Ramsey games near the critical threshold
David Conlon, Shagnik Das, Joonkyung Lee, Tam\'as M\'esz\'aros

TL;DR
This paper investigates the behavior of random graphs near the critical threshold for Ramsey properties, showing that even when not Ramsey, they are close to being so, especially for strictly 2-balanced graphs.
Contribution
It extends known results on the Ramsey properties of random graphs near the threshold to a broader class of graphs and addresses open questions about the necessity of strict 2-balance.
Findings
Random graphs near the threshold are nearly Ramsey even when not Ramsey.
The results generalize previous work from triangles to strictly 2-balanced graphs.
The theorems do not hold for graphs that are not strictly 2-balanced.
Abstract
A well-known result of R\"odl and Ruci\'nski states that for any graph there exists a constant such that if , then the random graph is a.a.s. -Ramsey, that is, any -colouring of its edges contains a monochromatic copy of . Aside from a few simple exceptions, the corresponding -statement also holds, that is, there exists such that whenever the random graph is a.a.s. not -Ramsey. We show that near this threshold, even when is not -Ramsey, it is often extremely close to being -Ramsey. More precisely, we prove that for any constant and any strictly -balanced graph , if , then the random graph a.a.s. has the property that every -edge-colouring without monochromatic copies of cannot be extended to an -free colouring after…
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Ramsey games near the critical threshold
David Conlon Department of Mathematics, California Institute of Technology, Pasadena, CA 91125, USA. E-mail: [email protected]. Research supported in part by ERC Starting Grant RanDM 676632.
Shagnik Das Institut für Mathematik, Freie Universität Berlin, 14195 Berlin, Germany. E-mail: [email protected]. Research supported by GIF grant G-1347-304.6/2016 and by the Deutsche Forschungsgemeinschaft (DFG) project 415310276.
Joonkyung Lee
Department of Mathematics, University College London, Gower Street, London WC1E 6BT, UK. E-mail: [email protected]. Research supported in part by ERC Consolidator Grant PEPCo 724903.
Tamás Mészáros Institut für Mathematik, Freie Universität Berlin, 14195 Berlin, Germany. E-mail: [email protected]. Funded by the DRS Fellowship Program and the Berlin Mathematics Research Center MATH+, Project “Learning hypergraphs”.
Abstract
A well-known result of Rödl and Ruciński states that for any graph there exists a constant such that if , then the random graph is a.a.s. -Ramsey, that is, any -colouring of its edges contains a monochromatic copy of . Aside from a few simple exceptions, the corresponding [math]-statement also holds, that is, there exists such that whenever the random graph is a.a.s. not -Ramsey.
We show that near this threshold, even when is not -Ramsey, it is often extremely close to being -Ramsey. More precisely, we prove that for any constant and any strictly -balanced graph , if , then the random graph a.a.s. has the property that every -edge-colouring without monochromatic copies of cannot be extended to an -free colouring after extra random edges are added. This generalises a result by Friedgut, Kohayakawa, Rödl, Ruciński and Tetali, who in 2002 proved the same statement for triangles, and addresses a question raised by those authors. We also extend a result of theirs on the three-colour case and show that these theorems need not hold when is not strictly -balanced.
1 Introduction
The study of sparse generalisations of combinatorial theorems has attracted considerable interest in recent years and there are now several general mechanisms [2, 3, 16, 17] that allow one to prove that analogues of classical results such as Ramsey’s theorem, Turán’s theorem and Szemerédi’s theorem hold relative to sparse random graphs and sets of integers. Much of this work is based, in one way or another, on the beautiful random Ramsey theorem of Rödl and Ruciński [14, 15] from 1995. This seminal result gives a complete answer to the question of when the binomial random graph is -Ramsey, that is, has the property that any -colouring of its edges contains a monochromatic copy of the graph .
To state the Rödl–Ruciński theorem precisely, we need some notation. For a graph , we write if has no edges, when and in the general case. We then write and call this quantity the -density of . Though we will not use these definitions immediately, we also say that is -balanced if and strictly -balanced if for all proper subgraphs of .
Theorem**.**
(Rödl–Ruciński, 1995) Let be a positive integer and let be a graph that is not a forest consisting of stars and paths of length . Then there are positive constants and such that
[TABLE]
There has been much work extending this result. We will not attempt an exhaustive survey, but refer the interested reader instead to some of the latest progress on hypergraphs [8], the asymmetric case [11], establishing sharp thresholds [18] and the equivalent problem in settings other than the binomial random graph [5, 13]. Our particular concern here will be with the following surprising result of Friedgut, Kohayakawa, Rödl, Ruciński and Tetali [6] regarding two-round Ramsey games against a random builder.
Theorem**.**
(Friedgut–Kohayakawa–Rödl–Ruciński–Tetali, 2003) Let be fixed and, for , let . Then, with high probability, the following statements hold:
- (a)
Let be an arbitrary monochromatic--free -edge-colouring of . If , then, with high probability, cannot be extended to a monochromatic--free -edge-colouring of .
- (b)
Let be an arbitrary monochromatic--free -edge-colouring of . If , then, with high probability, cannot be extended to a monochromatic--free -edge-colouring of .
When , the Rödl–Ruciński theorem implies that if for some sufficiently large , then every -edge-colouring contains a monochromatic triangle. Part (a) of the theorem above says that for any , no matter how small, if , then, even though there are -edge-colourings of containing no monochromatic , no such colouring can be extended to a monochromatic--free -edge-colouring after extra random edges are added. One interpretation of this result is that for any the random graph with is, with high probability, already extremely close to being -Ramsey. Part (b) gives a similar result for -edge-colourings, though in this case extra edges may be needed in the second round of colouring to guarantee a monochromatic triangle.
Addressing a problem raised by Friedgut, Kohayakawa, Rödl, Ruciński and Tetali [6], our main result says that a similar statement holds for all graphs containing an edge for which . In particular, the result applies when is strictly -balanced, since any edge works in this case.
Theorem 1.1**.**
Let be a graph and suppose that there is some edge whose removal decreases the -density, that is, . Let be fixed and, for , let . Then, with high probability, the following statements hold:
- (a)
Let be an arbitrary monochromatic--free -edge-colouring of . If , then, with high probability, cannot be extended to a monochromatic--free -edge-colouring of .
- (b)
Let be an arbitrary monochromatic--free -edge-colouring of . If , then, with high probability, cannot be extended to a monochromatic--free -edge-colouring of .
Observe that the densities and of the random graphs that must be added to create monochromatic copies of are best possible. Indeed, if , then with positive probability has no edges, so trivially extends to . Only slightly less trivially, if only uses two of the three colours on the edges of , then we can colour all the edges of with the third colour. If , then with positive probability is -free, thus giving a valid extension of . Finally, note that these results cannot be extended to colours, since the two random graphs and can be coloured independently with disjoint pairs of colours, so we can avoid creating a monochromatic copy of until the density of one of the two random graphs exceeds the random Ramsey threshold from Theorem Theorem.
2 The necessity of a condition
In Theorem 1.1, we impose the condition that there is some edge such that has a strictly lower -density than . While this condition covers, for example, strictly -balanced graphs (where the edge can be chosen arbitrarily), it is natural to ask whether it is necessary. In this section we show that Theorem 1.1 does not apply to all graphs , so some condition is indeed required.
2.1 Edge-rooted products of graphs
We first define the edge-rooted product of graphs.
Definition 2.1**.**
Let be a graph, let be a graph rooted at an edge and let . To build the -fold edge-rooted product , we start with a central copy of and then attach copies of to each edge such that is the root-edge in each copy of and all other vertices in each copy are new and distinct.
In other words, , induces a copy of and, for each and , induces a copy of with playing the role of . (Note that there is some slack in this definition, since we have not prescribed an orientation for each attached copy of . In practice, the particular choice of orientation makes no difference, so we will simply assume that some fixed choice has been made.)
The reduced -fold edge-rooted product, denoted , is the subgraph obtained by removing all the edges from the central copy of .
We have already defined , and stated what it means for a graph to be -balanced or strictly -balanced. In a similar fashion, we write if has no edges and otherwise. We then write and call this quantity the -density of . We say that is -balanced if and strictly -balanced if for all proper subgraphs of . Finally, write and , which we call the density of . We then say that is balanced if and strictly balanced if for all proper subgraphs of . We will make repeated use of the following simple lemma in what follows.
Lemma 2.2**.**
If is -balanced with , then is strictly -balanced and strictly balanced.
Proof.
Suppose that is not strictly -balanced and let be a subgraph with . That is, or, by multiplying the expression out,
[TABLE]
Since is -balanced, we have , which implies that . Rearranging gives . Substituting (1), we get
[TABLE]
Cancelling the like terms gives , which in turn implies that , which can be rewritten as . However, this is a contradiction, since by assumption and . The argument in the strictly balanced case follows along almost exactly the same lines. ∎
The key observation for our purposes is that the edge-rooted product behaves well with respect to the various graph densities.
Lemma 2.3**.**
For any graphs and of density at least , any edge and any :
- (a)
if is strictly balanced, is -balanced and , then is strictly balanced,
- (b)
* and*
- (c)
if , then .
Proof.
- (a)
Let be a smallest (induced) subgraph maximising . We wish to show that . We start with a lower bound on the density of :
[TABLE]
where the inequality on the second line follows since either , in which case we have equality, or . Note that the final equality, , is an application of Lemma 2.2 above.
We also observe that is a convex combination of and :
[TABLE]
Now, for each and , let be the subgraph induced by the vertices of in the th copy of attached to the edge in the central copy of . Let be the subgraph induced by the vertices of in the central copy of .
By the minimality of the size of , we may assume that is connected, as otherwise its densest component would be a smaller subgraph attaining the maximum density. We cannot have since, by (2), the density of is at least , which is strictly larger than . Thus, must be non-empty for at least two pairs and, hence, to be connected, each non-empty must contain at least one vertex of .
Now suppose there was some such that contained only one of the two endpoints of . Then, by removing from , we lose edges and vertices. Since , the ratio is at most , which by (2) is at most . Removing would therefore not decrease the density of , contradicting the minimality of its size. Thus, if is non-empty, we must have . Hence,
[TABLE]
Thus,
[TABLE]
Since and is balanced, . Similarly, for each and , . We therefore have
[TABLE]
Comparing this to (3), since , for to hold we require
[TABLE]
Now , with equality if and only if for all and . Therefore,
[TABLE]
Thus, in order to satisfy the inequality of (5), . As is strictly balanced, it follows that and then, since for all and , we have , as required.
- (b)
Since , we immediately have . The proof of the upper bound follows the same lines as in part (a). Let be a smallest subgraph realising the -density, that is, .
Let and, for each , be defined as in part (a). We may assume that for at least two pairs , since otherwise and thus . By the minimality of the size of , we may further assume that is -connected, as otherwise one of the blocks of will satisfy (see, for instance, Lemma 8 of [12]). In particular, this implies that whenever .
The vertices and edges of can then be enumerated as in (4), so
[TABLE]
Since and for each , it follows that .
- (c)
The product is obtained by deleting the edges of the central copy of from the product . We show by following the argument of part (b). To start, let be a smallest subgraph attaining the -density.
As before, let be the subgraph of induced by the vertices of from the central copy of and, for and , let be the subgraph of induced by the vertices of in the th copy of attached to the edge . Note that, since the edges from the central copy of are deleted in , neither the edges of nor the edge in (if present) appear in . However, it will be convenient for us to include them in and for our calculations.
If is only non-empty for one pair of , then , so . Otherwise, since must be -connected, must be in whenever is non-empty. We can then compute the -density of as in part (b), arriving at an expression similar to (6), except the edges in do not appear in . Thus,
[TABLE]
since implies . ∎
2.2 Graphs requiring unusually many extra random edges
Part (c) of Lemma 2.3 shows the role played by the assumption of the existence of the edge in Theorem 1.1. We will show how to use this to prove Theorem 1.1 in the next section, but first we use the other parts of this lemma to construct graphs for which the conclusion of Theorem 1.1 does not hold.
Theorem 2.4**.**
Let be a -balanced graph containing a cycle, let be an arbitrary edge of and let . Let for , where is a sufficiently small constant. Then, with high probability, the following statements hold:
- (a)
There is a monochromatic--free -edge-colouring of such that if the colouring can with high probability be extended to a colouring of without monochromatic copies of .
- (b)
There is a monochromatic--free -edge-colouring of and such that if the colouring can with high probability be extended to a colouring of without monochromatic copies of .
Proof.
Since is -balanced and contains a cycle, we have and . Using Lemma 2.3(b), . Hence, if the constant is sufficiently small, the Rödl–Ruciński theorem implies that with high probability we can find a -colouring of without any monochromatic copy of . This is the edge-colouring we extend in both cases.
- (a)
In this case, extend to the edges of arbitrarily. Observe that consists of edge-disjoint copies of . Since there are no monochromatic copies of in , any monochromatic copy of in must contain at least edges from .
There are at most potential copies of and ways to distribute its edges between and . Since , the probability that a copy with at least edges from appears in is at most . Thus, by the union bound, the probability that there is a copy of in with at least edges from is at most
[TABLE]
where we used that and . As , this simplifies to , which is by our choice of . Hence, with high probability our arbitrary extension of to does not create a monochromatic copy of .
- (b)
The colouring uses two colours, say red and blue. This leaves us with one unused colour, say green, that we can use when extending to the edges of .
As is -balanced, Lemma 2.2 implies that it is also strictly balanced. By Lemma 2.3(a), it follows that is strictly balanced. As a consequence, any union of two copies of that share at least an edge must be strictly denser than itself. Indeed, the subgraph common to both copies of is a proper subgraph and therefore strictly sparser than . Hence, the vertices and edges added in the second copy in the union must increase the overall density.
There is thus some such that, whenever , intersecting copies of do not appear in . That is, the copies of appearing in are with high probability pairwise edge-disjoint. We shall choose our to be less than this .
We now order the edges of arbitrarily and process them one-by-one. We colour each edge green, unless that would create a green copy of , in which case we colour the edge red. When colouring in this fashion, if we create a monochromatic copy of , it clearly must be red.
Consider a red copy of in our colouring of . Since is an edge-disjoint union of copies of and the colouring of has no monochromatic copy of , each copy of in must contain at least one red edge from . An edge from is only red if it is the last edge of an otherwise green copy of , which must be wholly contained in . Moreover, for , the copies and of are edge-disjoint.
This gives us a subgraph of with at most vertices and edges, of which at least edges come from . Since and there are at most some constant ways of building such a subgraph and dividing its edges between and , the probability of finding such a structure is at most
[TABLE]
Since , this is at most . Now , since is strictly balanced. Thus the upper bound on the probability of the appearance of a red copy of is , since . Hence, if we choose , this probability is , so with high probability we can extend the colouring to the edges of without creating a monochromatic copy of . ∎
3 The proof of Theorem 1.1
Having shown in the previous section that some condition on the graph is necessary in Theorem 1.1, we now show that our condition is sufficient. We begin with a sketch of the proof and then recall several useful results before providing the details of the argument.
3.1 An overview of the proof
We shall assume the colours used are red, blue and, in the case of three-colourings, green. Our goal is to find structures in the first random graph, , that force the creation of a monochromatic copy of no matter how the edges of the second random graph, , are coloured. To that end, we make the following definitions.
Definition 3.1** (Colour-forced edges).**
A copy of in is supported on the pair if maps to the missing edge . We then call the base of the copy. Given an edge-colouring , we say is a red, blue or green base if it is the base of a monochromatic copy of of the corresponding colour. Finally, we say a pair is green-forced if it is both a red and a blue base simultaneously, with blue-forced and red-forced defined similarly.
In the two-colour case, observe that it is impossible to extend to a green-forced pair, since colouring it either red or blue would create a monochromatic copy of . For the first assertion of Theorem 1.1, we shall show that with high probability is such that every two-colouring admits quadratically many green-forced pairs. Then, again with high probability when , one of these pairs will be an edge of the second random graph , so any extension of to will create a monochromatic copy of .
When dealing with three colours, our goal will instead be to show that there is some colour, say green, such that the green-forced pairs in are sufficiently dense that, when , we will find a copy of in consisting solely of green-forced pairs. If any one of its edges is coloured red or blue, it will complete a monochromatic copy of with edges from . On the other hand, if all of its edges are coloured green, we obtain a green copy of instead.
To find these colour-forced structures, we consider the reduced graph of a regular partition of (with respect to the colouring or ). In this reduced graph we will find two colours, say red and blue, and a copy of such that for each (removed) edge from the central copy of , one of the attached copies of is monochromatic red and the other is monochromatic blue. By applying the sparse counting lemma, we will deduce the existence of many potential copies of consisting of green-forced edges, from which we will be able to draw the desired conclusion.
Although the proof can be simplified in the two-coloured setting, for the sake of brevity we shall present a single unified argument allowing for three colours throughout, and only differentiate between the two cases at the end of the proof.
3.2 Some preliminaries
Here we collect several results about random graphs and sparse regularity that we shall use in our proof.
3.2.1 Random graphs
The Hoeffding inequality shows that does not have any subgraphs that are far sparser or denser than expected with high probability.
Proposition 3.2**.**
Let be fixed and suppose . Then, with high probability, is such that the following holds for any disjoint sets of vertices with :
- (i)
* and*
- (ii)
.
A simple application of Markov’s inequality also shows that is unlikely to contain many more copies of any subgraph than expected.
Proposition 3.3**.**
Given any graph with vertices and edges and any , the probability that there are more than copies of in is at most .
In the other direction, we can use Chebyshev’s inequality to establish the existence of subgraphs in when is suitably large. More precisely, it follows from Theorem 4.4.5 in The Probabilistic Method by Alon and Spencer [1] that if , then the number of copies of in is concentrated around its expectation.
Proposition 3.4**.**
Given a graph on vertices and a constant , let be a collection of potential copies of . If with some , then the probability that does not contain a copy of from is at most .
If the edge probability is even larger, then the following result, a consequence of Theorem 3.29 from the book Random Graphs by Janson, Łuczak and Ruciński [9], shows that there will be many pairwise edge-disjoint copies of in .
Proposition 3.5**.**
For every graph with , there is a constant such that, given constants and setting , with high probability every induced subgraph of on at least vertices contains at least edge-disjoint copies of .
3.2.2 Sparse regularity and counting
Given an -vertex graph , two disjoint sets of vertices and form an -regular pair of density if and, for all with and with , we have , where denotes . This notion of regularity is inherited by induced and random subgraphs (see Lemma 4.3 in [7]).
Proposition 3.6**.**
Suppose that , is a graph and are disjoint vertex sets, both of size , with an -regular pair of density . Then the following is true:
- (i)
for and with , the pair is -regular with density at least and
- (ii)
for , the subgraph of obtained by choosing edges from uniformly at random forms a -regular pair with high probability.
An -regular partition of is a partition such that , and all but at most pairs , , are -regular. When the graph is edge-coloured, we say a partition is -regular if for all but at most pairs of parts the edges of each colour between the two parts form an -regular subgraph. If has density , we say it is -upper-uniform if, for all disjoint sets and of size at least , we have . With these definitions in place, we may state a version of the sparse regularity lemma, originally due to Kohayakawa and Rödl [10].
Theorem 3.7**.**
For all and , there are and such that every -colouring of the edges of an -upper-uniform graph of density on at least vertices has an -regular partition with parts for some .
The final ingredient we will need is a sparse counting lemma due to Conlon, Gowers, Samotij and Schacht [4]. Given a graph , integers and , and , we define the family to be all graphs obtained by replacing each vertex of by an independent set of size and replacing each edge of by an -regular bipartite graph with exactly edges. Given such a graph , let denote the number of canonical copies of in (by which we mean that each vertex of in the copy belongs to the corresponding independent set in ).
Theorem 3.8**.**
For every graph and every , there exist with the following property. For every , there is such that if , then, with high probability, for every , and every subgraph of in , .
3.3 The reduced graph
With these preliminaries in hand, we can proceed with the proof of Theorem 1.1. We begin by describing the (standard) construction of the reduced graph and proving it has some useful properties.
Let be defined such that , where is the constant from Proposition 3.5, and ‘’ means these parameters are sufficiently small for the subsequent calculations to hold.
Now consider a monochromatic--free -edge-colouring of the edges of (where, as in the case of , we may only be using two of the three colours) and let , and represent the red, blue and green subgraphs of , respectively. Given our choice of and and setting and , let and be as in Theorem 3.7. Proposition 3.2 shows that is with high probability -upper-uniform. Hence, there is an -regular partition , where .
We next define three graphs, , and , on the same vertex set . has an edge between and if and only if the bipartite induced subgraph forms an -regular pair of density at least , with and defined similarly with respect to and , respectively. The reduced (multi)graph is the coloured union of in red, in blue and in green. Given a vertex , we write , and for its neighbourhoods in , and , respectively, and write and for the sizes of these sets.
We first show that any induced subgraph of with linearly many vertices has a vertex with large degree in at least two of the colours.
Lemma 3.9**.**
Define . Suppose satisfies
[TABLE]
Then, with high probability, for any subset of vertices of the reduced graph , we can find a vertex , two disjoint sets of size at least and two distinct colours such that, for each , is adjacent to all vertices in with edges of colour .
Proof.
Let be the vertices in the parts of corresponding to the vertices of and note that . Hence, by Proposition 3.5, we may assume contains at least edge-disjoint copies of . Since there are no monochromatic copies of in the -edge-colouring of , each such copy must contain two edges of distinct colours. It easily follows that there are two colours, say red and blue, that each appear on at least edges of .
Using Proposition 3.2, we observe that all but at most red edges of are contained within dense -regular pairs. Indeed, at most edges can be contained within the parts , at most edges can be within irregular pairs and at most edges are within -regular pairs of density less than . From (7), it follows that there are at least red edges in that are contained in -regular pairs of density at least . Again by Proposition 3.2, each such pair can account for at most edges in , so there must be at least such pairs, each of which corresponds to an edge of . By symmetry, we also find at least edges in .
Now let . By summing the red degrees of vertices in , distinguishing between those in and those not, we have
[TABLE]
from which we deduce that . Defining , we similarly have . If , let . Since , we can find the required disjoint sets and of size of red and blue neighbours, respectively.
Otherwise, for every and , by Proposition 3.2, there are at least edges in between and , so one of the three colours appears on at least edges. Let be the colour that appears most commonly as the majority colour in these pairs. Ignoring the pairs that give rise to irregular pairs in , it follows that there are at least edges in between and . Provided is sufficiently large with respect to , (7) and our lower bound on imply this is at least edges.
If is not red, then take to be red and, by averaging, find some with a set of at least neighbours in in the colour . Since , we have , so we can find a disjoint set of red neighbours of , as required. Otherwise, if is red, we take to be blue. By the same argument, we can find some with a set of at least red neighbours in and, since , it has large enough degree in to guarantee a disjoint set of blue neighbours. ∎
Through repeated use of this lemma, we can build large multicoloured structures in .
Corollary 3.10**.**
Given , let and, for , let . Provided , there is with high probability a vertex of contained in two monochromatic -cliques of distinct colours.
Proof.
Applying Lemma 3.9 with and , we find a vertex with large degrees in two colours. Without loss of generality, let the colours be red and blue. In the first stage of this algorithm, we iterate within the red neighbourhood of , finding either a vertex in blue and green -cliques, in which case we are done, or a red -clique containing .
To start, apply Lemma 3.9 again, this time taking to be the set of red neighbours of . This gives a vertex with large degrees in two colours. While one of those colours is red, we repeat the process, giving us a sequence of vertices with large nested red neighbourhoods. If this sequence (including ) has length , by choosing an arbitrary vertex in the final red neighbourhood, we obtain a red -clique containing and can proceed to the second stage.
Otherwise, after some steps we obtain a vertex that has large blue and green neighbourhoods. In this case, we first iterate within the blue neighbourhood of . Each subsequent vertex has either a large red neighbourhood or a large blue neighbourhood within which we can proceed. Once we have obtained a sequence of vertices, there are either of them (including ) for which we iterated within a red neighbourhood or of them (including ) for which we iterated within a blue neighbourhood. In the first case, we choose an arbitrary vertex in the final neighbourhood to create a red -clique containing and can then proceed to the second stage.
In the second case, choosing an arbitrary vertex in the final neighbourhood gives a blue -clique containing . We can then return to the green neighbourhood of and repeatedly iterate, at each point proceeding with a red or green neighbourhood of the latest vertex. Once we reach a sequence of length (including the vertices between and ), we again either have vertices with green neighbourhoods or vertices with red neighbourhoods. In the first case, we can complete a green -clique containing that, together with the earlier blue -clique, completes the desired structure. In the second case, choosing a vertex in the final neighbourhood again completes a red -clique containing , with which we proceed to the second stage.
If we proceed to the second stage, we will have already found a red -clique containing . The second stage consists of mirroring the above process in the blue neighbourhood of . This results in a blue -clique containing or a vertex in the blue neighbourhood that is contained in both red and green -cliques; in either case, we are done. ∎
3.4 Building colour-forced structures
Let denote the family of all copies of in . Using the cliques from Corollary 3.10, we will prove the following key proposition.
Proposition 3.11**.**
There are positive constants and such that, for any , with probability at least , for every monochromatic--free -edge-colouring of , there is some colour with at least -forced copies of in .
Proof.
For convenience, we write and . Setting and applying Corollary 3.10, which holds with high probability, we find a vertex in the reduced graph that is in, say, both a red and a blue . Let be the other vertices from the red clique and be the other vertices from the blue clique. Consider the corresponding parts in the graph . We know that, for all , the pairs and are all -regular pairs of density at least . This situation is illustrated below in the case .
Partition the part into equal-sized subsets, , letting denote the size of these sets. Define by , noting that , where we recall that is the number of parts in the -regular partition of . For each , let and be arbitrary subsets of size . Let , and . By Proposition 3.6(i), it follows that the pairs and are all -regular of density at least .
Next consider the graph and note that it has precisely vertices, with one central copy of , whose edges are deleted, and each deleted edge supporting two otherwise vertex-disjoint copies and of . We can build a bijection such that:
- •
and
- •
for all , and .
That is, for each edge , we send one of the attached copies of to the red parts and the other copy to the blue parts .
Let and consider an edge . If for some , then the pair is an -regular pair of density at least . Define to be the subgraph obtained from this pair by choosing edges uniformly at random. By Proposition 3.6, is -regular with high probability. Otherwise, for some , in which case we define to be the subgraph obtained by selecting edges uniformly at random from . We again have, with high probability, that is -regular.
Now define the subgraph to be the union of all these subgraphs , that is,
[TABLE]
From the above discussion, it is clear that , where this family of graphs is as defined before Theorem 3.8. Since and, by Lemma 2.3(c), , we can apply Theorem 3.8. This gives some constant such that there are with high probability at least copies of in , where each vertex comes from the set . To simplify this expression, we define and . Our lower bound on the number of copies of can then be written as . Note that is a constant, while, since , .
In each such copy of , each missing edge in the central copy of supports both a red copy of and a blue copy of . In particular, this means is green-forced and, as this holds for all edges , this shows that the central copy forms a green-forced copy of in .
However, we are not quite done, as these green-forced copies of may contribute to multiple copies of , in which case they will have been overcounted. To rectify this, and complete the proof, we now show that most of these copies of are not counted too often.
To this end, suppose we have found distinct green-forced central copies of above and enumerate them as . For each , let denote the number of copies of found above in which is the central copy . We have thus far established that
[TABLE]
while we wish to show that .
Now consider the quantity . This counts the number of ordered pairs of canonical copies of with as the central copy . Given such a pair, let . In , each edge is contained in a red copy of from and one from as well. The same holds true for the blue copies of . These attached copies of in are mostly disjoint outside the central , except that the two copies of the same colour supported on the same edge may share some vertices. We consider such a graph as a degenerate copy of . There are several isomorphism classes could belong to, depending on which vertices are shared by and .
For each edge , let be the subgraph of induced by the vertices shared between the red copies of in and supported on and define analogously for the blue copies. We include the edge in and , even though it does not appear in . Note that the union determines the isomorphism class of . Hence, there are at most possible isomorphism types, as for each vertex in , we can decide whether or not it belongs to the corresponding . Set .
We shall use Proposition 3.3 to show that, regardless of isomorphism type, there cannot be many copies of in . Indeed, we have
[TABLE]
and
[TABLE]
This gives
[TABLE]
Since and , we have for all . Thus, . Hence, by Proposition 3.3, with probability at least there are at most copies of in . Taking a union bound over all isomorphism classes, we find that with probability at least , there are at most of these degenerate copies of in .
We noted earlier that each pair of copies of counted by gives rise to a degenerate copy of . To reverse the correspondence, for each vertex in , we must decide how to assign the corresponding vertices of to and . Thus, there are at most pairs giving rise to the same .
Putting all this together, we have, with probability at least ,
[TABLE]
Define . It then follows from the above inequality that . Plugging this into (8), we obtain . As there are at most summands, each of which has size less than , we can conclude that
[TABLE]
Setting completes the proof. ∎
3.5 Finishing the proof
We begin with part (a). Suppose we have a monochromatic--free -edge-colouring of and for some . Set . By Proposition 3.11, with probability , there is some colour such that there are at least -forced copies of . As the colouring only has red and blue edges, the colour must be green.
Each edge can be in at most green-forced copies of , so there must be at least green-forced edges. If any of these edges were to appear in , we would not be able to extend the colouring , as colouring the edge red or blue creates a monochromatic copy of . Hence, the probability that extends to is at most
[TABLE]
as required.
Part (b) follows the same lines. We begin as before: given the colouring and some , where , we set . By Proposition 3.11, with probability , there is some colour with at least -forced copies of .
If any -forced copy of appears in , then cannot be extended. Indeed, colouring all of its edges with the colour clearly creates a monochromatic copy of , but since all the edges are -forced, using any other colour on an edge also completes a monochromatic copy. By Proposition 3.4, the probability that none of the -forced copies of appear in is at most
[TABLE]
as desired. This completes the proof of Theorem 1.1.
4 Concluding remarks
Our investigations point to several open problems, perhaps the most interesting of which is to classify all graphs for which Theorem 1.1 holds. We have shown that our condition, that there exists an edge such that , cannot be entirely dispensed with. However, there are also examples of graphs which do not satisfy this condition, but still satisfy some of the conclusions of Theorem 1.1.
Indeed, our proof of Theorem 1.1(a) readily generalises to the following statement.
Theorem 4.1**.**
Given a graph , suppose there are graphs , and a matching such that
- (i)
, with forming an independent set in and ,
- (ii)
,
- (iii)
* for all with and and*
- (iv)
for any partition of the matching , is a subgraph of or .
Let be fixed and, for , let . Then, with high probability, the following holds. Let be an arbitrary monochromatic--free -edge-colouring of . If , then, with high probability, cannot be extended to a monochromatic--free -edge-colouring of .
One of the simplest examples satisfying the conditions of Theorem 4.1 is the graph consisting of two triangles joined by a path of length . In this case we can take to have size three with the corresponding graphs and depicted below.
This example shows that the condition in Theorem 1.1 is not best possible, as Theorem 4.1 applies to a wider class of graphs. However, there is a subtle trade-off in finding appropriate forcing structures for Theorem 4.1 — we need them to be sparse enough to satisfy (ii) and (iii), but to have enough copies of for (iv).
Acknowledgements. Part of this work was carried out while the third author visited the second and fourth authors at FU Berlin and he is grateful for their hospitality.
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