Pairs of disjoint q-element subsets far from each other

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1 Pairs of disjoit q-elemet subsets far from each other Hikoe Eomoto Departmet of Mathematics, Keio Uiversity Hiyoshi, Kohoku-Ku, Yokohama, 223 Japa, Gyula O.H. Katoa Alfréd Réyi Istitute of Mathematics, HAS, Budapest P.O.B. 127 H-1364 Hugary July 27, 2001 Abstract Let ad q be give itegers ad X a fiite set with elemets. The followig theorem is proved for > 0 (q). The family of all q-elemet subsets of X ca be partitioed ito disjoit pairs (except possibly oe if ( q) is odd), so that A1 A 2 + B 1 B 2 q, A 1 B 2 + B 1 A 2 q holds for ay two such pairs {A 1,B 1 } ad {A 2,B 2 }. This is a sharpeig of a theorem i [2]. It is also show that this is a codig type problem, ad several problems of similar ature are posed. 1 Itroductio The followig theorem was proved i [2]. Theorem 1.1 Let X = ad 2 k>q. The family of all q-elemet subsets of X ca be partitioed ito uordered pairs (except possibly oe if ( ) q is odd), so that paired q-elemet subsets are disjoit ad if A 1,B 1 ad A 2,B 2 are two such pairs with A 1 A 2 k, the B 1 B 2 <k, provided > 0 (k, q). AMS Subject classificatio Primary 05B30, Secodary 05C45, 94B99. Keywords: desig, Hamiltoia cycle, code. The work was supported by the Japa Society for the Promotio of Sciece, Grat-i-Aid for Scietific Research (B), , the Hugaria Natioal Foudatio for Scietific Research grat umbers T029255, DIMACS ad UVO-ROSTE the electroic joural of combiatorics 8 (o. 2) (2001), #R7 1

2 The mai aim of the preset paper is to give a sharpeig of this theorem. Defie the closeess of the pairs {A 1,B 1 } ad {A 2,B 2 } by γ({a 1,B 1 }, {A 2,B 2 })=max{ A 1 A 2 + B 1 B 2, A 1 B 2 + B 1 A 2 } (1.1) It is obvious that A 1 A 2 k ad B 1 B 2 k imply γ((a 1,B 1 ), (A 2,B 2 )) 2k for sets satisfyig A 1 B 1 = A 2 B 2 =, therefore the followig theorem is really a sharpeig of Theorem 1.1. Theorem 1.2 Let X =. The family of all q-elemet subsets of X ca be partitioed ito disjoit pairs (except possibly oe if ( ) q is odd), so that γ({a1,b 1 }, {A 2,B 2 }) q holds for ay two such pairs {A 1,B 1 } ad {A 2,B 2 }, provided > 0 (q). The proof of Theorem 1.1 was based o a Hamiltoia type theorem. Here we will eed aother theorem of the same type. Two edge-disjoit (o-directed) simple graphs G 0 = (V,E 0 )adg 1 =(V,E 1 ) will be give o the same vertex set where V = N, E 0 E 1 =. Let r deote the miimum degree i G 0. The edges of the secod graph are labelled by positive itegers. The label of e E 1 is deoted by l(e). Deote the umber of edges of label i startig from the vertex v by d(v, i) = {e E 1 : v e, l(e) =i}. Let s be the maximum degree i G 1,thatis, s =max v V { i d(v, i)}. (1.2) Aother parameter t is defied by t = t(q) =max v V { i d(v, i)max { w V q+1 i j d(w, j)}}. (1.3) A 4-tuple (x, y, z, v) of vertices is called heavy C 4 iff (x, y), (z, v) E 0, (y,z), (x, v) E 1,l((y, z)) + l((x, v)) q + 1. After these defiitios we are able to formulate our theorem. Theorem 1.3 Suppose, that 2 r 4 t s 1 >N. (1.4) The there is a Hamiltoia cycle i G 0 such that if (a, b) ad (c, d) are both edges of the cycle, the (a, b, c, d) is ot a heavy C 4. Sectio 2 cotais the proofs. I Sectio 3 we will pose a geeral questio to fid the maximum umber of elemets whose paiwise distace is at least d i a fiite space furished with a distace. Theorem 1.2 is the solutio of this questio i a special case. the electroic joural of combiatorics 8 (o. 2) (2001), #R7 2

3 2 Proofs The proof of Theorem 1.3 is based o Dirac s famous theorem o a sufficiet coditio for existece of a Hamiltoia cycle ad o Lemma 2.2. Theorem 2.1 (Dirac [3]) If G is a simple graph o N vertices ad all degrees of G are at least N, the G has a Hamiltoia cycle. 2 Lemma 2.2 Let G 0, G 1, r, s, t ad N satisfy (1.4). Assume that there is a Hamiltoia path from a to b i G 0. The there exist c, c a, ad d, d b, adjacet vertices alog the path, such that c is betwee a ad d o the path, (a, d) E 0, (b, c) E 0, (a, d, b, c) is ot a heavy C 4, ad if (x, y) is a edge of the path, the either (a, d, x, y) or (b, c, x, y) is a heavy C 4. Proof of Lemma 2.2 We call a vertex x V a-bad (b-bad) if there exists a edge (y, z) of the Hamiltoia path such that (a, x, y, z) ((b, x, y, z), respectively) is a heavy C 4. Let t a be the umber of a-bad vertices ad t b be that of the b-bad vertices. Now, t a is bouded from above by the umber of four-tuples (a, z, y, x) such that (y,z) isaedge of the path, (a, z), (y,x) E 1 ad l((a, z)) + l((y,x)) q + 1 holds. There are d(a, i) choices for (a, z) oflabeli. The vertex y ca be chose i two differet ways alog the path, fially the umber of choices for (y, x) withlabelj is d(y,j). Therefore the umber of these paths ca be upperbouded by (see (1.3)). We obtaied 2{ i d(a, i)max { y V q+1 i j d(y, j)}} =2t t a,t b 2t. (2.1) The umber of pairs {c, d} (a c, d b) which are eighbours alog the path, c is betwee a ad d is N 3. At least r 2 of these pairs satisfy (a, d) E 0 ad at least r 2 of them satisfy (c, b) E 0. (The umber of edges i E 0 startig from a (b) isatleast r, three of these edges do ot cout here: the two edges alog the path ad evetually {a, b}.) Cosequetly, there are at least 2r N 1 pairs havig both of the edges i E 0. The pair {c, d} satisfies the coditios of the lemma if it is chose from the above 2r N 1 oes,d is ot a-bad, c is ot b-bad ad (d, b) E 1. (This last coditio implies that (a, d, b, c) isotaheavyc 4.) The umber of pairs {c, d} for which at least oe of these coditios does ot hold is at most t a + t b + s. Therefore if 2r N 1 >t a + t b + s holds the the existece of the pair i the lemma is proved. By (2.1) this is reduced to (1.4). Proof of Theorem 1.3 Let us suppose idirectly, that 2 r 4 t s 1 >N,but the required Hamiltoia cycle does ot exist. We say that K cotais a heavy C 4 if there exists a heavy C 4 whose E 0 edges are edges of K, wherek stads for a path or a cycle i G 0. the electroic joural of combiatorics 8 (o. 2) (2001), #R7 3

4 If E 1 =, thet ad s are zero, the coditio of Dirac s theorem holds for G 0,thusit cotais a Hamiltoia cycle. Furthermore o heavy C 4 could exist. So, we may assume that E 1 is o-empty. Let us drop edges oe-by-oe from E 1 util a required Hamiltoia cycle appears. Cosider the last dropped edge (u, v). Droppig it, a Hamiltoia cycle cotaiig o heavy C 4 appears. This meas, that there was a Hamiltoia cycle C i G 0 before, which cotaied such heavy C 4 s oly that used the edge (u, v) E 1.Letthe eighbours of v alog C be w ad z. A heavy C 4 usig the edge (u, v) must use either (w, v) or(z, v). Thus, the path of N 1 vertices from w to z obtaied by deletig the vertex v from C cotais o heavy C 4. Lemma 2.2 ca be applied for the Hamiltoia path obtaied from C by deletig the edge (z, v), takig a = v ad b = z. Replacig the edges (c, d) (provided by Lemma 2.2) ad (z, v) withedges(v, d) ad(z, c) a ew Hamiltoia cycle C is obtaied, which ca cotai a heavy C 4 oly if that heavy C 4 uses the edge (w, v). Now, a secod applicatio of Lemma 2.2 with a = w ad b = v gives a Hamiltoia cycle C cotaiig o heavy C 4, eve without droppig the edge (u, v), a cotradictio. Proof of Theorem 1.2 We costruct graphs G 0 =(V,E 0 )adg 1 =(V,E 1 )thatsatisfytherequiremetsof Theorem 1.3. The vertex set V cosists of the q-elemet subsets of X, V = ( ) q = N. Two q-elemet subsets are adjacet i G 0 if their itersectio is empty, while two q-elemet subsets are adjacet i G 1 if they have a o-empty itersectio. The label of the edge (A, B) isl((a, B)) = A B. G 0 is regular with degree r = ( ) q q = 1 q! q + O( q 1 ). I G 1 we have ( )( ) q q d(v, i) =d(i) = (1 i<q). (2.2) i q i By (1.2) ad (2.2) we have q 1 s = i=1 ( q i )( q q i O the other had (1.3) ad (2.2) imply t = q+1 i+j d(i)d(j) = ) = q+1 i+j q (q 1)! q 1 + O( q 2 ). (2.3) ( q i )( q q i )( q j )( q q j q 1 ( )( ) = q 1 q q 1 1 i=2 i q +1 i (q i)! (i 1)! + O(q 2 ). It is easy to check that 2 r 4 t s 1 >N= ( ) q = 1 q! q + O( q 1 ), provided > 0 (q). Accordig to Theorem 1.3, there is a Hamiltoia cycle H i G 0 that does ot cotai two disjoit edges that spa a heavy C 4. Now the required partitio of the q-elemet subsets ito disjoit pairs ca be obtaied by goig aroud H, every other edge will form a good pair. The coditio γ({a 1,B 1 }, {A 2,B 2 }) q ca be deduced from (1.1) ad from the fact that H cotais o heavy C 4. ) = the electroic joural of combiatorics 8 (o. 2) (2001), #R7 4

5 3 Geeralized codig problems Defie δ({a 1,B 1 }, {A 2,B 2 })=2q γ({a 1,B 1 }, {A 2,B 2 }). This is a distace i the space of all disjoit pairs of q-elemet subsets of X. Theorem 1.2 aswers a codig type questio, how may elemets ca be chose from this space with large pairwise distaces. I geeral, let Y be a fiite set ad δ(x, y) 0 a real-valued symmetric (δ(x, y) = δ(y, x)) fuctio defied o the pairs x, y Y. Let 0 <dbe a fixed iteger. A subset C = {c 1,...,c m } Y is called a code of distace d if δ(c i,c j ) d holds for i j. The followig (probably too geeral) questio ca be asked. Problem 3.1 Let Y, δ(x, y) ad the real d be give. Determie the maximum size C of a d-distace code. δ(x, y) is called a distace if δ(x, y) =0iffx = y ad the triagle iequality holds: δ(x, y) δ(x, z)+δ(z, y) for ay 3 elemets of Y. Problem 3.1 ca be asked for δ(x, y) ot possessig these coditios, but it is really more atural for distaces. The best kow fiite set with a distace is whe Y is the set of all sequeces of legth, the elemets take from a fiite set, the distace is the Hammig distace. Problem 3.1 leads to traditioal codig theory. There are may kow results of this type i geometry, but there Y is ifiite. Our case whe Y = Y 1 is the set of all disjoit pairs of q-elemet subsets of X ca also be cosidered as a set of sequeces, however the distace is ot a Hammig distace. Still, it is a distace. Propositio 3.2 Let Y 1 be the set of all disjoit pairs {A, B} of q-elemet subsets of a -elemet X. δ 1 ({A 1,B 1 }, {A 2,B 2 })=2q γ({a 1,B 1 }, {A 2,B 2 }) (3.1) is a distace. Proof of Propositio 3.2 It is easy to see that δ 1 ({A 1,B 1 }, {A 2,B 2 })=0iff{A 1,B 1 } = {A 2,B 2 }. So we really have to prove oly the triagle iequality. By (3.1) ad (1.1) this is reduced to Two cases will be distiguished. max{ A 1 A 3 + B 1 B 3, A 1 B 3 + B 1 A 3 }+ +max{ A 2 A 3 + B 2 B 3, A 2 B 3 + B 2 A 3 } 2q +max{ A 1 A 2 + B 1 B 2, A 1 B 2 + B 1 A 2 }. (3.2) the electroic joural of combiatorics 8 (o. 2) (2001), #R7 5

6 Case 1. Either the first or the secod value is larger (or equal) i both terms of the left had side of (3.2). By symmetry it ca be supposed that the first values are the larger oes. The the left had side of (3.2) is A 1 A 3 + B 1 B 3 + A 2 A 3 + B 2 B 3. (3.3) Observe that A 1 A 3 + A 2 A 3 A 3 + A 1 A 2 = q + A 1 A 2. The same holds for the Bs, therefore (3.3) is at most 2q + A 1 A 2 + B 1 B 2,provig (3.2) for this case. Case 2. The first value is larger i the first term, the secod oe i the secod term, or vice versa, o the left had side of (3.2). By symmetry we ca suppose that the left had side of (3.2) is A 1 A 3 + B 1 B 3 + A 2 B 3 + B 2 A 3. (3.4) All these itersectios are subsets of A 3 B 3. Usig the fact that A i B i =, oly the first ad the fourth, the secod ad the third, resp., itersectios ca be o-disjoit, the other pairs are disjoit. Therefore o elemet is i more tha two of the itersectios i (3.4) ad these elemets are all either i A 1 B 2 A 3 or i B 1 A 2 B 3.Thisgivesa upper boud o (3.4): A 3 + B 3 + A 1 B 2 A 3 + B 1 A 2 B 3 2q + A 1 B 2 + B 1 A 2, provig (3.2) for this case, too. The followig special case of Problem 3.1 arises ow aturally. Problem 3.3 Let Y 1, δ 1 (x, y) be the space with distace defied above. Determie the maximum size C of a q-distace code. Ufortuately, Theorem 1.2 is ot a solutio, sice the coditio o the distace permits the existece of a pair {A, B 1 }, {A, B 2 },B 1 B 2 =, which is excluded i Theorem 1.2 by the uique usage of every q-elemet subset. Let us see some other possible special cases of Problem 3.1. Problem 3.4 Let Y be set of all permutatios of elemets ad suppose that δ is the umber of iversios betwee two permutatios (umber of pairs beeig i differet order). Give the iteger 0 <d, determie the largest set of permutatios with pairwise distace at least d. Problem 3.5 Let Y be the set of matrices over a fiite field F ad suppose that the distace δ betwee two such matrices is the rak of the differece (etry by etry) of the matrices. Give the iteger 0 <ddetermie the largest set of matrices with pairwise distace at least d. Fially, let Y be the set of all simple graphs G =(V,E) o the same vertex set V, V =. The distace δ(g 1,G 2 ) betwee the graphs G 1 =(V,E 1 ),G 2 =(V,E 2 )isthesizeofthe largest complete graph i (V,E 1 E 2 )where is the symmetric differece. Some results o this problem will be published i a forthcomig paper [1]. the electroic joural of combiatorics 8 (o. 2) (2001), #R7 6

7 Refereces [1] N. Alo, G.O.H. Katoa, Codes amog graphs, paper uder preparatio. [2] J. Demetrovics, G.O.H. Katoa ad A.Sali, Desig type problems motivated by database theory J. Statist. Plaig ad Iferece 72 (1998) [3] G.A. Dirac, Some theorems o abstract graphs, Proc. Lodo Math. Soc., Ser. 3, 2 (1952), the electroic joural of combiatorics 8 (o. 2) (2001), #R7 7