Rainbow Pancyclicity in Graph Systems

Let $G_1,...,G_n$ be graphs on the same vertex set of size $n$, each graph having minimum degree $\delta(G_i)\geq \frac{n}{2}$. A recent conjecture of Aharoni asserts that there exists a rainbow Hamiltonian cycle i.e. a cycle with edge set $\{e_1,...,e_n\}$ such that $e_i\in E(G_i)$ for $1\leq i \leq n$. This can be seen as a rainbow variant of the well-known Dirac theorem. In this paper, we prove this conjecture asymptotically, namely, we show that for every $\varepsilon>0$, there exists an integer $N>0$, such that when $n>N$ for any graphs $G_1,...,G_n$ on the same vertex set of size $n$, each graph having $\delta(G_i)\geq (\frac{1}{2}+\varepsilon)n$, there exists a rainbow Hamiltonian cycle. Our main tool is the absorption technique. We also prove that with $\delta(G_i)\geq \frac{n+1}{2}$, one can find rainbow cycles of length $i$ where $3\leq i \leq n-1$.


Introduction
Let t be a positive integer and G i (i = 1, ..., t) be the t graphs on the same vertex set V of size n. We denote E(G i ) for the edge set of G i where 1 ≤ i ≤ t. Let S be an edge set which is a subset of ∪ t i=1 E(G i ), S is rainbow if each of whose edges is chosen from different E(G i ) i.e. we can label the edges in S with {e 1 , ..., e k } for some k such that there exist distinct i 1 , ..., i k with e j ∈ E(G ij ) for 1 ≤ j ≤ k. One of the motivations to study such rainbow structures in a graph system is the following famous conjecture due to Aharoni and Berger [1]: Conjecture 1. Let M i (i = 1, ..., n) be n matchings of size at least n+1 on the same vertex set V = X ∪Y where X and Y are disjoint and all edges of M i are between X and Y , then there exists a rainbow matching of size n.
The Aharoni-Berge conjecture generalizes the Brualdi-Stein Conjecture which asserts that every n × n Latin square has a partial transversal of size n − 1. Recall that a Latin square of order n is an n × n array filled with n different symbols, where no symbol appears in the same row or column more than once. A partial transversal of size m in a Latin square is a set of m entries such that no two entries are in the same row, same column, or have the same symbol. Now consider a Latin square S whose set of symbols is {1, ..., n} with the (i, j) symbol S i,j . We construct an edge-colouring of K n,n with colours {1, ..., n}. Set V (K n,n ) = {x 1 , ..., x n ; y 1 , ..., y n } and we colour the edge x i y j with colour S i,j . Notice that this colouring is proper i.e. adjacent edges are coloured with different colours. We call a matching M rainbow if any two different edges of it are coloured with distinct colours. Now it is immediate to see that S contains a Latin Square of order m if and only if K n,n contains a rainbow matching of size m. Thus Brualdi-Stein Conjecture can be restated as every proper edge-coloured K n,n using n colours contains a rainbow matching of size n − 1. This is a special case of Conjecture 1 for the reason that when are n edge-disjoint matchings defined in K n,n , there is a rainbow matching of size n − 1. The Aharoni-Berge conjecture has been confirmed asymptotically by Pokrovskiy [11]. For more details about this topic, see [14].
If we insist that the size of each matching is n, then how many such matchings can guarantee a rainbow matching of size n? Drisko [7] proved that 2n − 1 such matchings suffice and this bound is sharp.
Barát, Gyárfás and Sárközy [5] proposed the conjecture that for n even any 2n matchings of size n on the same vertex set have a rainbow matching of size n, and for n odd any 2n − 1 matchings of size n on the same vertex set have a rainbow matching of size n, which can be seen as a corresponding version of this problem in the case of general graphs. Recently, Aharoni et al. [2] showed that 3n − 2 matchings of size n on the same vertex set have a rainbow matching of size n, which is the best known bound up to now.
In [3], Aharoni et al. initiated the study of "rainbow extremal graph theory". They proved that if G 1 , G 2 , G 3 are graphs on a common vertex set of size n and |E(G i )| > 1+τ 2 4 n 2 for 1 ≤ i ≤ 3 where τ = 4− √ 7 9 , then there exists a rainbow triangle. They also showed that τ 2 cannot be replaced by any ε > 0 with ε < τ 2 , this asymptotically answered a question of D.Mubayi [8] about a three-colored version of Mantel's theorem. They further asked for every positive integer r, what is the smallest δ r such that whenever G 1 , ..., G ( r 2 ) are graphs on a common set of n vertices with |E(G i )| ≥ δ r n 2 for every 1 ≤ i ≤ r 2 there exists a rainbow K r . In the same paper, Aharoni gave the following elegant conjecture which is a natural generalization of Dirac's theorem to rainbow case: Conjecture 2. Given graphs G 1 , ..., G n on the same vertex set of size n, each graph having minimum degree at least n 2 , there exists a rainbow Hamiltonian cycle.
There have been some other generalizations of Dirac's theorem to rainbow case. Let G be an edgecoloured graph, G is k-bounded if no colour appears more than k times on its edges. A recent result of Coulson and Perarnau in [6], which asserts that there exists µ > 0 and positive integer n 0 such that if n ≥ n 0 and G is a µn-bounded edge-coloured graph on n vertices with minimum degree at least n 2 , then G contains a rainbow Hamiltonian cycle. This theorem generalized a previous result in [4] proving the same conclusion when G is replaced by a complete graph.
Our first result shows that given n graphs G i (i = 1, ..., n) where each one has minimum degree at least n+1 2 , we can find rainbow cycles with arbitrary lengths except the the one with length n i.e. the rainbow Hamiltonian cyle: Theorem 3. Given graphs G 1 , ..., G n on the same vertex set of size n, each graph having minimum degree at least n+1 2 , there exist rainbow cycles with size i for 3 ≤ i ≤ n − 1.
The lower bound of Theorem 3 is tight since one can take n copies of K n 2 , n 2 when n is even and there does not exist any odd rainbow cycle in such a system. The main result of this paper is to prove an asymptotical version of Conjecture 2 by showing that the minimum degree ( 1 2 + o(1))n is sufficient.
Theorem 4. For every ε > 0, there exists an integer N > 0, such that when n > N for any graphs G 1 , ..., G n on the same vertex set of size n, each graph having minimum degree at least ( 1 2 + ε)n, there exists a rainbow Hamiltonian cycle.
To prove Theorem 3, we first find a rainbow Hamiltonian path and a rainbow cycle of length n − 1 following the classical proof of Dirac's theorem. Then we obtain a rainbow cycle of length n − 2 or n − 3 and use it to build cycles of other lengths. In order to prove Theorem 4, we use the absorbing method introduced by Rödl, Ruciński and Szemerédi [12]. We first construct a short rainbow cycle C such that for any rainbow path P = v 1 ...v p disjoint from C, and any colour s, we can absorb P into C, that is, Next, we find a rainbow Hamiltonian path P on V (G)\V (C) and absorb P into C by the property of C (such approach was used by Lo [10]).

Preliminaries and Notation
Let G 1 , ..., G n be n graphs on the same vertex set V where |V | = n. Let δ(G i ) be the minimum degree of each G i where 1 ≤ i ≤ n. Without loss of generality, we identify this graph system with an edge-coloured multigraph G where E(G) is the disjoint union of E(G i ) for i = 1, ..., n and each edge in E(G i ) is coloured by i for 1 ≤ i ≤ n. For any subgraph H of G, let Col(H) be the set of colours used by the edges of H.
A subgraph H of G is rainbow if any two different edges of E(H) are coloured with distinct colours. It is easily to see that to find a rainbow graph in the graph system is equivalent to find a corresponding rainbow subgraph in G. For every vertex v ∈ V (G) and any colour 1 ≤ c ≤ n, let N c (v) be the set of neighbours of v which are adjacent to v by an edge coloured by c. Let S be any subset of V , we denote be the set of colours used in the edges between v 1 and v 2 . For three real numbers x 1 , x 2 and x 3 , if We first prove the following useful lemma: Lemma 5. Let P = v 1 ...v p be a rainbow path and let c, c ′ be two colours not used on P . If d c (v 1 , V (P )) + d c ′ (v p , V (P )) ≥ p, then there is a rainbow cycle of length p.
must be an rainbow cycle where the colours of v 1 v i+1 and v p v i are chosen to be c and c ′ . Thus we get Our first result is a weaker version of Conjecture 2, we show that a rainbow Hamiltonian path exists under a slightly weaker condition: Proposition 6. Given graphs G 1 , ..., G n on the same vertex set of size n, where δ(G i ) ≥ n−1 2 for 1 ≤ i ≤ n, then there exists a rainbow Hamiltonian path.
Proof. Suppose not, let P = v 1 ...v k , where k ≤ n − 1, be the rainbow path in G with the maximum length. Thus there exist at least two colors c, c ′ which are not used by the edges in P . Now consider the Suppose that the colour c ′′ is not used by this cycle. Since the monochromatic graph coloured by c ′′ is connected, at least one edge e 0 coloured by c ′′ is between V (C) and V (G)− V (C). Therefore, V (C)∪{e 0 } contains a rainbow path with length k + 1, a contradiction.
The lower bound here is still best possible. One can take n copies of K n 2 −1, n 2 +1 when n is even and there does not exist a rainbow Hamiltonian path since K n 2 −1, n 2 +1 does not contain a Hamiltonian path.

Proof of Theorem 3
Claim 1. G contains a rainbow cycle of size n − 1.
Proof. By Proposition 6, we firstly find a rainbow Hamiltonian path P = v 1 v 2 ...v n . Without loss of generality, suppose the colour of edge v i v i+1 is i for 1 ≤ i ≤ n − 1 and the only colour that does not appear in P is n. Now consider the subpath by Lemma 5 we can find a rainbow cycle of size n − 1.
Claim 2. G contains either a rainbow cycle of size n − 2 or a rainbow cycle of size n − 3.
Proof. Suppose that G neither contain a cycle of size n − 3 nor n − 2. By Proposition 6, we can find a rainbow path P 1 = v 1 v 2 ...v n−3 whose order is n − 3 and without loss of generality, suppose the colour of edge v i v i+1 is i for 1 ≤ i ≤ n−4 and the set of colours which are not used in P 1 is S = {n−3, n−2, n−1, n}.
One can easily deduce that N n−1 (v 1 )∩N n (v n−3 )∩(V (G)−V (P 1 )) = ∅ since else we already find a rainbow cycle of size n − 2, a contradiction. Now we get by Lemma 5, we can find a rainbow cycle of size n − 3, a contradiction.
Let C be a rainbow cycle of size n − 2 or n − 3. We give it an orientation to make it be a directed Lemma 7. Let C = v 1 ...v p v 1 be a rainbow cycle of length p, and let c, c ′ be two colours not used on C.
If d c (x, V (C)) + d c ′ (x, V (C)) ≥ p for some x / ∈ V (C) and there is no rainbow cycles of length i for some Proof. Suppose that d c (x, V (C)) + d c ′ (x, V (C)) ≥ p and there is no rainbow cycles of length i for some x is a rainbow cycle of size i by choosing the colours of xv j and xv j+i−2 to be c and c ′ .
Therefore, we get S 1 ∩ S 2 = ∅ by definition. However, since |S 1 | + |S 2 | ≥ p and S 1 ∪ S 2 ⊆ V (C) it follows that V (C) is partitioned into S 1 and S 2 , which completes the proof.
Proof. Suppose the set V (G) − V (C) = {v n−1 , v n } and the colours not used by C are n − 1 and n.
Suppose that for some 3 ≤ j ≤ n − 1, there does not exist a rainbow cycle of size j in G. By Lemma This is a contradiction. Therefore, G contains rainbow cycles of all size 3 ≤ j ≤ n − 1.
Proof. Suppose the set V (G) − V (C) = {v n−2 , v n−1 , v n } and the colours not used by C are n − 2, n − 1 and n. Suppose that for some 3 ≤ i ≤ n − 2, there does not exist a rainbow cycle of size i in G. We know that d n−1 (v n , V (C)) + d n (v n , V (C)) ≥ 2( n+1 2 − 2) ≥ n − 3, thus by Lemma 7 we get d n−1 (v n , V (C)) + d n (v n , V (C)) = n − 3 which implies that

By symmetry we now suppose that
, using an analogue of the proof of Lemma 7, we can get that T 1 ∩ T 2 = ∅ since otherwise suppose v j+i−3 ∈ T 2 for some j, thus v n−1 v n−2 v j v j+1 ...v j+i−3 v n−1 is a rainbow cycle with length i by choosing the colours of v n−1 v n−2 , v n−2 v j and v j+i−3 v n−1 to be n, n − 2 and n − 1, which is a contradiction. We actually get: thus all the inequalities above must be equalities and we get which implies that V (C) is partitioned into T 1 and T 2 . Since all colours and vertices are symmetric, the similar conclusion follows by considering T 1 and N a (v b , V (C)) for any n − 1 ≤ a, b ≤ n. Thus, we ). Therefore, we get the similar conclusion that T 2 = N a (v b , V (C)) for any n − 2 ≤ a ≤ n − 1 and n − 1 ≤ b ≤ n by considering T ′ 1 and T 2 . This implies T 2 = N a (v b , V (C)) for any n − 2 ≤ a ≤ n and n − 1 ≤ b ≤ n. Now since all the vertices are symmetric, we actually get get v j0−1 ∈ T 1 and there exists some v j1 ∈ N n−2 (v n−2 , V (C)) such that j 0 − 1 = j 1 + i − 3 which implies

Proof of Theorem 4
For any pair two not necessarily distinct vertices x 1 , x 2 ∈ V (G) and four distinct colours 1 ≤ s, i, j, k ≤ n, we define the absorbing paths set A s,i,j,k (x 1 , x 2 ) to be the family of edge-coloured 3-paths P which satisfy the following conditions: (ii) The edges v 1 v 2 , v 2 v 3 , v 3 v 4 are coloured respectively by i, j, k; (iii) s ∈ C(x 1 , v 2 ) and j ∈ C(x 2 , v 3 ).
For every path P in A s,i,j,k (x 1 , x 2 ), we say that P is an absorbing path for (x 1 , x 2 ) with colour pattern (s, i, j, k). In practice, we always choose x 1 and x 2 to be two endpoints of some path Q and thus P indeed absorbs Q.
Claim 3. For each pair (x 1 , x 2 ) and four distinct colours s, i, j, k, we have |A s,i,j,k (x 1 , x 2 )| ≥ εn 4 8 when n is sufficiently large.
Note that the total number of such v 1 , v 4 is at least ( n 2 + εn − 3)( n 2 + εn − 4) and hence we derive that there exist at least absorbing paths for (x 1 , x 2 ) when n is sufficiently large.
We will use the following version of Chernoff's bound, see [13].
Lemma 8. Let X = t 1 + ... + t n where t i are independent boolean random variables for 1 ≤ i ≤ n. Then for any ε > 0, Especially, when ε = 1 2 , we conclude that Lemma 9. There exists a rainbow cycle C with size at most εn 10 5 such that for every rainbow path P with V (P ) ∩ V (C) = ∅ and Col(P ) ∩ Col(C) = ∅, if s is a colour that is not used by C and P , there exists a Proof. Let ℓ = ⌈ εn 10 6 ⌉ ± 1 such that ℓ is divisible by 3. For simplicity, we assume that ℓ = εn 10 6 . We fix ℓ/3 groups of colours C i = {3i − 2, 3i − 1, 3i} where i = 1, ..., ℓ/3. Let P Ci be the set of all the paths P . Now we randomly and uniformly select an element from each P Ci where each element of P Ci is chosen by probability 1 |PC i | for every 1 ≤ i ≤ ℓ/3. Thus we get a random set with size ℓ/3, we denote it by W . For any colour s and i=1 X i and all these X ′ i s are independent. Using Claim 3, we get By the Chernoff's bound, we see that Now let Y be the number of pairs of 3-paths in W which are intersecting with each other. For some distinct 1 ≤ i, j ≤ ℓ/3, let Y i,j be the indicative variable of the event that the path we choose in A Ci intersects with the path we choose in A Cj . Thus we have Y = i,j Y i,j . We claim that the size of set {{P 1 , P 2 } | P 1 ∈ A Ci , P 2 ∈ A Cj , P 1 , P 2 are intersecting with each other} is at most 16n 7 for fixed i, j.
Since the number of P 1 is at most n 4 , and when P 1 is fixed, the number of P 2 that we can choose is at most 16n 3 . Besides, it is obvious that when P 1 , P 2 are fixed, the probability that we have chosen P 1 , P 2 together is 1 |A(Ci)||A(Cj)| ≤ 1 (n 4 /8) 2 since |A Ci |, |A Cj | ≥ n 4 8 when n is sufficiently large. Therefore, we get Using Markov's inequality, we get that ≤ ε 2 n 10 10 ε 2 n 10 9 = 1 10 .
Thus with positive possibility, for each s and any pair (x 1 , x 2 ) the following is true: (i) |A s (x 1 , x 2 )| ≥ εℓ 48 ≥ ε 2 n 48×10 6 ; (ii) Y < ε 2 n 10 9 . Fix such W , for each intersecting pair of W , we delete one 3-paths in it, suppose that the remaining path family is W ′ . Thus W ′ is a family containing mutually disjoint 3-paths and for every s and any pair Let W ′ = {P 1 , ..., P t } be the path family we found before and let S be the set of the colours that do not appear in any path in 4 for 1 ≤ i ≤ t. Now for P 1 , P 2 , it is obvious that we can find a vertex u 1 ∈ V ′ such that u 1 v  are two edges coloured with distinct colours in S. Delete these two colours from S and the vertex u 1 from V ′ . Repeat the above process for the path pair {P 2 , P 3 }, ..., {P t , P 1 }, and at last we find u 1 , ..., u t and a rainbow cycle C with size at most εn 10 5 which contains all the vertices in t i=1 V (P i ) and those u i where 1 ≤ i ≤ t. For every rainbow paths P ⊆ V (G) − V (C) such that the colour set of P is disjoint with the colour set of C, if x 1 , x 2 are two endpoints of P and s is a colour that does not appear in C and P , then the pair (x 1 , x 2 ) has at least one absorbing path in C with colour pattern (s, 3i − 1, 3i − 2, 3i) for some 1 ≤ i ≤ ℓ/3 since 9ε 2 n 10 9 ≥ 1 when n is sufficiently large. Therefore, we insert the path P into the cycle C which completes our proof.
Proof of Theorem 4. Let C be the absorbing cycle we get in Lemma 9, now let G ′ = G − V (C), thus for each maximal monochromatic graph H in G ′ we find that δ(H) ≥ 1+ε 2 |V (G ′ )|. Let S 1 be the set of colours that do not appear on any edge of C, thus one can construct a rainbow Hamiltonian path P 1 = v 0 v 1 ...v t in G ′ using exactly |S 1 | − 1 colours of S 1 by Proposition 6. Suppose s 1 is the unique colour in S 1 that is not used by P 1 . Now consider an absorbing path in ℓ/3 i=1 (A s,3i−2,3i−1,3i (v 1 , v t )) that is also contained in C, we can insert P 1 into C to be a rainbow Hamiltonian cycle which completes our proof.

Remark
Shortly after we finished this preprint, Joos and Kim [9] informed us that they have settled the Conjecture 2.

Acknowledgements
The first two authors are supported by the National Natural Science Foundation of China (11631014, 11871311) and Qilu Scholar award of Shandong University. The third author is partially supported by NSF grant DMS 1700622.