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. 2023 Aug 3;64(2):157–169. doi: 10.1002/rsa.21173

A lower bound for set‐coloring Ramsey numbers

Lucas Aragão 1, Maurício Collares 2, João Pedro Marciano 1, Taísa Martins 3, Robert Morris 1,
PMCID: PMC10952192  PMID: 38516561

Abstract

The set‐coloring Ramsey number Rr,s(k) is defined to be the minimum n such that if each edge of the complete graph Kn is assigned a set of s colors from {1,,r}, then one of the colors contains a monochromatic clique of size k. The case s=1 is the usual r‐color Ramsey number, and the case s=r1 was studied by Erdős, Hajnal and Rado in 1965, and by Erdős and Szemerédi in 1972. The first significant results for general s were obtained only recently, by Conlon, Fox, He, Mubayi, Suk and Verstraëte, who showed that Rr,s(k)=2Θ(kr) if s/r is bounded away from 0 and 1. In the range s=ro(r), however, their upper and lower bounds diverge significantly. In this note we introduce a new (random) coloring, and use it to determine Rr,s(k) up to polylogarithmic factors in the exponent for essentially all r, s, and k.

Keywords: probabilistic method, Ramsey theory, random graphs

1. INTRODUCTION

The r‐color Ramsey number Rr(k) is defined to be the minimum n such that every r‐coloring χ:E(Kn){1,,r} of the edges of the complete graph on n vertices contains a monochromatic clique of size k. These numbers (and their extensions to general graphs, hypergraphs, etc.) are among the most important and extensively‐studied objects in combinatorics, see for example the beautiful survey article [6].

In this paper, we will study the following generalization of the r‐color Ramsey numbers.

Definition 1.1

The set‐coloring Ramsey number Rr,s(k) is the least n such that every coloring χ:E(Kn)[r]s contains a monochromatic clique of size k, that is, a set SV(Kn) with |S|=k and a color i[r] such that iχ(e) for every eS2.

In other words, we assign a set χ(e)[r]={1,,r} of s colors to each edge of the complete graph, and say that a clique is monochromatic if there exists a color i[r] that is assigned to every edge of the clique. Note that when s=1 this is simply the usual r‐color Ramsey number Rr(k), for which the best known bounds are

2Ω(kr)Rr(k)rO(kr).

Both bounds have simple proofs: the upper bound follows from the classical method of Erdős and Szekeres [8], while the lower bound can be proved using a simple product coloring due to Lefmann [11]. Determining whether or not Rr(k)=2Θ(kr) is a major and longstanding open problem, see for example the recent improvements of the lower bound in [3, 13, 14].

The study of set‐coloring Ramsey numbers was initiated in the 1960s by Erdős, Hajnal and Rado [7], who conjectured that Rr,r1(k)2δ(r)k for some function δ(r)0 as r. This conjecture was proved by Erdős and Szemerédi [9] in 1972, who showed that in fact

2Ω(k/r)Rr,r1(k)rO(k/r).

The lower bound follows from a simple random coloring, while to prove the upper bound Erdős and Szemerédi showed that any 2‐coloring in which one of the colors has density at most 1/r contains a monochromatic clique of size c(r)logn, where c(r)=Ω(r/logr), and then applied this result to the color that is assigned to most edges.

For more general values of s, the first significant progress was made only recently, by Conlon, Fox, He, Mubayi, Suk and Verstraëte [4], who showed that

Rr,s(k)=2Θ(kr)

for every function s=s(r) such that s/r is bounded away from 0 and 1. More precisely, they used a clever generalization of the approach of Erdős and Szemerédi [9] to prove that

Rr,s(k)exp(ck(rs)2rlogrmin{s,rs}) (1)

for some absolute constant c>0, and defined a product colouring, which exploited an interesting and surprising connection with error‐correcting codes, to show that

Rr,s(k)exp(ck(rs)3r2) (2)

for some constant c>0. They also noted that a simple random coloring gives a lower bound of the form Rr,s(k)2Ω(k(rs)/r), which is stronger when rsr.

While the exponents in (1) and (2) differ by only a factor of logr when rs=Ω(r), they diverge much more significantly when (rs)/r0.1The main result of this paper is the following improved lower bound, which will allow us to determine Rr,s(k) up to a poly‐logarithmic factor in the exponent for essentially all r, s, and k.

Theorem 1.2

There exist constants C>0 and δ>0 such that the following holds. If r,s with srClogr, then

Rr,s(k)exp(δk(rs)2r) (3)

for every k(C/ε)logr, where ε=(rs)/r.

Note that the bound (3) matches the upper bound (1) on Rr,s(k), proved in [4], up to a factor of O(logr) in the exponent for all srClogr. When srClogr our method does not provide a construction, but in this case the bounds from [4] only differ by a factor of order (logr)2 in the exponent, the lower bound coming from a simple random coloring.

The lower bound on k in Theorem 1.2 is also not far from best possible, since if k1/ε then the most common color has density at least 11/k, and therefore Rr,s(k)k2, by Turán's theorem. We remark that the range in which Turán's theorem provides an optimal bound was investigated in detail by Alon, Erdős, Gunderson and Molloy [1] in the case s=r1. In Section 4, we will describe a simpler version of our construction which proves the lower bound

Rr,s(k)(ε(k1)e)εr/2 (4)

for all r, s, and k. Note that this bound matches the upper bound (1) up to a factor of order (logr)2 when 3/ε+1k(C/ε)logr. Moreover, as long as k(1+δ)/ε+1 for some constant δ>0, we can use the same construction to prove (with a slightly more careful calculation) a lower bound of the form

Rr,s(k)2Ω(εr). (5)

Thus, writing Θ˜ to hide poly‐logarithmic factors of r, we obtain the following corollary.

Corollary 1.3

Let r>s1 and δ>0, and set ε=(rs)/r. We have

Rr,s(k)=2Θ˜(ε2rk)

for every k(1+δ)/ε+1.

It would be interesting to determine the behavior of Rr,s(k) in the range εk=1+o(1), especially in light of the initial progress made in [1].

2. THE CONSTRUCTION

In this section, we will define the (random) coloring that we use to prove Theorem 1.2, and give an outline of the proof that it has the desired properties with high probability. The idea behind our construction, to let each color be a random copy of some pseudorandom graph, was introduced in the groundbreaking work of Alon and Rödl [2] on multicolor Ramsey numbers, and has been used in several recent papers in the area [10, 12, 13, 14]. However, our approach differs from that used in these previous works in several important ways; in particular, we will not count independent sets, and it will be important that our color classes are chosen (almost) independently at random. We will discuss the novel aspects of our construction in more detail at the end of this section.

Our first attempt at defining a coloring χ:E(Kn)[r]s with no monochromatic Kk is as follows: let the edges eE(Kn) with iχ(e) be a random blow‐up of the random graph G(m,p), where p=1Θ(ε), and m is chosen so that G(m,p) is unlikely to contain a copy of Kk. The problem with this coloring is that there will be some ‘bad’ edges which receive fewer than s colors,2and we will therefore need to modify the construction by giving these edges extra colors. The key idea is to only give them the colors for which they are ‘crossing’ edges, that is, not in the same part of the random blow‐up for that color.

We will next define the coloring precisely. Fix a sufficiently small3 constant δ>0, and set C=1/δ3. Recall that rs=εr, and define

m=2δ2εkandn=2δ4ε2rk.

Note that εmk, since k(C/ε)logr and ε1/r, and by our choice of C.

Set p=15δε, and for each color i[r], let

  1. Hi be an independently chosen copy of the random graph G(m,p), and

  2. ϕi:[n][m] be an independently and uniformly chosen random function.

Now define Gi to be the (random) graph with vertex set [n] and edge set

E(Gi)={uv:{ϕi(u),ϕi(v)}E(Hi)},

that is, a random blow‐up of Hi, with parts given by ϕi. Define a coloring χ of Kn by

χ(e)={i[r]:eE(Gi)},

and define the set of bad edges to be

B={eE(Kn):|χ(e)|<s}. (6)

We will also say that an edge e=uvE(Kn) is icrossing if ϕi(u)ϕi(v), and define

κ(e)={i[r]:eisicrossing}.

We can now define the colouring that we will use to prove Theorem 1.2.

Definition 2.1

For each eE(Kn), we define the set of colors χ(e)[r] by

χ(e)=χ(e)ifeB,κ(e)ifeB.

Our task is to show that with high probability |χ(e)|s for every eE(Kn), and moreover that χ contains no monochromatic copy of Kk. Before giving the details, let us briefly outline how we will go about proving these two properties.

The first property, that |χ(e)|s for every eE(Kn), is a relatively straightforward consequence of the definition of χ and our choice of m. Indeed, by Definition 2.1 and (6), it will suffice to show that |κ(e)|s for every eE(Kn), and this follows from a simple first moment calculation, using the fact that n=mδ2εr.

Proving that with high probability χ contains no monochromatic copy of Kk is more difficult, and will be the main task of Section 3. Cliques with few bad edges are easily dealt with using Chernoff's inequality, so let us focus here on cliques with many bad edges, where ‘many’ means at least t=δεk2. The difficulty in this case is that the events {eB} and {fB} are correlated, since the endpoints may be in the same part in some of the random partitions. In particular, if ϕi(u)=ϕi(v) for many colors i[r] and many pairs {u,v} of high‐degree vertices of F, the graph of bad edges in our monochromatic k‐clique, then Chernoff's inequality will not provide strong enough bounds on our large deviation events.

To deal with this issue, we will use the randomness of the partitions ϕ1,,ϕr to show (see Lemma 3.5) that there is not too much ‘clustering’ of the vertices of any graph FKn with k vertices and t edges. To do so, we will not be able to use a simple union bound over all graphs, since there will be too many choices for the low‐degree vertices of F; instead, we will need to find a suitable ‘bottleneck event’ for each F, and apply the union bound to these events. Roughly speaking, we will find an initial segment A of the vertices of F, ordered according to their degrees, with the following property: there are δεr|A| pairs vA and i[r] such that there exists an ‘earlier’ vertex uA with ϕi(u)=ϕi(v). We will then sum over the choices of A, using the fact that for each such pair vA and i[r], this event (conditioned on the choices of ϕi(u) for earlier u) has probability at most k/m.

On the other hand, when there is not too much clustering of the high‐degree vertices of F in the random partitions ϕ1,,ϕr, we will use the randomness in the choice of H1,,Hr to bound the probability that there are more bad edges than expected. More precisely, we will choose one vertex from each cluster in each color, and apply Chernoff's inequality, noting that the edges between these vertices are independent.

The proof outlined above diverges from the method of Alon and Rödl [2] (and the more recent constructions of [10, 12, 13, 14]) in several important ways. Of these new ideas, we note in particular that our choice of κ(e) for the bad edges is somewhat subtle (and perhaps surprising), since κ(e)=[r] for almost all edges eE(Kn), which seems very wasteful. The point is that there are very few bad edges, and by only including crossing colors in χ(e) we ensure that there is not too much dependence between the colors of different edges. It is also important that the random graphs H1,,Hr are chosen independently; if we used the same random graph for each color, then we would not be able to prove a sufficiently strong bound on the probability that a clique has too many bad edges.

3. THE PROOF

We begin by observing that with high probability |χ(e)|s for every eE(Kn).

Lemma 3.1

With high probability, |χ(e)|s for every eE(Kn).

Note first that, by the definition of B, if e is not a bad edge then χ(e)=χ(e) and |χ(e)|s. It will therefore suffice to show that

(|κ(e)|<s)1n3 (7)

for each edge eE(Kn). To do so, note that for each i[r] we have

(iκ(e))=1m,

all independently, by the definition of the functions ϕi. Recalling that rs=εr and that εmm=2δ2εk/2, since k(C/ε)logr and C=1/δ3, it follows that

(|κ(e)|<s)rεr(1m)εr(eεm)εr2δ3ε2rk,

Since n=2δ4ε2rk, we obtain (7), and so the lemma follows by Markov's inequality.

We are left with the (significantly more challenging) task of showing that, with high probability, χ contains no monochromatic copy of Kk. We will first deal with the (easier) case in which the clique has few bad edges. Recall that t=δεk2.

Lemma 3.2

With high probability, the colouring χ contains no monochromatic kclique with at most t bad edges.

Suppose χ contains a monochromatic clique S={v1,,vk} of color i[r] such that at most t of the edges eS2 are bad. For each j[k], let wj=ϕi(vj)V(Hi), and observe that the set W={w1,,wk} has size k, since by Definition 2.1, and the fact that χ(e)κ(e), every edge eE(Kn) such that iχ(e) is i‐crossing.

Now, if e=vjvS2 is not a bad edge, then iχ(e)=χ(e), and hence wjwE(Hi). Since there are at most t bad edges in S2, it follows that

e(Hi[W])k2t>pk2+δεk2,

since p=15δε and t=δεk2. Since Hi[W]G(k,p), it follows from Chernoff's inequality4 that this event has probability at most eδ2εk2. Therefore, taking a union bound over colors i[r] and sets WV(Hi) of size k, the probability that χ contains a monochromatic clique with at most t bad edges is at most

rmkeδ2εk2r(2δ2εk·eδ2εk)k. (8)

Since δ3εklogr, the right‐hand side of (8) tends to zero as k, as required.

It remains to prove the following lemma, which is not quite so straightforward.

Lemma 3.3

With high probability, every kclique contains at most t bad edges.

In other words, our task is to show that, with high probability, there does not exist a graph F in the family

={FKn:v(F)=kande(F)=t}

such that FB.5We will not be able to prove this using a simple first moment argument, summing over all graphs F, since the probability of the event {FB} is not always sufficiently small. Instead, we will need to identify a ‘bottleneck event’ for each F.

To do so, let us first choose an ordering F on the vertices of F satisfying

uFvdF(u)dF(v).

In other words, we order the vertices according to their degrees in F, breaking ties arbitrarily. Now, define

Qi(F)={vV(F):uV(F)withuFvsuch thatϕi(u)=ϕi(v)}

to be the set of vertices which share a part of ϕi with another vertex of F that comes earlier in the order F. We remark that if uFv, then uv.

We will bound the probability in two different ways, depending on the size of

XF=i=1rvQi(F)dF(v).

When XF is large, we will find an initial segment A of the order F such that there are at least δεr|A| pairs i[r] and vQi(F)A. First, however, we will deal with the case in which XF is small, where we can use a simple union bound.

Lemma 3.4

With high probability, there does not exist

F

with

XFεrt2andFB.

We first reveal the random functions ϕ1,,ϕr, and therefore the sets Qi(F) (and hence also the random variable XF) for each F. To prove the lemma we will only need to use the randomness in the choice of H1,,Hr. More precisely, we will consider the set

Y={(uv,i)E(F)×[r]:u,vQi(F)}

of pairs (e,i)E(F)×[r] such that neither endpoint of e is contained in Qi(F), and

Z=(e,i)Y1[eE(Gi)],

the number of such pairs for which iχ(e). Note that |Y|rt, and that

ZBin(|Y|,1p),

since the events {eE(Gi)} for (e,i)Y are independent, and correspond to the appearance of certain edges in the graphs H1,,Hr. Indeed, for each i[r], the graph {e:(e,i)Y} is contained in a clique with at most one vertex in each part of ϕi.

Thus, given |Y| (which is determined by ϕ1,,ϕr) the random variable Z is a binomial random variable with expectation

𝔼[Z||Y|](1p)rt=5δεrtεrt4.

Now, note that if FB, then for each edge eE(F), there are at least εr colors i[r] such that eE(Gi). Thus

i=1reE(F)1[eE(Gi)]εrt,

and moreover, if XFεrt/2, then

Zεrt2,

since for each vertex vQi(F) we remove at most dF(v) edges from Y. By Chernoff's inequality, it follows that for a fixed F we have6

(XFεrt/2andFB)(Zεrt/2)eδεrt.

Therefore, taking a union bound over F, and recalling that t=δεk2, it follows that the probability that there exists F with XFεrt/2 and FB is at most

nkk2teδεrtenkeδεδεkeδ2ε2rkk0,

as claimed, where in the final step we used our choice of n=2δ4ε2rk, the bound

ε=rsrClogrr,

which holds by our assumption that srClogr, and our choice of C=1/δ3.

Finally, we will use the randomness in ϕ1,,ϕr to show that XF is always small.

Lemma 3.5

With high probability,

XFεrt2

for every F.

For each graph F, and each j{1,,log2k}, define a set

Aj(F)={vV(F):2jkdF(v)<2j+1k}

and a random variable

sj(F)=i=1r|Aj(F)Qi(F)|.

Note that the random functions ϕ1,,ϕr determine Q1(F),,Qr(F), and hence sj(F). The key step is the following claim, which provides us with our bottleneck event.

Claim 3.6

If XFεrt/2, then there exists {1,,log2k} such that

s(F)>δεrj=1|Aj(F)|. (9)

Observe that

XF=i=1rvQi(F)dF(v)i=1rj=1log2kk2j1·|Aj(F)Qi(F)|=j=1log2kk2j1·sj(F),

and therefore if XFεrt/2, then

j=1log2ksj(F)2jεrt4k. (10)

Note also that

j=1log2k|Aj(F)|2j1kvV(F)dF(v)=2tk.

Thus, if (9) fails to hold for every {1,,log2k} then, by (10), we have

t4δk1δεr=1log2ks(F)2=1log2k12j=1|Aj(F)|=j=1log2k|Aj(F)|=jlog2k12j=1log2k|Aj(F)|2j14tk.

Since δ<24, this is a contradiction, and so the claim follows.

Fix {1,,log2k} such that  (9) holds, and set

A:=j=1Aj(F)anda:=|A|.

Now, if we reveal ϕi for the vertices of F one vertex at a time using the order F, then for each vertex vQi(F) we must choose ϕi(v) to be one of the (at most k) previously selected elements of [m]. The expected number of sets A such that (9) holds is thus at most7

a=1knaarδεar(km)δεara=1k(n·(eδε·km)δεr)a0

as k, as required, since n=2δ4ε2rk and εm/km=2δ2εk/2.

We remark that a simple union bound over graphs F would fail by a factor of Θ(ε) (in the exponent), since XFεrt/2 only implies that there are at least Ω(ε2rk) pairs (v,i) with i[r] and vQi(F), whereas ||=nΩ(k)=mΩ(εrk).

Combining Lemmas 3.4 and 3.5, we obtain Lemma 3.3.

Proof of Lemma 3.3

By Lemma 3.5, with high probability we have XFεrt/2 for every F. By Lemma 3.4, it follows that with high probability FB for every F. Therefore, with high probability every k‐clique contains at most t bad edges, as claimed.

We can now easily put together the pieces and prove our main theorem.

Proof of Theorem 1.2

With high probability, the random coloring χ satisfies:

  1. |χ(e)|s for every eE(Kn), by Lemma 3.1;

  2. χ contains no monochromatic Kk with at most t bad edges, by Lemma 3.2;

  3. every k‐clique contains at most t bad edges, by Lemma 3.3.

Thus Rr,s(k)>n=2δ4ε2rk, as required.

4. A SIMPLER CONSTRUCTION FOR SMALL K

To conclude, we will prove the alternative lower bounds (4) and (5), promised in the introduction, which hold for smaller values of k. Together with Theorem 1.2 and the results of [4], these bounds will allow us to prove Corollary 1.3.

Theorem 4.1

Let r>s1, and set ε=(rs)/r. We have

Rr,s(k)(ε(k1)e)εr/2

for every k.

Note that Theorem 4.1 only gives a non‐trivial bound when ε(k1)>e. However, the same construction (together with a slightly more careful calculation) in fact gives an exponential lower bound almost all the way down to the Turán range k1/ε.

Theorem 4.2

Let r>s1 and δ>0, and set ε=(rs)/r. We have

Rr,s(k)exp(c(δ)εr)

for some c(δ)>0 and every k(1+δ)/ε+1.

The construction is a simpler version of the one we used to prove Theorem 1.2: instead of taking blow‐ups of a random graph for our colors, we take blow‐ups of the complete graph with k1 vertices (i.e., we take complete (k1)‐partite graphs). In particular, this means that we do not need to worry about crossing edges.

To define the coloring precisely, let ϕ1,,ϕr be independent uniformly chosen functions from [n] to [k1], and for each i[r] define Gi to be the (random) graph with vertex set [n] and edge set

E(Gi)={uv:ϕi(u)ϕi(v)},

that is, a random complete (k1)‐partite graph, with parts given by ϕi. Define a coloring χ of the edges of Kn by

χ(e)={i[r]:eE(Gi)}.

Since each color class is (k1)‐partite, χ contains no monochromatic copy of Kk. Our task is to show that, with positive probability, every edge receives at least s colors.

Proof of Theorem 4.1

Let e=uvE(Kn). Since iχ(e) only if ϕi(u)=ϕi(v), which occurs with probability 1/(k1), independently for each color, it follows that

(|χ(e)|<s)rεr(1k1)εr(eε(k1))εr. (11)

Thus, if n<(ε(k1)/e)εr/2, then the expected number of edges eE(Kn) such that |χ(e)|<s is less than 1, and hence there exists a choice of the functions ϕ1,,ϕr such that every edge receives at least s colors, as required.

To prove an exponential bound when ε(k1)<e, we simply replace the union bound in (11) by an application of Chernoff's inequality.

Proof of Theorem 4.2

We may assume that δ2, since otherwise the claimed bound (with c(δ) an absolute constant) follows from Theorem 4.1. Let eE(Kn), and recall that the event iχ(e) occurs with probability 1/(k1), independently for each color i[r]. Since ε(k1)1+δ, it follows by Chernoff's inequality that

(|χ(e)|<s)=(Bin(r,1/(k1))εr)ec(δ)εr

for some constant c(δ)=Ω(δ2). Thus, if n<ec(δ)εr/2, then the expected number of edges eE(Kn) such that |χ(e)|<s is less than 1, and hence there exists a choice of the functions ϕ1,,ϕr such that every edge receives at least s colors, as required.

We can now easily deduce Corollary 1.3.

Proof of Corollary 1.3

The upper bound holds by (1), which was proved in Theorem 1.1 of [4], and holds for all r>s1 and all k3. When srClogr and k(C/ε)logr, the lower bound follows from Theorem 1.2 (though note that in the case rs=Ω(r) it was first proved in [4]), and when srClogr it follows from a simple random coloring (see Section 2 of [4]) that

Rr,s(k)2εk/6.

Finally, when (1+δ)/ε+1k(C/ε)logr, the lower bound follows from Theorem 4.2.

5. CONCLUDING REMARKS

We expect that determining logRr,s(k) up to a constant factor for all s=ro(r) will be extremely difficult (though it does not seem unreasonable to hope that doing so in this range might prove to be easier than in the case s=1). On the other hand, the upper and lower bounds in Corollary 1.3 differ by a factor of roughly (logr)2 in the exponent when s=rlogr, and also when εk=logr. We expect that the lower bound can be improved by a factor of logr in these cases.

Conjecture 5.1

There exists a constant δ>0 such that

Rr,s(k)exp(δk(rs)2r) (12)

for every r>s1 and k3/ε, where ε=(rs)/r.

It seems plausible that (12) could be proved in the case εk=O(logk) using a modified version of the construction in Section 4 (taking Turán graphs with fewer parts, and assigning the set [r] to edges with |χ(e)|<s), since the density of bad edges will be small, and the correlation between them not too large. The obstruction when srlogr is potentially more serious, since in this case the density of bad edges will be larger than ε, so one would not expect the conclusion of Lemma 3.3 to hold. It therefore seems unlikely that the construction from Section 2 can be used to prove (12) in this range.

ACKNOWLEDGMENTS

The research that led to this paper started at WoPOCA 2022. We thank the workshop organizers for a productive working environment. This research was funded in whole, or in part, by the Austrian Science Fund (FWF) P36161. For the purpose of open access, the author has applied a CC BY public copyright licence to any Author Accepted Manuscript version arising from this submission. This study was also financed in part by the Coordenação de Aperfeiçoamento de Pessoal de Nível Superior, Brasil (CAPES), Finance Code 001. This research was supported by: (Maurício Collares) the Austrian Science Fund (FWF P36131); (Taísa Martins) CNPq (Proc. 406248/2021‐4); (Robert Morris) FAPERJ (Proc. E‐26/200.977/2021) and CNPq (Proc. 303681/2020‐9).

Aragão L., Collares M., Marciano J. P., Martins T., Morris R., A lower bound for set‐coloring Ramsey numbers, Random Struct. Alg. 64 (2024), 157–169. 10.1002/rsa.21173

Footnotes

1

We remark that the range s=ro(r) was of particular interest to the authors of [4], who were motivated by an application to hypergraph Ramsey numbers, see [5].

2

Note that for edges that receive more than s colors, we can take an arbitrary subset of size s.

3

In fact taking δ=26 would suffice, but we will not make any attempt to optimize the constants in Theorem 1.2.

4

Indeed, the probability that e(G(k,1p))2μ/3, where μ=5δεk2, is at most eμ/18, and δ26.

5

Here, and below, we abuse notation slightly by treating the set of bad edges B as a graph.

6

Indeed, since Z is a binomial random variable with expected value at most εrt/4, the event Zεrt/2 has probability at most eεrt/12.

7

Indeed, there are at most na choices for the set AV(Kn), and at most arδεar choices for the ‘first’ (in some arbitrary order) δεar pairs (v,i) with i[r] and vA(F)Qi(F). For each such pair, we must have ϕi(v)=ϕi(u) for some uV(F) with uFv, and each of these choices is made independently.

DATA AVAILABILITY STATEMENT

Data sharing is not applicable to this article as no new data were created or analyzed in this study.

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Data Availability Statement

Data sharing is not applicable to this article as no new data were created or analyzed in this study.


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