On distinct sums and distinct distances
Post on 02-Jul-2016
Embed Size (px)
http://www.elsevier.com/locate/aimAdvances in Mathematics 180 (2003) 275289
On distinct sums and distinct distances
Renyi Institute, Realtanoda utca 13-15, 1053 Budapest, Hungary
Received 28 August 2001; accepted 2 August 2002
Communicated by Laszlo Lovasz
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 s2nsums 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 1=cJs=2n is dened in the paper and tends to e1 as s goes to innity. Here e is the base ofthe natural logarithm. As an application we improve the SolymosiToth bound on an old
Erd +os problem: we prove that n distinct points in the plane determine On 4e5e1e distinctdistances, where e40 is arbitrary. Our bound also nds applications in other related results indiscrete 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
For an n by s matrix A aij we dene SA faij aikj1pipn; 1pjokpsg theset of pairwise sums of entries from the same row. Let fsn be the minimum sizejSAj 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 fsn; especially for
large constant values of s:
ARTICLE IN PRESS
E-mail address: firstname.lastname@example.org.
0001-8708/$ - see front matter r 2003 Elsevier Science (USA). All rights reserved.
The motivation for this problem comes from the breakthrough paper of Solymosi
and Toth . They proved an On6=7 bound for an old problem of Erd +os , theminimum number distinct distances n points determine in the plane. This resultsubstantially improved earlier works of Moser , Chung , Chung et al. , andSzekely . See  for the background of this intriguing old Erd +os problem and forfurther references.
Solymosi and Toth implicitly use f3n On1=3 in their proof, and a closer lookreveals that any stronger bound for fsn; 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 f2n 1; f3n Yn1=3 and f4n Yn1=3 but for f5n andabove the correct order of magnitude is unknown. The best current bounds for f5and f6 are
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 fsn; i.e., the best construction is due to Ruzsa , he
fsn On12 12s2
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 f2k1npf2kn but for kX3 we have no other indication for thesefunctions being close to each other.
Theorem 1. For an integer kX2 we have
where for kp14 we have
i! 1k 1k!;
while for kX14 we have
3 7k2 20k 40k4 8k3 26k2 46k 40k!:
Notice, that both denitions of ck give the same value for c14:
ARTICLE IN PRESSG. Tardos / Advances in Mathematics 180 (2003) 275289276
It is easy to see, that the limit of the values ck as k goes to innity is e; the base ofthe natural logarithm. Thus we have the following.
Corollary 2. For every e40 we have a positive integer s se with
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 SolymosiToth 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 rst
non-trivial case of Theorem 1: we prove that f5n On4=11: We close the paperwith concluding remarks and open problems in Section 6.
2. The proofreduction to a linear program
Let us x 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 jSAj:Let I f1; 2; 3;y; sg be the set of column indices. For subsets U ; VDI and for an
s-tuple R a1;y; as we dene the UV pattern pU ;V R of R to be a sequence ofreal numbers consisting of the differences ai aj for i; jAU and for i; jAV and thesums ai aj for iAU and jAV : We dene
HU ; V HpUV R;
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 HU ; V is crucial for
Lemma 3. Let U ; U 0; V ; V 0DI : We have
(a) HU ; V HV ; U;(b) HU ; VpHU 0; V 0 if UDU 0 and VDV 0;(c) HU ; V 0 if U | and jV j 1;(d) HU ; VplogjSAj if UaV and jU j jV j 1;(e) HU ; V log n if U-Va| and jU,V j41;(f) HU,U 0; V,V 0 HU-U 0; V-V 0pHU ; V HU 0; V 0 if
U-U 0,V-V 0a|:
ARTICLE IN PRESSG. Tardos / Advances in Mathematics 180 (2003) 275289 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 HFplog k withequality if F is distributed uniformly.Part (c) follows since pU ;V R is constant in that case.Part (d) also follows since pU ;V R consists of a single value from SA in that
case.Part (e) of the lemma also follows from the above property. The pattern pU ;V R
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 Awe have that pU ;V R 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 HFXHG:Part (b) of the lemma follows as the pattern pU 0;V 0 R contains all entries of the
pattern pU ;V R:Part (a) of the lemma also follows as the patterns pU ;V R and pV ;UR 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 HG1 HG2pHF1 HF2:For part (f) of the lemma we use the submodularity of entropy as stated above.
Clearly the pattern pU-U 0;V-V 0 R is determined by either one of pU ;V R andpU 0;V 0 R: We need to show an entry ai7aj in pU,U 0;V,V 0 R is determined by thetwo patterns pU ;V R and pU 0;V 0 R: Indeed, the term ai7aj in the former pattern canbe expressed as a sum or difference of the terms ai7ak and aj7ak in the latterpatterns if kAU-U 0,V-V 0: &
Lemma 3 contains linear constraints on the entropies HU ; V and logjSAj; thussolving them as a linear program provides a bound on jSAj: This is indeed theroute we will take. The rest of the proof of the lower bound of jSAj uses solelyLemma 3. We remark here that the linear program dened by Lemma 3 has a uniqueoptimal solution for all values of s except for s 27 or s 28 where the optimalsolutions are the convex combinations of two extremal optimal solutions.Our rst step is to use averaging to decrease the exponential number of variables
to less than s2 of them.For integers i; jX0; 1pi jps we dene
Hi;j 1 1log nn
HU ; V;
where the summation extends over all nini
j pairs of disjoint subsets U and V of I
with jU j i and jV j j: (We consider the values Hi;j to form a matrix H with some
ARTICLE IN PRESSG. Tardos / Advances in Mathematics 180 (2003) 275289278
entries of this matrix missing. We will only use the values Hi;j satisfying 0pi; jpkand 1pi jp2k 1 where k Js=2n:)
Lemma 4. For i; j non-negative integers with 1pi jps we have:(a) (symmetry) Hi;j Hj;i;(b) (monotonicity) Hi;jXHi1; j if i jps 1;(c) H0;1 1;(d) H1;1X1 logjSAj=log n;(e) (convexity) Hi1; j Hi1; jX2Hi;j if iX1 and 2pi jps 1;(f) Hi;jXHi1; j Hi;j1 if i 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 jU 0j i; jU-U 0j i 1;V V 0; jV j j and U,U 0-V |: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\U,V 0: Applying Lemma 3(f) for U ;U 0 U,fkg; V V 0,fkg; and V 0 one gets
HU ; V 0 HU 0; VpHU ; V HU 0; V 0:
Here Lemma 3(e) applies and yields HU 0; V log n; thus, we haveHU ; V 0 log npHU ; V HU 0; V 0:
Part (f) of the lemma follows from averaging over all pairs of disjoint subsetsU ; V 0DI with jU j i and jV 0j j and for all possible indices k: &
We remark that the linear program dened 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 SA: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 jSAj we assume s 2k 1 forsome integer kX3:
ARTICLE IN PRESSG. Tardos / Advances in Mathematics 180 (2003) 275289 279
Lemma 5. Hk2;k1p 3k1 H0;k1:
Proof. By Lemma 4(e) the columns of the matrix H are convex, therefore we have
H0;k1 Hk2;k1k 2 XHk3;k1 Hk2;k1X
Using parts (f), (b) and (a) of Lemma 4 we get
Hk3;k1 Hk2;k1XHk3;kXHk1;k Hk;k1:
Combining the last two displayed inequalities we get
H0;k1 Hk2;k1k 2 X
XHk2;k1 Hk3;k1 Hk2;k1
yielding the claimed statement by rearrangement. &
Lemma 6. Suppose we have Hj1; jpaH0;j for some 3pjok and a40: If j 3ap2then we also have Hj2;j1pbH0;j1 for b 2 a=j a40 and j 4bp2 is alsosatisfied.
Proof. We consider the following four inequalities:
H0;j H1; j1pH0;j1;
by Lemma 4(f);
pH0;j1 Hj2;j1j 2 ;
by the convexity of column j 1 (Lemma 4(e));Hj2;j1 Hj;j1
2pH1; j1 Hj2;j1
j 3 ;
for j43 by the convexity of the same column; nally
by assumption and symmetry (Lemma 4(a)). We sum these inequalities with the non-negative coefcients a; 2 j 3a; j 3a; and 1, respectively, and rearrange toget the inequality Hj2;j1pbH0;j1 as claimed in the lemma. Notice that for j 3the third inequality is not valid but we use it with zero coefcient. Simple calculationyields the claimed bound on b: &
ARTICLE IN PRESSG. Tardos / Advances in Mathematics 180 (2003) 275289280
We need a closed form for the continued fraction in the next lemma. Note that asa consequence of the lemma, the corresponding innite continued fraction evaluatesto e; the base of the natural logarithm.
Lemma 7. For an integer kX1 and real xok2=k 1 we have
3 14 2
i! k xk2 k 1xk!:
For k 1; 2; 3;y the left-hand side of the equation in the lemma is understood tobe
3 14 x;
3 14 2
Proof. The proof is by induction on k: The k 1 case is trivial. For k41 we use theinductive hypothesis for k0 k 1 and x0 k 1=k 2 xok 12=k 2;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 2k 1 real matrix with no twodistinct rows sharing more than a single entry. Then we have
with ck Pk
Proof. Previously in this section we assumed kX3 so the k 2 s 3 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 3 case was already discussed in .For kX3 we prove a bound Hj1; jpajH0;j by reverse induction on j k 1;y; 2:
We use Lemma 5 to get ak1 3=k 1 as the bases of our induction. We useLemma 6 for the inductive step to get aj1 2 aj=j aj: Notice that the
ARTICLE IN PRESSG. Tardos / Advances in Mathematics 180 (2003) 275289 281
j 3ajp2 condition is satised at j k 1 because of the kp14 assumption andthis condition is preserved by Lemma 6. Rewriting the recursion to 1 aj1 j 2=j 1 1 aj and writing 1 ak1 k 2=k 1 we get thefollowing continued fraction expansion for a2:
1 a2 14 2
We further have
2H1;1 H0;1pH2;1 H1;2pa2H0;2pa2H0;1 H1;1;
by Lemma 4(e), Lemma 4(a), the statement above, and Lemma 4(f), respectively. Byrearrangement, and using H0;1 1 (Lemma 4(c)) we get the bound H1;1p1 a2=2 a2: By Lemma 4(d) we get
log n=logjSjp 11 H1;1p2 a2 3
4 25 3
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 j 3ap2 does not hold for j k 1 and a 3=k 1: The nextlemma provides the base j k 2 case for large k:
Lemma 9. If kX14 we have Hk3;k2p 2k3k2k4 H0;k2:
Proof. We combine six inequalities to get the desired bound. We use
k 1 H0;k1
The former inequality is provided by Lemma 5, while the latter inequality can beproven the same way. We also use
ARTICLE IN PRESSG. Tardos / Advances in Mathematics 180 (2003) 275289282
provided by Lemma 4(f). Finally, we also use
pH1;k2 Hk3;k2k 4
pH2;k2 Hk3;k2k 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 coefcients k 12k 3; kk 14; 32k 3; 3k 14; 3k 4k 17; and 3k 5k 14; inthis order, and rearrange to obtain the bound of the lemma. Note, that all thesecoefcients are non-negative if kX14: &
Theorem 10. Let kX14; nX1; and let A be an n by 2k 1 real matrix with no twodistinct rows sharing more than a single entry. Then we have
with ck Pk
Proof. We copy the proof...