On the Ramsey numbers for complete distance graphs with vertices in 

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Sbornik : Mathematics 200:12 1789–1806 c© 2009 RAS(DoM) and LMS

Matematicheskiı Sbornik 200:12 63–80 DOI 10.1070/SM2009v200n12ABEH004059

On the Ramsey numbers for completedistance graphs with vertices in {0, 1}n

K.A. Mikhaılov and A. M. Raıgorodskiı

Abstract. A new problem of Ramsey type is posed for complete distancegraphs in Rn with vertices in the Boolean cube. This problem is closelyrelated to the classical Nelson-Erdos-Hadwiger problem on the chromaticnumber of a space. Several quite sharp estimates are obtained for certainnumerical characteristics that appear in the framework of the problem.

Bibliography: 15 titles.

Keywords: Ramsey numbers, distance graphs, chromatic number.

§ 1. Introduction

1.1. The Ramsey numbers. One of the classical problems of combinatorics isthe problem of finding the Ramsey numbers. In order to state it accurately, we giveseveral definitions and introduce some notation in graph theory (see [1]).

Let G = (V,E) be some graph. If H = (W,F ) is a subgraph of G, then we writeH ⊆ G. If, in addition, H is a spanning subgraph of G (that is, W = V ), then,wishing to emphasize this fact, we write H � G. We say that H is an induced orgenerated subgraph of G if F = E

∣∣W

, that is, a pair of vertices x, y in W forms anedge {x, y} that belongs to F if and only if {x, y} ∈ E.

We denote by KN = (V,E) the complete graph on N vertices (it has |V | = Nand |E| = C2

N ). If G = (V, F ) � KN (that is, G is an arbitrary graph on Nvertices), then its complement (to the complete graph) is defined to be the graph[G] = (V, F ′) � KN in which {x, y} ∈ F ′ if and only if {x, y} /∈ F .

We fix arbitrary positive integers s, t. We define the Ramsey number (withparameters s, t) to be the quantity R(s, t) equal to the least N ∈ N such that forany colouring of the edges of KN in red and blue, either there exists H ⊆ KN

that is isomorphic to Ks all of whose edges are red, or there exists H ⊆ KN thatis isomorphic to Kt all of whose edges are blue. The quantity R(s, s) is calleda diagonal Ramsey number.

One can give an equivalent definition of the Ramsey numbers. Namely, R(s, t)is the minimal N ∈ N such that for any G � KN either G contains an isomorphiccopy of Ks or [G] contains an isomorphic copy of Kt.

This research was supported by the Russian Foundation for Basic Research (grantno. 06-01-00383), by the Programme of the President of the Russian Federation for Supportof Young Scientists (grant no. МД-5414.2008.1), by the Programme for Support of Leading Sci-entific Schools of the Russian Federation (grant no. НШ-691.2008.1), and by the “Dinastiya”Foundation.

AMS 2000 Mathematics Subject Classification. Primary 05C55; Secondary 05C15, 05C80.

1790 K. A. Mikhaılov and A.M. Raıgorodskiı

Since every induced subgraph of the graph KN is an isomorphic copy of some Ks,one can also say as follows: R(s, t) is the minimal N ∈ N such that for any G � KN

either G contains some generated subgraph of the graph KN on s vertices or [G]contains some generated subgraph of the graph KN on t vertices.

There is extensive literature concerning the Ramsey numbers (see, for example,[2]–[5]). Here we only point out that these numbers are always well defined andthat, in particular,

(√

2 + o(1))s 6 R(s, s) 6 (4 + o(1))s.

More precisely,√

2e

(1 + o(1))s2s/2 6 R(s, s) 6 e−γ(ln2 s)/(ln ln s) · 4s, γ > 0.

1.2. Chromatic numbers and distance graphs. Yet another classical problemof combinatorics consists in finding the chromatic number of a space, that is, thequantity χ(Rn) equal to the minimal number of colours required for colouring allthe points in Euclidean space Rn so that no two points of the same colour are atdistance 1.

An equivalent statement of the problem is again given in terms of graph theory(see [1]). Recall that the chromatic number of a graph is defined as the minimalnumber of colours needed to colour all the vertices of the graph so that the endpointsof any of its edges are of different colours. We define the graph Gn = (Vn,En) bysetting

Vn = Rn, En ={{x,y} : x,y ∈ Rn, |x− y| = 1

},

where |x − y| is the Euclidean distance between vectors. It is easy to see thatχ(Gn) = χ(Rn). Moreover,

χ(Gn) = maxG

χ(G),

where the maximum is taken over all the finite subgraphs of the graph Gn. Thisnon-obvious fact is a consequence of one the theorems of Erdos-de Bruijn (see [6]).

It is much easier to see that if instead of 1 we take an arbitrary a > 0 in thedefinition of the set of edges of the graph Gn, then the chromatic number does notchange because of that. Therefore often for convenience not only (finite) graphsG ⊂ Gn are considered but also their homothetic copies with positive homothetycoefficients.

The graph Gn, as well as any subgraphs of it and their homothetic copies withpositive homothety coefficients are customarily called distance graphs. If G is a gen-erated subgraph of Gn (or a homothetic copy of such a subgraph), then we say thatG is a complete distance graph (on its vertex set all the edges of fixed length aredrawn whichever could possibly be drawn).

There are numerous papers devoted to the chromatic number of a space and todistance graphs (see [5], [7]–[9]).

The following series of complete distance graphs is of special significance forfinding the chromatic number of a space. Let n be an arbitrary positive integer of

On the Ramsey numbers for complete distance graphs 1791

the form n = 4k. We set N = Cn/2n and consider GN = (VN ,EN ), where

VN ={x = (x1, . . . , xn) : xi ∈ {0, 1}, x1 + · · ·+ xn = 2k =

n

2

},

EN ={{x,y} ∈ VN × VN : |x− y| =

√n

2

}.

Thus, the complete distance graph GN ⊂ Gn has as vertices points in {0, 1}n andit has exactly N of these vertices.

Note that if k = pα, where p is a prime number and α ∈ N, then the chromaticnumber of the corresponding GN can be estimated much better than in the case ofan arbitrary n (see [5], [7], [8]). In other words, it makes sense to study separatelythe sequence of all the GN , and separately the sequence {G prime

N } ⊂ {GN}∞k=1 ofthose GN for which k = pα with some p and α.

In conclusion we mention that for the moment the strongest estimates for thechromatic number have the form

(1.239 . . .+ o(1))n 6 χ(Rn) 6 (3 + o(1))n

(see [5], [7], [8]).

§ 2. Statement of the problem

We now define analogues of the Ramsey numbersR(s, t) for the complete distancegraphs GN and G prime

N . But before that we introduce a couple of natural auxiliarynotions.

Let G � GN . Then its complement (to the complete distance graph) is definedto be the graph [G]dist � GN any two of whose vertices are connected by an edge ifand only if they are not connected by an edge in G but are connected by an edgein GN . For example, for G = (VN ,∅) we have [G]dist = GN .

For given s, t ∈ N we set Rdist(s, t) to be equal to the minimal N ∈ N such thatthe graph GN is well defined and for any G � GN either G contains some generatedsubgraph of GN on s vertices or [G]dist contains some generated subgraph of GN ont vertices.

In other words, the quantity Rdist(s, t) is completely analogous to the quantityR(s, t), since we replace KN and the complement in it by GN and the complementin it.

In exactly the same fashion we introduce Rprimedist (s, t) as the minimum of all the

N ∈ N for which the graph GN ∈ {G primeN } is well defined and for any G � GN

either G contains some generated subgraph of GN on s vertices or [G]dist containssome generated subgraph of GN on t vertices.

Given the evident similarity of the classical and new definitions, there are obvi-ously also substantial differences between them. The main difference is in the factthat while earlier a generated subgraph of the complete graph KN always had thesame structure as the complete graph itself (the subgraph was isomorphic to Ks),now generated subgraphs of the ‘complete’ graphs GN do not at all have to beisomorphic to any of the Gs. In particular, such subgraphs can even turn out

1792 K. A. Mikhaılov and A.M. Raıgorodskiı

to be ‘empty’ (that is, free of edges); as we shall see below, in the graphs GNthere are very large independent sets of vertices — sets whose elements are pairwisenon-adjacent. In the case of KN , a similar property was enjoyed exclusively by K1.

In the next section we state new results concerning the quantities Rdist(s, t)and Rprime

dist (s, t). They will be substantially different from the classical ones.

§ 3. Statements of the results

First of all we observe that the following holds.

Proposition 1. The quantities Rdist(s, t) and Rprimedist (s, t) are well defined for all

positive integers s and t.

The following theorem gives an explicit upper estimate for the diagonal Ramseynumbers.

Theorem 1. Let

c =4

33/4, ξ =

ln 2ln c

, b =35π

42=

24316

π.

Then for any β > 0 for all sufficiently large s ∈ N we have the inequality

Rdist(s, s) 6 16

√2π

(ln c)1/2−ξbξsξ(ln s)ξ−1/2(1 + β).

In other words, while the classical Ramsey number had exponential growth, the‘distance’ one, in essence, is bounded above by a polynomial.

The situation is somewhat worse for the numbers Rprimedist (s, s).

Theorem 2. There exists a function ϕ such that ϕ(s) = o(1) as s→∞ and

Rprimedist (s, s) 6 sξ+ϕ(s).

Of course, the form of the function ϕ in Theorem 2 is subject to concretization,but we shall give a more accurate formulation and comment on it only after we proveTheorem 2. In any case it will turn out that the expression sϕ(s) has a much fastergrowth than any power of the logarithm and therefore the estimate in Theorem 2is significantly weaker than the estimate in Theorem 1. This is related to thepeculiarities of the distribution of prime numbers among positive integers.

We now discuss the lower estimates.

Theorem 3. Let

c =4

33/4, ξ =

ln 2ln c

, d =

√2π· 4e · 33/2

,

θ1 =1ξ, θ2 =

12ξ

− 32, θ3 =

d

( 2π )θ1/2(ln 2)θ2

, θ4 =(

ln 22θ3ξθ2(ξ − 1)

)ξ.

Then for any β > 0 there exist infinitely many positive integers s such that

Rprimedist (s, s) > θ4s

ξ(ln s)(ξ−1)/2(1− β).

On the Ramsey numbers for complete distance graphs 1793

Theorem 3 is a kind of ‘omega-result’. It would be interesting to understandwhat happens if “infinitely many positive integers s” was replaced, for example, by“all sufficiently large s”. Unfortunately, this can be done only at the expense ofrather significant losses in the quality of the estimate — losses similar to those thatwe encountered when passing from Theorem 1 to Theorem 2. This is also relatedto the peculiarities of the distribution of prime numbers among positive integers.

Theorem 4. There exists a function ψ such that ψ(s) = o(1) as s→∞ and

Rprimedist (s, s) > sξ+ψ(s).

Clearly, the function ψ takes, more likely, negative values. But it is more conve-nient to concretize it after the proof of Theorem 4.

The following theorem appears to be unexpectedly weak.

Theorem 5. Let η = (ln 2)/ln 1.99. There exists a function µ such that µ(s) = o(1)as s→∞ and

Rdist(s, s) > sη+µ(s).

In other words, the gap between the estimates of the number Rdist(s, s) fromabove and from below has the order of a power of s. In this connection there is evenno particular sense in writing out accurately the function µ. This trouble is causedby the fact that some properties of an arbitrary graph GN have been studied toa much lesser extent than their analogues for a graph in the sequence {G prime

N }.We shall prove Theorem 1 in § 4. Section 5 is devoted to proving Theorem 2 and

discussing its refinements. Theorem 3 will be proved in § 6. Section 7 is devotedto substantiating Theorem 4 and refining it. In § 8 we substantiate Theorem 5.Finally, in § 9 we briefly summarize the information accumulated.

§ 4. Proof of Theorem 1

4.1. Auxiliary lemmas.

Lemma 1. Let N = Cn/2n , n = 4k. Then

N =

√2π· 2n√

n(1 + δ1(n)), δ1(n) → 0 as n→∞, (1)

n =lnNln 2

(1 + δ2(N)), δ2(N) → 0 as N →∞. (2)

Proof. To substantiate (1) we apply Stirling’s formula:

N =n!

(n2 !)2=

√2πnnne−n(√

2π n2 (n2 )n/2e−n/2)2 (1 + δ1(n)) =

√2πnnne−n

πn(n2 )ne−n(1 + δ1(n))

=

√2π· 2n√

n(1 + δ1(n)).

By taking the logarithm of the resulting equation we obtain

ln√πn+ lnN = (ln

√2 + n ln 2)(1 + δ3(n)), δ3(n) → 0 as n→∞,

1794 K. A. Mikhaılov and A.M. Raıgorodskiı

whencen =

lnNln 2

(1 + δ2(N)).

The asymptotics (2), and with it also Lemma 1, is proved.

In order to state the next lemma we need one more definition in graph theory.We define the independence number of a graph G = (V,E) as the cardinality α(G)of a largest (with respect to cardinality) independent set of its vertices.

Lemma 2. We have the inequality

α(GN ) >4n+2

πn33n/4+5(1 + δ(n)), δ(n) → 0 as n→∞.

Proof. Consider the set

F ={x = (x1, . . . , xn) ∈ VN :

n/2∑i=1

xi =[n

8

]− 1

}.

Clearly, F is an independent set of vertices of the graph GN .We put

q =n∑

i=n/2+1

xi =n

2−

([n

8

]− 1

).

Since [n/8] = n/8− ε for ε ∈ [0, 1), it follows that

q =n

2−

([n

8

]− 1

)=n

2− n

8+ 1 + ε =

3n8

+ 1 + ε.

We have|F | = C

[n/8]−1n/2 Cqn/2 = (C [n/8]−1

n/2 )2.

By applying Stirling’s formula and denoting ε1 = 1 + ε we now obtain that

α(GN ) > |F | =( n

2 !([n8 ]− 1)! q!

)2

=( n

2 !(n8 − ε1)! ( 3n

8 + ε1)!

)2

=( √

πn(n2 )n/2e−n/2√( n4 − 2ε1)π(n8 − ε1)n/8−ε1e−n/8+ε1

√( 3n

4 + 2ε1)π( 3n8 + ε1)3n/8+ε1e−q

)2

× (1 + δ4(n)) =πn(n2 )n

316n

2π2(n8 − ε1)n/4−2ε1( 3n8 + ε1)3n/4+2ε1

(1 + δ5(n))

=(n2 )n

316πn(n8 )n/4−2ε1(1− 8ε1

n )n/4−2ε1( 3n8 )3n/4+2ε1(1 + 8ε1

3n )3n/4+2ε1(1 + δ5(n))

=( 12 )n

316πn( 1

8 )n/4−2ε1( 38 )3n/4+2ε1

(1 + δ(n)) =4n+2

πn33n/4+1+2ε1(1 + δ(n))

>4n+2

πn33n/4+5(1 + δ(n)).

Here all the quantities of δ type tend to zero as n increases. The lemma is proved.

On the Ramsey numbers for complete distance graphs 1795

4.2. Completion of the proof. Suppose that for a given positive integer sand for some N the graph GN is well defined and α(GN ) > s. Then for anyG = (VN , E) � GN we also have α(G) > s. Let W , |W | = s, be any of the cor-responding independent sets in VN . Therefore the graph H = (W,E

∣∣W

) = (W,∅)is a generated subgraph of GN and simultaneously a subgraph of G. Thus, in thedescribed situation both G and [G]dist contain an induced subgraph of the graphGN on s vertices, and this is even more than required.

It remains to show that for every sufficiently large s there exists N = Cn/2n with

n = 4k and

N 6 16

√2π

(ln c)1/2−ξbξsξ(ln s)ξ−1/2(1 + β), α(GN ) > s.

We set

h(s) =ln bs(ln s)(1+β′)

ln c

ln c, 0 < β′ < β.

Clearly, for large s the quantity h(s) is nonnegative and, consequently, there isa positive integer n of the form n = 4k that does not exceed h(s)+4 and is greaterthan or equal to h(s). By Lemma 2, for the corresponding N we have

α(GN ) >4n+2

πn33n/4+5(1 + δ(n)) =

1b· cn

n(1 + δ(n)) =

1b· e

n ln c

n(1 + δ(n))

>1b· e

h(s) ln c

h(s) + 4(1 + δ(n)).

As s increases, n also increases, and therefore we can write δ(n) = γ(s), whereγ(s) → 0 as s→∞. Thus,

α(GN ) >1b· e

h(s) ln c

h(s) + 4(1 + γ(s)) =

1b· bs(ln s)(1 + β′)

(ln c)( ln (bs(ln s)(1+β′)/ ln c)

ln c + 4) (1 + γ(s))

=s(ln s)(1 + β′)

ln c4bs(ln s)(1+β′)ln c

(1 + γ(s)) =s(ln s)(1 + β′)

(ln s)(1 + γ′(s))(1 + γ(s)),

γ′(s) → 0 as s→∞.

Since β′ > 0, we finally obtain (for sufficiently large s) that α(GN ) > s.We now use Lemma 1 to obtain the following chain of inequalities:

N =

√2π· 2n√

n(1 + δ1(n)) 6

√2π· 2h(s)+4√

h(s)(1 + δ1(n))

= 16

√2π· 1√

h(s)· 2(ln(bs(ln s)(1+β′)/(ln c)))/(ln c)(1 + δ1(n))

= 16

√2π· 1√

h(s)·(bs(ln s)(1 + β′)

ln c

)ξ(1 + δ1(n)).

1796 K. A. Mikhaılov and A.M. Raıgorodskiı

We observe that h(s) > (ln s)/(ln c) for large s. Consequently,

N 6 16

√2π· 1√

ln sln c

·(bs(ln s)(1 + β′)

ln c

)ξ(1 + δ1(n))

6 16

√2π

(ln c)1/2−ξbξsξ(ln s)ξ−1/2(1 + β)

as soon as β′ and s are such that

(1 + β′)ξ(1 + δ1(n)) 6 1 + β.

The theorem is proved.

§ 5. Proof of Theorem 2 and its refinement

5.1. Proof. We proceed practically in the same fashion as in Subsection 4.2.Keeping the same notation, we set

h(s) =ln

( 2bs(ln s)ln c

)ln c

.

As in Subsection 4.2, we in fact wish to find the least possible positive integern > h(s) that has the form n = 4pα. For that we need a little digression into thearea of analytic number theory. Namely, let f be any function such that for everyx > 0 the closed interval [x, x + f(x)] contains a number that is a power of someprime number. For the moment, the only thing that matters for us is that thereexists f with the property f(x) = o(x) (see [10]).

The number n that we are interested in clearly does not exceed h(s)+4f(h(s)/4);furthermore, f(h(s)/4) = o(ln s). This circumstance and a series of calculationsthat repeat almost word-for-word those carried out in Subsection 4.2 mean thatthe inequality α(GN ) > s holds for a sufficiently large s and an appropriate N .

At the same time, as before,

N =

√2π· 2n√

n(1 + δ1(n)) 6

√2π· 2h(s)+4f(h(s)/4)√

h(s)(1 + δ1(n)) = sξ+ϕ(s)

with a suitable function ϕ = o(1).For small s the assertion of the theorem is obvious. The theorem is proved.

5.2. Refinements. It is now already clear how an absolutely precise assertion ofTheorem 2 must look (cf. Subsection 4.2).

Theorem 2′. Let c, ξ, b, β be the same as in Theorem 1, and let β′ > 0 be suchthat (1 + β′)ξ < 1 + β. We set

h(s) =ln

( bs(ln s)(1+β′)ln c

)ln c

.

On the Ramsey numbers for complete distance graphs 1797

Next, let f be any function in Subsection 5.1. Then for all sufficiently large s wehave the inequality

Rprimedist (s, s) 6 16f(h(s)/4)

√2π

(ln c)1/2−ξbξsξ(ln s)ξ−1/2(1 + β).

At present it is only known that as f one can take a function of the formf(x) = O(xr), where r ∈ (0.5, 1). There is also the conjecture: there exists fof the form f(x) = O(ln2 x). Unfortunately, even in the case where f has the hypo-thetical order of growth, the quantity 16f(h(s)/4) increases faster than any powerof the logarithm. Thus, the difference between the results of Theorems 1 and 2 isfundamental.

§ 6. Proof of Theorem 3

6.1. Auxiliary lemmas. In Lemma 2 we have established an asymptotic lowerestimate for the independence number of the graph GN for any N . It turns outthat it is much more difficult to obtain an upper estimate of the same number.Nevertheless, this was done in [11] by using the linear-algebraic method (see [5]).However, only for GN ∈ {G prime

N }. Namely, it was shown that α(GN ) 6 2Cn/4−1n−1

for n = 4pa.

Lemma 3. For n = 4pa we have the estimate

α(GN ) 6

√2 · 4n+1

e√πn3 · 27(n+2)/4

(1 + δ0(n)),

where δ0(n) → 0 as n→∞.

Proof. In view of Stirling’s formula we have

α(GN ) 6 2Cn/4−1n−1 =

2(n− 1)!(n4 − 1)! ( 3n

4 + 1)!

=2√

2π(n− 1)(n− 1)n−1e−n+1√π n−4

2 (n4 − 1)n/4−1e−n/4+1√π 3n+4

2 ( 3n4 + 1)3n/4+1e−3n/4+1

(1 + ∆1(n))

=4√

2(n− 1)(n− 1)n−1√π(n− 4)(3n+ 4)(n4 − 1)n/4−1( 3n

4 + 1)3n/4+1(1 + ∆1(n))

=4√

2 · 4nnn−1/2

e√π · nn+13(3n+6)/4

(1 + δ0(n)) =√

2 · 4n+1

e√πn3 · 27(n+2)/4

(1 + δ0(n)).

The lemma is proved.

Together Lemmas 2 and 3 mean that the independence number of the graphGN ∈ {G prime

N } has the form (4/33/4 + o(1))n, and only the function o(1) has notbeen fully concretized.

We rewrite the result of Lemma 3 in terms of N .

1798 K. A. Mikhaılov and A.M. Raıgorodskiı

Lemma 3′. Letc, ξ, d, θ1, θ2, θ3

be the same as in the statement of Theorem 3. Then for n = 4pa the estimate

α(GN ) 6 θ3(lnN)θ2Nθ1(1 + δ′0(N))

holds, where δ′0(N) → 0 as N →∞.

Proof. By Lemma 1 we have

N =

√2π· 2n√

n(1 + δ1(n)).

Therefore, lnN ∼ n ln 2. Thus,

θ3(lnN)θ2Nθ1 ∼ θ3nθ2(ln 2)θ2

(2π

)θ1/2 cn

n1/(2ξ)=

dcn

n3/2.

It is easy to see that, in turn, by Lemma 3,

α(GN ) 6dcn

n3/2(1 + δ0(n)).

The completion of the proof is obvious.

The following lemma is of a less technical nature than Lemmas 3 and 3′.

Lemma 4. Let GN be a complete distance graph in the sequence {G primeN }, and

G = (W,F ) an arbitrary generated subgraph of the graph GN that has s vertices.Let α = α(GN ) and suppose that s > (n + 1)α. Then there exists a function σ(n)such that σ(n) → 0 as n→∞ and the following inequality holds :

|F | >s2 − nsα+ 1

2n2α2

α(1 + σ(n)).

By Lemma 3 the assumption s > (n + 1)α makes sense for all sufficiently largen of the appropriate form. Thus, Lemma 4 is formally valid only for n > n0.

If s ∼ nα, then the estimate in Lemma 4 transforms into the inequality|F | > (s2/2α)(1 + σ′(n)). But if nα = o(s), then the estimate becomes in essencetwice as sharp: |F | > (s2/α)(1 + σ′′(n)). In the general case it is much easier toprove the first version of the inequality, but we prove the second one straight away.

Proof of Lemma 4. We begin with the fact that in W there is an independent set Aof vertices of the graph G that has maximal cardinality. We set β = |A|. SinceG is an induced subgraph of GN , it follows that β 6 α. Furthermore, since A ismaximal, every vertex in W \A is connected by an edge (which belongs to F ) withat least one vertex in A.

We divide W \A into two disjoint parts L and K. Here K is the set of verticessuch that for any vertex x ∈ K there exists exactly one vertex y ∈ A with theproperty {x, y} ∈ F . Formally,

K ={x ∈W \A : |{y ∈ A : {x, y} ∈ F}| = 1

},

L = W \A \K ={x ∈W \A : |{y ∈ A : {x, y} ∈ F}| > 2

}.

On the Ramsey numbers for complete distance graphs 1799

We claim that |K| 6 nβ. Suppose the opposite. Then by Dirichlet’s principlethere exists a vertex x ∈ A such that∣∣{y ∈ K : {x, y} ∈ F}

∣∣ > n+ 1.

Consider arbitrary

y1, . . . , yn+1 ∈{y ∈ K : {x, y} ∈ F

}.

We choose from them any two different vertices yi yj . Let M = (A\{x})∪{yi, yj}.Since |M | > |A|, in M there are edges of the graph G. Our construction hasthe property that necessarily {yi, yj} ∈ F . Therefore, the vertices y1, . . . , yn+1

of the distance graph in Rn are pairwise connected by edges, that is, they forma regular n-dimensional simplex. However, (0,1)-vectors, which we are dealingwith, obviously cannot form full-dimensional simplexes; we obtain a contradiction.

Thus, |K| 6 nβ 6 nα. Next, clearly,

F ={{x, y} : x ∈ K, y ∈ A

}∪

{{x, y} : x ∈ L, y ∈ A

}∪

{{x, y} : x ∈W \A, y ∈W \A

}.

In other words, setting

F1 ={{x, y} : x ∈W \A, y ∈W \A

},

we have

|F | =∣∣{{x, y} : x ∈ K, y ∈ A}

∣∣ +∣∣{{x, y} : x ∈ L, y ∈ A}

∣∣ + |F1|> 2(s− β)− |K|+ |F1| > 2(s− β)− nβ + |F1|> 2(s− α)− nα+ |F1| = 2s− α(n+ 2) + |F1|.

We now effect the same scheme of actions replacing W by W1 = W \A, A by A1

(an independent set of vertices in W1 that has maximal cardinality), β by β1 =|A1| 6 α, L and K by L1 and K1, so that W1 \A1 = L1 ∪K1, and so on. Settings1 = |W1| > s− α and

F2 = {{x, y} : x ∈W1 \A1, y ∈W1 \A1},

we have

|F1| = |{{x, y} : x ∈ K1, y ∈ A1}|+ |{{x, y} : x ∈ L1, y ∈ A1}|+ |F2|> 2(s1 − β1)− |K1|+ |F2| > 2(s1 − β1)− nβ1 + |F2|> 2(s− 2α)− nα+ |F2| = 2s− α(n+ 4) + |F2|.

We iterate the procedure described above k = [(s− nα)/α] times. As a resultwe obtain the estimate

|F | >k∑i=1

(2s− α(n+ 2i)) + |Fk|,

1800 K. A. Mikhaılov and A.M. Raıgorodskiı

whereFk =

{{x, y} : x ∈Wk−1 \Ak−1, y ∈Wk−1 \Ak−1

}.

All the steps of the procedure are well defined, since

|Wi| = |Wi−1 \Ai−1| > s− iα > s− kα > nα

for each i ∈ {1, . . . , k}, and therefore every time the consideration of the set Li giv-ing ‘doubled contribution’ into the estimate of the number of edges of the graph Gis justified in view of the inequality |Ki| 6 nβi 6 nα. Here, formally, W0 = W ,A0 = A, and so on. Moreover, due to the condition s > (n+ 1)α we obtain k > 1,which implies that we shall certainly realize at least one step of the procedure andthus confirm that the summation over i from 1 to k is well defined.

It remains to estimate the quantity |Fk|. This is, in essence, the number ofedges of the graph G on the set of vertices Wk. As before, we distinguish in Wk

a subset Ak that has maximal cardinality among all subsets in Wk that are freefrom edges of G. Clearly, every vertex in Wk \Ak is connected by an edge with atleast one vertex in Ak. Thus, the consideration of Ak gives a contribution of sizeat least |Wk| − |Ak| > s− kα− α to the quantity |Fk|.

We again iterate the procedure described above. Every time we remove at mostα vertices from Wk and every time we add at least s − kα − iα (where i is theiteration number) to the estimate of the quantity |Fk|. Taking into account that|Wk| > nα we arrive at the conclusion that we can carry out at least n iterations.As a result,

|Fk| >n∑i=1

(s− kα− iα).

Consequently,

|F | >k∑i=1

(2s− α(n+ 2i)) +n∑i=1

(s− kα− iα)

>k∑i=1

(2s− α(n+ 2i)) +n∑i=1

(nα− iα) = k(2s− nα)− k(k + 1)α+n(n− 1)

2α

>

(s− nα

α− 1

)(2s− nα)− s− nα

α

(s− nα

α+ 1

)α+

n(n− 1)2

α

=2s2 − 3snα+ n2α2

α− (2s− nα)− (s− nα)2 + (s− nα)α

α+α2(n2 − n)

2α

=s2 − nsα+ 0.5n2α2 + 0.5nα2 − sα

α− (2s− nα)

=s2 − nsα+ 0.5n2α2

α(1 + σ(n)).

The lemma is proved.

6.2. Completion of the proof. We fix all the parameters occurring in the state-ment of the theorem, including β > 0. We arrange in ascending order all numbersof the form 4pa, where p is a prime and a ∈ N. We obtain the set

P = {2, 3, 4, 5, 7, 8, 9, 11, 13, 16, . . . }.

On the Ramsey numbers for complete distance graphs 1801

For each q ∈ P we find N by the known formula. Preserving the order of thenumbers we form the set

N = {70, 924, 12870, 184756, . . . }.

For each N ∈ N we find s′ from the relation

N =[θ4(s′)ξ(ln s′)(ξ−1)/2(1− β′)

], β′ =

β

2.

We set s = ds′e. There arises an infinite sequence S of the positive integers sarranged in ascending order. Here every s ∈ S uniquely corresponds to N = N(s)and vice versa. Finally,

N(s) ∼ θ4sξ(ln s)(ξ−1)/2(1− β′),

that is,N(s) > θ4s

ξ(ln s)(ξ−1)/2(1− β)

for sufficiently large s.We now show that, starting from some s0, all the s ∈ S are such that

Rprimedist (s, s) > N = N(s),

and this will suffice for completing the proof of Theorem 3.Thus, let s be sufficiently large. For the corresponding N = N(s), consider the

graph GN . We need to verify that there exists G � GN such that neither itself norits complement contains generated subgraphs of the graph GN that have s vertices.We use the probabilistic method (see [12]–[14]). Namely, we construct a (random)graph G � GN by ‘putting’ into it every edge of the complete distance graph withprobability 0.5 independently of the other edges.

For any W ⊂ VN , |W | = s, we denote by l(W ) the number of edges of thecomplete distance graph on the vertices in W . Let AW be the event consisting inthat either the random graph G or its complement contains the induced subgraphof the graph GN with vertex set W . Clearly, P (AW ) = 2(1/2)l(W ) and, in view ofthe simplified version of Lemma 4, we have l(W ) > s2(1 + σ′(n))/2α. Here it isimportant that s > α = α(GN ).

As a result we obtain that

P

( ⋃W⊂VN

AW

)6 2CsN

(12

)s2(1+σ′(n))/2α

.

If we show that

2CsN

(12

)s2(1+σ′(n))/2α

< 1, (3)

then this will mean that

P

( ⋃W⊂VN

AW

)> 0,

and this is precisely what we need.

1802 K. A. Mikhaılov and A.M. Raıgorodskiı

Since CsN < Ns/sse−s, in order to prove (3) it is sufficient to verify the validityof the estimate

2Ns

sse−s

(12

)s2(1+σ′(n))/2α

< 1.

Next, from Lemma 3′ we know that

α 6 θ3(lnN)θ2Nθ1(1 + δ′0(N)).

Consequently,

2Ns

sse−s

(12

)s2(1+σ′(n))/2α

62Ns

sse−s

(12

)s2(1+σ(2)(n))/2θ3(lnN)θ2Nθ1

,

σ(2)(n) = o(1),

and our problem reduces to establishing the inequality

2Ns

sse−s

(12

)s2(1+σ(2)(n))/2θ3(lnN)θ2Nθ1

< 1.

We take the logarithm of the left-hand side of the last inequality and verify thatthe result is a negative number. We observe straight away that

N 6 θ4sξ(ln s)(ξ−1)/2(1− β′), lnN ∼ ξ ln s.

Thus,

ln 2 + s lnN − s ln s+ s− s2(ln 2)(1 + σ(2)(n))2θ3(lnN)θ2Nθ1

6 ln 2 + s ln(θ4(1− β′)) + ξs ln s+ξ − 1

2s ln ln s− s ln s+ s

− s2(ln 2)(1 + σ(3)(n))2θ3(ξ ln s)θ2θθ14 s

ξθ1(ln s)θ1(ξ−1)/2(1− β1)θ1

= (ξ − 1)(1 + σ(4)(n))s ln s

−(

ln 22θ3ξθ2θθ14

)· (1 + σ(3)(n))

(1− β1)θ1· s(ln s)3/2−1/2ξ−1/2+1/2ξ

= (ξ − 1)s ln s(

(1 + σ(4)(n))− (1 + σ(3)(n))(1− β1)θ1

)< 0

for all sufficiently large s.Strictly speaking, so far we have only proved that

Rprimedist (s, s) 6= N.

The fact is that the probabilistic technique described above might not have workedfor an arbitrary N ′ < N such that GN ′ ∈ {G prime

N }. Actually, this would be thecase if Lemma 3′ was valid only for N but not for other numbers of similar form.However, there are no problems whatsoever with Lemma 3′ and we really have theestimate

Rprimedist (s, s) > N.

The theorem is proved.

On the Ramsey numbers for complete distance graphs 1803

§ 7. Proof of Theorem 4 and its refinement

7.1. Proof. We fix an arbitrary β > 0 and any s ∈ N. Consider the maximalpositive integer N not exceeding

M = θ4sξ(ln s)(ξ−1)/2(1− β)

and having the form N = Cn/2n with some n = 4pa. Of course, for sufficiently

large s such N exists, and it is easy to see that all the technology of the precedingsection can be applied to such a pair s,N , so that

Rprimedist (s, s) > N

and it only remains to show that N = sξ+ψ(s) with a suitable ψ = o(1).Suppose that s is sufficiently large (otherwise everything is trivial). Consider the

largest positive integer n′ of the form n′ = 4k such that N ′ = Cn′/2n′ 6 M . Clearly,

n′ ∼ lnMln 2

∼ ξ ln sln 2

.

We know that there exists a function f ′ satisfying f ′(x) = o(x) such that for anyx ∈ R, x > 2, the closed interval [x− f ′(x), x] contains at least one prime number(cf. Subsection 5.1). Let p be some prime number contained in the closed interval[n′/4− f ′(n′/4), n′/4]. We set n = 4p. Obviously, n ∼ n′, that is,

n =ξ ln sln 2

(1 + ψ1(s)), ψ1 = o(1).

Therefore, for N = Cn/2n , on the one hand, we have N 6 M , and on the other

hand,

N ∼√

2π· 2n√

n= sξ+ψ2(s), ψ2 = o(1).

Thus, the sought-for N is contained between N and M , and

N > N = sξ+ψ(s), ψ = o(1).

The theorem is proved.

7.2. Refinements. Based on the preceding section, we can give the followingformulation.

Theorem 4′. Let β > 0 be an arbitrary number, and n′ the maximal positiveinteger divisible by 4 and satisfying the relation

Cn′/2n′ 6 θ4s

ξ(ln s)(ξ−1)/2(1− β).

Further, let f ′ be any function in Subsection 7.1. Then we have the inequality

Rprimedist (s, s) >

√2π· 2n

′−4f ′(n′/4)

√n′

(1 + δ′1(n′)), δ′1 = o(1).

1804 K. A. Mikhaılov and A.M. Raıgorodskiı

The estimate in Theorem 4′ is of a rather implicit nature. First, it is difficultto concretize the asymptotics n′ ∼ (ξ ln s)/ln 2, the form of the remainder term ofwhich substantially affects the difference between the quantities 2n

′and sξ. Second,

as before, the situation is bad with regard to estimating the quantity f ′(n′/4). Wecan use nothing better than the hypothetical (and with the current technique,unachievable) estimate

f ′(n′

4

)= O(ln2 n′) = O((ln ln s)2).

But 2−(ln ln s)2 decreases faster than any power of the logarithm. This means thatthe gap between the results of Theorems 4′ and 3 is certainly very big.

§ 8. Proof of Theorem 5

At first glance, it is completely unclear what is the difference between Theorems 4and 5. It would seem that we choose exactly the same N as in Subsection 7.1, andeverything will be all right: we obtain the estimate

Rdist(s, s) > N = sξ+ψ(s).

However, the subtlety is in the same thing that we discussed at the end of Sub-section 6.2. Namely, in the proofs of Theorems 3 and 4 the equivalence of theconditions

Rdist(s, s) 6= N and Rdist(s, s) > N

was related to the fact that Lemma 3′ is valid for all the n = 4pa with which weworked at that moment.

But taking N ′ < N we now have no right to claim that

α(GN ′) 6 θ3(lnN ′)θ2(N ′)θ1(1 + δ′0(N′)).

We need some universal estimate for the independence number of the graph GNthat would not appeal to the specific arithmetic nature of the number n. Butamazingly, if the strange condition n = 4pa is discarded, nothing better than

α(GN ) 6 (1.99 + ν(n))n, ν(n) = o(1)

has been invented so far (see [15]). This is related to the fact that the ‘simplicity’ ofthe quantity n/4 allows one to engage the powerful linear-algebraic method, whichfundamentally cannot be extended to ‘complicated’ numbers n/4 (see [7]).

The further arguments are standard and we do not carry them out. We simplytake c = 1.99 instead of c = 4/33/4 and consider η = (ln 2)/ln c instead of ξ =(ln 2)/ln c.

The theorem is proved.

On the Ramsey numbers for complete distance graphs 1805

§ 9. Comparison of the results obtained

Thus, we have the following facts.

1. We have the asymptotics

ln(Rprimedist (s, s)) ∼ ξ ln s = 1.2326229 . . . · ln s.

2. For infinitely many s we have the estimate

Rprimedist (s, s) > 17.735578 · s1.2326229...(ln s)0.1163114....

3. We have the estimate

1.0072842 . . . · (ln s)(1 + o(1)) = η(ln s)(1 + o(1)) 6 ln(Rdist(s, s))6 ξ(ln s)(1 + o(1)) = 1.2326229 . . . · (ln s)(1 + o(1)).

4. We have the inequality

Rdist(s, s) 6 2282.1524 · s1.2326229...(ln s)0.7326229....

Thus, the smallest gap is observed between the estimates in parts 2 and 4, andthe worst situation holds for the quantity Rdist(s, s), for which at the moment thereare even no asymptotics of the logarithm.

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1806 K. A. Mikhaılov and A.M. Raıgorodskiı

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K. A. Mikhaılov

Faculty of Mechanics and Mathematics,

Moscow State University

E-mail : airdesign@mail.ru

A. M. Raıgorodskiı

Faculty of Mechanics and Mathematics,

Moscow State University

E-mail : mraigor@yandex.ru

Received 4/JUN/08Translated by E. KHUKHRO