Constructions of Generalized MSTD Sets in Higher Dimensions
aa r X i v : . [ m a t h . N T ] S e p CONSTRUCTIONS OF GENERALIZED MSTD SETS IN HIGHERDIMENSIONS
ELENA KIM AND STEVEN J. MILLER
Abstract.
Let A be a set of finite integers, define A + A = { a + a : a , a ∈ A } , A − A = { a − a : a , a ∈ A } , and for non-negative integers s and d define sA − dA = A + · · · + A | {z } s − A − · · · − A | {z } d . A More Sums than Differences (MSTD) set is an A where | A + A | > | A − A | . It wasinitially thought that the percentage of subsets of [0 , n ] that are MSTD would go to zeroas n approaches infinity as addition is commutative and subtraction is not. However, in asurprising 2006 result, Martin and O’Bryant proved that a positive percentage of sets areMSTD, although this percentage is extremely small, about 10 − percent. This result wasextended by Iyer, Lazarev, Miller, ans Zhang [ILMZ] who showed that a positive percentageof sets are generalized MSTD sets, sets for { s , d } 6 = { s , d } and s + d = s + d with | s A − d A | > | s A − d A | , and that in d -dimensions, a positive percentage of sets are MSTD.For many such results, establishing explicit MSTD sets in 1-dimensions relies on the specificchoice of the elements on the left and right fringes of the set to force certain differences tobe missed while desired sums are attained. In higher dimensions, the geometry forces a morecareful assessment of what elements have the same behavior as 1-dimensional fringe elements.We study fringes in d -dimensions and use these to create new explicit constructions. We provethe existence of generalized MSTD sets in d -dimensions and the existence of k -generationalsets, which are sets where | cA + cA | > | cA − cA | for all 1 ≤ c ≤ k . We then prove that undercertain conditions, there are no sets with | kA + kA | > | kA − kA | for all k ∈ N . Contents
1. Introduction 22. Proof of Existence of 2-Dimensional Generalized MSTD Sets 33. 2-Dimensional Chains of Generalized MSTD Sets 94. Other 2-Dimensional Constructions 145. D -Dimensional Constructions 166. Further Remarks 18Appendix A. Corrections to “Generalized More Sums than Differences” 19References 21 Date : September 9, 2020.We thank John Haviland for his help in the creation of Theorem 2.2 and his contribution of Figures 7 and8. Additionally, we thank John Lentfer and Fernando Trejos Su´arez for their detailed feedback on drafts ofthis paper. Finally, we thank members of the SMALL 2020 REU at Williams College, especially Ph´uc Lˆamand those formerly mentioned, for enlightening conversations. This research was supported by NSF grantsDMS1947438 and DMS1561945 and by Williams College. Introduction
Given a finite set A ⊂ Z , we define the sumset A + A and the difference set A − A by A + A = { a + a : a , a ∈ A } ,A − A = { a − a : a , a ∈ A } . (1.1)It is natural to compare the sizes of A + A and A − A as we vary A over a family of sets.As addition is commutative while subtraction is not, a pair of distinct elements a , a ∈ A generates two differences a − a and a − a but only one sum a + a . We thus expect thatmost of the time, the size of the difference set is greater than that of the sumset—that is, weexpect most sets A to be difference-dominant . A set whose sumset has the same number ofelements as its difference set is called balanced . It is possible, however, to construct sets whosesumsets have more elements than their difference sets, sets known as sum-dominant or MoreSums Than Differences (MSTD) sets. The first example of an MSTD set was discovered byConway in the 1960s: { , , , , , , , } . While there are numerous constructions of suchsets and infinite families of such sets, one expects sum-dominant sets to be rare; however, in2006, Martin and O’Bryant [MO] proved that a positive percentage of sets are sum-dominant.They showed the percentage is at least 2 · − , which was improved by Zhao [Zh2] to at least4 . · − (Monte Carlo simulations suggest the true answer is about 4 . · − ). For a set A ⊂ [0 , n ], one can note that each nonzero k ∈ A + A has about n/ − | n − k | / A . This number of representations is large exceptwhen k is close to 0 or 2 n . In these cases, k can only be made from a select few pairs ofelements close to 0 or n . Thus a clever choice of the “fringe,” elements near the ends of theset, determines the size of its sumset and difference set, not the choice of the middle of the set.We can extend the idea of sumsets and difference sets by defining sA − dA = A + · · · + A | {z } s − A − · · · − A | {z } d . (1.2)Given s , d , s , d with s + d = s + d ≥
2, a generalized MSTD set is a set of integers A such that | s A − d A | > | s A − d A | . Let { x j } kj =1 , { y j } kj =1 , { w j } kj =1 , and { z j } kj =1 besequences of non-negative integers, such that x j + y j = w j + z j = j and { x j , y j } 6 = { w j , z j } for every 2 ≤ j ≤ k . We then define a chain of generalized MSTD sets as an A such that | x j A − y j A | > | w j A − z j A | for every 2 ≤ j ≤ k . A k -generational set is a set A suchthat for a specific positive integer k we have A, A + A, . . . , kA are all sum-dominant. Iyer,Lazarev, Miller, and Zhang [ILMZ] proved, through explicit constructions, that generalizedMSTD sets, chains of generalized MSTD sets, and k -generational sets exist. The authorsadditionally showed that these sets represent a positive percentage of subsets of [0 , n ] as n → ∞ .A natural question then arises: are such sets similarly common in higher-dimensionalspaces? This paper seeks to answer this question as a sequel to [ILMZ]. The previous work onhigher-dimensional MSTD sets has been conducted by Do, Kulkarni, Miller, Moon, Wellens,and Wilcox [DKMMWW] who looked at sets created by dilating d -dimensional polytopeswith vertices in Z d . The authors then used a probabilistic argument that was similar to thatof Martin and O’Bryant, not explicit constructions, to prove a positive percentage of MSTDsets in d -dimensions. In this paper, we extend ideas from [ILMZ] and [DKMMWW] to look ONSTRUCTIONS OF GENERALIZED MSTD SETS IN HIGHER DIMENSIONS 3 at generalized MSTD sets, chains of generalized MSTD sets, and k -generational sets in higherdimensions through explicit constructions. Interesting new features and complications arisein higher dimensions. Whereas on the line it is natural to consider subsets of the integers in agrowing interval, in d -dimensions we construct our sets as subsets of various parallelograms.Additionally, higher dimensions requires a clever generalization of the fringes.In Theorem 2.1, Theorem 3.1, and Theorem 3.1, we prove, respectively, the existence of2-dimensional generalized MSTD sets, chains of generalized MSTD sets, and k -generationalsets with 2-dimensional generalizations of the sets in [ILMZ].In Lemma 3.4, we show that for a 2-dimensional set, A , with certain elements and forsufficiently large N , the amount of missing sums in kA for k ≥ N grows linearly.The proof of this lemma examines elements, ( x, y ) in kA , by using a 1-dimensional resultby Nathanson [Na1] on x and y . We can then determine whether ( x, y ) is or is not in kA .We can then use this lemma to prove Theorem 3.5 that these sets cannot be k -generationalfor every k, or in other words cannot have | kA + kA | > | kA − kA | for all k ∈ N . We then extend our 2-dimensional work to d -dimensions by extending our explicit construc-tions for generalized MSTD sets into d -dimensions. As our work builds on that of [ILMZ],we end with an appendix which corrects some mistakes in that paper.2. Proof of Existence of -Dimensional Generalized MSTD Sets We begin with a brief survey of the proofs for 1-dimensional results from [ILMZ] thatwe extend to 2-dimensions. To prove the existence of generalized MSTD sets and chains ofgeneralized MSTD sets, the authors construct explicit sets with these properties. See Figure1 and Figure 2, which we use with permission from the authors, for depictions of these sets.Their sets have nearly symmetric left and right fringes, L and R , with R slightly longer than L . Specifically these fringes are defined as: L = { , , , , . . . , k − , k, k + 1 , k + 1 } = [0 , k + 1] \ ( { } ∪ [ k + 2 , k ]) R = { , , , , , . . . , k, k + 1 , k + 2 , k + 2 } = [0 , k + 2] \ ( { } ∪ [ k + 3 , k + 1]) . (2.1)The authors then create a filled-in middle interval M for their set. Let n > k + 1) . When s > s , the middle is M = [2 k + 1 − d , n − (2 k + 1 − d )] , (2.2)and when s > s , the middle is M = [2 k + 1 − s , n − (2 k + 1 − s )] . (2.3)Then the complete set is A = L ∪ M ∪ ( n − R ) . (2.4)As A is added and subtracted, the middle section and the fringes begin to overlap. Thisoverlap controls the amount of elements in sA − dA. We then adapt the 1-dimensional set to prove the following theorem.
ELENA KIM AND STEVEN J. MILLER
Figure 1. A , A + A , A + A + A , and A + A + A + A . The saw tooth meansall elements are present in that range. Figure 2. A , A + A , A + A − A , and A + A − A − A . The saw tooth meansall elements are present in that range. Theorem 2.1.
Let s , d , s , d be non-negative integers such that { s , d } 6 = { s , d } and s + d = s + d ≥ . Then there exists a finite, non-empty set A in a sufficiently large n × n integer lattice such that | s A − d A | > | s A − d A | . Let k = s + d = s + d . We begin by defining two fringe sets B and B , where B isslightly larger than B : B = ([0 , k + 1] × [0 , k + 1]) \ [ { (2 , , (0 , } ∪ { ( x, , (0 , y ) : k + 2 ≤ x, y ≤ k } ] B = ([0 , k + 2] × [0 , k + 2]) \ [ { (3 , , (0 , } ∪ { ( x, , (0 , y ) : k + 3 ≤ x, y ≤ k + 1 } ] . (2.5)We now must prove a lemma characterizing the behavior of these fringe sets. We claimthat for any a, b ∈ N , aB + bB has the same structure as B and B . The proof of thislemma relies on an analogous lemma in 1-dimensions.
ONSTRUCTIONS OF GENERALIZED MSTD SETS IN HIGHER DIMENSIONS 5
Figure 3.
The B fringe set for k = 4. Lemma 2.1 (ILMZ) . Let L = [0 , k + 1] \ ( { } ∪ [ k + 2 , k ]) and R = [0 , k + 2] \ ( { } ∪ [ k + 3 , k + 1]) . Then for all a, b ∈ N , we have aL + bR = [0 , k ( a + b ) + ( a + 2 b )] \ (2.6)( { k ( a + b −
1) + ( a + 2 b + 1) } ∪ [ k (2 a + 2 b −
1) + ( a + 2 b + 1) , k ( a + b ) + ( a + 2 b − a , then on b . We can then proveour 2-dimensional version of this lemma. Lemma 2.2.
For all a, b ∈ N , aB + bB , we have aB + bB = ([0 , k ( a + b ) + ( a + 2 b )] × [0 , k ( a + b ) + ( a + 2 b )]) \ [ { (2 k ( a + b −
1) + ( a + 2 b + 1) , , (0 , k ( a + b −
1) + ( a + 2 b + 1)) } ∪{ ( x, , (0 , y ) : k (2 a + 2 b −
1) + ( a + 2 b + 1) ≤ x, y ≤ k ( a + b ) + ( a + 2 b − } ] . (2.7) Proof.
One can note that our B and B have copies of the 1-dimensional fringe sets from[ILMZ] on their left and bottom edges. Addition between points of the form ( x,
0) only resultsin sums of the same form. Similarly, addition between points of the form (0 , y ) only resultsin sums of the same form. Adding of a point of the form ( x,
0) to a point of the form (0 , y )only creates a point that already was in the fringe set. Thus we can use Lemma 2.1 on theedges to prove our lemma. (cid:3)
We then use the fringes to construct a set A that lives in a sufficiently large n × n latticesuch that | s A − d A | > | s A − d A | . Let n > k + 1) . In order for our fringe sets to have the proper orientation in our set, we have to slightlymodify the fringe definitions. Let B = B , and B = B , and let B , = ([0 , k + 1] × [0 , k + 1]) \ [ { ((2 k + 1) − , , (0 , } ∪ { ((2 k + 1) − x, , (2 k + 1 , y ) : k + 2 ≤ x, y ≤ k } ] B , = ([0 , k + 2] × [0 , k + 2]) \ ELENA KIM AND STEVEN J. MILLER
Figure 4.
The corresponding generalized MSTD set for k = 4, n = 130 and s = 4, d = 0 and s = 2, d = 2.[ { (3 , k + 2) , (0 , (2 k + 2) − } ∪ { ( x, k + 2) , (0 , (2 k + 2) − y ) : k + 3 ≤ x, y ≤ k + 1 } ] . (2.8)Since for any s, d ∈ N , we have | sA − dA | = | dA − sA | , we assume without loss of generalitythat s ≥ d and s ≥ d . We first examine the case where s > s , which implies that d > d . We construct a set M that fills the middle of our set. M = { ( x, y ) : 2 k + 1 − d ≤ x ≤ n − (2 k + 1 − d ) , ≤ y ≤ n } ∪{ ( x, y ) : 2 k + 1 − d ≤ y ≤ n − (2 k + 1 − d ) , ≤ x ≤ n } . (2.9)We then define A = B , ∪ [( n, n ) − B , ] ∪ [(2 k + 1 , n ) − B , ] ∪ [( n, k + 2) − B , ] . (2.10)In other words, A is an n × n lattice where B has been placed in the left corners, B hasbeen placed in the right corners, and M is a filled-in cross that goes in between (but doesn’ttouch) the fringes. This can be seen in Figure 4.One can see that as M is the union of filled-in rectangles that we have | s M − d M | = | s M − d M | .We then examine the fringes. We begin with the left two fringes of s A − d A . Theseare identical up to rotation, so without loss of generality, we just examine one. The portionwhich does not lie in M is (up to translation) ( s B + d B ) ∩ ([0 , k − d ] × [0 , k − d ]) . We then compare 2 k − d to 2 k ( s + d −
1) + ( s + 2 d + 1).By Lemma 2.2, we know that the the smallest missing points in the fringe are ( x,
0) and(0 , y ) where x, y = 2 k ( s + d −
1) + ( s + 2 d + 1) and the smallest points in the missingintervals are ( x,
0) and (0 , y ) where x, y = k (2 s + 2 d −
1) + ( s + 2 d + 1). We then comparethese points to 2 k − d , to see how many points are missing from the portion of the fringe ONSTRUCTIONS OF GENERALIZED MSTD SETS IN HIGHER DIMENSIONS 7 that does not lie in M for s A − d A . When s > d , we have2 k − d ≥ k − s + 1= 2 k ( s + d ) − k + d + 1= s (2 k + 1) + d (2 k + 2) − k + 1= 2 k ( s + d −
1) + ( s + 2 d + 1) (2.11)and 2 k − d < k + d + 1= k (2 k −
1) + ( k + d + 1)= k (2 s + 2 d −
1) + ( s + 2 d + 1) . (2.12)Therefore, we know that( s B + d B ) ∩ ([0 , k − d ] × [0 , k − d ]) = ([0 , k − d ] × [0 , k − d ]) \ ((2 k ( s + d −
1) + ( s + 2 d + 1) , ∪ (0 , k ( s + d −
1) + ( s + 2 d + 1))) . (2.13)Thus we are missing 4 elements, 2 in each fringe.When s = d , then we have2 k − d < k − s + 1= 2 k ( s + d ) − k + d + 1= s (2 k + 1) + d (2 k + 2) − k + 1= 2 k ( s + d −
1) + ( s + 2 d + 1) . (2.14)We then know that( s B + d B ) ∩ ([0 , k − d ] × [0 , k − d ]) = [0 , k − d ] × [0 , k − d ] . (2.15)Thus we are not missing any elements.We then examine the right two fringes of s A − d A . These are identical up to a rotation,so without loss of generality, we just examine one. The portion which does not lie in M is(up to translation) ( d B + s B ) ∩ ([0 , k − d ] × [0 , k − d ]) . Again by Lemma 2.2, weknow that the the smallest missing points in the fringe are ( x,
0) and (0 , y ) where x, y =2 k ( d + s −
1) + ( d + 2 s + 1). Thus we then compare 2 k ( d + s −
1) + ( d + 2 s + 1) to2 k − d , to show that no points are missing from the portion of the fringe that does not liein M . We have 2 k − d < k − d + 1= 2 k − k + s + 1= 2 k ( d + s −
1) + ( d + 2 s + 1) . (2.16)Therefore, we know that( d B + s B ) ∩ ([0 , k − d ] × [0 , k − d ]) = [0 , k − d ] × [0 , k − d ] . (2.17)Thus these fringes do not have any missing elements.We then turn our attention to s A − d A. We first examine the two left fringes of s A − d A .These are identical up to a rotation, so without loss of generality we just examine one. Theportion which does not lie in M is (up to translation) ( s B + d B ) ∩ ([0 , k − d ] × [0 , k − d ]) . ELENA KIM AND STEVEN J. MILLER
By Lemma 2.2, we know that the the smallest missing points in the fringe are ( x,
0) and (0 , y )where x, y = 2 k ( s + d −
1) + ( s + 2 d + 1) and the smallest points in the missing intervalsare ( x,
0) and (0 , y ) where x, y = k (2 s + 2 d −
1) + ( s + 2 d + 1). We then compare thesepoints to 2 k − d , to see how many points are missing from the portion of the fringe thatdoes not lie in M . We note that s > s ≥ d > d . Therefore2 k − d ≥ k s + 1= 2 k ( s + d ) − k + d + 1= s (2 k + 1) + d (2 k + 2) − k + 1= 2 k ( s + d −
1) + ( s + 2 d + 1) (2.18)and 2 k − d < k + d + 1= k (2 k −
1) + ( k + d + 1)= k (2 s + 2 d −
1) + ( s + 2 d + 1) . (2.19)Therefore, we know that( s B + d B ) ∩ ([0 , k − d ] × [0 , k − d ]) = ([0 , k − d ] × [0 , k − d ]) \ ((2 k ( s + d −
1) + ( s + 2 d + 1) , ∪ (0 , k ( s + d −
1) + ( s + 2 d + 1))) . (2.20)Thus we are missing 4 elements, 2 in each fringe.We then, finally, examine the right two fringes of s A − d A . These are identical up toa rotation, so without loss of generality, we just examine one. The portion which does notlie in M is (up to translation) ( d B + s B ) ∩ ([0 , k − d ] × [0 , k − d ]) . By Lemma2.2, we know that the the smallest missing points in the fringe are ( x,
0) and (0 , y ) where x, y = 2 k ( d + s −
1) + ( d + 2 s + 1) and the smallest points in the missing intervals are( x,
0) and (0 , y ) where x, y = k (2 d + 2 s −
1) + ( d + 2 s + 1). We then compare these pointsto 2 k − d , to see how many points are missing from the portion of the fringe that does notlie in M . We have 2 k − d ≥ k − d + 1= 2 k − k + s + 1= 2 k ( d + s −
1) + ( d + 2 s + 1) (2.21)and 2 k − d < k + s + 1= k (2 k −
1) + ( k + s + 1)= k (2 d + 2 s −
1) + ( d + 2 s + 1) . (2.22)Therefore, we know that d B + s B ∩ ([0 , k − d ] × [0 , k − d ]) = ([0 , k − d ] × [0 , k − d ]) \ ((2 k ( d + s −
1) + ( d + 2 s + 1) , ∪ (0 , k ( d + s −
1) + ( d + 2 s + 1)) . (2.23)Therefore these two fringes have 4 missing elements. ONSTRUCTIONS OF GENERALIZED MSTD SETS IN HIGHER DIMENSIONS 9
Thus we have shown for s > s that | s A − d A | > | s A − d A | . We then consider the case where s > s . We define M = { ( x, y ) : 2 k + 1 − s ≤ x ≤ n − (2 k + 1 − s ) , ≤ y ≤ n } ∪{ ( x, y ) : 2 k + 1 − s ≤ y ≤ n − (2 k + 1 − d ) , ≤ x ≤ n } . (2.24)Thus A is constructed in the same manner as for s > s . The proof is identical to show | s A − d A | > | s A − d A | . (cid:3) We claim that we can create more constructions of generalized MSTD sets by applyinginjective linear transformations to the sets we just created.
Theorem 2.2.
All MSTD sets are preserved under injective linear transformations.Proof.
We first show that the size of the sumsets and difference sets are preserved underinjective linear transformation. Let A ⊆ Z d and T : Z d → Z d be an injective linear transfor-mation. As T in injective, we know that | A + A | = | T ( A + A ) | and | A − A | = | T ( A − A ) | . Let a , a ∈ A. As T is a linear transformation, we have T ( a + a ) = T ( a ) + T ( a ) and T ( a − a ) = T ( a ) − T ( a ). Thus | T ( A + A ) | = | T A + T A | and | T ( A − A ) | = | T A − T A | .We then conclude that | A + A | = | T A + T A | and | A − A | = | T A − T A | . Induction can showthat the the size of sA + dA is preserved under T. (cid:3)
3. 2 -Dimensional Chains of Generalized MSTD Sets
In this section we first prove the following theorem on the existence of chains of generalizedMSTD sets. We use the same sets that we constructed in the proof of Theorem 2.1 in theprevious section.
Theorem 3.1.
Let { x j } kj =1 , { y j } kj =1 , { w j } kj =1 , and { z j } kj =1 be finite sequences of non-negativeintegers of length k such that x j + y j = w j + y j = j, and { x j , y j } 6 = { w j , z j } for every ≤ j ≤ k. There exists a -dimensional set A that satisfies | x j A − y j A | > | w j A − x j A | forevery ≤ j ≤ k. To do this we first prove a series of lemmas.
Lemma 3.1.
For the set A constructed in the previous section, we have | s A − d A | = | s A − d A | for any s + d = s + d such that s + d = k. Proof. As M is the union of two filled-in rectangles, we first note that s M − d M = s M − d M . Thus for this proof we only consider the fringes. Since | sA − dA | = | dA − sA | , wesuppose without loss of generality that s ≥ d .We begin with the case where s + d = s + d > k or in other words s + d = s + d = k + c, where c ≥ s A − d A, rotating and translating them to lie in the bottomleft corner, the two empty points on the edges closest to the corner have the coordinates(2 k ( s + d −
1) + ( s + 2 d + 1) ,
0) and (0 , k ( s + d −
1) + ( s + 2 d + 1)). For the rightfringes these points are (2 k ( d + s − d +2 s +1) ,
0) and (0 , k ( d + s − d +2 s +1)) . Similarly, we note for the left fringes in s A − d A , rotating and translating them to liein the bottom left corner, the two empty points on the edges closest to the corner have the coordinates (2 k ( s + d − s +2 d +1) ,
0) and (0 , k ( s + d − s +2 d +1)). For the rightfringes these points are (2 k ( d + s − d +2 s +1) ,
0) and (0 , k ( d + s − d +2 s +1)) . In the cases where s > s , recall that M = { ( x, y ) : 2 k + 1 − d ≤ x ≤ n − (2 k + 1 − d ) , ≤ y ≤ n } ∪{ ( x, y ) : 2 k + 1 − d ≤ y ≤ n − (2 k + 1 − d ) , ≤ x ≤ n } . (3.1)We then compare 2 k + 1 − d , the smallest x or y can be for ( x, ∈ M or (0 , y ) ∈ M , with2 k ( s + d −
1) + ( s + 2 d + 1), the x or y coordinate for the smallest missing point in theleft fringe of s A − d A :2 k ( s + d −
1) + ( s + 2 d + 1) = 2 k ( k + c −
1) + ( k + c + d + 1) > k + ( k + d + 1) > k + 1 − d . (3.2)Thus the left fringe of s A − d A is not missing any points. This same inequality can beused for (2 k ( d + s −
1) + ( d + 2 s + 1)) , (2 k ( s + d −
1) + ( s + 2 d + 1)), and (2 k ( d + s −
1) + ( d + 2 s + 1)) . Therefore the fringes are not missing any points.In the case where s > s and thus M = { ( x, y ) : 2 k + 1 − s ≤ x ≤ n − (2 k + 1 − s ) , ≤ y ≤ n } ∪{ ( x, y ) : 2 k + 1 − s ≤ y ≤ n − (2 k + 1 − d ) , ≤ x ≤ n } , (3.3)the same argument shows that the fringes are not missing any points.We then examine the case where s + d = c < k . We show that the fringes do notintersect the middle. Up to translation, the end points for the left fringes of s A − d A are(2 k ( s + d ) + ( s + 2 d ) , , k ( s + d ) + ( s + 2 d )), the end points for the right fringesof s A − d A are (2 k ( d + s ) + ( d + 2 s ) ,
0) and (0 , k ( d + s ) + ( d + 2 s )), the end points forthe left fringes of s A − d A are (2 k ( s + d ) + ( s + 2 d ) ,
0) and (0 , k ( s + d ) + ( s + 2 d )) , and finally the end points for the right fringes of s A − d A are (2 k ( d + s ) + ( d + 2 s ) , , k ( d + s ) + ( d + 2 s )) . As the argument for all of these points is the same, without loss of generality, we justexamine the point (2 k ( s + d ) + ( s + 2 d ) , s > s , the point in the middle section that is closest to (2 k ( s + d ) +( s + 2 d ) ,
0) is (2 k + 1 − d , k ( s + d ) + ( s + 2 d ) = 2 kc + c + d ≤ k ( k −
1) + ( k −
1) + d ≤ k − k − − d = 2 k − s − ≤ k − d − ≤ k + 1 − d . (3.4)Thus the fringes do not intersect the middle. The argument for the the middle sectionwhen s > s is practically identical. ONSTRUCTIONS OF GENERALIZED MSTD SETS IN HIGHER DIMENSIONS 11 As | s M − d M | = | s M − d M | , it now suffices to show that2 | s B − d B | + 2 | d B − s B | = 2 | s B − d B | + 2 | d B − s B | . (3.5)We have that | s B − d B | = 4 k ( s + d ) − k , | d B − s B | = 4 k ( d + s ) − k , | s B − d B | = 4 k ( s + d ) − k , and | d B − s B | = 4 k ( d + s ) − k . Therefore, the fringe setsare all the same size, which implies that | s A − d A | = | s A − d A | . (cid:3) The next lemma defines the base expansion method in 2-dimensions and proves a propertyabout sets created through base expansion.
Lemma 3.2.
Fix a positive integer k . Let A, B ⊂ N and chose m > k · max( { a : ( a, y ) or ( x, a ) ∈ A } ) . Let C = A + m · B (where m · B represents the usual scalar multiplication). Then | sC − dC | = | sA − dA || sB − dB | whenever s + d ≤ k. Proof.
By the definition of C , we have | C | ≤ | A || B | . We claim that each element of C = A + mB can be written uniquely as some ( a , a ) + m ( b , b ) for ( a , a ) ∈ A and ( b , b ) ∈ B .If ( a , a ) + m ( b , b ) = ( a ′ , a ′ ) + m ( b ′ , b ′ ) then we have ( a − a ′ , a − a ′ ) = m ( b ′ − b , b ′ − b ).Since m > k · max( { a : ( a, y ) or ( x, a ) ∈ A } ), we must have ( a , a ) = ( a ′ , a ′ ) and ( b , b ) =( b ′ , b ′ ) . Therefore we know that | C | = | A || B | . We know that | C ± C | = | A ± A || B ± B | . We now claim each element of C ± C isuniquely written as ( ˜ a , ˜ a ) ± m ( ˜ b , ˜ b ) where ( ˜ a , ˜ a ) ∈ A ± A and ( ˜ b , ˜ b ) ∈ B ± B . If( ˜ a , ˜ a ) ± m ( ˜ b , ˜ b ) = ( ˜ a ′ , ˜ a ′ ) ± m ( ˜ b ′ , ˜ b ′ ) , then we have ( ˜ a − ˜ a ′ , ˜ a − ˜ a ′ ) ∓ m ( ˜ b ′ − ˜ b , ˜ b ′ − ˜ b ) . Again, by the size of m, this can only happen when ( ˜ a , ˜ a ) = ( ˜ a ′ , ˜ a ′ ) and when ( ˜ b , ˜ b ) =( ˜ b ′ , ˜ b ′ ) . Therefore, we know | C ± C | = | A ± A || B ± B | . (cid:3) The next lemma is a further generalization.
Lemma 3.3.
Fix a positive integer k . Say that A , . . . , A k ⊂ N . Choose some m > k · max( { a : ( a, y ) or ( x, a ) ∈ A i for ≤ i ≤ k } ) . Let C = A + m · A + · · · + m k − · A k (where m · A j is the usual scalar multiplication). Then | sC − dC | = Q kj =1 | sA j − dA j | whenever s + d ≤ k .Proof. This can be proved using induction on the previous lemma. (cid:3)
With these lemmas, we can now prove Theorem 3 . Theorem 3.1.
Let x j , y j , w j , x j be finite sequences of non-negative integers of length k suchthat x j + y j = w j + y j = j, and { x j , y j } 6 = { w j , z j } for every 2 ≤ j ≤ k. There exists a2-dimensional set A that satisfies | x j A − y j A | > | w j A − x j A | for every 2 ≤ j ≤ k. Proof.
For each i , choose a set A i such that | x i A i − y i A i | > | w i A i − z i A i | , and for j = i , | x j A i − y j A i | = | w j A i − z j A i | . We know such a set exists, because of Theorem 2.1 andLemma 3.1. Next, choose some m > k · max( { a : ( a, y ) or ( x, a ) ∈ A k for 1 ≤ i ≤ k } ). Define A = A + mA + m A + · · · + m k − A k . We have that for each 2 ≤ j ≤ k | x j A − y j A | = k Y i =1 | x j A i − y j A i | = | x j A j − y j A j | · Y i = j | x j A i − y j A i | = | x j A j − y j A j | · Y i = j | w j A i − z j A i | > | w j A j − z j A j | · Y i = j | w j A i − z j A i | = | w j A − z j A | . (3.6) (cid:3) A corollary on the existence of k -generational sets immediately follows from this theorem. Corollary 3.1.
For each k ∈ N there exists a k -generational set. That is, for each k, thereexists a set A such that | cA + cA | > | cA − cA | for all ≤ c ≤ k. This result begs the question: Are there any sets such that | kA + kA | > | kA − kA | for all k ∈ N ? We begin to answer this question by proving that sets with certain properties cannothave | kA + kA | > | kA − kA | for all k ∈ N .We begin with a 1-dimensional theorem proved by Nathanson [Na1] that is used in our2-dimensional proof. Theorem 3.2 (Nathanson) . Let A = { a , a , . . . , a k } be a finite set of integers with a = 0 | kA − kA | for all k ∈ N does not necessitate the lowest possible bound on k . Thus, as we know the classicalresult by Nathanson to be true, we choose to rely on it, instead of the similar theorem from[ILMZ]. Lemma 3.4.
Let A = { ( a , b ) , ( a , b ) , . . . , ( a m , b m ) } where a i , b i are non-negative integers.Let a be the smallest non-zero a i , a ′ be the largest a i , b be the smallest non-zero b i and b ′ be the largest b i , and N = max { a ′ , b ′ } . If a and a ′ are coprime, b and b ′ are coprime,and { (0 , , ( a, , (0 , b ) , ( a ′ , , (0 , b ′ ) , ( a, b ) , ( a, b ′ ) , ( a ′ , b ) , ( a ′ , b ′ ) } ⊂ A, then for k ≥ N andfor some constants ℓ, ℓ , ℓ , we have | kA | = k a ′ b ′ − ℓ − ℓ a ′ k − ℓ b ′ k . Remark.
Figure 5 demonstrates what these sets look like for k ≥ N . ONSTRUCTIONS OF GENERALIZED MSTD SETS IN HIGHER DIMENSIONS 13
Figure 5. kA where A = { (0 , , (3 , , (0 , , (5 , , (0 , , (3 , , (5 , , (3 , , (5 , } and k = 30 Proof.
First, we use Nathanson’s theorem to say that for A = { , a, a ′ } , there exists integers c and d and sets C ⊂ [0 , c −
2] and D ⊂ [0 , d −
2] such that for k ≥ a ′ , we have k A = C ∪ [ c , ka ′ − d ] ∪ ka ′ − D . Additionally, for B = { , b, b ′ } , we know there existsintegers c and d and sets C ⊂ [0 , c −
2] and D ⊂ [0 , d −
2] such that for k ≥ b ′ , wehave kB = C ∪ [ c , kb ′ − d ] ∪ ( kb ′ − D ) . For this proof, we use the concept of fringes and a middle, however, they look differentthan in previous sections. The fringes are a frame around the set and and the middle is arectangle in middle of the set. More specifically, we define the fringes to be F = kA \ M = kA ∩ [ { ( x, y ) : x ∈ C ∪ D } ∪ { ( x, y ) : y ∈ C ∪ D } ] , (3.7)and we define the middle of kA as M = kA ∩ [ { ( x, y ) : x ∈ [ c , ka ′ − d ] } ] ∪ [ { ( x, y ) : y ∈ [ c , kb ′ − d ] } ] . (3.8)Let N = max { a ′ , b ′ } . We first claim that M is completely filled-in for k ≥ N . We relyon Nathanson’s proof that showed for A ′ = { , a, a ′ } and for k ≥ a ′ , the middle intervalof kA ′ , [ c , ka ′ − d ], is completely filled. Thus as { (0 , , ( a, , ( a ′ , } ⊂ A , we know that { ( x,
0) : x ∈ [ c , ka ′ − d ] } ⊂ k a A where k a ≥ a ′ . We can use the same argument to showthat { (0 , y ) : y ∈ [ c , kb ′ − d ] } ⊂ k b A where k b ≥ b ′ . Let k ≥ N . Thus we know thatfor x ∈ [ c , ka ′ − d ], ( x, ∈ kA and for y ∈ [ c , kb ′ − d ], (0 , y ) ∈ kA . We then examine( x, y ) = ( x,
0) + (0 , y ) where x ∈ [ c , ka ′ − d ] and y ∈ [ c , kb ′ − d ]. We know that ( x, , y ) can be made in most k sums. As { ( a, b ) , ( a, b ′ ) , ( a ′ , b ) , ( a ′ b ′ ) } ⊂ A , we know we canreplace the elements ( a ′ , a,
0) and (0 , b ′ ), (0 , b ) which are summed to create ( x,
0) and(0 , y ), respectively with the elements of { ( a, b ′ ), ( a ′ , b ), ( a ′ , b ′ ), ( a, b ) } . Using this substitution,it is clear that the number of elements of A needed to sum to create ( x, y ) is at most k. Thus,for k ≥ N , the middle of kA is not missing any elements.We then examine the fringes of kA to show that they are always missing a set numberof points, C , and an amount of points that linearly depends on k , ℓ ka ′ + ℓ kb ′ . Suppose ( x, ∈ kA for k ≥ N where x ≤ c − x ≥ ka ′ − ( d − x ∈ (2 a ′ ) A , then x ∈ kA for k > a ′ . Similarly, if x / ∈ (2 a ′ ) A , then x / ∈ kA for k > a ′ . Thus ( x, ∈ kA for k > a ′ if and only if ( x, ∈ (2 a ′ ) A . The same argumentcan be applied to (0 , y ) for y ≤ c − y ≥ kb ′ − ( d − x ∈ (2 a ′ ) A , for x ≤ c − x ≥ ka ′ − ( d − x ∈ C ∪ D . Similarly, if y ∈ (2 y ′ ) A for y ≤ c − y ≥ kb ′ − ( d − y ∈ C ∪ D .We then have that for ( x, y ), if x ∈ C ∪ D and y ∈ C ∪ D then for k ≥ N , ( x, y ) isalways in kA or never in kA . Let the total number of these missing elements be ℓ ′ . If wehave x ′ ∈ C ∪ D and y / ∈ C ∪ D , then x ′ is either always or never present and y is alwayspresent as it is in the middle. If x ′ is never present, then we are missing a column defined by { ( x ′ , y ) : y ∈ [ c , kb ′ − d ] } . If x ′ is present, then that column is in kA. The length of of thiscolumn is kb ′ − d − c . We can examine x / ∈ C ∪ D and y ∈ C ∪ D to find that there mayalso be missing rows. The length of these rows is ka ′ − d − c . Let ℓ be the total number ofmissing columns and ℓ the missing rows. Then let ℓ = ℓ ′ − ℓ ( d + c ) − ℓ ( d + c ). Thenwe note that the number of missing elements in kA for k ≥ N is ℓ + ℓ ka ′ + ℓ kb ′ . (cid:3) We then use the previous lemma to prove a result about k -generational sets. Lemma 3.5.
Let A = { ( a , b ) , ( a , b ) , . . . , ( a m , b m ) } where a i , b i are non-negative integers.Let a be the smallest non-zero a i , a ′ be the largest a i , b be the smallest non-zero b i and b ′ bethe largest b i and N = max { a ′ , b ′ } . If a and a ′ are coprime, b and b ′ are coprime, and { (0 , , ( a, , (0 , b ) , ( a ′ , , (0 , b ′ ) , ( a, b ′ ) , ( a ′ , b ) , ( a ′ , b ′ ) , ( a, b ) } ⊂ A, then for k ≥ N we have | kA − kA | ≥ | kA + kA | .Proof. Let k ≥ N. We then know that | kA | = k a ′ b ′ − ℓ − ℓ a ′ k − ℓ b ′ k , or in other words, kA has a filled-in middle and a total of ℓ missing columns on left and right sides of theset and a total of ℓ missing rows on top and bottom edges of the set. We know both2 kA and kA − kA are subsets of an integer lattice with 4 k a ′ b ′ points and we note that | kA | = 4 k a ′ b ′ − ℓ − ℓ (2 a ′ k ) − ℓ (2 b ′ k ). We then examine how many points kA − kA ismissing. kA is the union of a rectangular filled-in middle and filled-in rows and columnswhere rows and columns overlap in the corners of the set. The rectangular middle, therows, and the columns are all preserved under subtraction. Thus kA − kA has at least4 k a ′ b ′ − ℓ − ℓ (2 a ′ k ) − ℓ (2 b ′ k ) points. However, under subtraction, points from different rowsand columns and the middle may interact to add new points to the set kA − kA . Therefore | kA − kA | ≥ k a ′ b ′ − ℓ − ℓ (2 a ′ k ) − ℓ (2 b ′ k ). We conclude that | kA − kA | ≥ | kA + kA | . (cid:3) Other -Dimensional Constructions The square lattice construction that was used in the previous two sections can also begeneralized to polygons that have integer vertices and are locally point symmetric. Fromthe work of [DKMMWW], we know that the shape of locally point symmetric polygons ispreserved under subtraction. The only shapes that satisfy these two conditions are parallel-ograms. Thus, we generalize our construction to rectangles, then to all parallelograms withinteger vertices. We first qualitatively describe the sets, then explicitly write their formulas.Figure 6 demonstrates the shape of the sets. The fringe sets and middle sets for these setsmirror the construction of the square lattice case. The fringe sets are created by placing the1-dimensional fringes along the edges by each vertex and then filling in the rest of the corner.In the rectangle case, there are almost completely filled-in squares in each of the corners, and
ONSTRUCTIONS OF GENERALIZED MSTD SETS IN HIGHER DIMENSIONS 15
Figure 6.
The generalized MSTD set for k = 4, n = 130 , s = 4, d = 0, s = 2, and d = 2 that has been sheared with slope m = 1.in the parallelogram case, the squares are appropriately sheared. We place a slightly largerfringe set in the right corners than the left corners. The middle set is created with a filled-inwide cross in the middle of the set, and sheared appropriately in the parallelogram.For a rectangle construction, we have side lengths n , n ∈ N where n , n > k + 1) .B , , B , , B , , and B , are identically constructed as in the square lattice case. In the casewhere s > s ,M = { ( x, y ) : 2 k + 1 − d ≤ x ≤ n − (2 k + 1 − d ) , ≤ y ≤ n } ∪{ ( x, y ) : 2 k + 1 − d ≤ y ≤ n − (2 k + 1 − d ) , ≤ x ≤ n } . (4.1)In the case where s > s ,M = { ( x, y ) : 2 k + 1 − s ≤ x ≤ n − (2 k + 1 − d ) , ≤ y ≤ n } ∪{ ( x, y ) : 2 k + 1 − s ≤ y ≤ n − (2 k + 1 − d ) , ≤ x ≤ n } . (4.2)In both cases we then construct our rectangle, A, as follows: A = B , ∪ [( n , n ) − B , ] ∪ [(2 k + 1 , n ) − B , ] ∪ [( n , k + 2) − B , ] . (4.3)For the parallelogram case, let n , n , n ∈ N . We let n represent the length of the bottomand top edges, n the height, and n the length from the bottom left vertex to where a per-pendicular line dropped from top left vertex reaches the bottom. We need to make n , n , n large enough so that both sides have more than 4(2 k + 1) integer points along them, so wepass to a parallelogram scaled by 4(2 k + 1) and relabel. We note the slope of the diagonaledge is n /n = m. To create this set, we can use a mapping to shear the square and rectangle lattice sets thatwe already created.We define a map ϕ : Z → Z by ϕ ( x, y ) = ( x + my, y ) . We know by Theorem 2.2 that if A is a set such that | s A − d A | > | s A − d A | , ϕ ( A ) has the same property. We can also explicitly describe the generalized MSTD sets that are subsets of parallelograms. We takethe same B , , B , , B , , B , as before, but then we shear them: B , = { ( x + my, , (0 , y ) : 0 ≤ x, y ≤ k + 1 } \ [ { (2 , , (2 m, } ∪ { ( x, , ( my, y ) : k + 2 ≤ x, y ≤ k } ] , (4.4) B , = { ( x + my, , (0 , y ) : 0 ≤ x, y ≤ k + 2 } \ [ { (3 , , (3 m, } ∪ { ( x, , ( my, y ) : k + 3 ≤ x, y ≤ k + 1 } ] , (4.5) B , = { (2 k + 1 − ( x + my ) , , (2 k + 1 , y ) : 0 ≤ x, y ≤ k + 1 } \ [ { (2 k + 1 − , k + 1) , (2 k + 1 − m, } ∪{ (2 k + 1 − x, k + 1) , (2 k + 1 − my, y ) : k + 2 ≤ x, y ≤ k } ] , (4.6) B , = { ( x + my, k + 2) , (0 , k + 2 − y ) : 0 ≤ x, y ≤ k + 2 } \ [ { (3 , k + 2) , (3 m, k + 2 − } ∪{ ( x, k + 2) , ( my, k + 2 − y ) : k + 3 ≤ x, y ≤ k + 1 } ] . (4.7)We then similarly shear the middle set. In the case where s > s , M = { ( x + my, y ) : 2 k + 1 − s ≤ x ≤ n − (2 k + 1 − d ) , ≤ y ≤ n } ∪{ ( x + my, y ) : 2 k + 1 − s ≤ y ≤ n − (2 k + 1 − d ) , ≤ x ≤ n } . (4.8)For s > s . M = { ( x + my, y ) : 2 k + 1 − s ≤ x ≤ n − (2 k + 1 − s ) , ≤ y ≤ n } ∪{ ( x + my, y ) : 2 k + 1 − s ≤ y ≤ n − (2 k + 1 − s ) , ≤ x ≤ n } . (4.9)Finally, we define our complete set: A = B , ∪ [( n + mm , n ) − B , ] ∪ [(2 k +1+ mn , n ) − B , ] ∪ [( n + m (2 k +2) , k +2) − B , ] . (4.10)5. D -Dimensional Constructions We now extend our rectangular construction of generalized MSTD sets to d -dimensions.These are d -dimensional rectangles with non-negative coordinates and one corner at (0 , , . . . , d -dimensional cubes with the 1-dimensional fringesplaced along their edges attached to the vertex, as seen in Figure 7. B = { ( x , , . . . , , . . . , (0 , , . . . , x d ) : 0 ≤ x , . . . , x d ≤ k + 1 } \ [ { (2 , , . . . , , (0 , , . . . , , . . . , (0 , , . . . , } ∪{ ( x , , . . . , , (0 , x , . . . , , . . . , (0 , , . . . , x d ) : k + 2 ≤ x , x , . . . x d ≤ k } ] .B = { ( x , , . . . , , . . . , (0 , , . . . , x d ) : 0 ≤ x , . . . , x d ≤ k + 2 } \ [ { (3 , , . . . , , (0 , , . . . , , . . . , (0 , , . . . , } ∪{ ( x , , . . . , , (0 , x , . . . , , . . . , (0 , , . . . , x d ) : k + 3 ≤ x , x , . . . x d ≤ k + 1 } ] . (5.1) ONSTRUCTIONS OF GENERALIZED MSTD SETS IN HIGHER DIMENSIONS 17
Figure 7.
One of the fringe sets in 3-dimensions. The black dots representthe included points on the edges. The rest of the cube is filled-in.However, to actually put these fringes into corners of our set, we need to change theirorientation. We introduce new notation. Let the side lengths be n , n , . . . , n d , where theside of length n j lies parallel to the j -th axis. We denote the fringes as B i ,i ,...,i d , where i j ∈ { , } . If i j = 0, then the j th coordinate of the corner B i ,i ,...,i d is in is n j . If i j = 1, thenthe j th coordinate of the corner B i ,i ,...,i d lies in is 0. We then define maps to create each fringewith proper orientation. The map ϕ i ,i ,...,i d corresponds to B i ,i ,...,i d . Let ϕ ,i ,...,i d : Z d → Z d be defined by ϕ ,i ,...,i d ( x , x , . . . , x d ) = ( x , (2 k + 1) i − x , . . . , (2 k + 1) i d − x d ) , (5.2)and let ϕ ,i ,...,i d : Z d → Z d be defined by ϕ ,i ,...,i d ( x , x , . . . , x d ) = (2 k + 2 − x , (2 k + 2) i − x , . . . , (2 k + 2) i d − x d ) . (5.3)Then we have ϕ ,i ,...,i d ( B ) = B ,i ,...,i d and ϕ ,i ,...,i d ( B ) = B ,i ,...,i d . We then define our middle section of the set. It is a d -dimensional rectangle with a d -dimensional cube missing from each corner, as seen in Figure 8.We define for s > s , M = d [ i =1 { ( x , x , . . . , x d ) : 2 k +1 − d ≤ x j ≤ n j − (2 k +1 − d ) for j = i, ≤ x i ≤ n i } . (5.4)For s > s we have, M = d [ i =1 { ( x , x , . . . , x d ) : 2 k +1 − s ≤ x j ≤ n j − (2 k +1 − s ) for j = i, ≤ x i ≤ n i } . (5.5) A = M ∪ (5.6) [ i ,...,i d ∈{ , } ,i =0 (( n , (1 − i ) n + i (2 k + 2) , . . . , (1 − i d ) n d + i d (2 k + 2)) − B ,i ,...,i d ) ∪ Figure 8.
The shape of the middle set in 3-dimensions. [ i ,...,i d ∈{ , } ,i =1 ((2 k + 1 , (1 − i ) n + i (2 k + 1) , . . . , (1 − i d ) n d + i d (2 k + 1)) − B ,i ,...,i d ) . We then can modify this construction to be a subset of a d -dimensional parallelogram byshearing. In the 2-dimensional case, only the edges parallel to the y -axis could be shearedin the x direction. For the 3-dimensional case, we are still able to shear the base, as inthe 2-dimensional case, but we can now also shear the edges parallel to the z -axis in the x and the y directions. Thus there are 3 ways to shear the fringe. More generally, thereare d ( d − / d ( d − / d ( d − / m , , m , , . . . , m ,d , m , , . . . , m ,d , . . . , m d − ,d that define each way to shear the fringe. Thesubscripts refer to the coordinates involved in the shear, for example m i,j is the j th axissheared in the i th direction.We then create a map that shears our rectangular constructions. We define ψ : Z d → Z d by ψ ( x , x , . . . , x d ) =( x + m , x + m , x + · · · + m ,d x d , x + m , x + · · · + m ,d x d , · · · , x d ) . (5.7)6. Further Remarks
We believe that there are several aspects of this paper that can be extended further in thefuture. First, work on 1-dimensions in [ILMZ] has shown positive percentages for generalizedMSTD sets, chains of generalized MSTD sets, and k -generational sets using the methods of[MO]. [DKMMWW] extended the probabilistic methods of [MO] into d -dimensions to find apositive percentage of d -dimensional MSTD sets. We believe that the same methods couldbe used to find positive percentages of generalized MSTD sets, chains of MSTD sets, and k -generational sets in d -dimensions.Second, we believe that the work on proving that for a set A with specific properties, | kA + kA | > | kA − kA | does not hold for all k ∈ N can be extended to sets with lessrestrictions. Consider slightly different conditions from those in Lemma 3.4: Let A = ONSTRUCTIONS OF GENERALIZED MSTD SETS IN HIGHER DIMENSIONS 19 { ( a , b ) , ( a , b ) , . . . , ( a m , b m ) } where a i , b i are non-negative integers. Let a be the smallestnon-zero a i , a ′ be the largest a i , b be the smallest non-zero b i and b ′ be the largest b i . Suppose a and a ′ are coprime, b and b ′ are coprime, and { (0 , , ( a, , (0 , b ) , ( a ′ , , (0 , b ′ ) , ( a ′ , b ′ ) } ⊂ A. For this set of conditions where fewer elements are required to be in A , the growth of | kA | is still linear, but the set is more complicated. Instead of having rows and columns missingin the fringes, the right and top parts of the fringes have a repeated jagged pattern. Thewidth of the right fringe and the height of the top fringe remain constant as k grows. Thusfor large k , this set also has linear growth for | kA | . Therefore, we believe it would be possibleto come up with a similar lemma to Lemma 3.4 with fewer conditions. In fact, we thinkthat it would be possible to prove that for sufficiently large k , for any 2-dimensional set, theamount of the points missing from kA has linear growth. A lemma about linear growth forany 2-dimensional set A would allow a general theorem that | kA + kA | ≤ | kA − kA | for large k . We additionally hope that we can generalize a statement about the growth of | kA | into d -dimensions. Appendix A. Corrections to “Generalized More Sums than Differences”
In this paper, many of our proof methods followed directly from those in [ILMZ]. However,in closely going through these authors’ work, we found some mistakes which must be corrected,especially as our work builds directly on theirs. • In Case 1 in the proof of Theorem 2 . M should be defined as M = [ kr − k + 1 − d , n − ( kr − k + 1 − d )] instead of M = [ kr − k + 1 − d , kn − ( kr − k + 1 − d )]. • The proof of Theorem 2 . s > d and thecase where s = d . The s > d case gives the original result that | s A − d A | = | s A − d A | + 1, but the s = d case gives | s A − d A | = | s A − d A | + 2 . • In the proof for Lemma 3 .
1, the authors state “[2 m − u, m − − ⊂ U + U ”. Thisshould be “[2 m − u, m − − ⊂ U + U .” • Based on its proof, Theorem 1 . s , d ≥
1, instead of s , d ≥ • In Lemma 4 .
1, equation (4 .
3) should have ( n − ( kr − k + 1 − d ) − R ) instead of( n − (2 kr − k + 1 − d ) − R ) • For Lemma 5 .
1, the authors claim that for k ≥ a m , the middle of the set is filled-in.However this does not always hold. For example for the set A = { , , } , we have a m = 8. Thus for k = a m , 54 should be in kA . However the only way to get 54 withsums of the elements in A is 6(5) + 3(8) = 54 . This would require k = 9. • For Lemma 5 .
1, the right fringe R is defined incorrectly. It should be R = kA ∩ [( k − a ) a m , ka m ], instead of R = kA ∩ [( k − a m , ka m ]. Then this fringe is symmetric tothe left fringe, which the authors’ proof needs to be true. • For Lemma 5 .
1, a lemma needed to show that there are no k -generational sets forall k , the authors define A = { , a , a , . . . , a m } ⊂ [0 , n −
1] to be a set of integerswhere a < a < · · · < a m and assume gcd( a , . . . , a m ) = 1. They then define B = { a ′ , . . . , a ′ m } where a i = a i mod a and claim that gcd( a ′ , . . . , a ′ m ) = 1 . Howeverwe can find a counterexample to the statement gcd( a ′ , . . . , a ′ m ) = 1 . Let a ∈ N and1 < x < a such that a and x are coprime. Suppose A = { , a , a , . . . , a m } where a < a < . . . < a m and a i = c ,i a + c ,i x where 1 ≤ c ,i , 0 ≤ c ,i x ≤ a . Then for if B = { a ′ , . . . , a ′ m } as defined above, we have gcd( a ′ , . . . , a ′ m ) = x or = 0 . To fix this claim, we prove in the following theorem that for A , | A − A | ≥ | A + A | and thereforeif we are studying k -generational sets, it doesn’t even make sense to consider thesesets. Theorem A.1.
Let a ∈ N and < x < a such that a and x are coprime. Suppose A = { , a , a , . . . , a m } is a set of integers where a < a < . . . < a m and a i = c ,i a + c ,i x where ≤ c ,i , ≤ c ,i x ≤ a . Then | A − A | ≥ | A + A | .Proof. To prove this theorem, we first show that | A − A | ≥ | A + A | is true in two specificcases. We then show that every perturbation that can create a set with the properties statedin the theorem still gives us | A − A | ≥ | A + A | .If every element greater than a is of the form a i = c ,i a , we then have an arithmeticprogression, which is balanced.We then examine the case where every element in A that is not a or 0 is of the form a i = c ,i a + c ,i x where c ,i = 0 . We start with the sub-case where the elements in A that arenot a or 0 come from a the intersection of intervals with arithmetic progressions A ℓ of theform a i = c i a + ℓx where 1 < ℓx < a i . In other words, each A ℓ is a section of an arithmeticprogression that is not missing any points with common difference a with each element equalto ℓx mod a . Let B = S A ℓ and let A = { , a } ∪ B . We note that the union of arithmeticprogressions is sum-difference balanced, thus we can let n = | B − B | = | B + B | . We then have A + A = ( B + B ) ∪ (0 + B ) ∪ (0 + 0) ∪ (0 + a ) ∪ ( a + a ) ∪ ( a + a m ).This can be verified to be a disjoint union as a does not divide a m , therefore | A + A | = n + ( m −
1) + 1 + 1 + 1 + 1 = n + m + 3.Additionally, A − A = ( B − B ) ∪ (0 − B ) ∪ ( B − ∪ (0 − a m ) ∪ ( a m − . This toocan be verified to be a disjoint union, therefore | A − A | = n + 2( m −
1) + 1 + 1 = n + 2 m .When m >
2, we have | A + A | ≤ | A − A | . We note m > A . Suppose the length of an arithmetic progression remains the same, but a “skip”is introduced into it. For example, if { , , , } becomes { , , , } .It can be verified that | A + A | grows by 2, and | A − A | grows by 4. If we add to the lengthof the skip (still while keeping the number of elements the same), we then have | A + A | growsby 1 and and | A − A | grows by 2. Thus for A ℓ which are subsets of arithmetic progressionswith common difference a with each element equal to ℓx mod a and A = { , a } S A ℓ , wehave | A − A | ≥ | A + A | .We then see what happens when we introduce points that are equal to 0 mod a to ourset or, in other words, add points of the form a i = c ,i a . We claim that introducing a pointof this form either increases the deficit between | A + A | and | A − A | or keeps it the same. Let A ′ be A ∪ a n where a n = 0 mod a . We then have A ′ + A ′ = ( A + A ) ∪ ( A + a n ) ∪ ( a n + a n )and A ′ − A ′ = ( A − A ) ∪ ( A − a n ) ∪ ( a n − A ). In the extremal case, ( A − a n ) and ( a n − A )differ by one point, an − a and a − a n respectively, thus | A ′ + A ′ | ≤ | A ′ − A ′ | . One can seethat adding in more points that are equal to 0 mod a has a similar effect.Through these perturbations, any set that satisfies the requirements of the theorem can becreated. Therefore any set that satisfies the characteristics outlined in the theorem cannotbe MSTD. (cid:3) ONSTRUCTIONS OF GENERALIZED MSTD SETS IN HIGHER DIMENSIONS 21
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E-mail address : [email protected] Department of Mathematics, Pomona College Claremont, CA 91711
E-mail address : [email protected], [email protected]@williams.edu, [email protected]