Maximizing the number of nonnegative subsets

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Maxmzng the number of nonnegatve subsets Noga Alon Hao Huang December 1, 213 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 2 + + + 1 by establshng and applyng a weghted verson of Hall s Theorem. Ths settles a problem of Tsukerman. 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 < 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 1 + + + 1. 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 1 + + + 1 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 } 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 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, 854, 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. Insttute for Advanced Study, Prnceton, NJ 854 and DIMACS at Rutgers Unversty. Emal: huanghao@math.as.edu. Research supported n part by the IAS-DIMACS postdoctoral fellowshp. 1

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 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 + 1. Ths s tght for all admssble values of t, k and n. 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 n t 1 n t 1 2 t 1 ( + + + 1 2 t ( + + + 1 k t k t 1 ( ( ( ( ( ( n t n t n t 1 n t 1 n t 1 n t 1 = 2 t 1 [ + + 1] k t k t 1 k t 1 ( ( ( ( n t n t 1 n t n t 1 = 2 t 1 [ + 1] k t k t 1 ( n t 1 = 2 t 1 [ 1]. k t The last quantty s strctly postve for all t < k < n 1 (and s zero f k = n 1. 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 =, 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. In the last secton, we dscuss some open problems and further research drectons. 2 The Man result The followng lemma can be regarded as a strengthenng of the fundamental theorem of Hall [3]. 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], [l], n the nduced bpartte graph G[A, B ] all the vertces n A have equal degrees and all the vertces n B have equal degrees too. Defne an auxlary bpartte graph H on the same vertex set, and replace every nonempty G[A, B ] by a complete bpartte graph. Then G contans a perfect matchng f and only f H contans a perfect matchng. 2

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 =1 B. 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 = Y B, and defne two sequences of numbers {x }, {y } so that X = x A, Y = y B. Consder the pars (, such that G[A, B ] s nonempty. In ths nduced bpartte subgraph suppose every vertex n A has degree d 1, and every vertex n B has degree d 2. Double countng the number of edges gves d 1 A = d 2 B. On the other hand, we also have d 1 X d 2 Y, snce every vertex n X has exactly d 1 neghbors n Y, and every vertex n Y has at most d 2 neghbors n X. Combnng these two nequaltes, we have y x for every par (, such that G[A, B ] s nonempty. We clam that these nequaltes mply that Y X,.e. l B y =1 k A x. (1 =1 To prove the clam t suffces to fnd d, defned on every par (, wth nonempty G[A, B ], such that l k d, (y x = B y A x., =1 In other words, the condtons for Hall s Theorem would be satsfed f the followng system has a soluton: d, = B ; d, = A ; d, ; d, = f G[A, B ] =. (2 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 s connected to the snk t wth capacty B. For every par (,, there s an edge from a to b. Its capacty s + f G[A, B ] s nonempty and otherwse. Then (2 s feasble f and only f there exsts a flow of value A = B. Now we consder an arbtrary cut n ths network: (s {a } U1 {b } U2, t {a } [k]\u1 {b } [l]\u2. Its capacty s fnte only when for every U 1, [l]\u 2, G[A, B ] s empty. Therefore n the auxlary graph H, f we take Z = U1 A, then the degree condton N H (Z Z mples that U 2 B U 1 A and thus the capacty of ths cut s equal to =1 A + B [k]\u 1 U 2 A + A = [k]\u 1 U 1 k A. =1 3

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, to (2, whch verfes the Hall s condton (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 adacent to B f G[A, B ] s nonempty. If for every subset X A, N H (X B X A, then G has a perfect matchng. For the case of parttonng A and B nto sngletons, ths s exactly Hall s Theorem. 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 adacent 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 ] nduced by A B. Note that when + r or + > m, G[A, B ] s empty, whle when r + 1 + mn{2r, m}, every vertex n A has degree ( m and every vertex n B has degree ( m. 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 adacent to the vertces {r + 1,, r} n B. The only nequaltes we need to verfy are: for every 1 t r, r =r+1 t B t =1 A. Note that r =r+1 t B = 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 adacent to vertces from r + 1 to mn{r, m }. More precsely, f 1 m r, then s adacent to {r + 1,..., r} n B, and f m r + 1 r, then s adacent to {r + 1..., 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 =r+1 t = r =1 r t =1 r =1 r =t+1 = For the second case, N H (X = {r + 1 t,, m s}, and snce m r + 1, m s =r+1 t = Ths concludes the proof of the corollary. m r+t 1 =s t =s. t =1. 4

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 1 + + x k+1 <. Suppose there are t k nonnegatve numbers,.e. x 1 x t and x t+1,, x n <. If t = 1, then every nonempty subset of nonnegatve sum must contan x 1, whch gves at most ( ( k 1 + + + 1 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 adacent to B T f and only f S, T {t + 1,, n}, S T = and S + T k t + 1. Note that by ths defnton, two adacent 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 + 1. 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 + 1. 3 Concludng remarks A conecture 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 2 + + (. 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. The followng strengthenng may also be true. Queston 3.1. Let k n 1, and let F 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. Is t true that F ( ( k 1 + ( k 2 + +? A postve answer to ths queston wll provde an alternatve proof of Theorem 1.1. Ths can be seen by takng F = {F : = F {1,, n}, F x }. The famly F satsfes the condtons 5

n Queston 3.1 snce f A, B F, then A x, B x. If moreover A B =, then A B x and t follows that A B k. It s not too dffcult to show that the statement n the queston ndeed holds f k s small wth respect to n, the challenge s to decde whether or not t holds for all k n 1. Another ntrgung queston s the followng: can one fnd 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 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. References [1] N. Alon, H. Huang and B. Sudakov, Nonnegatve k-sums, fractonal covers, and probablty of small devatons, J. Combnatoral Theory, Ser. B 12 (212, 784 796. [2] P. Erdős, C. Ko and R. Rado, Intersecton theorems for systems of fnte sets. Quart. J. Math. Oxford Ser., 12 (1961, 313 318. [3] P. Hall, On representatves of subsets, J. London Math. Soc., 1 (1 (1935, 26 3. [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, 385 392. [5] N. Manckam and N. M. Sngh, Frst dstrbuton nvarants and EKR theorems, J. Combnatoral Theory, Seres A 48 (1988, 91 13. [6] E. Tsukerman, Prvate communcaton. 6