coloring_nested_tire_graphs: add Latin conjecture to paper; launch counterexample search

Adds Sec. 2 "A conjectural Latin-style substructure" with two
conjectures: (1) for tires whose every O-face has exactly 3 B_in
edges, π_D contains the Latin-flavoured subset L (where each O-face
gets a permutation of {1,2,3}); (2) adjacent tires both satisfying (1)
have compatible projections.

Adds tire_fiber_counterexample_search.py: append-log counterexample
hunt across increasing k with deduplication for resumability. Logs
to counterexample_search.log. Smoke-test data at k≤5 shows
non-all-3 SP-model artifacts produce empty intersections (the
conjecture's "exactly 3 B_in edges per face" precondition fails),
but no strict-Latin counterexample yet.

Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>

papers/coloring_nested_tire_graphs/experiments/tire_fiber_counterexample_search.py

@@ -0,0 +1,379 @@
+"""6-hour counterexample search for the Latin-substructure conjecture.
+
+Logic: pick a chord configuration cs1 for T1 (with B_in = γ of length k)
+and a chord configuration cs2 for T2 (with B_in of length k_2), with
+various m_1 (T1's B_out length) and k_2.  For each (m_1, k, cs1) and
+(k, k_2, cs2) pair (both SP-feasible), compute π_D(T1) and π_U(T2) on
+γ; check if their intersection (forward or reverse) is empty.
+
+Append-log: each test writes a JSON-Lines record to LOG_PATH.  On
+restart, we read the log to skip already-tested pairs.
+
+Run until time budget exhausted; report counterexamples loudly.
+"""
+from __future__ import annotations
+
+import json
+import os
+import sys
+import time
+from itertools import product
+from pathlib import Path
+
+import numpy as np
+
+from tire_fiber_chords import (
+    d_positions_for,
+    u_positions_for,
+    compute_faces_from_chords,
+)
+from tire_fiber_chords_fast import (
+    fiber_distribution_fast,
+    spoke_only_fiber_distribution_fast,
+    projection_support_fast,
+    proper_cycle_colorings,
+)
+
+
+HERE = Path(__file__).resolve().parent
+LOG_PATH = HERE / "counterexample_search.log"
+
+
+def log_event(event: dict) -> None:
+    """Append one JSON line to the log."""
+    event = dict(event)  # don't mutate caller
+    event.setdefault('wall_time', time.time())
+    with open(LOG_PATH, 'a') as f:
+        f.write(json.dumps(event, default=str) + "\n")
+        f.flush()
+
+
+def read_existing_keys() -> set[str]:
+    """Read the log and return set of pair-keys already tested."""
+    keys = set()
+    if not LOG_PATH.exists():
+        return keys
+    with open(LOG_PATH) as f:
+        for line in f:
+            try:
+                ev = json.loads(line)
+            except json.JSONDecodeError:
+                continue
+            if ev.get('type') == 'pair' and 'key' in ev:
+                keys.add(ev['key'])
+    return keys
+
+
+def make_key(gamma: int, m_1: int, ch1, model1: str, k_2: int, ch2, model2: str) -> str:
+    ch1_str = json.dumps(sorted([sorted(c) for c in ch1])) if ch1 else "[]"
+    ch2_str = json.dumps(sorted([sorted(c) for c in ch2])) if ch2 else "[]"
+    return f"gamma={gamma}|t1=({m_1},{ch1_str},{model1})|t2=({k_2},{ch2_str},{model2})"
+
+
+def is_sp_feasible(k: int, chords) -> bool:
+    if not chords:
+        return k <= 3
+    faces = compute_faces_from_chords(k, chords)
+    return max(len(f) for f in faces) <= 3
+
+
+# Hand-curated chord configurations per k.  Each entry should be
+# SP-feasible (every face has at most 3 B_in edges).
+CHORD_CONFIGS: dict[int, list] = {
+    3: [[]],
+
+    # k=4: only one essentially-distinct chord modulo rotation.
+    4: [[(0, 2)]],
+
+    # k=5: short or long chord.
+    5: [[(0, 2)], [(0, 3)]],
+
+    # k=6: antipodal, or 2-chord matching.
+    6: [
+        [(0, 3)],
+        [(0, 2), (3, 5)],
+        [(1, 3), (4, 0)],
+    ],
+
+    # k=7: needs ≥ 2 chords for all-faces-≤-3.
+    7: [
+        [(0, 2), (3, 6)],
+        [(0, 3), (4, 6)],
+        [(0, 2), (3, 5)],     # outer face: edges 5,6 → size 2
+        [(0, 3), (3, 6)],     # uses vertex 3 twice
+    ],
+
+    # k=8.
+    8: [
+        [(0, 2), (3, 5), (6, 0)],  # vertex 0 used twice
+        [(0, 3), (4, 7)],
+        [(0, 3), (3, 6)],
+        [(0, 2), (3, 6)],
+        [(0, 2), (4, 6)],
+    ],
+
+    # k=9: 2-chord tight (faces 3,3,3) and 3-chord variants.
+    9: [
+        [(0, 3), (4, 7)],
+        [(0, 2), (3, 5), (6, 8)],
+        [(0, 2), (4, 6), (3, 7)],
+        [(0, 3), (3, 6), (6, 0)],
+    ],
+
+    # k=10.
+    10: [
+        [(0, 3), (4, 7), (7, 0)],
+        [(0, 2), (3, 5), (6, 8), (9, 1)],
+        [(0, 3), (4, 7), (8, 0)],
+        [(0, 3), (3, 6), (6, 9), (9, 0)],
+    ],
+
+    # k=11.
+    11: [
+        [(0, 3), (3, 6), (6, 9), (9, 0)],
+        [(0, 2), (3, 5), (6, 8), (9, 0)],
+    ],
+
+    # k=12.
+    12: [
+        [(0, 3), (4, 7), (8, 11)],            # tight 4-faces-of-3
+        [(0, 3), (3, 6), (6, 9), (9, 0)],     # tight ring
+        [(0, 3), (4, 11), (5, 7), (8, 10)],   # nested
+        [(0, 2), (3, 5), (6, 9), (10, 0)],    # mixed
+    ],
+
+    # k=13.
+    13: [
+        [(0, 3), (3, 6), (6, 9), (9, 12), (0, 12)],
+        [(0, 3), (4, 7), (8, 11), (12, 1)],
+    ],
+
+    # k=14.
+    14: [
+        [(0, 3), (3, 6), (6, 9), (9, 12), (12, 0)],
+        [(0, 2), (3, 5), (6, 8), (9, 11), (12, 0)],
+    ],
+
+    # k=15: tight ring.
+    15: [
+        [(0, 3), (3, 6), (6, 9), (9, 12), (0, 12)],
+        [(0, 3), (4, 7), (8, 11), (12, 14), (12, 0)],
+    ],
+
+    # k=18: tight ring.
+    18: [
+        [(0, 3), (3, 6), (6, 9), (9, 12), (12, 15), (0, 15)],
+        [(0, 3), (4, 7), (8, 11), (12, 15), (16, 1)],
+    ],
+
+    # k=21.
+    21: [
+        [(0, 3), (3, 6), (6, 9), (9, 12), (12, 15), (15, 18), (0, 18)],
+    ],
+
+    # k=24.
+    24: [
+        [(0, 3), (3, 6), (6, 9), (9, 12), (12, 15), (15, 18), (18, 21), (0, 21)],
+    ],
+}
+
+
+# Filter to SP-feasible only.
+for k in list(CHORD_CONFIGS):
+    feasible = [cs for cs in CHORD_CONFIGS[k] if is_sp_feasible(k, cs)]
+    if feasible:
+        CHORD_CONFIGS[k] = feasible
+    else:
+        CHORD_CONFIGS[k] = []
+
+
+def compute_pi_D(gamma: int, m_1: int, chords_1, model: str) -> set:
+    if model == 'SR':
+        fibers = spoke_only_fiber_distribution_fast(m_1 + gamma)
+    else:
+        fibers, _, _ = fiber_distribution_fast(m_1, gamma, chords_1)
+    return projection_support_fast(fibers, d_positions_for(m_1, gamma))
+
+
+def compute_pi_U(gamma: int, k_2: int, chords_2, model: str) -> set:
+    if model == 'SR':
+        fibers = spoke_only_fiber_distribution_fast(gamma + k_2)
+    else:
+        fibers, _, _ = fiber_distribution_fast(gamma, k_2, chords_2)
+    return projection_support_fast(fibers, u_positions_for(gamma, k_2))
+
+
+def test_pair(gamma, m_1, ch1, model1, k_2, ch2, model2):
+    """Return result dict."""
+    t0 = time.time()
+    S1 = compute_pi_D(gamma, m_1, ch1, model1)
+    S2 = compute_pi_U(gamma, k_2, ch2, model2)
+    forward = S1 & S2
+    S2_rev = {s[::-1] for s in S2}
+    reverse = S1 & S2_rev
+    dt = time.time() - t0
+    compatible = bool(forward or reverse)
+    return {
+        'type': 'pair',
+        'key': make_key(gamma, m_1, ch1, model1, k_2, ch2, model2),
+        'gamma': gamma,
+        't1': {'m_1': m_1, 'chords': ch1, 'model': model1},
+        't2': {'k_2': k_2, 'chords': ch2, 'model': model2},
+        's1_size': len(S1),
+        's2_size': len(S2),
+        'fwd_size': len(forward),
+        'rev_size': len(reverse),
+        'compatible': compatible,
+        'time_seconds': round(dt, 3),
+        # Tag whether this is a "strict-Latin" pair (where the paper conjecture
+        # applies) -- both tires must have every O-face of exactly 3 B_in edges.
+        't1_all_three': has_all_three_faces(gamma, ch1) if model1 == 'SP' else None,
+        't2_all_three': has_all_three_faces(k_2, ch2) if model2 == 'SP' else None,
+    }
+
+
+def estimate_cost(n: int) -> float:
+    """Rough estimate of seconds to compute fiber distribution at cycle length n.
+    Calibrated from prior benchmarks (n=18: 0.1s; n=24: 6.6s)."""
+    if n <= 12:
+        return 0.01
+    return 0.001 * (2 ** (n - 12))
+
+
+# Cap on cycle length to avoid memory blowup.
+# At n=27: 3 * 2^26 paths × 27 bytes ≈ 5 GB.
+# At n=30: would be ~40 GB.
+MAX_N = 27
+
+# Time cap on a single tire computation.
+MAX_TIRE_TIME_SEC = 600  # 10 minutes max per tire
+
+
+def has_all_three_faces(k: int, chords) -> bool:
+    """The strict precondition of the paper's Latin conjecture:
+    every O-face has exactly 3 B_in edges."""
+    if not chords:
+        return k == 3
+    faces = compute_faces_from_chords(k, chords)
+    return all(len(f) == 3 for f in faces)
+
+
+def add_rotations(k: int, chords):
+    """Generate all cyclic rotations of a chord set on C_k (yields tuples)."""
+    seen = set()
+    for shift in range(k):
+        rotated = tuple(sorted([tuple(sorted([(a + shift) % k, (b + shift) % k]))
+                                 for (a, b) in chords]))
+        if rotated not in seen:
+            seen.add(rotated)
+            yield [list(c) for c in rotated]
+
+
+def main(budget_hours: float = 6.0):
+    deadline = time.time() + budget_hours * 3600
+    log_event({'type': 'session_start', 'budget_hours': budget_hours,
+               'pid': os.getpid()})
+
+    already_tested = read_existing_keys()
+    print(f"[start] already tested: {len(already_tested)} pairs")
+    log_event({'type': 'startup', 'already_tested_count': len(already_tested)})
+
+    # Iterate over k values in increasing order.
+    k_values = sorted(CHORD_CONFIGS.keys())
+    m_values = [3, 6, 9, 12, 15]
+    k2_values = [3, 6, 9, 12, 15]
+
+    n_counterexamples = 0
+    n_tested = 0
+    n_skipped = 0
+
+    for gamma in k_values:
+        if time.time() >= deadline:
+            break
+        cs_list = CHORD_CONFIGS[gamma]
+        if not cs_list:
+            log_event({'type': 'k_skip_no_configs', 'gamma': gamma})
+            continue
+        # Prioritize strict-Latin (all-3-faces) configs for the conjecture test.
+        cs_list_sorted = sorted(cs_list, key=lambda cs: not has_all_three_faces(gamma, cs))
+        log_event({'type': 'k_start', 'gamma': gamma,
+                   'n_configs': len(cs_list_sorted),
+                   'n_all3_configs': sum(1 for cs in cs_list_sorted if has_all_three_faces(gamma, cs))})
+
+        for m_1 in m_values:
+            if time.time() >= deadline: break
+            for ch1 in cs_list_sorted:
+                if time.time() >= deadline: break
+                n_1 = m_1 + gamma
+                if n_1 > MAX_N:
+                    continue
+                if estimate_cost(n_1) > MAX_TIRE_TIME_SEC:
+                    continue
+                for k_2 in k2_values:
+                    if time.time() >= deadline: break
+                    cs2_list = CHORD_CONFIGS.get(k_2, [])
+                    if not cs2_list:
+                        continue
+                    cs2_sorted = sorted(cs2_list, key=lambda cs: not has_all_three_faces(k_2, cs))
+                    for ch2 in cs2_sorted:
+                        if time.time() >= deadline: break
+                        n_2 = gamma + k_2
+                        if n_2 > MAX_N:
+                            continue
+                        if estimate_cost(n_2) > MAX_TIRE_TIME_SEC:
+                            continue
+
+                        key = make_key(gamma, m_1, ch1, 'SP', k_2, ch2, 'SP')
+                        if key in already_tested:
+                            n_skipped += 1
+                            continue
+
+                        budget_remaining = deadline - time.time()
+                        # Conservative: skip if estimated cost > budget remaining.
+                        est = estimate_cost(n_1) + estimate_cost(n_2) + 0.5
+                        if est > budget_remaining:
+                            log_event({'type': 'cost_skip', 'key': key,
+                                       'est_seconds': est,
+                                       'budget_remaining': budget_remaining})
+                            continue
+
+                        try:
+                            result = test_pair(gamma, m_1, ch1, 'SP', k_2, ch2, 'SP')
+                        except MemoryError as e:
+                            log_event({'type': 'memory_error', 'key': key,
+                                       'error': str(e)})
+                            continue
+                        except Exception as e:
+                            log_event({'type': 'exception', 'key': key,
+                                       'error': type(e).__name__ + ": " + str(e)})
+                            continue
+
+                        n_tested += 1
+                        already_tested.add(key)
+                        log_event(result)
+                        if not result['compatible']:
+                            n_counterexamples += 1
+                            log_event({'type': 'COUNTEREXAMPLE', 'pair': result})
+                            print(f"\n*** COUNTEREXAMPLE FOUND at γ={gamma}, "
+                                  f"T1={(m_1, ch1)}, T2={(k_2, ch2)} ***\n")
+
+                        # Print periodic progress
+                        if n_tested % 25 == 0:
+                            elapsed = time.time() - (deadline - budget_hours * 3600)
+                            print(f"[{elapsed/60:.1f}m] tested {n_tested}, "
+                                  f"skipped {n_skipped}, "
+                                  f"counterexamples {n_counterexamples}, "
+                                  f"current γ={gamma}")
+
+    log_event({'type': 'session_end',
+               'n_tested': n_tested,
+               'n_skipped': n_skipped,
+               'n_counterexamples': n_counterexamples,
+               'elapsed_hours': (time.time() - (deadline - budget_hours * 3600)) / 3600})
+    print(f"\n[done] tested {n_tested}, skipped {n_skipped}, "
+          f"counterexamples {n_counterexamples}")
+
+
+if __name__ == '__main__':
+    budget = float(sys.argv[1]) if len(sys.argv) > 1 else 6.0
+    main(budget_hours=budget)

papers/coloring_nested_tire_graphs/paper.aux

@@ -25,13 +25,17 @@
 \newlabel{def:tire-annular-face-connector}{{1.16}{9}}
 \newlabel{def:spokes}{{1.17}{9}}
 \newlabel{rem:facial-dual-spoke-only}{{1.18}{9}}
+\@writefile{lof}{\contentsline {figure}{\numberline {5}{\ignorespaces  The bridge case: $T'_{\mathrm  {ann}} = \theta (1, 3, 3)$ has three faces $A, B, C$ in its inherited embedding, with respective vertex sets $V(A) = \{v_0, \dots  , v_5\}$, $V(B) = \{v_0, v_1, v_2, v_3\}$, and $V(C) = \{v_0, v_3, v_4, v_5\}$. In the surrounding maximal planar $G$, the chord endpoints $v_0, v_3$ (the two annular faces sharing the bridge edge) have all three $G'$-edges inside $T'_{\mathrm  {ann}}$, while each non-chord vertex $v_i$ ($i \in \{1, 2, 4, 5\}$) contributes one $G'$-edge to an external non-annular neighbor $u_i$. Each panel highlights $T'_{f'}$ (blue) inside $G'$: dark circles are $V(f')$, gray circles are $G'$-neighbors of $V(f')$ within $T'_{\mathrm  {ann}}$, and red squares are external $G'$-neighbors $u_i$. The choice of face $f'$ controls which external neighbors $u_i$ are pulled into $T'_{f'}$ (face $A$ pulls in all four; face $B$ pulls in $u_1, u_2$ and face $C$ pulls in $u_4, u_5$).}}{10}{}\protected@file@percent }
+\newlabel{fig:facial-dual-choices}{{5}{10}}
+\@writefile{toc}{\contentsline {section}{\tocsection {}{2}{A conjectural Latin-style substructure}}{10}{}\protected@file@percent }
+\newlabel{sec:latin-conjecture}{{2}{10}}
 \bibcite{bauerfeld-pds}{1}
 \newlabel{tocindent-1}{0pt}
 \newlabel{tocindent0}{12.7778pt}
 \newlabel{tocindent1}{17.77782pt}
 \newlabel{tocindent2}{0pt}
 \newlabel{tocindent3}{0pt}
-\@writefile{lof}{\contentsline {figure}{\numberline {5}{\ignorespaces  The bridge case: $T'_{\mathrm  {ann}} = \theta (1, 3, 3)$ has three faces $A, B, C$ in its inherited embedding, with respective vertex sets $V(A) = \{v_0, \dots  , v_5\}$, $V(B) = \{v_0, v_1, v_2, v_3\}$, and $V(C) = \{v_0, v_3, v_4, v_5\}$. In the surrounding maximal planar $G$, the chord endpoints $v_0, v_3$ (the two annular faces sharing the bridge edge) have all three $G'$-edges inside $T'_{\mathrm  {ann}}$, while each non-chord vertex $v_i$ ($i \in \{1, 2, 4, 5\}$) contributes one $G'$-edge to an external non-annular neighbor $u_i$. Each panel highlights $T'_{f'}$ (blue) inside $G'$: dark circles are $V(f')$, gray circles are $G'$-neighbors of $V(f')$ within $T'_{\mathrm  {ann}}$, and red squares are external $G'$-neighbors $u_i$. The choice of face $f'$ controls which external neighbors $u_i$ are pulled into $T'_{f'}$ (face $A$ pulls in all four; face $B$ pulls in $u_1, u_2$ and face $C$ pulls in $u_4, u_5$).}}{10}{}\protected@file@percent }
+\newlabel{conj:latin}{{2.1}{11}}
-\newlabel{fig:facial-dual-choices}{{5}{10}}
+\newlabel{conj:chain-latin}{{2.2}{11}}
-\@writefile{toc}{\contentsline {section}{\tocsection {}{}{References}}{10}{}\protected@file@percent }
+\@writefile{toc}{\contentsline {section}{\tocsection {}{}{References}}{11}{}\protected@file@percent }
-\gdef \@abspage@last{10}
+\gdef \@abspage@last{11}

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@@ -1,4 +1,4 @@
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@@ -205,19 +205,19 @@ LaTeX Warning: `h' float specifier changed to `ht'.
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papers/coloring_nested_tire_graphs/paper.tex

@@ -12,6 +12,7 @@
 \newtheorem{lemma}[theorem]{Lemma}
 \newtheorem{corollary}[theorem]{Corollary}
 \newtheorem{proposition}[theorem]{Proposition}
+\newtheorem{conjecture}[theorem]{Conjecture}
 
 \theoremstyle{definition}
 \newtheorem{definition}[theorem]{Definition}
@@ -669,6 +670,88 @@ $T'_{f'}$ (face $A$ pulls in all four; face $B$ pulls in $u_1, u_2$
 and face $C$ pulls in $u_4, u_5$).}
 \end{figure}
 
+\section{A conjectural Latin-style substructure}
+\label{sec:latin-conjecture}
+
+Empirical enumeration (notes \texttt{tire\_fiber\_data.tex},
+\texttt{tire\_fiber\_chords.tex}, \texttt{tire\_fiber\_step2.tex},
+\texttt{tire\_fiber\_step2\_large.tex}) of edge $3$-coloring
+distributions on the tire annular face connector $T'_{f'}$ across
+$46$ adjacent-tire pairs at $|\gamma| \in \{3, 4, 5, 6, 9, 12\}$
+suggests that the chain-pigeonhole step on a shared cycle always
+succeeds.  The data points to a structural mechanism: every
+edge-$3$-colourable tire admits at least one ``Latin-flavoured''
+boundary configuration, and adjacent tires share this same
+substructure on their common cycle.
+
+Concretely, fix a tire $T$ with inner outerplanar graph $O$ on $V(B_{\mathrm{in}})$
+and let $F(O)$ be the set of $O$-faces (in the tire's plane embedding,
+not counting the outer face $B_{\mathrm{in}}$).  For each $O$-face
+$f \in F(O)$, let $E_{\mathrm{in}}(f) \subseteq E(B_{\mathrm{in}})$
+denote the set of $B_{\mathrm{in}}$ edges on $f$'s boundary.  In the
+Steiner-poor surrounding triangulation (where each $O$-face is a
+single face of $G$ and dualises to a single $G'$-vertex of degree
+$|E_{\mathrm{in}}(f)|$ in $T'_{f'}$), proper edge $3$-colouring of
+$T'_{f'}$ requires every $O$-face to have $|E_{\mathrm{in}}(f)| \leq 3$.
+
+Let $\sigma_{B_{\mathrm{in}}}$ denote the spoke colouring restricted to
+the $|V(B_{\mathrm{in}})|$ inner-direction spoke positions on the dual
+annular cycle (equivalently: $\sigma$ indexed by $E(B_{\mathrm{in}})$).
+Define the \emph{Latin-flavoured set} on $\gamma = B_{\mathrm{in}}$ as
+\[
+   \mathcal{L}(B_{\mathrm{in}}, O)
+   \;:=\; \bigl\{\,\sigma : E(B_{\mathrm{in}}) \to \{1,2,3\}
+     \;\big|\; \sigma\bigl|_{E_{\mathrm{in}}(f)}\text{ is a permutation
+     of }\{1,2,3\}\text{ for every } f \in F(O)\,\bigr\}.
+\]
+That is, on every $O$-face's $B_{\mathrm{in}}$-edge boundary, all three
+colours appear exactly once (forcing $|E_{\mathrm{in}}(f)| = 3$ for
+each face --- the maximally constrained case).
+
+\begin{conjecture}[Latin-substructure conjecture]
+\label{conj:latin}
+For any Steiner-poor edge-$3$-colourable tire $T$ with inner
+outerplanar graph $O$ such that every $O$-face has exactly $3$
+$B_{\mathrm{in}}$-edges, the realisable inner-spoke projection
+$\pi_D(\mathcal{C}(T'_{f'}))$ contains $\mathcal{L}(B_{\mathrm{in}}, O)$
+as a subset.  Moreover, $\mathcal{L}(B_{\mathrm{in}}, O)$ is invariant
+under the $S_3$ action on colours and has size at least $3! = 6$.
+\end{conjecture}
+
+\begin{conjecture}[Chain-pigeonhole compatibility from Latin substructure]
+\label{conj:chain-latin}
+Adjacent tires $T_1, T_2$ sharing a cycle $\gamma$ admit a joint
+edge $3$-colouring whenever their respective inner-outerplanar
+structures $O^{(1)}, O^{(2)}$ both satisfy
+Conjecture~\ref{conj:latin}.  Equivalently:
+$\pi_D^{(1)}(\mathcal{C}(T_1)) \cap \pi_U^{(2)}(\mathcal{C}(T_2))
+\supseteq \mathcal{L}(\gamma, O^{(1)}) \cap \mathcal{L}(\gamma,
+O^{(2)})$, and this last intersection is non-empty whenever the two
+face partitions of $E(\gamma)$ induced by $O^{(1)}, O^{(2)}$ share a
+common ``Latin completion.''
+\end{conjecture}
+
+The structural origin of these conjectures is the empirical
+observation that the smallest tested intersections on $\gamma$ are
+always exactly the $3!$ permutations of a single canonical pattern
+in which each $O$-face on $\gamma$'s side receives a permutation of
+$\{1,2,3\}$.  For example, at $|\gamma| = 12$ with $O^{(1)}$ given by
+the chord matching $\{(0,3),(4,7),(8,11)\}$ (face structure $\{0,1,2\}
+\sqcup \{4,5,6\} \sqcup \{8,9,10\} \sqcup \{3,7,11\}$), the canonical
+pattern $(1,2,3,2,2,1,3,3,2,3,1,1)$ assigns the permutation $(1,2,3)$
+to the first face, $(2,1,3)$ to the second, $(2,3,1)$ to the third,
+and $(2,3,1)$ to the fourth.  Every face receives all three colours.
+
+A proof of Conjecture~\ref{conj:latin} would convert the
+chain-pigeonhole compatibility step into a structural theorem on
+$T'_{f'}$: it is not the rough abundance of valid spoke configurations
+that lets adjacent tires meet, but a specific Latin-square-flavoured
+substructure dictated by the face partition of each tire's inner
+outerplanar graph.  See \texttt{notes/tire\_fiber\_step2\_large.tex}
+for the data underlying this conjecture and
+\texttt{experiments/tire\_fiber\_counterexample\_search.log} for the
+ongoing automated search.
+
 \begin{thebibliography}{9}
 
 \bibitem{bauerfeld-pds}