On matchings in hypergraphs

Transcription

1 O matchigs i hypergraphs Peter Frakl Tokyo, Japa peter.frakl@gmail.com Tomasz Luczak Adam Mickiewicz Uiversity Faculty of Mathematics ad CS Pozań, Polad ad Emory Uiversity Departmet of Mathematics ad CS Atlata, USA tomasz@amu.edu.pl Katarzya Mieczkowska Adam Mickiewicz Uiversity Faculty of Mathematics ad CS Pozań, Polad kaska@amu.edu.pl Submitted: Mar 30, 2012; Accepted: Ju 3, 2012; Published: Ju 13, 2012 Mathematics Subject Classificatios: 05C35, 05C65, 05C70. Abstract We show that if the largest matchig i a k-uiform hypergraph G o vertices has precisely s edges, ad > 3k 2 s/2 log k, the H has at most ( ( k s k edges ad this upper boud is achieved oly for hypergraphs i which the set of edges cosists of all k-subsets which itersect a give set of s vertices. A k-uiform hypergraph G = (V, E is a set of vertices V N together with a family E of k-elemet subsets of V, which are called edges. I this ote by v(g = V ad e(g = E we deote the umber of vertices ad edges of G = (V, E, respectively. By a matchig we mea ay family of disjoit edges of G, ad we deote by µ(g the size of the largest matchig cotaied i E. Moreover, by ν k (, s we mea the largest possible umber of edges i a k-uiform hypergraph G with v(g = ad µ(g, ad by M k (, s we deote the family of the extremal hypergraphs for this problems, i.e. H M k (, s if v(h =, µ(h, ad e(h = ν k (, s. I 1965 Erdős [2] cojectured that, uless = 2k ad s = 1, all graphs from M k (, s are either cliques, or belog to the family Cov k (, s of hypergraphs o vertices i which the set of edges cosists of Partially supported by the Foudatio for Polish Sciece ad NSF grat DMS the electroic joural of combiatorics 19(2 (2012, #P42 1

2 all k-subsets which itersect a give subset S V, with S. This cojecture, which is a atural geeralizatio of Erdős-Gallai result [3] for graphs, has bee verified oly for k = 3 (see [5] ad [8]. For geeral k there have bee series of results which state that M k (, s = Cov k (, s for g(ks, (1 where g(k is some fuctio of k. The existece of such g(k was show by Erdős [2], the Bollobás, Dayki ad Erdős [1] proved that (1 holds wheever g(k 2k 3 ; Frakl ad Füredi [6] showed that (1 is true for g(k 100k 2 ad recetly, Huag, Loh, ad Sudakov [7] verified (1 for g(k 3k 2. The mai result of this ote slightly improves these bouds ad cofirms (1 for g(k 2k 2 /log k. Theorem 1. If k 3 ad the M k (, s = Cov k (, s. > 2k2 s log k, (2 I the proof we use the techique of shiftig (for details see [4]. Let G = (V, E be a hypergraph with vertex set V = {1, 2,..., }, ad let 1 i < j. The hypergraph sh i,j (G is obtaied from G by replacig each edge e E such that j e, i / e ad e ij = e \ {j} {i} / E, by e ij. Let Sh(G deote the hypergraph obtaied from G by the maximum sequece of shifts, such that for all possible i, j we have sh ij (Sh(G = Sh(G. It is well kow ad ot hard to prove that the followig holds (e.g. see [4] or [8]. Lemma 2. G M k (, s if ad oly if Sh(G M k (, s. Lemma 3. Let G M k (, s ad 2k + 1. Sh(G Cov k (, s. The G Cov k (, s if ad oly if Thus, it is eough to show Theorem 1 for hypergraphs G for which Sh(G = G. Let us start with the followig observatio. Lemma 4. If G is a hypergraph o vertex set [] such that Sh(G = G ad µ(g, the G A 1 A 2 A k, where for i = 1, 2,..., k. A i = {A [] : A = k, A {1, 2,..., i(s + 1 1} i}, Proof. Note that the set e 0 = {s + 1, 2s + 2,..., ks + k} is ot a edge of G. Ideed, i such a case each of the edges {i, i + s + 1,..., i + (k 1(s + 1}, i = 1, 2,..., s + 1, belogs to G due to the fact that G = Sh(G ad, clearly, they form a matchig of size s + 1. Now it is eough to observe that all sets which do ot domiate e 0 must belog to k A i. The followig umerical cosequece of the above result is crucial for our argumet. the electroic joural of combiatorics 19(2 (2012, #P42 2

3 Lemma 5. Let G be a hypergraph with vertex set {1, 2,..., } such that Sh(G = G ad µ(g, where k(s+1 1. The all except at most s(s+1 ( 1 2 k 2 edges of G itersect {1, 2,..., s}. Proof. Let A = k A i. Observe first that A (, for k(s Ideed, it follows from a easy iductio o k, ad the o. For k = 1 it is obvious. For k 1 ad = k(s we have clearly A = ( ( k. Now let k 2, k(s + 1 ad split all the sets of A ito those which cotai ad those which do ot. The, the iductioal hypothesis gives ( ( ( 1 1 A + s. k 1 k 1 Observe also that ( k = s ( i ( k 1 + s k, which is a direct cosequece of the idetity ( ( k = 1 ( k k. Thus, usig Lemma 4 ad the above observatio, the umber of edges of G which do ot itersect {1, 2,..., s} ca be bouded i the followig way. ( [( ( ] s G G A 1 A A 1 k 1 k k [ ( ( ] i s + k 1 k 1 s i j j=1 (i 1 i=2 ( 1 = (s i + 1 i ( s(s =. 2 Proof of Theorem 1. Let us assume that (2 holds for G M k (, s. The, by Lemma 2, the hypergraph H = Sh(G belogs to M k (, s. We shall show that H Cov k (, s which, due to Lemma 3, would imply that G Cov k (, s. Our argumet is based o the followig two observatios. Here ad below by the degree deg(i of a vertex i we mea the umber of edges cotaiig i, ad by V ad E we deote the sets of vertices ad edges of H respectively. Claim 6. If s 2, the {1, ks + 2, ks + 3,..., ks + k} E. the electroic joural of combiatorics 19(2 (2012, #P42 3

4 Proof. Let us assume that the assertio does ot hold. We shall show that the H has fewer edges tha the graph H = (V, E whose edge set cosists of all k-subsets itersectig {1, 2,..., s}. Let E i = {{i} e : e {ks + 2,..., }, e = k 1}, i [s] ad observe that the sets E i are pairwise disjoit ad E i = ( ks 1 k 1 for every i [s]. Moreover, sice H = Sh(H ad {1, ks + 2, ks + 3,..., ks + k} / E, E 1 E =, ad so E i E = for every i [s]. Thus, ( ks 1 E \ E s k 1 s( 1 ( k 1, (3 1 (k 1! k + 1 while from Lemma 5 we get Thus, E \ E s(s s( 1 k 1 (k 1! ( 1 ( 1 k 1 (k 1! ks k + 1. (s + 1(k 1 2( k + 1 e(h e(h s( 1 ( ( 1. (k 1! k + 1 k + 1 Let x = ks/( k + 1. It is easy to check that for all k 3 ad x (0, 0.7 log k/k we have (1 x k 1 > x. Thus, e(h e(h > 0 provided k 2 s < 0.7 log k( k + 1, which holds wheever 2sk 2 / log k. Thus, sice clearly µ(h, we arrive at cotradictio with the assumptio that H M k (, s. Claim 7. If s 2 the deg(1 = ( 1. I particular, the hypergraph H, obtaied from H by deletig the vertex 1 together with all edges it is cotaied i, belogs to M k ( 1, s 1. Proof. Let us assume that there is a k-subset of V, which cotais 1 ad is ot a edge i H. The, i particular, e = {1, k + 2,..., } / E. Let us cosider hypergraph H obtaied from H by addig e to its edge set. Sice H M k (, s, there is a matchig of size s + 1 i H cotaiig e. Hece, as H = Sh(H, there exists a matchig M i H such that M {2,..., ks + 1}. Note however that, by Claim 6, f = {1, ks + 2, ks + 3,..., ks + k} E. But the M = M {f} is a matchig of size s + 1 i H, cotradictig the fact that H M k (, s. Hece, we must have deg(1 = ( 1. Sice ks, the secod part of the assertio is obvious. Now Theorem 1 follows easily from Claim 7 ad the observatio that, sice s 1 1 s, if (2 holds the it holds also whe is replaced by 1 ad s is replaced by s 1. Thus, we ca reduce the problem to the case whe s = 1 ad use Erdős-Ko-Rado theorem (ote that the > 2k 2 / log k > 2k + 1. (4 the electroic joural of combiatorics 19(2 (2012, #P42 4

5 Refereces [1] B. Bollobás, E. Dayki, ad P. Erdős, Sets of idepedet edges of a hypergraph, Quart. J. Math. Oxford Ser. (2, 27:25 32, [2] P. Erdős, A problem o idepedet r-tuples, A. Uiv. Sci. Budapest. Eötvös Sect. Math., 8:93 95, [3] P. Erdős ad T. Gallai, O maximal paths ad circuits of graphs, Acta Math. Acad. Sci. Hugar. 10: , [4] P. Frakl, The shiftig techique i extremal set theory. I Surveys i Combiatorics, volume 123 of Lod. Math. Soc. Lect. Note Ser., pages Cambridge, [5] P. Frakl, O the maximum umber of edges i a hypergraph with give matchig umber, arxiv: [6] P. Frakl ad Z. Füredi, upublished. [7] H. Huag, P. Loh, ad B. Sudakov, The size of a hypergraph ad its matchig umber, Combiatorics, Probability & Computig, 21: , [8] T. Luczak, K. Mieczkowska, O Erdős extremal problem o matchigs i hypergraphs, arxiv: the electroic joural of combiatorics 19(2 (2012, #P42 5