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coloring_nested_tire_graphs: add Latin conjecture to paper; launch counterexample search

Browse Source 
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>
This commit is contained in:
didericis
2026-05-26 03:26:17 -04:00
parent 030ca67afb
commit 374fca98a5
6 changed files with 2976 additions and 37 deletions
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papers/coloring_nested_tire_graphs/experiments/counterexample_search.log
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papers/coloring_nested_tire_graphs/experiments/tire_fiber_counterexample_search.py
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"""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
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@@ -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{fig:facial-dual-choices}{{5}{10}}
\@writefile{toc}{\contentsline {section}{\tocsection {}{}{References}}{10}{}\protected@file@percent }
\gdef \@abspage@last{10}
\newlabel{conj:latin}{{2.1}{11}}
\newlabel{conj:chain-latin}{{2.2}{11}}
\@writefile{toc}{\contentsline {section}{\tocsection {}{}{References}}{11}{}\protected@file@percent }
\gdef \@abspage@last{11}
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\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}