d7e9b6af2f
NEW PROOF STRATEGY for Conjecture 5.1 (face-monochromatic-pair):
1. NEW Conjecture (Deciding face): For every chord-apex+Kempe
colouring φ of every reduced dual, the reduced dual has a face f
with ∂f ⊆ V(K_b) ∪ V(K_c) and |f| ≢ 0 (mod 3).
2. NEW Theorem: Deciding-face conjecture implies Conj 5.1.
Proof: contradiction. Assume no clauses-(1)-(3) witness for some
chord-apex+Kempe φ. By Lemma 5.3, h_φ ≡ ε ∈ {±1} on V(K_b) ∪ V(K_c).
By the deciding-face conjecture, ∃ face f with ∂f ⊆ V(K_b) ∪ V(K_c),
|f| ≢ 0 (mod 3). Heawood's face-sum identity (Heawood 1898) gives
Σ_{v ∈ ∂f} h_φ(v) = ε|f| ≡ 0 (mod 3). Since gcd(|f|, 3) = 1, we get
ε ≡ 0 (mod 3), but ε ∈ {±1} — contradiction.
3. EMPIRICAL: Conjecture (Deciding face) verified on 142,812 / 142,812
chord-apex+Kempe colourings of reduced duals up to |V(G)| ≤ 20 --
matching the full coverage of check_constancy_obstruction.py.
Face-length distribution:
|f| = 4: 13,074
|f| = 5: 102,498 (most common)
|f| = 7: 18,570
|f| = 8: 7,752
|f| = 10: 846
|f| = 11: 72
(All ≢ 0 mod 3.)
New scripts:
- check_kb_kc_coverage.py: |V(K_b) ∪ V(K_c)| / |V(Ĝ')| distribution.
73.87% of colourings have V(K_b) ∪ V(K_c) = V (full coverage); the
remaining 26% have coverage ≥ 70%, mostly ≥ 90%.
- check_deciding_face.py: existence of deciding face across all
colourings; 100.00% / 142,812.
Why this is the right reduction:
- It uses ALL THREE pieces of chord-apex+Kempe structure: Lemma 5.3
(constancy from no-witness), forced colour-equality at merged/spike,
and forced Kempe-cycle containment of merged + spike + side edges
(the latter two enter via V(K_b) ∪ V(K_c) covering specific
structural vertices).
- It uses Heawood's face-sum identity, which is the classical 3-fold
parity constraint on cubic plane 3-edge-colourings.
- The C28 counterexample to Conjecture 5.5 is not affected: it's not
a chord-apex+Kempe colouring of a reduced dual, so the deciding-face
structure doesn't apply.
Remaining work: prove the deciding-face conjecture structurally (likely
via the specific F_01 / F_12 "flank face" of the reduced dual, whose
length n_0 - 1 from the adjacent G'-face of length n_0 ≥ 5 is ≢ 0 mod 3
exactly when n_0 ≢ 1 mod 3, plus boundary-in-V(K_b) ∪ V(K_c) which
follows from Lemma 5.X kempe-spike + colour analysis at A_i).
Paper grows from 17 to 18 pages.
Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>
133 lines
4.9 KiB
Python
133 lines
4.9 KiB
Python
"""Check what fraction of V(Ĝ'_{v,i}) is covered by V(K_b) ∪ V(K_c)
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across all chord-apex+Kempe colourings of all reduced duals up to
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|V(G)| ≤ 18.
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This is the key prerequisite for the Heawood-face-sum proof attempt:
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if V(K_b) ∪ V(K_c) = V(reduced dual) in many cases, then under
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constancy on V(K_b) ∪ V(K_c) every face of the reduced dual must
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satisfy ε * |f| ≡ 0 (mod 3); since ε = ±1, this forces |f| ≡ 0 mod 3
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for every face -- and any reduced dual with a face of length 4, 5, or
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7 (etc., not multiple of 3) would yield an immediate contradiction.
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Run with: sage experiments/check_kb_kc_coverage.py
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"""
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import os
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import sys
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import time
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from sage.all import Graph
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from sage.graphs.graph_generators import graphs
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HERE = os.path.dirname(os.path.abspath(__file__))
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sys.path.insert(0, HERE)
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from check_conj_3_8_scaled import (
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apply_reduction,
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proper_3_edge_colorings,
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matches_chord_apex_kempe,
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kempe_cycle_set,
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edge_idx,
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)
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from check_heawood_on_kempe import (
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dual_of, heawood_numbers, vertices_of_kempe,
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)
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def test_one(D):
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D.is_planar(set_embedding=True)
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n_col = 0
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coverage_dist = {} # |V(K_b)∪V(K_c)| / |V| as ratio (rounded) -> count
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full_coverage = 0
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face_lengths_at_full = {} # face length distribution when full coverage
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for face in D.faces():
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if len(face) != 5: continue
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for i_red in range(5):
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res = apply_reduction(D, face, i_red, 9999)
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if res is None: continue
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H = res['H']; named = res['named']
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H.is_planar(set_embedding=True)
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edges, colorings = proper_3_edge_colorings(H)
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cand = [c for c in colorings
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if matches_chord_apex_kempe(edges, c, named)]
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n_v = H.order()
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face_lens = sorted([len(f) for f in H.faces()])
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for col in cand:
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n_col += 1
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merged_idx = edge_idx(edges, named['merged'])
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a = col[merged_idx]
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bs = [c for c in range(3) if c != a]
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kc_b = kempe_cycle_set(edges, col, merged_idx, (a, bs[0]))
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kc_c = kempe_cycle_set(edges, col, merged_idx, (a, bs[1]))
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V_b = vertices_of_kempe(edges, kc_b)
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V_c = vertices_of_kempe(edges, kc_c)
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cov = len(V_b | V_c)
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cov_ratio = cov / n_v
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bucket = round(cov_ratio * 20) / 20 # bucket in 5% increments
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coverage_dist[bucket] = coverage_dist.get(bucket, 0) + 1
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if cov == n_v:
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full_coverage += 1
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key = tuple(face_lens)
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face_lengths_at_full[key] = (
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face_lengths_at_full.get(key, 0) + 1)
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return n_col, coverage_dist, full_coverage, face_lengths_at_full
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def main(max_n=18, time_budget_per_n=300):
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print(f"V(K_b) ∪ V(K_c) coverage / V(Ĝ'_{{v,i}}) "
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f"on chord-apex+Kempe colourings, n_G ∈ [12, {max_n}]\n")
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grand_col = 0
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grand_dist = {}
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grand_full = 0
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grand_full_faces = {}
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for n in range(12, max_n + 1):
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start = time.time()
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try:
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triangulations = list(graphs.triangulations(n, minimum_degree=5))
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except Exception as ex:
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print(f"n={n}: cannot enumerate ({ex})")
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continue
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n_col_n = 0
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n_full_n = 0
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for tri_idx, G in enumerate(triangulations):
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if time.time() - start > time_budget_per_n:
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print(f" n={n}: timeout at tri {tri_idx}/{len(triangulations)}")
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break
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G.is_planar(set_embedding=True)
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D = dual_of(G)
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ni, dist, full, full_faces = test_one(D)
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n_col_n += ni
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n_full_n += full
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for k, v in dist.items():
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grand_dist[k] = grand_dist.get(k, 0) + v
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for k, v in full_faces.items():
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grand_full_faces[k] = grand_full_faces.get(k, 0) + v
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elapsed = time.time() - start
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print(f"n={n}: {n_col_n} col., {n_full_n} full-coverage "
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f"({100 * n_full_n / max(n_col_n, 1):.1f}%) [{elapsed:.0f}s]")
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sys.stdout.flush()
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grand_col += n_col_n
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grand_full += n_full_n
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print()
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print("=" * 70)
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print(f"Grand totals: {grand_col} chord-apex+Kempe colourings")
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print(f" full coverage (V(K_b)∪V(K_c) = V): {grand_full} "
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f"({100 * grand_full / max(grand_col, 1):.2f}%)")
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print()
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print(" Coverage ratio distribution "
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"(|V(K_b)∪V(K_c)| / |V(Ĝ')|):")
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for r in sorted(grand_dist):
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c = grand_dist[r]
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pct = 100 * c / max(grand_col, 1)
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bar = '#' * max(1, int(pct))
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print(f" {r:5.2f}: {c:>7} ({pct:5.2f}%) {bar[:50]}")
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if grand_full > 0:
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print()
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print(" Face-length distributions among full-coverage cases:")
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for key in sorted(grand_full_faces, key=lambda k: -grand_full_faces[k]):
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print(f" {key}: {grand_full_faces[key]}")
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if __name__ == '__main__':
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main()
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