Discrepancy in Arithmetic Progressions

Discrepancy in Arithmetic ProgressionsAuthor(s): Jiří Matoušek and Joel SpencerSource: Journal of the American Mathematical Society, Vol. 9, No. 1 (Jan., 1996), pp. 195-204Published by: American Mathematical SocietyStable URL: http://www.jstor.org/stable/2152845 .

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JOURNAL OF THE AMERICAN MATHEMATICAL SOCIETY Volume 9, Number 1, January 1996

DISCREPANCY IN ARITHMETIC PROGRESSIONS

JIRI MATOUSEK AND JOEL SPENCER

1. RESULTS AND HISTORY

Let A be a family of subsets of a finite set Q. By a two-coloring of Q we shall mean a map X: Q -) {-1, +1}. For any X C Q we define X(X) = ZXGX x(x). The discrepancy of A is defined by

(1) disc (A) = min Amax I x(A)| x AEA Let Q = { 1,... , n}, which we denote by [n]. Let A denote the set of arithmetic progressions on [n]. The discrepancy of this set system was investigated in 1964 by K. F. Roth [7]. If we define the function ROTH(n) = disc(A), his result can be written

ROTH(n) > cnl/4

for c a positive absolute constant. That is, for any two-coloring X of the first n integers there will be an arithmetic progression A on which the "imbalance" IX(A) is at least cnr/4

It is interesting that Roth himself did not believe his result to be best possible and speculated that perhaps ROTH(n) = nl/2O(l). Indeed a bound ROTH(nt) = 0(vn-lnn) follows by elementary probabilistic considerations. In the early 1970s Sarkozi (see [3]) showed ROTH(n) < nl/3+o(l). A breakthrough was given in 1981 by Beck [2] who showed ROTH(n) < cnl/4 ln5/2 n. Here we show

Theorem 1.1. ROTH(n) < Cn1/4 with C an absolute constant. In words, we show the existence of a two-coloring X of the first n integers so

that all arithmetic progressions A have imbalance IX(A)I <On14. We remark that the proof ,does not give a construction of X in the usual sense and is indeed not satisfactory from an algorithmic point of view. The methods of ?2 (see comments in [5]) are such that we have not been able to obtain an algorithm that would output this coloring X in time polynomial in n. Our proof involves variants of the probabilistic method; we give [1] as a general reference. The technique of our proof combines methods of [2], [5], [6].

Throughout the paper, we'll use the symbols c, c', etc. generically for denoting absolute constants, and in order to limit the number of symbols, we reuse them freely.

Received by the editors February 18, 1994 and, in revised form, December 29, 1994. 1991 Mathematics Subject Classification. Primary 11B25, 11N37. The first author was supported by Charles University grant No. 351 and Czech Republic

Grant GACR 201/93/2167. Part of this research was done during a visit to Princeton University supported by DIMACS.

?)1996 American Mathematical Society

195



196 JIRI MATOUSEK AND JOEL SPENCER

2. ENTROPY

Let A1,. Av C Q. A partial coloring is a map X: Q -1,0,+1}. When x(x) = 0 we call x uncolored, otherwise x is called colored. We define, for A C Q, X(A) = EXEA X(X). Our object will be to give a general condition under which there exists a partial coloring X with the IX(Ai) I "small" and "few" x E Q uncolored.

For any positive integer b define the b-roundoff function Rb(x) as that i so that 2bi is the nearest multiple of 2b to x. In case of ties take the larger. Thus

Rb(X) = O if and only if -b < x < b,

(2) Rb(X) > i if and only if x > (2i - 1)b, Rb(X) < -i if and only if x < -(2i - 1)b.

Let X be any discretely valued random variable. We use the standard definition of the entropy function H(X):

H(X) = E-Pilog2(Pi)

where pi = Pr[X = i], the summation is over the possible values of X, and 0 log2 0 is interpreted as 0. We shall use the following well-known facts about entropy:

* Entropy is subadditive, i.e., if X = (XI,... , Xv), then H(X) < >3> H(Xi). * When X takes on at most K values it has entropy at most log2 K, the extreme

case being a uniformly chosen value from a K-set. Moreover Ei.1 -Pi log2 (Pi)

< log2 Ill for any subset of values of X. * When X has entropy less than K it takes on some value with probability at

least 2-K

Let Sn, as standard, denote the sum of n independent random variables, each uniform on {-1,+1}. When X: Q -* {-1,+1} is uniform and A C Q,IAI n, then x(A) has distribution Sn. Now we come to a key definition:

ENT(n,b) = H(Rb(Sn))

With this definition we give our general criterion.

Lemma 2.1. Let Si,...,S, C Q with IQ = n and ISil = ni. Suppose bi,e and -y < 1 are such that -2

v

ZENT(ni, bi) < en i=l

and -yn

(3) S <~ <2(1) j=O 3

Then there is a partial coloring X of Q with

IX(Si)I < bi for all i and more than 2-yn points x E Q colored.

Proof. Consider the uniform probability space of all X: Q - {-1, +1} and define the random variable

L(x) = (Rb1 (X(Sl)), * * ,Rbv (X(Sv)))



DISCREPANCY IN ARITHMETIC PROGRESSIONS 197

By subadditivity of entropy v v

H(L) < H(Rb (X(Si))) = ZENT(ni,bi) < en.

Hence some value of L has probability at least 2-,n of being achieved. As all X have probability 2-n there is a set F of at least 2n(l-) colorings X so that if Xl, X2 E r, then L(Xi) = L(X2).

We naturally associate such colorings X with points on the Hamming Cube {-1, +1}n. (With Q = {l, ... ,n} associate X with (x(1), ... .,x(n)).) A theorem of Kleitman [4] (basically an isoperimetric inequality) states that any r c {-1, t1 }n of size bigger than Z1=o (n) with 1 < n contains two points at Hamming distance (i.e., the number of different coordinates) at least 21. (This is "best possible" as r may be the set of all sequences with at most 1 coordinates +1.) Thus there are Xl, X2 E F at Hamming distance at most 2yn. Set

x(x) =Xi (X) - X2 (X) for all x E C 2

Then X is a partial coloring. The number of colored points is precisely the Hamming distance which is at least 2-yn. For each i the values XI (Si), x2 (Si) have the same bi-roundoff and therefore lie in a common open interval of length less than 2b. Thus

Ix(S)l = Xl(Si) - X2(Si) < bi 2 as desired. O

We note that (3) holds for, say, 'y = - and e = 0.2; this value will suffice for our purposes. Also, we shall always use a bound on IX(S)I dependent only on jSj. We'll use the lemma in the following simpler form.

Corollary 2.2. Let A be a family of subsets of an n-set Q consisting of at most f (s) sets of size s. If b(s) satisfies

Z (s)ENT(s, b(s)) < 5'

then there is a partial coloring X with tX(S)I < b(jSj) for all S E A and fewer than half the points of Q uncolored.

In applying Corollary 2.2 we need upper bounds on ENT(n, b). The correct parameterization is b = A . Roughly Sn is like V/'iN where N is standard normal so that ENT(n, AV/;i) should be like g(A) = H(R\(N)). Analysis gives that g(A) = 8(A2e- 2/2) for A large (i = ?I giving the dominant terms) while g(A) = e(ln(A-1)) as A -* 0, the major contribution being pi = E)(A-1) for i = O(A-1). The following results are somewhat weaker and certainly not best possible but have the advantage of holding for all n,A.

Lemma 2.3. There is an absolute constant c so that ENT(n, AV/;i) < G(A) where we define

[ce-\2 /9 if A > 10, (4) G(A) c if 0.1 < A < 10,

cln(A-1) if A < 0.1.




Proof (outline). We employ the universal bound

Pr[S, ? r>v] < e72/2.

Set gi = exp(-A2(2i - 1)2/8), i > 1, and go = 1 - 2exp(-A2/8). From (2) pi,p-i < gi and po > go. On [0,1] the function -xlog2x increases to x = e-1 and then decreases. When A > 10, gi < e-1 for all i > 1 and go > e-1 so

00

ENT(n, An) <-go lg2 go + 2 E-gi lg2 gi i=1

This is a continuous function of A which is O(A2e- 2/8) or, giving ground, O(e- A2/9). When 0.1 < A < 10 set I = {-100, .. , +100}. The contribution to ENT(n, b) from i E I is at most log2 Ill <8. For i ~ I certainly gi <e-1 so

00

ENT(n, A6) < 8 + 2 , -gi 10g2 gi < 9- i=101

For A < 0.1 set I = {i: jil <A -20}. Again for i ? I we have gi < e-1 and

ENT(n, AV/;Y) < log2(2A-2 + 1) + 2 E -gi log2 gi < 40 ln(A-) lil >A-20

by computation. D1

We may now further reexpress Corollary 2.2.

Corollary 2.4. Let A be a family of subsets of an n-set Q consisting of at most f (s) sets of size s. If b(s) = a(s)\s where u(s) satisfies

(5) Z (s)G(a (s)) < ? S~~~~~

then there is a partial coloring X of Q with IX(A)l < b(jAI) for all A E A and at least half the points of Q colored. D

With these bounds we can already give a result which is interesting in its own right and may give significant insight into the somewhat technical computations to come.

Theorem 2.5. There is an absolute constant c so that the following holds for all n, s. If A1l . I An C Q, IQI = n and all IAi Ij s, then there is a partial coloring X of Q with less than half the points of Q uncolored and with

IX(Ai)l < cvs

for all 1 < i < n.

Proof. From Lemma 2.3 we may pick c so that G(c) < 0.2. Now apply Corollary 2.4. U

The monotonicity of G allows a further generalization of Corollary 2.4. Suppose A is a family of subsets of an n-set Q which breaks into subfamilies consisting of at most f(s) sets of size at most s. When (5) holds, the conclusion of Corollary 2.4 then holds. In particular, given any A1, ... , An C {1, ... , n} we have n sets of size at most n, and we may pick c so that G(c) < 0.2. Then there exists a partial coloring X of {1, . . ., n} with all IX(Ai)I < cfi- and at least half the points colored. This result was the core of [5].




3. THE FIRST PARTIAL COLORING

Let A denote the family of arithmetic progessions contained in Q = {1,.. . , n}. Here we show:

Lemma 3.1. There is a partial coloring of Q so that JX(A)J < cn1/4 for all A E A and at least half the points of Q are colored.

3.1. The decomposition. Let X = {xi,. . ., xl} be any set of integers with xl < * < xi. Define INT(X) to be the family of intervals i.e., all sets {xu,: i < u < j} where 1 < i < j K 1. Now define CINT(X) (the canonical intervals on X) by taking, for all powers of two, s = 2i < 1, all sets {x(j l)s+.1 . . , xjs} with js < 1. That is, we split X into consecutive intervals of length s = 2i, ignoring the "extra". The following observation is standard:

Lemma 3.2 (Decomposition lemma). Any A E INT(X) can be written as A = B\C with C c B and with B and C both decomposable into disjoint unions of sets in CINT(X) of different sizes.

Proof. With A = {xlt: i < u < j} set B {xu: 1 < u < j} and C = {xf,: 1 < u < i-1}t Take the binary expansion j 2bi + 2b2 +? , b, > b2 > ..., Of j Decompose B into the first 2bI elements of X union the next 2b2 elements of X,..., and do likewise with C. O

We can think of any arithmetic progression as a subinterval of an entire residue class so that

A= U U INT[{x E [n]: x i mod d}]. 1<d<in O<i<d

We define the "canonical arithmetic progressions"

(6) C=Cn= U U CINT[{xE[n]:x zi modd}]. 1<d<n O<i<d

Lemma 3.3. If X is a partial coloring of [n] so that

x(X) ? b(QXI) for all X E C, then

(7) x(A) < 2 E b(s) s;s=2i <n

for all XC A.

3.2. The coloring. For s = 2' < n how many s-sets are in Cn? We restrict 1 < d < n-I (otherwise the residue classes have fewer than s elements) and for - 8-1 each d the s-sets are disjoint so there are at most a of them, giving an upper bound of n n1 of them. For s = 1 there are only n distinct singletons. Ignoring s(s-1) asymptotically insignificant terms we'll say that Cn has at most n2s-2 sets of size s.

Remark. For s -'n we have - n sets of size s and Corollary 2.4 gives a partial coloring with I (A) I < cn1/4 for all such sets. We need to simultaneously color the larger and smaller sets. To avoid a logarithmic term in applying (7) we'll need a slightly better bound on JX(A)J when JAI is not near .




We parameterize s = T so that we have nT-2 sets of size s. We'll assume for convenience that Zn is a power of two so that T = 2i, i integral. We set

b(T\/ ) = vTv(Tx) where T) =cTlif T><l

We claim that, for an appropriately large constant c', (5) is now satisfied. We need to show

(8) E5T-2G(clT-l) + E T-2G(c'T-0.1) < ,r>l tr<l

where T in the sums runs over integral powers of 2 and G is given by (4). We will insist that c' > 1 so that G(c'y) < G(y). Both T-2G(T-1) = O(T-2 ln(T)) (T large) and T-2G(T-0.1) = O(T-2 exp(-_-0.2/9)) (T small) give convergent sums so we find an absolute constant T for which

E T-2G(T-1) + 5 T-2G(T-0 1) < 0.1. ,r>T r<T-1

As limxO,0 G(x) = 0 we may now select c' > 1 sufficiently large so that the finite sum

E T-2G(CT-0 1) < 0.1, T-1<r<T

yielding (8). Hence by Lemma 3.3 and Corollary 2.4 there is a partial coloring of [n] with at least half of the points colored and with

(9) Jx(A)I < 2 Z b(Tv?/;) < 2cn m/4 [ T 1-/2 + E To4 1T _1 i<1

for all A E A. As the bracketed sums both converge this gives Lemma 3.1.

4. NUMBER THEORY

Let X C {1, ... , n} be an m-element set. Let s be an integer, 1 < s < n. For an integer d, let U(d) denote the set of all x C X in residue classes modulo d for which at least s elements of X lie in that residue class. We are interested in the quantity

U= EIU(d)l. d

We can clearly restrict ourselves to the range 1 < d < n/s (for larger d, U(d) 0). Also, for each d, JU(d)l < m, and thus we get U < nm/s. This is tight for s = 1 but, for large enough s, the following theorem gives an improvement. The intuition behind it is that while for some individual value of d, the members of X can be distributed among very few residue classes modulo d only, such a distribution cannot occur for too many values of d at once.

Set p = m/n. We have

Proposition 4.1. Suppose that 5Vm < s < m. Then

U< nm ; Kc ap

for an absolute constant c.




Lemma 4.2. For any pair d, d' of distinct natural numbers, we have

JU(d)nU(d') U < S2 lcm(d, d')

Proof. A number x E U(d) n U(d') can be specified by giving the number r Lx/lcm(dd')] plus the residue classes of x modulo d and modulo d', by the Chinese Remainder Theorem. The number r can be chosen in at most Fn/lcm(d, d')] ways, and we note that U(d) may intersect at most JU(d)l/s residue classes modulo d, and similarly for U(d'). F1

Lemma 4.3. Let I C {1,2,...,n} be a set such that d > do for all d c I, and gcd(d, d') < M for all distinct d, d' c I. Suppose that d2/n < M < s2d2/(9mn). Then

S U(d)l< 2m. deI

Proof. If not, add indices to I one by one until the sum first gets over 2m. Stopping then would give a set I with the same assumptions where x = EdeI IU(d) satisfies 2m < x < 3m. We use Inclusion-Exclusion:

(10) m > U U(d) > E ? U(d)l- I U(d) n U(d') . deI deI d,d'eI,d<d'

By Lemma 4.2 and by the assumptions on I, we have

E U(d) n U(d') I < E IU(d)l I *U(d')l FnM1 d<d' d<d' 0

22~~~~~~ -<s ( U(d))

I nM +1)

The assumption on M implies nM/d2 > 1. Thus, from (10), we further get

x2 2nM 9m2nM m > x-22 d- > 2m- 2 > 2m-m = m 2s2 d 2 s2d~

(using the upper bound on M in the assumption of the lemma), a contradiction. D

Proof of Proposition 4.1. We may suppose that m, n, s, p- are all sufficiently large (otherwise the claim is satisfied trivially). We fix a parameter E = 5A/p. We let J be the interval

J= n,-n

(we may also suppose that Emn/s is an integer). We note that the d lying outside the interval J only contribute at most Emn/s = 5(nm/s)p to U. Hence it suffices to bound EdeJ U(d)l.

We want to partition the interval J into consecutive intervals I1, 12,... ,Ik, in such a way that Lemma 4.3 can be applied to each of them, giving the bound EdeI U(d)I < 2m. It remains to calculate how small k can be made. If we denote 1i = [d i, d?i+,), then we have gcd(d, d') < d?i+ I- di for any two distinct numbers




d, d' E hi. Thus, in order to apply Lemma 4.3, it is enough to have

sd? (12) di+, di :5 n '

9mn The upper bound (12) suggests we define the di's by the initial condition d1 = en/s and by the recurrence

di= di + L 2i2 One may check that with our choice of parameters, s2d2/(9mn) > 2, and therefore di+, ? di + s2d?/(18mn). We need to check the validity of (11), but this follows by calculation from the assumption s > 5A/1.

It remains to estimate the smallest k such that dk?1 > n/s. Set a = s2/(18mn). Then di+ I > di (1 + aedi). Given i, let us estimate the number j of steps needed so that di+j ? 2di. We have di+j ? di(1 + adi)i > di(jadi), so j ? 1/(2adi) suffices for the doubling. Therefore, the first doubling (from d1 to at least 2di) needs

1 9m = 2dia S

steps. Then the successive doubling times decrease geometrically, until the ra- tio of two successive members of the sequence of the di's exceeds 2. The number of remaining steps needed for reaching n/s after this happens is at most log2 ((n/s)/di) = 1og2 (1/E). Therefore k = O((n/s)Vp- + log(1/p)) = ((n1s)/), and EdEJ JU(d)l O((mn/s)fp-) as claimed. C]

Rernark. The set S {1,.,m} gives a value U = - p. Finding the maximal value of U U(n, rn, s) is an intriguing problem we do not pursue here but we conjecture that our Proposition 4.1 is not best possible.

5. THE END OF THE HUNT

Let X C [n], IXI = m= pn with n-3/4 < p ? 1. Our object is to find a partial coloring X of X so that I(X n A) I is small for all A E A and at least half the points of X are colored. Once successful, we'll apply this process iteratively beginning with X = [n] (which we did in ?3), resetting X to be the uncolored points, until

XI < n'/4 at which time the remaining points may be colored arbitrarily. Following (6) set

C=Cx= U U CINT[{x E X: x-i mod d}].

1<d<n O<i<d

For any A c A we may, as in ?3.1, decompose A n x = B\C with C C B and B, C both disjoint unions of sets of Cx of different cardinalities. Lemma 3.3 now generalizes.

Lemma 5.1. If X is a partial coloring of X so that JX(Y)l ? b(|Y|) for all Y E Cx, then

JX(A n X) I < 2 E b(s) s=2i

for all A E A.




Let f (m, s) denote the number of s-sets in Cx so that f (m, s) < s1U with U as in ?4. We first apply the elementary bound

f(m,s) < mn

Lemma 5.2. There is a partial coloring X of X with

Jx(A n X)I < cnl/4

for all A E A and with more than half the points of X colored.

Proof. We follow the proof of Lemma 3.1 precisely. For s = Tfi- we set

b(TV ) = V1TV?(Tx) with a(TVr) = {cT if r >1

Again we need (8) and the bound (9) is the same. [1

We iterate this result, beginning at X = [n], resetting X to be the uncolored points at each iteration, stopping when IXI < pon, with po a sufficiently small (as determined later) absolute constant. This is a constant number of iterations (recall the number of uncolored points is at least halved at each iteration) so together we have a partial coloring X with JX(A)I < cnl/4 for all A c A and a set X of fewer than pon points uncolored.

Remark. Continuing this process until IXI < n1/4 and then coloring the remaining points arbitrarily would give a full coloring with all IX(A) I < cn1/4 ln n. Our "slight" improvement of ?4 will allow a slight improvement as X becomes smaller so that the sum converges to 0(nl/4).

Now fix X with IXI = pn, n-3/4 < p < po. We set b(,Txj) = TxJa(,Txj)

with

(r-0.1 ifT 7< p/5,

(13) O(TV?) p if p1/5 < T <

Tr-Ip if I < r,

and set

fmT 2 if'T< p 1/5

f(Tji) = cmT-2 V/ if T> pl/5

which, by a slight weakening of Proposition 4.1, is an upper bound on the number of Trxj-sets in Cx.

We first claim that for p appropriately small

(14) m ,1 S f (T,T)G((TV)) <0.2 T

(we recall the convention from ?3-,T in summation runs through integral powers




of 2). We split the sum by the ranges of (13). As functions of p

E T G(T0?1) = O(p-2/5eP 1/30) T<pl/5

E CT-2 fpG(p) = 0(pl/2-2/5 In2 (p-1)) pl/5<T<1

SCTf-2VG(T-1p) < C Ew-2v/p(lnT + ln(p-1)) = 0(V/ ln(p-1)) 1<T 1<T

which are all o(p) so that (14) holds when po is picked sufficiently small. We bound

2n-4 E b(Tni) < 5 T04+ 5 T1 2p+EST1/2p T T<p1/5 P1/5<T<1 1<r

The first sum dominates and this is 0(p4/25) as p -* 0. We have shown:

Lemma 5.3. There are absolute positive constants po, c so that if IXI = pn, p < po, then there exists a partial coloring X so that IX(A n X)I < cn1/4p4/25 for all A c A and with at least half the points of X colored.

The exponent 4 clearly could be improved by more careful calculation but it does not matter. We are done. Begin with X = [n]. Apply Lemma 3.1 and then Lemma 5.1 until {XI < pon, then apply Lemma 5.3 until IXI < n1/4 and then color the remaining points arbitrarily. The final coloring X has

00

Ix(A)I < cn1/4 + E cInl/4(po2-i)4/25 + n1/4 < c*nl/4 i=o

for all A C A and has no points uncolored.

REFERENCES

1. N. Alon and J. Spencer, The probabilistic method, Wiley, New York, 1992. MR 93h:60002 2. J. Beck, Roth's estimate on the discrepancy of integer sequences is nearly sharp, Combina-

torica 1 (1981), 319-325. MR 83i:05040 3. P. Erd8s and J. Spencer, Probabilistic methods in combinatorics, Academic Press, New York,

1974. MR 52:2895 4. D. Kleitman, On a combinatorial problem of Erdo5s, J. Combin. Theory 1 (1966), 209-214.

MR 34:78 5. J. Spencer, Six standard deviations suffice, Trans. Amer. Math. Soc. 289 (1985), 679-706.

MR 86k:05004 6. J. Matousek, Tight upper bounds for the discrepancy of halfspaces, Discrete Comput. Geom.

(to appear). 7. K. F. Roth, Remark concerning integer sequences, Acta Arith. 9 (1964), 257-260. MR 29:5806

ABSTRACT. It is proven that there is a two-coloring of the first n integers for which all arithmetic progressions have discrepancy less than const.rn1/4. This shows that a 1964 result of K. F. Roth is, up to constants, best possible.

DEPARTMENT OF APPLIED MATHEMATICS, CHARLES UNIVERSITY, MALOSTRANSKE NAM. 25, 118 00 PRAHA 1, CZECH REPUBLIC

E-mail address: [email protected]. cuni. cz

COURANT INSTITUTE OF MATHEMATICAL SCIENCES, 251 MERCER STREET, NEW YORK, NEW

YORK 10012 E-mail address: spencerIcs.nyu.edu



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Discrepancy in Arithmetic Progressions