The equidistant dimension of graphs: NP-completeness and the case of lexicographic product graphs
Let $ V(G) $ be the vertex set of a simple and connected graph $ G $. A subset $ S\subseteq V(G) $ is a distance-equalizer set of $ G $ if, for every pair of vertices $ u, v\in V(G)\setminus S $, there exists a vertex in $ S $ that is equidistant to $ u $ and $ v $. The minimum cardinality among the...
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Published in | AIMS mathematics Vol. 9; no. 6; pp. 15325 - 15345 |
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Main Authors | , |
Format | Journal Article |
Language | English |
Published |
AIMS Press
01.01.2024
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Subjects | |
Online Access | Get full text |
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Summary: | Let $ V(G) $ be the vertex set of a simple and connected graph $ G $. A subset $ S\subseteq V(G) $ is a distance-equalizer set of $ G $ if, for every pair of vertices $ u, v\in V(G)\setminus S $, there exists a vertex in $ S $ that is equidistant to $ u $ and $ v $. The minimum cardinality among the distance-equalizer sets of $ G $ is the equidistant dimension of $ G $, denoted by $ \xi(G) $. In this paper, we studied the problem of finding $ \xi(G\circ H) $, where $ G\circ H $ denotes the lexicographic product of two graphs $ G $ and $ H $. The aim was to express $ \xi(G\circ H) $ in terms of parameters of $ G $ and $ H $. In particular, we considered the cases in which $ G $ has a domination number equal to one, as well as the cases where $ G $ is a path or a cycle, among others. Furthermore, we showed that $ \xi(G)\le \xi(G\circ H)\le \xi(G)|V(H)| $ for every connected graph $ G $ and every graph $ H $ and we discussed the extreme cases. We also showed that the general problem of finding the equidistant dimension of a graph is NP-hard. |
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ISSN: | 2473-6988 2473-6988 |
DOI: | 10.3934/math.2024744 |