2020
DOI: 10.48550/arxiv.2008.08541
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A characterization of always solvable trees in Lights Out game using the activation numbers of vertices

Ahmet Batal

Abstract: Lights out is a game that can be played on any simple graph G. A configuration assigns one of the two states on or off to each vertex. For a given configuration, the aim of the game is to turn all vertices off by applying a push pattern on vertices, where each push switches the state of the vertex and its neighbors. If every configuration of vertices is solvable, then we say that the graph is always solvable. We introduce a concept which we call the activation numbers of vertices and we prove several character… Show more

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Cited by 2 publications
(5 citation statements)
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“…Remark 2.4. Definitions of half activated, always activated and never activated vertices coincide with the definitions in [5]. Indeed, by ( O 3) we see that, for a vertex u, {u} is not solvable if and only if ℓ(u) = x {u} • ℓ = 1 for some null pattern ℓ.…”
Section: Introductionmentioning
confidence: 74%
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“…Remark 2.4. Definitions of half activated, always activated and never activated vertices coincide with the definitions in [5]. Indeed, by ( O 3) we see that, for a vertex u, {u} is not solvable if and only if ℓ(u) = x {u} • ℓ = 1 for some null pattern ℓ.…”
Section: Introductionmentioning
confidence: 74%
“…How the nullity of a graph change when a vertex is removed or what the nullity of the emergent graph becomes when two graphs are joined together by a single or multiple edges was investigated by several authors [3], [10], [8], [4], [5]. However, as far as we know, until now there has been no study which investigates how nullity changes when an edge is added to or removed from a graph.…”
Section: Introductionmentioning
confidence: 99%
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“…We form the join graph H := G 1 uwG 2 by connecting the vertices u of G 1 and w of G 2 by an edge. In [ [10], Theorem 3.3], we stated what the nullity of H and the activation numbers of u and w with respect to H become when we join two graphs with specific activation numbers of u and w with respect to G 1 and G 2 , respectively. Moreover, we explained the structure of the component sets of any odd dominating set of H in graphs G 1 and G 2 .…”
Section: Resultsmentioning
confidence: 99%
“…It turns out that a vertex u is in all, exactly half, or none of the odd dominating sets (see [[3], Lemma 1] and [ [10], Lemma 2.4]. We say u is always, half, or never activated if the first, the second, or the third case holds, respectively.…”
Section: Introductionmentioning
confidence: 99%