n ). This is tight for all admissible values of t, k and n. k t + + n t

Transcription

1 MAXIMIZING THE NUMBER OF NONNEGATIVE SUBSETS NOGA ALON, HAROUT AYDINIAN, AND HAO HUANG Abstract. Gven a set of n real numbers, f the sum of elements of every subset of sze larger than k s negatve, what s the maxmum number of subsets of nonnegatve sum? In ths note we show that the answer s ( ( k 1 + ( k , settlng a problem of Tsukerman. We provde two proofs, the frst establshes and apples a weghted verson of Hall s Theorem and the second s based on an extenson of the nonunform Erdős-Ko-Rado Theorem. Key words. Nonnegatve subsets, Hall s theorem, Erdős-Ko-Rado. AMS subject classfcatons. 05D05, 05C70 1. Introducton. Let {x 1,, x n } be a sequence of n real numbers whose sum s negatve. It s natural to ask the followng queston: What s the maxmum possble number of subsets of nonnegatve sum t can have? One can set x 1 = n 2 and x 2 = = x n = 1. Ths gves n =1 x = 1 < 0 and 2 nonnegatve subsets, snce all the proper subsets contanng x 1, together wth the empty set, have a nonnegatve sum. It s also not hard to see that ths s best possble, snce for every subset A, ether A or ts complement {x 1,, x n }\A must have a negatve sum. Now a new queston arses: suppose t s known that every subset of sze larger than k has a negatve sum, what s the maxmum number of nonnegatve subsets? Ths queston was rased recently by Emmanuel Tsukerman [6]. The prevous problem s the specal case when k = n 1. A smlar constructon x 1 = k 1, x 2 = = x n = 1 yelds a lower bound ( ( k In ths note we prove that ths s also tght. Theorem 1.1. Suppose that every subset of {x 1,, x n } of sze larger than k has a negatve sum, then there are at most ( ( k subsets wth nonnegatve sums. One can further ask whether the extremal confguraton x 1 = k 1, x 2 = = x n = 1 s unque, n the sense that the famly F = {U : U x 0} s unque up to somorphsm. Note that when k = n 1, an alternatve constructon x 1 = n, x 2 =, x n = 1 also gves 2 nonnegatve subsets, whle the famly F t defnes s non-somorphc to the prevous one. More generally, for k = n 1 any sequence X = {x 1,..., x n } of n ntegers whose sum s 1 contans exactly 2 nonnegatve subsets, as for any subset A of X, exactly one of the two sets A and X A has a nonnegatve sum. However, for every k < n 1, we can prove the unqueness by the followng result n whch the number of nonnegatve elements n the set s also taken nto account. Theorem 1.2. Let 1 t k < n 1 be ntegers, and let X be a set of real numbers {x 1,, x n }, n whch there are exactly t nonnegatve numbers. Suppose that the sum of elements of every subset of sze greater than k s negatve, then the number of nonnegatve subsets s at most 2 t 1 ( ( ( n t k t + + n t Ths s tght for all admssble values of t, k and n. Sackler School of Mathematcs and Blavatnk School of Computer Scence, Tel Avv Unversty, Tel Avv 69978, Israel and Insttute for Advanced Study, Prnceton, New Jersey, 08540, USA. Emal: nogaa@tau.ac.l. Research supported n part by an ERC Advanced grant, by a USA-Israel BSF grant, by an ISF grant, by the Israel I-Core program and by the Smony Fund. Department of Mathematcs, Unversty of Belefeld, Germany. Emal: ayd@math.un-belefeld.de Insttute for Advanced Study, Prnceton, NJ and DIMACS at Rutgers Unversty. Emal: huanghao@math.as.edu. Research supported n part by the IAS-DIMACS postdoctoral fellowshp. 1

2 For every fxed k and n wth k < n 1, the expresson n Theorem 1.2 s strctly decreasng n t. Indeed, f 1 t < t + 1 k n, then, usng Pascal s dentty: (( ( (( ( n t n t 2 t t k t 0 k t 1 0 [ k t ( n t k t ( k t 1 ( ] = 2 t =0 =1 =0 [( ( ( ( ] n t n t = 2 t k t k t [( ] = 2 t 1 1. k t The last quantty s strctly postve for all t < k < (and s zero f k =. Therefore, the above theorem mples Theorem 1.1 as a corollary and shows that t s tght for k < n 1 only when there s exactly one nonnegatve number. The bound s Theorem 1.2 s also tght by takng x 1 = k t, x 2 = = x t = 0, x t+1 = = x n = 1. In ths example, the sum of any k + 1 elements s negatve, and a subset s nonnegatve f and only f t s ether of the form {x 1 } S T, where S s an arbtrary subset of {x 2,, x t } and T s a subset of {x t+1,, x n } havng sze at most k t, or when t s a subset of {x 2,, x t }. The rest of ths short paper s organzed as follows. In Secton 2 we prove a Hall-type theorem and deduce from t the exstence of perfect matchngs n certan bpartte graphs. Ths enables us to obtan Theorem 1.2 as a corollary. Secton 3 ncludes a strengthenng of the non-unform verson of the Erdős-Ko-Rado theorem, whch leads to an alternatve proof of Theorem 1.1. In the last secton, we dscuss some further research drectons. 2. The Man result. The followng lemma s useful n provng the exstence of a perfect matchng when combned wth the fundamental theorem of Hall [3]. It can also be reformulated and vewed as a strengthenng of Hall s theorem tself. Lemma 2.1. In a bpartte graph G wth two parts A and B, suppose there exst parttons A = A 1 A k and B = B 1 B l, such that for every [k], j [l], n the nduced bpartte graph G[A, B j ] all the vertces n A have equal degrees and all the vertces n B j have equal degrees too. Defne an auxlary bpartte graph H on the same vertex set, and replace every nonempty G[A, B j ] by a complete bpartte graph. Then G contans a perfect matchng f and only f H contans a perfect matchng. Proof. The only f part s obvous snce G s a subgraph of H. In order to prove the f part, note frst that f H contans a perfect matchng, then k =1 A = l j=1 B j. We wll verfy that the graph G satsfes the condtons n Hall s Theorem: for any subset X A, ts neghborhood has sze N G (X X. Put Y = N G (X, and X = X A, Y j = Y B j, and defne two sequences of numbers {x }, {y j } so that X = x A, Y j = y j B j. Consder the pars (, j such that G[A, B j ] s nonempty. In ths nduced bpartte subgraph suppose every vertex n A has degree d 1, and every vertex n B j has degree 2

3 d 2. Double countng the number of edges gves d 1 A = d 2 B j. On the other hand, we also have d 1 X d 2 Y j, snce every vertex n X has exactly d 1 neghbors n Y j, and every vertex n Y j has at most d 2 neghbors n X. Combnng these two nequaltes, we have y j x for every par (, j such that G[A, B j ] s nonempty. We clam that these nequaltes mply that Y X,.e. (2.1 l k B j y j A x. j=1 =1 To prove the clam t suffces to fnd d,j 0 defned on every par (, j wth nonempty G[A, B j ], such that d,j (y j x =,j l B j y j j=1 k A x. In other words, the condtons for Hall s Theorem would be satsfed f the followng system has a soluton: (2.2 d,j = B j ; d,j = A ; d,j 0; d,j = 0 f G[A, B j ] =. j The standard way to prove that there s a soluton s by consderng an approprate flow problem. Construct a network wth a source s, a snk t, and vertces a 1,, a k and b 1,, b l. The source s s connected to every a wth capacty A, and every b j s connected to the snk t wth capacty B j. For every par (, j, there s an edge from a to b j. Its capacty s + f G[A, B j ] s nonempty and 0 otherwse. Then (2.2 s feasble f and only f there exsts a flow of value A = j B j. Now we consder an arbtrary cut n ths network: (s {a } U1 {b j } j U2, t {a } [k]\u1 {b j } j [l]\u2. Its capacty s fnte only when for every U 1, j [l]\u 2, G[A, B j ] s empty. Therefore n the auxlary graph H, f we take Z = U1 A, then the degree condton N H (Z Z mples that j U 2 B j U 1 A and thus the capacty of ths cut s equal to A + B j [k]\u 1 j U 2 =1 A + A = [k]\u 1 U 1 k A. Therefore the mnmum cut n ths network has capacty at least k =1 A, and there s a cut of exactly ths capacty, namely the cut consstng of all edges emanatng from the source s. By the max-flow mn-cut theorem, we obtan a maxmum flow of the same sze and ths provdes us wth a soluton d,j to (2.2, whch verfes the Hall s condton (2.1 for the graph G. Remark. Lemma 2.1 can also be reformulated n the followng way: gven G wth the propertes stated, defne the reduced auxlary graph H on the vertex set A B, where A = [k], B = [l], such that A s adjacent to j B f G[A, B j ] s nonempty. If for every subset X A, j N H (X B j X A, then G has a perfect matchng. For the case of parttonng A and B nto sngletons, ths s exactly Hall s Theorem. Moreover f we let k = 1 and l = 1 n Lemma 2.1, t mmedately mples the well-known result that every d-regular bpartte graph has a perfect matchng. 3 =1

4 Corollary 2.2. For m r + 1, let G be the bpartte graph wth two parts A and B, such that both parts consst of subsets of [m] of sze between 1 and r. S A s adjacent to T B ff S T = and S + T r + 1. Then G has a perfect matchng. Proof. For 1 r, let A = B = ( [m],.e. all the -subsets of [m]. Let us consder the bpartte graph G[A, B j ] nduced by A B j. Note that when + j r or + j > m, G[A, B j ] s empty, whle when r j mn{2r, m}, every vertex n A has degree ( m j and every vertex n Bj has degree ( m j. Therefore by Lemma 2.1, t suffces to check that the reduced auxlary graph H satsfes the condtons n the above remark. We dscuss the followng two cases. Frst suppose m 2r, note that n the reduced graph H, A = B = [r], every vertex n A s adjacent to the vertces {r + 1,, r} n B. The only nequaltes we need to verfy are: for every 1 t r, r j=r+1 t B j t =1 A. Note that r j=r+1 t B j = t =1 r t + t =1. The last nequalty holds because the functon ( m k s ncreasng n k when k m/2. Now we consder the case r + 1 m 2r 1. In ths case every vertex n A s adjacent to vertces from r + 1 to mn{r, m }. More precsely, f 1 m r, then s adjacent to {r +1,..., r} n B, and f m r +1 r, then s adjacent to {r , m } n B. It suffces to verfy the condtons for X = {1,, t} when t r, and for X = {s,, t} when m r s t r. In the frst case N H (X = {r + 1 t,, r}, and the desred nequalty holds snce r j=r+1 t = j r =1 r t =1 r =1 r =t+1 = t =1. For the second case, N H (X = {r + 1 t,, m s}, and snce m r + 1, m s j=r+1 t = j m r+t 1 =s t =s. Ths concludes the proof of the corollary. We are now ready to deduce Theorem 1.2 from Corollary 2.2. Proof. of Theorem 1.2: Wthout loss of generalty, we may assume that x 1 x 2 x n, and x x k+1 < 0. Suppose there are t k nonnegatve numbers,.e. x 1 x t 0 and x t+1,, x n < 0. If t = 1, then every nonempty subset of nonnegatve sum must contan x 1, whch gves at most ( ( k nonnegatve subsets n total, as needed. Suppose t 2. We frst partton all the subsets of {1,, t} nto 2 t 1 pars (A, B, wth the property that A B = [t], A B = and 1 A. Ths can be done by parng every subset wth ts complement. For every, consder the bpartte graph G wth vertex set V,1 V,2 such that V,1 = {A S : S {t + 1,, n}, S k t} and V,2 = {B S : S {t + 1,, n}, S k t}. Note that f a nonempty subset wth ndex set U has a nonnegatve sum, then U {t + 1,, n} k t, otherwse U {1,, t} gves a nonnegatve subset wth more than k elements. Therefore every nonnegatve subset s a vertex of one of the graphs G. Moreover, we can defne the edges of G n a way that A S s adjacent to B T f and only f S, T {t+1,, n}, 4

5 S T = and S + T k t + 1. Note that by ths defnton, two adjacent vertces cannot both correspond to nonnegatve subsets, otherwse S T {1,, t} gves a nonnegatve subset of sze larger than k. Applyng Corollary 2.2 wth m = n t, r = k t, we conclude that there s a matchng saturatng all the vertces n G except A and B. Therefore the number of nonnegatve subsets n G s at most ( n t k t + + ( n t Note that ths number remans the same for dfferent choces of (A, B, so the total number of nonnegatve subsets s at most 2 t 1 ( ( ( n t k t + + n t A strengthenng of the non-unform EKR theorem. A conjecture of Manckam, Mklós, and Sngh (see [4], [5] asserts that for any ntegers n, k satsfyng n 4k, every set of n real numbers wth a nonnegatve sum has at least ( k 1 k- element subsets whose sum s also nonnegatve. The study of ths problem (see, e.g., [1] and the references theren reveals a tght connecton between questons about nonnegatve sums and problems n extremal fnte set theory. A connecton of the same flavor exsts for the problem studed n ths note, as explaned n what follows. The Erdős-Ko-Rado theorem [2] has the followng non-unform verson: for ntegers 1 k n, the maxmum sze of an ntersectng famly of subsets of szes up to k s equal to ( ( k 1 + ( k The extremal example s the famly of all the subsets of sze at most k contanng a fxed element. Ths result s a drect corollary of the unform Erdős-Ko-Rado theorem, together wth the obvous fact that each such famly cannot contan a set and ts complement. In ths secton we show that the followng strengthenng s also true. It also provdes an alternatve proof of Theorem 1.1. Theorem 3.1. Let 1 k n 1, and let F 2 [n] be a famly consstng of subsets of sze at most k, where / F. Suppose that for every two subsets A, B F, f A B =, then A + B k. Then F ( ( k 1 + ( k Proof. Denote ( [n] k = {A [n] : A k}. Let us frst observe that f F s an upset n ( ( [n] k (that s A F mples that {B [n] k : B A} F then F s an ntersectng famly, and hence the bound for F holds. Suppose there exst A, B F, such that A B =, thus A + B k. Snce k and F s an upset, there exsts a C F such that A C, C B =, and C + B > k whch s a contradcton. Next let us show that applyng so called pushng up operatons S (F, we can transform F to an upset F ( [n] k of the same sze, wthout volatng the property of F. Ths, together wth the observaton above, wll complete the proof. For [n] we defne S (F = {S (A : A F}, where { A, f A {} F or A = k S (A = A {}, otherwse. It s clear that S (F = F and applyng fntely many operatons S (F, [n] we come to an upset F ( [n] k. To see that S (F does not volate the property of F let F = F 0 F 1, where F 1 = {A F : A}, F 0 = F \ F 1. Thus S (F 1 = F 1. What we have to show s that for each par A, B F the par S (A, S (B satsfes the condton n the theorem as well. In fact, the only doubtful case s when A, B F 0, A B =, A + B = k. The subcase when S (A = A {}, S (B = B {} s also clear. Thus, t remans to consder the stuaton when S (A = A (or S (B = B. In ths case (A {} F, snce A, B k 1. Moreover, (A {} B = and A {} + B = k + 1, a contradcton. 5

6 To see that Theorem 3.1 mples Theorem 1.1, take F = {F : F {1,, n}, F x 0}. The famly F satsfes the condtons n Theorem 3.1 snce f A, B F, then A x 0, B x 0. If moreover A B =, then A B x 0 and t follows that A B k. 4. Concludng remarks. We have gven two dfferent proofs of the followng result: for a set of n real numbers, f the sum of elements of every subset of sze larger ( than k s negatve, then the number of subsets of nonnegatve sum s at most ( k The connecton between questons of ths type and extremal problems for hypergraphs that appears here as well as n [1] and some of ts references s nterestng and deserves further study. Another ntrgung queston motvated by the frst proof s the problem of fndng an explct perfect matchng for Corollary 2.2 wthout resortng to Hall s Theorem. When r s small or r = m 1, one can construct such a perfect matchngs, but t seems that thngs get more complcated when r s closer to m/2. Acknowledgment We thank Emmanuel Tsukerman for tellng us about the problem consdered here, and Benny Sudakov for frutful dscussons, useful suggestons and helpful deas. Note added n proof: Alexey Pokrovsky nformed us that he has found an alternatve proof of Theorem 3.1. Hs proof s very dfferent from the one gven here, combnng the Kruskal-Katona Theorem wth certan EKR type nequaltes. REFERENCES [1] N. Alon, H. Huang and B. Sudakov, Nonnegatve k-sums, fractonal covers, and probablty of small devatons, J. Combnatoral Theory, Ser. B 102 (2012, [2] P. Erdős, C. Ko and R. Rado, Intersecton theorems for systems of fnte sets. Quart. J. Math. Oxford Ser., 12 (1961, [3] P. Hall, On representatves of subsets, J. London Math. Soc., 10 (1 (1935, [4] N. Manckam and D. Mklós, On the number of non-negatve partal sums of a non-negatve sum, Colloq. Math. Soc. Janos Bolya 52 (1987, [5] N. Manckam and N. M. Sngh, Frst dstrbuton nvarants and EKR theorems, J. Combnatoral Theory, Seres A 48 (1988, [6] E. Tsukerman, Prvate communcaton. 6