Computing the Partition Function for Cliques in a Graph

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www.theoryofcomputing.org

Computing the Partition Function

for Cliques in a Graph

Alexander Barvinok

∗

Received June 24, 2014; Revised February 17, 2015; Published December 4, 2015

Abstract: We present a deterministic algorithm which, given a graphGwithnvertices and an integer 1<m≤n, computes innO(lnm)time the sum of weightsw(S)over all m-subsetsSof the set of vertices ofG, wherew(S) =exp{γmσ(S) +O(1/m)}provided exactly

m 2

σ(S)pairs of vertices ofSspan an edge ofGfor some 0≤σ(S)≤1, andγ>0 is an

absolute constant. This allows us to tell apart the graphs that do not havem-subsets of high density from the graphs that have sufficiently manym-subsets of high density, even when the probability to hit such a subset at random is exponentially small inm.

ACM Classification:F.2.1, G.1.2, G.2.2, I.1.2 AMS Classification:15A15, 68C25, 68W25, 60C05

Key words and phrases:algorithm, clique, partition function, subgraph density, graph

1 Introduction and main results

1.1 Density of a subset in a graph

LetG= (V,E)be an undirected graph with setV of vertices and setEof edges, without loops or multiple edges. LetS⊂V be a subset of the set of vertices ofG. We define thedensityσ(S)ofSas the ratio of the number of the edges ofGwith both endpoints inSto the maximum possible number of edges spanned by vertices inS:

σ(S) =|{u,v

} ∈E:u,v∈S|

|S| 2

.

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Hence

0 ≤ σ(S) ≤1 for all S⊂V.

In particular,σ(S) =0 if and only ifSis anindependent set, that is, no two vertices ofSspan an edge of Gandσ(S) =1 if and only ifSis aclique, that is, every two vertices ofSspan an edge ofG.

In this paper, we suggest a new approach to testing the existence of subsetsSof a given sizem=|S|

with high densityσ(S). Namely, we present a deterministic algorithm, which, given a graphGwithn

vertices and an integer 1<m≤ncomputes innO(lnm)time (within a relative error of 0.1, say) the sum

Densitym(G) =

∑

S⊂V |S|=m

exp{γmσ(S)−ε(m)}

where 0 ≤ε(m) ≤ 0.1

m−1

(1.1)

andγ>0 is an absolute constant: we can chooseγ=0.07 and ifn≥8mandm≥10 then we can choose γ=0.27.

Let us fix two real numbers 0≤σ<σ0≤1. If there are nom-subsetsSwith densityσ or higher then

Densitym(G) ≤

n m

exp{γmσ−ε(m)}.

If, however, there are sufficiently manym-subsetsSwith densityσ0or higher, that is, if the probability to

hit such a subset at random is at least 2 exp{−γ(σ0₋

σ)m}, then

Densitym(G) ≥ 2

n m

exp{γmσ−ε(m)}

and we can distinguish between these two cases innO(lnm)time. Note that “many subsets" still allows for the probability to hit such a subset at random to be exponentially small inm.

It turns out that we can compute innO(lnm)time anm-subsetS, which is almost as dense as the average under the exponential weighting of theformula (1.1), that is, anm-subsetSsatisfying

expγmσ(S) ≥ 1

2

n m

−1

Densitym(G), (1.2)

cf.Remark 3.7.

We compute theexpression (1.1)using partition functions.

1.2 The partition function of cliques

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For an integer 1<m≤n, we define a polynomial

Pm(W) =

∑

S⊂{1,...,n}

|S|=m

∏

{i,j}⊂S i6=j

wi j,

which we call thepartition functionof cliques (note that it is different from what is known as the partition function of independent sets, see [20] and [2]). Thus ifW is the adjacency matrix of a graphGwith set V={1, . . . ,n}of vertices and setEof edges, so that

wi j=

(

1 if{i,j} ∈E,

0 otherwise,

thenPm(W) is the number of cliques inGof size m. Hence computing Pm(W) is at least as hard as counting cliques. Let us choose an 0<α <1 and modify the weightsW as follows:

wi j=

(

1 if{i,j} ∈E,

α otherwise.

Then Pm(W) counts all subsets ofmvertices in the complete graph on nvertices, only that a subset spanning exactlyknon-edges ofGis weighted down by a factor ofαk. In other words, anm-subsetS⊂V is counted with weight

exp

1−σ(S)

m 2

lnα

.

In this paper, we present a deterministic algorithm, which, given ann×nsymmetric matrixW= (wj) such that

w_{i j}−1

≤

γ

m−1 for all 1≤i6= j≤n (1.3)

computes the value ofPm(W) within relative errorε innO(lnm−lnε) time. Hereγ >0 is an absolute constant; we can chooseγ=0.07 and ifn≥8mandm≥10, we can chooseγ=0.27.

1.3 The idea of the algorithm

LetJ denote then×n matrix filled with 1s. Given ann×nsymmetric matrixW, we consider the univariate function

f(t) =lnPm J+t(W−J)

. (1.4)

Clearly,

f(0) =lnPm(J) =ln

n m

and f(1) =lnPm(W).

Hence our goal is to approximate f(1)and we do it by using the Taylor polynomial expansion of f at t=0:

f(1)≈ f(0) +

`

∑

k=1 1 k!

dk dtk f(t)

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It turns out that the right hand side of theapproximation (1.5)can be computed innO(`)time. We present the algorithm inSection 2. The quality of theapproximation (1.5)depends on the location of complexzeros ofPm.

Lemma 1.1. Suppose that there exists a realβ>1such that

Pm J+z(W−J)

6=0 for all z∈_C satisfying |z| ≤β.

Then the right hand side of theformula(1.5)approximates f(1)within an additive error of

m(m−1)

2(`+1)β`_(β₋₁₎.

In particular, for a fixed β >1, to ensure an additive error of 0<ε <1, we can choose `=

O(lnm−lnε), which results in the algorithm for approximating Pm(W) within relative error ε in

nO(lnm−lnε)time. We proveLemma 1.1inSection 2.

Thus we have to identify a class of weightsW for which the numberβ >1 ofLemma 1.1exists. We

prove the following result.

Theorem 1.2. There is an absolute constantω>0(one can chooseω=0.071, and if m≥10and n≥8m, one can chooseω=0.271) such that for any positive integers1<m≤n, for any n×n symmetric complex

matrix Z= (zi j), we have

Pm(Z)6=0

as long as

zi j−1

≤

ω

m−1 for all 1≤i6= j≤n.

We proveTheorem 1.2inSection 3. Theorem 1.2implies that if theinequality (1.3)holds, where 0<γ<ωis an absolute constant, we can chooseβ=ω/γinLemma 1.1and obtain an algorithm which

computesPm(W)within a given relative errorε innO(lnm−lnε)time. Thus we can chooseγ=0.07 and

β=71/70 and ifm≥10 andn≥8m, we can chooseγ=0.27 andβ =271/270.

1.4 Weighted enumeration of subsets

Given a graphGwith set of verticesV={1, . . . ,n}and setE of edges, we define weightsW by

wi j= (

1+ γ

m−1 if {i,j} ∈E, 1− γ

m−1 otherwise,

(1.6)

whereγ>0 is the absolute constant in theinequality (1.3). Thus we can chooseγ=0.07 and ifm≥10

andn≥8m, we can chooseγ=0.27. We define thefunctional (1.1)by

Densitym(G) =eγ m

1+ γ

m−1 −(m₂)

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Then we have

Density_m(G) =

_∑

S⊂V |S|=m

w(S),

where

w(S) = eγm

1+ γ

m−1

(σ(S)−1)(m₂) 1− γ

m−1

(1−σ(S))(m₂)

= exp{γmσ(S)−ε(m)} where 0 ≤ ε(m) ≤ 0.1

m−1

(in the upper bound forε(m)we use thatγ<0.3). Thus we are in the situation described inSection 1.1.

1.5 Comparison with results in the literature

The topic of this paper touches upon some very active research areas, such as finding dense subgraphs in a given graph (see, for example, [7] and references therein), computational complexity of partition functions (see, for example, [10] and references therein) and complex zeros of partition functions (see, for example, [20] and references therein). Detecting dense subsets is notoriously hard. It is anNP-hard problem to approximate the size|S|of the largest cliqueSwithin a factor of|V|1−ε _{for any fixed 0}_<

ε≤1

[14], [23]. In [11], a polynomial time algorithm for approximating the highest density of anm-subset within a factor ofO(n1/3−ε₎_{for some small}

ε>0 was constructed. The paper [8] presents an algorithm

ofnO(1/ε)complexity which approximates the highest density of anm-subset within a factor ofn1/4+ε and an algorithm ofO nlnn complexity which approximates the density within a factor ofO(n1/4), the current record (note that [11], [7] and [8] normalize density slightly differently than we do). It has also been established that modulo some plausible complexity assumptions, it is hard to approximate the highest density of anm-subset within a constant factor, fixed in advance, see [7].

We note that while searching for a densem-subset in a graph onnvertices, the most interesting case to consider is that of a growingmsuch thatm=o(n), since for any fixedε >0 one can compute in

polynomial time the maximum number of edges ofGspanned by anm-subsetSof vertices ofGwithin an additive error ofεn2[12] (the complexity of the algorithm is exponential in 1/ε). Of course, for any

constantm, fixed in advance, all subsets of sizemcan be found in polynomial time by an exhaustive search.

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inSection 1.3is also new. On the other hand, one can view our method as inspired by the intuition from statistical physics. Essentially, the algorithm computes the partition function similar to the one used in the “simulated annealing" method, see [1], for the temperature that is provably higher than the phase transition threshold (the role of the “temperature" is played by 1/γm, whereγ is the constant in

thedensity functional (1.1)). The algorithm uses the fact that the partition function has nice analytic properties at temperatures above the phase transition temperature.

The corollary asserting that graphs without densem-subsets can be efficiently separated from graphs with many densem-subsets is vaguely similar in spirit to the property testing approach [13].

The result requiring most work isTheorem 1.2bounding the complex roots of the partition function away from the matrix of all 1s. Studying complex roots of partition functions was initiated by Lee and Yang [22], [18], continued by Heilmann and Lieb [15], and it is a classical topic by now, because of its importance to locating the phase transition, see, for example, [20] and Section 8.5 of [19].

The general approach of this paper was first applied by the author [3] to approximate permanents of matrices that are sufficiently close to the matrix of all 1s (and related expressions, such as hafnians of symmetric matrices and multidimensional analogues of permanents of tensors). After the first version of this paper was written, P. Soberón and the author applied this approach to computing the partition function of graph homomorphisms [5] and its refinement [4]. In each case, the main work consists in obtaining a version ofTheorem 1.2, which bounds the complex roots of the polynomial in question. It is the easiest in the case of permanents [3] and the most cumbersome in the case of graph homomorphisms with multiplicities [4]. The most general framework under which the method of this paper works is still at large.

Open Question 1.3. Is it true that foranyγ>0, fixed in advance, thedensity functional (1.1)can be

computed (within a relative error of 0.1, say, and someε(m) =O(1/m)) innO(lnm)time?

2 The algorithm

2.1 The algorithm for approximating the partition function

Given ann×nsymmetric matrixW= (wi j), we present an algorithm that computes the right hand side of theapproximation (1.5)for the function f(t)defined byformula (1.4).

Let

g(t) =Pm J+t(W−J)

=

_∑

S⊂{1,...,n} |S|=m

∏

{i,j}⊂S i6=j

1+t(wi j−1), (2.1)

so f(t) =lng(t). Hence

f0(t) =g

0₍_t₎

g(t) and g

0₍

t) =g(t)f0(t).

Therefore, fork≥1 we have

dk dtkg(t)

t=0= k−1

∑

j=0

k−1 j

dj dtjg(t)

t=0

dk−j dtk−j f(t)

t=0

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(we agree that the 0-th derivative ofgisg). We note thatg(0) = _mn

. If we compute the values of

dk dtkg(t)

t=0 for k=1, . . . , ` , (2.3)

then theformulas (2.2)fork=1, . . . , `provide a non-degenerate triangular system of linear equations that allows us to compute

dk dtk f(t)

t=0 for k=1, . . . , ` . Hence our goal is to compute thevalues (2.3).

We have

dk dtkg(t)

t=0=_S⊂{1

∑

_,...,_n} |S|=m

∑

I=({i1,j1},...,{ik,jk}) {i1,j1},...,{ik,jk}⊂S

(wi1j1−1)· · ·(wikjk−1)

where the inner sum is taken over all ordered setsI ofkdistinct pairs{i1,j1},. . .,{ik,jk}of points from S.

For an ordered set

I= ({i1,j1}, . . . ,{ik,jk})

ofkpairs of points from the set{1, . . . ,n}, let

ρ(I) =

k [

s=1

{is,js}

be the total number of distinct points amongi1,j1, . . . ,ik,jk. Since there are exactly

n−ρ(I)

m−ρ(I)

m-subsetsS⊂ {1, . . . ,n}containing the pairs{i₁,j₁},. . .,{ik,jk}, we can further rewrite

dk dtkg(t)

t=0=

∑

I=({i1,j1},...,{ik,jk}) {i1,j1},...,{ik,jk}⊂{1,...,n}

n−ρ(I)

m−ρ(I)

(wi1j1−1)· · ·(wikjk−1),

where the sum is taken over all ordered setsIofkdistinct pairs{i₁,j₁}, . . . ,{ik,jk}of points from the set

{1, . . . ,n}. The algorithm consists in computing the latter sum. Since the sum contains not more than n2k=nO(`)terms, the complexity of the algorithm is indeednO(`), as claimed.

2.2 Proof ofLemma 1.1

The functiong(z)defined by theequation (2.1)is a polynomial inzof degreed≤ m 2

withg(0) = _mn

6=0, so we factor

g(z) =g(0)

d

∏

i=1

1− z αi

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whereα1, . . . ,αd are the roots ofg(z). By the condition ofLemma 1.1, we have

|αi| ≥ β >1 for i=1, . . . ,d.

Therefore,

f(z) =lng(z) =lng(0) +

d

∑

i=1 ln

1− z

αi

provided |z| ≤1, (2.4)

where we choose the branch of lng(z)that is real atz=0. Using the standard Taylor expansion, we obtain

ln

1− 1 αi

=−

`

∑

k=1 1 k 1 αi k +ζ`, where |ζ_`|= +∞

∑

k=`+1 1 k 1 αi k ≤ 1

(`+1)β`_(β₋₁₎.

Therefore, from theequation (2.4)we obtain

f(1) =f(0) +

`

∑

k=1

−1

k d

∑

i=1 1 αi k! +η`, where

|η_`| ≤ m(m−1)

2(`+1)β`_(β₋₁₎.

It remains to notice that

−1

k d

∑

i=1 1 αi k = 1 k! dk dtk f(t)

t=0.

3 Proof of

Theorem 1.2

3.1 Definitions

Forδ >0, we denote byU(δ)⊂Cn(n−1)/2the closed polydisc of weights

U(δ) =nZ= (zi j) :zi j−1

≤δ for all 1≤i6= j≤n o

.

For a subsetΩ⊂ {1, . . . ,n}where 0≤ |Ω| ≤m, we define

PΩ(Z) =

∑

S⊂{1,...,n}

|S|=m S⊃Ω

∏

{i,j}⊂S i6=j

zi j.

In words:PΩ(Z)is the restriction of the sum defining the clique partition function to the subsetsSthat contain a given setΩ. In particular,

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and if|Ω|<mthen

P_Ω(Z) = 1

m− |Ω|_i:

∑

_i_∈Ω_/ PΩ∪ {i}(Z). (3.1)

We note that the natural action of the symmetric groupSnon{1, . . . ,n}induces the natural action ofSn on the space of polynomials inZwhich permutes the polynomialsP_Ω(Z).

First, we establish a simple geometric lemma.

Lemma 3.1. Let u1, . . . ,un∈R2be non-zero vectors such that for some0≤α<2π/3the angle between any two vectors ui and uj does not exceedα. Let u=u1+· · ·+un. Then

kuk ≥ cosα 2

n

∑

i=1

kuik.

Proof. We observe that the origin does not lie in the convex hull of any three vectorsui,uj,uk since otherwise the angle between some two of these three vectors is at least 2π/3. The Carathéodory Theorem then implies that the convex hull of the whole set of vectorsu₁, . . . ,undoes not contain the origin and hence the vectors lie in an angle at mostα with vertex at the origin. The length of the orthogonal projection of each vectoruionto the bisector of the angle is at leastkuikcos(α/2)and hence the length of the orthogonal projection ofu=u₁+· · ·+un onto the bisector of the angle is at least

(ku1k+· · ·+kunk)cos(α/2). Since the length of the vectoru is at least as large as the length of its orthogonal projection, the proof follows.

Lemma 3.1and its proof was suggested by B. Bukh [9]. It replaces an earlier weaker bound used by the author, see Lemma 3.1 of [3].

We proveTheorem 1.2by reverse induction on|Ω|, usingLemma 3.1and the following two lemmas.

Lemma 3.2. Let us fix real0<τ <1, realδ >0 and an integer 1≤r≤m. Suppose that for all Ω⊂ {1, . . . ,n}such that|Ω|=r, for all Z∈U(δ), we have PΩ(Z)6=0and that for all i=1, . . . ,n we have

|P_Ω(Z)| ≥ τ

m−1 _j:

∑

_j6₌_i z_{i j}

∂

∂zi j P_Ω(Z)

.

Then, for any pair of subsets

Ω1,Ω2⊂ {1, . . . ,n} such that |Ω1|=|Ω2|=r and |Ω14Ω2|=2,

and for any Z∈U(δ), the angle between the complex numbers P_Ω1(Z)and P_Ω2(Z), interpreted as vectors inR2=C, does not exceed

θ=4δ(m−1) τ(1−δ)

and the ratio of|P_Ω1(Z)|and|P_Ω2(Z)|does not exceed

λ=exp

4δ(m−1) τ(1−δ)

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Proof. Let us choose anΩ∈ {1, . . . ,n}such that|Ω|=r. SincePΩ(Z)6=0 for allZ∈U(δ), we can choose a branch of lnPΩ(Z)that is a real number whenZ=J. Then

∂

∂zi jlnPΩ(Z) =

∂

∂zi jPΩ(Z)

. P_Ω(Z).

Sincezi j

≥1−δ for allZ∈U(δ), we have

∑

j: j6=i

∂

∂zi j

lnPΩ(Z)

≤ m−1

τ(1−δ) for i=1, . . . ,n. (3.2)

Without loss of generality, we assume that 1∈Ω1, 2∈/Ω1andΩ2=Ω1\ {1} ∪ {2}. GivenA∈U(δ), letB∈U(δ)be the weights defined by

b1j=a2j and b2j=aj1 for all j6=1,2

and

bi j=ai j for all other i,j.

Then

P_Ω2(A) =P_Ω1(B)

and

|lnP_Ω1(A)−lnP_Ω2(A)|=|lnPΩ1(A)−lnPΩ1(B)|

≤ sup

Z∈_U(δ)j:

∑

j6=1,2

∂

∂z1j

lnPΩ1(Z)

+

_∑

j: j6=1,2

∂

∂z2j

lnPΩ1(Z)

!

× max j: j6=1,2

n

a₁j−b1j ,

a₂j−b2j o

.

Sinceai j−bi j

≤2δ for allA,B∈U(δ), by theinequality (3.2)we conclude that the angle between the complex numbersP_Ω1(A)andP_Ω2(A)does not exceedθand that the ratio of their absolute values does

not exceedλ.

Lemma 3.3. Let us fix realδ >0, real0≤θ <2π/3, realλ>1and an integer1<r≤m. Suppose that for allΩ⊂ {1, . . . ,n}such that m≥ |Ω| ≥r and for all Z∈U(δ), we have PΩ(Z)6=0and that for any pair of subsets

Ω1,Ω2⊂ {1, . . . ,n} such that m ≥ |Ω1|=|Ω2| ≥r and |Ω14Ω2|=2,

and for any Z∈U(δ), the angle between two complex numbers P_Ω1(Z)and P_Ω2(Z)considered as vectors in_R2=Cdoes not exceedθ while the ratio of|PΩ1(Z)|and|PΩ2(Z)|does not exceedλ. Let

τ=cos

2_(θ/₂₎

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Then, for any subsetΩ⊂ {1, . . . ,n}such that|Ω|=r−1, all Z∈U(δ)and all i=1, . . . ,n, we have

|PΩ(Z)| ≥

τ

m−1 _j:

∑

_j6₌_i zi j ∂

∂zi jPΩ(Z) . (3.3) In addition, if n m ≥ λ

cos(θ/2),

theinequality(3.3)holds with

τ=cos(θ/2).

Proof. Let us choose a subsetΩ⊂ {1, . . . ,n}such that|Ω|=r−1. Let us define

Ωj=Ω∪ {j} for j∈/Ω.

Suppose first thati∈Ω. Then, for j∈Ω\ {i}we have

∂

∂zi j

P_Ω(Z) =z−1_{i j} P_Ω(Z)

and hence

zi j

∂

∂zi jPΩ(Z)

=|P_Ω(Z)| provided i,j∈Ω, i6=j.

For j∈/Ω, we have

∂

∂zi j

PΩ(Z) =

∂

∂zi j

PΩj(Z) =z −1 i j PΩj(Z)

and hence zi j ∂

∂zi jPΩ(Z)

=P_Ωj(Z)

provided i∈Ω and j∈/Ω.

Summarizing,

∑

j: j6=i zi j ∂

∂zi jPΩ(Z)

= (r−2)|PΩ(Z)|+

∑

j∈Ω/

P_Ω_j(Z)

. (3.4)

On the other hand, by theequation (3.1), we have

PΩ(Z) = 1

m−r+1 _j:

∑

_j_∈Ω_/ PΩj(Z).

ByLemma 3.1, we have

|P_Ω(Z)| ≥ cos(θ/2)

m−r+1_j:

∑

_j_∈Ω_/ P_Ω_j(Z)

(3.5)

and

(m−1)|PΩ(Z)| ≥ (r−2)|PΩ(Z)|+

cosθ 2

∑

j: j∈Ω/ P_Ω_j(Z)

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and theinequality (3.3)withτ =cos(θ/2)follows by theformula (3.4).

Suppose now thati∈/Ω. Then, for j∈Ω, we have

∂

∂zi jPΩ(Z) =

∂

∂zi jPΩi(Z) =z −1 i j PΩi(Z)

and hence

zi j

∂

∂zi jPΩ(Z)

=|P_Ωi(Z)| provided i∈/Ω and j∈Ω. (3.6)

Ifr=mthen

∂

∂zi j

PΩ(Z) =0 for any j∈/Ω

and

∑

j: j6=i zi j ∂

∂zi j PΩ(Z)

= (m−1)|PΩi|.

By theequation (3.1), we have

PΩ(Z) =

∑

j∈Ω/

PΩj(Z)

and hence byLemma 3.1

|PΩ(Z)| ≥

cosθ 2

∑

j∈Ω/ P_Ωj(Z)

≥ cosθ 2

|PΩi(Z)|,

which proves theinequality (3.3)in the case ofr=mwithτ=cos(θ/2).

Ifr<m, then for j∈/Ωi, letΩi j=Ω∪ {i,j}. Then, for j∈/Ωi, we have

∂

∂zi jPΩ(Z) = ∂

∂zi jPΩi j(Z) =z

−1 i j PΩi j(Z)

and hence zi j ∂

∂zi jPΩ(Z)

=P_{Ωi j}(Z)

provided i∈/Ω and j∈/Ωi.

From theformula (3.6),

∑

j: j6=i z_{i j}

∂

∂zi j P_Ω(Z)

= (r−1)|P_Ωi(Z)|+

_∑

j∈Ωi/ P_{Ωi j}(Z)

if r<m. (3.7)

By theequation (3.1), we have

P_Ωi(Z) = 1

m−r _j:

∑

_j_∈Ωi_/ PΩi j(Z)

and hence byLemma 3.1, we have

|PΩi(Z)| ≥

cos(θ/2)

m−r _j:

∑

_j_∈Ωi_/ P_{Ωi j}(Z)

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Comparing this with theequation (3.7), we conclude as above that

|PΩi(Z)| ≥

cos(θ/2)

m−1 _j:

∑

_j6₌_i zi j

∂

∂zi j PΩ(Z)

. (3.8)

It remains to compare|PΩi(Z)|and|PΩ(Z)|. Since the ratio of any two P_Ω_j

1(Z) ,

P_Ωj

2(Z)

does not exceedλ, from theinequality (3.5)we conclude that

|P_Ω(Z)| ≥ cos(θ/2) λ

n−r+1

m−r+1|PΩi(Z)| (3.9)

≥cos(θ/2)

λ |PΩi(Z)| for all i∈/Ω. (3.10)

The proof of theinequality (3.3)withτ=cos

2_(θ/₂₎

λ follows by theinequality (3.8)andinequality (3.10).

In addition, if

n m ≥

λ

cos(θ/2)

then

n−r+1 m−r+1 ≥

λ

cos(θ/2)

and the proof of theinequality (3.3)withτ=cos(θ/2)follows by theinequality (3.8)and the

inequal-ity (3.9).

3.2 Proof ofTheorem 1.2

One can see that for a sufficiently smallω>0, the equation

θ= 4ωexp{θ} (1−ω)cos2_(θ_/₂₎

has a solution 0<θ<2π/3. Numerical computations show that one can chooseω=0.071, in which case

θ≈0.6681776075.

Let

λ=exp{θ} ≈1.950679176 and τ =cos

2_(θ/₂₎

exp{θ} ≈0.4575206588.

Let

δ = ω

m−1.

Forr=m, . . . ,1 we prove the followingStatement 3.4,Statement 3.5andStatement 3.6:

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Statement 3.5. LetΩ⊂ {1, . . . ,n}be a set such that|Ω|=r. Then for any Z∈U(δ), Z= (zi j), and any i=1, . . . ,n, we have

|PΩ(Z)| ≥

τ

m−1 _j:

∑

_j6₌_i zi j ∂

∂zi j PΩ(Z)

.

Statement 3.6. LetΩ1,Ω2⊂ {1, . . . ,n}be sets such that|Ω1|=|Ω2|=r and|Ω14Ω2|=2. Then the angle between the complex numbers PΩ1(Z) and PΩ2(Z), considered as vectors in R2=C, does not exceedθ, whereas the ratio of|PΩ1(Z)|and|PΩ2(Z)|does not exceedλ.

Suppose thatr=mand letΩ⊂ {1, . . . ,n}be a subset such that|Ω|=m. Then

P_Ω(Z) =

_∏

{i,j}⊂Ω i6=j

zi j,

soStatement 3.4clearly holds forr=m. Moreover, fori∈Ω, we have

∑

j: j6=i zi j ∂

∂zi j PΩ(Z)

= (m−1)|PΩ(Z)|,

while fori∈/Ω, we have

∑

j: j6=i zi j ∂

∂zi jPΩ(Z)

=0,

soStatement 3.5holds forr=mas well.

Lemma 3.2implies that ifStatement 3.4andStatement 3.5hold for setsΩof cardinalityrthen for any two subsetsΩ1,Ω2⊂ {1, . . . ,n}such that|Ω1|=|Ω2|=rand|Ω14Ω2|=2 the angle between the two (necessarily non-zero) complex numbersP_Ω1(Z)andP_Ω2(Z)does not exceed

4δ(m−1) τ(1−δ) =

4ω

1− ω m−1

τ

= 4ωexp{θ}

1− ω m−1

cos2_(θ_/₂₎ ≤

4ωexp{θ}

(1−ω)cos2_(θ/₂₎ =θ

while the ratio of|PΩ1(Z)|and|PΩ2(Z)|does not exceed

exp

4δ(m−1) τ(1−δ)

= exp (

4ωexp{θ}

1− ω m−1

cos2_(θ_/₂₎ )

≤ exp

4ωexp{θ} (1−ω)cos2_(θ_/₂₎

= exp{θ} = λ

and henceStatement 3.6holds for setsΩof cardinalityr.

Lemma 3.3implies that ifStatement 3.4and Statement 3.6hold for setsΩof cardinality r then Statement 3.5holds for sets Ω of cardinality r−1. Formula (3.1), Lemma 3.1, Statement 3.4and Statement 3.6for setsΩof cardinalityrimplyStatement 3.4for setsΩof cardinalityr−1. Therefore, we conclude thatStatement 3.4andStatement 3.6hold for setsΩsuch that|Ω|=1.Formula (3.1)and Lemma 3.1imply then that

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as desired.

A more careful analysis establishes that ifm≥10 then forω =0.271 the equation

θ= 4ω

1− ω m−1

cos(θ/2)

has a solution

0<θ<1.626699443

with

λ=exp{θ}<5.1 and cos(θ/2)>0.68 so that λ

cos(θ/2) <7.5.

Then, ifn≥8m, we let

δ = ω

m−1, τ=cos(θ/2) and the proof proceeds as above.

Remark 3.7. As follows from Statement 3.4, we haveP_Ω(Z)6=0 for any complex weightsZ= (zi j) satisfying the conditions ofTheorem 1.2. The algorithm ofSection 2extends in a straightforward way to computingPΩ(W) for every weightsW = (wi j)satisfying the inequality (1.3). This allows us to compute innO(lnm)time by successive conditioning anm-subsetSwhich satisfies theinequality (1.2). Namely, we define weightsW by theformula (1.6)and construct subsetsS0, . . . ,Sm⊂ {1, . . . ,n}, where S0=/0 and|Si|=ias follows. Assuming thatSi−1is constructed, for each j∈ {1, . . . ,n} \Si−1, we let Ωj=Si−1∪ {j}and computePΩj(W)within a relative error of 1/10m. We then select jwith the largest computed value ofPΩj(W)and letSi=Ωj. The setS=Smsatisfies theinequality (1.2).

Acknowledgments

I am grateful to Alex Samorodnitsky and Benny Sudakov for several useful remarks, to anonymous referees for their comments and critique, in particular, for pointing out to [7] and to Boris Bukh for suggestingLemma 3.1.

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AUTHOR

Alexander Barvinok Professor of Mathematics

University of Michigan, Ann Arbor, MI barvinok umich edu

http://www.math.lsa.umich.edu/~barvinok

ABOUT THE AUTHOR