i , u j are adjacent if and only if j − i ≡ 1 ( mod 2 ) .
every closed walk of length at most 6 is a cycle of D .

Let CB 5 be a bipartite tournament, which has V ( CB 5 ) = { u 1 , u 2 , u 3 , u 4 , u 5 } and A ( CB 5 ) = { ( u 1 , u 2 ) , ( u 2 , u 3 ) , ( u 3 , u 4 ) , ( u 4 , u 5 ) , ( u 4 , u 1 ) , ( u 5 , u 2 ) } . Let TB 4 be a bipartite tournament, which has V ( TB 4 ) = { u 1 , u 2 , u 3 , u 4 } and A ( TB 4 ) = { ( u 1 , u 2 ) , ( u 2 , u 3 ) , ( u 3 , u 4 ) , ( u 1 , u 4 ) } (Figure 5).

$Figure 5 Two bipartite tournaments CB 5 {{CB}}_{5} and T B 4 T{B}_{4} .$

Figure 5

Two bipartite tournaments CB 5 and T B 4 .

Lemma 6.10

Let D = ( X , Y ) be an m -arc-coloured bipartite tournament with min { ∣ X ∣ , ∣ Y ∣ } ⩾ 3 . If all 4-cycles, 6-cycles and induced subdigraphs CB 5 in D are rainbow, and all induced subdigraphs TB 4 in D are properly coloured, then for any pair of distinct vertices u , v ∈ V ( D ) satisfying there exists a rainbow ( u , v ) -path and no rainbow ( v , u ) -path in D , at least one of the following conditions holds:

u → v ;
there exists a ( u , v ) -path of length 2.

Proof

Let P = ( u = u 0 , u 1 , u 2 , … , u n = v ) be the shortest rainbow ( u , v ) -path in D . The result holds clearly for n ⩽ 2 . Now assume n ⩾ 3 .

If n is odd, by Lemma 6.9 ( a ), we have u 0 , u n are adjacent. Since there is no rainbow ( v , u ) -path in D , we have u 0 → u n . The result holds. So we assume that n is even.

Also by Lemma 6.9 ( a ), we have u 1 , u n are adjacent. If u 1 → u n , then ( u = u 0 , u 1 , u n = v ) is a ( u , v ) -path of length 2 and the result holds. So we assume u n → u 1 .

If n = 4 , then P = ( u = u 0 , u 1 , u 2 , u 3 , u 4 = v ) . For u 3 → u 0 , we see that D [ u 0 , u 1 , u 2 , u 3 , u 4 ] is an induced rainbow CB 5 , which implies ( u 4 , u 1 , u 2 , u 3 , u 0 ) is a rainbow ( v , u ) -path, which is a contradiction. For u 0 → u 3 , we have ( u 0 , u 3 , u 4 ) is a ( u , v ) -path of length 2. So we assume n ⩾ 6 .

If u 0 → u i 0 → u n for some i 0 ∈ { 3 , 5 , … , n − 1 } , then ( u = u 0 , u i 0 , u n = v ) is a ( u , v ) -path of length 2 and the result holds. So we assume that either u i → u 0 or u n → u i for each i ∈ { 3 , 5 , … , n − 1 } .

Claim 1

For each i ∈ { 3 , 5 , … , n − 3 } , u 0 → u i and u n → u i .

Proof

It is sufficient to show that u 0 → u i . We process by induction on i . For i = 3 , suppose to the contrary that u 3 → u 0 . If u 3 → u n , then D [ u 0 , u 1 , u 2 , u 3 , u n ] is an induced rainbow CB 5 . It follows that ( u n , u 1 , u 2 , u 3 , u 0 ) is a rainbow ( v , u ) -path in D , which is a contradiction. If u n → u 3 , then D [ u n , u 3 , u 0 , u 1 ] is TB 4 which is properly coloured. It follows that ( u n , u 3 , u 0 ) is a rainbow ( v , u ) -path in D , which is a contradiction. Thus, ( u 0 , u 3 ) ∈ A ( D ) .

Assume that the claim holds for i < n − 3 . We consider the case i = n − 3 .

Suppose to the contrary that u n − 3 → u 0 . By the induction hypothesis, we have u 0 → u n − 5 and u n → u n − 5 . If u n − 3 → u n , then D [ u 0 , u n − 5 , u n − 4 , u n − 3 , u n ] is an induced rainbow CB 5 . It follows that ( u n , u n − 5 , u n − 4 , u n − 3 , u 0 ) is a rainbow ( v , u ) -path in D , which is a contradiction. If u n → u n − 3 , then D [ u n , u n − 3 , u 0 , u n − 5 ] is TB 4 which is properly coloured. It follows that ( u n , u n − 3 , u 0 ) is a rainbow ( v , u ) -path in D , which is a contradiction. So ( u 0 , u n − 3 ) ∈ A ( D ) .□

Now we show u 0 → u n − 1 . Suppose to the contrary that u n − 1 → u 0 . By Claim 1, we have u 0 → u n − 3 and u n → u n − 3 . Then D [ u 0 , u n − 3 , u n − 2 , u n − 1 , u n ] is an induced rainbow CB 5 . It follows that ( u n , u n − 3 , u n − 2 , u n − 1 , u 0 ) is a rainbow ( v , u ) -path in D , which is a contradiction. So u 0 → u n − 1 .

Now ( u 0 , u n − 1 , u n ) is a ( u , v ) -path of length 2.□

Theorem 6.11

Proof

According to Theorem 2.3, it is sufficient to prove that each cycle of C r ( D ) has a symmetrical arc. Suppose to the contrary that there exists a cycle C in C r ( D ) containing no symmetrical arc. We will get a contradiction by showing that C has a symmetrical arc. Let C = ( x 0 , x 1 , … , x n , x 0 ) . Since C has no symmetrical arc, for each i ∈ { 0 , 1 , … , n } , there exists a rainbow ( x i , x i + 1 ) -path and no rainbow ( x i + 1 , x i ) -path in D . The following claim follows directly from Lemma 6.10.

Claim 1

For each i ∈ { 0 , 1 , … , n } , either ( x i , x i + 1 ) ∈ A ( D ) or there exists a ( x i , x i + 1 ) -path of length 2 in D .

Let

P i = ( x i , x i + 1 ) , ( x i , x i + 1 ) ∈ A ( D ) ; ( x i , u i , x i + 1 ) , ( x i , x i + 1 ) ∉ A ( D ) ,

and C ′ = P 0 P 1 … P n . Then C ′ is a closed walk in D .

We consider the following two cases.

Case 1. n = 2 .

Now C is a 3-cycle. Then not all arcs of C are in D since D is a bipartite tournament. W.l.o.g., assume that ( x 0 , x 1 ) ∉ A ( D ) . Then ℓ ( P 0 ) = 2 , ℓ ( P 1 ) ⩽ 2 and ℓ ( P 2 ) ⩽ 2 . Now C ′ is a closed walk with length at most 6. By Lemma 6.9 ( b ), C ′ is a cycle. Since all 4-cycles and 6-cycles are rainbow, we have C ′ is rainbow. Now P 1 P 2 is a rainbow ( x 1 , x 0 ) -path and hence ( x 1 , x 0 ) ∈ A ( C r ( D ) ) . Note that ( x 0 , x 1 ) ∈ A ( C ) , which contradicts C has no symmetrical arc.

Case 2. n ⩾ 3 .

In this case, we set C ′ = ( v 0 , v 1 , … , v k , v 0 ) where v 0 = x 0 and k ⩾ n . By Lemma 6.9 ( a ), k is odd since ( v k , v 0 ) ∈ A ( D ) . Also v 0 , v 3 are adjacent and v 0 , v k − 2 are adjacent in D .

If ( v 3 , v 0 ) ∈ A ( D ) , then ( v 0 , v 1 , v 2 , v 3 , v 0 ) is a rainbow 4-cycle. This implies that ( v 1 , v 2 , v 3 , v 0 ) is a rainbow ( v 1 , v 0 ) -path and ( v 2 , v 3 , v 0 ) is a rainbow ( v 2 , v 0 ) -path. Then { ( v 1 , v 0 ) , ( v 2 , v 0 ) } ⊆ A ( C r ( D ) ) . Note that either v 1 = x 1 or v 2 = x 1 . We have ( x 1 , x 0 ) ∈ A ( C r ( D ) ) . Note that ( x 0 , x 1 ) ∈ A ( C ) , which contradicts C has no symmetrical arc.

If ( v 0 , v k − 2 ) ∈ A ( D ) , then ( v 0 , v k − 2 , v k − 1 , v k , v 0 ) is a rainbow 4-cycle. This implies that ( v 0 , v k − 2 , v k − 1 , v k ) is a rainbow ( v 0 , v k ) -path and ( v 0 , v k − 2 , v k − 1 ) is a rainbow ( v 0 , v k − 1 ) -path. Then { ( v 0 , v k ) , ( v 0 , v k − 1 ) } ⊆ A ( C r ( D ) ) . Note that either v k = x n or v k − 1 = x n . We have ( x 0 , x n ) ∈ A ( C r ( D ) ) . Note that ( x n , x 0 ) ∈ A ( C ) , which contradicts C has no symmetrical arc.

If ( v 0 , v 3 ) ∈ A ( D ) and ( v k − 2 , v 0 ) ∈ A ( D ) , we have v 3 ≠ v k − 2 and hence k − 2 ⩾ 5 . Also there exists i ∈ 1 , 2 , … , k − 5 2 such that ( v 0 , v 2 i + 1 ) ∈ A ( D ) and ( v 2 i + 3 , v 0 ) ∈ A ( D ) . Let

j 0 = max i ∈ 1 , 2 , … , k − 5 2 ∣ ( v 0 , v 2 i + 1 ) ∈ A ( D ) , ( v 2 i + 3 , v 0 ) ∈ A ( D ) .

Then ( v 0 , v 2 j 0 + 1 , v 2 j 0 + 2 , v 2 j 0 + 3 , v 0 ) is a rainbow 4-cycle.

If v 2 j 0 + 1 ∈ V ( C ) , let v 2 j 0 + 1 = x j . Now ( v 2 j 0 + 2 , v 2 j 0 + 3 , v 0 , v 2 j 0 + 1 ) is a rainbow ( v 2 j 0 + 2 , v 2 j 0 + 1 ) -path and ( v 2 j 0 + 3 , v 0 , v 2 j 0 + 1 ) is a rainbow ( v 2 j 0 + 3 , v 2 j 0 + 1 ) -path. Then { ( v 2 j 0 + 2 , v 2 j 0 + 1 ) , ( v 2 j 0 + 3 , v 2 j 0 + 1 ) } ⊆ A ( C r ( D ) ) . Note that either v 2 j 0 + 2 = x j + 1 or v 2 j 0 + 3 = x j + 1 . We have ( x j + 1 , x j ) ∈ A ( C r ( D ) ) . Note that ( x j , x j + 1 ) ∈ A ( C ) , which contradicts C has no symmetrical arc.

If v 2 j 0 + 1 ∉ V ( C ) , by the definition of C ′ , we have v 2 j 0 , v 2 j 0 + 2 ∈ V ( C ) . Let v 2 j 0 + 2 = x j + 1 . By the choice of j 0 , we have ( v 2 j 0 + 5 , v 0 ) ∈ A ( D ) . This implies that ( v 0 , v 2 j 0 + 1 , v 2 j 0 + 2 , v 2 j 0 + 3 , v 2 j 0 + 4 , v 2 j 0 + 5 , v 0 ) is a rainbow 6-cycle. So ( v 2 j 0 + 3 , v 2 j 0 + 4 , v 2 j 0 + 5 , v 0 , v 2 j 0 + 1 , v 2 j 0 + 2 ) is a rainbow ( v 2 j 0 + 3 , v 2 j 0 + 2 ) -path and ( v 2 j 0 + 4 , v 2 j 0 + 5 , v 0 , v 2 j 0 + 1 , v 2 j 0 + 2 ) is a rainbow ( v 2 j 0 + 4 , v 2 j 0 + 2 ) -path. Then { ( v 2 j 0 + 3 , v 2 j 0 + 2 ) , ( v 2 j 0 + 4 , v 2 j 0 + 2 ) } ⊆ A ( C r ( D ) ) . Note that either v 2 j 0 + 3 = x j + 2 or v 2 j 0 + 4 = x j + 2 , we have ( x j + 2 , x j + 1 ) ∈ A ( C r ( D ) ) . Note that ( x j + 1 , x j + 2 ) ∈ A ( C ) , which contradicts C has no symmetrical arc.

In any case, we get a contradiction. Thus, C r ( D ) is a KP-digraph.□

By Observation 2.2 and Theorem 6.11, the following corollary is direct.

Corollary 6.12

Funding information: This research was partially supported by the Youth Foundation of Shanxi Province (201901D211197).
Conflict of interest: Authors state no conflict of interest.

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Received: 2020-04-04

Revised: 2020-11-02

Accepted: 2021-01-21

Published Online: 2021-05-10

This work is licensed under the Creative Commons Attribution 4.0 International License.

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https://doi.org/10.1515/math-2021-0019

Keywords for this article

arc-coloured digraphs; kernels; kernels by rainbow paths

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