The Generation of Minimal Triangle Graphs

Abstract

Let be a finite set of triples (sets of three distinct elements). We define G to be the 1-skeleton of the 2-complex formed by (i.e. the graph with elements of triples of as vertices and two vertices adjacent iff they are distinct and lie in a common member of T). is a minimal triangle set (MTS) of order n if (i) G has n vertices and is connected, and (ii) G { t} ) satisfies (i) for no t in T. In such a case G is a minimal triangle graph (MTG) of order n. Our problem is to generate all (nonisomorphic) MTS's of a given order or equivalently (by Theorem 1) all MTG's of that order. THEOREM 1. If G is an MTG, then G = G(T) for a utnique MTS T. Proof. Suppose and T' are MTS's with G = G(T) = G(T') and t C T, t f T'. Let t = {P1, P2, P3}. As t f T' and G = G(T'), the pairs {Pi, P2}, {Pi, P3}, and {P2, P3} are in some triples {P1 , P2, P4, {P1i, P3 , P5}, and {2P2 , P3 , P6 of T'. Then all of P1P4, P4P2, P2P6, P6P3, P3P6, and P5P1 are all edges of G not lying on triangle t. Hence PlP4P2P6P3P6Pl is a cycle in G { t} ) and G { t} ) is connected and has the same vertices as G (T), contradicting an MTS. If P is a vertex of a graph G, we let Gp denote the graph formed by removing P and incident edges from G. LEMMA 1. If G is a connected graph, then Gp is connected for some vertex P of G. Proof. A spanning tree of G is a tree subgraph of G containing all the vertices of G. As G is connected, from [2] we know that G has a spanning tree K. Let P be a vertex of K of valency one. Then Kp is a spanning tree for Gp ; hence Gp is connected. LEMMA 2. If G is connected, then G { t} ) is connected for some t in T. Proof. Form the graph H by taking members of as vertices and letting a pair of them be adjacent iff they are distinct and have an element in common. Then H is connected if G is, for there is a path between two vertices in G iff there is one in H between elements of containing the vertices. As G is connected, so is H(T); and, by Lemma 1, so is H(T)t = H(T {t}) for some t in T. Then G t} ) is connected also. We note that the same proof applies to any complex; in any connected complex there is some simplex t such that the 1-skeleton of { t} is connected and hence { t} is connected also. An alternate proof of Lemma 2 appears in [1]. THEOREM 2. Let be an MITS of order n > 4. Then there is an MTS T' of order n 1 or n 2 such that T' = {t} for some t C T. Proof. Let t C as in Lemma 2 and set T' = -{t}. As is an MTS and G (T') is connected, G (T') must have fewer vertices than G (T). As G is connected and T' # 0, G (T') cannot have three fewer vertices than G (T). Hence G(T') has n 1 or n 2 vertices. Finally suppose t' C T', T = T' {t'}, with G (T) connected and having the same vertices as G (T'). Then G (T u { t} ) = G { t'} ) has the same vertices as G and hence is not connected since is an MTS. Since G (T) is connected, this means that t has no vertex in common with