Plane-section recurrence for facet access: $A(n,d)\le\lfloor n/2\rfloor\,A(n-1,d-1)$

Let $P\subseteq\mathbb R^d$ ($d\ge2$) be a bounded simple H-polytope with $n$ rows, all with nonzero normal, let $u$ be a vertex and $i$ a row met by $P$. Suppose every bounded H-polytope in $\mathbb R^{d-1}$ with $n-1$ rows has facet access at most $A$: from any vertex, some vertex on any given nonempty row is reachable by a padded walk of length $A$. Then some vertex of $P$ on row $i$ is reachable from $u$ by a padded walk of length $\lfloor n/2\rfloor\cdot A$.

Proof idea. A generic 2-plane through $u$ and a relative interior point of the face of row $i$ cuts $P$ in a convex polygon with at most $n$ sides, each in a facet of $P$, with $u$ a vertex and an edge on row $i$. Consecutive sides lie in facets sharing a vertex of the polygon, hence (by simplicity) sharing a ridge. The shorter boundary arc gives a dual path of at most $\lfloor n/2\rfloor$ facets from a facet through $u$ to row $i$; lifting each dual step costs one $(d-1)$-dimensional facet access inside the current facet, via Hirsch.facet_walk.

This is a geometric recurrence unavailable in the connected-layer-family abstraction (it uses convexity of plane sections). It is sharp in dimension two and unfolds to $n^{O(d)}$, so it does not by itself give a uniform polynomial.

Formalization Note Simplicity is assumed through the tight-row count; the general case reduces to it by the platform's push/perturbation lemmas (Hirsch.spindle_prepare_row_simple).