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Advances in Mathematics 180 (2003) 275–289

On distinct sums and distinct distances

Gabor Tardos*

Renyi Institute, Realtanoda utca 13-15, 1053 Budapest, Hungary

Received 28 August 2001; accepted 2 August 2002

Communicated by Laszlo Lovasz

Abstract

The paper (Discrete Comput. Geom. 25 (2001) 629) of Solymosi and Toth implicitly raised

the following arithmetic problem. Consider n pairwise disjoint s element sets and form all ðs2Þn

sums of pairs of elements of the same set. What is the minimum number of distinct sums one

can get this way? This paper proves that the number of distinct sums is at least nds ; where

ds ¼ 1=cJs=2n is defined in the paper and tends to e�1 as s goes to infinity. Here e is the base of

the natural logarithm. As an application we improve the Solymosi–Toth bound on an old

Erd +os problem: we prove that n distinct points in the plane determine Oðn4e5e�1�eÞ distinct

distances, where e40 is arbitrary. Our bound also finds applications in other related results in

discrete geometry. Our bounds are proven through an involved calculation of entropies of

several random variables.

r 2003 Elsevier Science (USA). All rights reserved.

MSC: 52C10; 11B75

Keywords: Distinct distances; Entropy; Erd +os problems

1. Introduction

For an n by s matrix A ¼ ðaijÞ we define SðAÞ ¼ faij þ aikj1pipn; 1pjokpsg theset of pairwise sums of entries from the same row. Let fsðnÞ be the minimum sizejSðAÞj for a real n by s matrix with all its sn entries being pairwise distinct.The goal of this paper is to study the asymptotic behavior of fsðnÞ; especially for

large constant values of s:

ARTICLE IN PRESS

*Fax: +36-14-83-83-33.

E-mail address: [email protected].

0001-8708/$ - see front matter r 2003 Elsevier Science (USA). All rights reserved.

doi:10.1016/S0001-8708(03)00004-5

The motivation for this problem comes from the breakthrough paper of Solymosi

and Toth [10]. They proved an Oðn6=7Þ bound for an old problem of Erd +os [4], theminimum number distinct distances n points determine in the plane. This resultsubstantially improved earlier works of Moser [6], Chung [2], Chung et al. [3], andSzekely [12]. See [7] for the background of this intriguing old Erd +os problem and forfurther references.

Solymosi and Toth implicitly use f3ðnÞ ¼ Oðn1=3Þ in their proof, and a closer lookreveals that any stronger bound for fsðnÞ; with a constant s would improve theirresult. Section 4 has the details; Corollary 15 states the bound we get on the numberof distinct distances in the plane.

We have that f2ðnÞ ¼ 1; f3ðnÞ ¼ Yðn1=3Þ and f4ðnÞ ¼ Yðn1=3Þ but for f5ðnÞ andabove the correct order of magnitude is unknown. The best current bounds for f5and f6 are

n4=11pf5ðnÞpf6ðnÞ ¼ Oðn2=5Þ:

These bounds are special cases of a construction of Ruzsa and Theorem 1 below.This special case of Theorem 1 (with a worse constant factor) has a much simplerproof than the full theorem as shown in Section 5.The best upper bound on fsðnÞ; i.e., the best construction is due to Ruzsa [9], he

proves

fsðnÞ ¼ Oðn12� 12s�2Þ

for even s: The lower bound of the following theorem is stated for odd values of s: Itis interesting to note that both the known lower and upper bounds are identical forthe functions f2k�1ðnÞpf2kðnÞ but for kX3 we have no other indication for thesefunctions being close to each other.

Theorem 1. For an integer kX2 we have

f2k�1ðnÞXn1ck ;

where for kp14 we have

ck ¼Xk

i¼0

1

i!þ 1

ðk � 1Þk!;

while for kX14 we have

ck ¼Xk

i¼0

1

i!þ k3 � 7k2 þ 20k � 40

ðk4 � 8k3 þ 26k2 � 46k þ 40Þk!:

Notice, that both definitions of ck give the same value for c14:

ARTICLE IN PRESSG. Tardos / Advances in Mathematics 180 (2003) 275–289276

It is easy to see, that the limit of the values ck as k goes to infinity is e; the base ofthe natural logarithm. Thus we have the following.

Corollary 2. For every e40 we have a positive integer s ¼ sðeÞ with

fsðnÞXn1=e�e:

Note, that the limit of the exponent in the Ruzsa construction is 12; so the lower and

upper bounds are far apart.In Sections 2 and 3 we give the proof of Theorem 1. In Section 4, we apply it (or

rather Corollary 2) to get an improvement over the Solymosi–Toth bound on thenumber of distinct distances n point determine in the plane (see Corollary 15). Wealso give references to other related problems where Corollary 2 could be used indiscrete geometry. In Section 5, we give an elementary and simple proof of the first

non-trivial case of Theorem 1: we prove that f5ðnÞ ¼ Oðn4=11Þ: We close the paperwith concluding remarks and open problems in Section 6.

2. The proof—reduction to a linear program

Let us fix the positive integers s; n and an n by s real matrix A: Our proof does notuse in full generality the assumption that all entries of A are distinct. It is enough tomake the slightly weaker assumption that no two rows of A have two commonentries. Our goal is to prove a lower bound on jSðAÞj:Let I ¼ f1; 2; 3;y; sg be the set of column indices. For subsets U ;VDI and for an

s-tuple R ¼ ða1;y; asÞ we define the UV pattern pU ;V ðRÞ of R to be a sequence of

real numbers consisting of the differences ai � aj for i; jAU and for i; jAV and the

sums ai þ aj for iAU and jAV : We define

HðU ;VÞ ¼ HðpUV ðRÞÞ;

where H denotes the entropy and R is a uniformly distributed random row of A: Allentropies and all logarithms in this paper are binary.The next lemma stating linear constraints on the entropies HðU ;VÞ is crucial for

the proof.

Lemma 3. Let U ;U 0;V ;V 0DI : We have

(a) HðU ;VÞ ¼ HðV ;UÞ;(b) HðU ;VÞpHðU 0;V 0Þ if UDU 0 and VDV 0;(c) HðU ;VÞ ¼ 0 if U ¼ | and jV j ¼ 1;(d) HðU ;VÞplogjSðAÞj if UaV and jU j ¼ jV j ¼ 1;(e) HðU ;VÞ ¼ log n if U-Va| and jU,V j41;(f) HðU,U 0;V,V 0Þ þ HðU-U 0;V-V 0ÞpHðU ;VÞ þ HðU 0;V 0Þ if

ðU-U 0Þ,ðV-V 0Þa|:

ARTICLE IN PRESSG. Tardos / Advances in Mathematics 180 (2003) 275–289 277

Proof. We use the well-known properties of the entropy to prove this lemma.

A. Range: For a random variable F that has k possible values HðFÞplog k withequality if F is distributed uniformly.Part (c) follows since pU ;V ðRÞ is constant in that case.Part (d) also follows since pU ;V ðRÞ consists of a single value from SðAÞ in that

case.Part (e) of the lemma also follows from the above property. The pattern pU ;V ðRÞ

contains 2ai for the index iAU-V and thus it determines ai and with it all othervalues aj with jAU,V : As two entries uniquely determine the row of the matrix A

we have that pU ;V ðRÞ is different for all the n rows of A; thus it is uniformlydistributed among n possible values.B. Monotonicity: If the value of a random variable F uniquely determines the

value of another random variable G then we have HðFÞXHðGÞ:Part (b) of the lemma follows as the pattern pU 0;V 0 ðRÞ contains all entries of the

pattern pU ;V ðRÞ:Part (a) of the lemma also follows as the patterns pU ;V ðRÞ and pV ;UðRÞ contain the

same entries, so they mutually determine each other.C. Submodularity: Suppose that the value of either one of the random variables F1

and F2 determines the value of the random variable G1 and the values of the randomvariables F1 and F2 together determine the value of the random variable G2: In thiscase, we have HðG1Þ þ HðG2ÞpHðF1Þ þ HðF2Þ:For part (f) of the lemma we use the submodularity of entropy as stated above.

Clearly the pattern pU-U 0;V-V 0 ðRÞ is determined by either one of pU ;V ðRÞ andpU 0;V 0 ðRÞ: We need to show an entry ai7aj in pU,U 0;V,V 0 ðRÞ is determined by thetwo patterns pU ;V ðRÞ and pU 0;V 0 ðRÞ: Indeed, the term ai7aj in the former pattern can

be expressed as a sum or difference of the terms ai7ak and aj7ak in the latter

patterns if kAðU-U 0Þ,ðV-V 0Þ: &

Lemma 3 contains linear constraints on the entropies HðU ;VÞ and logjSðAÞj; thussolving them as a linear program provides a bound on jSðAÞj: This is indeed theroute we will take. The rest of the proof of the lower bound of jSðAÞj uses solelyLemma 3. We remark here that the linear program defined by Lemma 3 has a uniqueoptimal solution for all values of s except for s ¼ 27 or s ¼ 28 where the optimalsolutions are the convex combinations of two extremal optimal solutions.Our first step is to use averaging to decrease the exponential number of variables

to less than s2 of them.For integers i; jX0; 1pi þ jps we define

Hi;j ¼ 1� 1

log nðniÞðn�i

jÞXU ;V

HðU ;VÞ;

where the summation extends over all ðniÞðn�i

jÞ pairs of disjoint subsets U and V of I

with jU j ¼ i and jV j ¼ j: (We consider the values Hi;j to form a matrix H with some


entries of this matrix missing. We will only use the values Hi;j satisfying 0pi; jpk

and 1pi þ jp2k � 1 where k ¼ Js=2n:)

Lemma 4. For i; j non-negative integers with 1pi þ jps we have:

(a) (symmetry) Hi;j ¼ Hj;i;(b) (monotonicity) Hi;jXHiþ1; j if i þ jps � 1;(c) H0;1 ¼ 1;(d) H1;1X1� logjSðAÞj=log n;(e) (convexity) Hi�1; j þ Hiþ1; jX2Hi;j if iX1 and 2pi þ jps � 1;(f) Hi;jXHiþ1; j þ Hi;jþ1 if i þ jps � 1:

We could also state the non-negativity of these variables, but we will not use it.

Proof. Parts (a)–(d) of this lemma follow from the corresponding parts of Lemma 3by simple averaging.Part (e) follows from part (f) of Lemma 3, here the averaging is over the four-

tuples of sets U ;V ;U 0;V 0DI satisfying jU j ¼ jU 0j ¼ i; jU-U 0j ¼ i � 1;

V ¼ V 0; jV j ¼ j and ðU,U 0Þ-V ¼ |:Finally, for part (f) of this lemma consider two disjoint subsets U and V 0 of I (not

both the empty set) and an index kAI\ðU,V 0Þ: Applying Lemma 3(f) for U ;U 0 ¼ U,fkg; V ¼ V 0,fkg; and V 0 one gets

HðU ;V 0Þ þ HðU 0;VÞpHðU ;VÞ þ HðU 0;V 0Þ:

Here Lemma 3(e) applies and yields HðU 0;VÞ ¼ log n; thus, we have

HðU ;V 0Þ þ log npHðU ;VÞ þ HðU 0;V 0Þ:

Part (f) of the lemma follows from averaging over all pairs of disjoint subsetsU ;V 0DI with jU j ¼ i and jV 0j ¼ j and for all possible indices k: &

We remark that the linear program defined by Lemma 4 is already tractable bystandard linear programming methods for small values of s but as we will see,considering the cases sp28 only can be misleading.

3. Solving the linear program

In this rather technical section we combine the inequalities in Lemma 4 to provean upper bound on H1;1 and thus a lower bound on the size of SðAÞ:The optimal solution of the linear program in Lemma 4 is the same for an odd

number s and for the next even number (and is unique unless s is either 27 or 28).Since our goal is simply to prove a lower bound on jSðAÞj we assume s ¼ 2k � 1 forsome integer kX3:


Lemma 5. Hk�2;k�1p 3kþ1 H0;k�1:

Proof. By Lemma 4(e) the columns of the matrix H are convex, therefore we have

H0;k�1 � Hk�2;k�1k � 2

XHk�3;k�1 � Hk�2;k�1XHk�2;k�1 � Hk;k�1

2:

Using parts (f), (b) and (a) of Lemma 4 we get

Hk�3;k�1 � Hk�2;k�1XHk�3;kXHk�1;k ¼ Hk;k�1:

Combining the last two displayed inequalities we get

H0;k�1 � Hk�2;k�1k � 2

XHk�2;k�1 � Hk;k�1

2X

Hk�2;k�1 � ðHk�3;k�1 � Hk�2;k�1Þ2

XHk�2;k�1 �

H0;k�1�Hk�2;k�1k�2

2;

yielding the claimed statement by rearrangement. &

Lemma 6. Suppose we have Hj�1; jpaH0;j for some 3pjok and a40: If ðj � 3Þap2

then we also have Hj�2;j�1pbH0;j�1 for b ¼ ð2þ aÞ=ðj þ aÞ40 and ðj � 4Þbp2 is also

satisfied.

Proof. We consider the following four inequalities:

H0;j þ H1; j�1pH0;j�1;

by Lemma 4(f);

Hj�2;j�1 � Hj;j�12

pH0;j�1 � Hj�2;j�1

j � 2;

by the convexity of column j � 1 (Lemma 4(e));

Hj�2;j�1 � Hj;j�12

pH1; j�1 � Hj�2;j�1

j � 3;

for j43 by the convexity of the same column; finally

Hj;j�1paH0;j;

by assumption and symmetry (Lemma 4(a)). We sum these inequalities with the non-negative coefficients a; 2� ðj � 3Þa; ðj � 3Þa; and 1, respectively, and rearrange toget the inequality Hj�2;j�1pbH0;j�1 as claimed in the lemma. Notice that for j ¼ 3

the third inequality is not valid but we use it with zero coefficient. Simple calculationyields the claimed bound on b: &


We need a closed form for the continued fraction in the next lemma. Note that asa consequence of the lemma, the corresponding infinite continued fraction evaluatesto e; the base of the natural logarithm.

Lemma 7. For an integer kX1 and real xok2=ðk � 1Þ we have

3� 1

4� 2

5� 36�^

kþ1� k�1kþ2�x

¼Xk

i¼0

1

i!þ k � x

ðk2 � ðk � 1ÞxÞk!:

For k ¼ 1; 2; 3;y the left-hand side of the equation in the lemma is understood tobe

3� x;

3� 1

4� x;

3� 1

4� 25�x

;

etc.

Proof. The proof is by induction on k: The k ¼ 1 case is trivial. For k41 we use the

inductive hypothesis for k0 ¼ k � 1 and x0 ¼ ðk � 1Þ=ðk þ 2� xÞoðk � 1Þ2=ðk � 2Þ;and a simple calculation yields the lemma. &

Instead of Theorem 1 we prove the somewhat stronger statements of Theorems 8and 10.

Theorem 8. Let 2pkp14; nX1; and let A be an n by ð2k � 1Þ real matrix with no two

distinct rows sharing more than a single entry. Then we have

jSðAÞjXn1=ck ;

with ck ¼Pk

i¼01i! þ 1

ðk�1Þk!:

Proof. Previously in this section we assumed kX3 so the k ¼ 2 ðs ¼ 3Þ case must bedealt with separately. One can either solve the linear program of Lemma 4 (there areonly four distinct relevant variables in this case) or use direct reasoning as in thebeginning of Section 5. This s ¼ 3 case was already discussed in [10].For kX3 we prove a bound Hj�1; jpajH0;j by reverse induction on j ¼ k � 1;y; 2:

We use Lemma 5 to get ak�1 ¼ 3=ðk þ 1Þ as the bases of our induction. We useLemma 6 for the inductive step to get aj�1 ¼ ð2þ ajÞ=ðj þ ajÞ: Notice that the


ðj � 3Þajp2 condition is satisfied at j ¼ k � 1 because of the kp14 assumption and

this condition is preserved by Lemma 6. Rewriting the recursion to 1� aj�1 ¼ðj � 2Þ=ððj þ 1Þ � ð1� ajÞÞ and writing 1� ak�1 ¼ ðk � 2Þ=ðk þ 1Þ we get the

following continued fraction expansion for a2:

1� a2 ¼1

4� 2

5� 36�^

k�k�2kþ1

:

We further have

2H1;1 � H0;1pH2;1 ¼ H1;2pa2H0;2pa2ðH0;1 � H1;1Þ;

by Lemma 4(e), Lemma 4(a), the statement above, and Lemma 4(f), respectively. Byrearrangement, and using H0;1 ¼ 1 (Lemma 4(c)) we get the bound H1;1pð1þ a2Þ=ð2þ a2Þ: By Lemma 4(d) we get

log n=logjSjp 1

1� H1;1p2þ a2 ¼ 3� 1

4� 2

5� 36�^

k�k�2kþ1

:

Lemma 7 provides a closed form for the continued fraction of the above statementand thus proves the theorem. &

In order to use Lemma 6 recursively as in the proof of Theorem 8 we need a basecase. Lemma 5 cannot be used for this if k414 as the inequality required for Lemma6 to apply ððj � 3Þap2Þ does not hold for j ¼ k � 1 and a ¼ 3=ðk þ 1Þ: The nextlemma provides the base j ¼ k � 2 case for large k:

Lemma 9. If kX14 we have Hk�3;k�2p 2kþ3k2�kþ4 H0;k�2:

Proof. We combine six inequalities to get the desired bound. We use

Hk�2;k�1p3

k þ 1H0;k�1

and

Hk�2;k�1p3

kH1;k�1:

The former inequality is provided by Lemma 5, while the latter inequality can beproven the same way. We also use

H0;k�1 þ H1;k�2pH0;k�2


and

H1;k�1 þ H2;k�2pH1;k�2

provided by Lemma 4(f). Finally, we also use

Hk�3;k�2 � Hk�2;k�12

pH1;k�2 � Hk�3;k�2

k � 4

and

Hk�3;k�2 � Hk�2;k�12

pH2;k�2 � Hk�3;k�2

k � 5;

both coming from symmetry (Lemma 4(a)) and the convexity of column k � 2(Lemma 4(e)). We sum the above six inequalities with coefficients ðk þ 1Þð2k þ3Þ; kðk � 14Þ; 3ð2k þ 3Þ; 3ðk � 14Þ; 3ðk � 4Þðk þ 17Þ; and 3ðk � 5Þðk � 14Þ; inthis order, and rearrange to obtain the bound of the lemma. Note, that all thesecoefficients are non-negative if kX14: &

Theorem 10. Let kX14; nX1; and let A be an n by ð2k � 1Þ real matrix with no two

distinct rows sharing more than a single entry. Then we have

jSðAÞjXn1=ck ;

with ck ¼Pk

i¼01i! þ k3�7k2þ20k�40

ðk4�8k3þ26k2�46kþ40Þk!:

Proof. We copy the proof of Theorem 8. We prove a bound Hj�1; jpajH0;j by

reverse induction on j ¼ k � 2;y; 2: We use Lemma 9 for the base case j ¼ k � 2

and ak�2 ¼ ð2k þ 3Þ=ðk2 � k þ 4Þ: Lemma 6 gives aj�1 ¼ ð2þ ajÞ=ðj þ ajÞ since

ðj � 3Þajp2 for j ¼ k � 2 and so it is true for all values of j considered. As in the

proof of Theorem 8 we get a continued fraction expansion of a2; in this case it is

1� a2 ¼1

4� 2

5� 36�^

k�2� k�4

k�1�k2�3kþ1k2�kþ4

:

Just as in the proof of Theorem 8 we get

log n=logjSjp2þ a2:


Thus, we have

log n=logjSjp3� 1

4� 2

5� 36�^

k�2� k�4k�1�k2�3kþ1

k2�kþ4

:

Lemma 7 provides a closed form for the continued fraction of the above statementand thus proves the theorem. &

4. Distinct distances in the plane

As mentioned in the introduction, the problem considered in this paper is a by-product of the paper [10] by Solymosi and Toth. One of their lemmas can be stated inour notation as follows.

Lemma 11 (Solymosi and Toth [10, Lemma 5]). Let A ¼ ðaijÞ be an n by 3 real

matrix with all its entries pairwise distinct. Assume that ai1oai2oai3 for i ¼ 1;y; n;and assume also that ai3oaiþ1;1 for all but at most t � 1 indices i ¼ 1;y; n � 1: Then

jSðAÞj ¼ OðN=t2=3Þ:

We generalize this lemma as follows. We include the simple proof along the samelines.

Lemma 12. Let s; and tpn be positive integers and let A ¼ ðaijÞ be an n by s real

matrix with all the ns entries pairwise distinct. Assume that maxj aijominj aiþ1; j holds

for all but at most t � 1 of the indices i ¼ 1;y; n � 1: Then

jSðAÞjX n

2t

j k fsðtÞ:

Proof. We find I n2tm pairwise disjoint real intervals, each containing all entries of t

rows of A: This can be done from left to right on the real line using a greedy strategy.Let AI be the submatrix of A consisting of the t rows fully contained in the interval I :Clearly jSðAI ÞjXfsðtÞ and SðAI ÞDSðAÞ: Furthermore, SðAI Þ and SðAI 0 Þ are disjointif I and I 0 are disjoint, yielding the lemma. &

With the help of this lemma we get


Theorem 13. Let us be given n points in the plane such that from any one of the points

there are at most t distinct distances to the other points. Then

t5

fsðtÞXcn4;

where c ¼ cðsÞ is a positive constant depending on s:

Proof (Sketch). Solymosi and Toth in [10] prove that n points in the plane

determine Oðn6=7Þ distinct distances. Their proof can be considered the s ¼ 3 specialcase of our proof. The beautiful proof is based on the method of Szekely, uses thecrossing number theorem of Ajtai et al. [1] and Leighton [5], and the point–lineincidence theorem of Szemeredi and Trotter [13]. As most of the proof goes throughwithout a change we only sketch the differences and refer the reader to the originalproof for details.We consider the same incidences between points and circles as in [10]. We partition

the points incident to a circle into s-tuples (rather than triplets as in [10]). Weconstruct a topological graph by connecting at most a single pair of points along thecircle from every s-tuple, a pair with a bisector not rich, i.e., not going through more

than en2=t2 points. If no such pair exists we call the s-tuple bad. We deduce (as in[10]) from the crossing number theorem that most of the s-tuples are bad. Now wecontrast the upper bound of the Szemeredi–Trotter theorem on incidences betweenpoints and rich lines, to the lower bound obtained from Lemma 12 (in place ofLemma 11 in [10]). All calculations of the paper go through with this modificationand they yield Theorem 13. &

The next corollary is a straightforward consequence of Theorem 13, whileCorollary 15 below is a consequence of Corollaries 2 and 14.

Corollary 14. If for some positive integer s and positive real a we have fsðnÞ ¼ OðnaÞthen the following holds. Any set P of n points in the plane has an element from which

the number of distinct distances to the other points of P is

Oðn45�aÞ:

Corollary 15. For any e40 we have that any set P of n points in the plane has an

element from which the number of distinct distances to the other points of P is

Oðn4e5e�1�eÞ:

Subsequent to the paper [10], it turned out that the same proof technique can be usedto prove generalizations of the result in [10]. The bound on the number of appearancesof the k most frequent distances among n points in the plane in [11] and the bound onthe number of isosceles triangles n points in the plane determined in [8] both implyCorollary 15, and both heavily rely on Corollary 2, the main result of this paper.


5. An elementary proof

In this section, we consider fs for small values of s: Trivially, f1ðnÞ ¼ 0; f2ðnÞ ¼ 1:

We claim that f3ðnÞEð6nÞ1=3; more precisely,

f3ðnÞ ¼ min mm

3

!�� Xn

( ):

Indeed, each row of an n by 3 matrix A of all distinct entries determines three distinctsums in SðAÞ and these three sums in turn determine the entries and thus the row.Thus, ðf3ðnÞ

3ÞXn must hold. To see the claimed equality we construct for an arbitrary

integer mX3 a matrix AS of n ¼ ðm3Þ rows and three columns from a rationally

independent set S of m reals. Each row of AS is of the form

x þ y � z

2;x � y þ z

2;�x þ y þ z

2

�;

where fx; y; zg is a different three-element subset of S for each row. Notice that allentries of AS are different and SðASÞ ¼ S:It is easy to see, that f3ðnÞpf4ðnÞp2f3ðnÞ; thus f4 has the same order of magnitude

as f3 (namely n1=3). Indeed add an extra column to the matrix AS constructed abovemaking each row add up to zero. Notice that the modified matrix A0

S has still all

different entries and furthermore SðA0SÞ ¼ S,ð�SÞ: More precisely, we claim that

min mm

3

!�� X4n

( )pf4ðnÞp2 min m 2

m

3

!�� Xn

( ):

Indeed, each three-element subset of a row of an n by 4 matrix A of all differententries determine three different sums in SðAÞ and a moment notice verifies that all4n triplets obtained this way are distinct proving the lower bound on f4ðnÞ: For theupper bound, consider an m-element rationally independent set S and the ðm

3Þ by 4

matrix A0S constructed above. Notice that by using all rows of A0

S and �A0S we get a

2ðm3Þ by 4 matrix A00ðSÞ of all different entries with SðA00

SÞ ¼ SðA0SÞ ¼ S,ð�SÞ:

Notice that the lower and upper bounds differ by at most 3 (and this can be furtherreduced by simple observations).The observations above on f3 appeared already in the paper of Solymosi and Toth

[10] and were used to prove a lower bound on the number of distinct distances n

points determine in the plane. It was clear from their proof, that an improved lowerbound for fs even if for a higher constant value of s would improve their bound onthe number of distances. It was independently found by Gyula Karolyi and TiborSzabo that f4 has the same order of magnitude as f3 and thus it cannot be used in thismanner.Next we consider f5ðnÞ for which the actual order of magnitude is not known, but

Theorem 1 gives f5ðnÞXn4=11: Here we give an elementary proof of this result (with a


constant multiplicative factor in the bound). This proof can serve as a motivation forthe general result, the techniques of the proof of Theorem 1 were introduced togeneralize this elementary proof below. Note also that the lower bound for fsðnÞ inTheorem 1 is very close to the lower bound presented here: for arbitrary sX7 theimprovement in the exponent of n is only in the third digit after the decimal point.

Theorem 16. f5ðnÞ ¼ Oðn4=11Þ:

Proof. Let A be an n by 5 real matrix of all distinct entries with S ¼ SðAÞ havingminimal size jSj ¼ m ¼ f5ðnÞ: We call a real value x heavy if it can be expressed as

x ¼ u � v with u; vAS in at least m1=4 different ways and x is light otherwise. We calla row of A heavy it has two distinct entries b and c with b � c being heavy, otherwisethe row is called light.Our goal is to bound the number n of rows of A in terms of m and we do this

separately for heavy and light rows.

Clearly, there are only m2 ways to form a difference x ¼ u � v from values u; vAS;

so there are at most m2=m1=4 ¼ m7=4 heavy numbers. Given any value x it can beexpressed as the difference between two distinct entries of the same row of A in onlym different ways, as the sum of these two entries is a value in S and the sum and thedifference together determines the entries (we use here that all entries of A aredistinct). The last two statements together bound the number of heavy rows in A by

m7=4 m ¼ m11=4:Now consider a light row ða1; a2; a3; a4; a5Þ of A: We identify this row with

revealing limited information, and use this to bound the number of light rows. Wefirst reveal u ¼ a1 þ a2 and u0 ¼ a1 þ a3: Both numbers are from the m-element set S

so we have m2 choices for these values. With this we also identified u � u0 ¼ a2 � a3which must be a light number. Thus, writing u � u0 ¼ ða2 þ a4Þ � ða3 þ a4Þ ¼ða2 þ a5Þ � ða3 þ a5Þ are two of the less than m1=4 ways u � u0 can be expressed as the

difference of two values in S: So we have less than ðm1=4Þ2 ¼ m1=2 choices whenrevealing the values v ¼ a2 þ a4; v0 ¼ a2 þ a5: Just as above, the value v � v0 ¼a4 � a5 is now revealed, and it must be a light number, thus v � v0 ¼ ða1 þ a4Þ �ða1 þ a5Þ is one of the less than m1=4 ways v � v0 can be expressed as a difference of

values in S: So for revealing w ¼ a1 þ a4 we have less than m1=4 choices. At thispoint, a1 ¼ ðu þ w � vÞ=2 is also revealed, so the row itself is identified (as all entriesof A are distinct). This limits the number of light rows by m2 m1=2 m1=4 ¼ m11=4:

Thus, no2m11=4 and so f5ðnÞ ¼ m4ðn=2Þ4=11: &

6. Concluding remarks and open problems

As we have already mentioned, the orders of magnitude of the functions fsðnÞ areunknown for sX5: Improving either the lower or the upper bounds is a challenge.One could hope to improve the lower bounds by the methods of this paper, i.e.,

considering a uniformly distributed random row of a matrix with all distinct entries,


several linear functions on this row, and using inequalities on the (joint) entropies ofthese functions. There is room for improvement. In this paper, we restricted ourattention to specific collections of pairwise sums and differences only. One can tryusing different collections, like Hða þ b; c þ dÞ or even linear combinations of adifferent type, like Hða þ b þ c þ dÞ: Here a; b; c; and d represent distinct entries ofthe random row. Submodularity gives a lot of inequalities, like

Hða þ b; b þ c; c þ dÞ þ Hða þ b þ c þ dÞpHða þ b; c þ dÞ þ Hðb þ c; d þ aÞ;

but the author was unable to use these additional inequalities to obtain betterbounds on fs: One could also try to use information inequalities that are notconsequences of the basic inequalities of Shannon type (monotonicity andsubmodularity). Such inequalities were published by Zhang and Yeung [14] butthey seem to be too complicated to easily lend themselves to applications.The original motivation for the problem considered in this paper is its applicability

to the Erd +os problem of finding the minimum number of distinct distances n pointsdetermine in the plane as formulated in Corollary 15. The Ruzsa construction shows

that one cannot use this theorem directly to prove an Oðn8=9Þ bound on the numberof distinct distances. One may try to modify the lower bound proof on the number ofdistinct distances by letting the parameter s grow with n: Unfortunately, a simple

modification of the Ruzsa construction is still in the way of proving Oðn8=9þeÞ in thisway. Indeed, one can show that for every e40 there exists d40 with

fndðnÞ ¼ Oðn1=2þeÞ:

Note added in proof. The recent paper ‘‘Nets Hawk Katz: An Improvement of aLemma of Tardos’’ (unpublished) was able to improve on the techniques of Section 5

of this paper and prove f5ðnÞ ¼ On7=19�e for any e40: By Corollary 14 and since7=1941=e this result improves our bound Corollary 15 on the Erdos distanceproblem. Combining ideas of the above and the present paper the authors of thesetwo papers could further improve on this bound by proving tighter lower bounds forfsðnÞ for sX7:

Acknowledgments

The author is indebted to Janos Pach for enlightening discussions. Fruitfulconversations with Nati Lineal, Imre Ruzsa, Gabor Simonyi, and Csaba Toth arealso acknowledged. Research was partially supported by the grants OTKA T030059,FKFP 0607/1999 and AKP 2000-78 2.1.


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