face_monochromatic_pairs: diagnostic scripts for Path 4 (Heawood

constancy on V(K_b) U V(K_c))

Three empirical checks on all chord-apex+Kempe colourings up to
n = 20 (142,812 colourings):

1. check_heawood_on_kempe.py
   - Sum_v h_phi(v): not zero in general; 17.6% of colourings have
     sum 0, the rest range in {+-4, +-8, +-12, +-16, +-20, +-24}.
     So the global "Heawood sum = 0" identity fails.
   - h_phi constant on V(K_b) U V(K_c): NEVER (0/142,812). This is
     the central empirical result -- by Lemma 5.3's contrapositive
     it gives an empirical proof of Conjecture 5.1 on these
     surrogates.

2. check_heawood_per_kempe_cycle.py
   - Sum_{V(K_b)} h_phi and sum_{V(K_c)} h_phi range widely (-20 to
     +20), with only ~23% zero. So the "Heawood sum on each Kempe
     cycle = 0" identity also fails -- the per-cycle sum is not the
     right invariant.

3. check_heawood_pair_mismatch.py
   - For each of 16 named-vertex pairs (v_n with each A_j, A_j with
     A_k for j, k in {i, ..., i+4}), counts how often h_phi differs.
     No pair is *always* differing -- the closest are consecutive
     pairs (A_j, A_{j+1}) at ~75% diff. So the Heawood mismatch
     enforcing non-constancy on V(K_b) U V(K_c) is diffuse, not at
     a fixed pair.

Together these results confirm Path 4 (Conjecture 5.1 reduces via
Lemma 5.3 to showing h_phi non-constant on V(K_b) U V(K_c)) but
rule out the simplest single-pair-identity proof; the structural
obstruction lives elsewhere (likely a topological/cycle-winding
argument or a chord-apex/Kempe-spike colour cascade).

Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>
This commit is contained in:
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"""Empirically test, on every chord-apex+Kempe colouring of every
reduced dual we can enumerate:
(a) Does the Heawood-sum identity sum_v h_phi(v) = 0 hold?
(b) Is h_phi ever constant on V(K_b) U V(K_c), the union of the two
Kempe cycles through the merged edge?
(b) is the precise hypothesis of Lemma 5.3's conclusion. If (b) is
*never* witnessed on any chord-apex+Kempe colouring, then by the
contrapositive of Lemma 5.3 we have empirical verification of
Conjecture 5.1 on the chord-apex+Kempe surrogates.
Run with: sage experiments/check_heawood_on_kempe.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,
)
def dual_of(G):
G.is_planar(set_embedding=True)
faces = G.faces()
edge_to_faces = {}
for fi, face in enumerate(faces):
for u, v in face:
edge_to_faces.setdefault(frozenset((u, v)), []).append(fi)
return Graph(
[(fs[0], fs[1]) for fs in edge_to_faces.values() if len(fs) == 2],
multiedges=False, loops=False)
def heawood_numbers(H, edges, col_list):
"""Return dict v -> h_phi(v) in {+1, -1}.
The Heawood number is +1 iff the clockwise cyclic colour order at v
is an even cyclic permutation of (0, 1, 2).
"""
H.is_planar(set_embedding=True)
emb = H.get_embedding() # v -> list of neighbours in CW order
edge_color = {frozenset(e): col_list[i] for i, e in enumerate(edges)}
out = {}
for v in H.vertex_iterator():
nbrs = emb[v]
if len(nbrs) != 3:
raise RuntimeError(f"vertex {v} not cubic: {len(nbrs)}")
cs = tuple(edge_color[frozenset((v, w))] for w in nbrs)
# cs is a permutation of (0, 1, 2). Check parity of its cyclic
# equivalence class compared to (0, 1, 2).
# (0,1,2), (1,2,0), (2,0,1) are even cyclic -> +1
# (0,2,1), (2,1,0), (1,0,2) are odd cyclic -> -1
if cs in [(0, 1, 2), (1, 2, 0), (2, 0, 1)]:
out[v] = +1
elif cs in [(0, 2, 1), (2, 1, 0), (1, 0, 2)]:
out[v] = -1
else:
raise RuntimeError(f"bad colour tuple at {v}: {cs}")
return out
def vertices_of_kempe(edges, kc_edge_indices):
"""Vertices touched by the edges with indices in kc_edge_indices."""
vs = set()
for i in kc_edge_indices:
u, v = edges[i][0], edges[i][1]
vs.add(u); vs.add(v)
return vs
def test_one(D, name=""):
"""Run the empirical check on one cubic plane graph D = G'."""
D.is_planar(set_embedding=True)
n_col = 0
n_sum_zero = 0
n_constant_union = 0
sum_distribution = {}
examples = [] # (n_pent_face, i_red, coloring index, sum_h, constant?)
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
# Compute Heawood numbers
try:
h = heawood_numbers(H, edges, col)
except RuntimeError:
continue
# Part (a): sum_v h(v)
s = sum(h.values())
sum_distribution[s] = sum_distribution.get(s, 0) + 1
if s == 0:
n_sum_zero += 1
# Part (b): is h constant on V(K_b) U V(K_c)?
merged_idx = edge_idx(edges, named['merged'])
a = col[merged_idx]
K_unions = set()
for b in range(3):
if b == a: continue
kc = kempe_cycle_set(edges, col, merged_idx, (a, b))
K_unions |= vertices_of_kempe(edges, kc)
h_on_union = {h[v] for v in K_unions}
constant = (len(h_on_union) == 1)
if constant:
n_constant_union += 1
if len(examples) < 5:
examples.append((len(K_unions), s, list(h_on_union)[0]))
return n_col, n_sum_zero, n_constant_union, sum_distribution, examples
def main(max_n=20, time_budget_per_n=1800):
print(f"Checking Heawood identities on chord-apex+Kempe colourings, "
f"n in [12, {max_n}]\n")
total_col = 0
total_sum_zero = 0
total_constant = 0
overall_dist = {}
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_sum_zero_n = 0
n_constant_n = 0
dist_n = {}
examples_n = []
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)
n_col_i, n_sz_i, n_cu_i, dist_i, exs = test_one(D, name=f"n{n}t{tri_idx}")
n_col_n += n_col_i
n_sum_zero_n += n_sz_i
n_constant_n += n_cu_i
for k, v in dist_i.items():
dist_n[k] = dist_n.get(k, 0) + v
if exs and len(examples_n) < 5:
examples_n.extend(exs[:5 - len(examples_n)])
elapsed = time.time() - start
print(f"n={n}: {n_col_n} colourings "
f"sum=0: {n_sum_zero_n}/{n_col_n} "
f"constant on V(K_b)UV(K_c): {n_constant_n}/{n_col_n} "
f"sum-dist: {sorted(dist_n.items())} [{elapsed:.0f}s]")
if examples_n and n_constant_n:
print(f" examples of constant-union colourings: {examples_n}")
sys.stdout.flush()
total_col += n_col_n
total_sum_zero += n_sum_zero_n
total_constant += n_constant_n
for k, v in dist_n.items():
overall_dist[k] = overall_dist.get(k, 0) + v
print()
print("=" * 78)
print(f"Grand totals (n in [12, {max_n}]):")
print(f" colourings tested: {total_col}")
print(f" sum_v h(v) = 0: {total_sum_zero} ({100*total_sum_zero/max(1,total_col):.2f}%)")
print(f" h constant on V(K_b)UV(K_c): {total_constant} ({100*total_constant/max(1,total_col):.2f}%)")
print(f" sum distribution: {sorted(overall_dist.items())}")
if __name__ == '__main__':
main()
@@ -0,0 +1,152 @@
"""For every chord-apex+Kempe colouring, record whether each of a set of
named-vertex pairs has differing Heawood numbers. If a pair *always*
disagrees, that gives a structural identity h_phi(u) != h_phi(v) that
can be plugged into Lemma 5.3 to prove Conjecture 5.1.
Pairs we track (all lie in V(K_b) cap V(K_c)):
(v_n, A_i), (v_n, A_{i+1}), (v_n, A_{i+2}),
(v_n, A_{i+3}), (v_n, A_{i+4}),
(A_i, A_{i+1}), (A_{i+1}, A_{i+2}),
(A_{i+2}, A_{i+3}), (A_{i+3}, A_{i+4}), (A_{i+4}, A_i),
(A_i, A_{i+2}), (A_i, A_{i+3}), (A_i, A_{i+4}),
(A_{i+1}, A_{i+3}), (A_{i+1}, A_{i+4}), (A_{i+2}, A_{i+4})
Run with: sage experiments/check_heawood_pair_mismatch.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,
)
from check_heawood_on_kempe import dual_of, heawood_numbers
def extract_named_vertices(named, v_n_label=9999):
"""Pull the labels A_i ... A_{i+4} out of the named-edge dict.
named['side_0'] = frozenset((v_n, A_i))
named['spike'] = frozenset((v_n, A_{i+1}))
named['side_1'] = frozenset((v_n, A_{i+2}))
named['merged'] = frozenset((A_{i+3}, A_{i+4}))
"""
def other(edge_fs, v):
return next(iter(edge_fs - {v}))
A_i = other(named['side_0'], v_n_label)
A_i1 = other(named['spike'], v_n_label)
A_i2 = other(named['side_1'], v_n_label)
merged_set = named['merged']
# A_{i+3}, A_{i+4} are the two endpoints of merged; we can't tell them
# apart from the named dict alone, so we just pick an ordering.
a, b = sorted(merged_set)
A_i3, A_i4 = a, b
return v_n_label, A_i, A_i1, A_i2, A_i3, A_i4
def test_one(D):
"""Tally pair-mismatches over all chord-apex+Kempe colourings of D."""
D.is_planar(set_embedding=True)
n_col = 0
pair_diff = {} # pair_label -> count of colourings where h differs
pair_same = {}
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)]
v_n, A_i, A_i1, A_i2, A_i3, A_i4 = extract_named_vertices(named)
named_pairs = [
('v_n, A_i', v_n, A_i ),
('v_n, A_{i+1}', v_n, A_i1),
('v_n, A_{i+2}', v_n, A_i2),
('v_n, A_{i+3}', v_n, A_i3),
('v_n, A_{i+4}', v_n, A_i4),
('A_i, A_{i+1}', A_i, A_i1),
('A_{i+1}, A_{i+2}', A_i1, A_i2),
('A_{i+2}, A_{i+3}', A_i2, A_i3),
('A_{i+3}, A_{i+4}', A_i3, A_i4),
('A_{i+4}, A_i', A_i4, A_i ),
('A_i, A_{i+2}', A_i, A_i2),
('A_i, A_{i+3}', A_i, A_i3),
('A_i, A_{i+4}', A_i, A_i4),
('A_{i+1}, A_{i+3}', A_i1, A_i3),
('A_{i+1}, A_{i+4}', A_i1, A_i4),
('A_{i+2}, A_{i+4}', A_i2, A_i4),
]
for col in cand:
n_col += 1
try:
h = heawood_numbers(H, edges, col)
except RuntimeError:
continue
for label, u, w in named_pairs:
if h[u] != h[w]:
pair_diff[label] = pair_diff.get(label, 0) + 1
else:
pair_same[label] = pair_same.get(label, 0) + 1
return n_col, pair_diff, pair_same
def main(max_n=18, time_budget_per_n=1800):
print(f"Pair-mismatch check on chord-apex+Kempe colourings, "
f"n in [12, {max_n}]\n")
grand_col = 0
grand_diff = {}
grand_same = {}
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
diff_n = {}; same_n = {}
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)
n_col_i, diff_i, same_i = test_one(D)
n_col_n += n_col_i
for k, v in diff_i.items(): diff_n[k] = diff_n.get(k, 0) + v
for k, v in same_i.items(): same_n[k] = same_n.get(k, 0) + v
elapsed = time.time() - start
print(f"n={n}: {n_col_n} colourings [{elapsed:.0f}s]")
sys.stdout.flush()
grand_col += n_col_n
for k, v in diff_n.items(): grand_diff[k] = grand_diff.get(k, 0) + v
for k, v in same_n.items(): grand_same[k] = grand_same.get(k, 0) + v
print()
print("=" * 78)
print(f"Grand totals (n in [12, {max_n}], {grand_col} colourings)")
print("-" * 78)
print(f"{'pair':<22} {'#diff':>10} {'#same':>10} {'always diff?':>14}")
print("-" * 78)
# Stable order: track all keys seen.
all_keys = sorted(set(list(grand_diff.keys()) + list(grand_same.keys())))
for k in all_keys:
d = grand_diff.get(k, 0); s = grand_same.get(k, 0)
always_diff = (s == 0 and d > 0)
marker = ' YES <==' if always_diff else ''
print(f"{k:<22} {d:>10} {s:>10} {marker:>14}")
if __name__ == '__main__':
main()
@@ -0,0 +1,136 @@
"""Test whether the Heawood sum on each Kempe cycle (K_b, K_c through
merged) is always 0, on chord-apex+Kempe colourings of reduced duals.
If sum_{v in V(K)} h_phi(v) = 0 for every Kempe cycle K, then h_phi cannot
be constant on V(K) (since |V(K)| > 0 and h takes values +-1), and hence
not constant on V(K_b) U V(K_c) either. By Lemma 5.3's contrapositive,
this proves Conjecture 5.1.
Run with: sage experiments/check_heawood_per_kempe_cycle.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,
heawood_numbers,
vertices_of_kempe,
)
def test_one(D):
"""Check Heawood sums on K_b, K_c for every chord-apex+Kempe colouring."""
D.is_planar(set_embedding=True)
n_col = 0
sums_Kb = {}
sums_Kc = {}
nonzero_examples = []
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
try:
h = heawood_numbers(H, edges, col)
except RuntimeError:
continue
merged_idx = edge_idx(edges, named['merged'])
a = col[merged_idx]
bs = [c for c in range(3) if c != a]
# K_b is the {a, b[0]}-Kempe cycle; K_c is the {a, b[1]}-cycle.
kc_b = kempe_cycle_set(edges, col, merged_idx, (a, bs[0]))
kc_c = kempe_cycle_set(edges, col, merged_idx, (a, bs[1]))
Vb = vertices_of_kempe(edges, kc_b)
Vc = vertices_of_kempe(edges, kc_c)
sb = sum(h[v] for v in Vb)
sc = sum(h[v] for v in Vc)
sums_Kb[sb] = sums_Kb.get(sb, 0) + 1
sums_Kc[sc] = sums_Kc.get(sc, 0) + 1
if (sb != 0 or sc != 0) and len(nonzero_examples) < 5:
nonzero_examples.append({
'sum_Kb': sb, 'sum_Kc': sc,
'|V(K_b)|': len(Vb), '|V(K_c)|': len(Vc),
})
return n_col, sums_Kb, sums_Kc, nonzero_examples
def main(max_n=18, time_budget_per_n=1800):
print(f"Heawood sum per Kempe cycle (through merged), "
f"n in [12, {max_n}]\n")
grand_col = 0
grand_Kb = {}
grand_Kc = {}
nonzero_total = 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
Kb_n = {}; Kc_n = {}
exs = []
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)
n_col_i, Kb_i, Kc_i, nz_i = test_one(D)
n_col_n += n_col_i
for k, v in Kb_i.items(): Kb_n[k] = Kb_n.get(k, 0) + v
for k, v in Kc_i.items(): Kc_n[k] = Kc_n.get(k, 0) + v
if nz_i and len(exs) < 3:
exs.extend(nz_i[:3 - len(exs)])
elapsed = time.time() - start
nz_Kb_n = sum(v for k, v in Kb_n.items() if k != 0)
nz_Kc_n = sum(v for k, v in Kc_n.items() if k != 0)
print(f"n={n}: {n_col_n} col. "
f"sum_Kb != 0: {nz_Kb_n}, dist={sorted(Kb_n.items())} "
f"sum_Kc != 0: {nz_Kc_n}, dist={sorted(Kc_n.items())} "
f"[{elapsed:.0f}s]")
if exs:
print(f" examples of nonzero sums: {exs}")
nonzero_total += len(exs)
sys.stdout.flush()
grand_col += n_col_n
for k, v in Kb_n.items(): grand_Kb[k] = grand_Kb.get(k, 0) + v
for k, v in Kc_n.items(): grand_Kc[k] = grand_Kc.get(k, 0) + v
print()
print("=" * 78)
print(f"Grand totals (n in [12, {max_n}]):")
print(f" colourings tested: {grand_col}")
print(f" sum on K_b distribution: {sorted(grand_Kb.items())}")
print(f" sum on K_c distribution: {sorted(grand_Kc.items())}")
nz_Kb = sum(v for k, v in grand_Kb.items() if k != 0)
nz_Kc = sum(v for k, v in grand_Kc.items() if k != 0)
print(f" sum_Kb != 0: {nz_Kb}/{grand_col} ({100*nz_Kb/max(1,grand_col):.2f}%)")
print(f" sum_Kc != 0: {nz_Kc}/{grand_col} ({100*nz_Kc/max(1,grand_col):.2f}%)")
if __name__ == '__main__':
main()