% SLACK_FORCING_G.tex
\section{Slack-Forcing from Doeblin-Word Avoidance}\label{sec:slack-forcing}

This section isolates a quantitative principle that will be used later in the forcing dichotomy:
once a Doeblin word is certified on a tight aperiodic SCC and the bounded-gap recurrence certificate (RC1) holds for tight trajectories,
then any attempt to avoid that word over long time windows necessarily incurs a \emph{linear} amount of reduced-cost slack.

\subsection{Setup and constants}

Fix the calibrated potential $\phi$ and reduced costs $c_\phi(e)\ge 0$ as in \eqref{eq:reduced-cost-u}.
Recall the tight edge set $E^{(0)}=\{e: c_\phi(e)=0\}$ and the tight graph $G^{(0)}=(V^{(0)},E^{(0)})$.

Let
\begin{equation}\label{eq:alpha-min-positive-slack}
\alpha := \min\{c_\phi(e): e\in E\setminus E^{(0)}\},
\end{equation}
with the convention $\alpha=+\infty$ if $E^{(0)}=E$.
When $E^{(0)}\neq E$, we have $\alpha>0$ by finiteness of $E$.
This constant already appears in Proposition~\ref{prop:asymptotic-tight}.

\begin{definition}[Slack count and slack mass on a window]\label{def:slack-count-mass}
For a finite edge-word $\gamma=(e_0,\dots,e_{L-1})$, define
\[
\mathrm{SlackCount}(\gamma):=\big|\{0\le t<L:\ e_t\notin E^{(0)}\}\big|,
\qquad
\mathrm{SlackMass}(\gamma):=\sum_{t=0}^{L-1} c_\phi(e_t).
\]
\end{definition}

By construction, $\mathrm{SlackMass}(\gamma)\ge \alpha\cdot \mathrm{SlackCount}(\gamma)$ whenever $\alpha<\infty$.

\subsection{RC1 turns word-avoidance into a bound on tight-run lengths}

Let $H$ be a tight SCC of $G^{(0)}$ and let $w$ be a finite tight edge-word supported in $H$.
Assume that the recurrence certificate \textbf{(RC1)} holds for $(H,w)$ in the sense of Definition~\ref{def:rc1}.
Let $L_{\mathrm{syn}}(H,w)$ denote the explicit syndetic return modulus from Lemma~\ref{lem:avoidance-syndetic}.
Set
\begin{equation}\label{eq:L0-def}
L_0:=L_{\mathrm{syn}}(H,w)+|w|.
\end{equation}

\begin{lemma}[No long tight runs in $H$ can avoid $w$]\label{lem:long-tight-run-impossible}
Assume \textbf{(RC1)} holds for $(H,w)$ and let $L_0$ be as in \eqref{eq:L0-def}.
If $\gamma=(e_0,\dots,e_{L-1})$ is a finite path that stays inside $H$ and uses only tight edges $e_t\in E^{(0)}$,
then $\gamma$ contains an occurrence of $w$ provided $L\ge L_0$.
Equivalently, any tight path segment inside $H$ with no occurrence of $w$ has length at most $L_0-1$.
\end{lemma}

\begin{proof}
Because $H$ is strongly connected and contains a directed cycle, any finite tight path segment inside $H$ can be extended forward to an infinite tight path in $H$.
Apply Lemma~\ref{lem:avoidance-syndetic} to that infinite tight extension: it contains occurrences of $w$ with gaps between \emph{starting indices} bounded by $L_{\mathrm{syn}}(H,w)$.
In particular, there exists an occurrence of $w$ whose start index is at most $L_{\mathrm{syn}}(H,w)$.
Since an occurrence of $w$ occupies $|w|$ consecutive edges, the entire occurrence lies within the first $L_{\mathrm{syn}}(H,w)+|w|=L_0$ edges.
Thus any tight segment of length $L\ge L_0$ must contain an occurrence of $w$.
\end{proof}

\subsection{Linear slack forcing on word-avoiding windows}

We now turn Lemma~\ref{lem:long-tight-run-impossible} into a quantitative lower bound on slack.

\begin{proposition}[Avoiding $w$ for length $L$ forces linear slack]\label{prop:slack-forcing-window}
Let $H$ be a tight SCC of $G^{(0)}$ and let $w$ be a tight word supported in $H$ such that \textbf{(RC1)} holds for $(H,w)$.
Let $L_0$ be as in \eqref{eq:L0-def}.
Consider any finite path segment $\gamma=(e_0,\dots,e_{L-1})$ that stays inside $H$ and contains \emph{no} occurrence of $w$.
Then
\begin{equation}\label{eq:slackcount-lower}
\mathrm{SlackCount}(\gamma)\ \ge\ \max\!\left\{0,\ \left\lceil\frac{L-L_0}{L_0+1}\right\rceil\right\},
\end{equation}
and consequently, if $\alpha<\infty$,
\begin{equation}\label{eq:slackmass-lower}
\mathrm{SlackMass}(\gamma)\ \ge\ \alpha\cdot \max\!\left\{0,\ \left\lceil\frac{L-L_0}{L_0+1}\right\rceil\right\}.
\end{equation}
In particular, $\mathrm{SlackMass}(\gamma)$ grows at least linearly in $L$ once $L>L_0$.
\end{proposition}

\begin{proof}
If $\gamma$ contains no tight edges, then $\mathrm{SlackCount}(\gamma)=L$ and \eqref{eq:slackcount-lower} holds.
Assume $\gamma$ contains at least one tight edge.
Decompose $\gamma$ into maximal consecutive blocks of tight edges (tight runs) separated by slack edges.
Let $k$ be the number of tight runs and let $s=\mathrm{SlackCount}(\gamma)$ be the number of slack edges.
Then necessarily $k\le s+1$.

By Lemma~\ref{lem:long-tight-run-impossible}, every tight run inside $H$ that avoids $w$ has length at most $L_0-1$.
Hence the total number of tight edges in $\gamma$ is at most $k(L_0-1)$.
Therefore
\[
L = (\text{number of tight edges}) + s \ \le\ k(L_0-1) + s \ \le\ (s+1)(L_0-1) + s = s(L_0+1) + (L_0-1).
\]
Rearranging gives $s \ge (L-L_0)/(L_0+1)$.
Since $s$ is an integer and $s\ge 0$, we obtain \eqref{eq:slackcount-lower}.
Finally, \eqref{eq:slackmass-lower} follows from $\mathrm{SlackMass}(\gamma)\ge \alpha\,\mathrm{SlackCount}(\gamma)$.
\end{proof}

\begin{remark}[What this does and does not say]\label{rem:slack-forcing-scope}
Proposition~\ref{prop:slack-forcing-window} is a \emph{local forcing} statement:
it applies to windows that stay inside a fixed tight SCC $H$ where \textbf{(RC1)} holds for a chosen word $w$.
If \textbf{(RC1)} fails, Lemma~\ref{lem:avoidance-cycle-witness} provides a tight periodic witness that avoids $w$ with zero slack,
showing that no positive lower bound of the form \eqref{eq:slackmass-lower} can hold in that case.
Excursions that leave $H$ and later return are handled separately in Section~\ref{sec:return-cost}.
\end{remark}
