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APPEARED IN BULLETIN OF THE

arXiv:math/9304211v1 [math.NT] 1 Apr 1993APPEARED IN BULLETIN OF THEAMERICAN MATHEMATICAL SOCIETYVolume 28, Number 2, April 1993, Pages 329-333ON THE DISTRIBUTION OF SUMS OF RESIDUESJerrold R. GriggsAbstract. We generalize and solve the mod q analogue of a problem of Littlewoodand Offord, raised by Vaughan and Wooley, concerning the distribution of the 2nsums of the form Pni=1 εiai, where each εi is 0 or 1.

For all q, n, k we determinethe maximum, over all reduced residues ai and all sets P consisting of k arbitraryresidues, of the number of these sums that belong to P .1. IntroductionVaughan and Wooley [15] raised the problem of determining the maximum num-ber of the 2n sums of the form Pni=1 εiai, where each εi is 0 or 1, that are congruentto 0 mod q.

The maximum is over all residues a1, . .

. , an that are reduced, whichmeans that (ai, q) = 1 for all i.

Results about this problem have been applied tostudy the solutions of simultaneous additive equations.By using analytical tools, including exponential sums and classical inequalities,and by treating many cases, Vaughan and Wooley show that the maximum isn⌊n/2⌋provided that q > ⌈n/2⌉. This bound is sharp, since it is attained byletting ai be 1 for i ≤⌈n/2⌉and −1 for i > ⌈n/2⌉.

(To see this, observe that forthis choice of ai’s, we have Pni=1 εiai ≡0 precisely when an equal number of εi’sare 1 for i ≤⌈n/2⌉and −1 for i > ⌈n/2⌉. This happens if and only if the numberof indices i with i ≤⌈n/2⌉and εi = 1 plus the number with i > ⌈n/2⌉and εi = 0is ⌊n/2⌋, so that the choices correspond to the subsets of {1, .

. ., n} of size ⌊n/2⌋.

)When ⌈n/2⌉≥q, wraparound effectsmod q come into play. For example, withthe ai’s chosen as above, the sum Pqi=1 ai is also congruent to 0, so the answerexceedsn⌊n/2⌋.We solve the problem for arbitrary n and q, using an inductive argument that isinspired by the study of the extremal properties of the Boolean lattice Bn on thecollection 2[n] of all subsets of the n-set [n] = {1, .

. .

, n}, ordered by inclusion. Letus adopt the notationnsq:= |{A ⊆[n]: |A| ≡s}| =Xj≡sjs,for the mod q binomial coefficients in n. We shall see that for general n and q, themaximum number of sums congruent to 0 is the middlemod q binomial coefficient1991 Mathematics Subject Classification.

Primary 11P83; Secondary 11A07, 05A05, 06A07.Received by the editors September 2, 1992Research supported in part by NSA/MSP Grant MDA90-H-4028 and by a Visiting Professor-ship at Simon Fraser Universityc⃝1993 American Mathematical Society0273-0979/93 $1.00 + $.25 per page1

2J. R. GRIGGSn⌊n/2⌋q.

The maximum is attained as before by dividing the ai’s as evenly aspossible between 1 and −1. In general, the maximum number of sums congruent toany single residue isn⌊n/2⌋q.

Throughout the paper we maintain the conditionthat the residues ai be reduced. Without this restriction, one would select ai’s withcommon factors, in order to increase the number of sums congruent to 0.This problem we are considering is the analogue for residues of a famous prob-lem about the clustering of partial sums of a collection of complex numbers.

Inconnection with their study of roots of random polynomials, Littlewood and Offord[13] were led to consider the following question. For a1, .

. .

, an ∈C with ∥ai∥≥1for all i and for an open ball S ⊂C of unit diameter, how many of the 2n sumsof the form Pni=1 εiai, where each εi is 0 or 1, can belong to S? They sought themaximum over all choices of ai’s and S. In particular, if one selects ai to be 1 for alli and centers S at ⌊n/2⌋, one can packn⌊n/2⌋sums into S, and this was believedto be optimal.

While Erd¨os [3] soon proved this for the real ai case, it was twentyyears before the original complex case was solved, by Katona [8] and Kleitman [10]independently, from an appropriate extension of the theorem of Sperner [14] aboutthe maximum size of an antichain in the Boolean lattice.Although the usual Sperner method does not extend to higher dimensions, Kleit-man [11, also in 5] found a remarkable proof that the answer is stilln⌊n/2⌋in anydimension m (or indeed, in Hilbert space): For any vectors a1, . .

. , an ∈Rm oflength at least one, there is a partition of 2[n] into justn⌊n/2⌋blocks, such thatfor any sets I, J in the same block, the sums Pi∈I ai and Pj∈J aj are far apart(distance at least one).

Hence any open ball S of unit diameter contains at mostn⌊n/2⌋sums.The idea behind this construction is that for every n, the sizesof the blocks partitioning 2[n] exactly match the sizes of the chains in the famousinductive symmetric chain decomposition of Bn discovered by de Bruijn et al. [2].Erd¨os [3] considered the more general problem of maximizing the number ofsums of vectors inside an open ball in Rm of diameter d ≥1.He solved thisproblem for the real case (m = 1) using Sperner theory, and he found that thevalue attained when all ai = 1 is optimal for all d. This value is the sum of the ⌈d⌉middle binomial coefficients, P(n−⌈d⌉)/2≤j<(n+⌈d⌉)/2nj.

However, the problem ismore complicated when m ≥2 and completely solved only in some special cases.Using a variety of tools from extremal set theory, probability, and geometry, manyauthors have attacked this more general question, including Kleitman [12], Griggs[6], and Frankl and F¨uredi [4]. Also see the survey by Anderson [1].In marked contrast to previous results of the Littlewood-Offord type, the set-ting for the work of Vaughan and Wooley is the additive group Zq of integersmod q.

Nonetheless, as with the unit diameter problem above, we shall see thattheir theorem can be obtained by an inductive partition construction inspired bya particular chain partition of the Boolean lattice. The method yields the solutionto the extension of their problem to general n and q.More generally, we determine the maximum number of the 2n sums Pni=1 εiaicongruentmod q to any of k arbitrary residues ρj, for 1 ≤j ≤k, over all choicesof the residues ρj and the reduced residues ai.

The answer is the sum of the kmiddlemod q binomial coefficients in n. This bound is attained by selecting all

SUMS OF RESIDUES3ai to be 1 and selecting the k middle values for the residues ρj. Switching some aiin this solution to −1 has the effect of shifting the collection of all 2n sums downby 1.

Thus the bound is also attained by selecting ai to be 1 for i ≤⌈n/2⌉and −1for i > ⌈n/2⌉and by choosing the k initial values in the sequence 0, 1, −1, 2, −2, . .

.for the residues ρj.2. The main resultWe fix the integer q > 0 and work in Zq.Theorem 1.

Let a1, . .

. , an be reduced residues in Zq.

Let P ⊆Zq, where |P| = k.Then the number of the 2n sums Pni=1 εiai in P, where each εi is 0 or 1, is at mostthe sum of the k middlemod q binomial coefficients P(n−k)/2≤j<(n+k)/2njq, andthis bound is best possible.Proof. For S ⊆Zq and a ∈Zq, let S + a := {s + a: s ∈S} ⊆Zq.

For A ⊆2[n],define the sum setS(A) =(Xi∈Iai (mod q): I ⊂A).We say that A ⊆2[n] is a structure for a1, . .

. , an provided that the sums in S(A)are distinct.We shall partition 2[n] inton⌊n/2⌋q structures in such a way that the bound inthe theorem will follow for all k. The construction is carried out by induction on nfor a given sequence of reduced residues a1, a2, .

. .

. It starts at n = 0 with the singlestructure {∅}.

For the induction step, suppose we are given a partition of 2[n−1] intostructures Aj for a1, . .

. , an−1.

Then the structures Aj and A′j := {I∪{n}: I ∈Aj}for a1, . .

. , an partition 2[n], but they are not quite the ones we want.

Notice thatS(A′j) = S(Aj) + an. We require an easy fact.Lemma.

Let ∅̸= S ⊆Zq and a ∈Zq with (a, q) = 1. Then S + a = S if and onlyif S = Zq.If S(Aj) is Zq, then so is S(A′j), and we leave both structures alone.

However, ifS(Aj) ̸= Zq, then by the lemma there exists at least one element t ∈S(A′j)\S(Aj),say t = Pi∈I ai where I ∈A′j, so that we may replace Aj and A′j by the structuresBj = Aj ∪{I} and B′j = A′j\{I}. We have |Bj| = |Aj| + 1 and |B′j| = |Aj| −1.

Inthe case where |Aj| = 1, we discard B′j.Now denote the structures in this partition of 2[n] by Aj for j = 1, 2, . .

. .

Sincesets in a structure have distinct sums, it follows that(1)I ⊆[n]:Xi∈Iai ∈P ≤Xjmin(k, |Aj|) .It suffices to show that the sum on the right-hand side of inequality (1) is at mostthe sum of the k middle mod q binomial coefficients in n.Since the collection of structure sizes |Aj| depends in no way on the actual valuesof the ai’s, it is enough to consider the case where all ai = 1. One can verify byinduction on n that the sum set S(Aj) for each structure consists of all q residuesor else consists of values congruent to an interval x, x + 1, .

. .

, y ∈Z centered

4J. R. GRIGGSabout n/2, which means x + y = n. The number of structures in the partitionisn⌊n/2⌋q, because every structure contains a set with sum (i.e., cardinality)≡⌊n/2⌋.

For general k, we see that the sum on the right-hand side of (1) is thenumber of subsets of cardinality congruent to any of the k middle values aroundn/2.□When q > ⌈n/2⌉, we have thatn⌊n/2⌋q =n⌊n/2⌋, which implies the originalresult of Vaughan and Wooley [15]:Corollary 1. Let a1, .

. .

, an be reduced residues in Zq, where q > ⌈n/2⌉. Then thenumber of the 2n sums Pni=1 εiai congruent to 0, where each εi is 0 or 1, is at mostn⌊n/2⌋, and this bound is best possible.3.

Related remarksThe inspiration for the proof of the theorem is the inductive partition of theBoolean lattice Bn into saturated chains of size at most q, that is, into collectionsof at most q totally ordered subsets of consecutive sizes.Katona [9] used thisconstruction to determine the maximum number of subsets of {1, . .

. , n} containingno sets A ⊂B with 0 < |B\A| < q.

The author [7] later independently devisedthe same construction to obtain a maximum-sized collection of disjoint saturatedchains of size q in Bn. The collection of structure sizes |Aj| in our constructionexactly corresponds to the collection of chain sizes in Katona’s partition.By applying the theorem with k = q −1, it is also possible to determine theminimum number of sums in any residue class.Corollary 2.

Let a1, . .

. , an be reduced residues in Zq, where n ≥q−1.

Let ρ ∈Zq.Then the number of the 2n sums Pni=1 εiai congruent to ρ, where each εi is 0 or 1,is at leastn⌈(n−q)/2⌉q, and this bound is best possible.The bound in Corollary 2 is attained by taking all ai = 1 and ρ ≡⌈(n −q)/2⌉.For n < q −1, no sums are congruent to −1 when all ai equal 1. The asymptoticgrowth of the mod q binomial coefficients, studied in connection with saturatedchain partitions [7], implies that the lower bound in Corollary 2 approaches 2n/qas n →∞with fixed q.

(This remains true even if q grows with n, provided thatq = o(n1/2).) Hence, for any sequence {a1, a2, .

. . } of reduced residues mod q, thedistribution of the mod q sums of the first n residues is asymptotically uniform asn →∞.The Littlewood-Offord problem has an equivalent formulation that considers theconcentration of sums of the form Pni=1 δiai with each δi = 1 or −1, where as before∥ai∥≥1 for all i.

The analogous problem in Zq can be solved by a reduction tothe original problem of Theorem 1.Corollary 3. Let a1, .

. .

, an be reduced residues in Zq. Let P ⊆Zq, where |P| = k.Then the number of the 2n sums Pni=1 δiai in P, where each δi is 1 or −1, is atmost the sum of the k middle mod r binomial coefficients in n, where r is q whenq is odd and q/2 when q is even, and this bound is best possible.

SUMS OF RESIDUES5AcknowledgmentThe author is grateful to Oren Patashnik and Ted Sweetser for many suggestionsthat greatly improved the presentation of this paper.References1. I. Anderson, Combinatorics of finite sets, Clarendon Press, Oxford, 1987.2.

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Oxford (2) 42 (1991), 379–386.Department of Mathematics, University of South Carolina, Columbia, South Car-olina 29208E-mail address: griggs@math.scarolina.edu

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