Some dominant results have been obtained for maximal outer planar graphs (mops). The classical domination problem is $G – N[S]The graph $ obtained by removing the closed neighborhood of $S$ contains no vertices. In the proof of the Art Gallery theorem, Chv\'{a}tal is called the dominating number of $G$, and the minimum size represented by $\gamma(G)$ is at most $n/3$ showed that there is $G$ is the mop. Now consider the change by allowing $G – N.[S]The maximum degree of $ is at most $k$. Let $\iota_k(G)$ denote the size of the minimal set $S$ for which this is achieved. For $n \le 2k+3$ it is trivially $\iota_k(G) \leq 1$ . Let $G$ be a mop of $n \ge \max\{5,2k+3\}$ vertices, of which $n_2$ are of degree $2$. Upper bounds on $\iota_k(G)$ were obtained for $k = 0$ and $k = 1$. That is, $\iota_{0}(G) \le \min\{\frac{n}{4} ,\frac{n+n_2}{5},\frac{n-n_2}{3}\}$ and $\iota_1(G) \le \min\{\frac{n}{5},\frac{ n+n_2}{6},\frac{n-n_2}{3}\}$. {k}(G) \le \min\{\frac{n}{k+4},\frac{n+n_2}{k+5},\frac{n-n_2}{any $k \ge k+2}\}$ for 0$. In the original setting of the art gallery theorem, if an art gallery has exactly $n$ corners, and at least one of $k + 2$ consecutive corners must be visible to at least one guard, The argument presented goes like this: The number of guards required is at most $n/(k+4)$. Also, unless $\gamma(G) \le \frac{n – n_2}{2}$ ($n = 2n_2$, $n_2$ is odd, $\gamma(G) = \frac{n – n_2) prove) + 1}{2}$. Together with the inequality $\gamma(G)\le\frac{n+n_2}{4}$ obtained by Campos and Wakabayashi and independently obtained by Tokunaga, this is the boundary of his Chv\'{a}tal improve. Boundaries are sharp.

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