Monochromatic cycle partitions of $2$-coloured graphs with minimum degree $3n/4$

Balogh, Bar\'at, Gerbner, Gy\'arf\'as, and S\'ark\"ozy proposed the following conjecture. Let $G$ be a graph on $n$ vertices with minimum degree at least $3n/4$. Then for every $2$-edge-colouring of $G$, the vertex set $V(G)$ may be partitioned into two vertex-disjoint cycles, one of each colour. We prove that this conjecture holds for $n$ large enough, improving approximate results by the aforementioned authors and by DeBiasio and Nelsen.


History
While undergraduates in Budapest, Gerencsér and Gyárfás [14] proved the following simple result: for any 2-edge-colouring of the complete graph K n , there exists a monochromatic path of length at least 2n/3 . It is easy to see that this statement is sharp. In their paper, Gerencsér and Gyárfás observe that a weaker result, asserting the existence of a monochromatic path of length at least n/2, can be deduced from the following simple observation: for any red and blue colouring of K n , there is a Hamilton path which is the union of a red path and a blue path. The latter observation, simple as it is, inspired intensive research.
In a later paper, Gyárfás [12] proved that, in fact, more is true. He showed that for any red and blue colouring of K n the vertices may be covered by a red cycle and a blue one sharing at most one vertex. Lehel went even further: he conjectured that for every 2-colouring of K n the vertex set may be partitioned into two monochromatic cycles of distinct colours. This conjecture first appeared in [2], where it was proved for some special colourings of K n .
Almost twenty years after this conjecture was made, Luczak, Rödl and Szemerédi [22] proved it for large n, using the Regularity Lemma. Ten years later, Allen [1] proved it for large n avoiding the use of the Regularity Lemma. Finally, Lehel's conjecture was fully resolved by Bessy and Thomassé [5] with an elegant and short proof.

Conjectures and progress
In the hope of generalising the above result of Gerencsér and Gyárfás, Schelp [26] considered 2-colourings of graphs which are not necessarily complete. In particular, he conjectured that given a graph G on n vertices with δ(G) > 3n/4, there is a monochromatic path of length at least 2n/3. Benevides, Luczak, Skokan, Scott and White [4] and Gyárfás and Sárközy [17] proved approximate versions of this conjecture.
Inspired by Schelp's conjecture,Balogh,Barát,Gerbner,Gyárfás,and Sárközy [3] proposed the following conjecture: given a graph G on n vertices with minimum degree δ(G) > 3n/4, for every 2-colouring of G the vertex set can be partitioned into two monochromatic cycles of distinct colours. We remark that for the purpose of this conjecture, the empty set, a single vertex and an edge are considered to be cycles. We note that there are examples of 2-coloured graphs G on n vertices with δ(G) = 3n/4 − 1 which do not admit such a partition (see Section 2).
In [3], the authors prove the following approximate result of their conjecture. For every ε > 0 there exists n 0 such that for every 2-coloured graph G on n ≥ n 0 vertices with minimum degree δ(G) ≥ (3/4 + ε)n, there exist vertex-disjoint monochromatic cycles of distinct colours covering all but at most εn of the vertices.
Recently, DeBiasio and Nelsen [8] proved the following stronger approximate result of the latter conjecture: for every ε > 0 there exists n 0 such that, for every 2-coloured graph G on n ≥ n 0 vertices and δ(G) ≥ (3/4 + ε)n, the vertex set may be partitioned into two monochromatic cycles of distinct colours.

The main result
Our main aim is to prove that the conjecture of Balogh et al. [3] holds if n is large enough. Theorem 1.1. There exists n 0 such that if a graph on n ≥ n 0 vertices and minimum degree at least 3n/4 is 2-coloured then its vertex set may be partitioned into two monochromatic cycles of different colours.
In [21], Luczak introduced a technique that uses the Regularity Lemma to reduce problems about paths and cycles into problems about connected matchings, which are matchings that are contained in a connected component. This technique, which we shall describe in more detail in Section 6, has become fairly standard by now and can be used to prove the approximate result of Balogh et al. [3]. The second result by DeBiasio and Nelsen [8], requires further ideas, most notably the "absorbing technique" of Rödl, Ruciński and Szemerédi (see [25] and [20]). Nevertheless, the stronger conditions on the minimum degree make their proof a great deal easier than ours. In order to prove Theorem 1.1, we use a variety of additional ideas and techniques.
We remark that Theorem 1.1 is sharp. Indeed, for every n ≥ 4, there exists a 2-coloured graph on n vertices with minimum degree 3n 4 − 1 admitting no partition into two monochromatic cycles of distinct colours. We give such extremal examples in the following section. These examples disprove the conjecture from [8], that a slightly stronger version of Theorem 1.1 may hold, namely that the conclusion holds for graphs with minimum degree at least 3n−3 4 . The following section consists of some extremal examples for Theorem 1.1. In Section 3 we give an overview of the proof as well as the structure of the rest of this paper.

Sharpness examples
Before we turn to the proof of Theorem 1.1, we give some extremal examples showing that the theorem is sharp. More precisely, we give examples of 2-coloured graphs on n vertices with minimum degree 3n 4 −1 admitting no partition into two monochromatic cycles of distinct colours. Figures (1,2,3,4) depict several families of such examples differing in the values of n (mod 4).
In these figures, we use black and grey for the colours of the edges. The areas coloured in either black or grey denote a complete (or complete bipartite) subgraph of the corresponding colour. Shaded areas denote complete (or complete bipartite) subgraphs which may be coloured arbitrarily. White areas are empty subgraphs. A small dot denotes a single vertex and a larger shape denote a cluster of vertices whose size is written in it. It is not hard to see that, indeed, each of these figures is an extremal example for Theorem 1.1. We leave the details to the reader. : Black and grey graphs of orders 4m + 2 and 4m + 1 (from left to right) with minimum degrees 3m + 1 and 3m respectively, admitting no partition into a black cycle and a grey one.
Other extremal examples may be formed by removing a vertex or two from the left-hand graph, or by removing the extra vertex from the right-hand graph.    Figure 4: A black and grey graph on 4m + 1 vertices and minimum degree 3m with no partition into a black cycle and a grey one. Any subgraph of this graph, obtained by removing one vertex, is also an extremal example.
We remark that there may exist other extremal examples. It may be possible to find all of them by carefully analysing the proof of Theorem 1.1, but we make no attempt to do so at this point.
3 Outline of the proof and structure of the paper In this section we describe the proof of our main theorem (1.1). We are given a graph G on n vertices and minimum degree 3n/4 with a red and blue colouring of the edges. In what follows, we give an outline of our proof that V (G) may be partitioned into a red cycle and a blue one.
Given an edge coloured graph, a monochromatic connected matching is a monochromatic matching which is contained in a connected component of the same colour. Similarly to the earlier results on our problem (by Balogh et al. [3] and DeBiasio and Nelsen [8]), as well as many other results in the area, one of the key tools is the technique of reducing problems about cycles to problems about connected matchings using the Regularity Lemma. This technique was introduced by Luczak [21] and since then has become fairly standard. In our setting, the basic idea, which is described in more detail in Section 6, is as follows. We are given a 2-coloured graph G and consider the reduced graph obtained by applying the Regularity Lemma. If the reduced graph has a perfect matching consisting of a red connected matching and a blue connected matching, we may use the blow-up lemma [18] (or in fact, a much simpler special case), to find two vertex-disjoint monochromatic cycles, a red one and a blue one, which cover almost all of the vertices.
The next ingredient is the "absorbing method" of Rödl, Ruciński and Szemerédi (see [25] and [20]). As in [8], in order to apply this method, we use a notion of "robust subgraphs", which are defined to be graphs with certain expansion properties (see Section 5 for the exact definition). Such graphs can be shown to contain short "absorbing paths", which are paths that can absorb small sets of vertices. We observe (see Section 6) that monochromatic connected components in the reduced graph (obtained from the Regularity Lemma) correspond to monochromatic robust subgraphs in the original graph. This observation allows us to obtain information about the rough structure, by using the Regularity Lemma and finding the corresponding robust subgraphs (see Section 8).
After a study of the some properties of the rough "robust structure" of the graph, we aim to apply the Regularity Lemma again, in order to find a suitable perfect matching, namely a perfect matching consisting of a connected red matching and a connected blue matching. This matching is used to find two vertex-disjoint cycles, a red one and a blue one, which cover most of the vertices and have the additional absorbing property implying that the leftover vertices can be inserted into one of these cycles (see Sections (9 -14)).
We should like to emphasize that as we prove the sharp result, namely that we only assume that the minimum degree is at least 3n/4, rather than (3/4 + ε)n as in the previous results, new difficulties arise, making our proof much harder. Firstly, we have to deal with several different cases for the rough robust structure, some of which do not arise when δ(G) ≥ (3/4 + ε)n. Interestingly, these cases require a variety of ideas and techniques, making the proof of the general theorem rather intricate. Secondly, when applying the Regularity Lemma we cannot guarantee that the minimum degree would be at least 3n/4, thus it may not be possible to find a suitable matching in the reduced graph directly.
The combination of the following two ideas helps us with these challenges. The first idea is to use stability versions of results promising a perfect matching. These enable us to conclude that if the reduced graph does not have the required perfect matching, it has some specific structure which we can further analyse to find the required monochromatic cycle partition.
The second idea is the following simple yet important observation: given two robust components of the same colour, if they can be connected with two vertex-disjoint paths, they may essentially be treated as one larger component. In several parts of the proof (see Sections (10,12,13)), we use this observation to conclude that either we may join two robust components to obtain a larger one, or the graph admits some restrictive structure, for which the desired partition may be found "by hand" (see Sections 15 and 16). We remark that even at this stage, the proof is rather hard due to the fact that our result is sharp.

Structure of the paper
In the next section, Section 4, we introduce the notation that will be used in this paper. In Section 5, we define our notion of robustness and prove some properties of robust components, most notably the existence of absorbing paths. In Section 6, we state the version of the Regularity Lemma that we use here. We also prove some results about the correspondence between connected components of the reduced graph and robust subgraphs of the original graph and describe the method of converting connect matchings in the reduced graph into cycles in the original graph. In Section 7, we list some results which will be used throughout the proof.
Sections (8 -16) are devoted to the proof of Theorem 1.1. In Section 8, we obtain some information about the rough structure and point out how to prove Theorem 1.1 using the results in subsequent lemmas. In each of Sections 9 -14, we consider one of the cases arising from the structural result in Section 8. These cases vary in difficulty and we have to use various techniques used to deal with them. In Sections 15 and 16, we prove Lemmas 7.13 and 7.14, which are used in earlier sections and prove the main theorem under certain restrictive conditions on the colouring and the structure of the graph. Finally, Section 17 is devoted to some concluding remarks.

Notation
We use mostly standard notation. Write |G| for the order of a graph G and δ(G) and ∆(G) for its minimum and maximum degrees respectively. The neighbourhood of a vertex x ∈ V (G) is denoted by N G (x) and its degree by d G (x) = |N G (x)|. Given A ⊆ V (G), we write N G (x, A) = N G (x) ∩ A and d G (x, A) = |N G (x, A)|. We will write, for example, d(x, A) for d G (x, A) if this is unambiguous. Given a set of vertices X ⊆ V (G), we write G[X] for the graph induced by G on X. Similarly, for disjoint subsets X, Y ⊆ V (G), we write G[X, Y ] for the bipartite graph with bipartition {X, Y } induced by G. We denote e G (X, Y ) = |E(G[X, Y ])|.
Given a graph G, we denote a 2-colouring of G by E(G) = E(G B ) ∪ E(G R ), where G B , G R are graphs on vertex set V (G) (in proper colouring, the graphs G B , G R are edge-disjoint). The edges of G B are called blue edges and the edges of G R are red edges. We sometimes use B or R for a subscript instead of G B or G R . For instance, N B (x) is a shorthand for N G B (x).
We denote by (u 1 u 2 . . . u k ) the path on vertices u 1 , . . . , u k taken in this order. We use the same notation to denote the cycle obtained by adding the edge (u k , u 1 ) to the given path. It should be clear from the context if we are dealing with a path or a cycle. Given paths P 1 , P 2 which share an end and are otherwise disjoint, we denote by (P 1 P 2 ) the concatenation of the two paths. Similarly, if the paths share both ends but are otherwise disjoint, the same notation denotes the cycle obtained by joining the two paths.
Throughout this paper we omit floors and ceilings whenever the argument is unaffected. The constants in the hierarchies used to state our results are chosen from right to left. For example, the claim that a statement holds for 0 < a, 1 n b c 1 means that there exist non-decreasing functions f, g : (0, 1] → (0, 1] and a constant c 0 such that the statement holds for all 0 < a, b, c ≤ c 0 and integers n with b ≤ f (c), a ≤ g(b) and n ≥ 1 g(b) . We normally do not specify the functions in question.

Robust components and absorbing paths
Similarly to the proof of DeBiasio and Nelsen [8], one of the main tools in our proof is the notion of robust subgraphs. As we shall see, these are graphs with certain expansion properties. The role of robust subgraphs in our proof is similar to their role in [8], but our definition is different and is often easier to apply. Nevertheless, the two definitions are in some sense equivalent, as can be seen in Lemma 5.4 in [8]. After defining a robust subgraph, we state and prove some simple properties of these components. Finally, we prove that robust subgraphs contain "absorbing paths", which may absorb small sets of vertices.

Definitions of robust subgraphs
We define two notions of robustness: strong and weak. The difference between the two is that strong robust subgraphs are far from being bipartite. It will be easier for our application, though not essential, to define a robust subgraph relative to a fixed ground graph. The precise definitions are as follows.
Given a graph G, vertices x, y ∈ V (G) and an integer l, denote by con G,l (x, y) the number of paths of length l + 1 in G between x and y.
As a shorthand, we often omit the parameters α and k when they are clear from the context. From now on, we use the term "robust" to signify either strongly robust or weakly robust with suitable parameters. We point out that in our context α and k are fixed and n tends to infinity. We remark that an (α, k) robust subgraph F of a graph G on n vertices has δ(F ) ≥ αn. In particular, robust subgraph are always dense.
Before discussing some properties of robust subgraphs, let us give a few examples. Any graph of minimum degree at least (1/2 + α/2)n is (α, 1) strongly robust, because any two vertices have at least αn common neighbours. Similarly, the random graph G(n, α), is w.h.p. (α 2 /2, 1) strongly robust. Furthermore, the blow-up of a path of length k ≥ 2, where every vertex is replaced by a complete graph on n/k vertices, is (α, k − 1) strongly robust for a suitable α. Similarly, if vertices of a path of length k are replaced by empty graphs, we obtain an (α, k − 1) weakly robust graphs.

Properties of robust subgraphs
We shall make use of some simple properties of robust graphs. The following lemma states that a robust subgraph remains robust after removing a small number of vertices.
Lemma 5.2. Given α > 0 and k an integer, the following holds for small enough β. Let G be a graph on n vertices and let F be an (α, k)-robust subgraph. Suppose that F is obtained from F by removing at most βn vertices. Then F is (α/2, k)-robust.
Proof. We prove the lemma under the assumption that F is strongly robust; the proof in case F is weakly robust is analogous. Let l ≤ k satisfy |con F,l (x, y)| ≥ αn l for every x, y ∈ V (F ). For every x, y ∈ V (F ), the number of paths of length l + 1 between x and y containing at least one vertex from V (F ) \ V (F ) is at most lβn l ≤ α 2 n l . It follows that | con F ,l (x, y)| ≥ α 2 n l for every x, y ∈ V (F ), i.e. F is (α/2, l) strongly robust.
The next lemma shows that a robust subgraph remains robust after removing a graph of small maximum degree. Lemma 5.3. Given α > 0 and k, the following holds for suitably small β and large n. Let G be a graph on n vertices and let F be an (α, k)-robust subgraph. Suppose that F is a subgraph of F such that for every vertex Proof. We prove this lemma for F strongly robust; the proof for F weakly robust is similar. Let l ≤ k be such that |con F,l (x, y)| ≥ αn l for every x, y ∈ V (F ). Fix some x, y ∈ V (F ). We consider the family of paths in F of length l + 1 between x and y which contain at least one edge outside of F . There are at most lβn l such paths, i.e. |con F ,l (x, y)| ≥ |con F ,l (x, y)| − lβn l ≥ α 2 n l (for small enough β). It follows that F is (α/2, l) strongly robust.
The following lemma states that a robust subgraph F remains robust after the addition of vertices which have a large neighbourhood in F . Lemma 5.4. Given α > 0 and k an integer, the following holds for large enough n. Let G be a graph on n vertices and let F be an (α, k)-robust component. Let F be a subgraph of G containing F , such that every vertex in V (F ) \ V (F ) has at least αn neighbours in F . Then F is (α 3 /2, k + 2)-robust.
Proof. We prove the statement assuming that F is strongly robust; the proof in case F is weakly robust is very similar and we omit the details. Let l ≤ k satisfy |con F,l (x, y)| ≥ αn l for every x, y ∈ V (F ). Fix some x, y ∈ V (F ). For every z, w ∈ V (F ) such that z ∈ N (x) and w ∈ N (y), we have |con F,l (w, z)| ≥ αn l . Thus, since every vertex in V (F ) has at least αn neighbours in F , we have that the number of walks between x and y in F with l + 2 inner vertices is at least α 3 n l+2 for large n. Since there are at most O(n l+1 ) such walks which are not paths, we have that |con F ,l (x, y)| ≥ α 3 2 n l+1 . It follows that H is (α 3 /2, k + 2)-strongly robust.
So far we listed and proved several simple properties of robust subgraphs. In the following subsection we state and prove a more interesting property.

Absorbing paths
The main reason robust subgraphs are so useful in our context, is the fact, which was proved by DeBiasio and Nelsen [8] that they contain short "absorbing paths". We conclude this section with a proof of this fact.
Lemma 5.5. Let 1 n ρ α, 1 k 1, let G be a graph on n vertices and let F be an (α, k)-robust subgraph of G. Then there exists a path Q in F satisfying the following conditions. 1. If F is strongly robust, for every set W ⊆ V (F ) \ V (Q) of size at most ρ 2 n, there exists a Hamilton path in F [V (Q) ∪ W ] with the same ends as Q.

If
contains a Hamilton path with the same ends as Q.
We follow the footsteps of DeBiasio and Nelsen in their proof of Lemma 5.6 from [8]. The main tool is absorbing method of Rödl, Ruciński and Szemerédi [25]. We shall use "gadgets", which we will define to be Hamiltonian graphs that are can absorb a single vertex under the condition that it is adjacent to some of the vertices in the gadget. By a simple application of the probabilistic method and the robustness of the given graph, we show that there exists a not too large collection of vertex-disjoint gadgets, such that every vertex may be absorbed by a rather large number of them. From there it will be easy to construct the required path Q.
Proof of Lemma 5.5. We start by proving the first part of Lemma 5.5. Suppose that F is (α, k)strongly robust. In particular, δ(F ) ≥ αn so we may apply the following claim.
Claim 5.6. Let p, 1 n α 1 and let F be a graph on at most n vertices with δ(G) ≥ αn. Then there exists a family F of disjoint pairs of vertices of V (F ) such that the following conditions hold.

• |F| ≤ pn
• For every u ∈ V (F ) there are at least 1 16 pαn 2 pairs (x, y) ∈ F such that x, y are neighbours of u.
This claim is a simple application of Chernoff's bound. We shall pick a family of pairs of vertices randomly and then delete a small number of pairs so as to ensure that the pairs are disjoint.
Proof of Claim 5.6. Let F be the family of pairs obtained by choosing each pair of vertices in V (G) independently with probability p n . By Chernoff's bound, we have that with high probability, the following properties hold.
The expected number of pairs of intersecting pairs in F is at most ( p n ) 2 n 3 ≤ p 2 n. It follows by Markov's inequality that with probability at least 1/2, the number of pairs of intersecting pairs in F is at most 2p 2 n. In particular, we may pick a family F which satisfies the above conditions and which has at most 2p 2 n pairs of intersecting pairs. We obtain a subfamily F of F containing no intersecting pairs by deleting at most 2p 2 n pairs from F. It is easy to verify that if p is suitably small, F satisfies the requirements of the claim.
Let F = {(x j , y j )} N j=1 be a family of pairs as in Claim 5.6 (so N ≤ pn). We use the following simple technical claim, to avoid divisibility issues. Claim 5.7. Let β be suitably small and n suitably large. Then for some 1 ≤ l ≤ k, there are at least βn 4l−2 paths of length 4l − 1 between each pair of vertices in F .
Proof. Since F is (α, k) strongly robust, there exists l ≤ k such that between every u, v ∈ V (F ) there are at least αn l−2 paths in F of length l − 1. We conclude that for every u, v ∈ V (F ) there are at least α 10 n 4l−2 walks of length 4l − 1 between u and v. Indeed, given u, v ∈ V (F ), there are at least (αn) 6 ways to pick edges e 1 , e 2 , e 3 ∈ E(F ), since we may pick one end of each edge in at least |F | ≥ αn ways, and then there are at least αn ways to pick a neighbour. Denote e i = (a i , b i ). There are (αn l−2 ) 4 ways to pick paths P 1 , P 2 , P 3 , P 4 in F of length l − 1 with ends u and a 1 , b 1 and a 2 , b 2 and a 3 , b 3 and v respectively. It follows that there are at least α 10 n 4l−2 walks in F between u and v. At most O(n 4l−3 ) of them are not paths, so for large enough n, there are at least α 10 2 n 4l−2 walks of length 4l − 2 between u and v.
We shall build vertex-disjoint paths Q j of length 8l 2 − 4l + 1 one by one for j = 1, . . . , N as follows. Suppose that Q 1 , . . . , Q j−1 are already defined. We would like to pick paths as follows.
It is easy to see, by the choice of l according to Claim 5.7, that if p is small enough, we may pick such paths to be vertex-disjoint of all previously defined paths and to have pairwise disjoint interiors.
Suppose that w ∈ V (G) is a neighbour of x j = u 1 and y j = u 2l . The following path is a path in F with vertex set V (Q j ) ∪ {w} and same ends as Q j (this path is illustrated in Figure 5 together with Q j ).
Finally, we let Q be a path which contains Q 1 , . . . , Q N by connecting the ends of the Q j 's with paths of length 4l − 1. Denote ρ = 8pk 2 , and note that we may pick p small enough such that ρ 2 ≤ 1 16 pα 2 . The length of Q is at most ρn and for every vertex z ∈ V (F ) \ V (Q), there are at least ρ 2 n values of j ∈ [N ] such that x j , y j ∈ N (z). We show that Q has the desired absorbing property. Let W be a set of at most with the same ends as Q j . By replacing the occurrence of Q j i by Q j i in the path Q, we obtain a path on vertex set V (Q) ∪ W with the same ends as Q.
We now turn to the proof of the second part of Lemma 5.5. Let F be an (α, k) weakly robust component with bipartition {X, Y }. The proof will use similar ideas, with some variations which take into account the bipartition of F . A similar argument as in Claim 5.6 implies that for small enough p we may find a family F of disjoint quadruples of vertices of F with the following properties. Figure 5: An illustration of the absorbing structure for l = 2. The path Q j is represented by the straight line between u 2 and u 8 which is marked in grey, and the path absorbing w is represented by the bold black path.

• |F| ≤ pn
• For every x ∈ X, y ∈ Y the number of quadruples (a, b, c, d) such that a, c ∈ N (x) and b, d ∈ N (y) is at least 1 16 pα 4 n.
. . , N we pick a path Q j as follows. As before, there exists 1 ≤ l ≤ k such that | con F,4l−2 (x, y)| ≥ βn 4l−2 for every x ∈ X, y ∈ Y . Assuming that Q 1 , . . . , Q j−1 were already chosen to be paths of length at most 10l 2 , we may pick paths (a 1 b 1 . . . a 2l b 2l ) a path with ends a 1 = a j , b 2l = b j (c 1 d 1 . . . c 2l d 2l ) a path with ends c 1 = c j , d 2l = d j which are vertex-disjoint of each other and of previously defined paths. We may further pick paths of length at most k P i a path with ends which are pairwise vertex-disjoint and are disjoint of previously defined paths. Let Suppose that x, y satisfy a j , c j ∈ N (x) and b j , c j ∈ N (y). Then the following path is a path in F with vertex set V (Q j ) ∪ {x, y} with the same ends as Q j (See figure 6).
We may proceed as in the previous part to complete the proof of Lemma 5.5. Figure 6: An illustration of the absorbing structure for l = 3. The black lines represent edges, whereas the grey ones represent paths. The dashed blue line represents the path Q j and the dotted red one represents the path obtained from Q j by absorbing w.
The proof of Lemma 5.5 concludes our introduction of the notion of robust subgraphs and their properties. In order to make use of the properties we established, we shall use Lemmas 6.3, 6.7 and 6.8 from the next section, Section 6. These lemmas establish the connection between connected components of the reduced graph (obtained by an application of the Regularity Lemma) and robust subgraphs.

The Regularity Lemma
In our proof, we shall use Szemerédi's Regularity Lemma extensively. Before stating the version we use, we introduce some notation. Let U, W be disjoint subsets of vertices of a graph G. The density d(U, W ) of edges between U and W is defined to be where e(U, W ) is the number of edges between U and W . A bipartite graph with bipartition U, W is said to be ε-regular if for every U ⊆ U and W ⊆ W with |U | ≥ ε|U | and |W | ≥ ε|W |, We use a variant of the so-called degree form of the Regularity Lemma (see [19]), which is applicable to 2-coloured graphs. Furthermore, it will be useful for our purpose to start with a cover of the vertices of a graph and require that the partition obtained by the lemma is a refinement of the initial cover.
Lemma 6.1. For every ε > 0 and integer l there exists M = M (l, ε) such that the following holds. Let G be a 2-coloured graph on n vertices, C a cover of V (G) with at most l parts and d > 0. Then there exists a partition {V 0 , . . . , V m } of V (G) and a subgraph G of G with vertex set V (G) \ V 0 , such that the following conditions hold.
is contained in one of the parts of C.
(R6) All pairs (V i , V j ) are ε-regular in both colours in G , with density in each colour either 0 or at least d.
It is often useful to work with the reduced graph, obtained from the partition given by the Regularity Lemma as follows. Given a 2-coloured graph G, and parameters ε, d, l, we define the (ε, d)-reduced graph Γ as follows. Let {V 0 , . . . , V m } be the partition obtained by an application of Lemma 6.1, and let G be the given subgraph of G. We take V (Γ) = {V 1 , . . . , V m }. A pair V i V j is a t-coloured edge in Γ if it has density at least d in colour t in G . Note that an edge of Γ can have more than one colour.
The following observation shows why it is useful to work with the degree form of the Regularity Lemma.
Observation 6.2. Let G be a 2-coloured graph on n vertices with δ(G) ≥ cn and let Γ be the (ε, d)-reduced graph obtained by applying Lemma 6.1.
The rest of this section is divided into two parts. In the first part, Subsection 6.1, we establish the connection between robust subgraph and connected components of a reduced graph. In the second part, Subsection 6.2, we describe the connection between connected matchings in a reduced graph and cycles in the original graph.

From connected components of the reduced graph to robust subgraphs
One of our main tools in the proof of Theorem 1.1 is the following lemma. It gives us the means to obtain a robust subgraph from a connected subgraph of the reduced graph.
Lemma 6.3. Let α, 1 k , 1 n ε, 1 l 1 and d ≥ 4ε. Let G be a graph on n vertices and let Γ be the (ε, d)-reduced graph obtained by an application of Lemma 6.1. Suppose that Φ is a connected subgraph of Γ. Then there exists a subgraph F of G with the following properties.

Let U be the set of vertices contained in clusters of
2. F is (α, k)-robust. If Φ is bipartite, F is weakly robust, otherwise it is strongly-robust.
In order to find the required robust component F , we consider the clusters represented by V (Φ), and for each of the clusters we remove vertices with low degree. The regularity of pairs of clusters which are connected by an edge, together with the choice of d, implies that the number of low degree vertices in each cluster is small. We show that the subgraph induced by the remaining vertices has the required expansion properties, using the regularity of the pairs of clusters.
Proof of Lemma 6.3.
Let G be the subgraph of G given by Lemma 6.1). Let Proof. Suppose otherwise. Recall that for every j ∈ I i , (V i , V j ) is an ε-regular pair in G with density at least 4ε. It follows by the definition of a regular pair that Note that F satisfies Property (1) in Lemma 6.3. It remains to show that Property (2) holds. We suppose that Φ is non-bipartite, the proof for the bipartite case follows similarly. We use the following simple claim.
Claim 6.5. Let G be a connected non-bipartite graph on n vertices. Then there exists k ≤ 3n such that between every two vertices of G there is a walk of length k.
Proof. We first show that between every two vertices of G there is a walk of odd length not exceeding 3n. Indeed, let x, y ∈ V (G). Let C be an odd cycle, and pick some z ∈ C. Pick some path from x to z and a path from z to y. Combining the two paths, we obtain a walk from x to y of length at most 2n. If this walk has even length, we add the cycle C to it. In any case we obtain an odd walk of length at most 3n. Let k be the length of the longest of these walks. By possibly adding 2-cycles to the given walks, we obtain walks of length k between each pair of vertices.
Fix some k ≤ 3n as in the previous claim. In the following claim, we show that the existence of a walk of length k in Φ implies the existence of many paths of the same length in F . Claim 6.6. Let k ≤ 3m. There exists β = β(l, ε) such that the following holds. Suppose that V i 1 , . . . , V i k is a walk in Φ. Let U 1 and U k be subsets of V i 1 ∩ V (F ) and V i k ∩ V (F ) respectively of size at least 2ε|V 1 |. Then there exists at least βn k+1 paths of length k in F between U 1 and U k .
is an ε-regular pair in H with density at least 4ε, we have e(A, U k ) ≥ 3ε|A||U k | ≥ 3ε 3 |V 1 | 2 . Each such edge completes the above walk into a distinct walk in H between v 1 and U k of length k − 1.
We conclude that the total number of walks of length k − 1 between U 1 and U k is at least where M is as in Lemma 6.1. Note that the number of such walks which are not paths (namely, a vertex appears more than once) is O(n k−1 ). It follows that indeed, there are at least βn k paths between U 1 and U k of length k − 1, where β is a suitable constant. Let x, y ∈ V (F ) and suppose that x ∈ V i and y ∈ V j . Note that the definition of F implies the existence of t, s ∈ I such that (V i , V t ), (V j , V s ) ∈ E(Φ) and x and y have at least 2ε|V 1 | neighbours in H in V t and V s respectively. It follows from Claim 6.6 that there are at least βn k paths from the neighbourhood of x in V t to the neighbourhood of y in V s . This shows that |con F,k (x, y)| ≥ β 2 n k for every x, y ∈ V (F ), implying that F is (β/2, m) strongly robust.
In fact, we need a stronger version of Lemma 6.3. In our application, since we deal with 2-coloured graph, we will typically have two collections of connected subgraphs of Γ, one for each colour, and it would be useful to obtain collections of robust components which preserve containment. For example, if in the reduced graph we have blue components Φ 1 , Φ 2 and red components Φ 3 , , we would like the corresponding robust subgraphs F 1 , F 2 , F 3 , F 4 to satisfy the corresponding equality, namely . This is achieved by the following lemma. We remark that the proof is very similar to the previous one, so we omit it.
Let Γ be the (ε, d)-reduced graph obtained by an application of Lemma 6.1 and let {V 0 , . . . , V m } be the corresponding partition of V (G). Let P B and P R be collections of disjoint connected subgraphs of Γ B and Γ R respectively. Then there exist subsets U i ⊆ V i for i ∈ [m] satisfying the following properties.
In our proof we shall find robust components and then apply the Regularity Lemma. Therefore we need the following result, stating that given a robust component F in G, the corresponding subgraph of F in the reduced graph Γ is connected.
Lemma 6.8. Let ε, 1 n α, 1 k , 1 l 1 and let G be a graph on n vertices with a 2-colouring were t ∈ {B, R} and let C be a cover of V (G) with at most l parts refining {V (F ), V (G) \ V (F )}. Let Γ be the (ε, d)-reduced graph obtained by an application of Lemma 6.1. Then the t-coloured subgraph Φ of Γ spanned by the clusters contained in V (F ) is connected.
Proof. Let G be the corresponding subgraph of G obtained by applying Lemma 6.1. Let U be the set of vertices of F which belong to sets Hence F is obtained by removing at most εn vertices of F and then removing a subgraph with maximum degree at most 9εn. By Lemmas 5.2, 5.3, we have that F is (α/4, k)-robust in G . In particular, F is connected and it follows that Φ (the t-coloured subgraph of Γ spanned by clusters contained in V (F )) is connected.

From connected matchings to long cycles
We shall use the technique of converting connected matchings in the reduced graph into cycles in the original graph. This was introduced by Luczak [21], and since then has become fairly standard (see [3], [8] and [15], [16], [17] and [22]).
For the sake of completeness, we prove the following result, stating that given a connected matching in the reduced graph, there exists a cycle in the original graph through most of the vertices in the clusters and few additional vertices. Lemma 6.9. Let ε > 0 and d ≥ 3ε and let n be suitably large. Let G be a graph on n vertices and let Γ be the (ε, d)-reduced graph obtained by an application of Lemma 6.1. Suppose that M is a connected matching in Γ and denote by U the set of vertices spanned by the clusters of M. Then G contains a cycle covering at least (1 − 6ε)|U | of the vertices of U .
Let us first sketch the proof. Using the fact that M is connected we can connect the matching edges by paths, following a cyclic ordering of the edges in M. We replace these paths by short vertex-disjoint paths in the original graph G between the cluster pairs associated to the edges of M. These connecting paths will be parts of the final cycle. To define the rest of the cycle, remove the internal vertices of these connecting paths, and in each cluster pair find a long path to close the connecting pairs to a cycle. There are various ways to find these long cycles, e.g. using the rotation-extension technique of Pósa [24] or as a very special case of the Blow-up Lemma [18]. We shall use a much simpler result to obtain these cycles.
Proof of Lemma 6.9. Denote by {V 0 , . . . , V m } the corresponding partition of V (G) and M = (V i 1 , V j 1 ), . . . , (V im , V jm ). It is easy to see that we may find paths P 1 , . . . , P m with the following properties.
• The paths P l are vertex disjoint.
• Each path P l contains at most one vertex from each cluster V i .
• Denote the ends of P l by x l , y l . Then Note that this is indeed possible because by the choice of x l , y l and since for large enough n, |U | ≤ ε|V 1 |. Pick sets We shall show that for each l, G[C l , D l ] contains a path Q l missing at most 2ε|V 1 | vertices from each set C l , D l . Assuming that we may find such paths, it is now easy to construct the required cycle. By the definition of the reduced graph, the density of edges between subsets of V i l and V j l of size at least ε|V 1 | is positive. Thus there is an edge between A l and the last 2ε|V 1 | vertices of Q l as well as between B l and the first 2ε|V 1 | vertices of Q l+1 . Thus by losing at most 4ε|V 1 | vertices from each path Q l we may use the paths P l to obtain a cycle C which misses at most 6ε|V 1 | vertices from each cluster in V (M).
It remains to show that such paths Q l may be found. We use the following simple but useful claim which was proved independently by Pokrovskiy [23] and Dudek and Pra lat [9]. For the sake of completeness, we include the proof here. Proposition 6.10. For every graph G there exist two disjoint sets of vertices U, W of equal size such that G has no edges between U and W and the graph G \ (U ∪ W ) has a Hamilton path.
Proof. In order to find sets with the desired properties, we apply the following algorithm, maintaining a partition of V (G) into sets U, W and a path P . Start with U = V (G), W = ∅ and P an empty path. At every stage in the algorithm, do the following. If |U | ≤ |W |, stop. Otherwise, if P is empty, move a vertex from U to P (note that U = ∅). If P is non-empty, let v be its endpoint. If v has a neighbour u in U , put u in P , otherwise move v to W . Note that at any given point in the algorithm there are no edges between U and W . Furthermore, the value |U | − |W | is positive at the beginning of the algorithm and decreases by one at every stage, thus at some point the algorithm will stop and will produce sets U, W with the required properties.
We need the following corollary of Proposition 6.10.
Corollary 6.11. Let G be a balanced bipartite graph on n vertices with bipartition V 1 , V 2 , which has no path of length k. Then there exist X i ⊆ V i such that |X 1 | = |X 2 | ≥ (n − k)/4 and G has no edges between X 1 and X 2 .
Proof. Let U, W be as in Proposition 6.10 and let P be a Hamilton path in G \ (U ∪ W ). Note that P must alternate between V 1 and V 2 , thus |V (P ) ∩ V 1 | = |V (P ) ∩ V 2 | (by the assumptions, the number of vertices in P is even).
The graph G[C l , D l ] is a balanced bipartite graph, satisfying that for every choice of sets C ⊆ C l , D ⊆ D l of size at least ε|V 1 | each, we have e(G[C , D ]) > 0. It follows from Corollary 6.11 that G[C l , D l ] contains a path missing at most 2ε|V 1 | vertices from each side. As explained previously, Lemma 6.9 follows.
In fact, we shall need a slight generalisation of Lemma 6.9. The usual setting in which we apply the described technique is as follows. We consider the reduced graph obtained by applying the Regularity Lemma. In the reduced graph, we find a perfect matching consisting of a connected blue matching and a connected red matching, and we use the described technique to find two disjoint cycles, one blue and one red, which together cover almost all of the vertices.
To obtain disjoint cycles, we set aside small sets of each cluster of the red component which will be used to connect the cluster pairs of the red connected matching. We then find a blue cycle as above covering most of the clusters of the blue matching, and then we find a red cycle similarly, using the sets we set aside.
We shall not write the exact statement of the modified version of Lemma 6.9, but let us remark that the constants change slightly. We need d ≥ 6ε and are able to cover (1 − 9ε)n of the vertices of each cluster in the matchings.
Finally, we point out that we often first find "absorbing paths" in the original graph and then apply the Regularity Lemma to the remaining graph. When building the cycles obtained by the connected matchings, we would like them to contain these predefined paths. This is obtained by the same method, enabling us to find a path (rather than a cycle) between the neighbourhoods of the two ends of the absorbing path.
This concludes the introduction of the tools we shall need for our proof of Theorem 1.1. To complete the preliminary material needed for our proof, we list several extremal results in the next section.

Extremal results
In this section we list a number of extremal results we shall use in our proofs. They concern mainly with the existence of matchings, paths and cycles in graphs with certain structural conditions.
The following is Chvátal's theorem [7] giving sufficient conditions on the degree sequence of a graph for containing a Hamilton cycle.
Theorem 7.1. Let G be a graph on n ≥ 3 vertices and let d 1 ≤ . . . ≤ d n be the degree sequence of G. Suppose that d i ≥ i + 1 or d n−i ≥ n − i for every i ≤ n/2. Then G contains a Hamilton cycle.
In some cases it is easier to use the following version of Chvátal's result for bipartite graphs.
Corollary 7.2. Let G be a balanced bipartite graph on 2n vertices with bipartition {X, Y }. Let x 1 ≤ . . . ≤ x n be the degree sequence of X and let y 1 ≤ . . . ≤ y n be the degree sequence of Y . Suppose that x i ≥ i + 1 or y n−i ≥ n − i + 1 for every i ∈ [n]. Then G contains a Hamilton cycle.
Proof. Consider the graph G obtained from G by adding all edges with both ends in X. By Theorem 7.1, G contains a Hamilton cycle C. As |X| = |Y |, the cycle C contains no edges with both ends in X, i.e. C is a Hamilton cycle in G.
The following is a simple result by Erdős and Gallai [10], giving an upper bound on the number of edges in a graph with no path of a given length. Theorem 7.3. Let G be a graph on n vertices with no paths of length at least l + 1, then e(G) ≤ nl/2.
A graph on n vertices is called pancyclic if for every l ≤ n, G contains a cycle on l vertices. The following result by Bondy [6] is a generalisation of Dirac's Theorem, asserting that graphs with large enough minimum degree are pancyclic.
Theorem 7.4. Let G be a graph on n vertices with δ(G) > n/2. Then G is pancyclic.
In the following two subsections, we use the following well known theorem of Tutte, giving a necessary and sufficient condition for having a perfect matching.
Theorem 7.5. Let G be a graph on an even number of vertices. Then G has a perfect matching if and only if for every set of vertices U , the number of odd components of G \ U is at most |U |.

Matchings in tripartite graphs
In section 14 we shall analyse conditions for certain tripartite graph to have a perfect matching. Here we describe the extremal results we shall need for the analysis. We use a stability version of the following lemma of DeBiasio and Nelsen [8]. For the sake of completeness, we prove it here. Lemma 7.6. Let n be even, and let G be a tripartite graph on n vertices with tripartition Then G has a perfect matching.
Proof. Suppose to the contrary that G has no perfect matching. Without loss of generality, we assume that G is a maximal counter example. It is easy to check that the graph obtained from the complete tripartite graph with partition {X 1 , X 2 , X 3 } by removing any edge has a perfect matching. Thus we may assume that G is not complete as a tripartite graph. Without loss of generality, suppose that Denote by M 1 the edges of M which have ends in X 1 and X 2 , and let M 2 = M \M 1 . It follows that for every e ∈ M 1 \{v 1 v 2 }, deg(v 1 , e)+deg(v 2 , e) ≤ 1 and for every e ∈ M 2 , deg(v 1 , e) + deg(v 2 , e) ≤ 2.
For the last equality, we used the fact that |M 2 | = |X 3 |. By the degree condition on G, we have The following lemma is a stability version of Lemma 7.6. We prove it by applying Tutte's theore, Theorem 7.5.
and x ∈ X i . Then one of the following holds.
• G has a perfect matching.
Proof. We assume that G has no perfect matching. By Tutte's theorem, Theorem 7.5, there exists a subset S ⊆ V (G) such that the number of odd components of G \ S is larger than S.
Denote G = G\S. We show first that δ(G ) ≤ εn. Suppose not. Then the number of components of G is at most 1/ε, thus |S| ≤ 1/ε. We show that G is connected, contradicting the choice of S. Given u, v ∈ X i , they are non-adjacent to at most (1/4 + ε)n vertices in V (G) \ X i . But |V (G) \ X i | ≥ (1/2 + 4ε), hence u, v have at least 2εn ≥ 1/ε common neighbours. It follows that indeed, G is connected.
By the assumptions on G, δ( Note that |S| ≤ n/2, because the number of odd component of G is at most n − |S|. Denote In particular, u is in a component of G order at least εn. Furthermore, every non isolated vertex of X 1 is adjacent to some vertex in X 2 ∪ X 3 , and thus is in a component of size at least εn. Since X 2 ∪ X 3 is non empty (e.g. because |S| ≤ n/2), it follows that G has at most (1/4 + ε)n isolated vertices, and the rest of the vertices are in components of order at least εn. Hence the number of odd components of G is at most (1/4 + ε)n + 1/ε ≤ |S|, a contradiction.
Their sum is n + 2|S| ≤ 2n. Without loss of generality, it follows that If in addition |X 1 | + |X 2 | ≥ (1/4 + 2ε)n, every vertex in X 3 is in a component of order at least εn. But the number of isolate vertices in X 2 s at most (1/4 + ε)n, so the number of odd components in G is at most (1/4 + ε)n + 1/ε ≤ |S|. We conclude that |X 1 The set Y obtained by picking one vertex from each component of G is an independent set in

Matchings in bipartite graphs
The following lemma is a simple consequence of Hall's theorem for perfect matchings in bipartite graphs.
Then one of the following conditions hold.
• G contains a perfect matching.
contains no edges.
Proof. Suppose that G contains no perfect matching. Then by Hall's Theorem there exists In Section 14, we shall also need the following stability result for graphs with a bipartition satisfying certain conditions. The proof is again an applications of Tutte's theorem, Theorem 7.5.
Lemma 7.9. Let 1 n ε 1, where n is even. Let G be a graph on n vertices, and suppose that , u ∈ X i . Then one of the following conditions holds.
• G has a perfect matching.

• There exists an independent set
• |X i | > |X 3−i | and X i contains an independent set of size at least (1/2 − ε)n.
Proof. Suppose that G has no perfect matching. It follows from Tutte's theorem, Theorem 7.5, that there exists a set S such that the number of odd components in G = G \ S is larger than S.
If δ(G ) ≥ εn, the number of components of G is at most 1/ε, implying that |S| ≤ 1/ε. In this case, the conditions on the graph G[X 1 , X 2 ] imply that either G is connected, contradiction our assumption on S, or G consists of two connected components, implying that |S| ≤ 1, so G is not 2-connected.
We now assume that δ(G ) ≤ εn. It follows that |S| ≥ (1/4 − 2ε)n. Suppose that |X 1 | ≥ (1/4 + 3ε)n and X 2 = ∅. Denote by A 1 the set of isolated vertices in X 1 . Then every vertex in It follows that the number of components is at most and the third conditions of the lemma holds. The case |X 2 | ≥ (1/4 + 3ε)n and X 1 = ∅ follows similarly.
Suppose now that X 2 = ∅. Note that |S| < n/2 because the number of component in G is at most n − |S|. By our assumption, consists of at least (1/2 − ε)n components, in particular it contains an independent set of size at least (1/2 − ε)n, and the fourth condition holds.

Hamilton cycles in bipartite graphs
The following result is a stability version of a special case of Chvátal's theorem, Theorem 7.1, which we shall use in Section 15, for the proof of Lemma 7.13 below.
Lemma 7.10. Let 1 n ε 1 and let G be a balanced bipartite graph on n vertices with bipartition {X 1 , X 2 }. Suppose that δ(G) ≥ (1/4 − ε)n and between every two subsets of X 1 and X 2 of size at least (1/4 − 3ε)n there are at least εn 2 edges. Then G is Hamiltonian. Furthermore, there is a Hamilton path between every pair of points We prove this result by a relatively simple application of the absorbing method of of Rödl, Ruciński and Szemerédi [22]. In fact, all we need for this proof is Lemma 5.5, asserting the existence of short absorbing paths in robust subgraphs, and the Regularity Lemma. A graph G as in Lemma 7.10 is (ε, 2)-weakly robust, thus by Lemma 5.5, it is possible to find an absorbing path P in G. We consider the reduced graph Γ, obtained from applying the Regularity Lemma on the graph G \ V (P ). We deduce from the conditions of the lemma that Γ has an almost perfect matching, implying that G contains a cycle extending P and spanning almost all vertices. The remaining vertices may be absorbed by P .
Proof of Lemma 7.10. It is easy to check from the conditions that G is (ε, 2) weakly robust with bipartition {X 1 , X 2 }. Indeed, let x 1 ∈ X 1 , x 2 ∈ X 2 . Recall that | con G,2 (x 1 , x 2 )| is the number of paths of length three in G between x 1 and x 2 . Then | con G,2 (x 1 , x 2 )| is the number of edges between N (x 1 ) \ {x 2 } and N (x 2 ) \ {x 1 }, which, by the assumptions in the lemma is at least εn 2 .
It follows from Lemma 5.5, that for ρ > 0 is small enough, there exists a ρ 2 n-absorbing path P in G of length at most ρn.
contains a Hamilton path with the same ends as P . Note that we may assume for convenience that P has one end in X 1 and the other in X 2 .
Pick η > 0 suitably small. Apply the regularity lemma, Lemma 6.1, with the graph G \ V (P ), the bipartition {X 1 , X 2 }, where X i = X \ V (P ) and parameter η. Let G be the subgraph of G promised by Lemma 6.1 and let {V 0 , . . . , V m } be the given partition. Denote by m i the number of parts V j (where j ≥ 1) which are contained in X i (so m = m 1 + m 2 ). Without loss of generality, we assume that m 1 ≥ m 2 .
Let Γ be the (η, 4η)-reduced graph defined by the m 2 clusters V i contained in X 2 and by some Proof. Both parts of the claim follow from Condition 4 in Lemma 6.1 and the definition of Γ as long as η and ρ are small enough.
It follows from Lemma 7.8, that Γ contains a perfect matching, which is a connected matching as Γ is connected. By Lemma 6.9, we obtain a cycle C in G, containing the path P and spanning all but at most 6ηn ≤ ρ 2 n vertices.
It follows from the absorbing property of P that the vertices of W may be absorbed into P and thus into C to obtain a Hamilton cycle.
It is easy to modify the proof to obtain a Hamilton path between any given vertices x i ∈ X i .

Monochromatic cycle partitions in 2-coloured graphs with the red graph almost disconnected
The following two lemmas, Lemmas 7.13 and 7.14, state that a 2-coloured graph G has the desired partition into a red cycle and a blue one if G admits some restrictive structural property. We shall use these results several times in the following sections and delay their proofs to the ends of the paper. We prove Lemma 7.13 in Section 15 and Lemma 7.14 in Section 16.
Lemma 7.13. Let 1 n ε 1 and let G be a graph on n vertices with δ(G) ≥ 3n/4 and a 2-colouring E(G) = E(G B ) ∪ E(G R ). Suppose that S, T ⊆ V (G) satisfy the following conditions.
Then V (G) may be partitioned into a red cycle and a blue one.
Lemma 7.14. Let 1 n ε 1 and let G be a graph on n vertices with δ(G) ≥ 3n/4 and a 2- with the following properties.
• The sets S and T belong to different components of G R \ X.
Then V (G) may be partitioned into a red cycle and a blue one.
This concludes the preliminary material needed for the proof of Theorem 1.1. We are now finally ready to turn to the heart of the proof.

Rough structure
In this section we make the first step towards our proof of Theorem 1.1. We use the Regularity Lemma, Lemma 6.1, to obtain information about the rough structure.
Then one of the following assertions holds, where a robust component refers to an (α, k)-robust component, possibly with the roles of red and blue reversed.
1. There exists a monochromatic strongly robust blue component on at least (1 − ε)n vertices.
2. There exists a weakly robust blue component of order at least (1 − ε/4)n and a red strongly robust component of order at least (1/2 + ε/2)n.

There exists a weakly robust blue component with bipartition
≤ εn 2 and one of the following holds.
4. There exist a blue strongly robust component and a red robust component, each has order at least (3/4 − ε)n and together the span all but at most εn of the vertices.
5. There exist sets We remark that in light of the variety of extremal examples for Theorem 1.1 (see Section 2), it should not be surprising that there is a large number of cases to consider for the rough structure. Furthermore, it is perhaps useful to note that many of the above cases describe the structure of the extremal examples we gave in Section 2. For example, the left-hand graph in Figure  This lemma, technical as it seems, is a simple application of Lemmas 6.3 and 6.7, which imply that monochromatic components in the reduced graph correspond to robust components in the original graph. Before proving Lemma 8.1, we give a brief overview of the proof. After applying the Regularity Lemma with suitable parameters, we obtain a reduced graph Γ, which has minimum degree close to 3m/4 where m = |Γ|. It is a routine check to verify that either there is a spanning monochromatic component, or there are two monochromatic components of size almost 3m/4 spanning the whole vertex set, or for each colour there are two almost half-sized components spanning the whole vertex set. In the case where there is a spanning monochromatic component, further analysis is needed to show that one of the cases (1, 2, 3) holds.
Suppose first that Φ 1 is a spanning subgraph of Γ. If it is in addition non-bipartite, by Lemma 6.3, there is a strongly robust blue component F 1 of order at least (1 − 2η)n, as in (1).
] is a red non bipartite component. By Lemma 6.7, we obtain a weakly robust blue component F 1 on at least (1 − 3η)n vertices and a red strongly robust component F 2 on at least (1/2 + 23η)n vertices, as in (2).
1. Γ i is connected in red and non-bipartite.
By the definition of a reduced graph and our choice of parameters, the number of blue edges of G which are not present in G is at most 14ηn 2 and similarly for the red edges. It follows from Lemma 6.7 that one of the conditions in (3) holds. Suppose that Condition (2) (3) holds.
We assume now that Φ 1 does not span Γ. Denote s = |Φ 1 |. Suppose first that s > (1/2 + 26η)m. Denote U = V (Γ)\V (Φ 1 ). Then every vertex u ∈ U , is incident to at least (3/4−13η)m−(m−s) > s/2 neighbours. It follows that every two vertices outside of V (Φ 1 ) have a common red neighbour, implying that U is contained in a red component Φ 2 of order at least (3/4 − 13η)m. Indeed, pick u ∈ U . Then the red neighbourhood of u in Γ is contained in Φ 2 as well as U .
Then for every x 1 ∈ X 1 , x 2 ∈ X 2 , the vertices x 1 , x 2 are non-adjacent in Γ. Thus, x 1 sends at least (3/4 − 13η)m − |X 1 | (blue) edges to Y and similarly x 2 sends at least (3/4 − 13η)m − |X 2 | red edges into Y . If both Φ 1 and Φ 2 are bipartite, it follows that Y contains a set A 1 of (3/4 − 13η)m − |X 1 | vertices spanning no blue edges and a set A 2 of (3/4 − 13η)m − |X 2 | spanning no red edges. It follows that Y contains an independent set of size at least |A 1 This is a contradiction to the minimum degree condition on Γ.
It remains to consider the case where s = |Φ 1 | ≤ (1/2+26η)m. An argument similar to a previous one shows that if s < (1/2−26η)m, every two vertices of Φ 1 have a common red neighbour outside of Φ 1 , contradicting the choice of Φ 1 as the largest monochromatic component. Thus we have that (1/2 − 26η)m ≤ s ≤ (1/2 + 26η)m. Note that we may find u 1 , u 2 ∈ V (Φ 1 ) which have no common red neighbour outside of Φ 1 (otherwise there is a red component of order larger than |Φ 1 | contradicting our choice of Φ 1 ). Denote by X i the set of red neighbours of u i outside of V (Φ 1 ). Let Y i be the red neighbourhood of X i in Φ 1 . It follows from the minimum degree condition and the order of Φ 1 that |X i |, |Y i | ≥ (1/4 − 39η)m. Furthermore, the sets X 1 , X 2 , Y 1 , Y 2 are disjoint, there are no red edges in between X 1 ∪ Y 1 and X 2 ∪ Y 2 and no blue edges between X 1 ∪ X 2 and Y 1 ∪ Y 2 . In particular, there are no edges between X i and Y 3−i . Considering the minimum degree conditions and the size of the various sets, it follows that the blue subgraph Φ Moreover, of the four components, there cannot be both a red and a blue bipartite component. Condition (5) follows.

Proof of the main theorem
We now prove Theorem 1.1, using Lemma 8.1 and other results which we shall state and prove in subsequent sections.
Theorem (1.1). There exists n 0 such that if a graph on n ≥ n 0 vertices and minimum degree at least 3n/4 is 2-coloured then its vertex set may be partitioned into two monochromatic cycles of different colours.
Proof of Theorem 1.1. Let 1 n ε 1 and let G be a graph on n ≥ n 0 vertices with minimum degree at least 3n/4 and a red and blue colouring of the edges. By Lemma 8.1, we may assume that one of the Cases (1 -5) from the statement of the lemma hold. It remains to conclude that in each of these cases, we may find a partition of V (G) into a red cycle and a blue one. We prove this for each of the above cases using lemmas appearing in Sections (9 -14).
We start by resolving Case (2), which is perhaps the easiest to deal with, in Lemma 9.1 in Section 9.
Case (3) is dealt with in Sections 10, 11. Lemmas 10.1, 10.2 and 10.3 share some similarities and are used to prove all possible combinations of Cases (3a, 3b, 3c) except for the case where Condition (3b) holds for both graphs in question. The proof of Theorem 1.1 in the latter case can be deduced from Lemma 11.1 and is of different nature.
Case (4) can be resolved by Lemma 12.1 in Section 12. It shares some ideas with the proofs in Section 10, but requires further analysis. Case (5) is dealt with by Lemma 13.1 in Section 13. Case (1) turns out to be hardest to deal with, thus we prove it last in Lemma 14.1 in Section 14.

Large weakly robust blue component
In this section we deal with Case (2) of Lemma 8.1. This is perhaps the easiest case to deal with. We recommend the reader to follow the proof here carefully, since the methods appearing here will be used in later sections, often in less detail.
In order prove Theorem 1.1 in Case (2), we prove the following lemma. Note that we make the further assumption that the given robust components cover all the vertices of G (rather than almost all of them). This can be easily be justified by Lemma 5.4, which states that given a robust subgraph F , the graph obtained by adding vertices of large degree into F remains robust.
Lemma 9.1. Let 1 n ε, α, 1 k 1 and let G be a graph of order n with δ(G) ≥ 3n/4 and a 2-colouring E(G) = E(G B ) ∪ E(G R ). Suppose that F 1 , F 2 satisfy the following conditions.
Then V (G) may be partitioned into a blue cycle and a red cycle.
This case is the most straightforward of the various cases arising from Lemma 8.1. We apply Lemma 5.5 to find vertex disjoint absorbing paths P i in F i for i ∈ [2]. Then, using the regularity lemma and the connected matching method, we find two vertex-disjoint monochromatic cycles containing the paths P 1 , P 2 and covering almost all of the vertices. Finally, we use the absorption property to insert the remaining vertices into the paths P 1 , P 2 so as to obtain the desired monochromatic cycle partition.
Proof of Lemma 9.1. We use Lemma 5.5 to build absorbing paths for F 1 and F 2 . More precisely, given a suitably small ρ > 0, there exists vertex disjoint paths P i ⊆ F i for i ∈ [2] satisfying the following conditions, where U = V (P 1 ) ∪ V (P 2 ).
has a Hamilton path with the same ends as P 1 .
• For every set W ⊆ V (F 2 ) \ U of size at most ρ 2 n, the graph F 2 [V (P 2 ) ∪ W ] has a Hamilton path with the same ends as P 2 .
Indeed, by Lemma 5.5, we may find a ρ 2 n-weakly-absorbing path P 1 in F 1 of length at most ρn. By Lemma 5.2, the subgraph F 2 = F 2 \ V (P 1 ) is robust (with suitable parameters), thus we may find a ρ 2 n absorbing path P 2 in F 2 of length at most ρn.
Using the regularity lemma, we shall find vertex-disjoint monochromatic cycles, containing the paths P 1 and P 2 and covering most vertices of G. We then cover the remaining vertices using the absorption properties of the paths P 1 and P 2 . However, the fact that P 1 is weakly-absorbing presents a technical difficulty which we overcome as follows.
Claim 9.2. Γ has a perfect matching.
We conclude from Theorem 7.1 that Γ has a Hamilton cycle and in particular a perfect matching.
Remark. In the last claim, we implicitly assumed that m is even. It is indeed possibly to make this further assumption in the Regularity Lemma. We shall make this assumption whenever convenient without stating so explicitly.
By Claim 9.2, Γ has a perfect matching consisting of a blue connected matching in Φ 1 and a red connecting matching in Φ 2 . Thus, we may use Lemma 6.9 to obtain a blue cycle C 1 and a red cycle C 2 which are disjoint, each C i contains the respective absorbing path P i and together they cover all but at most 7ηn ≤ ρ 2 n/4 vertices of V (G) \ A.
We now show how to absorb the leftover vertices into the cycles C 1 , C 2 . Let B be the set of vertices which are not contained in the cycles C 1 , C 2 or in the set A (so |B| ≤ ρ 2 n/4). Denote are disjoint of the cycles C 1 , C 2 and have sizes ρ 2 n/2 and ρ 2 n/4 respectively. It follows that ρ 2 n/4 ≤ |A 2 ∪ B 2 | ≤ ρ 2 n/2. Thus we may choose A 1 ⊆ A 1 such that A 1 ∪ B 1 and A 2 ∪ B 2 are of equal size which is most ρ 2 n/2. Recall that the path P 1 , which is contained in C 1 is ρ 2 nabsorbing in F 1 , so these sets can be absorbed by P 1 and thus by C 1 . We remain with the vertices (A 1 \ A 1 ) ∪ B 3 . There are at most ρ 2 n of them and they belong to F 2 , so we may absorb them into P 2 and thus into C 2 . This completes a partition of V (G) into a red and a blue cycle. Now that we have proved Theorem 1.1 in Case (2) of Lemma 8.1, we are ready to consider harder cases.

Large weakly robust blue graph with almost balanced bipartition
In this section we consider Condition (3)  The main idea in the various cases arising here is that if two red robust components may be joined by two vertex-disjoint red paths, then they can essentially be treated as one bigger component, at which case we may use the argument of the previous section to finish the proof. If that is not possible, we deduce that the red graph G R may be disconnected by removing a small number of vertices into two almost half-sized subgraphs. This case may be resolved by Lemmas 7.13 and 7.14.
The following lemma deals with the case where one of the red graphs in question satisfies condition (3a) and the other satisfies (3a) or (3b).
Then V (G) may be partitioned into a blue cycle and a red cycle.
To prove this lemma, we show that we may either join the two robust components F 2 , F 3 or we can finish using Lemma 7.13.
Proof of Lemma 10.1. By Menger's theorem, one of the following holds.
1. There exist two vertex-disjoint red paths P 2 , P 3 , each having one end in F 2 and the other in F 3 .

There exists
In Case (2), it is easy to deduce that the required monochromatic cycle partition exists from Lemma 7.14.
It remains to consider Case (1). The idea is to use the paths P 2 and P 3 so as to essentially connect the two components F 2 , F 3 into one large component. We achieve this as follows.
Note that we may assume that the internal vertices of P 2 , P 3 belong to V (G) \ (V (F 2 ) ∪ V (F 3 )). It follows that |P 2 |, |P 3 | ≤ 2εn and the components F 1 , F 2 , F 3 remain robust after removing the vertices of the paths P 2 , P 3 . Note that every vertex has either large (say, at least αn) blue degree into F 1 , or large red degree into either F 2 or F 3 . It follows that we may extend these components to cover the remaining vertices. Namely, using Lemma 5.4, we obtain (α/2, k+2)-robust components F 1 , F 2 , F 3 which extend the given components and cover V (G) \ (V (P 2 ) ∪ V (P 3 )). Denote by {X , Y } the bipartition of F 1 .
By Lemma 5.5, F i contains a ρ 2 n-absorbing path P i of length at most ρn for each i ∈ [3]. We may assume that the paths P 1 , P 2 , P 3 are vertex disjoint. For i ∈ {2, 3}, we may connect P i with P i in F i to a path Q i using at most k + 2 additional vertices. For convenience, we denote Q 1 = P 1 . To conclude, the paths Q 1 ⊆ G B and Q 2 , Q 3 ⊆ G R are vertex-disjoint paths, such that Q i is ρ 2 n-absorbing in F i . Furthermore, each of Q 2 , Q 3 has one end in F 2 and one in F 3 .
Recall that F 1 is weakly robust and that F 2 is strongly robust. It is perhaps not evident from the definitions that a strongly robust graph is weakly robust, but for our purpose, the place where the difference is important is in the definition of an absorbing path. But clearly, a weakly absorbing path is strongly robust. Thus, it suffices to consider the case where F 3 is weakly robust with bipartition {Z 1 , Z 2 }.
Similarly to the proof in Section 9, in order to overcome the technical issues arising when dealing with weakly robust components, we pick sets A 1 , A 2 , A 3 such that • |A 1 | = ρ 2 n/16, |A 2 | = ρ 2 n/4 and |A 3 | = ρ 2 n/2.
We now consider the (η, 6η)-reduced graph obtaining by applying the Regularity Lemma, Lemma 6.1, to the graph obtained from G by removing the vertices in the Q i 's and A i 's (with the cover . The reduced graph Γ consists of a large blue component Φ 1 (containing almost all vertices) and two disjoint almost half-sized connected red subgraphs Φ 2 , Φ 3 . It is easy to verify, similarly to the proof of Claim 9.2, using Theorem 7.1, that Γ has a perfect matching consisting of edges in Φ 1 , Φ 2 and Φ 3 .
By Lemma 6.9, there exist a blue cycle C 1 and a red cycle C 2 such that • They cover all but at most 4ηn vertices of V (G) \ A.
• C 1 contains the path Q 1 and C 2 contains the paths Q 2 , Q 3 .
Let us elaborate slightly more on how to obtain the required cycles by pointing out that C 2 may be obtained by connecting the ends of Q 2 , Q 3 by two paths, one in F 2 and the other in F 3 .
Let B be the set of vertices which do not belong to the cycles C 1 , C 2 or to A (so |B| ≤ 7ηn ≤ ρ 2 n/16) and denote We perform the following steps in order to absorb B.
The following lemma deals with the case where one of the red graphs in question satisfies Condition (3c) in Lemma 8.1 and the other satisfies one of the other two conditions. Lemma 10.2. Let 1 n ε, α, 1 k 1 and let G be a graph of order n with δ(G) ≥ 3n/4 and a 2-colouring E(G) = E(G B ) ∪ E(G R ). Suppose that F 1 , F 2 , F 3 , F 4 satisfy the following assertions.
Then V (G) may be partitioned into a blue cycle and a red cycle.
The proof of this lemma is similar to the previous one. Either F 2 may be joined to one of F 3 , F 4 or we may finish using Lemma 7.13.

Proof of Lemma 10.2.
Similarly to the previous case, two possibilities arise.
1. There exist vertex-disjoint red paths P 1 , P 2 with one end in F 2 and either both have the other end in F 3 or both have the other end in F 4 .

There exists a set U of size at most 2 such that
In Case (2), the required monochromatic cycle partition exists by Lemma 7.13. Indeed, we may find S ⊆ X \ U and T ⊆ Y \ U such that the three conditions in the lemma hold, for some parameter η = η(ε). In particular, the third condition holds because most vertices in Y have degree at most (1/4 + √ ε)n in G[Y ], and thus have degree at least (1/2 − √ ε)n into X.
Case (1) may be dealt with similarly to the previous proof of Lemma 10.1, we omit further details.
The following lemma deals with the remaining case, where both red graphs satisfy condition (3c) in Lemma 8.1.
Then V (G) may be partitioned into a blue cycle and a red cycle.
To prove this lemma we follow similar ideas to the previous results in this section. We show that either at least three of the four components F 2 , F 3 , F 4 , F 5 may be joined or we may finish using Lemma 7.13 or Lemma 7.14.
Proof of Lemma 10.3. We consider four cases. In order to be able to distinguish between them, we define a graph H on vertex set {2, 3, 4, 5} with an edge ≥ εn 2 (so H is a bipartite graph on four vertices with bipartition { [2,3], [4,5]}. Clearly, one of the following conditions.
1. H contains a path of length 2.
2. H consist of two vertex-disjoint edges.
3. H has exactly one edge.

H is the empty graph.
In Case (1), without loss of generality suppose that (2,4), (2,5)  We now suppose that Case (2) holds. Without loss of generality, E(H) = {(2, 4), (3,5)}. If there are two vertex-disjoint red paths P 1 , P 2 between V (F 2 ) ∪ V (F 4 ) and V (F 3 ) ∪ V (F 5 ), we conclude that the four component F 2 , F 3 , F 4 , F 5 may be joined (note that to join F 2 to F 4 and F 3 to F 5 we use edges between them which can be chose so as to not intersect the given paths P 1 , P 2 ). Otherwise, there is a vertex u ∈ V (G) such that G R \ {u} is disconnected, with F 2 , F 4 in one component and F 3 , F 5 in another. The proof of Lemma 10.3 can be completed by Lemma 7.14.
We now assume that Case (3) holds. Without loss of generality, If it contains a matching of size at least 5, then the one of F 2 , F 4 may be joined to one of F 3 , F 5 . Since there are many edges between F 2 and F 4 , it follows that we may form a red component using F 2 , F 4 and one of F 3 , F 5 . The proof may be completed as before. Thus we assume that the above graph has no matching of size 5. It follows that we may remove four vertices from G so as to disconnect F 2 and F 4 from F 3 and F 5 . Thus there exists a set U ⊆ V (G) of size at most 3εn such that F 2 and F 4 are disconnected from F 3 and F 5 . Lemma 10.3 now follows from Lemma 7.13 (note that most vertices in V (F 3 ) ∪ V (F 5 ) have red degree at most about n/4).

Finally, we consider Case (4). Denote
Since H is the empty graph, we have |U | ≤ 5 √ εn, thus we may complete the proof by Lemma 7.13.
In order to finish the proof of Theorem 1.1 under the assumption that Condition (3) from Lemma 8.1, we need to consider the case where both graphs in question satisfy Condition (3b). This is done in the next section, Section 11.

Almost balanced large weakly robust blue component and two half-sized weakly robust red components
In this section we consider Case (3) Lemma 11.1. Let 1 n ε 1 and let G be a graph of order n with δ(G) ≥ 3n/4 and with a 2-colouring E(G) = E(G B ) ∪ E(G R ). Suppose that there exists four disjoint sets Y i,j , i, j ∈ [2] with the following properties.
We notice that a graph with the given conditions has a rather specific structure. Namely, the sets Y i,j span few edges, whereas the graphs G R [Y i,1 , Y i,2 ] and G B [Y 1,j , Y 2,j ] are almost complete. By Lemma 7.13, we conclude that we may finish the proof unless say G R [Y 1,1 , Y 2,2 ] and G B [Y 1,2 , Y 2,1 ] are almost complete. In the latter case we construct the required partition into a red cycle and a blue one "by hand".
Proof of Lemma 11.1. The conditions imply that for some η = η(ε), we can find disjoint sets S 1 , S 2 , T 1 , T 2 such that the following holds. .
Proof. Suppose to the contrary that |S 1 ∩ S |, |T 1 ∩ T | ≥ 10 √ ηn. For every u ∈ S 1 , the number of vertices in T 1 which are not blue neighbours of u is at most 4ηn. If S ∩ S 1 and T ∩ T 1 both have size at least 10 √ η, we deduce e(G B [S , T ]) ≥ 90ηn, a contradiction.
By the above claim, without loss of generality, we may assume that |S 2 ∩ S |, |T 1 ∩ T | ≤ 10 √ ηn, so G B [S 1 , T 2 ] is almost empty, and G R [S 1 , T 2 ] is almost complete. Similar arguments imply that we may assume that G B [S 2 , T 1 ] is almost complete.
We deduce that there exists ρ = ρ(ε) such that we may find disjoint sets A 1 , A 2 , A 3 , A 4 satisfying the following conditions.
For the time being, we assume that n is even. We obtain a partition Note that every vertex will be added to one of the A i 's.
Denote m i = |A i |. We will find values k 1 , k 2 , k 3 and l 1 , l 2 , l 3 such that there exist a partition of V (G) into a blue cycle C 1 and a red cycle C 2 with the following properties.
To that end, we find blue paths P 1 , P 2 , P 3 , P 4 forming a blue cycle C 1 = P 1 P 2 P 3 P 4 such that the following assertions hold.
• P 1 ∈ G B [A 1 , A 2 ], its ends are in A 2 and it has k 1 vertices in A 1 .
• P 2 , P 4 ∈ G B [A 2 , A 3 ], both have one end in A 2 and the other in A 3 and together they have k 2 + 1 vertices in A 2 .
• P 3 ∈ G B [A 3 , A 4 ], its ends are in A 3 and it has k 3 vertices in A 4 .
Similarly, we find red paths Q 1 , Q 2 , Q 3 , Q 4 forming a red cycle C 2 = Q 1 Q 2 Q 3 Q 4 such that C 1 and C 2 partition V (G) and the following assertions hold.
• Q 1 ∈ G R [A 1 , A 3 ], its ends are in A 1 and it has l 1 vertices in A 3 .
• Q 2 , Q 3 ∈ G R [A 1 , A 4 ], both have one end in A 1 and the other in A 4 and together they have l 2 + 1 vertices in A 1 .
• Q 4 ∈ G R [A 2 , A 4 ], its ends are in A 2 and it has l 3 vertices in A 2 .
The values k 1 , k 2 , k 3 and l 1 , l 2 , l 3 clearly need to satisfy the following system of equations.
Which may be solved as follows.
Since n is even, if k 3 , l 3 are integers then so are k 1 , k 2 , l 1 , l 2 . Note that the following inequalities hold.
We pick l 3 = k 3 = 12ρn. It follows that 10ρn ≤ l 1 , k 1 ≤ 14ρn. We now choose the paths P 1 , P 2 , P 3 as follows. Let P 2 be any edge starting with u 2 and ending in some u 1 ∈ A 2 , such that P 1 contains k 1 vertices from A 1 . Similarly, let P 3 be a path in G B [S 3 , A 4 ] with k 3 vertices from A 4 and ends u 3 and u 4 where u 4 is some vertex in A 4 .
We now construct Q 1 , Q 2 , Q 3 as follows (we implicitly ensure that these paths are disjoint of the paths P 1 , P 2 , P 3 ). Let Q 2 be any edge containing all vertices A 1 \ A 1 , with ends v 2 and v 1 ∈ A 1 , and with l 1 vertices from A 3 . We remark that such a path exists. Indeed, we have |A 1 \ A 1 | ≤ 5ρn and any two vertices in A 1 \ A 1 may be connected using at most three additional vertices from A 1 ∪ A 3 . Thus we may find a path with at most 10ρn vertices from A 1 starting with v 2 and containing the vertices A 1 \ A 1 . We extend it arbitrarily to the desired length. Similarly, pick a path Q 3 ∈ G R [A 2 , A 4 ] containing the vertices A 4 \ A 4 , with ends v 3 and v 4 ∈ A 4 and with l 3 vertices in A 2 .
Denote by U the set of inner vertices in the paths P 1 P 2 P 3 and Q 1 Q 2 Q 3 and denote A i = A i \ U for i ∈ [4]. By the definition of the paths P i and Q i , i ∈ [3] we obtain the following equalities.
It is easy to verify that G B [A 2 , A 3 ] contains a Hamilton path P 4 with ends u 1 , u 4 and that G R [A 1 , A 4 ] contains a Hamilton path Q 4 with ends v 1 , v 4 .
It remains to consider the case where n is odd. If there exists a vertex u which has blue neighbours v 1 ∈ A 1 ∪ A 3 and v 2 ∈ A 2 ∪ A 4 , we consider the graph G \ {u} and partition it into a red cycle and a blue path with ends v 1 , v 2 . This may be done by the same arguments as when n is even. Thus we assume that no vertex has blue neighbour in both A 1 ∪ A 3 and A 2 ∪ A 4 . By symmetry, we may also assume that no vertex has red neighbours in both A 1 ∪ A 2 and A 3 ∪ A 4 . It follows that every vertex has neighbours in at most three of the sets A 1 , A 2 , A 3 , A 4 , so it sends at least (1/4 − 2ρ)n edges to each of these three sets. We extend the sets A 1 , A 2 , A 3 , A 4 as follows. For It is easy to verify that the following claim holds.
Claim 11.3. The sets A 1 , A 2 , A 3 , A 4 partition V (G) and satisfy the following conditions.
Without loss of generality, suppose that |A 1 | ≥ n/4. Then in fact, |A 1 | > n/4, because n is odd. Hence G[A 1 ] contains an edge uv. If it is blue, u has blue neighbours in both A 1 and A 2 . If it is red, u contains red neighbours in both A 1 and A 4 . We may continue as before, to partition G \ {u} into a monochromatic cycle and a monochromatic path.
The proof of Lemma 11.1 completes the proof of our main Theorem under the assumption that Case (3) from Lemma 8.1 holds.
12 Two monochromatic robust components of size almost 3n/4 In this section, we consider Case (4) from Lemma 8.1. Similarly to Section 9, we may assume that the given robust components cover V (G).
Lemma 12.1. Let 1 n ε, α, 1 k 1 and let G be a graph of order n with δ(G) ≥ 3n/4 and a 2-colouring E(G) = E(G B ) ∪ E(G R ). Suppose that F 1 , F 2 satisfy the following assertions.
Then V (G) may be partitioned into a blue cycle and a red cycle.
We proceed as before, building absorbing paths, and considering the reduced graph on the remaining vertices, where we have a blue component and a red one, each with almost 3/4 of the vertices. If a perfect matching can be found using the edges in these component, we continue as before to obtain the required partition into cycles. If no such perfect matching exists, we conclude that the graph G satisfies some structural conditions which enable us to either find the required partition "by hand", or to join two components in a similar way to previous cases.
Proof of Lemma 12.1. Let Q 1 , Q 2 be disjoint ρ 2 n-absorbing paths in F 1 , F 2 respectively of length at most ρn. As pointed out in Section 10 we may assume that F 2 is weakly robust. Denote its bipartition by {X, Y }.
Note that, assuming ρ, η are small enough, we have δ(Γ) ≥ (3/4 − ε)m, where m = |Γ|. Furthermore, there is a blue component Φ 1 and a red component Φ 2 which cover V (Γ) and are of order at least (3/4 − 2ε)m each. We consider the subgraph Γ of Γ spanned by the blue edges in Φ 1 and the red ones in Φ 2 .
Consider the following claim.
Claim 12.2. One of the following conditions holds.
2. The following holds for some η, β, l depending only on ε, α, k. There exist subsets are (β, l)-robust, with at least one of them being strongly-robust. Let W 0 be the intersection of the blue neighbourhood of W 1 in U 0 and the red neighbourhood of W 2 in U 0 . By the definition of the sets U i , there are no edges between U 1 and U 2 , there are no red edges between U 1 and U 0 and no blue edges between U 2 and U 0 . It follows that the each vertex in U 1 has at least (1/2 − 10ε)m blue neighbours in U 0 , and by the analogous argument for are connected and it is not hard to see that at least one of them is non-bipartite. It follows from Lemma 6.7 that condition (2) holds.
Similarly to the proofs in Section 10, one of the following holds.
1. There are two vertex-disjoint blue paths between V 2 and V 1 ∪ V 0 .
2. There are two vertex-disjoint red paths between V 1 and V 2 ∪ V 0 .
3. There exists a set X of size at most 2 such that the sets V 2 and V 1 ∪ V 0 are not connected in G B \ X and the sets V 1 and V 2 ∪ V 0 are not connected in G R \ X.
In the first two cases we proceed as in Section 10 to join say the two blue component to form an almost spanning component, which together with the large red component (V 2 ∪ V 0 ) may be used to find the required cycle partition. Thus we assume that the Case (3) holds.
Claim 12.3. There exists θ = θ(ε) such that W may be partitioned into sets W 1 , W 2 satisfying the following conditions.
• G B [W 1 ] contains a blue path of length k.
• Pick any partition {W 1 , W 2 } with the following properties.
Such a partition satisfies the required conditions of Claim 12.3 with θ = √ η.

Four half-sized robust components
In this section we consider Case (5) 3,4].
Then V (G) may be partitioned into a blue cycle and a red one.
We follow similar ideas to previous sections. If there exist four vertex-disjoint paths, two of which are blue and connect F 1 with F 2 and two are red and connect F 3 with F 4 , then we may continue as in previous sections, by essentially having two robust components, one red and one blue, which both span almost all the vertices. The main effort in this case goes into showing that if such paths do not exist, the desired partition may be found by Lemma 7.14.
Proof of 13.1. We extend the components F i as follows. For every vertex v not in any of the components, if there exist i ∈ {1, 2}, j ∈ {3, 4} such that v sends at least αn blue edges to F i and at least αn red edges to F j , we add v to F i and F j . Note that if no such i, j exist, then v either blue degree at least (3/4 − 3ε)n or red degree at least (3/4 − 3ε)n.
Note that the obtained components satisfy the conditions above (though with relaxed parameters α, k in the definition of robustness and with say 2ε instead of ε). We abuse notation by denoting the modified components by F 1 , F 2 , F 3 , F 4 . So in addition to the above conditions, we have that every vertex not in V (F 1 ) ∪ V (F 2 ) has either blue degree or red degree at least (3/4 − 3ε)n.
We claim that one of the following assertions holds.
1. There exist vertex-disjoint paths P 1 , P 2 , P 3 , P 4 such that P 1 , P 2 are blue paths from F 1 to F 2 and P 3 , P 4 are red paths from F 3 to F 4 .
2. There exist two vertices u, v such that F 1 , F 2 belong to different connected components of G B \ {u, v}. Furthermore, v sends at most εn blue edges to either F 1 or F 2 .
3. There exist two vertices u, v such that F 3 , F 4 belong to different connected components of G R \ {u, v}. Furthermore, v sends at most εn red edges to either F 3 or F 4 .
Condition (1) implies that we may connect F 1 and F 2 using P 1 , P 2 to obtain a large robust blue component, and similarly we may connect F 3 and F 4 using the paths P 3 , P 4 to obtain a large strongly robust red component. We may continue as in Section 10 to obtain the desired partition of V (G) into a red cycle and a blue one.
If one of Conditions (2, 3) holds, we may find the desired partition into a red cycle and a blue one by Lemma 7.14.
It remains to prove that indeed, one of the above three cases holds. We call a vertex blue if it sends at least αn blue edges to both F 1 and F 2 . Similarly, a vertex is red if it sends at least αn red edges to both F 3 and F 4 . We show that either one of the above three conditions holds, or there are at least four vertices which are either blue or red.
If true, we conclude that one of the three conditions above holds. It is easy to verify that if there exist four vertices, two of them red and two blue, we may find paths as Condition (1). Thus we may assume that there is at most one red vertex. This implies that either Condition (3) holds, or there are three vertex disjoint red paths P 1 , P 2 , P 3 between F 3 and F 4 . We may assume that the inner vertices of these path are not in V (F 3 )∪V (F 4 ). If there exist four blue vertices u 1 , u 2 , u 3 , u 4 , without loss of generality, u 1 , u 2 / ∈ V (P 1 ) ∪ V (P 2 ) and we may find two vertex-disjoint blue paths between F 1 and F 2 which are disjoint of P 1 , P 2 using the blue vertices u 1 , u 2 (we can take them to be of length 2 and be centred at u 1 and u 2 ). If there exist three blue vertices u 1 , u 2 , u 3 and a different red vertex v, without loss of generality, the path P 1 has length 2, is centred at v and avoids u 1 , u 2 , u 3 . We may further assume that P 2 does not contain u 1 , u 2 . It follows that we may find two vertex-disjoint blue paths, disjoint of P 1 , P 2 , between F 1 and F 2 .
It remains to show that indeed, if Conditions (1 -3) do not hold, there are at least four vertices which are either red or blue. Clearly, each vertex v ∈ V (G) \ (F 1 ∪ F 2 ) is either red or blue. Let l be the number of these vertices. Denote Consider the following claim.
Claim 13.2. Suppose that |A i | ≥ n/4 − 1 + k where 1 ≤ k ≤ 4. Then either one of the Conditions (1 -3) from above holds or A i contains at least k vertices which are either blue or red.
Using the claim, if the above conditions do not hold and there are at most three vertices which are either blue or red, we have l ≤ 3 and the total number of vertices is at most 4(n/4−1)+(3−l)+l ≤ n − 1, a contradiction. It remains to prove the claim.
Proof of Claim 13.2. Without loss of generality, i = 1. Note that if G[A 1 , A 4 ] has a matching of size 7, Condition (1) holds. Indeed, let M be such a matching. If in such a matching at least two edges are red and at least two are blue, condition (1) holds, since each edge connects F 1 and F 2 as well as F 3 and F 4 . Thus we may assume that at most one edge is red.
Recall that there exist vertex-disjoint red paths P 1 , P 2 between F 3 and F 4 (otherwise Condition (3) holds). We assume that the inner vertices of P 1 and P 2 do not belong to V (F 3 ∪ F 4 ) = V (F 1 ∪ F 2 ), so each of the paths P 1 and P 2 intersects at most two edges of M . It follows that there exist two blue edges e 1 , e 2 ∈ M which are disjoint of P 1 and P 2 , and Condition (1) holds. It remains to consider the case where G[A 1 , A 4 ] has no matching of size 7.
We deduce that there is a set U ⊆ A 1 ∪ A 4 of at most six vertices which intersects each of the edges of G[A 1 , A 4 ]. Note that by the minimum degree conditions, every vertex in A 4 has at least k neighbours in A 1 . Hence, We conclude that at least k vertices in U ∩ A 1 have at least n/25 neighbours in A 4 . Indeed, otherwise we have Each of the vertices in A 1 with at least n/25 neighbours in A 4 is either red or blue. E.g. if it sends at least αn blue edges to A 4 is blue.
14 Large strongly robust blue component In this section, we resolve Case (1) from Lemma 8.1, which is the last remaining case. Lemma 14.1. Let 1 n ε, α, 1 k 1 and let G be a graph of order n with δ(G) ≥ 3n/4 and a 2-colouring E(G) = E(G B ) ∪ E(G R ). Suppose that F is a blue (α, k)-strongly robust component on at least (1 − ε)n vertices. Then V (G) may be partitioned into a blue cycle and a red cycle.
In our proof of Lemma 14.1, we extend F to include all vertices which send a fairly large number of blue edges into F , and denote the set of remaining vertices by Z. We consider two cases, according to the size of Z.
The case where Z is relatively small turns out to be harder. We apply the Regularity Lemma on F and prove a structural result on the reduced graph, focusing on ways to obtain perfect matchings. In each of the cases for the structure of the reduced graph, we can partition almost all of the vertices into a red cycle and a blue one. The vertices of Z may be covered using their large degree and the leftover vertices of F may be absorbed as usual.
In the case where Z is large, the reduced graph can easily be seen to have a perfect matching consisting of a connected blue matching and a connected red matching. We have to be slightly more careful than usual when obtaining cycles from the connected matching so as to cover Z. The leftover vertices of F can be absorbed as usual.
Proof of Lemma 14.1. Let F 1 be the graph obtained by adding to F the vertices in G with at least αn blue neighbours in F . By Lemma 5.4, F 1 is (α 3 /2, k + 2) strongly robust. Denote It follows that every two vertices in Z have at least (1/2 − 4ε)n red neighbours in common. We consider two cases according to the size of Z. Pick β > 0 by Lemma 5.2 so that F 1 remains (α 3 /4, k + 2) strongly robust after removing at most βn vertices.

Case 1: |Z| ≥ βn
Let Q be a ρ 2 n-absorbing path in F 1 of length at most ρn. Pick a suitably small η, and let G be the subgraph obtained by applying Lemma 6.1 with the graph G \ V (Q), the partition {V (F 1 ), Z} and parameters η and d = 6η. Let Γ be the corresponding reduced graph. As usual, Let Φ 1 be the blue subgraph spanned by the clusters which are contained in V (F 1 ). Then |Φ 1 | ≥ (1 − 2ε)m and Φ 1 is connected by Lemma 6.8. The vertices in clusters contained in Z have red degree at least (3/4 − 4ε)m in Γ (note that since |Z| ≥ βn ≥ 2ηn, there are such clusters). In particular, Γ contains a red component Φ 2 of order at least (3/4 − 4ε)m. It is easy to check that Φ 1 ∪ Φ 2 has a perfect matching, e.g. by Theorem 7.1.
Let U 1 be the set of vertices in G belonging to clusters of the blue matching and let U 2 be the set of vertices in G belonging to clusters of the red matching. We obtain the required partition into a blue cycle and a red cycle as follows.
Suppose first that |U 2 | ≤ 3n/8. Fix two vertices z 1 , z 2 ∈ Z. Let W be a set of vertices in G containing 3η of the vertices of each cluster in V (F 1 ). Clearly, z 1 , z 2 each have many neighbours in the set of vertices belonging to the clusters of Φ 2 . Thus, by Lemma 6.9, there is a red path P 1 in G between z 1 and z 2 spanning at least (1 − 6η) of the vertices of U 2 \ W and using at most m 2 other vertices.
Denote by Z the set of vertices in Z which are not covered by this path. Note that |Z | ≤ 9ηn. Furthermore, every two vertices of Z ∪ {z 1 , z 2 } have at least n/16 common red neighbours in V (G ) \ U 2 . It is thus possible to find a path in G \ (V (P 1 ) ∪ U 1 ) between z 1 and z 2 , containing Z and using at most 200η of the vertices of each cluster. Indeed, such a path can be constructed greedily. Suppose Z = {z 3 , . . . , z t }. For i = 2, . . . , t we find a common red neighbour of z i , z i+1 (where subscripts are taken modulo t) which was not used before and which does not belong to a cluster in U 2 with at least 200η of its vertices already used. Note that this would give the required red path, completing P 1 to a red cycle C. In each step, there are at least n/16 possible choices, out of which at most 9ηn were already used, and at most 9ηn/200η < n/20 belong to clusters for which at least 200η of its vertices are used. Thus it is possible to choose a suitable vertex.
We now construct a blue cycle which is disjoint of C, contains Q and misses at most 210η of the vertices vertices of each cluster in U 1 , using Lemma 6.9 (we use the vertices of W to connect cluster pairs, as described after the statement of Lemma 6.9). The missing vertices may be absorbed by Q, completing the desired cycle partition.
Suppose now that |U 2 | ≥ 3n/8. By Lemma 6.9, there exist a blue cycle C 1 and a red cycle C 2 which are disjoint and C i covers all but at most 9η of the vertices of U i for i ∈ [2]. In particular, the red cycle has length at least 5n/16. Let Z be the set of vertices of Z which are not covered by either of the cycles. We show how to obtain a red cycle To that end, we claim that the vertices of Z can be inserted one by one, such that in each stage at most 20 of the original vertices of C are removed, none of them from Z. If z cannot be inserted as explained, the number of red neighbours of z in the cycle is at most 40|Z| + n/20. But every vertex z ∈ Z has at least (1/16 − 3ε)n red neighbours in the cycle obtained from C 2 , as long as it has length at least (5/16 − 60η)n, implying that z may indeed be inserted. There are at most 20|Z | + 9ηn ≤ ρ 2 n vertices missing from V (C 1 ) ∪ V (C 2 ), all of them from V (F 1 ). They can be absorbed by Q.
We shall use the following proposition. Proposition 14.2. One of the following assertions holds.
1. Γ B has a perfect matching.
2. There exists a red component Φ on at least (1/2 − 50ε)m vertices such that Γ B ∪ Φ has a perfect matching.
3. There exist disjoint subsets X 1 , 4. There exist disjoint subsets X 1 , X 2 , Y 1 , Y 2 of size at least (1/4 − 500ε)m such that Before proving Proposition 14.2, we show how to complete the proof of Lemma 14.1 in this case using the proposition.

Γ B has a perfect matching
Suppose that Γ B has a perfect matching. Recall that Γ B is connected, so this matching is connected. We complete Z ∪ V (P ) to a red cycle C 1 using at most four additional vertices of F 3 . By Lemma 6.9, there exists a blue cycle C 2 , disjoint of C 1 which extends the absorbing path Q and contains all but at most 6ηn ≤ ρ 2 n vertices of F 3 . The remaining vertices can be absorbed by Q to obtain a blue cycle C 2 . The cycles C 1 , C 2 form the required cycle partition.
An almost half-sized red component whose union with Γ B has a perfect matching Suppose that Φ is a red component of size at least (1/2 − 50ε)m and Γ B ∪ Φ has a perfect matching. Define a red path P as follows. If |Z| ≥ 2, extend P to a path P containing Z with ends z 1 , z 2 ∈ Z (we may do this using at most three additional vertices). If |Z| = 1, take P to be the path (z) where Z = {z}. If Z = ∅, take P to be the empty path. Note that every vertex of Z sends many red edges to the set of vertices contained in the clusters defined by V (Φ). By Lemma 6.9, there exist vertex-disjoint cycles C 1 , C 2 such that C 1 is blue and contains the absorbing path Q and C 2 is red and contains the path P . Furthermore, the cycles C 1 , C 2 cover all but at most ρ 2 n vertices belonging to F 3 , which may be absorbed by Q, completing the desired partition into a red cycle and a blue one.
Two almost quarter-sized red components whose union with Γ B has a perfect matching Let X 1 , X 2 ⊆ V (Γ) be disjoint sets of size at least (1/4−500ε)m satisfying the following conditions.
• X 1 ∪ X 2 is independent in Γ B .
• Γ R [X i ] is connected and Γ B ∪ Γ R [X i ] has a connected matching for i ∈ [2].
Note that we may assume that Γ R [X 1 ∪ X 2 ] is not connected, since otherwise we may proceed as in the previous case.
Let U i be the set of vertices contained in the clusters in X i . Note that |U i | ≥ (1/4 − 501ε)n.
We define a path P as follows. If |Z| ≥ 3, without loss of generality, z 1 , z 2 send at least 2ηn red edges into U 1 . Extend P to a path P containing Z with ends z 1 , z 2 (using at most three additional vertices). If |Z| = 1, denote Z = {z} and take P = (z), and suppose without loss of generality that z has at least 2ηn red neighbours in U 1 . If Z = ∅, take P to be the empty path. As before, since Γ B ∪ Γ R [X 1 ] has a perfect matching, Lemma 6.9 implies that there exist disjoint cycles C 1 , C 2 such that C 1 is blue and contains Q and C 2 is red and contains P , and together they cover all but at most ρ 2 n vertices of F 3 , which may be absorbed by C 1 .
It remains to consider the case where |Z| = 2. Denote Z = {z 1 , z 2 }. If for some i ∈ [2], both z 1 , z 2 have at least 2ηn red neighbours in U i , we may continue as before by taking P to be a path of length 2 connecting z 1 , z 2 . Thus we assume that Recall that deg R (z i ) ≥ (3/4 − 4ε)n for i ∈ [2]. It follows that |U 1 |, |U 2 | ≤ (1/4 + 5ε)n. Let W be the set of common red neighbours of z 1 and z 2 in V (G) \ (U 1 ∪ U 2 ). Then |W | ≥ (1/2 − 20ε)n.
Suppose that there exists a vertex w 1 ∈ W with at least 4ηn red edges into U 1 ∪ U 2 . Without loss of generality, w 1 has at least 2ηn red neighbours in U 1 . Take P = (w 1 z 2 w 2 z 1 ) for some w 2 ∈ W and continue as before (when |Z| = 2) to conclude that the desired partition exists.
We now assume that every vertex in W has at most εn red neighbours in U 1 ∪U 2 . Furthermore, by the definition of the reduced graph, every vertex in U 1 ∪ U 2 has at most 9ηn ≤ εn red neighbours in W . Note that since e(Γ[X 1 , X 2 ]) = 0, the graph G B [U 1 ∪ U 2 , W ] is almost complete. It follows that we may apply Lemma 7.13 with parameter 1002ε, to conclude that V (G) may be partitioned into a red cycle and a blue one.

Four half-sized monochromatic components
Suppose that X 1 , X 2 , Y 1 , Y 2 are disjoint sets of size at least (1/4 − 500ε)m with the following properties.
• Γ R [X i ∪ Y i ] is connected and non-bipartite for i ∈ [2].
By Lemma 13.1, V (G) may be partitioned into a red cycle and a blue one. This completes the proof of Lemma 14.1, under the assumption that Proposition 14.2 holds. We prove it in the following subsection.

Proof of Proposition 14.2
In this subsection, we prove Proposition 14.2. We shall consider four cases according to the sizes of the red components in Γ. In each of these cases, we apply Lemmas 7.9 or 7.7 to gain additional structural information about the graph Γ in case Γ B has no perfect matching (otherwise we are done). This information will enable us to show that one of the conditions in the proposition holds.
Claim 14.3. One of the following conditions holds.
3. There exists a tripartition {X 1 , X 2 , X 3 } of V (Γ) such that |X i | ≤ (1/2 − 24ε)m and every red component is contained in one of the red sets X i .
We prove Proposition 14.2 in each of the four cases in Claim 14.3. The second case, where there is a large red component, turns out to be hardest and we leave it to the end of the proof.

No large red components
In Case (1) of the previous claim, we have deg B (Γ) ≥ m/2, implying that Γ B has a perfect matching.

Tripartition
In Case (3) of Claim 14.3, there exists a tripartition {X 1 , X 2 , X 3 } of V (Γ) such that |X i | ≤ (1/2 − 24ε)m and every red component is contained in one of the red sets X i . We assume that Γ B contains no perfect matching. By Lemma 7.7, without loss of generality, there exist subsets It follows that In particular, Γ R [Y i ] is connected. We show that Γ B ∪ Γ R [Y i ] has a perfect matching for i ∈ [2]. Suppose to the contrary that Φ = Γ B ∪ Γ R [Y 1 ] has no perfect matching. By Lemma 7.7, it follows that there exist subsets Z i ⊆ X i and Z j ⊆ X j of size at least (1/4 − 10ε)n such that Z 1 ∪ Z 2 is independent in Φ, for some 1 ≤ i = j ≤ 3.
Suppose first that say i = 2. Then the intersection of Y 2 and Z 2 is non-empty. Let u ∈ Y 2 ∪ Z 2 .

Almost equipartition into four parts
In Case (4)  Claim 14.4. One of the following assertions holds.
1. Γ B has a perfect matching.

For some
has a perfect matching for l ∈ {i, j}.
Proof. Consider Lemma 7.9 with the graph Γ B , the bipartition {X 1 ∪X 2 , X 3 ∪X 4 } and parameter 72ε. Assuming that Γ B has no perfect matching, it is easy to verify that the first three conditions cannot hold. It follows that Γ B has an independent set Y of size at least (1/2 − 432ε)m. This is the union of two red connected subgraphs of order at least (1/4 − 434ε)m. Without loss of generality, , by Lemma 7.9. Otherwise, Γ B [X 1 ∪ X 2 , X 3 ∪ X 4 ] is almost complete, and by Lemma 7.9,if Γ B has no perfect matching, X 3 ∪ X 4 contains an independent set Y of size at least (1/2 − 432ε)m. 3,4], we conclude as before that ] has a perfect matching for i ∈ [3,4].

Large red component
We now consider Case (2) of Claim 14.3, which the last remaining case. Let Φ 1 be a red component on at least (1/2 − 50ε)m vertices and denote The following claim is a simple application of Theorem 7.1.
Claim 14.5. One of the following assertions holds.
Proof. Suppose that Γ 1 does not have a perfect matching. In particular, Γ 1 has no Hamilton cycle. Note that It follows from Theorem 7.1 that either there are at least m/2 vertices of degree at most (1/2 + 50ε)m or there are at least (1/4 + 2ε)m vertices of degree at most (1/4 + 2ε)m.
Suppose that the former holds and let Y be the set of vertices of degree at most (1/2 + 50ε)m in Γ 1 . Then deg R (u, X 2 ) ≥ (1/4 − 52ε)m for every u ∈ Y . It follows that either Γ R [Y ] is connected, or it consists of two connected components of order at least (1/4 − 52ε)m.
Suppose now that the latter holds, and let Y be the set of vertices of degree at most (1/4 + 2ε)m in Γ 1 . Then deg R (u, X 2 ) ≥ (1/2 − 4ε)m for every u ∈ Y , and in particular, X 2 contains a red component of order at least (1/2 − 4ε)m.
In order to prove Proposition 14.2, we may assume that Γ 1 has no perfect matching. We consider first the third case of Claim 14.5, where |X 2 | ≥ m/2 and X 2 contains two red components of size at least (1/4 − 52ε)m. Denote by Y 1 and Y 2 the vertex sets of these red component. Since Γ 1 = Γ B ∪ Γ R [X 1 ] has no perfect matching, it follows from Lemma 7.9 that X 2 contains an independent set of size at least (1/2 − 50ε)m, implying that Γ[Y 1 , Y 2 ] is almost empty and Γ[X 1 , X 2 ] is almost complete. It is easy to conclude from Lemma 7.9 that Γ B ∪ Γ R [Y i ] has a perfect matching for i ∈ [2].
In order to complete the proof of Proposition 14.2 in this case, it remains to consider the case where Γ 1 has no perfect matching and X 2 contains a red component of order at least (1/2 − 4ε)m. The following claim completes the proof of Proposition 14.2.
Claim 14.6. One of the following conditions holds.

There exist disjoint sets
Proof. Without loss of generality, we assume that |X 1 | ≥ |X 2 |. We assume that both Γ 1 and Γ 2 have no perfect matchings and make use of Lemma 7.9 (with parameter 50ε). We start by considering Γ 1 . It is easy to verify that Γ 1 is 2-connected. Furthermore, since the vertices of X 1 have degree at least (3/4 − 2ε), it follows that the two last conditions cannot hold (because they imply that there are many vertices in X 1 of degree at most approximately m/2). We conclude that the following holds.
There exists an independent set A 2 ⊆ X 2 of size at least (1/4 − 200ε) Similarly, if Γ 2 has no perfect matching, one of the two following conditions holds.
2. There exists an independent set A 1 ⊆ X 1 of size at least (1/4−200ε) Suppose that Condition (1) above holds. Then every vertex in A 2 has at least (1/2 − 152ε)m red neighbours (in Γ) in X 2 . Denote It is easy to verify that Condition 2 from Claim 14.6 holds, completing the proof in this case.
We now suppose that Condition (2) holds. So we have sets A i ⊆ X i of size at least (1/4 − 200ε) In particular, A 2 N (A 1 ), so by the same argument, It is easy to check that Condition (2) in Claim 14.6 holds.
). There are no edges in Γ between A 1 and B 2 or between A 2 and B 1 . It is easy to conclude from here that Condition (2) in Claim 14.6 holds.
The proof of Lemma 14.1 concludes our proof of Theorem 1.1. We remind the reader that Lemmas 7.13 and 7.14 were used several times in the proof, and we have yet to proved them. The next two sections, Sections 15 and 16 are devoted to the proofs of Lemmas 7.13 and 7.14 respectively.
15 Proof of Lemma 7.13 In this section, we prove Lemma 7.13. Before turning to the proof, we remind the reader of the statement.
Lemma (7.13). Let 1 n ε 1 and let G be a graph on n vertices with δ(G) ≥ 3n/4 and a 2-colouring E(G) = E(G B ) ∪ E(G R ). Suppose that S, T ⊆ V (G) satisfy the following conditions.
Then V (G) may be partitioned into a red cycle and a blue one.
The main tool we use in the proof is Lemma 7.10, which is a stability version of a special case of Chvátal's theorem, Theorem 7.1. Our aim would be to find a short red cycle C and a short blue path P , whose removal from G leaves a balanced bipartite graph. We then apply Lemma 7.10 to deduce that P may be extended to a blue cycle with vertex set V (G) \ V (C).
Proof of Lemma 7.13. We start by modifying the sets S, T as follows.
Remark. The vertices in X have red degree at least (3/4 − 50ε)n in G.
We will find a red cycle and a blue path with one end in S and one in T , which are disjoint, cover X and their removal from G leaves a balanced bipartite graph with a large number of vertices. We will then use the following claim to obtain the required partition into a blue cycle and a red one.
Hamiltonian. Furthermore, for every s ∈ S and t ∈ T , the graph G B [S , T ] contains a Hamilton path with ends s and t.
Proof. Denote G = G B [S , T ] and Y = (S ∪ T ) \ (S ∪ T ). We claim that there exists a path P of length at most 12εn whose vertex set contains Y . Indeed, we may construct P greedily, by adding a vertex of Y one at a time. Suppose that we want to add the vertex y 1 ∈ Y to a path P in G of length at most 12εn, one of whose ends is y 2 ∈ Y . We may pick z 1 , z 2 ∈ ((S ∪ T ) ∩ V (G )) \ V (P ) such that z i is a neighbour of y i in G . We have N G (z i ) ≥ (1/4 − 11ε)n. In particular, by the third assumption of the lemma, there exists a path of length at most 2 between N G (z 1 ) and N G (z 2 ). Thus we may add y 1 to P using at most five additional vertices. Using this process, we obtain the desired path P , containing the vertices of Y . Denote by s, t the ends of P and assume that s ∈ S, t ∈ T (we may need to extend P slightly).
Let G be the graph obtained from G by removing the inner vertices of P and denote n = |G | and η = 25ε. Then δ(G ) ≥ (1/4 − η)n . Let S ⊆ S, T ⊆ T be subsets of size at least (1/4 − 3η)n ≥ (1/4 − 75ε)(1 − 20ε)n ≥ (1/4 − 80ε)n. Then by the assumptions of Lemma 7.13, e(G [S , T ]) ≥ ηn 2 . By Lemma 7.10, it follows that G is Hamiltonian. The same argument may be used to show that G contains a Hamilton path with ends s, t for every s ∈ S , t ∈ T .
We consider several cases, depending on the size of X and the behaviour of the vertices in X.
Case 1: X = ∅ Without loss of generality, |S 1 | ≥ |T 1 |. Denote k = |S 1 | − |T 1 |. Suppose first that k is even. We use the following claim. We show that G R [U ] contains a cycle of length k. We may assume k ≥ 4 because this assertion is trivial for k = 0, 2. Pick u ∈ U and denote has a path of length k − 2, together with the vertex u it forms a red cycle in S of length k.
Thus, we assume that G R [A] contains no path of length k − 2. It follows from Theorem 7.3 that e(G R [A]) ≤ |A| · εn. We deduce that at most |A|/2 vertices in A have red degree at least 4εn in A. In particular, we may pick a set B of k/2 vertices in A with red degree at most 4εn in A. For every v ∈ B we have deg R (v, U \ A) ≥ (1/4 − 4ε − η)n. It follows that every two vertices in B have at least say n/8 common red neighbours in U \ A. In particular, if B = {b 1 , . . . , b k/2 }, we may pick distinct c 1 , . . . , c k/2 ∈ U \ A such that (b 1 , c 1 , . . . , b k/2 , c k/2 ) is a red cycle in S of length k.
By Claim 15.2, either S 1 contains a blue path P of length k or it contains a red cycle C of length k. In the first case, it is easy to verify that P may be extended to a Hamilton cycle of G B by Claim 15.1. Indeed, consider the bipartite graph G B [S 1 \U, T 1 ] where U is the set of inner vertices of P 1 . This graph is almost balanced (namely the first set has one more vertex than the other), so it contains a Hamilton path whose ends are the ends of P . In the second case, we may conclude directly from Claim 15.1 that the graph G B \ V (C) is Hamiltonian.
We now suppose that k is odd. If S 1 contains a blue path P of length k, we continue as before.
Otherwise, if S 1 contains a blue edge uv, we may find a red cycle C in S 1 \ {u, v} of length k − 1, by the argument of Claim 15.1. It follows that the graph G B \ V (C) is Hamiltonian. Finally, if S 1 has no blue edges, we have δ(G R [S 1 ]) ≥ |S 1 | − n/4 > |S 1 |/2, since |S 1 | ≥ (n + 1)/2. It follows from Bondy's theorem (7.4) that G R [S 1 ] is pancyclic, in particular it contains a cycle C of length k. We proceed as before to conclude that G B \ V (C) is Hamiltonian.
Suppose first that k = 2. Let u, v ∈ S 1 be two blue neighbours of x in S 1 . Let C 1 be a (red) cycle consisting of a single vertex in S 1 \ {u, v}. We may find a blue cycle spanning V (G) \ V (C 1 ) by Claim 15.1.
Suppose now that k = 4. We have |S 1 | = n/2 + 3/2. Thus x has at least three blue neighbours u, v, w ∈ S 1 . Pick an edge ab in S 1 such that a and b are distinct from u, v, w. Since deg R (a, S 1 ) ≥ |S 1 | − n/4 ≥ n/4 + 3/2, we conclude that a and b have a common red neighbour c ∈ S 1 . Let C 1 be the red triangles (abc). Without loss of generality c = u, v. We proceed as before, to show that the graph G B \ {a, b, c} is Hamiltonian.
It remains to consider the case k ≥ 6. Fix u, v to be blue neighbours of x in S 1 and denote S 2 = S 1 \ {u, v}. Note |S 2 | ≥ n/2 + 1/2 and δ(G R [S 2 ]) ≥ |S 2 | − n/4 > |S 2 |/2. It follows from Theorem 7.4, that G R [S 2 ] is pancyclic. In particular, it contains a red cycle of length k − 1. We proceed as before.
Claim 15.5. The graph G[S 2 ] either contains a blue path of length 5εn or for every 2 ≤ l ≤ 5εn it contains a red path of length l − 1 whose one end is a red neighbour of x 1 and the other is a red neighbour of x r .
Similarly, the graph G[T 1 ] either contains a blue path of length 5εn or for every 2 ≤ l ≤ 5εn it contains a path of length l − 1 a red neighbour of x 1 as one end and a red neighbour of x 2 as the other end.
Without loss of generality, we assume that |S 2 | ≥ |T 2 |. Denote k = |S 2 | − |T 2 |, so k ≤ 4εn. It is easy to conclude from Claim 15.5 that we may find vertex-disjoint paths P 1 ∈ G B [S 2 ] and Q 1 ∈ G R [S 2 ] such that the following holds.
• Either P 1 has length k and Q 1 is a singleton, or P 1 is the empty path and Q 1 has length k.
• One end of Q 1 is a red neighbour of x 1 and the other end is a red neighbour of x r (if Q 1 is a singleton, then it is a common red neighbour of x 1 , x r ).
Indeed, if G[S 2 ] contains a blue path of length 5εn, we may find such P 1 , Q 1 where P 1 has length k − 1 and Q 1 is any common red neighbour of x 1 , x r in S 2 (note that such a common neighbour exists). Otherwise, G[S 2 ] contains a red path of length k − 1 whose ends are a red neighbour of x 1 and a red neighbour of x 2 .
Similarly, we may pick P 2 , Q 2 to be a blue and a red path in G[T 2 ] as follows.
• Either P 2 is the empty path and Q 2 has length 1 or P 2 has length 1 and Q 2 is a singleton.
• Q 2 has one end which is a red neighbour of x 1 and the other is a red neighbour of x 2 .
We take C to be the red cycle (Q 1 x 1 Q 2 x 2 y 2 x 3 . . . x l ). The paths P 1 , P 2 may be extended to a Hamilton cycle of G B \ V (C) by Claim 15.1.
Without loss of generality, we may now assume the following.
Case 4: some x 1 , x 2 ∈ X have at least 3ηn common red neighbours in T We proceed similarly to the previous case. For 2 ≤ i ≤ r − 1, pick y i to be a common red neighbour of x i and x i+1 such that the y i 's are distinct and do not belong to X. (2). Consider the set D of common red neighbours of x 1 , x r in S . Then |D| ≥ (1/2 − 23η)n. Clearly, D has either a blue path or a red one of length 5εn.
Claim 15.6. The graph G[T 2 ] either has a blue path of length 5εn or it contains a red path of length l with ends which are neighbours of x 1 , x 2 respectively, for every even 2 ≤ l ≤ 5εn.
Proof. Suppose that G[T 2 ] has no blue path of length 5εn. Denote by U the set of vertices of T 2 with at most ηn blue neighbours. Then |T 2 \ U | ≤ ηn and δ(G R [U ]) ≥ (1/4 − 3η)n. Denote A = N R (x 1 , U ) and B = N R (x 2 , U ). Note that |A|, |B| ≥ (1/4 − 2η)n and U ∩ A ∩ B = ∅, by the assumptions on x 1 and x 2 , thus we may pick u 1 ∈ U ∪ A ∪ B.
If |A ∪ B| ≥ (1/4 + 10η)n, we may find a red path of length l as follows. Greedily pick a red path P = (u 1 , . . . , u l ) in U . As in Claim 15.5, there exists u l+1 ∈ (A ∪ B)\V (P ) which is a red neighbour of u l . The path (u 1 , . . . , u l+1 ) satisfies the requirements.
If |A ∪ B| ≤ (1/4 + 10η)n, it follows that |A ∩ B| ≥ |A| + |B| − |A ∪ B| ≥ (1/4 − 14η)n. If A ∩ B contains a red path of length l, we are done. Otherwise, we may continue as in Claim 15.2 to conclude that the graph G R [A ∩ B, U \ (A ∩ B)] has a path of length l with ends in A ∩ B.
Denote k = |S 2 | − |T 2 |. Pick 2 ≤ l 1 , l 2 ≤ 5εn such that l 2 is even and l 1 − l 2 = k. Similarly to the previous case, we may pick vertex-disjoint paths with the following properties.
• Q 1 has ends which are a red neighbour of x 1 and a red neighbour of x r . Similarly, Q 2 has ends which are a red neighbour of x 2 and a red neighbour of x 1 .
• Either P i = ∅ and Q i has length l i or P i has length l i and Q i is a singleton, for i ∈ [2].
As before, we take C to be the red cycle (Q 1 x 1 Q 2 x 2 y 2 x 3 . . . x r ). The paths P 1 , P 2 may be extended to a Hamilton cycle in G B \ V (C).
Note that if |X| ≥ 3, one of Cases 3 and 4 holds (perhaps with the roles of S and T reversed). Thus we may assume that |X| = 2, and denote X = {x 1 , x 2 }.
either contains a blue path P or a red path Q of length k − 2. In the former case, pick C to be a 4-cycle consisting of x, y and two vertices in D \ V (P ). In the latter case, extend Q to a cycle C through x, y using an additional vertex of D. For convenience denote P = ∅. In both cases, the path P may be extended to a Hamilton cycle of G B \ V (C). From now on, we may assume the following.
Suppose that G R [T 1 ] contains a path P with one end in A and the other in B, of length l, where 0, k ≤ l ≤ 3εn (by a path of length 0 we mean a single vertex). Then we may find the desired cycle partition as follows. G[D] contains a path Q of length l − k which is either red or blue. If Q is red, take C to be the red cycle (x 1 P x 2 Q), and the remaining graph G B \ V (C) is Hamiltonian. If Q is blue, take C = (x 1 P x 2 u) where u ∈ D \ V (Q). The leftover graph G B \ V (C) has a Hamilton path extending Q. Thus from now on we assume the following.
G R [T 1 ] has no path of length l, where 0, k ≤ l ≤ 3εn, with ends in A and B.
In particular, e(G R [A, B \ A]) ≤ εn 2 , implying that G R [A], G R [B] are almost complete (recall that A ∩ B have a small intersection). Suppose that x 1 has two blue neighbours u, v ∈ T 1 . It follows from the previous assumption that G R [B] contains a path P on k + 1 vertices. Form a red cycle C by adding the vertex x 2 to P . Then the graph G B \ V (C) is Hamiltonian, since the graph G B \ (V (C) ∪ {x 1 }) has a Hamilton path with ends u, v. Thus we may assume that both x 1 and x 2 have at most one blue neighbour in T 1 . In particular, |A|, |B| ≥ |T 1 | − n/4.
We can now finish the proof in case |S 1 | − |T 1 | ∈ {0, 1}. Note that we have e(G B [T 1 ]) ≤ 2 by Assumption (3). It is easy to conclude, using Assumption (14) that one of the following conditions holds.
• There exists a vertex u ∈ T 1 which has red neighbours in both A and B.
It follows that there exists a red path in T 1 with one end in A and the other in B of length at most 2, contradiction Assumption (4). Thus the desired monochromatic cycle partition exists.
Denote A = A \ Y and B = B \ Y . If there exists a path of length at most 2 in G R [T 1 ] with one end in A and the other in B , this path may be extended to a red path between A and B of length k + 2, contradicting assumption (4). Thus we assume the following.
No vertex in T 1 has red neighbours in both A and B .
If |Y | ≤ (k − 1)/2, it follows that |A |, |B | ≥ n/4 − 1/2. By the minimum degree condition, we have that δ(G[A , B ]) ≥ 1, and it is easy to deduce that G[A , B ] has a (blue) path forest on at least four edges. By (10), we may complete this into a blue path forest in T 1 with at least k + 2 edges, a contradiction to (3).
16 Proof of Lemma 7.14 In this section, we prove Lemma 7.14. We first remind the reader of the statement.
Lemma ( 7.14). Let 1 n ε 1 and let G be a graph on n vertices with δ(G) ≥ 3n/4 and a 2-colouring E(G) = E(G B ) ∪ E(G R ). Suppose that there exists a partition {S, T, X} of V (G) with the following properties.
• The sets S and T belong to different components of G R \ X.
Then V (G) may be partitioned into a red cycle and a blue one.
The idea of the proof is as follows. By Lemma 7.13, we may assume that there exist subsets S ⊆ S, T ⊆ T of size almost n/4 such that G[S , T ] is close to being empty, implying that the subgraphs G B [S , T \ T ] and G B [S \ S , T ] are almost complete. We aim, similarly to the proof of Lemma 7.13 to find a red cycle C, and two blue paths P 1 , P 2 , whose removal from G leaves two balanced bipartite subgraphs of the aforementioned graphs. We then find Hamilton paths in the remainder subgraphs which together with P 1 , P 2 form a blue cycle with vertex set V (G) \ V (C). We remark that we run into some technical difficulties when X is non-empty.
Proof of Lemma 7.14. We abuse notation and slightly change the definition of S, T and X as follows. Denote (if |X| < 2, add vertices to X arbitrarily). Recall that by the conditions of Lemma 7.14, we may assume that deg R (x, S) ≤ εn or deg R (x, T ) ≤ εn. If the former holds, we move x from X to T , otherwise we move x from X to S. Similarly, if deg B (y, S) ≥ n/32 we move y from X to T and otherwise, if deg R (y, T ) ≤ n/32, we put y in S. After the modification, we have either X = ∅, or X = {y} and deg B (y) ≤ n/16. Furthermore, the only red edges in G[S, T ] are adjacent to x or y.
We may assume that there exist subsets S ⊆ S, T ⊆ T of size at least (1/4 − 100ε)n such that e(G[S , T ]) ≤ 25εn 2 , because otherwise the proof can be completed immediately by Lemma 7.13. It is easy to deduce the following claim (we omit the exact details of the proof).
Claim 16.1. The following holds for some η = η(ε) ≥ ε. There exist partitions {S 1 , S 2 } of S and {T 1 , T 2 } of T with the following properties.
• All but at most ηn vertices of S 2 ∪ T 1 have degree at most ηn in G B [S 2 , T 1 ]. Let F be the graph (V (G), E(M 1 )∪E(M 2 )). Note that by the choice of H , F contains no cycles(in fact it contains no paths of length larger than 3), hence it is a collection of vertex-disjoint paths.
We obtain paths Q 1 , Q 2 as follows. Start with the collection of paths in F . In each stage, if there are two paths with an end in S 2 , connect them using a path in G B [S 2 , T 2 ] of length at most 4, such that it remains vertex-disjoint of all other paths constructed so far. Similarly, if there are two paths with an end in T 1 we join them using a path of length at most 4 in G B [S 1 , T 1 ]. Note that whenever there are at least three paths, we may continue the process. We stop when exactly two paths remain.
By the definition of Q 1 , Q 2 , the set of edges of G B [S 2 , T 1 ] which are contained in one of Q 1 , Q 2 is exactly E(F ), so in particular, it has even size. It follows that either both paths Q 1 , Q 2 have one end in S 2 and the other in T 1 or one of them has both ends in S 2 and the other has both ends in T 1 . Recall that by the way the paths Q 1 , Q 2 were constructed, the first and last edges in each of them are in G B [S 2 , T 1 ]. Hence, by either removing the first and last edges from one of the paths or by removing the first edge from each of them, we obtain paths Q 1 , Q 2 satisfying the requirements of Claim 16.4.
Let Q 1 , Q 2 be paths as in Claim 16.4. Denote by s i ∈ S 2 , t i ∈ T 1 the ends of Q i , let U be the set of inner vertices of these paths, and let S i = S i \ U and T i = T i \ U . It is easy to verify that G B [S 1 , T 1 ] has a Hamilton path with ends t 1 , t 2 and that G[S 2 ∪ T 2 ] may be partitioned into a blue path with ends s 1 , s 2 and a red cycle (note that |S 2 | ≥ |T 2 |, and we use the fact that G[S 2 ] is almost complete).
Case 1.4: |S 2 | < |T 2 | Here we have |S 1 | ≥ |T 1 | + 2. If G[S 1 , T 2 ] contains a blue path of length 10ηn, we may proceed as in Condition (2) from Case 1.1. Otherwise, G[S 1 , T 2 ] has few edges, so G[S 1 ] contains many edges and we may proceed as in the previous case.
We may thus assume that G R [S 1 ] is almost complete. If e(G B [S 2 ]) ≥ 10ηn 2 , pick u, v ∈ A 2 such that the set D = N R (u, S 1 ) ∩ N R (v, S 1 ) has size at least n/8. Denote S 2 = S 2 \ {u, v}. It is easy to verify that G B [S 2 , T 2 ] has a Hamilton path with ends s 1 , s 2 . We form a red cycle by picking a red path P 2 in D of length |S 1 | − |T 1 | and joining it to (uyv) (note that this is indeed a cycle since |S 1 | − |T 1 | ≥ 1). The graph G B [S 1 \ V (P 2 ), T 1 ] has a Hamilton path with ends t 1 , t 2 , completing the required partition.
Finally, if e(G B [S 2 ]) ≤ 10ηn 2 , we have that G R [S] is almost complete, hence we may pick a red cycle C in S ∪ {y} with the following properties.
Case 2.5: |S 1 | < |T 1 | We proceed as in Case 1.3, partitioning the graph remaining from G[S 2 ∪ T 2 ] into a blue path with suitable ends and a red path contained in A 2 .
Case 2.6: |S 2 | ≤ |T 2 | We may continue as in the last part of Case 2.1. We pick u, v ∈ A 2 such that |N R (u, v)∩S 1 | ≥ n/8 and proceed to partition G[S ∪T ] (where S = S\{u, v}) into a blue cycle and a red path contained in the red neighbourhood of u, v in S 1 , using an analogue of Claim 16.4.
The proof of Lemma 7.14 concludes the proof of our main Theorem, Theorem 1.1. We finish this paper with some concluding remarks.

Concluding Remarks
More than two colours -results and future work As a further line of research, one may consider colourings of K n with more than two colours. Gyárfás [13] conjectured that for every r-colouring of K n , the vertex set may be partitioned into at most r monochromatic paths. Erdős, Gyárfás and Pyber [11] considered partitions into monochromatic cycles rather than paths. The defined c(r) to be the smallest t such that whenever a complete graph is r-coloured, it may be partitioned into c(r) monochromatic cycles. They proved that c(r) is bounded, and furthermore c(r) ≤ cr 2 log r for some constant c. Note that Bessy and Thomassé's result [5], mentioned in the introduction, implies that c(2) = 2.
Gyárfás, Ruszinkó, Sárközy and Szeméredi [15] proved that c(r) ≤ cr log r, which is the best known upper bound on c(r) so far. The same authors [16] proved an approximate result of the last conjecture for r = 3. Furthermore, they showed that for large enough n, if K n is 3-coloured, it may be partitioned into 17 monochromatic cycles. However, it turns out that the full conjecture is false, even for r = 3, as shown by Pokrovskiy [23]. Nevertheless, in the same paper, he proved Gyárfás' conjecture for r = 3, so it may still be the case that Gyárfás' conjecture holds in general.
In addition, the counter examples given in [23] are 3-colourings of K n for which all but one vertex may be covered by vertex-disjoint monochromatic paths. This raises the following question: is it true that for every r-colouring of K n all but at most c = c(r) vertices may be covered by r vertex-disjoint monochromatic paths?
Finally, it is natural to consider the Schelp-type version of these problems, namely for graphs with large minimum degree rather than for complete graph. An example for a concrete question of this type is: what is the smallest value of c such that any 3-coloured graph G on n vertices and minimum degree δ(G) ≥ cn can be partitioned into three monochromatic paths?
We believe that the methods we have employed for the proof of Theorem 1.1 may prove useful in resolving the latter questions, as well as many others, regarding partitions of r-coloured graphs into paths and cycles.