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By Peter L. Hammer (Eds.)

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Extra resources for Combinatorics 79

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What is a hypermap Let [ n ] denote the subset {1,2,. . , n } of positive integers. To verify this condition one can construct a graph with vertices {1,2,. . The graph is connected if (a,a) is a hypermap. The underlying hypergraph of (a,a) is obtained as follows: take as set of vertices the orbits of a ; each orbit of a determines then an edge consisting of all the vertices intersecting it. Example. Let 2 = (a,a) be defined as follows: a = ( 1 , 2 ) ( 3 , 4 , 5 ) ( 6 , 7 ,8)(9, 10)(11, 12, 13), a = (1,5)(2, 11,8,4)(3,6,9)(7,12, 10)(13).

Soit G = ( X , E ) un mark ayant exactement 16 sommefs, alors il ne posstde pas de sous-gruphe isomorphe a H (Fig. 1). Fig. 1. Preuve. La preuve est simple, il suffit en effet de compter le nombre d’arttes possibles dans G X - H , puis d’inventorier les sous-graphes possibles pour G X - H(il y en a 5 ) : chaque cas nous menant a un graphe 3-coloriable: contradiction. 2. Soit G = ( X , E ) un mark ayant exuctemenf 16 sommets alors il ne possbde pas de sous-graphe isomorphe a K (Fig. 2). Fig. 2. -L.

Then a K ” ( A ,B ) is any graph consisting of a complete graph on A, together with a pair of edges connecting each point of B with a different pair of points of A. Furthermore, if F is a K ” ( A ,B ) , define HF to be the graph with A as its vertices, with a pair of vertices joined in HF if they are joined in F through a point of B. Moreover, call a K”(A,B ) in G maximal in a given graph if there exists no K”(A,B , ) in the graph with lBll > (Bl. We will now prove a series of facts about G, leading finally to a contradiction.

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