228

ISSN 1064–5624, Doklady Mathematics, 2007, Vol. 75, No. 2, pp. 228–230. © Pleiades Publishing, Ltd., 2007.Original Russian Text © I.M. Shitova, 2007, published in Doklady Akademii Nauk, 2007, Vol. 413, No. 2, pp. 178–180.

This paper considers the classical problem of com-binatorial geometry, which goes back to Nelson, Erdösand Hadwiger and consists in finding the chromaticnumber of a metric space with some distances forbid-den. By the chromatic number we mean the quantity

χ

((

X

,

ρ

),

A

)

, which is defined as follows. Given a metricspace

X

with metric

ρ

and a (possibly infinite) set

A

ofpositive real numbers, the chromatic number

χ

((

X

,

ρ

),

A

)

of the metric space

X

with metric

ρ

and forbiddendistance set

A

is the minimum number of colorsrequired to color

X

so that any

x

,

y

and

x

∈

X

,

y

∈

X

forwhich

ρ

(

x

,

y

)

∈

A

are of different colors.

One of the cases that have been studied most thor-oughly is that in which

X

=

R

n

,

ρ

=

l

2

, where

l

2

is theusual Euclidean metric, and

A

is a finite set. For exam-ple, it is known that

(1)

The lower bound is due to Raigorodskii (see [1]), andthe upper one is due to Larman and Rogers (see [2]).

For a set

A

of cardinality 2, the same authorsobtained the estimates

(2)

(the lower bound was obtained in [3], and the upperbound was obtained in [2]).

In general, for a set

A

of cardinality

|

A

|

=

k

, we have

(3)

where

c

1

and

c

2

are some positive constants (see [3]).

1.239 o 1( )+( )n χ Rn l2,( ) a{ },( ) 3 o 1( )+( )n≤ .a

max≤

1.439 o 1( )+( )n χ Rn l2,( ) a1 a2,{ },( )a1 a2,{ }max≤

≤ 9 o 1( )+( )n

c1k( )c2n

χ Rn l2,( ) A,( )A: A k=max 3 o 1( )+( )kn,≤ ≤

The important case of

X

=

R

n

and

ρ

=

l

1

has alsobeen studied fairly well. It is known that, in this case,

(4)

(see [4, 5]). As in the case of

l

2

, an estimate for the chro-matic number of

R

n

with more than one forbidden dis-tances can be obtained.

In this paper, we improve the lower bound in (2) andobtain some other estimates. In the next section, westate the main results.

STATEMENT OF THE RESULTS

Theorem 1.

The following estimate holds

:

(5)

Theorem 2.

The following estimate holds

:

(6)

Remark.

Generally, the method of proof of Theo-rems 1 and 2 makes it possible to obtain estimates forsets of forbidden distances of any finite cardinality. Asan example, we give several more estimates below.

Theorem 3.

The following estimates are valid

:

(7)

(8)

Theorem 4.

The following estimates are valid

:

(9)

(10)

1.365 o 1( )+( )n χ Rn l1,( ) a{ },( )a

max≤

≤ 5 o 1( )+( )n

χ Rn l2,( ) a1 a2,{ },( ) 1.465 o 1( )+( )n.≥a1 a2,{ }max

χ Rn l1,( ) a1 a2,{ },( ) 1.691 o 1( )+( )n.≥a1 a2,{ }max

χ Rn l2,( ) A,( )A: A 3=max 1.664 o 1( )+( )n,≥

χ Rn l2,( ) A,( )A: A 4=max 1.836 o 1( )+( )n.≥

χ Rn l1,( ) A,( )A: A 3=max 2.000 o 1( )+( )n,≥

χ Rn l1,( ) A,( )A: A 4=max 2.250 o 1( )+( )n.≥

MATHEMATICS

On the Chromatic Number of a Space with Several Forbidden Distances

I. M. ShitovaPresented by Academician V.V. Kozlov June 20, 2006

Received July 24, 2006

DOI: 10.1134/S1064562407020147

Moscow State University, Leninskie gory, Moscow, 119992 Russiae-mail: irshitova@yandex.ru

DOKLADY MATHEMATICS Vol. 75 No. 2 2007

ON THE CHROMATIC NUMBER OF A SPACE WITH SEVERAL FORBIDDEN DISTANCES 229

PROOF OF THEOREM 1Preliminary remarks. In essence, the proofs of

Theorems 1–4 are very similar. Below, we present someconsiderations concerning the proof of Theorem 1 and,thereby, clarify the main ideas on which the proofs ofTheorems 2–4 are based.

Suppose that a dimension n and a set A = {a1, a2} offorbidden distances are given. To these n and A weassign the infinite geometric graph defined by

where

Thus, � is the graph whose vertex set coincides withthe set of all points in Rn; two vertices are joined by anedge if the distance between the corresponding pointsof space equals ai for some i ∈ {1, 2}. The chromaticnumber of the space (Rn, l2) with the set of forbiddendistances A = {a1, a2} is the usual chromatic number ofthe graph �.

It is known that the chromatic number of � is finite(see [2]). According to the Erdös–de Bruijn theorem(see [6]), if some graph has finite chromatic number,then it has a finite subgraph with the same chromaticnumber; therefore, in finding the chromatic numbers ofthe metric spaces under consideration, it is useful toconsider finite geometric graphs, which are, by defini-tion, finite subgraphs of the corresponding (infinite)geometric graphs �. The point is to choose a suitablesubgraph.

Frankl and Wilson (see [7]) and Raigorodskii (see[1, 3, 8]) showed that it is convenient to deal with finitegeometric graphs for which the coordinates of verticescan take only finitely many values. These authors stud-ied the graphs � = (V, E), where V ⊆ {0, 1}n and V ⊆{–1, 0, 1}n. The graphs which we consider in this paperhave the form G = (V, E), where

or

These graphs have already been studied to a certainextent (see [8]), but the problem is difficult, and no newresults have been obtained. We use them to improvelower bounds. The chromatic number of a graph � isbounded below as

� � Rn A,( ) V E,( ),= =

V Rn,=

E x y,( ): x V , y V , l2 x y,( ) A∈ ∈ ∈{ },=

V 0 1 2 3, , ,{ }n,⊆

E x y,( ): x V , y V , l2 x y,( ) A∈ ∈ ∈{ },=

V 0 1 2 3 4, , , ,{ }n,⊆

E x y,( ): x V , y V , l2 x y,( ) A∈ ∈ ∈{ }.=

χ �( ) Vα �( )-------------,>

where α(�) is the independence number of the graph �(see [9]). Consider the case of V = {0, 1, 2, 3}n.

Sketch of the proof. Below, we outline the proof ofTheorem 1. It consists of the following steps.

Step 1. Take nonnegative integers v0, v1, v2, and v3for which v0 + v1 + v2 + v3 = n. Consider the set G =G(v0, v1, v2, v3) of graphs G = (V, E) such that pre-cisely v0 coordinates of points from V equal 0, v1 coor-dinates equal 1, v2 coordinates equal 2, and v3 coordi-nates equal 3. Clearly,

Step 2. Let s and S denote, respectively, the mini-mum and maximum scalar product of vectors from V.For each set of v0, v1, v2, and v3, the values of s and Sare linear combinations of these numbers. We set q =

. A well-known theorem from number theory says

that, for any sufficiently large q, there is a prime

between q and q + O (see [10]). We assume that

vi ~ · n for i = 0, 1, 2, 3, where the are constants.The choice of the vi ensures that q is sufficiently large.

We also set a1 = and a2 = .

Step 3. Using the linear-algebraic method of combi-natorics (see [11]), we can prove the following lemma.

Lemma 1. For any family of vectors Q = {x1, x2, …,xs} ⊂ V such that l2(xi, xj) ∉ A for any i ≠ j, the quantity|Q| satisfies the inequality

where M is the set determined by the system

(11)

Step 4. It follows from the lemma that, for any � ∈G,

Vn!

v 0!v 1!v 2!v 3!---------------------------------.=

S s–3

-----------

q38/61( )v i

0v i

0

2 p 4 p

Q Cns1Cn s1–

s2 Cn s1– s2–s3

s0 s1 s2 s3, , ,( ) M∈∑≤

= n!

s0!s1!s2!s3!--------------------------,

s0 s1 s2 s3, , ,( ) M∈∑

s0 0,≥s1 0,≥s2 0,≥s3 0,≥

s0 s1 s2 s3+ + + n,=

s1 2s2 3s3 p 1.–≤+ +

α �( ) n!s0!s1!s2!s3!--------------------------.

s0 s1 s2 s3, , ,( ) M∈∑≤

230

DOKLADY MATHEMATICS Vol. 75 No. 2 2007

SHITOVA

Therefore,

Thus,

and analytically, the problem reduces to finding the bestestimate among those given above.

The sum over the set M is bounded above by its larg-est term multiplied by the cardinality of M and belowby this largest term itself. We analytically estimate theasymptotics of the denominator and then numerically(with the aid of a computer) find the best estimate withthis denominator.

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(2000).2. D. G. Larman and C. A. Rogers, Mathematika 19, 1–24

(1972).3. A. M. Raigorodskii, Usp. Mat. Nauk 56 (1), 107–146

(2001).4. A. M. Raigorodskii, Usp. Mat. Nauk 59 (5), 161–162

(2004).5. J.-H. Kang and Z. Füredi, Theor. Comp. Sci. 319 (1/3),

357–366 (2004).6. N. G. de Bruijn and P. Erdös, Proc. Konink. Ned. Acad.

Wettensch. Ser A 54 (5), 371–373 (1951).7. P. Frankl and R. Wilson, Combinatorica 1, 357–368

(1981).8. A. M. Raigorodskii, Mat. Sb. 196 (1), 123–156 (2005).9. F. Harary, Graph Theory (Addison-Wesley, Reading,

Mass., 1969; Mir, Moscow, 1973).10. K. Prachar, Primzahlverteilung (Springer-Verlag, Ber-

lin, 1957; Mir, Moscow, 1967).11. L. Babai and P. Frankl, Linear Algebra Method in Com-

binatorics (Depart. Computer Sci., Univ. Chicago, Chi-cago, 1992), Preliminary Version 2, Part 1.

χ G( )V

α �( )------------≥ =

n!v 0!v1!v2!v3!------------------------------

n!s0!s1!s2!s3!------------------------

s0 s1 s2 s3, , ,( ) M∈∑

= 1

v 0!v 1!v 2!v 3!---------------------------------

n!s0!s1!s2!s3!--------------------------.

s0 s1 s2 s3, , ,( ) M∈∑

χ Rn l2,( ) A,( ) χ G( ),G G∈max≥