Discrepancy of Sums of Three Arithmetic Progressions AleˇsPˇr´ıvˇetiv´y ∗ Department of Applied Mathematics of Charles University Malostransk´en´am. 25, 11800 Praha, Czech Republic privetivy@kam.mff.cuni.cz Submitted: Jul 24, 2005; Accepted: Dec 5, 2005; Published: Jan 25, 2006 Mathematics Subject Classification: 11K38 Abstract The set system of all arithmetic progressions on [n] is known to have a discrep- ancy of order n 1/4 . We investigate the discrepancy for the set system S 3 n formed by all sums of three arithmetic progressions on [n] and show that the discrepancy of S 3 n is bounded below by Ω(n 1/2 ). Thus S 3 n is one of the few explicit examples of systems with polynomially many sets and a discrepancy this high. 1 Introduction Let (X, F) be a set system on a finite set. The discrepancy problem is to color each point of X either red or blue, in such a way that any of the sets of F has roughly the same number of red points and blue points. The maximum deviation from an even splitting, over all sets of F, is the discrepancy of F, denoted by disc(F). Formally disc(F)= min χ:X→{−1,1} max S∈F     x∈S χ(x)    . For further information see Beck and S´os [BS95], Chazelle [Cha00], and Matouˇsek [Mat99]. Let n be a positive integer and let [n]denotetheset{0, 1, ,n−1}. For any a ∈ Z and d 1 ,n 1 ∈ N we define the arithmetic progression AP (a, d 1 ,n 1 )astheset{a+id 1 : i ∈ [n 1 ]}. The set system formed by all arithmetic progressions on [n]wedenoteby([n], S n )where S n = {AP (a, d 1 ,n 1 ) ∩ [n]:a, d 1 ,n 1 ∈ N}. The lower bound Ω(n 1/4 ) on the discrepancy of arithmetic progressions S n proved by Roth [Rot64] was one of the early results in combinatorial discrepancy. In 1974, S´ark˝ozy (see [ES74]) established an O(n 1/3+ ) upper bound. This was improved by Beck [Bec81], ∗ Supported by Czech Science Foundation grant 201/05/H014. the electronic journal of combinatorics 13 (2006), #R5 1 who obtained the near-tight upper bound of O(n 1/4 log 5/2 n), inventing the powerful partial coloring method for that purpose. The asymptotically tight upper bound O(n 1/4 )was finally proved by Matouˇsek and Spencer [MS96]. Discrepancies of related set systems were also studied. One possible extension of the original problem is to consider set systems formed by sums of arithmetic progressions, where a sum of k arithmetic progressions AP k (a, d 1 , ,d k ,n 1 , ,n k ) is defined for a ∈ Z and d 1 , ,d k ,n 1 , ,n k ∈ N as the set {a + i 1 d 1 + + i k d k : i l ∈ [n l ],l =1, ,k}. The corresponding set system of all sums of k arithmetic progressions on [n]isthen ([n], S k n ), where S k n = {AP k (a, d 1 , ,d k ,n 1 , ,n k ) ∩[n]:a ∈ Z; d 1 , ,d k ,n 1 , ,n k ∈ N}. Hebbinghaus [Heb04] proved that disc(S k n )=Ω(n k 2k+2 ). Here we show that for k ≥ 3, disc(S k n )=Ω(n 1/2 ). Thus S 3 n is one of the few explicit examples of systems with polynomially many sets and a discrepancy this high. For a fixed k ≥ 3, the lower bound on S k n is nearly tight since the random coloring lemma [AS92] provides the upper bound O(n 1/2 log 1/2 n). In the case k =2,thereis still a considerable gap, Ω(n 1/3 )versusO(n 1/2 log 1/2 n), and estimating the correct bound remains still open. We start in Section 2 with recalling the eigenvalue bound method and then we show how it can be used for wrapped set systems. In Section 3 we discuss how to construct suitable wrapped set systems and illustrate this approach on the system of arithmetic progressions on [n] (this version of proof is attributed to Lov´asz). Then we construct a wrapped set system for our main result. 2 Preliminaries In this section we recall some basic facts. We start with some definitions from discrepancy theory; for more definitions see [Mat99]. Let (X, F) be a set system on a finite set. Let us enumerate the elements of X as x 1 ,x 2 , ,x n and the sets of F as S 1 ,S 2 , ,S m in some arbitrary order. The incidence matrix of (X, F)isthem × n matrix A, with columns corresponding to points of X and rows corresponding to sets of F, whose element a ij is given by a ij =  1ifj ∈ S i 0 otherwise. As we will see, it is useful to reformulate the definition of the discrepancy of F in terms of the incidence matrix. Now let us regard a coloring χ : X →{−1, +1} as the column vector (χ(x 1 ),χ(x 2 ), ,χ(x n )) T ∈ R n . Then the product Aχ is the row vector (χ(S 1 ),χ(S 2 ), ,χ(S n )) ∈ R m , where we extend the coloring χ for sets as χ(S)=  x∈S χ(x). Therefore, the definition of the discrepancy of F can be written as disc(F)= min x∈{−1,1} n Ax ∞ . the electronic journal of combinatorics 13 (2006), #R5 2 For many lower bound techniques, it is easier to consider the L 2 -discrepancy instead of the worst-case discrepancy. In our case, this means replacing the max-norm . ∞ by the usual Euclidean norm .. Namely, we have disc(F) ≥ disc 2 (F)=min χ  1 m m  i=1 χ(S i ) 2  1/2 = 1 √ m · min x∈{−1,1} n Ax. To obtain a lower bound on the L 2 -discrepancy for a set system, we can use the following eigenvalue lower bound: Theorem 2.1 (Eigenvalue bound, see [BS95]) Let (X, F) be a system of m sets on an n-point set, and let A denote its incidence matrix. Then we have disc(F) ≥ disc 2 (F) ≥  n m · λ min , where λ min denotes the smallest eigenvalue of the n ×n matrix A T A. The computation of eigenvalues becomes much easier when the matrix A T A is a circu- lant matrix. A circulant matrix is an n ×n matrix whose rows are composed of cyclically shifted copies of the first row. Namely, for an n-dimensional vector (a 0 ,a 1 , ,a n−1 )we define the n × n circulant matrix C(a 0 ,a 1 , ,a n−1 ) by putting c ij = a (j−i)modn , i.e. C(a 0 ,a 1 , ,a n−1 )=       a 0 a 1 a 2 a n−1 a 1 a 2 a 3 a 0 a 2 a 3 a 4 a 1 ···· · a n−1 a 0 a 1 a n−2       . Let ζ 0 ,ζ 1 , ,ζ n−1 denote the n-th roots of the unity, which are defined as roots of the cyclotomic equation x n = 1. All the roots lie on the unit circle and we can order them according to the sequence of visiting them if we go around the unit circle counterclockwise starting at 0, namely we put ζ k = e 2πi n k . This simplifies the following operations: • ζ j ζ k = ζ (j+k)modn • ζ k j = ζ (jk)modn For convenience, we will consider all operations +,.on indices reduced modulo n and thus we will later omit the mod n suffix. We define the complex argument as usual by arg(x + iy) = arctan( y x ) and restrict its range to the interval (−π, +π]. The complex argument of the n-th root of unity is then as follows arg(ζ k )=  2πk n if 0 ≤ k ≤ n 2 2π(k− n) n if n 2 <k<n. the electronic journal of combinatorics 13 (2006), #R5 3 Let B be a circulant matrix C(a 0 ,a 1 , ,a n−1 ). It can be easily verified that z i = (1,ζ 1 i ,ζ 2 i , ,ζ n−1 i ) T is an eigenvector of B and thus the eigenvalues λ 0 ,λ 1 , ,λ n−1 of B are λ i = a 0 + a 1 ζ i + a 2 ζ 2 i + + a n−1 ζ n−1 i . Let A be an incidence matrix for the set system (X, F)andletB denote A T A.Then the element b ij counts the number of sets S i ∈Fcontaining both elements x i and x j . Moreover, if the matrix B is a circulant matrix C(a 0 ,a 1 , ,a n−1 ), we can derive a more useful expression for the eigenvalues of B: nλ k = n n−1  i=0 a i ζ i k = n−1  i=0 n−1  j=0 a (i−j)modn ζ (i−j)modn k = n−1  i=0 n−1  j=0 b ij ζ (i−j) k = =  S∈F  x i ∈S  x j ∈S ζ (i−j) k =  S∈F     x i ∈S ζ i k    2 . And thus λ k = 1 n  S∈F     x i ∈S ζ i k    2 . Let (X, F) be a set system, where X =[n]andF contains exactly mn sets enumerated as F = {S 0 ,S 1 , ,S mn−1 }. We say that a set system (X, F)iswrapped if for every i ∈ [m] and j ∈ [n]thesetS in+j is the set S in cyclically translated by j, i.e. S in+j = {(k + j)modn : k ∈ S i }. The incidence matrix A of a wrapped set system (X, F)iscomposedofm square n×n circulant matrices A 0 ,A 1 , ,A m−1 stacked up vertically, one on top of the other: A =      A 0 A 1 . . . A m−1      . By the definition of a wrapped set system every A i is a circulant and thus every A T i A i is a circulant too. Note that although A is not a circulant itself, the matrix B = A T A is equal to  m−1 i=0 A T i A i and therefore B is a circulant. Alternatively, we can observe that the (i, j) entry of A T A is the number of sets from F that contain both elements i and j. Since the sets forming F are invariant under cyclic shifts, the entries (i, j)and(i + k, j + k)ofA T A are the same for an arbitrary shift by k and thus A T A is a circulant. the electronic journal of combinatorics 13 (2006), #R5 4 Lemma 2.2 Let (X, F) be a wrapped set system, where |X| = n and |F| = mn, and let A be its incidence matrix. Then the n × n matrix B = A T A is a circulant and its eigenvalues a re λ k = m−1  i=0     j∈S in ζ j k    2 . Proof. Since for each set S in ,wehaveitsn − 1 translates in F that give the same contribution, we may just count n-times the contribution of the set S in and thus λ j = 1 n  S∈F     k∈S ζ k j    2 = m−1  i=0     k∈S in ζ k j    2 .  3 Lower b ounds In this section we will prove the lower bound for the sums of three arithmetic progressions. For this purpose we will use following lemma: Lemma 3.1 Let ([n], F) be a wrapped set system, where |F| = mn, and let ζ 0 ,ζ 1 , ,ζ n−1 be the n-th roots of unity. If there are real constants c, α > 0 such that for each j ∈ [n] there is S in ∈Fsuch that     k∈S in ζ k j    ≥ cn α holds, then disc(F) ≥ cn α √ m . Proof. To invoke the eigenvalue bound for an L 2 -discrepancy we need to lowerbound the value of smallest eigenvalue λ min . Since our set system is wrapped, we know that all eigenvalues are given by the expression λ j = m−1  i=0     k∈S in ζ k j    2 . We know that for each j there is a set S in that makes the eigenvalue ‘large’ and hence for every eigenvalue we know that λ j = m−1  i=0     k∈S in ζ k j    2 ≥     k∈S in ζ k j    2 ≥ c 2 n 2α . Thus disc(F) ≥ disc 2 (F) ≥  n mn c 2 n 2α = cn α √ m .  the electronic journal of combinatorics 13 (2006), #R5 5 If we want to obtain a good lower bound from Lemma 3.1, the number of sets forming F has to be small and F has to contain for each j ∈ [n]asetB j , such that    k∈B j ζ k j   is large. Our goal is to ensure that for each j ∈ [n]thereisaB j ∈Fsuch that all ζ k j for k ∈ B j are concentrated in one part of the unit circle. Namely, if all k ∈ B j satisfy − π 3 ≤ arg ζ k j ≤ π 3 , then Re ζ k j ≥ 1 2 for all k ∈ B j , and the value of     k∈B j ζ k j    will be at least |B j |/2. 0 0 n −1 n −1 B j B  j = {jk mod n : k ∈ B j } k → jk mod n 2 3 π 0  n 6   5 6 n k → ζ k j k → ζ k Figure 1: The relation of the set B  j and the sum  k∈B j ζ k j For convenience, we define for every B j ⊆ [n]asetB  j as the set {jk mod n : k ∈ B j }. The set B  j is actually the set of indices i of ζ i = ζ k j that participate in the sum  k∈B j ζ k j (see figure 1). The condition |arg ζ k |≤ π 3 for all k ∈ B j is thus equivalent to the condition B  j ⊆  0, ,  1 6 n  ∪  5 6 n  , ,n− 1  . Moreover, if n is a prime and 0 <j<n, the mapping k → jk mod n is a bijection and the cardinalities of B j and B  j are the same. Now let us apply this method to prove the Ω(n 1/4 ) lower bound for the set system of arithmetic progressions on [n]. For this purpose we construct a small auxiliary wrapped set system F that is suitable for Lemma 3.1 and disc(S n ) is asymptotically bounded below by disc(F). We will show, that for each j ∈ [n] we can find a positive integer d j = O( √ n), such that |arg ζ d j j | = O(n −1/2 ). Let us take as B j an arithmetic progression with difference d j having Ω( √ n) elements, such that |arg ζ k j |≤ π 3 for all k ∈ B j .ForsuchaB j ,weget     k∈B j ζ k j    =Ω( √ n). Since there are only O( √ n) possible choices of d j , it suffices us to put only O( √ n) different sets B j into F. With each inserted B j ,wealsohavetoput into F its n −1 wrapped translates, and thus F has size O(n 3/2 ). The following theorem the electronic journal of combinatorics 13 (2006), #R5 6 summarizes our discussion. This version of the proof of the lower bound for S n was first suggested by Lov´asz and can be found in [BS95]. Theorem 3.2 For any n ∈ N we put k =   n/6 and m =6k. Let us consider the following set system ([n], F), where F = {S 0 ,S 1 , ,S mn−1 } and the sets S dn+j for d ∈ [m] and j ∈ [n] are given as S dn+j = {(di + j)modn : i ∈ [k]}. Then disc(F) ≥ cn 1/4 . Proof.Forafixedj ∈ [n], there is by the Pigeonhole Principle a positive integer c 0 , 1 ≤ c 0 ≤ m such that − 2π m ≤ arg(ζ c 0 j ) ≤ 2π m . Then Re ζ ic 0 j ≥ 1/2 for 0 ≤ i ≤ k −1, and hence     i∈S c 0 n ζ i j    ≥  Re  i∈S c 0 n ζ i j  ≥ k/2. From this and Lemma 3.1 it immediately follows that disc(F) > 1 10 n 1/4 .  Since every S ∈Ffrom theorem 3.2 is a disjoint union of two arithmetic progressions, we get the following corollary. Corollary 3.3 For n ∈ N,let([n], S n ) be a set system formed by all arithmetic progres- sions on [n]. Then disc(S n )=Ω(n 1/4 ). For the set system ([n], S n )istheΩ(n 1/4 ) lower bound tight. We would like to show that the set systems ([n], S k n ) for k ≥ 2 have their discrepancy bounded below by Ω(n 1/2 ). Unfortunately, we are able to prove this only for k ≥ 3, while for k = 2 the currently known best lower bound is Ω(n 1/3 ); see [Heb04]. As we have seen in the proof of Theorem 3.2, for each j ∈{1, ,n−1} we can find two positive integers 0 <c 1 ≤ √ n and 0 <d 1 ≤ √ n, such that |arg ζ c 1 j | = 2πd 1 n . Without loss of generality, let us assume that arg ζ c 1 j is positive and thus d 1 = c 1 j mod n, the other case can be handled in the same way. Let A j be the set {ic 1 : i ∈ [n 1 ]}.Ifn 1 ≤ min{ n c 1 , n 6d 1 }, then A j is an arithmetic progression on [n]andA  j is an arithmetic progression on [n/6] (see figure 2). Although is n 1 is at least Ω( √ n), we cannot generally expect a greater value. the electronic journal of combinatorics 13 (2006), #R5 7 0 0 k → jk mod n n −1 n −1 A j A  j Figure 2: A j with difference c 1 goes to A  j with difference d 1 Our goal is to find a B j such that B  j covers a constant fraction of {0, , 1 6 n} ∪ { 5 6 n, ,n− 1}. In next two steps we will schematically (and possibly misleadingly) show how to achieve this. In the first step we extend the arithmetic progression A  j to a longer arithmetic progression B  j with the same difference. This is done in such a way that B  j consists of several copies of A  j and thus B j is taken as a sum of two arithmetic progressions (see figure 3). In this way we can have Ω(n/d 1 )elementsinB j . 0 n −1 A j A  j 0 n −1 A  j + d 2 A  j +2d 2 A j + c 2 A j +2c 2 k → jk mod n Figure 3: B j (resp. B  j ) composed from copies of A j (resp. A  j ) In the last step we take a suitable sum of three arithmetic progressions for C j such that C  j is composed of Ω(d 1 ) interlaced copies of B  j that are mutually disjoint (see figure 4), and thus C  j has Ω(n)elements. B j B  j B  j + d 3 B j + c 3 0 n −1 0 n −1 k →jk mod n Figure 4: C  j is composed from interlaced copies of B  j the electronic journal of combinatorics 13 (2006), #R5 8 The following lemma provides, for each j ∈{1, ,n−1}, a precise and more careful construction of the set C j . This construction requires n to be a prime. Lemma 3.4 Let n be a prime. For each j ∈{1 n−1} there exists a set C j such that • C j is a sum of three a rithmetic progressions on [n] • Re ζ k j ≥ 1/2 for every k ∈ C j •|C j |≥ 1 5000 n. Proof.Forafixedj we find integer constants c 1 ,c 2 ,c 3 ,d 1 ,d 2 ,d 3 ,n 1 ,n 2 and n 3 as follows: 1. Let c 1 be the k ∈{1  √ n} for which the value Re ζ k j is maximum. We put d 1 =min{jc 1 mod n, −jc 1 mod n} and n 1 =  n 12 max{c 1 ,d 1 } . 2. If c 1 ≤ 12d 1 , then we put c 2 =1,d 2 =1andn 2 = 1, otherwise we put c 2 = n mod c 1 , d 2 = d 1  n c 1  and n 2 =  c 1 30d 1 . 3. If d 1 < 6, then we put c 3 =1,d 3 =1andn 3 = 1, otherwise we put c 3 to be the k ∈{1  2n d 1 } for which the value Re ζ k j is maximum. We put d 3 = min{jc 3 mod n, −jc 3 mod n} and n 3 =  d 1 12 . 4. We put C j = {i 1 c 1 + i 2 c 2 + i 3 c 3 : i k ∈ [n k ],k =1, 2, 3}. We have chosen c 1 as the k ∈{1  √ n} for which the value of Re ζ k j is maximum, i.e. as the k ∈{1  √ n} for which the value of |arg ζ k j | is minimum. By the Pigeonhole Principle − 2π  √ n ≤ arg ζ c 1 j ≤ 2π  √ n , and since |arg ζ c 1 j | =argζ d 1 = 2πd 1 n , we conclude that d 1 ≤ √ n. Similarly we arrive at c 3 ≤ 2n d 1 and d 3 ≤ d 1 2 . Claim A: C j is a sum of three arithmetic progressions on [n]. By construction the set C j is a sum of three arithmetic progressions. The largest element of C j is bounded by max C j = c 1 (n 1 − 1) + c 2 (n 2 − 1) + c 3 (n 3 − 1) ≤ ≤ n 12 + n 30 + n 6 ≤ n 2 , and thus C j ⊆ [n]. the electronic journal of combinatorics 13 (2006), #R5 9 Claim B:Reζ k j ≥ 1/2 for every k ∈ C j . We show that for every k ∈ C j the value of |arg ζ k j | is less than π/3 and this already implies the claim. max k∈C j |arg ζ k j | =max (i 1 ,i 2 ,i 3 )∈[n 1 ]×[n 2 ]×[n 3 ] |arg ζ c 1 i 1 j +arg ζ c 2 i 2 j +arg ζ c 3 i 3 j |≤ ≤ max i 1 ∈[n 1 ] |arg ζ c 1 i 1 j | +max i 2 ∈[n 2 ] |arg ζ c 2 i 2 j | +max i 3 ∈[n 3 ] |arg ζ c 3 i 3 j | = = 2π n (d 1 (n 1 − 1) + d 2 (n 2 − 1) + d 3 (n 3 − 1)). Inthecasethatc 1 ≤ 12d 1 max k∈C j |arg ζ k j | = 2π n (d 1 (n 1 − 1) + d 2 (n 2 − 1) + d 3 (n 3 − 1)) ≤ ≤ 2π n  d 1 n 12d 1 + d 2 · 0+ d 1 2 d 1 12  ≤ π 3 , otherwise max k∈C j |arg ζ k j | = 2π n (d 1 (n 1 − 1) + d 2 (n 2 − 1) + d 3 (n 3 − 1)) ≤ ≤ 2π n  d 1 n 144d 1 + d 1  n c 1  c 1 30d 1 + d 1 2 d 1 12  ≤ π 3 . Claim C: |C j |≥ 1 5000 n. We put D = {i 1 d 1 + i 2 d 2 : i 1 ∈ [n 1 ],i 2 ∈ [n 2 ]}. From the fact that d 2 = d 1  n c 1  and d 2 >n 1 d 1 we deduce that D is a subset of an arithmetic progression with difference d 1 and |D| = n 1 n 2 . The set E = {i 1 d 1 + i 2 d 2 + i 3 d 3 : i 1 ∈ [n 1 ],i 2 ∈ [n 2 ],i 3 ∈ [n 3 ]} is a union of n 3 shifted copies of D. Our goal is to show that those n 3 shifted copies of D are mutually disjoint. If there were two intersecting copies of D, then there has to exist a k ∈{1, ,n 3 }, such that d 1 |kd 3 . Let it be so and let k, l be the integers demonstrating this case, i.e. ld 1 = kd 3 . Since d 1 >d 3 ,wehavel<k.Thusl<n 3 ≤ n 1 and the preimage of ld 1 under the mapping f j (k)=jk mod n is f −1 j (ld 1 )=lc 1 and similarly f −1 j (kd 3 )=kc 3 . The mapping f j is a bijection and therefore ld 1 = kd 3 implies lc 1 = f −1 j (ld 1 )=f −1 j (kd 3 )=kc 3 .But since we also have c 1 <c 3 , we arrive at the contradiction l>k.Thusalln 3 shifted copies of D are mutually disjoint and |C j | = |E| = n 1 n 2 n 3 ≥ n/5000.  Theorem 3.5 For each prime n there exists a wrapped set system ([n], F n ), where F n = {S 0 ,S 1 , ,S n 2 −1 } such that each S i ∈F n is a union of two sums of three arithmetic progressions and disc(F n ) > 1 10000 n 1/2 . the electronic journal of combinatorics 13 (2006), #R5 10 [...]... of Integer Sequences is Nearly Sharp Combinatorica 1(4):319-325, 1981 [BS95] J Beck and V S´s Discrepancy Theory In Handbook of Combinatorics, pages o 1405-1446 North-Holland, Amsterdam, 1995 [Cha00] B Chazelle The Discrepancy Method Cambridge University Press, 2000 [ES74] P Erd˝s and J Spencer Probabilistic Methods in Combinatorics Academic Press, o 1974 [Heb04] N Hebbinghaus Discrepancy of Sums of. .. Sums of Arithmetic Progressions In Electronic Notes in Discrete Mathematics 17C, pages 185-189, 2004 [Mat99] J Matouˇek Geometric Discrepancy Springer, 1999 s [MS96] J Matouˇek and J.Spencer Discrepancy in Arithmetic Progressions Journal of s the American Mathematic Society, 9:195-204, 1996 [Rot64] K F Roth Remark Concerning Integer Sequences Acta Arithmetica 9:257-260, 1964 the electronic journal of combinatorics...Proof For a fixed n we construct Fn = {S0 , S1 , , Sn2 −1 } as follows: For S0 just take the set {0, 1, n/5000 } and for 0 < j < n we put Sjn = Cj as constructed in Lemma 3.4 Since for all 0 ≤ j < n we know that 1 k k n, ζj ≥ Re ζj ≥ 10000 k∈Sjn k∈Sjn from lemma 3.1 it immediately follows that disc(S) > 1 n1/2 10000 Corollary 3.6 For n ∈ N, let ([n], Sn ) be a set system formed by all sums of three. .. three arithmetic progressions on [n] Then disc(Sn ) = Ω(n1/2 ) Acknowledgements I would like to thank Jiˇ´ Matouˇek for introducing me into the area and for help with rı s writing this paper, and Lisa Bailey for reading the paper and correcting grammatical errors References [AS92] N Alon and J Spencer The Probabilistic Method J Wiley and Sons, New York, NY, 1992 [Bec81] J Beck Roth’s Estimate of the . Discrepancy of Sums of Three Arithmetic Progressions AleˇsPˇr´ıvˇetiv´y ∗ Department of Applied Mathematics of Charles University Malostransk´en´am. 25, 11800. set system of all arithmetic progressions on [n] is known to have a discrep- ancy of order n 1/4 . We investigate the discrepancy for the set system S 3 n formed by all sums of three arithmetic. [MS96]. Discrepancies of related set systems were also studied. One possible extension of the original problem is to consider set systems formed by sums of arithmetic progressions, where a sum of k arithmetic