1. Introduction
In 1978, Caccetta and Häggkvist [4] made the following conjecture.
Conjecture 1 (CaccettaHäggkvist).
For all positive integers , every simple vertex digraph with minimum outdegree at least contains a directed cycle of length at most .
A digraph is simple if for all there is at most one arc from to . The CaccettaHäggkvist conjecture has proven to be a notoriously difficult problem. See Sullivan [15] for a summary of partial results. Although there has been a lot of progress on approximate versions, it is known to hold exactly for only a few values of . For example, the case has received considerable attention [4, 5, 3, 14, 6, 12, 10, 8], but still remains open.
On the other hand, Conjecture 1 is known to hold for small values of . The case was actually proved by Caccetta and Häggkvist [4].
Theorem 2 ([4]).
Every simple vertex digraph with minimum outdegree at least contains a directed cycle of length at most .
Given a graph and a colouring of , we say that a subgraph of is rainbow if no two edges of are of the same colour. Aharoni (see [1]) recently proposed the following strengthening of the CaccettaHäggkvist conjecture.
Conjecture 3 (Aharoni).
Let be a simple vertex graph and be a colouring of with colours, where each colour class has size at least . Then contains a rainbow cycle of length at most .
In fact, we now show that the following weakening of Aharoni’s conjecture implies the CaccettaHäggkvist conjecture.
Conjecture 4.
Let be a simple vertex graph and be a colouring of with colours, where each colour class has size at least . Then contains a cycle of length at most such that no two incident edges of are of the same colour.
Proof of Conjecture 1, assuming Conjecture 4.
Let be a simple digraph of order and minimum outdegree at least . Let be the graph obtained from by forgetting the orientations of all arcs. Let and colour with colour if . Clearly, this colouring uses colours. Moreover, since has minimum outdegree at least , each colour class has size at least . Therefore, by Conjecture 4, contains a properly edgecoloured cycle of length at most . Let be the subdigraph of corresponding to . We claim that is a directed cycle. If not, then there exists such that and . Thus, the two edges of incident to vertex are the same colour. This contradicts that is properly edgecoloured. ∎
Our main theorem is that Aharoni’s conjecture holds for .
Theorem 5.
Let be a simple vertex graph and be a colouring of with colours, where each colour class has size at least . Then contains a rainbow cycle of length at most .
The rest of the paper is organized as follows. In Section 2, we prove our main theorem. We show that our bound is tight in Section 3, and that there is a sharp increase in the ‘rainbow girth’ as the number of colours decreases from . In Section 4, we show that the natural matroid generalization of Theorem 5 holds for cographic matroids, but fails for binary matroids. We conclude with some open problems in Section 5.
2. Proof of the Main Theorem
Given an edgecoloured graph , a transversal of is a spanning subgraph of which contains exactly one edge from each colour class. In particular, a transversal is a rainbow subgraph (which may contain isolated vertices). A theta is a graph consisting of internallydisjoint paths , between two vertices and .
Proof of Theorem 5.
Suppose the theorem is false and let be a counterexample with minimum. By minimality, each colour class contains exactly two edges. We claim that contains a vertex such that all edges incident to have different colours. If not, then at each vertex, there is at least one colour that appears twice. Since there are only colours, there is exactly one colour that appears twice at each vertex. For each vertex , let and be the two edges incident to that have the same colour. For each , we orient and away from and apply Theorem 2 to find a directed cycle of length at most . This corresponds to a rainbow cycle in , which contradicts the fact that is a counterexample.
Let be a transversal of . Since has edges and vertices, it follows that contains a rainbow cycle. Moreover, since cannot contain two rainbow cycles that meet in at most one vertex, contains exactly one nontrivial block . Suppose has an ear decomposition with at least two ears. In this case, , since contains at most edges. Since contains at least two ears, contains two cycles meeting in at most two vertices, or contains a subdivision of . In either case, contains a cycle of length at most . Thus, contains at most one ear. That is, is a cycle or a theta. It follows that is either a connected graph with exactly one cycle, or has exactly two components, one of which is a tree and the other of which is a theta with trees attached.
Since contains a vertex such that all edges incident to have different colours, there is a transversal of such that is an isolated vertex in . It follows that the other component of contains a theta. In particular, contains a rainbow theta. The rest of the proof only uses the fact that contains a rainbow theta.
Let be a rainbow theta in with minimum. Note that , since every rainbow subgraph of contains at most edges. Let be the union of three paths , where for all . We claim that at most two edges of are chords of . Let be a chord of . If has both endpoints on , then by the minimality of , contains two rainbow cycles and with . One of or has length at most , since . Therefore, we may assume that the ends of are on and . Suppose is coloured red. If does not contain a red edge, then there exist rainbow cycles such that each edge of is contained in exactly two of . Since , at least one has length at most . It follows that some edge of is red. By the minimality of , contains a triangle with two red edges. For , let be the second vertex of . If contains more than two chords, then by symmetry we may assume that and are red, and that and are blue. But then contradicts the minimality of .
Since contains a transversal, there is a rainbow cycle that is edgedisjoint from . Let , , and . Since is a counterexammple, . Let be the number of chords of . Observe that . Therefore, , and
since . Observe that if a theta has vertices, where , then it contains a cycle of length at most . Therefore, if is congruent to , or , it is easy to check that contains a cycle with at most vertices. The remaining case is if , , , and . In particular, we are done unless and each of , and contains exactly vertices. Let be a chord of and be the edge of of the same colour as . By replacing by , we may assume that the number of vertices of , and are and . But now has vertices, as required. ∎
3. Tightness of the Bound
We now show that our bound is tight, and that there is a dramatic change of behaviour as we decrease the number of colours from . To be precise, define the rainbow girth of an edgecoloured graph , denoted , to be the length of a shortest rainbow cycle in . If does not contain a rainbow cycle, then . Let
Theorem 6.
For all and ,
Proof.
By Theorem 5, . For the corresponding lowerbound, let be a graph with vertex set and edges and for all . Colour both and with colour for all . See Figure 1. It is easy to check that the shortest rainbow cycle in this graph has length .
We now show . For the upperbound, let be a graph with , and let be a colouring of such that each colour class has size . Since there are only colours, there is a vertex of such that all edges incident to are coloured differently. Therefore, there is a transversal of such that is an isolated vertex in . Since contains vertices and edges, contains a cycle of length at most . For the corresponding lowerbound, let be the wheel graph on vertices. Let be a colouring of such that each colour class is a path with two edges, one of which is incident to the hub vertex. See Figure 2. Observe that no rainbow cycle of can use the hub vertex. Therefore, the shortest rainbow cycle in has length .
By deleting two edges of the same colour from we obtain a graph on vertices and edges that does not contain a rainbow cycle. Therefore, , for all . ∎
We have determined exactly for all . What happens for ? The best general upper bound we can prove follows from a theorem of Bollobás and Szemerédi [2]. To state their result, we need some definitions. The girth of a graph , denoted , is the length of a shortest cycle in . Define
Bollobás and Szemerédi prove the following.
Theorem 7 ([2]).
For all and ,
As a corollary, we obtain the following.
Theorem 8.
For all and ,
Proof.
Let be a simple vertex graph, with and be a colouring of where each colour class has size . Let be a transversal of . Note that has vertices and edges. By Theorem 7, contains a cycle of length at most . Since this cycle is necessarily rainbow, we are done. ∎
4. Matroid Generalizations
In this section, we assume basic familiarity with matroids, and we follow the notation of Oxley [11]. An attractive feature of Aharoni’s conjecture as opposed to the CaccettaHäggkvist conjecture, is that there is a natural matroid generalization.
Conjecture 9.
Let be a simple rank matroid and be a colouring of with colours, where each colour class has size at least . Then contains a rainbow circuit of size at most .
Unfortunately, it is easy to see that Conjecture 9 is false, since the uniform matroid does not contain any circuits of size less than . On the other hand, Theorem 5 asserts that Conjecture 9 holds for graphic matroids. We now prove that Conjecture 9 also holds for cographic matroids.
Theorem 10.
Let be a simple rank cographic matroid and be a colouring of with colours, where each colour class has size at least . Then contains a rainbow circuit of size at most .
Proof.
For a graph , we let and denote the graphic and cographic matroids of , respectively. Let be a connected graph such that . By deleting elements from , we may assume that each colour class has size exactly . Therefore, . Since has rank , has rank . Therefore, . Since is simple, has minimum degree at least . For a vertex , let be the set of edges of incident to . Since only uses colours, there are at least two distinct vertices and of such that and are both rainbow. If or , then contains a rainbow circuit of size at most . Thus, we may assume that . It follows that . Therefore, some vertex has degree at most , which contradicts that is simple. ∎
We finish this section by giving an infinite family of binary matroids for which Conjecture 9 fails.
Theorem 11.
For each even integer , there exists a simple rank binary matroid on elements, and a colouring of where each colour class has size , such that does not contain a rainbow circuit of size at most .
Proof.
Let be even. Instead of specifying via vectors in , we specify via subsets of , where binary vector addition is replaced by symmetric difference of sets. We represent by the following sets

,

, and

.
We now specify the colouring, which is just a pairing of these sets. For each , we pair and (subscripts are read modulo ). The last two pairs are and . To illustrate, the case is given by the following matrix.
Let be a subset of these sets containing at most one set from each pair and such that . Here, denotes the symmetric difference of all the sets in . We will prove that .
We first consider the case or . Since and are paired, exactly one of them, which we call , is in . Since and are the only other sets in that contain , exactly one of them, which we call , is in . Let . In all four cases, note that is either or . Since is even, and all sets in have size at most , . Therefore, .
The remaining case is contains neither nor . The only other sets that contain are and . Since and are paired, both cannot appear in . Therefore, neither nor appears in . If all of the sets are in , then . Thus, we may assume that for some , but (subscripts are read modulo ). Since , we must also have . But, and are paired, so they cannot both be in . This contradiction completes the proof. ∎
5. Open Problems
Note that in proving Theorem 8, we only use one fixed transversal. By considering multiple transversals, we suspect that the bound in Theorem 8 can be improved.
Problem 12.
Determine for .
Since the CaccettaHäggkvist conjecture is known to hold for , another possible direction is to prove Aharoni’s conjecture for .
Recall that our proof of Aharoni’s conjecture for uses Theorem 2 as a blackbox. It would be interesting to find a proof of Theorem 5 that avoids using Theorem 2.
Finally, by Theorems 5 and 10, Conjecture 9 holds for both graphic and cographic matroids. Therefore, we suspect there is a proof of Conjecture 9 for regular matroids via Seymour’s regular matroid decomposition theorem [13].
Conjecture 14.
Let be a simple rank regular matroid and be a colouring of with colours, where each colour class has size at least . Then contains a rainbow circuit of size at most .
Acknowledgements.
References
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