The Distinguishing Index of Mycielskian Graphs

Let $V(K_{1,m})=\{v_{0},v_{1},v_{2},\dots,v_{m}\}$ be the vertex set of $K_{1,m}$ such that $\deg(v_{0})=m$. Let $u_{0},u_{1},u_{2},\dots,u_{m}$ be the shadow vertices in $\mu(K_{1,m})$, and let $w$ be the root. Additionally, let $I=\{1,2,\dots,m\}$. Then we have the vertex set

and the edge set

We consider two important subsets of the edges - let $L_{1}$ and $L_{2}$ be sets of ordered pairs of edges such that $L_{1}=\{(u_{0}v_{i},v_{i}v_{0}):i\in I\}$ and $L_{2}=\{(wu_{i},u_{i}v_{0}):i\in I\}$. Note that each edge in $E(\mu(K_{1,m}))$ is in an ordered pair in $L_{1}$ or $L_{2}$ or is the edge $u_{0}w$. In Figure 2, $L_{1}$ represents the "elbows" drawn on the left, and $L_{2}$ represents the "elbows" drawn on the right.

Choose $r$ least such that $r^{2}\geq m+1$. We will color $\mu(K_{1,m})$ with $r$ colors. Given $r$ colors, there are $r^{2}$ ordered pairs of colors. Let $P=\{p_{1},p_{2},\dots p_{r^{2}}\}$ be the set of all ordered pairs of $r$ colors. First, we color the edges in $L_{1}$. For $i\in I$, we assign the color pair $p_{i}$ to the ordered pair $(u_{0}v_{i},v_{i}v_{0})$ so that $u_{0}v_{i}$ gets the first color in the ordered pair $p_{i}$ and $v_{i}v_{0}$ gets the second color in the ordered pair $p_{i}$. Similarly, for the $i^{th}$ ordered pair in $L_{2}$, $(wu_{i},u_{i}v_{0})$, we assign the color pair $p_{i+1}$. Since $r^{2}\geq m+1$, there are enough color pairs to do so. Finally, color the edge $u_{0}w$ any of the $r$ colors.

We will now show this coloring is distinguishing. Suppose $\phi$ is a color-preserving automorphism on $\mu(K_{1,m})$. By Lemma 3, $\phi(w)\in\{w,u_{0}\}$. By construction, every pair $p_{j}$ for $1\leq j\leq r^{2}$ is distinct so $\phi(v_{i})\neq v_{\alpha}$ for any $\alpha\neq i$ and $\phi(u_{i})\neq u_{\alpha}$ for any $\alpha\neq i$. Because the color pair $p_{1}$ colors an edge pair in $L_{1}$ and does not color any edge pair in $L_{2}$, there is no $\beta\in I$ such that $\phi(v_{1})=u_{\beta}$. So $\phi(v_{i})=v_{i}$ and $\phi(u_{i})=u_{i}$ for every $i\in I$. It follows that $\phi(w)=w$ and $\phi(u_{0})=u_{0}$. So we have a distinguishing edge coloring for $\mu(K_{1,m})$ with $r$ colors and $\operatorname{Dist^{\prime}}(\mu(K_{1,m}))\leq r$.

To show that $\operatorname{Dist^{\prime}}(\mu(K_{1,m}))=r$, suppose that we have a distinguishing coloring with $s$ colors such that $s<r$. Let $P^{\prime}=\{p_{1},p_{2},\dots p_{s^{2}}\}$ be the set of all ordered pairs of $s$ colors. Since $s<r$, and $r$ is least such that $r^{2}\geq m+1$, $s^{2}<m+1$ and $P^{\prime}$ contains at most $m$ color pairs. Note that for every $i\in I$, $N(v_{i})=\{u_{0},v_{0}\}$ and hence for all $i,j\in I$, $v_{i}$ and $v_{j}$ are twins. By Lemma 2, every edge pair in $L_{1}$ must be colored with a different color pair. Since there are $m$ edge pairs in $L_{1}$ we must have $s^{2}\geq m$. Since $s^{2}\leq m$ by assumption, suppose $s^{2}=m$. In this case all color pairs must be used to color the edge pairs of $L_{1}$ and all color pairs must be used to color the edge pairs of $L_{2}$. Because $L_{1}$ and $L_{2}$ use all of the color pairs of $P^{\prime}$, we can construct a non-trivial automorphism $\phi$ that is color preserving where $\phi(w)=u_{0}$. So we have found that we cannot make a distinguishing edge coloring with fewer than $r$ colors and $\operatorname{Dist^{\prime}}(\mu(K_{1,m}))=r$.

Note that $\operatorname{Dist}^{\prime}(K_{1,m})=m$ for if any two edges were the same color, their degree $1$ vertices could be switched in a color preserving automorphism. For $m\geq 2$, we know $m^{2}\geq m+1$ and hence $r=m$ satisfies $r^{2}\geq m+1$. Since $r$ is least such that $r^{2}\geq m+1$, $r\leq m$ and we conclude

∎

Define the vertices of $\mu(G)$ to be $v_{1},\dots,v_{n}$, their shadows to be $u_{1},\dots,u_{n}$ and the root to be $w$. Let $c$ be a distinguishing coloring of $E(G)$ with colors $1,\dots,q$. We define a coloring $\overline{c}$ of $E(\mu(G))$ that “mimics" the coloring on $E(G)$. If $xy\in\{v_{j}v_{k},v_{j}u_{k},v_{k}u_{j}\}$ for any $1\leq k,j\leq n$ then $\overline{c}(xy)=c(v_{j}v_{k})$. Lastly, $\overline{c}(wu_{i})=1$ for all $1\leq i\leq n$.

Let $\phi$ be a color-preserving automorphism. Since $G\neq K_{1,m}$ we know by Lemma 3 that for the root $w$ of $\mu(G)$, $\phi(w)=w$. For all $1\leq i\leq n$, $d(u_{i},w)=1$ and $d(v_{i},w)=2$. Therefore, for all $1\leq i\leq n$, $\phi(u_{i})=u_{j}$ for some $1\leq j\leq n$ and $\phi(v_{i})=v_{j}$ for some $1\leq j\leq n$. That is, the levels of $\mu(G)$ are fixed set-wise. Moreover, $\overline{c}$ fixes $v_{i}\in V(\mu(G))$ for $1\leq i\leq n$ because $\overline{c}$ restricts to a distinguishing edge coloring of $G$. Let $u_{k}$ be the shadow of a non-twin vertex in $\mu(G)$. Then $u_{k}$ has a unique and fixed neighborhood, and $\phi(u_{k})=u_{k}$ for all such $u_{k}$.

We will show that twins $u_{i},u_{j}\in V(\mu(G))$ are fixed using a proof by contradiction. So, we assume that $u_{i},u_{j}$ are not fixed by $\phi$, so $\phi(u_{i})=u_{j}$ and $\phi(u_{j})=u_{i}$. Since $u_{i},u_{j}$ are twins, there are twins $v_{i}$ and $v_{j}$ in $G$ where $N(v_{i})=N(v_{j})$, so by Lemma 2, there exists some $v_{0}\in N(v_{i})=N(v_{j})$ such that $c(v_{k}v_{i})\neq c(v_{0}v_{j})$. Since $c(v_{k}v_{i})=\overline{c}(v_{k}u_{i})$ and $c(v_{k}v_{j})=\overline{c}(v_{k}u_{j})$ we know $\overline{c}(v_{k}u_{i})\neq\overline{c}(v_{k}u_{j})$. Then $\phi$ cannot be a color-preserving automorphism on $\mu(G)$ which is a contradiction. Thus $u_{i}$ and $u_{j}$ are fixed as desired. ∎

Let $V(\mu_{t}((K_{1,m}))=\{u^{j}_{0},u^{j}_{1},u^{j}_{2},\dots,u^{j}_{m}:0\leq j\leq t\}\cup\{w\}$ where $u^{0}_{0}$ is the center of $K_{1,m}$ and $w$ is the root.

To define the edge set, we start by defining $L_{i}$ for $0\leq i\leq t$ as sets of ordered pairs of edges. Let $L_{0}=\{(u^{1}_{0}u^{0}_{i},u^{0}_{i}u^{0}_{0}):i\in I\}$, $L_{t}=\{(wu^{t}_{0},u^{t}_{i}u^{t-1}_{0}):i\in I\}$, and

where $1\leq\alpha<t$. The edges in ordered pairs in each $L_{\alpha}$ plus $u_{0}^{t}w$ make up $E(\mu_{t}(K_{1,m}))$. In Figure 4 the sets $L_{i}$ for $0\leq i\leq 5$ are shown for $\mu_{5}(K_{1,3})$.

Choose $r$ least such that $r^{2}\geq m+1$. Given $r$ colors, there are $r^{2}$ ordered pairs of colors. Let $P=\{p_{1},p_{2},\dots p_{r^{2}}\}$ be the set of all ordered pairs of $r$ colors. For each $L_{\alpha}$ with $1\leq\alpha<t$, we assign the color pair $p_{i}$ to the $i^{th}$ ordered pair in $L_{\alpha}$, $(u^{\alpha+1}_{0}u^{\alpha}_{i},u^{\alpha}_{i}u^{\alpha-1}_{0})$ so that $u^{\alpha+1}_{0}u^{\alpha}_{i}$ gets the first color in the ordered pair $p_{i}$ and $u^{\alpha}_{i}u^{\alpha-1}_{0}$ gets the second color in the ordered pair $p_{i}$. Similarly, for the $i^{th}$ ordered pair in $L_{0}$, $(u_{0}^{1}u^{0}_{i},u^{0}_{i}u^{0}_{0})$, we assign the color pair $p_{i}$. Finally, for the $i^{th}$ ordered pair in $L_{t}$, $(wu^{t}_{i},u^{t}_{i}u^{t-1}_{0})$, we assign the color pair $p_{i+1}$. Since $r^{2}\geq m+1$, there are enough color pairs to do so. Finally, color the edge $u_{0}w$ any of the $r$ colors. Examples of this coloring on $\mu_{5}(K_{1,3})$ and $\mu_{2}(K_{1,5})$ are shown in Figure 5.

We’ll now prove this coloring is distinguishing. Suppose $\phi$ is a color-preserving automorphism on $\mu_{t}(K_{1,m})$ and suppose $\phi(w)=u_{0}^{t}$. Since $u_{m}^{t}$ has degree $2$ and is distance $1$ from $w$, $\phi(u_{m}^{t})$ must have degree $2$ and be distance $1$ from $\phi(w)=u_{0}^{t}$. Thus $\phi(u_{m}^{t})=u_{i}^{t-1}$ for some $1\leq i\leq t$. Similarly, by distance and degree, $\phi(u_{0}^{t-1})=u_{0}^{t-2}$. Note that $(wu_{m}^{t},u_{m}^{t}u_{0}^{t-1})\in L_{t}$ has color pair $p_{m+1}$, and no pair of edges in $L_{t-1}=\{(u^{t}_{0}u^{t-1}_{i},u^{t-1}_{i}u^{t-2}_{0}):1\leq i\leq n\}$ has this color pair. So $u_{m}^{t}$ can’t map to $u_{i}^{t-1}$ because then the pair of edges $(wu_{m}^{t},u_{m}^{t}u_{0}^{t-1})$ would map to the pair of edges $(u_{0}^{t}u_{i}^{t-1},u_{i}^{t-1}u_{0}^{t-2})$ and this wouldn’t be color-preserving. So $\phi(w)\neq u_{0}^{t}$. Therefore, by Lemma 7, $\phi(w)=w$.

Since $w$ is fixed, by distance and degree, the levels of $\mu_{t}(K_{1,m})$ are fixed setwise by $\phi$. The construction of our coloring means that on a given level $j$ with $0\leq j\leq t$, we used a distinct color ordered pair for the edges incident to any degree $2$ vertex. So, $\phi(u^{j}_{i})\neq u^{j}_{k}$ for any $k\neq i$. Thus, $\phi(u^{j}_{i})=u^{j}_{i}$ for $0\leq j\leq t$ and $1\leq i\leq m$. Since we also know $\phi(u^{j}_{0})=u^{j}_{0}$ for $0\leq j\leq t$ and $\phi(w)=w$, we conclude that $\phi$ is the trivial automorphism and $\operatorname{Dist^{\prime}}(\mu_{t}(K_{1,m}))\leq r$.

To show that $\operatorname{Dist^{\prime}}(\mu(K_{1,m}))=r$, we assume for the sake of a contradiction that there’s a distinguishing coloring with $s$ colors such that $s<r$. Let $P^{\prime}=\{p_{1},p_{2},\dots p_{s^{2}}\}$ be the set of all ordered pairs of $s$ colors. Since $s<r$, and $r$ is least such that $r^{2}\geq m+1$, we know a$s^{2}\leq m$.

First we’ll show $s^{2}\geq m$. For each $1\leq j\leq t$ and $1\leq i\leq n$, $N(u^{j}_{i})=\{u^{j+1}_{0},u^{j-1}_{0}\}$. Similarly, for $1\leq i\leq n$, $N(u^{0}_{i})=\{u^{1}_{0},u^{0}_{0}\}$ and $N(u^{t}_{i})=\{w,u^{t-1}_{0}\}$. Hence, for a fixed $j$, and all $1\leq a,b\leq n$, $u_{a}^{j}$ and $u_{b}^{j}$ are twins. By Lemma 2, every edge pair in $L_{j}$ must be colored with a different color pair. There are $m$ edge pairs in $L_{t}$ and $s^{2}$ color pairs, so $s^{2}\geq m$.

Suppose $s^{2}=m$. In this case all color pairs must be used to every color the edge pairs of each $L_{j}$ for $0\leq j\leq t$. Let $\mathcal{L}_{j}$ be the set of vertices $\{u^{j}_{i}\mid 1\leq i\leq n\}$. for every $j$, each vertex in $\mathcal{L}_{j}$ is the same distance from $w$ so we define $d(\mathcal{L}_{j},w)$ to be this distance. We will consider the cases for $t$ even and $t$ odd separately.

When $t$ is odd, for each $\mathcal{L}_{j}$ that is distance $d$ from $w$ there is exactly one other set, call it $\mathcal{L}_{j}^{\prime}$, that is distance $d$ from $u^{t}_{0}$. Let $\psi$ be the automorphism which swaps $w$ and $u^{t}_{0}$ and maps the $i^{th}$ vertex in $\mathcal{L}_{j}$ to the vertex in $\mathcal{L}_{j}^{\prime}$ that has the appropriate colors on its incident edges and complete the automorphism so that distances are preserved. Then $\psi$ is a color preserving automorphism.

If $t$ is even, then, as seen in Figure 5, $\mathcal{L}_{0}=\mathcal{L}_{0}^{\prime}$. Since all $m$ colorings were used, for any edge pair $\epsilon$ in $L_{0}$ with an ordered pair of different colors, the reverse ordered pair is also assigned to an edge pair $\delta$ in $L_{0}$. Let $\psi$ be the automorphism that swaps $w$ and $u^{t}_{0}$ and for $j\geq 1$, maps the $i^{th}$ vertex in $\mathcal{L}_{j}$ to the vertex in $\mathcal{L}_{j}^{\prime}$ that has the appropriate colors on its incident edges. Let $\psi^{\prime}$ be the automorphism which swaps the vertices in $\mathcal{L}_{0}$ such that $\epsilon$ maps to $\delta$ and vice versa. Then $\psi^{\prime}\psi$ is a color preserving automorphism.

So we have shown that we can always construct a non-trivial automorphism $\phi$ that is color preserving when $s^{2}=m$. Therefore, we cannot make a distinguishing edge coloring with fewer than $r$ colors and $\operatorname{Dist^{\prime}}(\mu(K_{1,m}))=r$.

For $m\geq 2$, we know $m^{2}\geq m+1$ and hence $r=m$ satisfies $r^{2}\geq m+1$. Since $r$ is least such that $r^{2}\geq m+1$, $r\leq m$ and we conclude that

∎

Let $|V(G)|=n$ and define $V(\mu_{t}((G))=\{u^{j}_{1},u^{j}_{2},\dots,u^{j}_{n}:0\leq j\leq t\}\cup\{w\}$ where $w$ is the root. Let $c$ be a distinguishing coloring of $E(G)$ with colors $1,\dots,q$. We define a coloring of $E(\mu_{t}(G))$ called $\overline{c}$. For an edge $u_{i}^{0}u_{k}^{0}\in E(G)$, we let $\overline{c}(u_{i}^{0}u_{k}^{0})=\overline{c}(u_{i}^{j}u_{k}^{j-1})=\overline{c}(u_{k}^{j}u_{i}^{j-1})=c(u_{i}^{0}u_{k}^{0})$ for $1\leq j\leq t$. That is, we mimic the coloring $c$ so that edges between layers are colored with the same color as the edge that produced them.