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http://dx.doi.org/10.4134/BKMS.2013.50.2.469

A CORRECTION TO A PAPER ON ROMAN κ-DOMINATION IN GRAPHS  

Mojdeh, Doost Ali (Department of Mathematics University of Tafresh / School of Mathematics Institute for Research in Fundamental Sciences (IPM))
Moghaddam, Seyed Mehdi Hosseini (Department of Mathematics University of Tafresh)
Publication Information
Bulletin of the Korean Mathematical Society / v.50, no.2, 2013 , pp. 469-473 More about this Journal
Abstract
Let G = (V, E) be a graph and k be a positive integer. A $k$-dominating set of G is a subset $S{\subseteq}V$ such that each vertex in $V{\backslash}S$ has at least $k$ neighbors in S. A Roman $k$-dominating function on G is a function $f$ : V ${\rightarrow}$ {0, 1, 2} such that every vertex ${\upsilon}$ with $f({\upsilon})$ = 0 is adjacent to at least $k$ vertices ${\upsilon}_1$, ${\upsilon}_2$, ${\ldots}$, ${\upsilon}_k$ with $f({\upsilon}_i)$ = 2 for $i$ = 1, 2, ${\ldots}$, $k$. In the paper titled "Roman $k$-domination in graphs" (J. Korean Math. Soc. 46 (2009), no. 6, 1309-1318) K. Kammerling and L. Volkmann showed that for any graph G with $n$ vertices, ${{\gamma}_{kR}}(G)+{{\gamma}_{kR}(\bar{G})}{\geq}$ min $\{2n,4k+1\}$, and the equality holds if and only if $n{\leq}2k$ or $k{\geq}2$ and $n=2k+1$ or $k=1$ and G or $\bar{G}$ has a vertex of degree $n$ - 1 and its complement has a vertex of degree $n$ - 2. In this paper we find a counterexample of Kammerling and Volkmann's result and then give a correction to the result.
Keywords
dominating set; Roman k-dominating function; correction;
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Times Cited By KSCI : 1  (Citation Analysis)
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