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Determining Number and Cost of Generalized Mycielskian Gr...
Debra Boutin, Sally Cockburn, Lauren Keough, Sarah Loeb, K. E. P · 2020-07-30 · via math.CO updates on arXiv.org

A set $S$ of vertices is a determining set for a graph $G$ if every automorphism of $G$ is uniquely determined by its action on $S$. The size of a smallest determining set for $G$ is called its determining number, $Det(G)$. A graph $G$ is said to be $d$-distinguishable if there is a coloring of the vertices with $d$ colors so that only the trivial automorphism preserves the color classes. The smallest such $d$ is the distinguishing number, $Dist(G)$. If $Dist(G) = 2$, the cost of 2-distinguishing, $ρ(G)$, is the size of a smallest color class over all 2-distinguishing colorings of $G$. The Mycielskian, $μ(G)$, of a graph $G$ is constructed by adding a shadow master vertex $w$, and for each vertex $v_i$ of $G$ adding a shadow vertex $u_i$ with edges so that the neighborhood of $u_i$ in $μ(G)$ is the same as the neighborhood of $v_i$ in $G$ with the addition of $w$. That is, $N(u_i)=N_G(v_i)\cup\{w\}$. The generalized Mycielskian $μ^{(t)}(G)$ of a graph $G$ is a Mycielskian graph with $t$ layers of shadow vertices, each with edges to layers above and below, and $w$ only adjacent to the top layer of shadow vertices. A graph is twin-free if it has no pair of vertices with the same set of neighbors. This paper examines the determining number and, when relevant, the cost of 2-distinguishing for Mycielskians and generalized Mycielskians of simple graphs with no isolated vertices. In particular, if $G \neq K_2$ is twin-free with no isolated vertices, then $Det(μ^{(t)}(G)) = Det(G)$. Further, if $Det(G) = k \geq 2$ and $t \ge k-1$, then $Dist(μ^{(t)}(G))=2$, and $Det(μ^{(t)}(G)) = ρ(μ^{(t)}(G))= k$. For $G$ with twins, we develop a framework using quotient graphs with respect to equivalence classes of twin vertices to give bounds on the determining number of Mycielskians. Moreover, we identify classes of graphs with twins for which $Det(μ^{(t)}(G)) = (t{+}1) Det(G)$.