Estimating the chromatic numbers of Euclidean space by convex minimization methods

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Sbornik : Mathematics 200:6 783–801 c© 2009 RAS(DoM) and LMS

Matematicheskiı Sbornik 200:6 3–22 DOI 10.1070/SM2009v200n06ABEH004019

Estimating the chromatic numbers of Euclideanspace by convex minimization methods

E. S. Gorskaya, I.M. Mitricheva, V.Yu. Protasov and A. M. Raıgorodskiı

Abstract. The chromatic numbers of the Euclidean space Rn with k for-bidden distances are investigated (that is, the minimum numbers of coloursnecessary to colour all points in Rn so that no two points of the same colourlie at a forbidden distance from each other). Estimates for the growthexponents of the chromatic numbers as n→∞ are obtained. The so-calledlinear algebra method which has been developed is used for this. It reducesthe problem of estimating the chromatic numbers to an extremal problem.To solve this latter problem a fundamentally new approach is used, whichis based on the theory of convex extremal problems and convex analysis.This allows the required estimates to be found for any k. For k 6 20these estimates are found explicitly; they are the best possible ones in theframework of the method mentioned above.

Bibliography: 18 titles.

Keywords: chromatic number, distance graph, convex optimization.

§ 1. Introduction

Given a system of different positive numbers a1, . . . , ak, by the chromatic numberof the Euclidean space Rn with the set of forbidden distances A = {a1, . . . , ak} wemean the quantity

χ(Rn,A ) = min{χ : Rn = V1 t · · · t Vχ, ∀ i ∀x, y ∈ Vi |x− y| /∈ A

}.

In other words, χ(Rn,A ) is the minimum number of colours required for colouringall points in the space so that no two points of the same colour have the distancebetween them lying in the set A .

The problem of finding the chromatic number χ(Rn,A ) goes back to Nelson,Hadwiger and Erdos. It was posed in the late 1940s or early 1950s and is nowadays

The research of V.Yu. Protasov was carried out with the support of the Russian Foundationfor Basic Research (grant no. 08-01-00208) and in the framework of the Programmes of Supportof Young Doctors of Science of the RF (grant no. МД-2195.2008.1) and of Support of LeadingScientific Schools of the RF (grant no. НШ-3233.2008.1); the research of A. M. Raıgorodskiı wascarried out with the support of the Russian Foundation for Basic Research (grant no. 06-01-00383),the Programmes of Support of Young Doctors of Science of the RF (grant no. МД-5414.2008.1) andof Support of Leading Scientific Schools of the RF (grant no. НШ-691.2008.1), and the DynastyFoundation; the research of I. M. Mitricheva was supported by the Russian Foundation for BasicResearch (grant no. 06-01-00383).

AMS 2000 Mathematics Subject Classification. Primary 52C10, 51K05, 05C15; Secondary05C12.

784 E. S. Gorskaya, I. M. Mitricheva, V.Yu. Protasov and A.M. Raıgorodskiı

a classical problem of combinatorial geometry (see [1]–[5]). Following Erdos, mostauthors consider the quantity

χ(Rn; k) = maxA :|A |=k

χ(Rn,A ).

In this paper we investigate the asymptotic behaviour of χ(Rn; k) as n→∞ fora fixed value of k. In this direction the following results are known. For k = 1,

(ζ1 + o(1))n 6 χ(Rn; 1) 6 (3 + o(1))n, ζ1 = 1.239 . . . . (1)

The lower bound in (1) is due to Raıgorodskiı [6] and the upper one to Larman andRogers [7]. For k = 2, 3, 4 the following estimates hold:

(ζ2 + o(1))n 6 χ(Rn; 2) 6 (9 + o(1))n, ζ2 = 1.465 . . . , (2)

(ζ3 + o(1))n 6 χ(Rn; 3) 6 (27 + o(1))n, ζ3 = 1.664 . . . , (3)

(ζ4 + o(1))n 6 χ(Rn; 4) 6 (81 + o(1))n, ζ4 = 1.836 . . . . (4)

The lower bounds in (2)–(4) are due to Shitova (Mitricheva) [8], [9] and the upperones are obvious consequences of a similar estimate in (1). Finally, for an arbitrary kwe have

(c1k)c2n 6 χ(Rn; k) 6 (3 + o(1))kn, c1 > 0, c2 > 0, (5)

as proved by Raıgorodskiı in [1] (see also [2] and [10]).In this paper we obtain lower asymptotic bounds for the chromatic numbers as

n → ∞, that is, we calculate the ζk for all k ∈ N. To do this we use a construc-tion given in [1], [8] and [9], which reduces the calculation of the exponents ζk toa certain extremal problem. However, in this paper, to solve it we use a funda-mentally new approach: we reduce the problem to a family of convex extremalproblems of a special form (minimization problems for a convex function on theboundary of a polyhedron), and to solve the latter we develop an effective method(which is perhaps of independent interest). All this enables us to:

1) calculate the constants ζk, k = 1, . . . , 20 (the values that we find are bestpossible for our method; in particular, for k = 3, 4 we have improved the previouslyknown estimates (3) and (4));

2) obtain analytic expressions for the bounds ζk for all k > 20 and to makea conjecture that these estimates are also optimal for the method selected.

The structure of the paper is as follows. In § 2 we describe a method for find-ing estimates for the chromatic numbers, which reduces the task to solving someextremal problem. In § 3 we transform the latter into a family of convex extremalproblems and develop a method for their solution. We also prove in that sec-tion several auxiliary results in convex geometry and the theory of extremal valuesof functions. In § 4 we present an algorithm for the numerical realization of ourmethod. In § 5 we collect our main numerical results to give a table of values of theoptimal values of the ζk and all the auxiliary parameters for k = 1, . . . , 20. Finally,in § 6 we find asymptotic estimates for arbitrary ζk with k > 20 and make a con-jecture that these estimates cannot be improved in the framework of our method.

Estimating the chromatic numbers 785

§ 2. The method for finding estimates. The extremal problem

First of all note that the problem § 1 can be conveniently reformulated in graph-theoretic terms. Recall that the chromatic number of a graph G = (V,E) withvertex set V and edge set E is the minimum number of colours required for colouringits vertices so that vertices of the same colour are not connected by an edge.

Let G be a graph such that

V ⊂ Rn, |V | <∞, E ={{x, y} : x, y ∈ V, |x− y| ∈ A

},

where A is a set of positive real numbers. Then, obviously, χ(Rn,A ) > χ(G). Infact, considering finite graphs (called (finite) distance graphs) implies no restric-tions. By the de Bruijn-Erdos theorem [11] there exists a finite distance graph withchromatic number equal to the corresponding chromatic number of the space Rn.

Thus, to find the best lower bound for χ(Rn; k) we must pick V ⊂ Rn anda1, . . . , ak such that the corresponding graph G = (V,E) has the maximum chrom-atic number χ(G).

In turn, the simplest way to obtain a lower estimate for χ(G) is by means of theelementary inequality χ(G) > |V |/α(G), where

α(G) = max{|W | : W ⊆ V, ∀x, y ∈W {x, y} /∈ E

}is the independence number of the graph G. Note that we definitely do not losevery much by taking this approach: ‘almost every’ graph has chromatic numberasymptotically equal to |V |/α(G) as |V | → ∞ (see [12]).

Among the known upper bounds for the independence number of a distancegraph the sharpest rely on the so-called linear algebra method in combinatorics [2].As shown in [9], this results in a certain class of distance graphs, which we definebelow.

Thus, fix n, k and an integer d > 3. Furthermore, let

v0, . . . , vd−1 ∈ (0, 1), v0 + · · ·+ vd−1 = 1.

We set

w0 = bv0nc, w1 = bv1nc, . . . , wd−1 = n− w0 − w1 − · · · ,

where btc is the integer part of t. Thus, w0+· · ·+wd−1 = n, and as the dimension ngrows, we obtain the asymptotic behaviour w0 ∼ v0n, . . . , wd−1 ∼ vd−1n. Let

V = V ({0, . . . , d− 1};w0, . . . , wd−1)

={x = (x1, . . . , xn) : xi ∈ {0, 1, . . . , d− 1} ∀ i, |{i : xi = j}| = wj ∀ j

}.

Consider the quantities

r = r({0, . . . , d− 1};w0, . . . , wd−1) = maxx,y∈V

(x, y),

r = r({0, . . . , d− 1};w0, . . . , wd−1) = minx,y∈V

(x, y),

786 E. S. Gorskaya, I. M. Mitricheva, V.Yu. Protasov and A.M. Raıgorodskiı

where (x, y) is the standard scalar product in Rn. Asymptotically as n→∞, thesequantities behave like Mn and mn, respectively, where M = M(v) and m = m(v)are ‘constants’ (that is, they do not depend on n) each of which is a function ofv = (v0, . . . , vd−1). We also set h(x) = M(x)−m(x).

Let p′ be the smallest prime such that r − (k + 1)p′ < r. Then p′ ∼ pn, wherep = h(v)/(k + 1) (cf. [13]). We set

a1 =√

2p′, a2 =√

4p′, . . . , ak =√

2kp′,

E ={{x, y} : |x− y| ∈ {a1, . . . , ak}

}.

Thus, the class of graphs mentioned above consists of all possible graphs G =(V,E) parameterized by quantities v0, . . . , vd−1 ∈ (0, 1) such that v0+· · ·+vd−1 = 1.That is, the values of the parameters form the simplex

∆ ={x ∈ Rd : x > 0, (x, e) = 1

},

where e = (1, . . . , 1) and the inequality x > 0 is taken to hold for each coor-dinate. Note that the simplex ∆ is closed even though we were only discussingopen intervals before. However, this is not essential. Also, as shown in [9], foreach v = (v0, . . . , vd−1) ∈ ∆ the independence number α(G) of the correspondinggraph G has the following estimate:

α(G) 6∑

(q0,...,qd−1)∈A

n!q0! · · · qd−1!

,

A ={q = (q0, . . . , qd−1) : qi ∈ N ∪ {0} ∀ i,q0 + · · ·+ qd−1 = n, q1 + 2q2 + · · ·+ (d− 1)qd−1 6 p′ − 1

}.

Note that the condition q1 + 2q2 + · · · + (d − 1)qd−1 6 p′ − 1 can be conciselyexpressed as (q,b) 6 p′ − 1, where b = (0, 1, . . . , d− 1).

Hence the best estimate for χ(Rn; k) that can be derived from [9] is as follows:

χ(Rn; k) > maxd>3

maxv∈∆

|V |α(G)

> maxd>3

maxv∈∆

n!/(w0! · · ·wd−1!)∑(q0,...,qd−1)∈A n!/(q0! · · · qd−1!)

. (6)

Here we do not explicitly indicate that the right-hand side in the chain of inequalitiesdepends on k, but the edge sets of the graph depend on k, therefore so does thedomain of summation A . That the interior maxima depend on d is obvious. Nowlet f(x) =

∑di=1 xi lnxi be the entropy function. It is defined as a continuous

function on Rd+ = {x ∈ Rd | x > 0}; here we set xi lnxi = 0 for xi = 0. In terms of

the entropy function we can express the right-hand side of (6) as follows:

maxv∈∆

n!/(w0! · · ·wd−1!)∑(q0,...,qd−1)∈A n!/(q0! · · · qd−1!)

= (eSd,k + o(1))n as n→∞,

whereSd,k = max

v∈∆

(min

s∈∆, (s,b)6h(v)/(k+1)f(s)− f(v)

)(7)

Estimating the chromatic numbers 787

(see [9]). As a result the estimate

χ(Rn; k) >(max

deSd,k + o(1)

)n

, (8)

which is optimal in the framework of this method, is a solution of the extremalproblem of finding the quantity Sd,k for each d and maximizing the result over d.

We see that to find bounds for χ(Rn; k) we must solve the following problem:F (v) =(

mins∈∆, (s,b)6h(v)/(k+1)

f(s)− f(v))→ max,

v ∈ ∆.(9)

For d = 3, 4, 5 the problem was solved numerically in [9], and this led to theestimates (2)–(4). Estimate (1) is slightly simpler; it is sufficient to consider the cased = 3. The fact that for d > 3 the lower estimate in (1) cannot be refined showsthe limitations of this method. However, we know of no method providing betterresults.

Note that the value of the target function in (9) is easy to calculate for each v∈∆.Indeed, we find the minimum of the convex function f(s) on a polyhedron (theintersection of the simplex ∆ and the hyperplane {s ∈ Rd | (s,b) 6 h(v)/(k + 1)}).Hence this minimum can be explicitly calculated using the method of Lagrangemultipliers [14] (we present the corresponding argument in Lemma 2). Were thetarget function concave in v, we could calculate its maximum Sd,k effectively fromits values [15], but it is not concave and, moreover, can have many local maxima. Itis always fairly difficult to find the global maximum of such a function: all the knownalgorithms are inefficient and in fact imply examining the values of the function atthe nodes of some net on the simplex ∆ systematically, and reducing the meshof the grid repeatedly. This method was used to obtain the estimates (2)–(4)in [9]. However, as d increases the time required to make such an exhaustivesearch increases sharply (‘the curse of dimensionality’), which makes it impossibleto derive a sharp bound even for k = 5. For this reason in our paper we putforward an entirely new approach to this problem, which consists in the following.Problem (9) is reduced to a family of convex extremal problems of a special kindin Theorem 1. We have developed an efficient method to solve them and haveobtained an analytic expression for the maxima (see Theorem 2). This enables usto solve the original problem and to find optimal values (for our method) of theconstants Sd,k and ζk for all k 6 20, while for other values of k we can deduceestimates for ζk.

§ 3. Solving the extremal problem

Recall our notation: e = (1, . . . , 1) ∈ Rd, c = (1/d)e = (1/d, . . . , 1/d),b = (0, 1, . . . , d − 1), (x, y) =

∑di=1 xiyi is the standard scalar product in Rd,

∆ ={x ∈ Rd | x > 0, (x, e) = 1

}is the unit simplex in Rd. For fixed λ > 0 we set

[λ] =1∑d−1

j=0 λj(1, λ, . . . , λd−1) ∈ Rd.

788 E. S. Gorskaya, I. M. Mitricheva, V.Yu. Protasov and A.M. Raıgorodskiı

If a = (a1, . . . , ad) is a vector with nonnegative integer components, then

[λ]a =1∑d−1

j=0 λaj+1

(λa1 , . . . , λad).

Thus, [λ]a∈∆ for each a. In particular, [λ]b =[λ], [1]=c. Now, f(x)=∑d

i=1 xi lnxi

is the entropy function; here we assume that xi lnxi = 0 for xi = 0. The func-tion f is strictly convex, therefore it has a unique minimum point on each convexcompact set; f attains its minimum on the simplex ∆ at a point c. We also seth(x) = M(x)−m(x), x ∈ ∆, where M(x) and m(x) are the constants defined in § 2.

We start with a simple auxiliary result on the position of the minimum pointsof the entropy function f .

Lemma 1. If G is a convex compact subset of ∆ containing at least one interiorpoint of the simplex ∆, then the minimum point of f on G is also an interior pointof ∆.

Proof. Let x ∈ G be a point in the interior of ∆. Assume that f takes its minimumon G at a point z ∈ ∂∆. Without loss of generality suppose that zi = 0, i = 1, . . . , q,and zi > 0, i = q + 1, . . . , d. The partial derivatives fvi

= ln vi + 1, i = 1, . . . , q,reduce to −∞ at v = z. The derivative of f in the direction of the vector x − zis equal to

∑dj=1(xi − zi)fvi

. Since xi − zi > 0 for i = 1, . . . , q, this derivativealso reduces to −∞ at z. Hence at points v ∈ [z, x] sufficiently close to z we havef(v) < f(z), which contradicts our assumption.

Now we find a point at which the interior minimum in the problem (9) is attained.

Lemma 2. For each p > (d− 1)/2 the function f attains its minimum on{s ∈ ∆ | (s,b) 6 p} at the point c, and for each p ∈ (0, (d− 1)/2] it attainsthe minimum at [λ], where λ is the unique positive zero of the polynomial Pb(z) =∑d−1

j=0(j − p)zj .

Proof. Since (c,b) = (d− 1)/2, for p > (d− 1)/2 the point c lies in the setunder consideration and therefore gives the minimum of f . For p 6 (d− 1)/2 theminimum is attained on the boundary of this set, but by Lemma 1 not onthe boundary of the simplex ∆. Hence the minimum point lies in the plane(s,b) = p. The necessary condition for the minimum is fulfilled at this point:Ls = 0, where L = f(s) + λ1((s,b)− p) + λ2((s, e)− 1) is the Lagrangian. Thus,ln sn +1+λ1(n−1)+λ2 = 0, n = 1, . . . , d. Hence sn = Cλn−1, n = 1, . . . , d, whereλ and C are constants. From the condition (s, e) = 1 we obtain C =

(∑d−1j=0 λ

j)−1,

so s = [λ]. From the condition (s,b) = p we see that λ is a zero of Pb(z). Thispolynomial cannot have another positive root λ since then the point [λ] would alsosolve the equation Ls = 0 and, by the Karush-Kuhn-Tucker Theorem [14], it wouldalso deliver the minimum of f for s ∈ ∆, (s,b) = p. But this is impossible since fis strictly convex.

Now we consider problem (9); when we solve it we shall find the quantities Sd,k

given in (7). For each v ∈ ∆, using Lemma 2 we can calculate the value of thetarget function. The main difficulty, as we already mentioned, is as follows: gen-erally speaking this function is not concave and can have several local maximum

Estimating the chromatic numbers 789

points. For this reason finding its maximum on the simplex ∆ to within 0.01, say,can be impossible for all practical purposes even for d = 5. To find a way outof this situation we use some special properties of the function under considera-tion. The key point here is Proposition 1 below, which means that problem (9)reduces to a family of convex extremal problems depending smoothly on a param-eter p. To state it we require the quantity r(d, k), which is equal to the minimumof (c,b) and h(c)/(k − 1). Note that (c,b) =

∑d−1j=0 j/d = (d− 1)/2. Now using

the well-known formula 12 + · · ·+m2 = m(m+ 1)(2m+ 1)/6 we obtain

h(c) =d−1∑j=0

j2

d−

d−1∑j=0

j(d− 1− j)d

=d2 − 1

6.

Thus,

r(d, k) = min{d− 1

2,d2 − 1

6(k + 1)

}. (10)

Proposition 1. For any d and k, Sd,k > 0 and furthermore,

Sd,k = max0<p6r(d,k)

(min

s∈∆, (s,b)=pf(s)− min

v∈∆, h(v)=(k+1)pf(v)

). (11)

Proof. First we show that the solution Sd,k of problem (9) is positive for any kand d. It is sufficient to demonstrate the existence of v ∈ ∆ such that F (v) > 0,where

F (v) = mins∈∆, (s,b)6h(v)/(k+1)

f(s)− f(v)

(see the notation in problem (9)). We set vλ = [λ]. As λ → +0, we obtain(vλ,b) = λ + o(λ) and h(vλ) = λ + o(λ). Since h(c) > 0, it follows for all small λthat

1k + 1

h(vλ) < (vλ,b) <1

k + 1h(c).

Since h is continuous, there exists a point v on the interval [vλ, c] at whichh(v)/(k + 1) = (vλ,b). Obviously, v is different from vλ and c. By Lemma 2,for small λ (when (vλ,b) 6 (d− 1)/2) the function f attains its minimum on{s ∈ ∆ | (s,b) 6 (vλ,b)

}at the point vλ. Hence

mins∈∆, (s,b)6h(v)/(k+1)

f(s) = f(v).

Thus, F (v) = f(vλ) − f(v). Since f is a strictly convex function attaining itsminimum at c, it follows that f(v) < max

{f(vλ), f(c)

}= f(vλ). Hence F (v) > 0,

so that Sd,k > 0.Now we show that the maximum value in (9) remains the same after passing to

the equality in (s,b) 6 h(v)/(k + 1):

Sd,k = maxv∈∆

(min

s∈∆, (s,b)=h(v)/(k+1)f(s)− f(v)

). (12)

Assume the contrary: the inner minimum and the outer maximum in (7) areattained at points s, v ∈ ∆ for which (s,b) < h(v)/(k + 1). This inequality will

790 E. S. Gorskaya, I. M. Mitricheva, V.Yu. Protasov and A.M. Raıgorodskiı

also hold in sufficiently small neighbourhoods of s and v. In such a neighbour-hood the values of f(s) do not decrease since otherwise s is not a minimum point.Applying Lemma 1 to the entire simplex G = ∆ we obtain s ∈ int∆, and therefores = c. Similarly, in a sufficiently small neighbourhood of v the values of f(v) do notdecrease for otherwise the outer maximum would increase. Thus, v = c. But thenSd,k = 0, which leads to a contradiction and completes the proof of formula (12).It remains to set p = (s,b) in it and to show that (s,b) < r(d, k) at the mini-mum point s. As a result, we will obtain (11). If (s,b) > (d− 1)/2, then alsoh(v)/(k + 1) > (d− 1)/2 = (c,b), so that the inner minimum in (7) is equal tof(c). Thus, Sd,k 6 0, which again leads to a contradiction. On the other hand, if(s,b) > (d2 − 1)/(6(k + 1)) = h(c)/(k + 1), then replacing the maximum point vby c in (7) we reduce f(v), but do not reduce the inner minimum (since we shrinkthe set over which the minimum is taken). Hence the outer maximum will increase,which is impossible. Thus,

p = (c,b) 6 min{d− 1

2,d2 − 1

6(k + 1)

}.

Now, to find Sd,k we must find the maximum over p ∈ [0, r(d, k)] of the functionin parentheses in (11). This function is the difference between the minima of thestrictly convex function f on two sets. The first set is a hyperplane, and we findthe minimum on it using Lemma 2. The second set is the surface in Rd describedby the equation h(v) = (k+1)p. We shall show in Proposition 2 that h(v) is a con-cave piecewise linear function, which therefore defines the surface of a polyhedron.In this way we arrive at the following extremal problem.

Problem 1. Find the minimum of a convex function on the boundary of a poly-hedron in Rd described by a system of linear inequalities.

The standard approach to the solution of this problem consists in listing all the(d− 1)-faces, finding an approximate value of the minimum of the function on eachface by solving the corresponding convex problem with linear constraints, and com-paring the minimum values over all faces. The convex problem with constraintscan be solved, for instance, using the interior point method [16], [17]. However, wecannot actually use this approach in our case, not only because it requires largecomputational resources, but also because we are interested not in an approximatevalue of the minimum for each p, but in an analytic expression for the minimum,which we can differentiate with respect to p to find the maximum of the func-tion (11). We point out that the minimum can be attained on a lower-dimensionalface and that to examine all the faces of different dimensions requires immense com-putational resources. For these reasons we put forward another approach, based onthe following simple result.

Lemma 3. Let ϕ : Rd → R be a convex function and G ⊂ Rd a polyhedron boundedby hyperplanes {αi}1 and containing the global minimum point of ϕ. Then the mini-mum of ϕ on the boundary of G is equal to its minimum on the union of thehyperplanes {αi}.

1Some αi can be disjoint from the polyhedron.

Estimating the chromatic numbers 791

Proof. Assume that there exists a point a ∈ αi such that ϕ(a) < minx∈∂G ϕ(x).Then a /∈ G. Hence the line interval joining a to the global minimum point of ϕintersects the boundary of G at a point b. Since ϕ is convex, we have ϕ(b) 6 ϕ(a),which is a contradiction.

Thus, if the minimum point of a smooth function lies in the interior of a poly-hedron, then to minimize the function over the boundary of the polyhedron it issufficient to find its minimum on each hyperplane bounding the polyhedron (oneach of them this is a problem without constraints) and to select the smallest valueamong those obtained. We see that to solve the problem stated above it is sufficientto show that the set under consideration is a polyhedron and to find hyperplanesbounding it.

Proposition 2. The function h(v) is concave on ∆ and the inequality h(v) > q forany q is equivalent to a system of at most 2d−1 linear inequalities (v,ai) > q withnonnegative integer vectors ai independent of q.

Proof. Let

ψj(t) =

{0, t < v0 + · · ·+ vj ,

1, t > v0 + · · ·+ vj ,

and ψ =∑d−2

j=0 ψj . Thus, ψ(x) = min{j > 0 | x <

∑ji=0 vi

}is a piecewise constant

nondecreasing right-continuous function assuming integer values lying between zeroand d− 1 inclusive. Then

M(v) =∫ 1

0

ψ2(t) dt, m(v) =∫ 1

0

ψ(t)ψ(1− t) dt.

Obviously, M(v) =∑d−1

j=1 j2vj . Thus, M(v) is a linear function. We claim that

m(v) is convex. For any x we set x+ = max{x, 0}. For arbitrary i and j,∫ 1

0

ψi(t)ψj(1− t) dt =( ∑

r>i+1

vr +∑

s>j+1

vs − 1)

+

=( ∑

r>i+1

vr −∑s6j

vs

)+

. (13)

Note also that ∫ 1

0

ψi(t)ψj(1− t) dt =∫ 1

0

ψj(t)ψi(1− t) dt

since the second integral is obtained from the first by substituting t by 1− t. Hence

m(v) =∫ 1

0

ψ(t)ψ(1− t) dt =∑i,j

∫ 1

0

ψi(t)ψj(1− t) dt

=∑

i

∫ 1

0

ψi(t)ψi(1− t) dt+∑i 6=j

∫ 1

0

ψi(t)ψj(1− t) dt

=∑

i

∫ 1

0

ψi(t)ψi(1− t) dt+ 2∑i>j

∫ 1

0

ψi(t)ψj(1− t) dt.

792 E. S. Gorskaya, I. M. Mitricheva, V.Yu. Protasov and A.M. Raıgorodskiı

Using equality (13) we obtain

m(v) =∑

i

( ∑r>i+1

vr −∑s6i

vs

)+

+ 2∑i>j

( ∑r>i+1

vr −∑s6j

vs

)+

. (14)

The right-hand side of this equality contains (d − 1)2 terms, each being thenonnegative part of a linear function. Hence the function m(v) is convex; accord-ingly, h(v) = M(v) −m(v) is concave. For each q > 0 the inequality h(v) > q isequivalent to a system of 2(d−1)2 linear inequalities (ai, v) > q corresponding to thevarious ways in which the maxima can actually be realized in the (d− 1)2 terms ofthe form x+ = max{x, 0} in (14). Each of these inequalities has integer coefficients(of the variables vj) because the coefficients in each term are ±1. For any j thereexist precisely j2 terms in (14) in which the coefficient of vj is +1. Hence for anyrealization of the maxima in (14) the coefficient of vj does not exceed j2. Since inM(v) the same variable has coefficient j2, in each inequality (ai, v) > q the coeffi-cient ai

j (of vj) is nonnegative. In a similar way we can show that aij 6 (d − 2)2.

Thus, all the coefficients in the inequalities (ai, v) > q are nonnegative integers notexceeding (d− 2)2.

It remains to show that the number of meaningful inequalities in the system doesnot exceed 2d−1, which means that there exist at most 2d−1 ways of realizing themaxima in (14); the other possibilities cannot occur. For each j 6 d − 2 we settj =

∑s6j vs. The way in which the maxima in (14) are realized is determined by

the signs of the quantities (1 − ti) − tj for the various pairs i, j ∈ {0, . . . , d − 2}.The number of different cases is equal to the number of possible positions of thepoints (1− t0), . . . , (1− td−2) on [0, 1] relative to the points t0, . . . , td−2. If exactlyl points among t0, . . . , td−2 lie to the left of 1/2, then we must put (d− l) points onthe interval [0, 1/2], where the l points are fixed (on [1/2, 1] we will have a symmetricarrangement of points). There are

(d−1

l

)ways of doing this. Hence the total number

of possible arrangements of the points is∑d−1

l=0

(d−1

l

)= 2d−1.

Remark 1. The function h is only defined on the simplex ∆. Proposition 2 allows usto extend the solution set of the inequality h(x) > q by including in it all the pointsx ∈ Rd as follows: this inequality is described by a system of 2d−1 inequalities(ai, x) > q. Thus, for each q > 0 the inequality h(x) > q defines a polyhedron withat most 2d−1 faces in Rd. In the next section we present an algorithm for listing allthe 2d−1 vectors ai and, accordingly, all the faces of the polyhedron.

Proposition 3. For each p ∈ (0, r(d, k)],

minv∈∆, h(v)=(k+1)p

f(v) = minj=1,...,2d−1

minv∈∆, (v,ai)=(k+1)p

f(v).

Proof. We set f(x) = f(x) for x ∈ ∆ and f = +∞ for x ∈ Rd \∆. Then we obtaina convex function with a minimum at c. Since p 6 r(d, k) 6 (d2 − 1)/(6(k + 1)),it follows that (k + 1)p 6 (d2 − 1)/6 = h(c). Hence the point c giving the globalminimum of f is contained in the polyhedron G =

{x ∈ Rd | h(x) > (k + 1)p

}. By

Lemma 3 the minimum of f on the boundary of G is equal to the minimum of fon the union of the planes (x,ai) = (k + 1)p bounding G.

Estimating the chromatic numbers 793

Combining Propositions 1 and 3 we obtain the following result.

Theorem 1. For any d and k,

Sd,k = maxi=1,...,2d−1

max0<p6r(d,k)

(min

s∈∆, (s,b)=pf(s)− min

v∈∆, (v,ai)=(k+1)pf(v)

). (15)

The next result lets us find the minimum of f on each of the 2d−1 hyperplanes{v ∈ Rd | (v,ai) = (k + 1)p

}. Its proof is the same as Lemma 2.

Lemma 4. Let a = (a1, . . . , ad) ∈ Rd be a vector with nonnegative integer compo-nents and assume that the set

{v ∈ ∆ | (v,a) = (k + 1)p

}contains at least one

interior point of the simplex ∆. Then f attains its minimum on this set at thepoint [µ]a, where µ is the unique positive zero of the polynomial

Pa(z) =d−1∑j=0

(aj+1 − (k + 1)p)zaj+1 .

Combining the results we have obtained gives the following theorem.

Theorem 2. For all d and k,

Sdk = maxi=1,...,2d−1

max0<p6r(d,k)

(f([λ])− f([µ]ai)

), (16)

where for any i and p, λ is the unique positive zero of the polynomial

Pb(z) =d−1∑j=0

(j − p)zj ,

and µ is the unique positive zero of the polynomial

Pai(z) =d−1∑j=0

(aij+1 − (k + 1)p)zai

j+1 .

Theorem 2 enables us to find an analytic expression for the quantity Sd,k; wewill do this in § 4.

§ 4. An algorithm for the numerical realization of the method

In this section we describe an algorithm for the implementation of our method.By Theorem 2, to find Sd,k we must find the maximum of the right-hand sideof (15) over all p ∈ (0, r(d, k)] and all the planes

{v ∈ ∆ | (v,ai) = (k + 1)p

},

i = 1, . . . , 2d−1. We split this problem into two:1) finding the inner maximum in (16) over p ∈ (0, r(d, k)] for each i;2) listing all the faces and the corresponding vectors ai, i = 1, . . . , 2d.We start with the first problem. Differentiating the equalities

d−1∑j=0

(j − p)λj = 0,d−1∑j=0

(aij+1 − (k + 1)p)µai

j+1 = 0,

794 E. S. Gorskaya, I. M. Mitricheva, V.Yu. Protasov and A.M. Raıgorodskiı

we obtain

( d−1∑j=0

j(j − p)λj−1

)dλ =

( d−1∑j=0

λj

)dp,

( d−1∑j=0

aij+1(a

ij+1 − (k + 1)p)µai

j+1−1

)dµ = (k + 1)

( d−1∑j=0

µaij+1

)dp,

so that

dλ

dp=

∑d−1j=0 λ

j∑d−1j=0 j(j − p)λj−1

,dµ

dp=

(k + 1)∑d−1

j=0 µai

j+1∑d−1j=0 a

ij+1(a

ij+1 − (k + 1)p)µai

j+1. (17)

To find the point p delivering the maximum of f([λ])−f([µ]ai), i = 1, . . . , 2d−1, wedifferentiate with respect to p:

0 =d

dp

(f([λ])− f([µ]ai)

)=∂f([λ])∂λ

dλ

dp− ∂f([µ]ai)

∂µ

dµ

dp.

Calculating the derivatives of the functions f([λ]) and f([µ]ai) and substituting dλdp

and dµdp from (17) in the last equation we obtain an equation for p.

Now we describe these transformations in greater detail. Since

f([λ]) =d−1∑j=0

λj∑d−1j=0 λ

jln

λj∑d−1j=0 λ

j=

1∑d−1j=0 λ

j

d−1∑j=0

(λj lnλj − λj ln

d−1∑j=0

λj

)

=lnλ

(∑d−1j=0 jλ

j)∑d−1

j=0 λj

− ln( d−1∑

j=0

λj

),

it follows that

∂f([λ])∂λ

= lnλ

∑d−1j=1 j

2λj−1∑d−1

j=0 λj − λ

(∑d−1j=1 jλ

j−1)2(∑d−1

j=0 λj)2 .

In a similar way,

f([µ]) =lnµ

(∑d−1j=0 a

ij+1µ

aij+1

)∑d−1j=0 µ

aij+1

− ln( d−1∑

j=0

µaij+1

),

∂f([µ])∂µ

= lnµ

∑d−1j=0(ai

j+1)2 µai

j+1−1 ∑d−1j=0 µ

aij+1 − µ

(∑d−1j=0 a

ij+1 µ

aij+1−1

)2(∑d−1j=0 µ

aij+1

)2 .

Estimating the chromatic numbers 795

Finally, using the expressions for dλdp and dµ

dp in (17), for each i = 1, . . . , 2α−1 weobtain the following system of equations for p, λ and µ:

lnλ

∑d−1j=1 j

2λj−1∑d−1

j=0 λj − λ

(∑d−1j=1 jλ

j−1)2∑d−1

j=0 λj∑d−1

j=0 j(j − p)λj−1

= (k+1) lnµ

∑d−1j=0(ai

j+1)2µai

j+1−1 ∑d−1j=0 µ

aij+1 − µ

(∑d−1j=0 a

ij+1µ

aij+1−1

)2∑d−1j=0 µ

aij+1

∑d−1j=0 a

ij+1(a

ij+1 − (k + 1)p)µai

j+1,

d−1∑j=0

(j − p)λj = 0,

d−1∑j=0

(aij+1 − (k + 1)p)µai

j+1 = 0.

(18)This system contains three equations for three unknowns. Solving it numerically

for each i we find the optimal parameters p, λ and µ. To solve problem 1) we mustexamine 2d−1 faces of the polyhedron; for each we face must find the corresponding pand the quantity f([λ])− f([µ]ia) and then must take the maximum of those found.

Now we will work on solving the second problem: for fixed d find the equationsof all the 2d−1 faces of the polyhedron. To this end we express m(v) as follows:

m(v) =d−2∑j=0

(1− 2v0 − · · · − 2vj)+

+ 2[(1− 2v0 − v1)+ + (1− 2v0 − v1 − v2)+ + · · ·+ (1− 2v0 − 2v1 − v2)+

+ (1− 2v0 − 2v1 − v2 − v3)+ + · · ·+ (1− 2v0 − · · · − 2vd−3 − vd−2)+].

The equations of all the faces are obtained by realizing all possible ways ofremoving the parentheses in the expression for m. We write the terms of thisexpression as a table:

(1− 2v0)+2 (1− 2v0 − v1)+ (1− 2v0 − 2v1)+2(1− 2v0 − v1 − v2)+ 2(1− 2v0 − 2v1 − v2)+ (1− 2v0 − 2v1 − 2v2)+. . .

The nth row of this table contains the following terms:

2(1− 2v0 − v1 − v2 − · · · − vn−1)+, 2(1− 2v0 − 2v1 − v2 − · · · − vn−1)+, . . . ,

(1− 2v0 − 2v1 − 2v2 − · · · − 2vn−1)+.

If some term is positive, then since vi > 0, the terms to its left or above it arealso positive.

796 E. S. Gorskaya, I. M. Mitricheva, V.Yu. Protasov and A.M. Raıgorodskiı

Now consider the grid of size N (see Fig. 1; there are N nodes on the sides)with a zero or one written at each node. We call an arrangement of zeros and ones‘regular’ if it satisfies the following conditions: if one is written in some node, thenones are also written in all the nodes attainable by moving left and upwards.

Figure 1

Note that each ‘regular’ arrangement of zeros and ones in a grid of size d − 1corresponds to a face of the polyhedron and vice versa. Hence for fixed d it issufficient to list all the ‘regular’ arrangements of zeros and ones.

Suppose we can solve this problem for i = 1, . . . , N . We shall show how to solveit for i = N +1. Consider the grid of size N +1 and let b1, . . . , bN+1 be the integersat the nodes of the bottom row. Then the required number of ‘regular’ arrange-ments is

∑N+1j=0 Aj , where Aj is the number of ‘regular’ arrangements satisfying the

conditions bi = 1, i = 1, . . . , j, and bi = 0, i = j + 1, . . . , N + 1.

Also note that if bi = 1, i = 1, . . . , j, then we also have ones at all the nodesabove them, while the nodes above the bi, i = j + 1, . . . , N + 1, may carry any‘regular’ arrangement of zeros and ones of a size less than N , which we alreadyknow how to find.

Thus, to solve the problem for the grid of size N+1 we must consider successivelythe following cases:

1) bi = 0, i = 1, . . . , N + 1; here we use the solution of the problem for the tableof size N ;

2) b1 = 1, bi = 0, i = 2, . . . , N + 1; then the first column contains only ones andto find the number of ‘regular’ arrangements in the rest of the table we use oursolution of the problem for the table of size N − 1;

· · ·j) bi = 1, i = 1, . . . , j, and bi = 0, i = j + 1, . . . , N + 1; then the first j columns

contain only ones and to find the number of ‘regular’ arrangements in the rest ofthe table we use the solution for the table of size N − j;

· · ·N) bi = 1, i = 1, . . . , N , and bN+1 = 0;

N+1) bi = 1, i = 1, . . . , N + 1, that is, the table contains only ones.

Estimating the chromatic numbers 797

§ 5. Numerical results. Estimates for chromatic numbers

We calculate the constants Sd,k for all pairs d, k by the same scheme. Using thealgorithm from § 4 we examine 2d−1 vectors ai, for each vector solve the system ofequations (18) numerically, find λ and µ, after which we find Sd,k as the largestof the numbers

f([λ])− f([µ]ai), i = 1, . . . , 2d−1.

This algorithm can be implemented on a PC for small values of d (for instance,d 6 20).

We have tabulated our calculations. For each k = 2, . . . , 20 we find Sd,k,d = 2, . . . , 20, and put the maximum of these integers into the table. However,our calculations have shown that for each k the function Sd,k increases in d andattains its maximum for d = 20. So we put in the table only the integers S20,k. Onthe other hand, for d close to 10 the growth of Sd,k is very weak and the value virtu-ally stabilizes. This allows us to assume that for the values of k under considerationthe values of S20,k and ζk are close to optimal. In the table we also present thevalues of the parameter p delivering the maximum in (16) and the correspondingparameters λ and µ of the minimum points for these p.

k p λ µ S20,k ζk = eS20,k

2 0.298312 0.229769 0.612487 0.382448 1.4658693 0.323872 0.244640 0.703286 0.511303 1.6674624 0.338967 0.253156 0.759762 0.614183 1.8481465 0.348944 0.258679 0.798228 0.699665 2.0130786 0.356030 0.262553 0.826097 0.772739 2.1656907 0.361325 0.265422 0.847212 0.836531 2.3083468 0.365430 0.267630 0.863759 0.893129 2.4427609 0.368701 0.269380 0.877075 0.943979 2.57018910 0.371360 0.270797 0.888019 0.990147 2.69163011 0.373540 0.271954 0.897170 1.032414 2.80783612 0.375332 0.272903 0.904932 1.071391 2.91943713 0.376778 0.273667 0.911594 1.107542 3.02690814 0.377904 0.274260 0.917369 1.141252 3.13068615 0.379727 0.275219 0.922529 1.172885 3.23130316 0.380802 0.275783 0.927027 1.202605 3.32877717 0.381754 0.276282 0.931031 1.230658 3.42348218 0.382601 0.276726 0.934618 1.257222 3.51564019 0.383356 0.277120 0.937850 1.282445 3.60544520 0.384066 0.277491 0.940780 1.306459 3.693075

798 E. S. Gorskaya, I. M. Mitricheva, V.Yu. Protasov and A.M. Raıgorodskiı

In the last column we present the required bounds ζk for the exponents of theasymptotic growth of the chromatic numbers χ(Rn, k) as n→∞. Even for k = 3, 4we see that the estimates (3) and (4) have been slightly improved, and for k > 5 weobtain new estimates, which cannot be improved in the framework of our method.In the next section we describe how the bounds ζk can be found for any k > 21 andconjecture that they also cannot be improved in the framework of our method.

On the basis of the data obtained we can pose many new problems. For instance,it would be interesting to know how ζk grows with k. Judging by our table we canconjecture that its growth is slower than linear, and it gets increasingly slow ask grows, that is, ζk must be increasing and concave in k.

§ 6. Conjectures and generalizations

The results in § 5 allow us to make conjectures about the functions S d,k and ζk.For instance, based on the fact that for each k = 2, . . . , 20 the function Sd,k isincreasing in d for d 6 20, we can put forward the following conjecture.

Conjecture 1. For each k the function Sd,k is increasing in d.

In particular, this would mean (that is, if Conjecture 1 holds) that the boundζk = eS∞,k , where S∞,k = limd→∞ Sd,k is optimal. However, it is difficult toestablish such an estimate since for each d we must examine 2d−1 planes to calcu-late Sd,k. Here Conjecture 2 may be helpful. It is based on the following fact: ascalculations demonstrate, for all d = 2, . . . , 20 and k = 2, . . . , 20 the maximum inproblem (16) is always attained on the same face, which has the direction vector awith components aj = (j − 1)j/2, j = 1, . . . , d.

Conjecture 2. For any d and k the required maximum is attained on the facecorresponding to the vector a.

In special cases we can confirm this conjecture by calculation. If Conjecture 2 istrue, then we need not examine all the 2d−1 planes for each d and can immediatelyfind S∞,k. Now we describe how we can calculate this quantity. (We point outthat we obtain some estimate of the chromatic number in any case. The meaningof Conjectures 1 and 2 is that this estimate is the best possible in the framework ofour method.)

Let α = (α1, α1, . . . , αd), G(t) =∑d

j=1 tαj . Assume also that, as d → ∞, the

function G1(t) corresponds to the system (0, 1, . . . , d− 1) and G2(t) corresponds tothe system (a1, . . . , ad−1), where aj = (j − 1)j/2.

Since µ is a root of the equation

d∑j=1

(j − 1)j2

t(j−1)j/2 = p(k + 1)d∑

j=1

t(j−1)j/2,

and λ is a root of the equation

d∑j=1

(j − 1)tj−1 = pd∑

j=1

tj−1,

Estimating the chromatic numbers 799

as d→∞, it follows that µ is a root of the equation

tG′2(t) = p(k + 1)G2(t), (19)

and λ is a root of the equation

tG′1(t) = pG1(t). (20)

As d→∞, we have

G2(z) =∞∑

j=1

t(j−1)j/2, G1(t) =1

1− t, tG′1(t) =

t

(t− 1)2.

Hence λ/(λ− 1)2 = p/(1−λ), so that λ = p/(p+ 1). From equation (19) we obtainp = µG′2(µ)/

((k + 1)G2(µ)

). Then

λ =µG′2(µ)/

((k + 1)G2(µ)

)µG′2(µ)/

((k + 1)G2(µ)

)+ 1

=µG′2(µ)

µG′2(µ) + (k + 1)G2(µ).

Thus,

p =µG′2(µ)

(k + 1)G2(µ), λ =

µG′2(µ)µG′2(µ) + (k + 1)G2(µ)

. (21)

Now, from (19) and (20) we deduce

dµ

dp=

(k + 1)(G2(µ))2

G2(µ)G′2(µ) + µG′′2(µ)G2(µ)− µ(G′2(µ))2,

dλ

dp= (1− λ)2.

Next, as d→∞,

f([λ]) =λ lnλG′1(λ)G1(λ)

− lnG1(λ) =λ lnλ1− λ

+ ln(1− λ).

Therefore,∂f([λ])∂λ

=lnλ

(1− λ)2,

∂f([λ])∂λ

dλ

dp= lnλ.

Furthermore, since

f([µ]) =µ lnµG′2(µ)G2(µ)

− lnG2(µ),

it follows that

∂f([µ])∂µ

= lnµµG2(µ)G′′2(µ) +G2(µ)G′2(µ)− µ(G′2(µ))2

(G2(µ))2,

which means that∂f([µ])∂µ

dµ

dp= (k + 1) lnµ.

800 E. S. Gorskaya, I. M. Mitricheva, V.Yu. Protasov and A.M. Raıgorodskiı

Hence, as d → ∞, the condition that the derivatives of f([λ]) and f([µ]a) withrespect to the parameter p must be equal can be written as lnλ = (k+1) lnµ. Nowusing the expression for λ(µ) in (21) we obtain an equation for µ:

lnµG′2(µ)

µG′2(µ) + (k + 1)G2(µ)= (k + 1) lnµ.

Thus, to find the quantity S∞,k and the corresponding bound ζk for fixed k we startby finding µ from the above equation, and then find p and λ from equations (21).Then by Theorem 2 we have S∞,k = f([λ]) − f([µ]a), where a is the vector withcomponents aj = (j − 1)j/2, j = 1, . . . , d.

Remark 2. The function

G2(z) =∞∑

j=1

z(j−1)j/2

is analytic in the disc |z| < 1. It is easy to show (using, for instance an approachsimilar to the one in [18], Ch. 7, § 7.8) that there is a dense subset of singular pointson the boundary of the convergence disc such that the function G2(z) approachesinfinity as its argument approaches one of these points along a radius. This seemsto indicate that (by contrast to G1) there exists no formula expressing G2(t) interms of elementary functions. Hence in calculating the parameters µ and λ wehave to replace G2(t) by a partial sum of the corresponding series.

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Estimating the chromatic numbers 801

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E. S. Gorskaya

Faculty of Mechanics and Mathematics,

Moscow State University

E-mail : gorskaja@zmail.ru

I.M. Mitricheva

Faculty of Mechanics and Mathematics,

Moscow State University

E-mail : irshitova@yandex.ru

V. Yu. Protasov

Faculty of Mechanics and Mathematics,

Moscow State University

E-mail : v-protassov@yandex.ru

A. M. Raıgorodskiı

Faculty of Mechanics and Mathematics,

Moscow State University

E-mail : mraigor@yandex.ru

Received 5/MAY/08 and 9/DEC/08Translated by IPS(DoM)