Ridge-visible access in the $n\le 2d+3$ two-leftover regime

Let $P\subseteq\mathbb R^d$ be a bounded simple H-polytope with $n\le 2d+3$ inequalities. Simplicity means every vertex is tight on exactly $d$ rows. Let $i$ be a describing row that is tight at two distinct vertices $z_1\neq z_2$. Then every vertex $u$ of $P$ is at graph distance at most $3$ from the ridge-visible set of row $i$.

Ridge-visible means $u$ is tight on row $i$, or tight on some other row that meets row $i$ inside $P$. The cases $n\le 2d$ (distance $0$), $n=2d+1$ and $n=2d+2$ (distance at most $1$) are already proved. The new layer is $n=2d+3$.

If $u$ is not ridge-visible, its $d$ tight rows are far from $i$, and $z_1,z_2$ contribute at least $d+1$ near rows. That leaves at most two leftover far inequalities $H,H_2$. The one-collision case (every leave-one ray from $u$ enters the same leftover) is the simplex argument of $n=2d+2$: $P$ is that simplex, all generators are strictly slack on $i$, contradiction.

When the two leftovers are mixed, the far rows are exactly the $d+2$ rows $t_U\cup\{H,H_2\}$. Non-ridge-visible vertices have tight sets that are $d$-subsets of these $d+2$ rows (Johnson graph $J(d+2,2)$). Vertices that share $d-1$ tight rows bound an edge. Consequently every far vertex is at graph distance at most $2$ from $u$.

A linear height $\psi$ uniquely minimized at $u$ (dual-basis cone) has a minimizer $x$ among the finitely many ridge-visible vertices. A decreasing neighbour $y$ of $x$ cannot be ridge-visible, hence is far, hence is at distance $\le 2$ from $u$. The path $u\rightsquigarrow y-x$ has length at most $3$ and ends in the ridge-visible set.

for simple bounded H-polytopes, on any row attained at two vertices.

This is the two-leftover layer of the ridge-visible recurrence. It does not treat $n\ge 2d+4$, and it does not by itself close the polynomial Hirsch conjecture.

Formalization Note Boundedness guarantees leave-one rays hit a new inequality. Dual-basis coordinates uniquely minimize the height at $u$. Extreme points inject into $d$-subsets of the $n$ rows, so the ridge-visible set is finite. Sharing $d-1$ tight rows produces an equality-face edge.