Ore- and P\'osa-type conditions for partitioning $2$-edge-coloured graphs into monochromatic cycles

In 2019, Letzter confirmed a conjecture of Balogh, Bar\'at, Gerbner, Gy\'arf\'as and S\'ark\"ozy, proving that every large $2$-edge-coloured graph $G$ on $n$ vertices with minimum degree at least $3n/4$ can be partitioned into two monochromatic cycles of different colours. Here, we propose a weaker condition on the degree sequence of $G$ to also guarantee such a partition and prove an approximate version. This resembles a similar generalisation to an Ore-type condition achieved by Bar\'at and S\'ark\"ozy. Continuing work by Allen, B\"ottcher, Lang, Skokan and Stein, we also show that if $\operatorname{deg}(u) + \operatorname{deg}(v) \geq 4n/3 + o(n)$ holds for all non-adjacent vertices $u,v \in V(G)$, then all but $o(n)$ vertices can be partitioned into three monochromatic cycles.

1. Introduction 1.1. Background. The initial spark of what has today become the sizeable field of research into monochromatic cycle covers can be found in a four-page paper by Gerencsér and Gyárfás [21] from 1967: In a seemingly innocent footnote, they mention that every 2-edge-coloured complete graph K n can be covered by two vertex-disjoint paths of different colours. Inspired by this simple observation, Lehel [3] conjectured that the statement would still hold replacing the term path with cycle; provided the latter includes edges, vertices and the empty set. 1 This conjecture remained unsolved for about 20 years, when it was finally confirmed for large n by Luczak, Rödl, and Szemerédi [41]. The restriction to graphs of large order came from the use of Szemerédi's regularity lemma, but could later be relaxed by Allen [1] and then completely removed by Bessy and Thomassé [8], both finding proofs not relying on regularity arguments.
Ensuing research modified the setting above in multiple directions. Firstly, Erdős, Gyárfás, and Pyber [18] varied the number of colours. In particular, they established O(r 2 log r) as an upper bound for the number of monochromatic cycles needed to partition an r-edge-coloured complete graph K n . Moreover, they conjectured that r colours might even suffice, which follows from Lehel's conjecture for r = 2, but was later refuted for all r 3 by Pokrovskiy [43]. So far, the best improvement of the upper bound is due to Gyárfás, Ruszinkó, Sárközy, and Szemerédi [25], who were able to lower it to O(r log r), provided n is large in terms of r. According to Conlon and Stein [13], Lang and Stein [36] as well as a recent paper by Sárközy [50], the same can be achieved for local r-edge-colourings, where the colour limit only applies to the incident edges of each vertex. Related areas of research also considered hypergraphs [10,11,20,26,39,48] and infinite graphs [9,17,46,56] as host graphs. Alternatively, one may look at not only partitions into cycles, but also into monochromatic paths [21,43], powers of cycles [49,51], regular graphs [52,53], graphs of bounded degree [14,23,24], or arbitrary connected graphs [4,18,19,22,28].
The second main modification was relaxing the completeness requirement on the host graph. Originally suggested in a posthumous paper by Schelp [54], imposing a lower bound on the minimum degree was considered as a replacement, which bears some resemblance to Dirac's theorem [16]. Indeed, Balogh, Barát, Gerbner, Gyárfás, and Sárközy [5] conjectured that for an n-vertex 2-edge-coloured host graph, a minimum degree above 3n/4 would still suffice to guarantee a partition into two monochromatic cycles of different colours. Constructions show that this would be optimal. In support of their conjecture, they proved an approximate version that required minimum degree 3n/4 + o(n) and only guaranteed that all but at most o(n) vertices could be covered. Since then, the two error terms have been gradually eliminated by DeBiasio and Nelsen [15] as well as Letzter [37], both using advanced absorbing techniques. One should note that other density measures for the host graph have also been examined, such as prescribing its independence number [5,47,53] or considering complete multipartite [27,35] as well as random graphs [4,32,33].

Main results.
In this work, we continue two developments initiated by [5]. As its natural Ore-type analogue, Barát and Sárközy [7] proved that the following holds for all large 2-edge-coloured graphs G on n vertices: If G satisfies deg(u) + deg(v) 3n/2+o(n) for all uv / ∈ E(G), then there are two vertex-disjoint and distinctly coloured monochromatic cycles in G, which together cover at least n − o(n) vertices.
We take this one step further and propose a Pósa-type condition, owing its name to a Hamiltonicity condition given by Pósa [45]: Every graph on n 3 vertices whose degree sequence d 1 . . . d n satisfies d i > i for all 1 i < n/2 contains a Hamilton cycle. 2 We conjecture that the following stronger condition guarantees a partition of any large 2-edge-coloured graph into two monochromatic cycles of different colours. Conjecture 1.1. There is n 0 such that the following holds for all 2-edge-coloured graphs G on n n 0 vertices: If the degree sequence d 1 . . . d n of G satisfies d i > i + n/2 for all 1 i < n/4, then there is a partition of V (G) into two distinctly coloured monochromatic cycles.
Unlike for Hamiltonicity, there is no easy link anymore between this Pósa-type condition and the Ore-type condition in [7]. We address this fact in Section 3, also providing a construction to show that each inequality required here is essentially tight. Our first main result is the following approximate version of Conjecture 1.1. Theorem 1.2. For every β > 0, there is n 0 (β) such that the following holds for all 2-edge-coloured graphs G on n n 0 (β) vertices: If the degree sequence d 1 . . . d n of G satisfies d i > i+(1/2+β)n for all 1 i < n/4, then there are two vertex-disjoint and distinctly coloured monochromatic cycles in G, which together cover at least (1 − β)n vertices.
For our second main result, we want to allow a third monochromatic cycle in the partition, but work with even smaller degrees in the 2-edge-coloured host graph. Here, Pokrovskiy [44] conjectured 2n/3 as a minimum degree threshold to guarantee a partition into three monochromatic cycles. This has recently been confirmed approximately by Allen, Böttcher, Lang, Skokan, and Stein [2]. Recalling the aforementioned results for partitions into two monochromatic cycles, we believe that this minimum degree condition might again be replaceable by its natural Ore-type analogue. We therefore propose the following conjecture, which would be best possible as indicated by the construction of Pokrovskiy [44].

Open problems.
Having derived a Pósa-type analogue of the Ore-type condition for two cycles with Theorem 1.2, it is only natural to ask whether a similar analogue also exists in the case of three cycles, so for Theorem 1.4. Here, the task of finding the optimal Pósa-type condition can be formulated as follows.
Problem 1.5. Determine the minimum x, y ∈ [0, 1] that satisfy: For every β > 0, there is n 0 (β) such that the following holds for all 2-edge-coloured graphs G on n n 0 (β) vertices: If the degree sequence d 1 . . . d n of G satisfies d i > i + (x + β)n for all 1 i < yn, then there are three pairwise vertex-disjoint monochromatic cycles in G, which together cover at least (1 − β)n vertices.
As any graph G with minimum degree δ(G) (x + y + β)n automatically satisfies such a Pósa-type condition, the construction of Pokrovskiy [44] for the sharpness of the minimum degree threshold immediately implies that x + y 2/3 must hold. In fact, any solution of Problem 1.5 with x + y = 2/3 would approximately generalise the result in [2].
Among all such solutions, the stronger statements arise from decreasing x and increasing y. Since any graph satisfying the Ore-type condition of Theorem 1.4 also satisfies the Pósa-condition above with (x, y) = (1/6, 1/2) 3 , the lowest achievable x is 1/6, which would fully generalise Theorem 1.4. As we will discuss in Section 3, however, such a full generalisation is not possible for two cycles, so it seems unlikely that it would hold for three cycles. Considering Theorem 1.2, we suggest (x, y) = (1/2, 1/6) as a sensible conjecture for further research.
1.4. Methodology. We briefly sketch the proof idea for our main results. As it is the same for Theorem 1.2 and Theorem 1.4, we focus on the former. The main tool is a colour version of Szemerédi's regularity lemma. Starting with a host graph G satisfying the Pósa-type condition, we first apply the regularity lemma (Lemma 4.2) to partition V (G) into a bounded number of clusters. We find that up to a negligible loss, the reduced graph R with these clusters as vertices inherits the Pósa-type condition of G. Using our structural lemma for two cycles (Lemma 5.2), we identify two distinctly coloured monochromatic components of R that are suitable for constructing the desired cycles. More precisely, the union H of these components does not contain a contracting set (for a formal definition, see Section 4.1). By a well-known analogue of Tutte's theorem (Lemma 4.1), this is equivalent to H having a perfect 2-matching. Leveraging regularity, this 2-matching can be used to lift each monochromatic component in H to one monochromatic cycle in G. The technical details are encapsulated in Lemma 4.3 and guarantee that the cycles are vertex-disjoint and approximately cover the same fraction of vertices as the 2-matching, as desired.
The proof of Theorem 1.4 is similar and only requires one minor adjustment. Here, we cannot completely exclude the occurrence of contracting sets, but only limit what we call their contraction. However, it turns out that this does not invalidate the approach above, although it complicates the proof of the respective structural lemma (Lemma 5.4). Nevertheless, we are still able to follow the line of argumentation from the proof of the corresponding structural lemma in [2] for the minimum degree case.
1.5. Organisation of the paper. The rest of this paper is organised as follows. The next section introduces some basic notation and definitions. In Section 3, we provide the aforementioned constructions showing that Conjectures 1.1 and 1.3 are essentially tight. Afterwards, Section 4 presents the necessary tools for embedding cycles, which will allow us to prove Theorems 1.2 and 1.4 in Section 5. Each proof relies on a structural lemma that is used as a black box. Finally, Sections 6 and 7 are dedicated to proving these structural lemmas, thereby completing the proofs of our two main results.

Notation
We write [k] = {1, . . . , k}. For a graph G, a function c : ) is a pair of a graph G and a 2-edge-colouring c of G although we generally suppress the latter in the notation. For the sake of simplicity, we denote the subgraphs of G retaining only the edges of colour class i as G i , but routinely refer to these colour classes as red and blue. Any connected component of such a G i is called a (red/blue) monochromatic component of G. In particular, isolated vertices of G i form monochromatic components with only one vertex and no edges.
For a subset U ⊆ V (G), we let the neighbourhood N G (v, U ) be the set of all vertices in U that are adjacent to v by an edge of G and omit U if U = V (G). This extends to neighbourhoods of subsets We denote by δ(G) and ∆(G) the minimum degree and the maximum degree of G, respectively. If G is a graph on n vertices v 1 , . . . , v n , ordered such that deg G (v 1 ) . . . deg G (v n ), then this non-decreasing sequence is called the degree sequence of G.
The union of two graphs H 1 , H 2 has vertex set V (H 1 )∪V (H 2 ) and edge set E(H 1 )∪ E(H 2 ). For a vertex subset U ⊆ V (G), the complement U := V (G)\U is always understood relative to the largest graph G in the context. We can then remove the vertices of U from G by considering G \ U , the induced subgraph on U . In conjunction with set-theoretical operands such as cardinality, subset, complement, union, intersection or set difference, we also use the symbol of a graph to refer to its vertex set. For example, if H 1 , H 2 are subgraphs of some common graph, then H 1 ∩ H 2 means V (H 1 ) ∩ V (H 2 ). In particular, this notation allows us to denote the number of vertices of a graph G as |G|.
In constant hierarchies, we write x ≪ y if for all y ∈ (0, 1], there is some x 0 ∈ (0, 1) such that the subsequent statement holds for all x ∈ (0, x 0 ]. Hierarchies with more than two constants are defined similarly and read from right to left. Furthermore, we will assume all constants to be positive real numbers and x to be a natural number if 1/x appears in such a hierarchy.

Constructions
Labeling conditions as Ore-or Pósa-type conditions stems from the well-known Hamiltonicity conditions given by Ore [42] and Pósa [45]. In fact, the latter generalises the former, which also carries over to stronger versions of both conditions. For example, the Ore-type condition conjectured by Barát and Sárközy [7] can be seen a special case of the following Pósa-type condition with x = 1/4. Proposition 3.1. Let x ∈ [0, 1/2) and G be a non-complete graph on n vertices such that deg G (u) + deg G (v) (1 + 2x)n holds for all uv ∈ E(G). Then the degree sequence d 1 . . . d n of G satisfies d j > j + xn for all 1 j < n/2.
Proof. Assume otherwise, so d j j + xn for some 1 j < n/2. Let v 1 , . . . , v n be an enumeration of V (G) in order of non-decreasing degree and define U := {v 1 , . . . , v j }. As deg G (u) + deg G (u ′ ) 2d j 2j + 2xn < (1 + 2x)n for all u, u ′ ∈ U , the induced subgraph G[U ] must be a clique by the Ore-type condition. Therefore, each u ∈ U satisfies deg G (u, U ) = |U | − 1 = j − 1 and So there are at most (xn + 1)|U | edges between U and U . Let v ∈ U be the vertex incident to the least number of these edges. Then as |U | > |U |, this vertex v must satisfy deg G (v, U ) (xn + 1)|U |/|U | < xn + 1. Now if all edges from U to U existed, δ(G) = d 1 = n − 1 would follow and imply that G is complete. As this is not the case by assumption, we can pick some u ∈ U \ N G (v). Thus, the obvious observation which contradicts the Ore-type condition.
However, as the following construction shows, this Pósa-type condition is too weak to guarantee that the graph can be covered by two vertex-disjoint monochromatic cycles. Indeed, we can show that the stronger Pósa-type condition of Conjecture 1.1 is essentially tight. This means that up to a constant number of vertices, every inequality required is necessary in order to ensure the existence of a red and a vertex-disjoint blue cycle covering the whole graph. In fact, this is still true even if cycles of the same colour are allowed. Construction 3.2. Let k < m and G k,m be a 2-edge-coloured graph on 4m vertices as follows. The vertex set of G k,m consists of one cluster U of k vertices, two clusters A 1 and A 2 of m vertices each, and one cluster B of 2m − k vertices. The only edges missing from G are edges from U to B and from A 1 to A 2 . The edges inside A 1 , from A 1 to B and from U to A 2 are red. Similarly, the edges inside A 2 , from A 2 to B and from U to A 1 are blue. The edges inside U and B have arbitrary colours (see Figure 1). Proposition 3.3. For k < m and n = 4m, the 2-edge-coloured graph G k,m satisfies both of the following: (1) The degree sequence d 1 . . . d n of G k,m satisfies d j > j + n/2 − 1 for all 1 j < n/4 except j = k.
(2) The vertices of G k,m cannot be covered by two vertex-disjoint monochromatic cycles.
Proof. The vertices with the smallest degree in G k,m are those in U . So the first k terms in the degree sequence of G k,m are d 1 = . . . = d k = k + 2m − 1, and (1) holds for all 1 j < k, but not for j = k. As every vertex v ∈ U satisfies deg G k,m (v) 3m − 1 > j + 2m − 1 for all j < m, (1) also holds for k < j < m.
It is easy to see that any monochromatic cycle intersecting U can only intersect either A 1 or A 2 , but cannot cover this A i completely. So as no monochromatic cycle can intersect all three of A 1 , A 2 and B, the graph G k,m satisfies (2).
Since the Pósa-type condition from Conjecture 1.1 is strictly stronger than the one obtained from Proposition 3.1, the question whether the former still generalises the Ore-type condition from Barát and Sárközy [7] arises naturally. The following proposition answers this in the negative.

Regularity.
A connected matching in a graph is a 1-regular subgraph contained in a single connected component. With the help of Szemerédi's regularity lemma [57], the task of finding large cycles in a dense graph G can be relaxed to finding large connected matchings in an appropriately defined reduced graph R. This idea was first used by Komlós, Sárközy, and Szemerédi [29] to prove an approximate version of the Pósa-Seymour conjecture and then transferred to monochromatic cycle covers by Luczak [40,41]. Ever since then, the method has become standard practice and fueled numerous advances [2,5,7,15,31,33,37,38], including many of the results mentioned in Section 1. We therefore limit ourselves to stating the necessary definitions and lemmas, mostly following the notation from Lang and Sanhueza-Matamala [34].
Let A, B be two non-empty vertex subsets of a graph G and denote the number of edges of G with one endpoint in A and the other in B as e G (A, B). The density of such a pair is then defined as Moreover, an ε-regular pair with density at least d is called (ε, d)-regular. Now let V = {V j } r j=1 be a family of r disjoint sets and R be a graph on [r]. We say is edgeless for every j ∈ [r], and jk is an edge of R whenever e G (V j , V k ) > 0. We call the sets V j of the partition its clusters and refer to R as the reduced graph of G or, more precisely, We will use the degree version of Szemerédi's regularity lemma, adapted for the use with 2-edge-coloured graphs [30, Theorems 1.10 and 1.18]. With the notation introduced above, it can be formulated as follows: graphs on n common vertices. Then there are r 0 r r 1 and a family V of r disjoint subsets of these vertices with the following properties: As already mentioned, the method introduced by Luczak allows us to lift large connected matchings in a reduced graph R i to large cycles in the corresponding G i . In fact, Christofides, Hladký, and Máthé [12] observed that the same is also true for fractional matchings and similarly, 2-matchings. Formally, the following statement holds: Lemma 4.3 (From connected matchings to cycles). Let 1/n ≪ 1/r ≪ ε ≪ d ≪ η ≪ β. Let G 1 , G 2 be graphs on n common vertices, V be a family of r disjoint subsets of these vertices, and R 1 , R 2 be graphs on [r].
. Let H be the union of m 1 components of R 1 and m 2 components of R 2 , and suppose that there is a 2-matching in H that covers at least

Proof of the main results
In this section, we show the main results detailed in Section 1.2.

5.1.
A Pósa-type condition for two cycles. Let us start with our first main result, Theorem 1.2. Its proof follows the argument outlined in Section 1.4 and uses the tools of Section 4 to simplify the problem to finding a large 2-matching in two monochromatic components of the reduced graph. For convenience, we introduce the following notation to abbreviate the Pósa-type condition.
The structural analysis is then encapsulated in the following lemma, whose proof we defer to Section 6.
Lemma 5.2 (Structural lemma for two cycles). Let 1/n ≪ γ and G be (n, γ)-Pósa. Suppose G is 2-edge-coloured. Then there are a red and a blue component of G whose union H is a spanning subgraph of G without contracting sets.
Consider an (n, β)-Pósa 2-edge-coloured graph G and apply Lemma 4.2. This yields r 0 r r 1 and a family V = {V j } r j=1 of r disjoint sets, together with (ε, δ/2)- . Each of the r clusters must then contain between (1 − ε)n/r and n/r vertices.
Since we want to apply Lemma 5.2 to the graph R := R 1 ∪ R 2 on [r], we need to check that R is (r, γ)-Pósa. For this, we note that as the Now enumerate the clusters of V by ascending maximum index of their vertices in the degree sequence d 1 . . . d n of G and pick 1 j < r/4.
Then for each j k r, the vertex v k ∈ V k of maximum such index h k must have a larger index than all vertices in V 1 , . . . , V k , so we get h k k(1 − ε)n/r j(1 − ε)n/r. By the choice of constants, we have εn/r εn/r 1 1, so there must be some integer j ′ with j(1−2ε)n/r j ′ j(1−ε)n/r h k . As j ′ < jn/r < n/4 and G is (n, β)-Pósa, it follows that Since G ′ is an (ε, δ)-approximation of G, this implies that deg G ′ (v k ) deg G (v k )−δn > jn/r + (1/2 + γ)n. As each cluster of V contains at most n/r vertices, the number of clusters containing a vertex in N G ′ (v k ) must thus be at least j + (1/2 + γ)r. Due to the (G ′ i , V)'s being R i -partitions, the indices of these clusters are then adjacent to k in R, so deg R (k) j + (1/2 + γ)r.
Applying this argument to all j k r, we find that the j-th entry of the degree sequence of R must be at least j + (1/2 + γ)r. As this holds for all 1 j < r/4, the graph R is (r, γ)-Pósa and thus satisfies all requirements of Lemma 5.2 with r playing the role of n. Hence, there are a red component H 1 ⊆ R 1 and a blue component H 2 ⊆ R 2 such that their union H is a spanning subgraph of G without contracting sets. By Lemma 4.1, there is a 2-matching in H that covers all r (1 − η)r vertices of R. But then all requirements of Lemma 4.3 are fulfilled, which guarantees the existence of two vertex-disjoint cycles C 1 ⊆ G 1 and C 2 ⊆ G 2 together covering at least (1 − β)n vertices of G. As the G i 's only contain edges of one colour, these cycles are monochromatic and of different colours, so we are done.

An
Ore-type condition for three cycles. Using the same approach, we also want to prove Theorem 1.4. In contrast to minimum degree or Pósa-type conditions, the Ore-type condition only approximately carries over to the reduced graph, which motivates the following abbreviation of our setting.
Excluding these exceptional edges E(X) from the Ore-type condition means that we cannot exclude the occurrence of contracting sets, but only limit their contraction to a fraction of the total number of vertices in R. But since we only aim to cover almost all vertices of G with few monochromatic cycles, this small loss is manageable and the following lemma suffices together with the tools of Section 4. Its proof is deferred to Section 7.
Consider an (n, β)-Ore 2-edge-coloured graph G and apply Lemma 4.2. This yields r 0 r r 1 and a family V = {V j } r j=1 of r disjoint sets, together with (ε, δ/4)- . Each of the r clusters must then contain between (1 − ε)n/r and n/r vertices.
Since we want to apply Lemma 5.4 to the graph R := R 1 ∪ R 2 on [r], we need to check that for some appropriately defined graph X on [r], the pair (R, X) is (r, δ, γ)-Ore. For this, we note that as the G ′ i are (ε, δ/4)-approximations of the G i , the graph Observe that every u ∈ V j loses at most δn/2 incident edges from G to G ′ . So there can be at most δn/(2(1 − ε)n/r) < δr clusters V k in V such that uv ∈ E(G) \ E(G ′ ) for all v ∈ V k . This shows that ∆(X) < δr.
. This together with G ′ being an (ε, δ/2)-approximation of the (n, β)-Ore graph G yields As each cluster of V contains at most n/r vertices, the number of clusters containing a vertex in N G ′ (u) plus the number of clusters containing a vertex in N G ′ (v) must thus be at least (4/3 + γ)r. Due to the (G ′ i , V)'s being R i -partitions, the indices of these clusters are then adjacent to j or k in R, so deg R (j) + deg R (k) (4/3 + γ)r as desired.
This shows that (R, X) is (r, δ, γ)-Ore and thus satisfies all requirements of Lemma 5.4 with r playing the role of n. Hence, there are three monochromatic components H 1 , H 2 , H 3 ⊆ R such that their union H covers at least (1 − η)r vertices of R and every stable set S of H satisfies c H (S) ηr. By Lemma 4.1, there is a 2-matching in H that covers at least |H| − ηr (1 − 2η)r vertices of R. But then all requirements of Lemma 4.3 are fulfilled, which guarantees the existence of three pairwise vertexdisjoint cycles in G 1 , G 2 together covering at least (1 − β)n vertices of G. As the G i 's only contain edges of one colour, these cycles are monochromatic and we are done.

Proof of the structural lemma for two cycles
In this section, we show Lemma 5.2, which completes the proof of Theorem 1.2. Recall from Definition 5.1 that the input graph G is a 2-edge-coloured graph on n vertices with d j > j + (1/2 + γ)n for all 1 j < n/4. We have to find a red and a blue component R, B of G such that their union H := R ∪ B is a spanning subgraph of G without contracting sets. Before we address any details, we collect a few general observations. Observation 6.1. Let 1/n ≪ γ and G be (n, γ)-Pósa. Then all of the following hold: Proof. The first statement follows directly from the Pósa-type condition when choosing j = 1. For (2) and (3), let U be non-empty and u ∈ U the vertex of maximum index in the degree sequence of G, which implies deg G (u) d |U | . If |U | n/4, we can require 1/n γ and apply the Pósa-type condition with j = ⌈n/4 − 1⌉ to find d |U | d j > j + (1/2 + γ)n 3n/4. Conversely, |U | < n/4 allows us to apply the Pósa-type condition directly and find d |U 6.1. Component structure. We first show that there are a red and a blue component R, B in G such that their union R ∪ B is spanning. Formally, we prove the following intermediate result.
Then there are a red and a blue component of G whose union H is a spanning subgraph of G.
Proof. By Observation 6.1(2), there is a vertex u 1 with deg G (u 1 ) > 3n/4. Let R and B 1 be its red and blue component, respectively. Now if |R ∩ B 1 | n/2, Observation 6.1 (1) implies that every vertex of G is adjacent to some vertex in R ∩ B 1 . Thus, R ∪ B 1 is a spanning subgraph of G and we are done. So we may assume |R ∩ B 1 | < n/2 for the remainder of this proof.
In particular, N G (u 1 ) \ (R ∩ B 1 ) contains at least n/4 vertices and can thus play the role of U in Observation 6.1 (2). Hence, there is a vertex u 2 ∈ N G (u 1 ) \ (R ∩ B 1 ) with deg G (u 2 ) > 3n/4. Without loss of generality, we can assume the edge u 1 u 2 ∈ E(G) to be red, otherwise swap colours. This implies u 2 ∈ R \ B 1 , and we denote the blue component of u 2 as B 2 .
If R is spanning already, there is nothing to show. Similarly, if |R| < n/4, we can apply Observation 6.1(3) with R as U to find v ∈ R with deg G (v, R) > n/2. As all these edges must be blue, the blue component B(v) of v satisfies |R ∩ B(v)| > n/2 and we are again done by the argument above. So we may assume |R| n/4. Here, Observation 6.1(2) applied with R as U guarantees that there is a vertex v 1 ∈ R with deg G (v 1 ) > 3n/4. We let B be the blue component of v 1 and show that R ∪ B is the desired spanning subgraph H.
Assuming otherwise, we will consider R ∪ B as U in either Observation 6.1(2) or 6.1(3) and arrive at a contradiction in both cases. For |R ∪ B| n/4, Observation 6.1(2) guarantees the existence of v 2 ∈ R ∪ B with deg G (v 2 ) > 3n/4. Now u 1 , u 2 , v 1 , v 2 all have degree above 3n/4 in G, so there must be a vertex w that is adjacent to all four of them. However, at least one of the edges wu 1 and wu 2 must be red as the blue components of u 1 and u 2 differ. Similarly, at least one of wv 1 and wv 2 must be red. But this would put some u j ∈ R with j ∈ [2] and some v k ∈ R with k ∈ [2] into the same red component, a contradiction.
We may therefore assume 0 < |R ∪ B| < n/4 and apply Observation 6.1(3) to find v ′ ∈ R ∪ B with deg G (v ′ , R ∪ B) > n/2. Denote its red and blue component as R ′ and B ′ , respectively. We immediately observe that But no vertex in R ∩ B ′ can be adjacent in G to v 1 ∈ R ∩ B with deg G (v 1 ) > 3n/4, so we get |R ∩ B ′ | < n/4. Similarly, we can choose j ∈ [2] such that B = B j and observe that no vertex in R ′ ∩ B can be adjacent in G to u j ∈ R ∩ B j with deg G (u j ) > 3n/4, so |R ′ ∩ B| < n/4. Combining these results yields the desired contradiction This proves that H := R ∪ B must indeed be a spanning subgraph of G.

6.2.
Proof of the structural lemma. The only thing missing to complete the proof of Lemma 5.2 and thus, Theorem 1.2 is to exclude the existence of contracting sets. Having established two suitable monochromatic components in Lemma 6.2, we can now prove the full statement.
Proof of Lemma 5.2. By Lemma 6.2, there are a red and a blue component R, B of G such that R ∪ B is a spanning subgraph of G. If one of R and B is already spanning on its own, we may freely choose the component of the other colour and do so by picking the largest one. However, even if neither R nor B is spanning, it is easy to see that they must be the largest components of their respective colour in G. Indeed, choose u ∈ R \ B and v ∈ B \ R. As R ∪ B is spanning and there is no edge between R \ B and B \ R, we have N G (u) ⊆ R and N G (v) ⊆ B. So both R and B must already contain n/2 vertices by Observation 6.1 (1).
For a proof by contradiction, fix some contracting set S in H := R ∪ B, which cannot be empty as c H (∅) = 0. We immediately observe that the stability of S implies N H (s) ⊆ N H (S) for all s ∈ S. By the contraction property, we additionally know that deg H (s) |N H (S)| < (|S| + |N H (S)|)/2 < n/2, which is less than deg G (s) by Observation 6.1(1). So every s ∈ S must lose incident edges from G to H and can therefore not be a vertex of R ∩ B. This implies S ⊆ (R \ B) ∪ (B \ R). Now let s 1 ∈ S be the vertex of maximum index in the degree sequence of G. Without loss of generality, we can assume s 1 ∈ R \ B, otherwise swap colours. Denote the blue component of s 1 as B 1 and recall that the largest blue component of G is B, so |B 1 | n/2 must hold.
We first want to show that |S| n/4. Assuming otherwise and using that G is (n, γ)-Pósa, we find that So s 1 must lose more than n/2 incident edges from G to H. As all of them are blue, |B 1 | > n/2 follows in contradiction to what we have observed above. So |S| n/4 must hold. In particular, applying Observation 6.1(2) with S as U yields deg G (s 1 ) > 3n/4 by choice of s 1 as the maximum degree vertex in S.
For the remainder of this proof, we partition B 1 into S 1 := S∩B 1 , N 1 := N H (S)∩B 1 , and W 1 := B 1 \ (S 1 ∪ N 1 ). Similarly, we also partition its complement B 1 into S ′ := S \B 1 , N ′ := N H (S)\B 1 , and W ′ = B 1 \(S ∪ N H (S)). Obviously, there can be no blue edges from S 1 ⊆ B 1 to S ′ ∪ W ′ ⊆ B 1 in G. But by stability of S in H and choice of W ′ , there can also be no red edges. So deg G (s 1 ) > 3n/4 implies that |S ′ ∪ W ′ | < n/4.
Moreover, S ′ ∪ W ′ cannot be empty as then B 1 = N ′ ⊆ N H (S) would imply that n/2 |B 1 | |N H (S)| or n/2 < |B 1 | hold, both of which we already know to be false. This allows us to apply Observation 6.1(3) to S ′ ∪ W ′ as U to find u ∈ S ′ ∪ W ′ with deg G (u, S ′ ∪ W ′ ) > n/2. Additionally, u cannot have an edge to S 1 by the argument above, so we conclude Using the contraction property of S together with |S 1 | + |N 1 | + |W 1 | = |B 1 | n/2, we get Together with |S ′ | |S ′ ∪ W ′ | < n/4, this allows us to apply Observation 6.1(3) to S ′ as U and obtain a vertex s ′ ∈ S ′ with deg G (s ′ , S ′ ) > n/2. By the same argument as above, none of these edges may go to S 1 ∪ W 1 and we observe that But now adding the inequalities (6.1) and (6.2) yields the desired contradiction So H = R ∪ B cannot have a contracting set.

Proof of the structural lemma for three cycles
In this section, we show Lemma 5.4, which completes the proof of Theorem 1.4. Recall from Definition 5.3 that G is a 2-edge-coloured graph on n vertices and X is another graph on the same vertices with bounded maximum degree. We try to find three monochromatic components of G such that their union H contains almost all vertices and has no stable sets with large contraction in H. 7.1. Component structure. Let us first shed some light on the structure of the monochromatic components of G. We find that two of them suffice to cover almost all vertices of G.
Proof. We will prove this by assuming otherwise and finding three vertices v 1 , v 2 , v 3 which are pairwise non-adjacent in X and lie in distinct red and blue components of G. These are then also non-adjacent in G, so as the graph G ∪ X is (n, γ)-Ore, we get |B j | 2n and contradicts 2 3 j=1 deg G (v j ) > 4n from above. It remains to show that if every pair of monochromatic components of G misses more than 6δn vertices, then there must be three vertices as described above. For this, let v 1 ∈ V (G) be arbitrary and denote its monochromatic components as R 1 and B 1 . As together they miss more than 6δn vertices and ∆(X) < δn, we can select v 2 ∈ V (G) \ (R 1 ∪ B 1 ∪ N X (v 1 )). We denote its monochromatic components as R 2 , B 2 and let R ′ : ) would complete a triple as described above, so we may assume |R ′ ∩ B ′ | 2δn for the remainder of this proof. As R 1 , R 2 together miss more than 6δn vertices, this implies that at least one of R ′ ∩ B 1 and R ′ ∩ B 2 must contain more than 2δn vertices. Similarly, at least one of R 1 ∩ B ′ and R 2 ∩ B ′ must contain more than 2δn vertices.
(1) A triple (G, X, R) is called plain (n, δ, γ)-Ore if R is a monochromatic component of G with |R| (1 − 10δ)n. (2) A quadruple (G, X, R, B) is called mixed (n, δ, γ)-Ore if (1) does not hold for any choice of R, and R, B are two monochromatic components of G with different colours as well as |R ∪ B| (1 − 8δ)n. (3) A quadruple (G, X, R 1 , R 2 ) is called split (n, δ, γ)-Ore if neither (1) nor (2) holds for any choice of R, B, and R 1 , R 2 are two monochromatic components of G with the same colour as well as |R 1 ∪ R 2 | (1 − 6δ)n.
The remainder of this chapter is dedicated to three separate proofs of Lemma 5.4, one for each of these three cases. That is, we show that the following three statements hold: Lemma 7.3. Let 1/n ≪ δ ≪ η ≪ γ and (G, X, R) be plain (n, δ, γ)-Ore. Then there are three monochromatic components of G whose union H contains at least (1 − η)n vertices and has no ηn-contracting sets.
With these at hand, the proof of Lemma 5.4 becomes trivial.
Proof of Lemma 5.4. Choosing one or two monochromatic components of G according to Lemma 7.1, we can extend (G, X) to a triple or quadruple that is either plain, mixed or split (n, δ, γ)-Ore. Thus, we are done by Lemmas 7.3 to 7.5.
While the first two cases are quite straightforward to solve (see Sections 7.2 and 7.3), the third one will require a more involved argument (see Sections 7.4 and 7.5). Before we address any details, we collect a few general observations that hold in all three cases.
Observation 7.6. Let 1/n ≪ δ ≪ γ and (G, X) be (n, δ, γ)-Ore. If S is a contracting set in the subgraph H of G and S ′ is a subset of S, then all of the following hold: (1) |S| c H (S).
(2) |N H (S)| < n/2. We first observe that S ′ is disjoint from B∪B ′ : Assuming otherwise, there would be a u ∈ S ′ that belongs to both a red and a blue component kept from G to H and therefore does not lose incident edges. By Observation 7.6(1) and our choice of constants, we have |S ′ | c H (S ′ ) 3δn and so there is v ∈ S ′ \(N X (u)∪{u}). Moreover, v is incident to more than n/3 edges lost from G to H by Observation 7.6(5). As v ∈ R and R is kept from G to H, all of these edges must be blue and belong to the blue component L of v with |L| > n/3. But then L is among the two largest blue components B, B ′ and thus also a subgraph of H, a contradiction.
Next, we want to show that there also exist two blue components B 1 , B 2 such that |S ′ \ (B 1 ∪ B 2 )| 2δn. For this, let s 1 ∈ S ′ be arbitrary and denote its blue component as B 1 . If |S ′ \ B 1 | 2δn, there is nothing to show, so we can assume there is some s 2 ∈ S ′ \ (B 1 ∪ N X (s 1 )). Denote its blue component as there is nothing to show, so assume otherwise and choose s 3 ∈ S ′ \ (B 1 ∪ B 2 ∪ N X (s 1 ) ∪ N X (s 2 )). Let B 3 be its blue component. The vertices s 1 , s 2 , s 3 ∈ S ′ are then pairwise non-adjacent in X and lie in different blue components of G, meaning there can be no blue edges between them. By stability of S ′ ⊆ R, there can also be no red edges, so the vertices are pairwise non-adjacent in G, as well. Thus, the Ore-type condition is applicable and adding the three inequalities yields 4n < 2 3 j=1 deg G (s j ). Now every incident edge of s j ∈ S ′ ⊆ R is either kept from G to H and hence an edge to N H (S ′ ), or lost and therefore blue. This shows deg Reordering yields |N H (S ′ )| > n/2 in contradiction to Observation 7.6(2). So there must have been two blue components Finally, we show for s 1 ∈ S ′ ∩ B 1 that its blue component B 1 must be among B, B ′ : As v is not in N H (S ′ ) and belongs to a different blue component than all of S ′ , there can be no edge from v to S ′ in G. It follows that deg G (v) n − |S ′ |. In particular, vs 1 / ∈ E(G ∪ X) by choice of v. Since G ∪ X is (n, γ)-Ore, we find 4n/3 < deg G (v) + deg G (s 1 ). Recall from above that deg G (s 1 ) |N H (S ′ )| + |B 1 \ N H (S ′ )|, which is smaller than |S ′ | + |B 1 | as S ′ is contracting. So we can deduce 4n/3 < n + |B 1 | and obtain |B 1 | > n/3. But then B 1 must be among the two largest blue components B, B ′ intersecting R and s 1 ∈ S ′ ∩ B 1 contradicts the disjointness of S ′ from B ∪ B ′ we have shown above. So H cannot have an ηn-contracting set. 7.3. Two monochromatic components of different colours. In a similar fashion, we can also prove Lemma 7.4. This lemma deals with the mixed case, that is when there are two monochromatic components R, B of different colours that only together cover almost all of G.
Proof of Lemma 7.4. We first note that if R and B covered exactly the same set of vertices, then (G, X, R) would be plain (n, δ, γ)-Ore, which is not the case by assumption. So there must be some vertex in V (R ∪ B) that belongs to only one of R and B. In particular, there must be a monochromatic component of G intersecting V (R ∪ B) that is neither R nor B. Without loss of generality, we can assume the largest such component to be red (otherwise swap colours) and denote it as R ′ . Let R be the red component among R, B. The graph H := R ∪ B ∪ R ′ then still covers at least (1 − 8δ)n (1 − η)n vertices by the choice of constants. It remains to show that H has no stable sets S with c H (S) > ηn. For a proof by contradiction, fix such a set and consider the still stable set S ′ := S ∩ (R ∪ B) with c H (S ′ ) > (η − 10δ)n 8δn by Observation 7.6(3) and the choice of constants. Now as R, B both miss more than 10δn vertices, but together miss at most 8δn vertices, we observe that |R \ B| > 2δn and similarly, |B \ R| > 2δn. So we can pick u ∈ R \ B and v ∈ B \ (R ∪ N X (u)), which share no monochromatic component. On the one hand, this implies that they cannot be adjacent, so the Ore-type condition yields deg G (u) + deg G (v) > (4/3 + γ)n and thus, |N G (u) ∩ N G (v)| > (1/3 + γ)n. On the other hand, their edges to some w ∈ N G (u) ∩ N G (v) must have different colours. If uw is red and vw is blue, we automatically get w ∈ R ∩ B. If uw is blue and vw is red, then w / ∈ R ∪ B, so there can be at most 8δn such vertices in N G (u) ∩ N G (v). This shows that |R ∩ B| > (1/3 + γ − 8δ)n.
Similar to the proof of Lemma 7.3, we can now observe that S ′ ⊆ R ∪ B is disjoint from R ∩ B and R ′ : Assuming otherwise, there would be some u ∈ S ′ that belongs to both a red and a blue component kept from G to H and therefore does not lose incident edges. By Observation 7.6(1), we can select some v ∈ S ′ \ (N X (u) ∪ {u}), which is incident to at least n/3 lost edges by Observation 7.6(5). As one of its monochromatic components is kept from G to H, this implies that all of these lost edges go to the same monochromatic component L with |L| > n/3. But this contradicts the choice of R ′ : Obviously, R ′ and L are disjoint from R ∩ B. The intersection R ′ ∩ L can only exist if L is blue and must then lie outside of R ∪ B, so it can contain at most 8δn vertices. Hence, we find that |R ′ | n − |R ∩ B| − |L| + |R ′ ∩ L| < (1/3 − γ + 16δ)n < n/3 < |L| by the choice of constants, although R ′ is supposed to be the largest such component. This contradiction proves that S ′ must indeed be disjoint from R ∩ B and R ′ . The next step is a general observation we will use multiple times in the following arguments.
Proof of the claim. We can apply the Ore-type condition to find that Recall from above that apart from at most 8δn vertices missed by R ∪ B, all common neighbours w of u and v in G must belong to R ∩ B, so the connecting edges uw, vw are kept from G to H. As at least one of u, v belongs to S ′ , this puts w into N H (S ′ ).
Now it is easy to see that S ′ must intersect with both R \ B and B \ R: Claim 7.7. But as v ∈ B \ R and S ′ ⊆ R \ B, the neighbourhood of v cannot intersect with S ′ and n/3 < |N G (u) ∩ S ′ | follows. By stability of S ′ , all these edges incident to u ∈ S ′ ⊆ R must be lost from G to H and therefore be blue. So the blue component L of u ∈ R \ B contradicts the choice of R ′ , exactly as above. Similarly, if S ′ were disjoint from R \ B, then S ′ ⊆ B \ R follows and we can choose v ∈ S ′ ∩ B \ R as well as u ∈ R \ (B ∪ N X (v)). Here, the neighbours of u cannot belong to S ′ , so Claim 7.7 implies n/3 < |N G (v) ∩ S ′ | with all these neighbours of v belonging to the red component L of v ∈ B \ R. However, as L contains v ∈ S ′ and R ′ is disjoint from S ′ , these two must be different red components and L being larger yields the same contradiction as above. So indeed, both intersections S ′ ∩ (R \ B) and S ′ ∩ (B \ R) must be non-empty.
According to Observation 7.6(1), there are more than 2δn vertices in S ′ . So we can choose s 1 in the smaller and s 2 / ∈ N X (s 1 ) in the larger set of S ′ ∩(R\B) and S ′ ∩(B \R). Using s 1 , s 2 as u, v in Claim 7.7, we obtain n/3 < |N G (s 1 ) ∩ S ′ | + |N G (s 2 ) ∩ S ′ |. So for some j ∈ [2], we have n/6 < |N G (s j ) ∩ S ′ |. By the stability of S ′ , all these vertices must belong to the lost component L of s j , which thereby contains more than n/6 vertices.
But then this component L is again larger than R ′ . Indeed, we already know that the sets S ′ , R ∩ B and R ′ ∩ B are pairwise disjoint. By definition, S ′ is also disjoint from N G (S ′ ) \ N H (S ′ ). The same holds for the other two sets because vertices in R ∩ B or R ′ ∩ B do not lose incident edges from G to H. This shows that by Observation 7.6(2). Together with |R ′ ∩ B| n − |R ∪ B| 8δn, we get The fact that G ∪ X is (n, γ)-Ore now yields (4/3 + γ)n deg G (s 1 ) + deg G (s 2 ). The right side deg G (s 1 ) + deg G (s 2 ) can be expressed as the sum of |N G (s 1 ) ∪ N G (s 2 )| |S ′ |+|N G (S ′ )| and |N G (s 1 )∩N G (s 2 )|. Again recall that apart from at most 8δn vertices outside of R∪B, all vertices in N G (s 1 )∩N G (s 2 ) must belong to R∩B. Taken together, we get Now adding the inequalities (7.2) and (7.3) yields |R ′ | (1/6 − γ + 16δ)n after simplification, which is less than n/6 by the choice of constants. As there is a larger monochromatic component L that intersects R ∪ B in s j , this contradicts the choice of R ′ . So H cannot have an ηn-contracting set. Proof. For a proof by contradiction, let S be such a stable set with c H (S) > ηn. We first show that S must be disjoint from the set T of all vertices for which both monochromatic components are in H. Assuming otherwise, there is a vertex u ∈ S ∩ T not losing incident edges from G to H, so by Observation 7.6(1) and the choice of constants, there is a v ∈ S \ (N X (u) ∪ {u}). The vertex v loses more than n/3 incident edges from G to H by Observation 7.6(5). But these lost edges cannot go to T , which contradicts |T | 2n/3. So S and T are indeed disjoint. Next, we let W := V (G) \ (S ∪ N H (S)) and partition V (G) into S, N H (S), W ∩ T and W \ T . Since S and W \ T are disjoint subsets of T and S is contracting, we get |S|+ |N H (S)|+ |W \T | < 2|T | 2n/3. This implies that the fourth set W ∩ T contains at least n/3 > δn vertices, so we can select some u ∈ S and v ∈ (W ∩ T ) \ N X (u). As vertices in T do not lose incident edges from G to H, every edge from S to T must go to N H (S). Consequently, there can be no edge between S and W ∩ T . So the Ore-type condition yields |N G (u) ∩ N G (v)| > n/3. But by the same argument, none of these joint neighbours can belong to S or W ∩ T . Hence, |N G (u) ∩ N G (v)| |N H (S)| + |W \ T | < |S| + |W \ T | |T | n/3 follows. This contradicts the previous inequality and we conclude that H cannot have an ηn-contracting set.
This means that if three monochromatic components double-cover G and already contain almost all vertices, we can immediately deduce Lemma 7.5 from Lemma 7.9. If that is not the case, however, we will show in Lemma 7.11 that we may assume the following setting.
Proof. Without loss of generality, we can assume R 1 , R 2 to be red. We distinguish between two cases. Case 1. Assume that |R k \ (B ∪ B ′ )| δn for every choice of blue components B, B ′ and k ∈ [2]. We show that this leads to a contradiction. Claim 7.12. There are six vertices u 1 , u 2 , u 3 ∈ R 1 and v 1 , v 2 , v 3 ∈ R 2 such that none of the u j 's and none of the v j 's share their blue component, and Proof of the claim. We distinguish two subcases: For the first subcase, assume that there is a blue component B 3 that intersects with R 1 , but not with R 2 . As no two blue components completely cover R 1 , we can pick two vertices u 1 , u 2 ∈ R 1 that lie in distinct blue components B 1 , B 2 other than B 3 . By assumption, |R 2 \B 1 | δn. Hence, we can select v 1 ∈ R 2 \(B 1 ∪N X (u 1 )) with blue component B(v 1 ). Similarly, we use the assumption For the second subcase, assume that every blue component intersecting with R 1 also intersects with R 2 . We start with any blue component ). Let B 2 be its blue component, pick any v 3 ∈ R 2 ∩B 2 and again use the assumption Now G ∪ X is (n, γ)-Ore, which can be applied to the pairs u j , v j returned by Claim 7.12. The three resulting inequalities add up to 3 j=1 |N G (u j ) ∩ N G (v j )| > n. Next, we show that the three sets on the left are pairwise disjoint. For this, consider . The edges u j w and u k w cannot both be blue as u j and u k lie in different blue components. The same is true for v j w and v k w. So w must be adjacent to some u ∈ R 1 and some v ∈ R 2 by red edges, which is obviously false. Hence, the sets on the left-hand side of the inequality above must indeed be pairwise disjoint. This yields the desired contradiction and concludes Case 1.
Case 2. Assume that |R k \(B 1 ∪B 2 )| < δn for some k ∈ [2] and two blue components Proof of the claim. Let j ∈ [2] be arbitrary and observe that as (G, X, It is not hard to see that most of this intersection belongs to R 3−k ∩ B j . Indeed, as u j and v do not share monochromatic components, their edges to some w ∈ N G (u j ) ∩ N G (v) must have different colours. If u j w is red and vw is blue, then w ∈ R k \ (B 1 ∪ B 2 ), of which there are less than δn vertices. So the remaining at least (1/3 + γ − δ)n > n/3 vertices w must have a blue edge to u j ∈ B j and a red edge to v ∈ R 3−k , putting them into R 3−k ∩ B j . But then R 3−k , B 1 , B 2 double-cover G. Moreover, R 3−k , B 1 , B 2 contain all of R 1 ∪ R 2 except for the fewer than δn vertices in R k \(B 1 ∪B 2 ), so in total at least |R 1 ∪R 2 |−δn (1−7δ)n vertices. Thus, R 3−k , B 1 , B 2 satisfy (1).
So if (1) does not hold, we may assume R 3−k ⊆ B 1 ∪B 2 . But then |R 3−k \(B 1 ∪B 2 )| = 0 < δn, so we can also apply Claim 7.13 to find R k ⊆ B 1 ∪ B 2 . In total, we observe R 1 ∪ R 2 ⊆ B 1 ∪ B 2 and (G, X, B 1 , B 2 ) is split (n, δ, γ)-Ore, thus proving (2). 7.5. Two monochromatic components of the same colour. The remainder of the proof now deals with the setting of Definition 7.10. Here, both two red components R 1 , R 2 and two blue components B 1 , B 2 of G together cover almost all vertices of G. This means that for each monochromatic component L among R 1 , R 2 , B 1 , B 2 , the union H L of the other three contains enough vertices to satisfy Lemma 7.5. So its statement can only be wrong if each of these H L 's contains a stable set S L of sufficient contraction in H L . Our first task will be to locate these sets with Lemmas 7.14, 7.15 and 7.17. We start by showing that accepting negligible losses in contraction, we may assume S L to belong to the intersection of L with only one component of the other colour. The proof is in two steps (Lemmas 7.14 and 7.15).
Being a subset of the (2η ′ n)-contracting set S ′ , the set S ′′ is still stable with c H (S ′′ ) > (2η ′ − 2δ)n η ′ n by Observation 7.6(3) and the choice of constants. In particular, this implies that S ′′ contains at least one vertex v by Observation 7.6(1). As S ′′ ⊆ R 2 ∩ B j , all of v's edges to R 1 must be blue and therefore go to Using |R 1 | |R 2 | + 6δn and assuring δ η/18 when choosing the constants, we get 4n/3 < |R 2 | + |B j | − 12δn. But then at least one of R 2 and B j would contain more than (2/3 + 6δ)n vertices and thus double-cover G together with the two components of the other colour. As this is not the case by assumption, there can be no u ∈ S ′ ∩ R 1 and S ′ ⊆ R 2 holds as claimed.
Lemma 7.15. Let 1/n ≪ δ ≪ γ and (G, X, R 1 , R 2 , B 1 , B 2 ) be evenly split (n, δ, γ)- Proof. Without loss of generality, we assume R 1 , R 2 to be red. Note that since S is stable, N H (S ∩ B 1 ) and N H (S ∩ B 2 ) partition N H (S). Indeed, all s ∈ S ⊆ R 2 lose their incident red edges from G to H = R 1 ∪ B 1 ∪ B 2 . So any vertex in N H (S ∩ B 1 ) ∩ N H (S ∩ B 2 ) would be adjacent to vertices in both S ∩ B 1 and S ∩ B 2 by blue edges, which is obviously impossible. This implies that Before we combine the results of Lemmas 7.14 and 7.15 to obtain Lemma 7.17, we note another general observation in Lemma 7.16. It will be used both to obtain additional information on the locations of the contracting sets in Lemma 7.17 as well as multiple times in the remainder of this third case. Lemma 7.16. Let 1/n ≪ δ ≪ η ≪ γ and (G, X, R 1 , R 2 , B 1 , B 2 ) be evenly split (n, δ, γ)-Ore. Suppose H := R 1 ∪ B 1 ∪ B 2 has an ηn-contracting set S ⊆ R 2 ∩ B j with j ∈ [2]. Then at least one of |(R 1 ∩ B j ) \ N H (S)| < δn and |R 2 ∩ B j | > n/3 hold, both of which imply |R 1 ∩ B j | < |R 2 ∩ B j |.
Proof. Without loss of generality, we can assume j = 1, otherwise exchange the labels of B 1 , B 2 . Assume that |(R 1 ∩ B 1 ) \ N H (S)| δn. So picking any v ∈ S, there must be some u ∈ (R 1 ∩ B 1 ) \ (N H (S) ∪ N X (v)). Then u has no edge to S or R 2 ∩ B 2 in H and furthermore does not lose incident edges from G to H, so deg G (u) = deg H (u) n − |S| − |R 2 ∩ B 2 |. In particular, the Ore-type condition is applicable to u, v and As at most 6δn vertices do not belong to B 1 ∪ B 2 , this immediately implies the desired |R 2 ∩ B 1 | |R 2 | − |R 2 ∩ B 2 | − 6δn > n/3 by the choice of constants.
The second statement is an easy observation: If |(R 1 ∩B 1 )\N H (S)| < δn holds, then |R 1 ∩ B 1 | < |N H (S)| + δn < |S| − (η − δ)n < |R 2 ∩ B 1 | by the ηn-contracting property of S ⊆ R 2 ∩ B 1 and the choice of constants. On the other hand, |R 2 ∩ B 1 | > n/3 and |R 1 ∩ B 1 | |R 2 ∩ B 1 | would immediately combine to R 1 , R 2 , B 1 double-covering G, which is not the case by assumption. Lemma 7.17. Let 1/n ≪ δ ≪ η ′ ≪ η ≪ γ and (G, X, R 1 , R 2 , B 1 , B 2 ) be evenly split (n, δ, γ)-Ore. For each choice of L ∈ {R 1 , R 2 , B 1 , B 2 }, denote the union of the other three as H L . Suppose all of these H L 's have an ηn-contracting set S L . Then for each L, there also is an η ′ n-contracting set Proof. For each lost component L ∈ {R 1 , R 2 , B 1 , B 2 }, Lemma 7.14 guarantees the existence of a (2η ′ n)-contracting set S ′ L in H L that only intersects with L and the two components C 1 , C 2 of the other colour. By Lemma 7.15, we may additionally assume S ′ L to only intersect with one of the C j 's, while at worst cutting its contraction in half to c H L (S ′ L ) > η ′ n. Now Lemma 7.16 applied with η ′ and S ′ L playing the roles of η and S implies that L ∩ C j contains more vertices than K ∩ C j , with K being the other component among R 1 , R 2 , B 1 , B 2 that has the same colour as L and is therefore kept from G to H L . Using this multiple times yields the desired locations of the contracting sets.
Firstly, there is some j ∈ and thereby enforces S ′ Assuming k = 2 and combining these four inequalities then leads to the contradiction So we must have k = 1 as claimed.
The remainder is a two-step argument about these diagonal intersections R 1 ∩ B j and R 2 ∩ B 3−j that contain two contracting sets each. We first bound the number of vertices in these intersections from above and then find vertices of small degree in them, which will later contradict the Ore-type condition.
Proof. Without loss of generality, we can assume R 1 , R 2 to be red and j = 1. We first observe that vertices of R 1 ∩ B 1 cannot be adjacent to vertices of R 2 ∩ B 2 in G. However, both intersections contain sets of contraction above ηn, so more than δn vertices by Observation 7.6(1) and the choice of constants. Thus picking u ∈ R 1 ∩ B 1 and v ∈ (R 2 ∩ B 2 ) \ N X (u), we can use that G ∪ X is (n, γ)-Ore to find that at least one of these intersections has a vertex of degree greater than 2n/3 in G. But then the other intersection must contain less than n/3 vertices. Again without loss of generality, we can assume |R 1 ∩ B 1 | < n/3, otherwise exchange the labels of all four components.
In particular, this assumption excludes the possibility of |B 1 ∩ R 1 | > n/3 when applying Lemma 7.16 to S B 1 ⊆ B 1 ∩ R 1 . So |B 2 ∩ R 1 | < |N H B 1 (S B 1 )| + δn < |S B 1 | − (η − δ)n < |S B 1 | − δn must hold by the choice of constants. Similarly, S R 1 ⊆ R 1 ∩ B 1 guarantees |R 2 ∩ B 1 | < |S R 1 | − δn. Moreover, the Ore-type condition implies that n/3 |N G (u) ∩ N G (v)|. So as each common neighbour of u and v in G must either belong to B 2 ∩ R 1 or R 2 ∩ B 1 , we get Now recall that S R 1 , S B 1 ⊆ R 1 ∩ B 1 , so these two must intersect in at least 2δn vertices. This allows us to pick u 1 ∈ S R 1 ∩ S B 1 and u 2 ∈ (S R 1 ∩ S B 1 ) \ N X (u 1 ). By stability of S R 1 in H R 1 , the edge u 1 u 2 cannot be blue and similarly by stability of S B 1 in H B 1 , it also cannot be red. So u 1 u 2 ∈ E(G ∪ X) and as G ∪ X is (n, γ)-Ore, at least one of the u k ∈ R 1 ∩ B 1 with k ∈ [2] has degree deg G (u k ) > 2n/3. But then |R 2 ∩ B 2 | < n/3 follows by the initial argument.
Lemma 7.19. Let 1/n ≪ δ ≪ η ≪ γ and (G, X, R 1 , R 2 , B 1 , B 2 ) be evenly split (n, δ, γ)-Ore. For each choice of L ∈ {R 1 , R 2 , B 1 , B 2 }, denote the union of the other three as H L . Suppose there is k, j ∈ [2] such that R k ∩ B j contains an ηn-contracting set S R k in H R k and an ηn-contracting set S B j in H B j , but satisfies |R k ∩ B j | < n/3. Then there are at least δn vertices v ∈ R k ∩ B j with deg G (v) < 2n/3.
Proof. Without loss of generality, we can assume R 1 , R 2 to be red. Consider the following two subsets of R k ∩ B j : T R := {v ∈ R k ∩ B j | v has more than |(R k ∩ B j ) \ S B j | red edges to R k ∩ B j } .
Taken together, T B and T R cover all vertices in R k ∩ B j with degree at least 2n/3 in G. Indeed, fix v ∈ (R k ∩ B j ) \ (T B ∪ T R ) and note that (7.8) Applying Lemma 7.16 with S R k ⊆ R k ∩ B j or S B j ⊆ B j ∩ R k playing the role of S and using our assumption of |R k ∩ B j | < n/3 to exclude the possibility of |R k ∩ B j | > n/3, we get Outside of (R 1 ∪ R 2 ) ∩ (B 1 ∪ B 2 ), the vertex v may have at most 12δn neighbours, so the three inequalities (7.8), (7.9) and (7.10) combine to Using |R k ∩ B j | < n/3 again, we observe that deg G (v) < 2n/3 by the contraction property of S R k and S B j as well as the choice of constants. This means that T B ∪ T R can only miss vertices in R k ∩ B j with degree below 2n/3 in G.
It remains to show that there are at least δn such vertices v ∈ R k ∩ B j that have deg G (v) < 2n/3. For a proof by contradiction, we assume otherwise and observe that every subset S ⊆ R k ∩ B j must satisfy |S \ (T B ∪ T R )| < δn. Now every vertex of T B must have a blue edge to S R k ⊆ R k ∩ B j by construction, and therefore cannot itself belong to S R k by stability of S R k in H R k . This shows that T B ⊆ N H R k (S R k ). In particular, |S R k \ T R | = |S R k \ (T B ∪ T R )| < δn holds by the disjointness of S R k and T B . Similarly, we also find T R ⊆ N H B j (S B j ) and |S B j \ T B | < δn. Using c H B j (S B j ) > ηn, this combines to |S R k | < |T R | + δn |N H B j (S B j )| + δn < |S B j | − (η − δ)n < |S B j | by the choice of constants. But the same way, we can also deduce |S B j | < |S R k | from c H R k (S R k ) > ηn and obtain the desired contradiction. So there must indeed be at least δn vertices v ∈ R k ∩ B j with deg G (v) < 2n/3.
The outcome of Lemma 7.19 will of course contradict the Ore-type condition, so we are finally able to prove Lemma 7.5, which is the last missing piece in the proof of Theorem 1.4.
Proof of Lemma 7.5. We choose the constants such that 1/n ≪ δ ≪ η ′ ≪ η ≪ γ satisfies the requirements of Lemmas 7.9, 7.11 and 7.17 to 7.19, in the last two cases with η ′ playing the role of η. Assuring δ η/7, it suffices to prove the statement of Lemma 7.5 with (1 − 7δ)n instead of (1 − η)n. Whenever G is double-covered by three monochromatic components together containing at least (1 − 7δ)n vertices, there is nothing to show as their union cannot contain ηn-contracting bad sets by Lemma 7.9. So by Lemma 7.11, we may assume the existence of two components B 1 , B 2 such that (G, X, R 1 , R 2 , B 1 , B 2 ) is evenly split (n, δ, γ)-Ore.
For a proof by contradiction, we assume that for every choice of L ∈ {R 1 , R 2 , B 1 , B 2 }, the union H L of the other three contains an ηn-contracting set S L . Then Lemma 7.17 yields η ′ n-contracting sets S ′ L in H L such that S ′ R 1 , S ′ B j ⊆ R 1 ∩ B j and S ′ R 2 , S ′ B 3−j ⊆ R 2 ∩ B 3−j for some j ∈ [2]. Applying Lemma 7.18 with η ′ and S ′ * playing the roles of η and S * , these intersections R 1 ∩ B j and R 2 ∩ B 3−j have fewer than n/3 vertices. Again using η ′ and S ′ * in place of η and S * , Lemma 7.19 guarantees that both intersections contain at least δn vertices of degree below 2n/3 in G. We can thus pick u ∈ R 1 ∩ B j with deg G (u) < 2n/3 and v ∈ (R 2 ∩B 3−j )\N X (u) with deg G (v) < 2n/3. However, this obviously contradicts the fact that G∪X is (n, γ)-Ore, thereby proving the lemma.