The general position problem on Kneser graphs and on some graph operations

Abstract

A vertex subset S of a graph G is a general position set of G if no vertex of S lies on a geodesic between two other vertices of S. The cardinality of a largest general position set of G is the general position number (gp-number) gp(G) of G. The gp-number is determined for some families of Kneser graphs, in particular for K(n, 2) and K(n, 3). A sharp lower bound on the gp-number is proved for Cartesian products of graphs. The gp-number is also determined for joins of graphs, coronas over graphs, and line graphs of complete graphs.

The general position problem and strong resolving graphs

“…Proof. From [2,Theorem 4.3], it can be noticed that gp(G ⊙ H) = n(G) i n i , and also that the union of the sets of vertices of every copy of H in G⊙ H form a gp-set S of G ⊙ H. Every two vertices belonging to one copy of H are MMD, as well as are MMD every two vertices belonging to two different copies of H. Hence S forms a complete subgraph of (G ⊙ H) SR . Thus we deduce the equality by Theorem 3.1.…”

“…If G = (V (G), E(G)) is a graph, then S ⊆ V (G) is a general position set if no triple of vertices from S lie on a common geodesic in G. The general position problem is to find a largest general position set of G, the order of such a set is the general position number gp(G) of G. A general position set of G of order gp(G) is shortly called gp-set. The general position problem has been further studied in a sequence of very recent papers [1,2,8,10].…”

On independent position sets in graphs

“…But the same concept has already been studied two years earlier in [5] under the name geodetic irredundant sets. Refer [6,7,8,9,10,11] to understand the recent developments on general position number.…”

On the general position problem on Kneser graphs

“…, n} and two k-subsets S and T are joined by an edge if and only if S ∩ T = ∅. Ghorbani et al [10] determined gp(Kn n,2 ) and gp(Kn n,3 ) for all n and showed that for any fixed k if n is large enough, then gp(Kn n,k ) = n−1 k−1 holds. Theorem 1.1 ([10]).…”