% Section: Preliminaries
\section{Preliminaries}\label{sec:prelim}

\subsection{Definitions}

Throughout, graphs are finite, simple and undirected; $N(w)$ is the open neighbourhood of $w$ and, for $X\subseteq V$, $\bd X=\{w\notin X: N(w)\cap X\neq\emptyset\}$ is the (vertex) boundary and $N[X]=X\cup\bd X$.

\begin{definition}[generalized Petersen graph]\label{def:gp}
For integers $k\ge 1$ and $n>2k$, $\PP{n}{k}$ has vertex set $\{u_i,v_i: i\in\mathbb Z_n\}$ and edge set $\{u_iu_{i+1},\ u_iv_i,\ v_iv_{i+k}: i\in\mathbb Z_n\}$. The pair $\{u_i,v_i\}$ is \emph{column} $i$; $u_i$ lies in the \emph{outer layer} and $v_i$ in the \emph{inner layer}. The rotation $\rho\colon u_i\mapsto u_{i+1},\ v_i\mapsto v_{i+1}$ and the reflection $\sigma\colon u_i\mapsto u_{-i},\ v_i\mapsto v_{-i}$ generate a dihedral group $D_n$ of order $2n$ acting on $V$.
\end{definition}

\begin{definition}[zero forcing]\label{def:zf}
Let $B\subseteq V$ be the set of black vertices. A black vertex $b$ with exactly one white neighbour $w$ may \emph{force} $w$, written $b\to w$, after which $w$ is black. A \emph{chronology} from $S$ is a sequence of forces, each legal at the colouring produced by its predecessors, starting from $B=S$; it is \emph{maximal} if it cannot be extended. $S$ is a \emph{zero-forcing set} if some maximal chronology from $S$ ends with $B=V$, and $\Z(G)$ is the minimum size of a zero-forcing set.
\end{definition}

\begin{definition}[forts, boundary minimum]\label{def:fort}
A \emph{fort} of $G$ is a non-empty $F\subseteq V$ such that every $w\notin F$ has $|N(w)\cap F|\neq 1$. For $p\ge1$ let $\beta_p(n)=\min\{|\bd X|: X\subseteq V(\PP{n}{5}),\ |X|=p\}$.
\end{definition}

\subsection{Order independence and the closure}

The following standard fact makes Definition~\ref{def:zf} independent of the choice of chronology; we include the proof because the certificates of Section~\ref{sec:framework} depend on replaying one particular chronology.

\begin{proposition}[order independence]\label{prop:order}
Let $S\subseteq V$. Every maximal chronology from $S$ ends in the same black set $\cl(S)$. Consequently $\cl$ is monotone and idempotent, and $\cl(\varphi S)=\varphi\,\cl(S)$ for every automorphism $\varphi$ of $G$.
\end{proposition}

\begin{proof}
\emph{Step 1 (monotonicity of the rule).} Let $b\in B\subseteq B'$ and suppose $b$ can force $w$ at $B$, i.e.\ $N(b)\setminus B=\{w\}$. If $w\notin B'$ then $\emptyset\ne N(b)\setminus B'\subseteq N(b)\setminus B=\{w\}$, so $b$ can force $w$ at $B'$.

\emph{Step 2.} Put $\Phi(B)=B\cup\{w\notin B:\exists b\in B,\ N(b)\setminus B=\{w\}\}$. By Step~1, $B\subseteq B'$ implies $\Phi(B)\subseteq\Phi(B')$. The final set $C$ of a maximal chronology satisfies $\Phi(C)=C$.

\emph{Step 3.} Let $D\supseteq S$ with $\Phi(D)=D$ and let $S=B_0\subset B_1\subset\dots\subset B_m=C$ be any chronology, $B_{i+1}=B_i\cup\{w_i\}$ with $b_i\to w_i$. By induction $B_i\subseteq D$: if $B_i\subseteq D$ then either $w_i\in D$ or, by Step~1, $b_i\in D$ forces $w_i$ at $D$, contradicting $\Phi(D)=D$.

\emph{Step 4.} If $C,C'$ are final sets of two maximal chronologies from $S$, both are fixed points of $\Phi$ containing $S$, so Step~3 gives $C\subseteq C'$ and $C'\subseteq C$. Thus $\cl(S)$ is the least fixed point of $\Phi$ above $S$. Monotonicity follows from Step~3, idempotence from $\Phi(\cl S)=\cl S$, and equivariance because an automorphism maps legal forces to legal forces and fixed points to fixed points.
\end{proof}

\begin{proposition}[fort characterization, {\citealp[Thm.~8]{brimkov2019computational}}]\label{prop:fort}
$S$ is a zero-forcing set of $G$ if and only if $S\cap F\neq\emptyset$ for every fort $F$.
\end{proposition}

\begin{proof}
($\Rightarrow$) Let $F$ be a fort disjoint from a zero-forcing set $S$ and consider, in a chronology from $S$ that colours $V$, the first force $b\to w$ with $w\in F$. Just before it no vertex of $F$ is black, so $b\notin F$ and all neighbours of $b$ other than $w$ are black, hence outside $F$; thus $|N(b)\cap F|=1$, contradicting the fort property.
($\Leftarrow$) If $\cl(S)\ne V$, let $W=V\setminus\cl(S)$. For $b\in\cl(S)$ the set $N(b)\cap W$ consists of the white neighbours of $b$ at a fixed point, so it has size $\ne1$. Hence $W$ is a fort disjoint from $S$.
\end{proof}

\begin{proposition}[orbit reduction]\label{prop:orbit}
Let $H\le\Aut(G)$. There is a zero-forcing set of size $s$ if and only if some representative of each $H$-orbit on $s$-subsets is zero forcing. Nothing requires an optimal set to be $H$-invariant.
\end{proposition}

\begin{proof}
By Proposition~\ref{prop:order}, $\cl(\psi S)=\psi\cl(S)$, so $S$ is zero forcing if and only if $\psi S$ is, and $|\psi S|=|S|$.
\end{proof}

\subsection{Graph facts and the isomorphism \texorpdfstring{$P(n,k)\cong P(n,l)$}{P(n,k) = P(n,l)}}

\begin{proposition}\label{prop:graph}
For $n\ge 11$, $\PP{n}{5}$ is simple and cubic with $2n$ vertices and $3n$ edges, connected, its inner subgraph is a disjoint union of $\gcd(n,5)$ cycles of length $n/\gcd(n,5)$, and $D_n$ acts by automorphisms.
\end{proposition}

\begin{proof}
The three edge families are pairwise disjoint. The outer edges form an $n$-cycle. An inner edge $v_iv_{i+5}$ is a loop iff $5\equiv0$, and $v_iv_{i+5}=v_jv_{j+5}$ with $i\ne j$ iff $10\equiv0\pmod n$; both are excluded by $n\ge 11$. So there are $3n$ distinct edges, every $u_i$ has the three distinct neighbours $u_{i\pm1},v_i$ and every $v_i$ has $v_{i\pm5},u_i$. The outer cycle is connected and every $v_i$ is joined to it by a spoke. The inner subgraph is the Cayley graph of $\mathbb Z_n$ with connection set $\{\pm5\}$; the orbit of $i$ under $i\mapsto i+5$ has $n/\gcd(n,5)\ge 11/5>2$ elements, so each orbit induces a genuine cycle. $\rho$ and $\sigma$ map each edge family to itself.
\end{proof}

\begin{proposition}[isomorphism]\label{prop:iso}
Let $\gcd(n,k)=1$ and $kl\equiv1\pmod n$. Then $\psi(v_{jk})=u_j$, $\psi(u_{jk})=v_j$ $(j\in\mathbb Z_n)$ is an isomorphism $\PP{n}{k}\to\PP{n}{l}$, and $\PP{n}{l}\cong\PP{n}{n-l}$. For $k=5$ and $n\ge 11$ the partner $l'=\min(l,n-l)$ satisfies $l'\le 4$ exactly when $n\in\{11,14,16,19,21\}$, giving $\PP{11}{5}\cong\PP{11}{2}$, $\PP{14}{5}\cong\PP{14}{3}$, $\PP{16}{5}\cong\PP{16}{3}$, $\PP{19}{5}\cong\PP{19}{4}$ and $\PP{21}{5}\cong\PP{21}{4}$.
\end{proposition}

\begin{proof}
Since $\gcd(n,k)=1$, $j\mapsto jk$ is a bijection of $\mathbb Z_n$, so $\psi$ is a bijection of the vertex set (swapping layers). Inner edge $v_{jk}v_{(j+1)k}\mapsto u_ju_{j+1}$; spoke $u_{jk}v_{jk}\mapsto v_ju_j$; outer edge $u_{jk}u_{jk+1}$: since $lk\equiv1$, $jk+1=(j+l)k$, so $u_{jk}u_{(j+l)k}\mapsto v_jv_{j+l}$, an inner edge of $\PP{n}{l}$. Thus $\psi$ maps $E(\PP{n}{k})$ injectively into $E(\PP{n}{l})$, and both have $3n$ edges. The map $i\mapsto-i$ on both layers sends $v_iv_{i+l}$ to $v_{-i}v_{-i-l}$, an edge of $\PP{n}{n-l}$. For $k=5$: $5l'\equiv\pm1\pmod n$ with $1\le l'\le4$ means $n\mid 5l'\mp1\in\{4,6,9,11,14,16,19,21\}$, and $n\ge 11$ leaves exactly the listed cases.
\end{proof}

The isomorphism of Proposition~\ref{prop:iso} is the well-known sufficiency direction of the classification of isomorphisms among generalized Petersen graphs; with \citet[Theorem~1.1]{arxiv2610_24059} ($\Z(\PP{n}{3})=8$ for $n\ge 13$, $\Z(\PP{n}{4})=10$ for $n\ge 18$) it yields $\Z(\PP{14}{5})=\Z(\PP{16}{5})=8$ and $\Z(\PP{19}{5})=\Z(\PP{21}{5})=10$, which we treat as known and use as reference tests.