Paint Cost and the Frugal Distinguishing Number

You are handed a graph with vertices in a neutral color and asked to color a subset of vertices with expensive paints in $d$ colors in such a way that only the trivial symmetry preserves the color classes. Your goal is to minimize the number of vertices needing this expensive paint. This paper addre...

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Zusammenfassung:You are handed a graph with vertices in a neutral color and asked to color a subset of vertices with expensive paints in $d$ colors in such a way that only the trivial symmetry preserves the color classes. Your goal is to minimize the number of vertices needing this expensive paint. This paper address the issues surrounding your choices. In particular, a graph is said to be $d$-distinguishable if there exists a coloring with $d$ colors so that only the trivial automorphism preserves the color classes. The distinguishing number of $G$, denoted ${\rm Dist}(G)$, is the smallest $d$ for which $G$ is $d$-distinguishable. We define the -paint cost of $d$-distinguishing, denoted $\rho^d(G)$, to be the minimum number of vertices that need to be painted to $d$-distinguish $G$. This cost varies with $d$. The maximum paint cost for $G$ is called the upper paint cost, denoted $\rho^u(G)$, and occurs when $d={\rm Dist}(G)$; the minimum paint cost is called the lower paint cost, denoted $\rho^\ell(G)$. Further, we define the smallest $d$ for which the paint cost is $\rho^\ell(G)$, to be the frugal distinguishing number, ${\rm Fdist}(G)$. In this paper we formally define $\rho^d(G)$, $\rho^u(G)$, $\rho^\ell(G)$, and ${\rm Fdist}(G)$. We also show that $\rho^u(G)$ and $\rho^\ell(G)$, as well as ${\rm Fdist}(G)$ and ${\rm Dist}(G)$, can be arbitrarily large multiples of each other. Lastly, we find these parameters for the book graph $B_{m,n}$, summarized as follows. For $n\geq 2$ and $m\geq 4$, we show $\bullet$ $\rho^\ell(B_{m,n}) = n-1;$ $\bullet$ $\rho^u(B_{m,n}) \geq (m-2) \left( n-k^{m-3} \right) +1$, where $k={\rm Dist}(B_{m,n});$ $\bullet$ ${\rm Fdist}(B_{m,n}) = 2+\left\lfloor \frac{n-1}{m-2} \right\rfloor.$
DOI:10.48550/arxiv.2103.06097