Searching for perfection on hypergraphs

Transcription

1 Available online at Electronic Notes in Discrete Mathematics 44 (2013) Searching for perfection on hypergraphs Deborah Oliveros-Braniff 1,2 Instituto de Matemáticas, Unidad Juriquilla Natalia Garcia-Colin 3 Instituto de Matemáticas Amanda Montejano 4 UMDI-Facultad de Ciencias Luis Montejano 5 Instituto de Matemáticas, Unidad Juriquilla Abstract In this paper we study the family of oriented transitive 3 hypergraphs that arise from cyclic permutations and intervals in the circle, in order to search for the notion of perfection on hypergraphs. Keywords: transitive oriented 3 hypergraphs, cyclic permutations, perfection on hypergraphs /$ see front matter 2013 Elsevier B.V. All rights reserved.

2 156 D. Oliveros-Braniff et al. / Electronic Notes in Discrete Mathematics 44 (2013) Introduction We begin by proving a curious fact about permutations. Theorem 1.1 Let P be a permutation of [n] ={1,..., n}. Then the minimum number of ascending subsequences of P that cover the whole sequence P is equal to the length of the largest descending subsequence of P. Proof. Define a directed graph D associated to the permutation P as follows. Let V (D) =[n] and, for 1 i<j n, the directed arc (i, j) from the vertex i to the vertex j, isind if and only if the subsequence (i, j) is a descending subsequence of P.NotethatDis actually a transitive oriented graph. Thus, the underlying graph G of D is a comparability graph, and so it is perfect. Hence χ(g) =ω(g). The statement follows since, by definition, there is a natural correspondence between the cliques of G and the maximal descending subsequences of P, as well as between the independent subsets of V (G) and the ascending subsequences of P. The main idea in the previous proof is to associate to every permutation a transitive oriented graph whose underlying graph is perfect. (Related works can be found in the literature, but they approach the problem mostly from an algorithmic point of view, see [2,3,4]). We are interested in studying cyclic permutations. In such case, the purpose of the research is to associate to every cyclic permutation a 3 hypergraph with a natural structure of transitivity. As we will show, the 3 hypergraphs that arise from cyclic permutations are an excellent family to analyse within the problem of searching for perfection on hypergraphs. 2 Cyclic permutations and transitive 3 hypergraphs A cyclic permutation is a cyclic ordering of [n]. This is, an equivalence class [φ] of the set of bijections φ :[n] [n], in respect to the cyclic equivalence relation. A cyclic permutation [φ] will be denoted by (φ(1) φ(2)... φ(n)). In resemblance to the transitive oriented graphs associated to linear permutations, we will now define the 3 hypergraphs associated to cyclic permu- 1 Research supported by CONACyT-México under Project dolivero@matem.unam.mx 3 pmnatz@gmail.com 4 montejano.a@gmail.com 5 luis@matem.unam.mx

3 D. Oliveros-Braniff et al. / Electronic Notes in Discrete Mathematics 44 (2013) tations and study some of its properties. We need first to introduce, as in [6], the notion of transitive orientation on a 3 hypergraph. Let H be a 3-hypergraph, that is, a pair of sets H =(V (H),E(H)) where V (H) is the vertex set of H, and the edge set of H is E(H) ( ) V (H) 3. Definition 2.1 Let H be a 3 hypergraph. An orientation of H is an assignment of exactly one of the two possible cyclic orderings to each edge. An orientation of a 3 hypergraph is called an oriented 3 hypergraph, and we will denote the oriented edges by O(H). Example 1 Let H = (V (H),E(H)) with V (H) = {a 1,a 2,a 3,a 4,a 5 } and E(H) ={{a 1,a 2,a 3 }, {a 1,a 3,a 4 }, {a 1,a 3,a 5 }} then a possible orientation of H could be O(H) ={(a 1,a 2,a 3 ), (a 1,a 4,a 3 ), (a 1,a 3,a 5 )} obtaining an oriented 3 hypergraph. a 1 a 2 a 5 a 4 a 3 Definition 2.2 An oriented 3 hypergraph H is transitive if whenever (uv z) and (zvw) O(H) then(uvw) O(H) (this implies (uwz) O(H)). Observe that in the example, since (a 1,a 5,a 2 )and(a 2,a 5,a 3 )arenotino(h) then H is not transitive. Let [φ] be a cyclic permutation. Three elements i, j, k [n], with i<j<k, are said to be in clock-wise order in respect to [φ] if there is ψ [φ] suchthat ψ 1 (i) <ψ 1 (j) <ψ 1 (k); otherwise the elements i, j, k are said to be in counter-clockwise order with respect to [φ]. Definition 2.3 The oriented 3 hypergraph H [φ] associated to a cyclic permutation [φ] is the hypergraph with vertex set V (H [φ] )=[n], whose edges are the triplets {i, j, k}, withi<j<k, which are in clockwise order in respect to [φ], and whose edge orientations are induced by [φ]. An oriented 3 hypergraph H

4 158 D. Oliveros-Braniff et al. / Electronic Notes in Discrete Mathematics 44 (2013) is called a cyclic permutation 3 hypergraph if there is a cyclic permutation [φ] such that H [φ] = H. It can be easily checked that, for any cyclic permutation [φ], its corresponding oriented 3 hypergraph H [φ] is transitive. Hence, in parallel to the study of perfection in transitive oriented graphs, using the following definitions of cromatic and clique numbers, the question below is very natural. For a cyclic permutation [φ], let the chromatic number χ([φ]) of [φ] be the minimum number of clock-wise subsequences of [φ] that cover the whole [φ], and the clique number ω([φ]) of [φ] be the length of the largest counterclockwise subsequence of [φ]. (1) It is true that χ([φ]) = w([φ])/2 for any cyclic permutation [φ]? This question is equivalent to asking if, for a cyclic permutation 3 hypergraph H, χ(h) = w(h)/2?, where ω(h) andχ(h) denote the clique number and the (weak) chromatic number of the unoriented hypergraph of H, respectively (see [5] for precise definitions). It is important to note that, in contrast to graphs where ω(g) χ(g), for 3 hypergraphs we have w(h)/2 χ(h). Another related question is: (2) It is true that any 3 hypergraph that can be transitively oriented, satisfies w(h)/2 = χ(h)? Unfortunately, as we will see later, both answers are negative. So, the following definition is a natural alternative. Definition 2.4 A 3 hypergraph H is perfect if every induced subhypergraph F of H, and every induced sub-hypergraph F of the complement of H, satisfy w(f )/2 = χ(f ). Now, the following question arises; (3) If both H and its complement can be transitively oriented, is it true that w(h)/2 = χ(h)? In the next section we will answer the latter question. 3 Self transitive 3 hypergraphs An oriented 3 hypergraph containing all possible oriented 3 edges is called a 3 hypertournament. LetTT 3 n be the oriented 3 hypergraph associated to the cyclic permutation ( n). Clearly TT 3 n is a transitive 3 hypertournament.

5 D. Oliveros-Braniff et al. / Electronic Notes in Discrete Mathematics 44 (2013) Theorem 3.1 There is just one transitive 3 hypertournament with n vertices, up to isomorphism. Theorem 3.1 allows us to hereafter refer to the transitive 3 hypertournament on n vertices. Although the proof is not difficult, we do not include it in this extended abstract. Besides, this fact is implied in the literature regarding cyclic orders [1]. For an oriented 3 hypergraph, it is not immediately clear how to define its oriented complement. However, for a spanning oriented subhypergraph of TTn, 3 H, we define its complement as the oriented 3 hypergraph H with V (H) =V (TTn)andO(H) 3 =O(TTn) 3 \ O(H). An oriented 3 hypergraph, H, such that it is a spanning subhypergraph of TTn 3 is called self-transitive if it is transitive and its complement is also transitive. In [6] the following theorem is proved. Theorem 3.2 H is an oriented cyclic permutation 3 hypergraph if and only if H is self transitive. Using Theorem 3.2 we show that not every self transitive 3 hypergraph is perfect, by exhibiting a cyclic permutation whose corresponding oriented 3 hypergraph is not perfect. Let [φ] be the cyclic permutation of {1, 2,...,7} given by ( ). The largest counter-clockwise subsequence of [φ] has length 4, but [φ] can not be decomposed into just 2 clockwise subsequences, hence showing that 2 = w(h [φ] )/2 <χ(h [φ] )=3. Nevertheless, we believe that a large number of cyclic permutation 3 hypergraphs are perfect. So, we propose the problem characterize perfect cyclic permutation 3 hypergraphs. 4 Families of intervals in the circle The intersection graph of a family of intervals in the line played an important role in the search for perfection in graphs. The same could also be possible for 3 hypergraphs. Let F be a finite family of closed intervals in the circle S 1, and let H = H(F) be the 3 hypergraph with V (H) =F such that a triple of intervals defines and edge of H if and only if the three intervals are pairwise disjoint. It is easy to see that H can be naturally associated to an oriented transitive 3 hypergraph. A sub-family of intervals I Fcorresponds to a clique of H if and only if I is a pairwise disjoint subset of intervals of F. So,w(H) is the cardinality of the largest pairwise disjoint subset of intervals in F. On the other hand, a subset of intervals I Fcorresponds to a (weak) independent set of vertices

6 160 D. Oliveros-Braniff et al. / Electronic Notes in Discrete Mathematics 44 (2013) in H if and only if it satisfies the property (3, 2), this is, for any set of three intervals in I, at least two of them intersect. A (weak) coloring of H is an assignment of colors to the intervals of F such that each chromatic class corresponds to a (weak) independent subset of H, that is, a set satisfying the property (3, 2). Let χ(h) be the minimum number of colors in a (weak) coloring of H so ω(h) 2 χ(h). Theorem 4.1 Let F be a finite family of closed intervals in the circle S 1 and let H be the 3 hypergraph associated to F. Then = χ(h). w(h) 2 Proof. We proceed by induction. If w(h) = 2, by definition, the family F satisfies the property (3, 2) which implies χ(h) =1. Now, assume that w(h) =2n +1. Let {I 0,I 1,..., I 2n } Fbe a pairwise disjoint set of intervals. Without loss of generality, we might assume that there is no other interval of F contained in I 0. Denote by G 0 the set of intervals of F that intersect I 0, and note that it satisfies the property (3, 2). Thus, G 0 corresponds to a independent set of H. On the other hand, F G 0 consists of all intervals contained in S 1 I 0, which may be regarded as the real line. The maximum number of pairwise disjoint intervals of F G 0 is 2n. Thus, we can pierce the intervals of F G 0 with 2n points, {v 1,..., v 2n }. For 1 i n, let G i be the subset of intervals of F G 0 that intersects {v 2i 1,v 2i }. Clearly, each G i is an independent set because it satisfies the (3, 2) property. Consequently, χ(h) n +1. Suppose now that w(h) = 2n. Let {I 1,I 2,..., I 2n } F be a pairwise disjoint set of intervals. Without loss of generality, suppose that the distance between I 1 and I 2 is the smallest possible between all choices of pairwise disjoint subsets of F. This implies two facts; firstly, that the collection G 0 of intervals of F which intersects {I 1,I 2 } satisfies the property (3, 2) and, secondly that the collection of intervals F G 0 may be regarded as contained in the real line. The maximum number of pairwise disjoint intervals of F G 0 is 2n 2, hence we can pierce the intervals of F G 0 with 2n 2points {v 1,..., v 2n 2 }. Let G i be the subset of intervals of F G 0 that intersects {v 2i 1,v 2i } for each i =1,...n 1. Clearly, each G i is independent because satisfies the property (3, 2). Consequently χ(h) n. This concludes the proof of the theorem. In the search for perfection on 3 hypergraphs, it will now be important to characterize the families F of intervals in the circle for which the complement of H satisfies the property that w(h)/2 = χ(h).

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