Infinite Families of Partitions into MSTD Subsets
aa r X i v : . [ m a t h . N T ] O c t INFINITE FAMILIES OF PARTITIONS INTO MSTD SUBSETSH`ung Viˆe.t Chu
Department of Mathematics, Washington and Lee University, Lexington, VA24450 [email protected]
Noah Luntzlara
Department of Mathematics, University of Michigan, Ann Arbor, MI 48109 [email protected]
Steven J. Miller
Department of Mathematics and Statistics, Williams College, Williamstown, MA01267 [email protected]
Lily Shao
Department of Mathematics and Statistics, Williams College, Williamstown, MA01267 [email protected]
Received: , Revised: , Accepted: , Published:
Abstract
A set A is MSTD (more-sum-than-difference) if | A + A | > | A − A | . Though MSTDsets are rare, Martin and O’Bryant proved that there exists a positive constant lowerbound for the proportion of MSTD subsets of { , , . . . , r } as r → ∞ . Later, Asadaet al. showed that there exists a positive constant lower bound for the proportionof decompositions of { , , . . . , r } into two MSTD subsets as r → ∞ . However, themethod is probabilistic and does not give explicit decompositions.Continuing this work, we provide an efficient method to partition { , , . . . , r } (for r sufficiently large) into k ≥ k . Next, let R ( k ) bethe smallest integer such that for all r ≥ R ( k ), { , , . . . , r } can be k -decomposedinto MSTD subsets. We establish rough lower and upper bounds for R ( k ). Lastly,we provide a sufficient condition on when there exists a positive constant lowerbound for the proportion of decompositions of { , , . . . , r } into k MSTD subsets as r → ∞ . The authors were supported by NSF grants DMS1659037 and DMS1561945, the FinnertyFund, University of Michigan, Washington and Lee University and Williams College.
NTEGERS: 19 (2019)
1. Introduction1.1. Background
Given a set A of integers, define A + A = { a i + a j | a i , a j ∈ A } and A − A = { a i − a j | a i , a j ∈ A } . Then A is said to be sum-dominant or MSTD (more-sums-than-differences) if | A + A | > | A − A | , balanced if | A + A | = | A − A | and difference-dominated if | A + A | < | A − A | ; see [2, 6, 10, 11, 13, 14, 15] for some history and earlyresults in the subject. Research on MSTD sets has made great progress in the lasttwenty years. In particular, Martin and O’Bryant [7] showed that with the uniformmodel, where each element is chosen with probability 1 /
2, the proportion of MSTDsubsets of { , , . . . , r } is uniformly bounded below by a positive constant for largeenough r . Zhao [18] showed that the proportion converges as r → ∞ and improvedthe lower bound to 4 . · − . On the other hand, Hegarty and Miller [3] provedthat with a sparse model, where each element is chosen with probability p ( r ) suchthat r − = o ( p ( r )) and p ( r ) → r → ∞ , almost all sets are difference-dominated.These two results do not contradict each other since the probability of being MSTDsubsets depends on which model we are using. In proving a lower bound for theproportion of MSTD subsets, Martin and O’Bryant used the probabilistic methodand did not give explicit constructions of MSTD sets. Later works gave explicitconstruction of large families of MSTD sets: Miller et al. [8] gave a family ofMSTD subsets of { , , . . . , r } with density Θ(1 /r ) , while Zhao [17] gave a denserfamily with density Θ(1 /r ), the current record.In [1], the authors used a technique introduced by Zhao [18] to show that theproportion of 2-decompositions (i.e., partitions into two sets) of { , , . . . , r } thatgives two MSTD subsets is bounded below by a positive constant. This result issurprising in view of the conventional method of constructing MSTD sets, which isto fix a fringe pair ( L, R ) of two sets containing elements to be used in the fringeof the interval and argue that all the middle elements appear with some positiveprobability. (The fringe pair ensures that some of the largest and smallest differencesare missed and that our set is MSTD.) However, the result in [1] seems to suggestthat we can find two (or more) disjoint fringe pairs ( L , R ) and ( L , R ) such that L ∪ L and R ∪ R cover a full set of left and right elements of { , , . . . , r } and( L , R ) , ( L , R ) are two fringe pairs for two disjoint MSTD sets. Previous researchhas focused on each fringe pair independently, so it is interesting to see that two (ormore) fringe pairs can complement each other nicely on both sides of { , , . . . , r } .Motivated by that, we provide a method to construct these fringe pairs and studypartitions of { , , . . . , r } into MSTD subsets more thoroughly. [5] showed that with slightly more work, the density is improved to Θ(1 /r ). NTEGERS: 19 (2019) Let [ a, b ] denote { ℓ ∈ Z | a ≤ ℓ ≤ b } and I r denote [1 , r ]. We use the idea of P n setsdescribed in [8]. A set A is said to be P n if the following conditions are met. Let a = min A and b = max A . Then A + A ⊇ [2 a + n, b − n ] (1) A − A ⊇ [( a − b ) + n, ( b − a ) − n ] . (2)A set A is P n with respect to sums ( SP n ) if condition (1) is satisfied, and P n withrespect to differences ( DP n ) if condition (2) is satisfied. Next, let [ a, b ] denote { ℓ ∈ Z | a ≤ ℓ ≤ b and ℓ − a is even } . Finally, a 2-decomposition of a set S is A ∪ A = S , where A , A are nonempty and A ∩ A = ∅ . We use the words decomposition and partition interchangeably. Our main result is the following. Theorem 1.1.
Let A and A be chosen such that both are MSTD, A is P n and A is P n − . Also,1. ( A , A ) partition [1 , n ] ,2. A i = L i ∪ R i with L i ⊆ [1 , n ] and R i ⊆ [ n + 1 , n ] for i = 1 , ,3. [1 , ∪ { n } ⊆ L and { n + 1 } ∪ [2 n − , n ] ⊆ R , and4. [5 , ⊆ L and [2 n − , n − ⊆ R .(See Remark 1.2 for an example of such sets A and A .)Pick k ≥ n/ and m ∈ N . Set R ′ = R + m + 4 k + 4 ,R ′ = R + m + 4 k + 4 ,O = { n + 4 } ∪ [ n + 5 , n + 2 k + 1] ∪ { n + 2 k + 2 } ,O = { n + m + 2 k + 3 } ∪ [ n + m + 2 k + 4 , n + m + 4 k ] ∪ { n + m + 4 k + 1 } ,O = [ n + 1 , n + 3] ∪ [ n + 6 , n + 2 k ] ∪ [ n + 2 k + 3 , n + 2 k + 5] ,O = [ n + m + 2 k, n + m + 2 k + 2] ∪ [ n + m + 2 k + 5 , n + m + 4 k − ∪ [ n + m + 4 k + 2 , n + m + 4 k + 4] . Let M ⊆ [ n + 2 k + 6 , n + m + 2 k − such that within M , there exists a sequence ofpairs of consecutive elements, where consecutive pairs in the sequence are not morethan k − apart and the sequence starts with a pair in [ n + 2 k + 6 , n + 4 k + 1] andends with a pair in [ n + m + 4 , n + m + 2 k − . Let M ⊆ [ n + 2 k + 6 , n + m + 2 k − NTEGERS: 19 (2019) such that within M , there exists a sequence of triplets of consecutive elements,where consecutive triplets in the sequence are not more than k + 5 apart and thesequence starts with a triplet in [ n + 2 k + 6 , n + 4 k + 5] and ends with a triplet in [ n + m, n + m +2 k − . Also, M ∩ M = ∅ and M ∪ M = [ n +2 k +6 , n + m +2 k − .Then A ′ = L ∪ O ∪ M ∪ O ∪ R ′ A ′ = L ∪ O ∪ M ∪ O ∪ R ′ are both MSTD and partition [1 , n + m + 4 k + 4] . Remark 1.2.
To show that our family is not empty, we need to show the existenceof at least one pair of A and A . Note that our technique is similar to many otherpapers [2, 7, 8, 9, 12] in the sense that we need a good fringe to start with. Arandom search yielded A = { , , , , , , , , , , , , , , , , , , , , } ,A = { , , , , , , , , , , , , , , , , , , } . We have A + A = [2 , A − A = [ − , \{± } and A + A = [10 , A − A = [ − , \{± } . Clearly, A is P and A is P . It can be easily checked that all conditionsmentioned in Theorem 1.1 are satisfied. These pairs of sets A and A are not hardfor computers to find: for n = 20, computer search shows that there are 48 suchpairs. Remark 1.3.
Our method of decomposing I r into two MSTD sets allows a lot offreedom in choosing the middle elements. This is because once the fringe elementsare chosen, the conditions placed on M and M are relatively weak.Next, we answer positively question (3) in [1], where the authors ask: Can wedecompose { , , . . . , r } into three sets which are MSTD? For any finite number k ,is there a sufficiently large r for which there is a k -decomposition into MSTD sets? Theorem 1.4.
Let k ∈ N ≥ be chosen.1. There exists the smallest R ( k ) ∈ N such that for all r ≥ R ( k ) , I r can be k -decomposed into MSTD subsets, while I R ( k ) − cannot be k -decomposed intoMSTD subsets.2. In particular, we find some rough bounds : We make no attempt to optimize these bounds. Finer analysis may give us better bounds.
NTEGERS: 19 (2019) (a) when k is even, k ≤ R ( k ) ≤ k ,(b) when k ≥ odd, k ≤ R ( k ) ≤ k − , and(c) when k = 3 , ≤ R ( k ) ≤ T + 24 ,where T = min { max A : | A + A | − | A − A | ≥ | A |} . We prove Theorem 1.4 using sets constructed by the base expansion method that helps generate an infinite family of MSTD sets from a given MSTD set. Themethod is a very powerful tool and has been used extensively in the literatureincluding [2],[4] and [5]. However, the base expansion method turns out to beinefficient in terms of our MSTD sets’ cardinality. Hence, we present a second,more efficient approach by using a particular family of MSTD sets. We presentboth proofs since they are of independent interest: the first proof is less technicalbut less efficient. Also, the second proof cannot resolve the case k = 3 while thefirst can.Lastly, we give a sufficient condition on when there exists a positive constantlower bound for the proportion of k -decompositions of I r into MSTD subsets. Thecondition offers an alternative proof of Theorem 1.4 in [1] ( k = 2). Due to thecondition, we make the following conjecture. Conjecture 1.5.
For any finite k ≥ , the proportion of k -decompositions intoMSTD subsets is bounded below by a positive constant. The outline of the paper is as follows. In Section 2, we provide an efficientmethod to decompose I r into two MSTD subsets; Section 3 presents two methodsto decompose I r into k ≥ k -decompositions into MSTD subsets inAppendix B. Appendix C gives a proof of Theorem 3.4. Appendix D contains manyexamples illustrating our lemmas and theorems.
2. Explicit 2-decomposition into MSTD Subsets
In this section, we show how we can decompose I r into two MSTD subsets. Webelieve that the method can be applied to the general case of k -decompositions,but the proof will be much more technical. However, for k ≥
4, we have a way todecompose I r into k MSTD subsets by simply using 2-decompositions, which willbe discussed later. We can generate an infinite family of MSTD sets from a given MSTD set through the baseexpansion method. Let A be an MSTD set, and let A k,m = { P ki =1 a i m i − : a i ∈ A } . If m issufficiently large, then | A k,m ± A k,m | = | A ± A | k and | A k,m | = | A | k . NTEGERS: 19 (2019) The following lemma is useful in proving many of our results.
Lemma 2.1.
Let A = L ∪ R be an MSTD, P n set containing and n , where L ⊆ [1 , n ] and R ⊆ [ n + 1 , n ] . Form A ′ = L ∪ M ∪ R ′ , where M ⊆ [ n + 1 , n + m ] and R ′ = R + m for some m ∈ N . If A ′ is a SP n set, then A ′ is MSTD.Proof. We prove that A ′ is MSTD by showing that the increase in the number ofdifferences is at most the increase in the number of sums. As shown in the proofof Lemma 2.1 in [8], the number of new added sums is 2 m . Because R ′ = R + m ,all differences in [ − (2 n + m − , − ( n + m )] can be paired up with differences in[1 − n, − n ] from L − R and differences in [ n + m, n + m −
1] can be paired up withdifferences in [ n, n −
1] from R − L . Because the set A is P n , A contains all numbersin [ − n + 1 , n − A ′ − A ′ contains all differences in [ − ( n + m ) + 1 , ( n + m ) − | A ′ − A ′ | − | A − A | = | [ − ( n + m ) + 1 , ( n + m ) − | − | [ − n + 1 , n − | = 2 m .This completes our proof. Corollary 2.2.
Let A = L ∪ R be an MSTD, P n − set containing and n − , where L ⊆ [5 , n ] and R ⊆ [ n + 1 , n − . Form A ′ = L ∪ M ∪ R ′ , where M ⊆ [ n + 1 , n + m ] and R ′ = R + m for some m ∈ N . If A ′ is a SP n − set, then A ′ is MSTD.Proof. In Lemma 2.1, we use n − n , then consider 4 + A . Lemma 2.3.
Let an MSTD, P n set A be chosen, where A = L ∪ R for L ⊆ [1 , n ] and R ⊆ [ n + 1 , n ] . Additionally, L and R must satisfy the following conditions: [1 , ∪ { n } ⊆ L and { n + 1 } ∪ [2 n − , n ] ⊆ R . Pick k ≥ n/ and m ∈ N .Form O = { n + 4 } ∪ [ n + 5 , n + 2 k + 1] ∪ { n + 2 k + 2 } O = { n + m + 2 k + 3 } ∪ [ n + m + 2 k + 4 , n + m + 4 k ] ∪ { n + m + 4 k + 1 } . Let M ⊆ [ n + 2 k + 3 , n + m + 2 k +2] be such that within M , there exists a sequence ofpairs of consecutive elements, where consecutive pairs in the sequence are not morethan k − apart and the sequence starts with a pair in [ n + 2 k + 3 , n + 4 k + 1] andends with a pair in [ n + m + 4 , n + m + 2 k + 2] . Denote A ′ = L ∪ O ∪ M ∪ O ∪ R ′ ,where R ′ = R + m + 4 k + 4 . Then A ′ is MSTD.Proof. We know that A ′ ⊆ [2 , n +2 m +8 k +8]. To prove that A ′ is MSTD, it sufficesto prove that A ′ is SP n . In particular, we want to show that [ n +2 , n +2 m +8 k +8] ⊆ A ′ + A ′ . Due to symmetry , it suffices to show that [ n +2 , n + m +4 k +5] ⊆ A ′ + A ′ . Due to symmetry, A ′ has the same structure as (2 n + m +4 k +5) − A ′ . If [ n +2 , n + m +4 k +5] ⊆ A ′ + A ′ , then [ n + 2 , n + m + 4 k + 5] ⊆ (2 n + m + 4 k + 5 − A ′ ) + (2 n + m + 4 k + 5 − A ′ ) =(4 n + 2 m + 8 k + 10) − ( A ′ + A ′ ) and so, [2 n + m + 4 k + 5 , n + 2 m + 8 k + 8] ⊆ A ′ + A ′ . NTEGERS: 19 (2019) n + 2 , n + 4] ⊆ A ′ + A ′ (because 2 , , , n ∈ A ′ )(1 + O ) ∪ (2 + O ) = [ n + 5 , n + 2 k + 4] O + O = [2 n + 8 , n + 4 k + 4] . Since n + 2 k + 4 ≥ n + 8, [ n + 2 , n + 4 k + 4] ⊆ A ′ + A ′ . Consider M + O . Inthe worst scenario (in terms of getting necessary sums), the two smallest elementsof M are n + 4 k and n + 4 k + 1, while the two largest elements of M are n + m + 4and n + m + 5. We have M + O ⊇ [2 n + 4 k + 4 , n + m + 2 k + 7]. We completethe proof by showing that [2 n + m + 2 k + 8 , n + m + 4 k + 5] ⊆ A ′ + A ′ . We have(( n + 4) + O ) ∪ (( n + 5) + O ) = [2 n + m + 2 k + 7 , n + m + 4 k + 6]. So, A ′ is SP n and thus, MSTD by Lemma 2.1. Lemma 2.4.
Let an MSTD, P n − set A be chosen, where A = L ∪ R for L ⊆ [5 , n ] and R ⊆ [ n +1 , n − . Additionally, L and R must satisfy the following conditions: [5 , ⊆ L and [2 n − , n − ⊆ R . Pick k ≥ n/ − and m ∈ N . Form O = [ n + 1 , n + 3] ∪ [ n + 6 , n + 2 k ] ∪ [ n + 2 k + 3 , n + 2 k + 5] ,O = [ n + m + 2 k, n + m + 2 k + 2] ∪ [ n + m + 2 k + 5 , n + m + 4 k − ∪ [ n + m + 4 k + 2 , n + m + 4 k + 4] . Let M ⊆ [ n + 2 k + 6 , n + m + 2 k − such that within M , there exists a sequence oftriplets of consecutive elements, where consecutive triplets in the sequence are notmore than k +5 apart and the sequence starts with a triplet in [ n +2 k +6 , n +4 k +5] and ends with a triplet in [ n + m, n + m + 2 k − . Denote A ′ = L ∪ O ∪ M ∪ O ∪ R ′ ,where R ′ = R + m + 4 k + 4 . Then A ′ is MSTD.Proof. We want to show that A ′ is SP n − . By Corollary 2.2, we know that A ′ isMSTD. In particular, we prove that [ n + 6 , n + 2 m + 8 k + 4] ∈ A ′ + A ′ . It sufficesto prove that [ n + 6 , n + m + 4 k + 5] ⊆ A ′ + A ′ . We have(5 + O ) ∪ (6 + O ) ∪ (7 + O ) = [ n + 6 , n + 2 k + 12] O + O = [2 n + 2 , n + 4 k + 10] . Because n + 2 k + 12 ≥ n + 2, A ′ + A ′ contains [ n + 6 , n + 4 k + 10]. Consider M + O . In the worst scenario (in terms of getting sums), the smallest elements in M are n + 4 k + 3 , n + 4 k + 4 and n + 4 k + 5, while the largest elements in M are n + m, n + m + 1 and n + m + 2. Then M + O ⊇ [2 n + 4 k + 4 , n + m + 2 k + 7]. Due to symmetry, A ′ has the same structure as (2 n + m +4 k +5) − A ′ . If [ n +6 , n + m +4 k +5] ⊆ A ′ + A ′ , then [ n + 6 , n + m + 4 k + 5] ⊆ (2 n + m + 4 k + 5 − A ′ ) + (2 n + m + 4 k + 5 − A ′ ) =(4 n + 2 m + 8 k + 10) − ( A ′ + A ′ ) and so, [2 n + m + 4 k + 5 , n + 2 m + 8 k + 4] ⊆ A ′ + A ′ . NTEGERS: 19 (2019) n + m + 2 k + 8 , n + m + 4 k + 5] ⊆ A ′ + A ′ . We have(( n + 1) + O ) ∪ (( n + 2) + O ) ∪ (( n + 3) + O ) = [2 n + m + 2 k + 1 , n + m + 4 k + 7].This completes our proof that A ′ is SP n − and thus, MSTD. We are now ready to prove Theorem 1.1. The proof follows from Lemma 2.3 andLemma 2.4.
Proof.
As indicated in Remark 1.2, there exist pairs of sets A and A such that allconditions in Theorem 1.1 are satisfied. Pick k ≥ n/ m sufficiently large.Set R ′ i = R i + m + 4 k + 4 for i = 1 ,
2. Form O = { n + 4 } ∪ [ n + 5 , n + 2 k + 1] ∪ { n + 2 k + 2 } ,O = { n + m + 2 k + 3 } ∪ [ n + m + 2 k + 4 , n + m + 4 k ] ∪ { n + m + 4 k + 1 } ,O = [ n + 1 , n + 3] ∪ [ n + 6 , n + 2 k ] ∪ [ n + 2 k + 3 , n + 2 k + 5] ,O = [ n + m + 2 k, n + m + 2 k + 2] ∪ [ n + m + 2 k + 5 , n + m + 4 k − ∪ [ n + m + 4 k + 2 , n + m + 4 k + 4] . We see that O ∪ O = [ n +1 , n +2 k +5] and O ∪ O = [ n + m +2 k, n + m +4 k +4].By Lemma 2.3 and Lemma 2.4, we know that A ′ = L ∪ O ∪ M ∪ O ∪ R ′ and A ′ = L ∪ O ∪ M ∪ O ∪ R ′ are MSTD sets and ( A ′ , A ′ ) partitions [1 , n + m + 4 k + 4],given that the following three conditions are satisfied:1. M ⊆ [ n + 2 k + 6 , n + m + 2 k −
1] such that within M , there exists a sequenceof pairs of consecutive elements, where consecutive pairs in the sequence arenot more than 2 k − n + 2 k +6 , n + 4 k + 1] and ends with a pair in [ n + m + 4 , n + m + 2 k − M ⊆ [ n + 2 k + 6 , n + m + 2 k −
1] such that within M , there exists a sequenceof triplets of consecutive elements, where consecutive triplets in the sequenceare not more than 2 k + 5 apart and the sequence starts with a triplet in[ n + 2 k + 6 , n + 4 k + 5] and ends with a triplet in [ n + m, n + m + 2 k − M ∪ M = [ n + 2 k + 6 , n + m + 2 k −
1] and M ∩ M = ∅ .For m sufficiently large, it is obvious that M and M exist. This completes theproof of the theorem. Remark 2.5.
Observe that our fringe pairs in this case are ( L ∪ O , R ′ ∪ O )and ( L ∪ O , R ′ ∪ O ). Though disjoint, the union of the two fringe pairs givesus a full set of left and right elements of [1 , n + m + 4 k + 4] and each is a fringepair for an MSTD set. NTEGERS: 19 (2019)
3. Explicit k -decomposition into MSTD Subsets3.1. Overview Theorem 1.1 gives us a way to partition I r into two MSTD subsets. Due to lineartransformations, we can partition any (long enough) arithmetic progressions intotwo MSTD subsets. If we can find an MSTD subset S of I r such that I r \ S is aunion of k arithmetic progressions ∪ kj =1 P j , then we can partition I r into 1 + 2 k MSTD subsets (because each P j can be partitioned into two MSTD subsets). Thisis the central idea in both methods we use to k -decompose I r into MSTD subsetspresented later. k ≥ We explicitly provide a way to k -decompose I r into MSTD subsets. First, we needto define a “strong MSTD” set. We call a set S a 10-strong MSTD set if | S + S | −| S − S | ≥ | S | . This type of MSTD set does exist. For example, using the baseexpansion method, we can construct such a set. Using ˜ S = { , , , , , , , } ,by the method, we can construct S such that | S | = | ˜ S | = 8 = 4096, | S + S | =26 = 456976 and | S − S | = 25 = 390625; then, | S + S | − | S − S | > | S | . Lemma 3.1. If S is a -strong MSTD set, then S ∪ { a , a , a , a } , where a >a > a > a > max S , is an MSTD set. Similarly, S ∪ { b , b , b , b } , where b < b < b < b < min S , is also an MSTD set.Proof. We want to show that S ∪{ a , a , a , a } is MSTD. Adding one more elementto a set S produces at least 0 new sums and at most 2 | S | new differences. So, | ( S ∪ { a } ) + ( S ∪ { a } ) | − | ( S ∪ { a } ) − ( S ∪ { a } ) | ≥ | S | + 0 − | S | = 8 | S | .So, S ∪ { a } is MSTD. Define S = S ∪ { a } with | S | = | S | + 1. Similarly, | ( S ∪ { a } ) + ( S ∪ { a } ) | − | ( S ∪ { a } ) − ( S ∪ { a } ) | ≥ | S | + 0 − | S | =8 | S | − | S | + 1) = 6 | S | −
2. Again, S ∪ { a } is MSTD. Repeating this argument,we can show that S = S ∪ { a , a , a , a } is an MSTD set. The proof is similarfor S ∪ { b , b , b , b } . Note that in the end, we reach the requirement that | S | > Remark 3.2.
The following fact will be useful later: An arithmetic progression ofintegers (assumed long enough) can contain an arbitrarily large number of disjoint10-strong MSTD sets.With the above remark, we are ready to prove the following.
Lemma 3.3.
There exists N ∈ N such that for all r ≥ N , I r can be partitionedinto exactly three MSTD subsets. We pick the number 10 just to be safe for our later arguments. We make no attempt to providean efficient way to decompose I r into k MSTD subsets.
NTEGERS: 19 (2019) Proof.
We use a pair of fringe elements described in [7]: L = { , , , , , , } and R = { r − , r − , r − , r − , r − , r − , r − , r } . We see that I r \ ( L ∪ R ) = { , , , } ∪ [12 , r − ∪ { r − , r − , r − } . We have L + L = [2 , \ { } L + R = [ r − , r + 11] R + R = [2 r − , r ] . Consider K = { ℓ | ≤ ℓ ≤ r − , ℓ is even } ∪ { r − } . We have ( { } ∪ K ∪ { r − } )+( { }∪ K ∪{ r − } ) = [22 , r − L ∪ K ∪ R )+( L ∪ K ∪ R ) = [2 , r ] \{ } .Because ± ( R − L ) lacks ± ( r − L ∪ K ∪ R is an MSTD set. It is not hard to seethat adding numbers in [12 , r − \ K to L ∪ K ∪ R still gives an MSTD set.Now, [12 , r − \ K contains an arithmetic progression of consecutive odd inte-gers. We can make this arithmetic progression arbitrarily large by increasing r . ByRemark 3.2, this arithmetic progression can contain two disjoint 10-strong MSTDsets, called S and S . We write [12 , r − \ K = S ∪ S ∪ M . By Lemma 3.1, S ∗ = S ∪ { , , , } and S ∗ = S ∪ { r − , r − , r − } are both MSTD. By whatwe say above, K ∗ = M ∪ L ∪ K ∪ R is also MSTD. Because S ∗ ∪ S ∗ ∪ K ∗ = I r , wehave completed the proof. Proof of item (1) of Theorem 1.4.
Let k ≥ k = 2 m + 3 m forsome m and m ∈ N . We can find N ∈ N such that I N = [1 , N ] = (cid:0) [1 , k ] ∪ [ k + 1 , k ] ∪ · · · ∪ [ k m − + 1 , k m ] (cid:1) ∪ (cid:0) [ k m + 1 , k m +1 ] ∪ [ k m +1 + 1 , k m +2 ] ∪ · · · ∪ [ k m + m − + 1 , N ] (cid:1) , such that each of the first m intervals are large enough to bepartitioned into two MSTD sets while the next m intervals are large enough to bepartitioned into three MSTD sets. So, I N can be partitioned in exactly k MSTDsets. This completes our proof. k ≥ We introduce a notation to write a set; this notation was first used by Spohn [16].Given a set S = { a , a , . . . , a n } , we arrange its elements in increasing order andfind the differences between two consecutive numbers to form a sequence. Supposethat a < a < · · · < a n ; then our sequence is a − a , a − a , a − a , . . . , a n − a n − .Then we represent S = ( a | a − a , a − a , a − a , . . . , a n − a n − ) . Take S = { , , , , } , for example. We arrange the elements in increasing orderto have 2 , , , ,
10 and form a sequence by looking at the difference between twoconsecutive numbers: 1 , , ,
1. So, we write S = (2 | , , , NTEGERS: 19 (2019) Lemma 3.4.
The following are MSTD sets for a given m ∈ N : (1 | , , , , . . . , | {z } m -times , , , , , (1 | , , , , . . . , | {z } m -times , , , , , , (1 | , , , , . . . , | {z } m -times , , , . We present the proof in Appendix C. k ≥ ) In our decomposition of I r into k MSTD subsets for k ≥
3, we use the base expansionmethod. However, the base expansion method is inefficient in terms of cardinalitiesof our sets. Is there a more efficient way to decompose? In answering this question,we present a method of decomposing I r into k MSTD subsets ( k ≥
4) that helpsreduce the cardinalities of sets. We use the infinite family of MSTD sets in Lemma3.4 to achieve this.We want to decompose I r for sufficiently large r into k ( k ≥
4) MSTD subsets.If k is even, we can simply write I r as the union of k/ k ≥ r mod 4. If r ≡ r = 4 m + 13 for some m ∈ N and consider (1 | , , , , . . . , | {z } m -times , , , , I r \{ , , , , , , , , , , . . ., m, m,
10 + 4 m,
11 + 4 m,
13 + 4 m } = { , , , , , . . . , m,
12 + 4 m } ∪ { , , , , . . . , m } . Notice that both { , , , , , . . . , m,
12 + 4 m } and { , , , , . . . , m } are arithmetic progressions and each of these sets can be decomposed into an evennumber of MSTD sets. So, our original sets I r = [1 ,
13 + 4 m ] can be decomposedinto exactly k MSTD sets. If r ≡ r = 4 m + 14 and consider(1 | , , , , . . . , | {z } m -times , 3 , , , , r ≡ r = 4 m + 11 and con-sider (1 | , , , , . . . , | {z } m -times , , , r ≡ r = 4 m + 12 and consider(2 | , , , , . . . , | {z } m -times , , , I r can be decomposed into exactly k MSTD sets. We prove the upper and lowerbounds for R ( k ) in Appendix A. NTEGERS: 19 (2019)
4. Future Work
We end with several additional questions to pursue.1. In [1], the authors show that there is a positive constant lower bound forthe percentage of decompositions into two MSTD sets. Is there a positiveconstant lower bound for the percentage of decompositions into k MSTD setsfor k ≥
3? In other words, is Conjecture 1.5 true? A method is to find afamily of sets ( A i ) ki =1 that satisfies the condition in Theorem B.4.2. Is there a method of k -decomposition that is of high density, for exampleΘ(1 /r c ) for small c ?3. For the 3-decomposition, we use the base expansion method, which is ineffi-cient. Can we find an efficient way to decompose I r into three MSTD subsets.4. Can we find some better bounds for R ( k ) in Theorem 1.4? There is a yawninggap between our upper and lower bounds.5. Suppose that I r can be decomposed into k MSTD subsets. Can we concludethat I r +1 can be decomposed into k MSTD subsets?
A. Lower and Upper Bounds for R ( k ) in Theorem 1.4 Given k ≥
2, the lower bound is obvious since by [2], the smallest cardinality of anMSTD set is 8. In [1], it is shown that for all r ≥ I r can be partitioned intotwo MSTD subsets. To decompose I r into k ≥ I r to be the union of k/ r ≥ k , I r can be decomposed into k MSTD subsets. Hence, R ( k ) ≤ k .In our method to decompose I r into k ≥ A = (1 | , , , , . . . , | {z } m -times , , , , ,A = (1 | , , , , . . . , | {z } m -times , , , , , ,A = (1 | , , , , . . . , | {z } m -times , , , ,A = (2 | , , , , . . . , | {z } m -times , , , . NTEGERS: 19 (2019) I max A i \ A i , i ∈ [1 ,
4] is the union of two arithmeticprogressions. Given m , I max A \ A gives a pair of arithmetic progressions of shortestlength, m + 2 and 2 m + 1, while max A = 4 m + 14 = max { max A i | i ∈ [1 , } . Weconsider two cases.1. k = 4 j + 1 ( j ≥ m + 2 can be partitioned into 2 j MSTD sets. Then m + 2 ≥ j and so, m ≥ j −
2, which also guarantees that all arithmetic progressions of lengthat least 2 m + 1 can be partitioned into 2 j MSTD sets. So, we find out that for r ≥ j −
2) + 14 = 20 k − I r can be partitioned into k MSTD subsets.Hence, R ( k ) ≤ k − k = 4 j + 3 ( j ≥ m + 2 can be partitioned into 2 j MSTD sets. Then m + 2 ≥ j and so, m ≥ j −
2, which also guarantees that all arithmetic progressions of lengthat least 2 m + 1 can be partitioned into 2 j + 2 MSTD sets. So, we find outthat for r ≥ j −
2) + 14 = 20 k − I r can be partitioned into k MSTDsubsets. Hence, R ( k ) ≤ k − r − + 1.Let T be min { max A : A is 10 − strong } . Then we require r − + 1 ≥ T or r ≥ T + 24. Hence, 24 ≤ R ( k ) ≤ T + 24. B. Sufficient Condition for a Positive Constant BoundLemma B.1.
Consider S ⊆ { , , . . . , r − } and S = L ∪ M ∪ R . Fix L ⊆ [0 , ℓ − and R ⊆ [ r − ℓ, r − for some fixed ℓ . Let M be a uniformly randomly chosensubset of [ ℓ, r − ℓ − . Then for any ε > , there exists sufficiently large r such that P ([2 ℓ − , r − ℓ − ⊆ S + S ) ≥ − −| L | + 2 −| R | ) − ε. (3) Proof.
We write P ([2 ℓ − , r − ℓ − ⊆ S + S ) = 1 − P ([2 ℓ − , r − ℓ − S + S )= 1 − P ([2 ℓ − , r − ℓ − ∪ [ r + ℓ − , r − ℓ − S + S or [ r − ℓ, r + ℓ − S + S ) ≥ − P ([2 ℓ − , r − ℓ − ∪ [ r + ℓ − , r − ℓ − S + S − P ([ r − ℓ, r + ℓ − S + S ) . (4)By Proposition 8 in [7], P ([2 ℓ − , r − ℓ − ∪ [ r + ℓ − , r − ℓ − S + S ) ≤ −| L | + 2 −| R | ) . (5) NTEGERS: 19 (2019) P ([ r − ℓ, r + ℓ − S + S ) ≤ P ([ r − ℓ, r + ℓ − M + M ) , (6)because [ r − ℓ, r − S + S implies [ r − ℓ, r + ℓ − M + M . By a linear shift of ℓ , we can consider M a subset of [0 , r − ℓ −
1] and P ([ r − ℓ, r + ℓ − M + M ) turnsinto P ([ r − ℓ, r − M + M ). Use the change of variable N = r − ℓ . We have: M ⊆ [0 , N −
1] and we estimate: P ([ r − ℓ, r − M + M ) = P ([ N, N + 2 ℓ − M + M ) ≤ P N +2 ℓ − k = N P ( k / ∈ M + M ). Lemma 7 in [7] shows that the last quantitytend to 0 as N goes to infinity. So, for any ε >
0, there exists sufficiently large r such that P ([ r − ℓ, r + ℓ − S + S ) < ε . This completes our proof. Lemma B.2.
Consider S ⊆ { , , . . . , r − } and S = L ∪ M ∪ R . Fix L ⊆ [0 , ℓ − and R ⊆ [ r − ℓ, r − for some fixed ℓ . Let M be a uniformly randomly chosen subsetof [ ℓ, r − ℓ − . Let a denote the smallest integer such that both [ ℓ, ℓ − a ] ⊆ L + L and [2 r − ℓ + a − , r − ℓ − ⊆ R + R . Then, for all ε > , there exists sufficientlylarge r such that P ([2 ℓ − a + 1 , r − ℓ + a − ⊆ S + S ) ≥ − ( a − − τ ( R ) + 2 − τ ( L ) ) − −| L | + 2 −| R | ) − ε, (7) where τ ( L ) = |{ i ∈ L | i ≤ ℓ − a + 1 }| and τ ( R ) = |{ i ∈ R | i ≥ r − ℓ + a − }| .Proof. We have: P ([2 ℓ − a + 1 , r − ℓ + a − ⊆ S + S )= P ([2 ℓ − , r − ℓ − ⊆ S + S and [2 ℓ − a + 1 , ℓ − ⊆ S + S and [2 r − ℓ, r − ℓ + a − ⊆ S + S )= 1 − P ([2 ℓ − , r − ℓ − S + S or [2 ℓ − a + 1 , ℓ − S + S or [2 r − ℓ, r − ℓ + a − S + S ) ≥ − P ([2 ℓ − a + 1 , ℓ − S + S ) − P ([2 r − ℓ, r − ℓ + a − S + S ) − P ([2 ℓ − , r − ℓ − S + S ) . By Lemma B.1, P ([2 ℓ − , r − ℓ − S + S ) ≤ −| L | + 2 −| R | ) + ε . We have P ([2 ℓ − a + 1 , ℓ − S + S ) ≤ ℓ − X k =2 ℓ − a +1 P ( k / ∈ S + S ) . (8)Let τ ( L ) = |{ i ∈ L | i ≤ ℓ − a + 1 }| and τ ( R ) = |{ i ∈ R | i ≥ r − ℓ + a − }| . Foreach value of k in [2 ℓ − a + 1 , ℓ − k / ∈ S + S , all pairs of numbersthat sum up to k must not be both in S . Take k = 2 ℓ − a + 1, for example. For a NTEGERS: 19 (2019) x ≤ ℓ − a + 1, the number y that when added to x gives 2 ℓ − a + 1 is atleast ℓ and y / ∈ S . So, P ( k / ∈ S + S ) ≤ − τ ( L ) and hence, P ([2 ℓ − a + 1 , ℓ − S + S ) ≤ ( a − − τ ( L ) . (9)Similarly, P ([2 r − ℓ, r − ℓ + a − S + S ) ≤ ( a − − τ ( R ) . (10)We have shown that P ([2 ℓ − a + 1 , r − ℓ + a − ⊆ S + S ) ≥ − ( a − − τ ( R ) + 2 − τ ( L ) ) − −| L | + 2 −| R | )) − ε. This completes our proof.
Corollary B.3.
Consider set A = L ∪ R , where L ⊆ [0 , n − and R ⊆ [ n, n − .Let m ∈ N be chosen. Set R ′ = R + m ⊆ [ n + m, n + m − . Let M ⊆ [ n, n + m − be chosen uniformly at random. Form S = L ∪ M ∪ R ′ . Let a be the smallest integersuch that [ n, n − a ] ⊆ L + L and [2 n + a − , n − ⊆ R + R . Then for all ε > ,there exists sufficiently large m such that P ([2 n − a + 1 , n + 2 m + a − S + S ) ≤ ( a − − τ ( R ) + 2 − τ ( L ) ) + 6(2 −| L | + 2 −| R | ) + ε, (11) where τ ( L ) = |{ i ∈ L | i ≤ n − a + 1 }| and τ ( R ) = |{ i ∈ R | i ≥ n + a − }| .Proof. The corollary follows immediately by setting r = 2 n + m and ℓ = n in LemmaB.2. Also, notice that R ′ is a linear shift of R .For conciseness, we denote f ( L, R ) = ( a − − τ ( R ) + 2 − τ ( L ) ) + 6(2 −| L | + 2 −| R | ) . Theorem B.4.
Suppose that there exist sets ( A i ) ki =1 , which are pairwise disjoint,MSTD, P n and ∪ ki =1 A i = [0 , n − . In particular, each A i = L i ∪ R i , where L i ⊆ [0 , n − and R i ⊆ [ n, n − . Let m ∈ N be chosen. Form R ′ i = R i + m and ( M i ) ki =1 ⊆ [ n, n + m − such that ( M i ) ki =1 are pairwise disjoint and ∪ ki =1 M i =[ n, n + m − . Then the proportion of cases where all S i = L i ∪ M i ∪ R ′ i are MSTDis bounded below by a positive constant if − k X i =1 f ( L i , R i ) > , (12) for m sufficiently large. In other words, there exists a positive constant lower boundfor the proportion of k -decompositions into MSTD subsets as m → ∞ . NTEGERS: 19 (2019) Proof.
Let a i be the corresponding a value (defined in Lemma B.2) for L i and R i .By Corollary B.3, for any ε > m sufficiently large, the probability P ( ∀ i, S i is MSTD) ≥ P ( ∀ i, S i + S i ⊇ [2 n − a i + 1 , n + 2 m + a i − ≥ − P ( ∃ i, S i + S i [2 n − a i + 1 , n + 2 m + a i − > − k X i =1 f ( L i , R i ) − ε > . (13)The first inequality is because [2 n − a i + 1 , n + 2 m + a i − ⊆ S i + S i guaranteesthat S i is SP n and thus, MSTD. By Lemma 2.1, S i is MSTD.Now, we prove Theorem 1.4 in [1] easily. Corollary B.5.
There exists a constant c > such that the proportion of 2-decompositions of [0 , r − into two MSTD subsets is at least c .Proof. Let L = { , , , , , , , , , , } ,R = { , , , , , , , , , } ,L = { , , , , , , , , } ,R = { , , , , , , , , , } . Notice that n = 20. We find that a = 12 and a = 4. From that we calculate τ ( L ) = 6 , τ ( R ) = 6 , τ ( L ) = 8 and τ ( R ) = 9. So, f ( L , R ) is less than 0.33,while f ( L , R ) is less than 0.03 for m sufficiently large. By Theorem B.4, we aredone. C. Proof of Lemma 3.4
We prove that for a fixed m ∈ N , S = (0 | , , , , . . . , | {z } m -times , , , ,
2) is MSTD. Theproof for other sets in the lemma follows similarly.Note that max S = 12 + 4 m . We will prove that | S + S | ≥
26 + 6 m . Since S contains 0, 1 and 2, if the difference between two numbers, say x < y , in S is lessthan or equal to 3, then S + S contains [ x, y ]. If a, b ∈ S and a − b = 4, then inthe worst case (in term of cardinality of the sum set), S + S does not contain a − ,
12 + 4 m ], S + S misses at most m − m differences of 4 and 8 = 4 + 4 ∈ S + S . Next, consider [13 + 4 m,
24 + 8 m ]and observe that S = { ℓ | ≤ ℓ ≤ m and ℓ ≡ } ⊆ S . Since 0 ∈ S , NTEGERS: 19 (2019) S ∈ S + S . We also have(12 + 4 m ) + S = { ℓ |
13 + 4 m ≤ ℓ ≤
21 + 8 m and ℓ ≡ } , (10 + 4 m ) + S = { ℓ |
11 + 4 m ≤ ℓ ≤
19 + 8 m and ℓ ≡ } , (9 + 4 m ) + S = { ℓ |
10 + 4 m ≤ ℓ ≤
18 + 8 m and ℓ ≡ } . Note that 16 + 4 m = (12 + 4 m ) + 4 ∈ S + S,
16 + 8 m = (8 + 4 m ) + (8 + 4 m ) ∈ S + S,
20 + 8 m = (10 + 4 m ) + (10 + 4 m ) ∈ S + S,
22 + 8 m = (10 + 4 m ) + (12 + 4 m ) ∈ S + S,
24 + 8 m = (12 + 4 m ) + (12 + 4 m ) ∈ S + S. On the interval [13 + 4 m,
24 + 8 m ], S + S misses at most the whole set { ℓ |
20 + 4 m ≤ ℓ ≤
12 + 8 m and ℓ ≡ } ∪ {
23 + 8 m } , which has m numbers. Therefore, intotal, S + S misses at most 2 m − S − S misses at least 2 m numbers by provingthat S − S contains none of the elements in { ℓ | ≤ ℓ ≤ m − } . We use proofby contradiction. Suppose that there exists 0 ≤ ℓ ≤ m − ℓ is in S − S . Then, there must exist a run within 1 , , , , . . . , | {z } m -times , , , , ℓ . Because 6 + 4 ℓ ≡ , , , , , ,
2. Consider the following two cases:1.
Case I : the run starts within 1 , , ,
1. Because 1 + 1 + 2 + 1 = 5 <
6, therun must end within 3 , , ,
2. Therefore, the run sums up to a number of theform a + 4 m + b , where the value of a and b depend on where the run startsand where it ends, respectively. Since a + 4 m + b = 6 + 4 ℓ ≤ m − a + b ≤
2. This is a contradiction because b ≥ Case II : the run ends within 3 , , ,
2. Because there is no run within 4 . . . , | {z } m -times , , , , ℓ , the run must start within 1 , , ,
1. Repeatingthe argument used in
Case I and we have a contradiction.Therefore, ( S − S ) ∩ { ℓ | ≤ ℓ ≤ m − } = ∅ and so, S − S misses at least 2 m elements. This completes our proof that S is MSTD. ✷ NTEGERS: 19 (2019) D. ExamplesD.1. Theorem 1.1
We use A and A mentioned in Remark 1.2. Pick k = 12 and m = 30. Set A ′ e = { , , , , , , , , , , }∪ { } ∪ [25 , ∪ { } ∪ { , , , }∪ { } ∪ [78 , ∪ { }∪ { , , , , , , , , , } ,A ′ e = { , , , , , , , , }∪ [21 , ∪ [26 , ∪ [47 , ∪ { , , , , , }∪ [74 , ∪ [79 , ∪ [100 , ∪ { , , , , , , , , , } . We have | A ′ e + A ′ e | − | A ′ e − A ′ e | = 243 −
241 = 2, | A ′ e + A ′ e | − | A ′ e − A ′ e | =227 −
225 = 2 and ( A ′ e , A ′ e ) partitions [1 , D.2. 5-decompositions
We do not give an example of a 3-decomposition because our method is inefficientand involves a large set arising from the base expansion method. Neither do wegive an example of a 4-decomposition because the method is straightforward. Weuse the efficient method to have a 5-decomposition into MSTD sets. Set M = (1 | , , , , . . . , | {z } , , , , { , , , } ∪ [6 , ∪ { , , , } . Observe that [1 , \ M = [4 , ∪ [7 , . Because [4 , is an arithmetic progression of length 122, we have M = 4 A ′ e and M = 4 A ′ e partition [4 , . Notice that in the example mentioned in D.1, we can pick m =147 and find A ′′ e (containing 1) and A ′′ e that partition [1 , M = 2 A ′′ e + 5 and M = 2 A ′′ e + 5partition [7 , . We have found ( M , M , M , M , M ) that partitions [1 , NTEGERS: 19 (2019) Acknowledgement.
We thank the referee for helpful comments on an earlier draft. We thank theparticipants from the 2018 SMALL REU program for many helpful conversations.
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