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face_monochromatic_pairs: retract n_i = 6 lemma as empirically false

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CRITICAL AUDIT FINDING: experiments/check_subcase_iib.py shows that
Lemma (Flank covering, n_i = 6) is empirically FALSE in full
generality, not just unproven:

  Across 142,812 chord-apex+Kempe colourings up to |V(G)| ≤ 20:
    -   9,228 (6.46%) reach sub-case (ii.B) of Case (b)
        (φ(A_i P_1) = c_1 AND φ(P_1 P_2) = c_0);
    -   1,314 (0.92%) of those have P_1 ∉ V(K_b) ∪ V(K_c),
        falsifying the lemma's conclusion ∂F_flank^♭ ⊆ V(K_b) ∪ V(K_c).

So the original n_i = 6 lemma cannot be saved by patching the proof;
the conclusion itself is wrong.

Paper changes:
  - Lemma (Flank covering, n_i = 6): retracted in full generality.
    Restated with a weakened conclusion (true only for Case (a) and
    Case (b) sub-case (i)), with explicit acknowledgement that the
    sub-case (b)(ii) configuration falsifies the lemma on 1,314
    colourings.
  - Proof of the lemma: rewritten to honestly stop at the proven sub-
    cases; sub-case (b)(ii) is identified as unprovable by local
    argument (and now demonstrated empirically false).
  - Theorem (Partial proof via flank): restricted from n_i ∈ {5, 6}
    to n_i = 5 only.
  - Theorem (Extended partial proof): cases relabelled (a'), (b'), (c)
    with a' = (n_i = 5), b' = (n_{i+1} = 5), c = (n_{i+2} = n_{i+4} = 5).
  - Empirical coverage remark: structural proof covers
    7,531 / 7,930 (94.97%) of (G, v, i) configurations up to
    |V(G)| ≤ 20. The other 399 (5.03%) have at least one n_k = 6 but
    no n_k = 5 in the right position; the flank face on the n_k = 6
    side is the natural candidate but is no longer a tight covering.
  - Deciding-face conjecture itself remains empirically true on all
    142,812 colourings; the proof's structural step is what's open
    on the 399 triples.

Lessons from the audit:
  - The "+P_2 ∈ V(K_b) ∪ V(K_c) implies P_1 ∈ V(K_b) ∪ V(K_c)"
    propagation in the original n_i = 6 proof was wrong: the cycle
    type (K_b vs K_c) matters in a way the proof glossed over, and
    specifically when φ(P_1 P_2) = c_0 the K_c cycle through P_2
    doesn't use that edge.
  - A correct n_i = 6 lemma would require a global K_b-walk argument
    showing the {c, c_0}-cycle through P_2 coincides with K_b in the
    bad sub-case. Empirically this is FALSE in 0.92% of colourings,
    so no such argument exists; the n_i = 6 covering must instead come
    from a different face entirely for those colourings.

Paper stays at 21 pages.

Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>
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papers/face_monochromatic_pairs/experiments/check_subcase_iib.py
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"""Check whether sub-case (ii.B) of Case (b) of the Flank-covering
lemma (n_i = 6) ever arises in chord-apex+Kempe colourings.
Sub-case (ii.B) is the configuration in which the structural
propagation argument in the partial proof has a real gap:
- n_i = 6: F_flank_{i, i+1}^♭ has 2 intermediates P_1, P_2 on the
F-arc from A_i to A_{i+1}.
- Case (b): varphi(A_i P_1) = c_1.
- Sub-case (ii): varphi(P_1 P_2) = c_0.
- Derived: varphi(A_{i+1} P_2) = c_1, varphi(P_2 O_2) = c,
varphi(P_1 O_1) = c, varphi(A_i Q) = c.
In this sub-case the lemma's local argument (that the cycle at P_2
passes from P_2 to P_1) needs P_2 ∈ V(K_b), which isn't forced from
the local data: the {c, c_0}-Kempe cycle through P_2 might not be K_b.
We check: across all chord-apex+Kempe colourings of all reduced duals
with |V(G)| ≤ 20, does sub-case (ii.B) ever occur? If yes, in how many
of those configurations is P_1 actually in V(K_b) ∪ V(K_c) (i.e., the
conclusion of the lemma still holds even though our proof gap is
real)?
Run with: sage experiments/check_subcase_iib.py
"""
import os
import sys
import time
from sage.all import Graph
from sage.graphs.graph_generators import graphs
HERE = os.path.dirname(os.path.abspath(__file__))
sys.path.insert(0, HERE)
from check_conj_3_8_scaled import (
apply_reduction,
proper_3_edge_colorings,
matches_chord_apex_kempe,
kempe_cycle_set,
edge_idx,
)
from check_heawood_on_kempe import dual_of, vertices_of_kempe
def find_flank_arc_intermediates(H, named, i, side):
"""Return the ordered F-arc from A_i (or A_{i+1}) along F's
boundary as a list of vertices.
side='lower': arc from A_i to A_{i+1} (= F_flank_{i, i+1}^♭).
side='upper': arc from A_{i+1} to A_{i+2} (= F_flank_{i+1, i+2}^♭).
We use the planar embedding of H to walk around the
face containing v_n + side_0 + spike (lower) or v_n + spike + side_1
(upper).
"""
H.is_planar(set_embedding=True)
spike = named['spike'] # frozenset({A_{i+1}, v_n})
side_0 = named['side_0']
side_1 = named['side_1']
# The flank face contains v_n + spike + side_0/side_1 + the arc edges
target_edges = set()
if side == 'lower':
target_edges = {side_0, spike}
else:
target_edges = {side_1, spike}
for face in H.faces():
# face is list of (u, v) edges (or tuples of vertex pairs)
face_edges_set = {frozenset(e) for e in face}
if target_edges.issubset(face_edges_set):
# Trace the boundary, return the vertices in cyclic order
verts = []
for e in face:
verts.append(e[0])
return verts
return None
def test_one(D):
D.is_planar(set_embedding=True)
n_col = 0
n_subcase_iib = 0
n_subcase_iib_p1_covered = 0 # of those, how many have P_1 ∈ V(K_b) ∪ V(K_c)
for face in D.faces():
if len(face) != 5: continue
for i_red in range(5):
res = apply_reduction(D, face, i_red, 9999)
if res is None: continue
H = res['H']; named = res['named']
H.is_planar(set_embedding=True)
edges, colorings = proper_3_edge_colorings(H)
cand = [c for c in colorings
if matches_chord_apex_kempe(edges, c, named)]
for col in cand:
n_col += 1
# We need to identify P_1, P_2, A_i, A_{i+1}, etc.
# Recover the named vertices from `named`.
v_n = 9999
# named['side_0'] = {A_i, v_n}, named['spike'] = {A_{i+1}, v_n}
# side_1 = {A_{i+2}, v_n}, merged = {A_{i+3}, A_{i+4}}
A_i = next(iter(named['side_0'] - {v_n}))
A_ip1 = next(iter(named['spike'] - {v_n}))
A_ip2 = next(iter(named['side_1'] - {v_n}))
A_ip3, A_ip4 = sorted(named['merged'])
# Get flank arc for lower (A_i → A_{i+1})
# The face F_flank_{i, i+1}^♭ has boundary v_n - A_i - ... - A_{i+1} - v_n
# Find this face:
target_edges = {named['side_0'], named['spike']}
arc_verts = None
for f in H.faces():
f_edges = {frozenset(e) for e in f}
if target_edges.issubset(f_edges):
arc_verts = [e[0] for e in f]
break
if arc_verts is None or len(arc_verts) != 5:
# Not the n_i = 6 case (length 5 flank face)
continue
# Identify P_1, P_2 along the arc
# The cycle visits v_n, A_i, P_1, P_2, A_{i+1} in some order
# (possibly reversed). Find the index of v_n.
if v_n not in arc_verts: continue
k = arc_verts.index(v_n)
cyc = arc_verts[k:] + arc_verts[:k] # rotate so v_n is first
# cyc should be [v_n, A_i, P_1, P_2, A_{i+1}] or
# [v_n, A_{i+1}, P_2, P_1, A_i]
if cyc[1] == A_i:
P_1, P_2 = cyc[2], cyc[3]
assert cyc[4] == A_ip1
elif cyc[1] == A_ip1:
P_2, P_1 = cyc[2], cyc[3]
assert cyc[4] == A_i
else:
continue
# Check sub-case (ii.B): φ(A_i P_1) = c_1, φ(P_1 P_2) = c_0
# c = color of merged/spike
merged_idx = edge_idx(edges, named['merged'])
c = col[merged_idx]
# side_0 has color c_0
side_0_idx = edge_idx(edges, named['side_0'])
c_0 = col[side_0_idx]
# side_1 has color c_1
side_1_idx = edge_idx(edges, named['side_1'])
c_1 = col[side_1_idx]
# Get edge colours of (A_i, P_1) and (P_1, P_2)
e_AiP1 = edge_idx(edges, frozenset((A_i, P_1)))
e_P1P2 = edge_idx(edges, frozenset((P_1, P_2)))
if e_AiP1 is None or e_P1P2 is None:
continue
if col[e_AiP1] != c_1 or col[e_P1P2] != c_0:
continue
# Sub-case (ii.B) detected
n_subcase_iib += 1
# Check if P_1 ∈ V(K_b) ∪ V(K_c)
a = c
other = [x for x in range(3) if x != a]
# K_b = {a, b_color} Kempe cycle through merged
kc_b = kempe_cycle_set(edges, col, merged_idx, (a, other[0]))
kc_c = kempe_cycle_set(edges, col, merged_idx, (a, other[1]))
V_b = vertices_of_kempe(edges, kc_b)
V_c = vertices_of_kempe(edges, kc_c)
if P_1 in V_b or P_1 in V_c:
n_subcase_iib_p1_covered += 1
return n_col, n_subcase_iib, n_subcase_iib_p1_covered
def main(max_n=20, time_budget_per_n=1800):
print("Check: how often does sub-case (ii.B) of Case (b) of\n"
"Lemma 'Flank covering, n_i = 6' actually arise?\n"
f"n_G in [12, {max_n}]\n")
grand_col = 0
grand_sub = 0
grand_sub_covered = 0
for n in range(12, max_n + 1):
start = time.time()
try:
triangulations = list(graphs.triangulations(n, minimum_degree=5))
except Exception as ex:
print(f"n={n}: cannot enumerate ({ex})")
continue
n_col_n = 0; n_sub_n = 0; n_cov_n = 0
for tri_idx, G in enumerate(triangulations):
if time.time() - start > time_budget_per_n:
print(f" n={n}: timeout at tri {tri_idx}/{len(triangulations)}")
break
G.is_planar(set_embedding=True)
D = dual_of(G)
ni, ns, ncov = test_one(D)
n_col_n += ni; n_sub_n += ns; n_cov_n += ncov
elapsed = time.time() - start
print(f"n={n}: {n_col_n} col., {n_sub_n} sub-case (ii.B) "
f"({100*n_sub_n/max(n_col_n, 1):.2f}% of col.); "
f"{n_cov_n} of those have P_1 ∈ V(K_b)∪V(K_c) "
f"[{elapsed:.0f}s]")
sys.stdout.flush()
grand_col += n_col_n
grand_sub += n_sub_n
grand_sub_covered += n_cov_n
print()
print("=" * 70)
print(f"Grand totals: {grand_col} chord-apex+Kempe colourings")
print(f" sub-case (ii.B) detected: {grand_sub} "
f"({100*grand_sub/max(grand_col, 1):.2f}%)")
if grand_sub > 0:
print(f" of those, P_1 ∈ V(K_b)∪V(K_c): {grand_sub_covered} "
f"({100*grand_sub_covered/max(grand_sub, 1):.2f}%)")
gap_open = grand_sub - grand_sub_covered
print(f" of those, P_1 NOT in V(K_b)∪V(K_c): {gap_open}")
else:
print(" Sub-case (ii.B) never arises empirically.")
print(" → the proof gap is structurally INACTIVE")
print(" → the n_i = 6 lemma's conclusion is empirically tight")
if __name__ == '__main__':
main()
papers/face_monochromatic_pairs/paper.pdf
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papers/face_monochromatic_pairs/paper.tex
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@@ -1041,21 +1041,45 @@ its colour is in $\{c_0, c_1\}$, so the corresponding Kempe cycle
placing $P$ in that cycle's vertex set.
\end{proof}
\begin{lemma}[Flank covering, $n_i = 6$; partial]
\begin{lemma}[Flank covering, $n_i = 6$; \textbf{empirically FALSE in full
generality}]
\label{lem:flank-covering-hex}
If $n_i = 6$ then $\partial F_{i, i+1}^{\flat} \subseteq V(K_b) \cup V(K_c)$.
\emph{Status:} The naive statement ``if $n_i = 6$ then
$\partial F_{i, i+1}^{\flat} \subseteq V(K_b) \cup V(K_c)$'' is
\emph{false} in full generality. An exhaustive check across the
$142{,}812$ chord-apex+Kempe colourings of reduced duals with
$|V(G)| \le 20$
(\texttt{experiments/check\_subcase\_iib.py}) finds:
\begin{itemize}
\item $9{,}228$ colourings ($6.46\%$) reach the configuration of
Case (b) sub-case (ii) below (with $\varphi(A_i P_1) = c_1$ \emph{and}
$\varphi(P_1 P_2) = c_0$);
\item of those $9{,}228$, exactly $1{,}314$ ($0.92\%$ of the full
$142{,}812$) actually have $P_1 \notin V(K_b) \cup V(K_c)$, falsifying
the naive lemma's conclusion.
\end{itemize}
\emph{Status:} The proof below establishes the conclusion except in a
specific sub-case (Case (b), sub-case~(ii) below) where the local
propagation argument from $P_2$ to $P_1$ requires the
$\{c, c_0\}$-Kempe cycle through $P_2$ to coincide with $K_b$; this
isn't forced by the local data and would require a global argument
through the rest of the graph. Empirically (audit of $7{,}930$
$(G, v, i)$ triples up to $|V(G)| \le 20$,
\texttt{experiments/audit\_tight\_coverage.py}) the gap does not
manifest --- $399$ of the $7{,}930$ triples rely on this lemma and
all of them have a deciding face. So the lemma's conclusion is
empirically robust; only the structural proof is incomplete.
The argument below correctly handles Case~(a) and Case~(b)
sub-case~(i), and so establishes only the weakened conclusion
\[
\partial F_{i, i+1}^{\flat} \;\subseteq\; V(K_b) \cup V(K_c)
\qquad\text{provided either}\qquad
\begin{cases}
\varphi(A_i P_1) = c, \quad\text{or} \\
\varphi(A_i P_1) = c_1 \;\text{and}\; \varphi(P_1 P_2) = c.
\end{cases}
\]
The remaining sub-case ($\varphi(A_i P_1) = c_1$ \emph{and}
$\varphi(P_1 P_2) = c_0$) cannot be settled by local argument: in
that sub-case the $\{c, c_0\}$-Kempe cycle through $P_2$ may or may
not coincide with $K_b$, depending on the global structure of the
colouring, and empirically the wrong outcome occurs on
$1{,}314 / 142{,}812$ colourings.
For those $1{,}314$ colourings, the deciding face for
Conjecture~\ref{conj:deciding-face} \emph{is} some other face of the
reduced dual (the empirical deciding-face count is still
$142{,}812 / 142{,}812$); identifying that face structurally is open.
\end{lemma}
\begin{proof}
@@ -1084,62 +1108,57 @@ $\varphi(P_1 P_2) = c_1$. Hence
\[
\varphi(P_1 P_2) \;\in\; \{c, c_0\}.
\]
Now $P_2 \in V(K_b) \cup V(K_c)$ as shown, and at $P_2$ the cycle that
contains $P_2$ uses two specific edges of $P_2$. If $P_2 \in V(K_b)$
the cycle uses $P_2$'s colour-$c$ and colour-$c_0$ edges; if
$P_2 \in V(K_c)$ it uses colour-$c$ and colour-$c_1$. In either case
the cycle at $P_2$ uses $P_2$'s colour-$c$ edge. By the
preceding paragraph $\varphi(P_1 P_2) \in \{c, c_0\}$, so the
$\{c, \varphi(P_2 P_1)\}$-Kempe cycle through $P_2$ passes from $P_2$
to $P_1$. If this cycle is $K_b$, $P_1 \in V(K_b)$; if it is $K_c$
(which requires $\varphi(P_1 P_2) = c$, since $K_c$ uses
$\{c, c_1\}$), then $P_1 \in V(K_c)$.
Case~(b) splits further on $\varphi(P_1 P_2)$.
In either case $P_1 \in V(K_b) \cup V(K_c)$.
\emph{Sub-case (i):} $\varphi(P_1 P_2) = c$. Then $P_2$'s colour-$c$
edge is $P_1 P_2$. Both $K_b$ and $K_c$ use the colour-$c$ edge
at any vertex they pass through, so whichever of $K_b, K_c$ contains
$P_2$ walks via $P_1 P_2$ to $P_1$, placing $P_1 \in V(K_b) \cup V(K_c)$.
\textbf{Audit note.} In Case~(b) sub-case~(ii) (where
$\varphi(A_i P_1) = c_1$ \emph{and} $\varphi(P_1 P_2) = c_0$), the
preceding paragraph's claim that ``the cycle at $P_2$ passes from
$P_2$ to $P_1$'' requires the $\{c, c_0\}$-Kempe cycle through $P_2$
to be $K_b$ (so that the $c_0$-edge $P_1 P_2$ is on the cycle).
We can rule out one configuration of this sub-case by properness at
$P_2$: $\varphi(A_{i+1} P_2)$ cannot be $c_0$ (since $\varphi(P_1 P_2)
= c_0$ already accounts for $P_2$'s colour-$c_0$ edge), nor $c$ (since
$A_{i+1}$'s only colour-$c$ edge is the spike). So
$\varphi(A_{i+1} P_2) = c_1$, and hence $P_2 \in V(K_c)$ via the
$K_c$-walk $A_{i+1} \to P_2$ along the $c_1$-edge.
\emph{Sub-case (ii):} $\varphi(P_1 P_2) = c_0$. Properness at $P_2$
forces $\varphi(A_{i+1} P_2) = c_1$ (the option $c_0$ would put two
colour-$c_0$ edges at $P_2$, and $c$ is impossible because $A_{i+1}$'s
only colour-$c$ edge is the spike). Hence $P_2 \in V(K_c)$ via $K_c$'s
walk $A_{i+1} \xrightarrow{c_1} P_2$. But $K_c$'s walk at $P_2$ uses
$P_2$'s colour-$c$ edge (which in this sub-case is $P_2 O_2$, not
$P_1 P_2$), and exits at $O_2$ rather than $P_1$. For $P_1$ to lie in
$V(K_b)$ we would need the $\{c, c_0\}$-cycle through $P_2$ (which
contains the $c_0$-edge $P_1 P_2$) to coincide with $K_b$ --- but
this isn't forced from the local data, and \emph{empirically} fails
on $1{,}314$ of the $142{,}812$ chord-apex+Kempe colourings tested.
However the further conclusion $P_2 \in V(K_b)$ --- needed for the
above argument to place $P_1 \in V(K_b)$ via the $c_0$-edge
$P_1 P_2$ --- is not forced by the local picture; whether the
$\{c, c_0\}$-cycle through $P_2$ coincides with $K_b$ depends on the
$\{c, c_0\}$-walk through the rest of the graph, which we have not
controlled here. This gap is empirically inactive on
all chord-apex+Kempe colourings tested
($142{,}812$ / $142{,}812$ colourings, $|V(G)| \le 20$, each with a
deciding face), but the structural proof of this sub-case remains
\emph{open}.
So the proof handles Case~(a) and Case~(b) sub-case (i) uniformly,
giving the weakened conclusion stated above. Case~(b) sub-case (ii)
is not settled by local argument.
\end{proof}
\begin{theorem}[Partial proof of
Conjecture~\ref{conj:deciding-face} via flank face]
\label{thm:deciding-face-partial}
If $n_i \in \{5, 6\}$ or $n_{i+1} \in \{5, 6\}$ for the reduction
index $i$, then $\widehat{G}'_{v,i}$ has a deciding face: one of
the two flank faces $F_{i, i+1}^{\flat}$ or $F_{i+1, i+2}^{\flat}$.
If $n_i = 5$ or $n_{i+1} = 5$ for the reduction index $i$, then
$\widehat{G}'_{v,i}$ has a deciding face: one of the two flank faces
$F_{i, i+1}^{\flat}$ or $F_{i+1, i+2}^{\flat}$.
\end{theorem}
\begin{proof}
Without loss of generality $n_i \in \{5, 6\}$ (the $n_{i+1}$ case is
symmetric, swapping side-$0$ with side-$1$). By
Lemma~\ref{lem:flank-length}, $|F_{i, i+1}^{\flat}|$ is $4$ (if
$n_i = 5$) or $5$ (if $n_i = 6$); both are $\not\equiv 0 \pmod 3$.
By Lemmas~\ref{lem:flank-covering-base}
and~\ref{lem:flank-covering-hex}, $\partial F_{i, i+1}^{\flat}
\subseteq V(K_b) \cup V(K_c)$. Hence $F_{i, i+1}^{\flat}$ is a
deciding face.
Without loss of generality $n_i = 5$ (the $n_{i+1}$ case is symmetric,
swapping side-$0$ with side-$1$). By Lemma~\ref{lem:flank-length},
$|F_{i, i+1}^{\flat}| = 4 \not\equiv 0 \pmod 3$. By
Lemma~\ref{lem:flank-covering-base},
$\partial F_{i, i+1}^{\flat} \subseteq V(K_b) \cup V(K_c)$. Hence
$F_{i, i+1}^{\flat}$ is a deciding face.
\end{proof}
\begin{remark}[$n_i = 6$ is not handled by the flank face alone]
\label{rem:n-i-6-flank-fails}
The natural extension to $n_i = 6$ via the flank face fails ---
Lemma~\ref{lem:flank-covering-hex} is empirically false in full
generality. So the flank face $F_{i, i+1}^{\flat}$ does \emph{not}
yield a structural proof in the $n_i = 6$ case for every
chord-apex+Kempe colouring, and that case must be handled by other
faces of the reduced dual.
\end{remark}
We extend the partial proof to handle configurations where neither
flank face qualifies (i.e., both $n_i, n_{i+1} \ge 7$). The candidate
is the \emph{outer face} inside $F$, on the merged side of $v_n$.
@@ -1207,13 +1226,13 @@ Conjecture~\ref{conj:deciding-face}]
\label{thm:deciding-face-partial-extended}
The deciding face exists for $\widehat{G}'_{v,i}$ whenever at least
one of the following holds:
\textbf{(a)} $n_i \in \{5, 6\}$;
\textbf{(b)} $n_{i+1} \in \{5, 6\}$; or
\textbf{(a$'$)} $n_i = 5$;
\textbf{(b$'$)} $n_{i+1} = 5$; or
\textbf{(c)} $n_{i+2} = n_{i+4} = 5$.
\end{theorem}
\begin{proof}
Cases (a) and (b) are covered by Theorem~\ref{thm:deciding-face-partial},
Cases (a$'$) and (b$'$) are Theorem~\ref{thm:deciding-face-partial},
yielding the flank face $F_{i, i+1}^{\flat}$ or $F_{i+1, i+2}^{\flat}$
as the deciding face. For case (c),
$|F_{\mathrm{outer}}^{\flat}| = 5 + 5 - 3 = 7 \equiv 1 \pmod 3$ by
@@ -1230,28 +1249,29 @@ Across all $1{,}586$ $(G, v)$ pairs underlying chord-apex+Kempe
colourings for $|V(G)| \le 20$, every cyclic neighbour-degree
sequence around $v$ contains at least one degree-$5$ entry (see
\texttt{experiments/check\_v\_neighbour\_degrees.py}). Of the
$7{,}930$ $(G, v, i)$ triples we have:
\begin{itemize}
\item $7{,}906$ ($99.70\%$) satisfy case (a) or (b);
\item the remaining $24$ ($0.30\%$) have $n_i, n_{i+1} \ge 7$, but
\emph{all $24$} have $(n_{i+2}, n_{i+3}, n_{i+4}) = (5, 5, 5)$, so
they satisfy case (c).
\end{itemize}
Combining
Theorems~\ref{thm:deciding-face-implies-conj-5-1}
and~\ref{thm:deciding-face-partial-extended}, this yields a
structural proof of
Conjecture~\ref{conj:face-monochromatic-pair-on-merged-kempe-cycle}
on every chord-apex+Kempe colouring of every reduced dual with
$|V(G)| \le 20$, \emph{modulo} the open sub-case in
Lemma~\ref{lem:flank-covering-hex}. A stricter accounting
(\texttt{experiments/audit\_tight\_coverage.py}) shows that
$7{,}531$ of the $7{,}930$ $(G, v, i)$ configurations
($94.97\%$) are covered by the \emph{tight} subset of cases
(\textbf{a$'$}~$n_i = 5$, \textbf{b$'$}~$n_{i+1} = 5$, \textbf{c}~$n_{i+2} = n_{i+4} = 5$),
which has no proof gap. The remaining $399$ ($5.03\%$) rely on the
$n_i = 6$ flank-covering lemma, and would only constitute a complete
structural proof once that lemma's Case (b) sub-case (ii) is closed.
$7{,}930$ $(G, v, i)$ triples,
Theorem~\ref{thm:deciding-face-partial-extended} (cases (a$'$),
(b$'$), (c)) gives a deciding face for $7{,}531$ ($94.97\%$); see
\texttt{experiments/audit\_tight\_coverage.py}.
The remaining $399$ triples ($5.03\%$) have $n_i \ne 5$ \emph{and}
$n_{i+1} \ne 5$ \emph{and} $(n_{i+2}, n_{i+4}) \ne (5, 5)$, but at
least one of $\{n_i, n_{i+1}\}$ equals $6$. The flank face on the
$n_k = 6$ side is the natural candidate (length $5 \not\equiv 0
\pmod 3$), but
Lemma~\ref{lem:flank-covering-hex} is empirically false in full
generality (boundary coverage fails on $0.92\%$ of colourings via
Case (b) sub-case (ii)). So Theorem~\ref{thm:deciding-face-partial-extended}
does \emph{not} extend to all $7{,}930$ triples structurally;
identifying a uniform deciding face for the remaining $399$ triples
is open.
Combined, Theorems~\ref{thm:deciding-face-implies-conj-5-1}
and~\ref{thm:deciding-face-partial-extended} yield a structural proof
of Conjecture~\ref{conj:face-monochromatic-pair-on-merged-kempe-cycle}
on $7{,}531 / 7{,}930$ ($94.97\%$) of chord-apex+Kempe configurations
up to $|V(G)| \le 20$. The deciding-face conjecture itself remains
empirically true on all $142{,}812$ colourings.
The structurally open remaining case is configurations where
\emph{both} $n_i, n_{i+1} \ge 7$ \emph{and} the merged-side