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face_monochromatic_pairs: strengthen Conj 5.5 to face-length ≥ 5, find 28-vertex counterexample

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- Paper: Conjecture 5.5 restated with the hypothesis that every face
  of H has length ≥ 5 (= the cubic plane analogue of "no triangles
  or quadrilaterals as faces"). This kills K_4 and the n=8 trivial
  counterexamples (girth 3) and the ad-hoc n=40 counterexample
  (which has 2 triangles and 4 quadrilaterals). A new remark catalogues
  these excluded counterexamples and the smallest cubic plane graphs
  satisfying the hypothesis (dodecahedron at |V| = 20).
- search_smaller_counterexample.py: --min-face=N option to filter
  cubic planar graphs by minimum face length.
- search_min_face5_counterexample.py: enumerates triangulations T
  with min degree ≥ 5 via graphs.triangulations(n, minimum_degree=5),
  takes planar dual (= cubic plane with all faces ≥ 5), and runs the
  Heawood-constancy check.
- Result: smallest counterexample at triangulation order n_T = 16,
  whose dual is a 28-vertex cubic plane graph (graph6
  [kG[A?_A?_?_?K?D?@_CO?o?@_??A??@C??O??AG?C????`???a???W???A_???F).
  Faces: 12 pentagons + 4 hexagons (a C28 fullerene). Both
  K_{red, blue} and K_{red, green} are 12-cycles sharing the
  colour-red edge (0, 1) and both have h_φ ≡ -1. 8 of 28 vertices
  lie outside V(K_0) ∪ V(K_1).
- verify_28_vertex_counterexample.py: reproduces the counterexample,
  verifies all properties, and renders figures/min-face-5-counterexample.png.

Note on the boundary: face-length ≥ 6 is impossible for cubic plane
graphs by Euler (6F = 6(V/2 + 2) > 3V = sum face lengths for V > 4).
So face-length ≥ 5 is the strongest face-length restriction admitting
any cubic plane graphs at all.

Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>
This commit is contained in:
didericis
2026-05-25 03:42:08 -04:00
parent 3aec31b3ac
commit 10a53e9de1
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papers/face_monochromatic_pairs/experiments/search_min_face5_counterexample.py
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"""Search for counterexamples to the strengthened Conjecture 5.5:
Let H be a cubic plane graph in which every face has length ≥ 5,
with a proper 3-edge-colouring φ. If K_0 = {a,b}-Kempe cycle and
K_1 = {a,c}-Kempe cycle share a colour-a edge, then h_φ cannot be
constant on both V(K_0) and V(K_1).
The graphs H satisfying the face-length-≥-5 hypothesis are exactly
the duals of triangulations of the sphere with minimum degree ≥ 5.
We enumerate these via plantri (through Sage's
`graphs.triangulations(n, minimum_degree=5)`), take the planar dual,
and run the same Heawood-constancy check as
`search_smaller_counterexample.py`.
For each triangulation order n, the dual has 2n - 4 cubic vertices.
The smallest case n = 12 is the icosahedron, whose dual is the
dodecahedron (20 vertices, all faces pentagonal).
Run with:
sage experiments/search_min_face5_counterexample.py # default max_n=14
sage experiments/search_min_face5_counterexample.py 16
sage experiments/search_min_face5_counterexample.py 16 --all
"""
import sys
import time
from sage.all import Graph
from sage.graphs.graph_generators import graphs
# Reuse the verification machinery from the cubic-search script.
import os
HERE = os.path.dirname(os.path.abspath(__file__))
sys.path.insert(0, HERE)
from search_smaller_counterexample import (
check_graph,
min_face_length,
)
def planar_dual(G):
"""Build the planar dual of a planar graph G with its embedding
already set via G.is_planar(set_embedding=True)."""
faces = G.faces()
D = Graph(multiedges=False, loops=False)
n_faces = len(faces)
D.add_vertices(range(n_faces))
# Map each (undirected) edge of G to the indices of the two faces
# that contain it.
edge_to_faces = {}
for i, face in enumerate(faces):
for e in face:
key = frozenset(e[:2])
edge_to_faces.setdefault(key, []).append(i)
for key, fs in edge_to_faces.items():
if len(fs) == 2 and fs[0] != fs[1]:
D.add_edge(fs[0], fs[1])
return D
def main():
args = [a for a in sys.argv[1:] if not a.startswith('--')]
max_n = int(args[0]) if args else 14
flag_all = '--all' in sys.argv[1:]
print(f"Searching for counterexamples to the face-length-≥-5 form "
f"of Conjecture 5.5.\n"
f"Iterating over triangulations with min degree ≥ 5 for "
f"n_T in [12, {max_n}].\n"
f"Each dual is cubic with all faces of length ≥ 5; the dual "
f"has 2n_T - 4 vertices.\n"
f"{'Continuing past first hit.' if flag_all else 'Stopping at first hit.'}\n")
first_found = None
for n_T in range(12, max_n + 1):
start = time.time()
count = 0
found = None
try:
gen = graphs.triangulations(n_T, minimum_degree=5)
except Exception as ex:
print(f"n_T={n_T:>3}: cannot enumerate ({ex})")
continue
for T in gen:
T.is_planar(set_embedding=True)
H = planar_dual(T)
n_H = H.order()
# Sanity: H should be cubic and planar; faces length ≥ 5.
if max(H.degree()) != 3 or min(H.degree()) != 3:
continue
if not H.is_planar(set_embedding=True):
continue
if min_face_length(H) < 5:
continue
count += 1
res = check_graph(H)
if res is not None:
found = (T.copy(), H.copy(), res, n_H)
if not flag_all:
break
elapsed = time.time() - start
n_H_min = 2 * n_T - 4
if found is None:
print(f"n_T={n_T:>3}: checked {count} triangulation duals "
f"(each with {n_H_min} cubic vertices), no counterexample "
f"[{elapsed:.1f}s]")
else:
T, H, (col, K0, K1, h0, h1, a, b, c, e), n_H = found
colour_name = {0: 'red', 1: 'blue', 2: 'green'}
print(f"n_T={n_T:>3}: COUNTEREXAMPLE in dual #{count} "
f"(|V(H)| = {n_H}) [{elapsed:.1f}s]")
print(f" triangulation canonical graph6 = "
f"{T.canonical_label().graph6_string()}")
print(f" dual (cubic) canonical graph6 = "
f"{H.canonical_label().graph6_string()}")
print(f" dual edges = {sorted(H.edges(labels=False))}")
print(f" colouring = {col}")
print(f" shared colour = {colour_name[a]} ({a}), edge {e}")
print(f" K_{{a,b}} = K_{{{colour_name[a]},{colour_name[b]}}} "
f"= {K0} (h={h0:+d}, |V|={len(K0)})")
print(f" K_{{a,c}} = K_{{{colour_name[a]},{colour_name[c]}}} "
f"= {K1} (h={h1:+d}, |V|={len(K1)})")
if first_found is None:
first_found = n_T
if not flag_all:
break
sys.stdout.flush()
if first_found is not None:
print(f"\nSmallest counterexample at triangulation order n_T "
f"= {first_found} (dual has {2 * first_found - 4} vertices).")
else:
print(f"\nNo counterexample found for triangulation orders ≤ "
f"{max_n} (dual orders ≤ {2 * max_n - 4}).")
if __name__ == '__main__':
main()
papers/face_monochromatic_pairs/experiments/search_smaller_counterexample.py
+37 -4
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@@ -146,19 +146,45 @@ def check_graph(G):
return None
def min_face_length(G):
"""Min face length of G in its current planar embedding."""
return min(len(f) for f in G.faces())
def main():
args = [a for a in sys.argv[1:] if not a.startswith('--')]
max_n = int(args[0]) if args else 18
flag_all = '--all' in sys.argv[1:]
# Parse --min-face N
min_face = None
for i, a in enumerate(sys.argv[1:]):
if a == '--min-face' and i + 1 < len(sys.argv) - 1:
min_face = int(sys.argv[i + 2])
break
if a.startswith('--min-face='):
min_face = int(a.split('=', 1)[1])
break
print(f"Searching for cubic plane graph counterexamples to "
f"Conjecture 5.5, n in [4, {max_n}] "
f"({'continuing past first hit' if flag_all else 'stopping at first hit'})\n")
f"({'continuing past first hit' if flag_all else 'stopping at first hit'})"
+ (f", min face length >= {min_face}" if min_face else "")
+ "\n")
# If filtering by min face length, skip small n where impossible.
# For all-faces-length->=L cubic plane: V - E + F = 2, E = 3V/2,
# sum face lengths = 3V, F = V/2 + 2, so min sum = L*(V/2 + 2)
# <= 3V gives V >= 2L*(L-3)/(3-L/2)... let's just set V >= 20 for L=5.
if min_face is not None and min_face >= 5:
n_start = max(4, 20)
else:
n_start = 4
first_found = None
for n in range(4, max_n + 1, 2): # cubic requires n even
for n in range(n_start, max_n + 1, 2): # cubic requires n even
start = time.time()
count = 0
skipped = 0
found = None
try:
gen = graphs.planar_graphs(
@@ -172,6 +198,11 @@ def main():
for G in gen:
if max(G.degree()) != 3:
continue # not cubic
if min_face is not None:
G.is_planar(set_embedding=True)
if min_face_length(G) < min_face:
skipped += 1
continue
count += 1
res = check_graph(G)
if res is not None:
@@ -180,8 +211,10 @@ def main():
break
elapsed = time.time() - start
if found is None:
print(f"n={n:>3}: checked {count} graphs, no counterexample "
f"[{elapsed:.1f}s]")
extra = (f", skipped {skipped} due to small face"
if min_face is not None else "")
print(f"n={n:>3}: checked {count} graphs, no counterexample"
f"{extra} [{elapsed:.1f}s]")
else:
G, (col, K0, K1, h0, h1, a, b, c, e) = found
colour_name = {0: 'red', 1: 'blue', 2: 'green'}
papers/face_monochromatic_pairs/experiments/verify_28_vertex_counterexample.py
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"""Verify the n=28 counterexample (smallest dual of min-degree-5
triangulation that violates the face-length-≥-5 form of Conjecture
5.5) and render a planar PNG of it.
The graph H is the planar dual of a 16-vertex triangulation with
min degree ≥ 5 (the 3rd one in sage's enumeration). It has 28
vertices, 42 edges, and all faces of length ≥ 5. The colouring shown
below makes K_{red, blue} and K_{red, green} both 12-cycles sharing
the colour-red edge (0, 1) with h_φ ≡ -1 on each.
Run with: sage experiments/verify_28_vertex_counterexample.py
"""
import os
import sys
import math
from sage.all import Graph
HERE = os.path.dirname(os.path.abspath(__file__))
sys.path.insert(0, HERE)
from search_smaller_counterexample import (
edge_key, heawood_numbers, trace_kempe,
)
OUT_PNG = os.path.join(HERE, '..', 'figures', 'min-face-5-counterexample.png')
# Dual edges and the discovered colouring (sorted-edge order).
EDGES_LIST = [
(0, 1), (0, 4), (0, 6), (1, 2), (1, 5), (2, 3), (2, 8), (3, 4),
(3, 11), (4, 13), (5, 7), (5, 9), (6, 7), (6, 15), (7, 17),
(8, 10), (8, 12), (9, 10), (9, 19), (10, 20), (11, 12), (11, 14),
(12, 22), (13, 14), (13, 16), (14, 23), (15, 16), (15, 18),
(16, 25), (17, 18), (17, 19), (18, 26), (19, 21), (20, 21),
(20, 22), (21, 27), (22, 24), (23, 24), (23, 25), (24, 27),
(25, 26), (26, 27),
]
COLOURING = (0, 1, 2, 2, 1, 1, 0, 2, 0, 0, 2, 0, 1, 0, 0, 1, 2, 2,
1, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 0, 2, 1, 1,
2, 1, 2, 0, 1, 2)
COLOUR_NAME = {0: 'red', 1: 'blue', 2: 'green'}
def build():
G = Graph(multiedges=False, loops=False)
for u, v in EDGES_LIST:
G.add_edge(u, v)
# Sage's sorted edges should align with EDGES_LIST.
edges_sorted = sorted([edge_key(u, v) for (u, v) in G.edge_iterator(labels=False)])
assert edges_sorted == EDGES_LIST, "edge order mismatch"
col_of_edge = {edges_sorted[i]: COLOURING[i] for i in range(len(edges_sorted))}
return G, col_of_edge
def main():
G, col_of_edge = build()
print(f"|V| = {G.order()}, |E| = {G.size()}")
assert G.is_planar(set_embedding=True)
print(f"face lengths: {sorted([len(f) for f in G.faces()])}")
assert all(len(f) >= 5 for f in G.faces()), "face length < 5"
# Verify proper 3-edge-colouring.
for v in G.vertex_iterator():
cs = sorted(col_of_edge[edge_key(v, u)] for u in G.neighbors(v))
assert cs == [0, 1, 2], f"vertex {v} colours {cs}"
print("proper 3-edge-colouring ✓")
h = heawood_numbers(G, col_of_edge)
plus = sum(1 for v in h if h[v] == +1)
minus = sum(1 for v in h if h[v] == -1)
print(f"global h_φ: {plus} (+1) / {minus} (-1)")
# Verify both Kempe cycles through (0, 1) are constant.
K0 = trace_kempe(G, col_of_edge, (0, 1), (0, 1)) # red+blue
K1 = trace_kempe(G, col_of_edge, (0, 1), (0, 2)) # red+green
h_K0 = [h[v] for v in K0]
h_K1 = [h[v] for v in K1]
print(f"K_{{red, blue}} (len {len(K0)}): {K0}, h = {h_K0[0]} (const? {len(set(h_K0))==1})")
print(f"K_{{red, green}} (len {len(K1)}): {K1}, h = {h_K1[0]} (const? {len(set(h_K1))==1})")
print(f"V(K0) ∪ V(K1)| = {len(set(K0) | set(K1))} / {G.order()}")
# Canonical graph6
print(f"canonical graph6 = {G.canonical_label().graph6_string()}")
# Render PNG with planar layout
import matplotlib
matplotlib.use('Agg')
import matplotlib.pyplot as plt
pos = G.layout_planar()
fig, ax = plt.subplots(figsize=(10, 10), dpi=160)
ax.set_aspect('equal')
ax.axis('off')
for (u, v) in G.edge_iterator(labels=False):
c = COLOUR_NAME[col_of_edge[edge_key(u, v)]]
x1, y1 = pos[u]; x2, y2 = pos[v]
ax.plot([x1, x2], [y1, y2], color=c, linewidth=2.0, solid_capstyle='round')
K0set = set(K0); K1set = set(K1)
for v, (x, y) in pos.items():
if v in K0set and v in K1set:
face_c = '#bbbbff' # on both cycles
elif v in K0set or v in K1set:
face_c = '#ddddff'
else:
face_c = 'lightgrey'
ax.plot(x, y, 'o', markersize=18, markerfacecolor=face_c,
markeredgecolor='black', markeredgewidth=1.0)
ax.text(x, y, str(v), ha='center', va='center', fontsize=8)
xs = [p[0] for p in pos.values()]
ys = [p[1] for p in pos.values()]
pad = 0.5
ax.set_xlim(min(xs) - pad, max(xs) + pad)
ax.set_ylim(min(ys) - pad, max(ys) + pad)
os.makedirs(os.path.dirname(OUT_PNG), exist_ok=True)
fig.tight_layout()
fig.savefig(OUT_PNG, dpi=160, bbox_inches='tight')
plt.close(fig)
print(f"\nWrote: {OUT_PNG}")
if __name__ == '__main__':
main()
papers/face_monochromatic_pairs/figures/min-face-5-counterexample.png
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papers/face_monochromatic_pairs/paper.tex
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@@ -814,65 +814,64 @@ that proof produces a clauses-(1)--(3) witness without ever needing to
inspect the other Kempe cycle.
\end{proof}
\begin{conjecture}[Constant Heawood on two edge-sharing Kempe cycles --- \textbf{FALSE}]
\begin{conjecture}[Constant Heawood on two edge-sharing Kempe cycles, large-face cubic plane graphs]
\label{conj:no-two-constant-kempe-cycles}
Let $H$ be a cubic plane graph with a proper $3$-edge-colouring
$\varphi$, fix a colour $a \in \{1, 2, 3\}$, and let $\{b, c\} =
\{1, 2, 3\} \setminus \{a\}$. Let $K_0$ be an $\{a, b\}$-Kempe cycle
of $\varphi$ and $K_1$ an $\{a, c\}$-Kempe cycle of $\varphi$ such that
Let $H$ be a cubic plane graph in which every face has length at
least $5$, with a proper $3$-edge-colouring $\varphi$. Fix a colour
$a \in \{1, 2, 3\}$ and let $\{b, c\} = \{1, 2, 3\} \setminus \{a\}$.
Let $K_0$ be an $\{a, b\}$-Kempe cycle of $\varphi$ and $K_1$ an
$\{a, c\}$-Kempe cycle of $\varphi$ such that
$E(K_0) \cap E(K_1) \neq \emptyset$ (equivalently, $K_0$ and $K_1$
share at least one colour-$a$ edge). If $h_\varphi$ is constant on
$V(K_0)$, then $h_\varphi$ is \emph{not} constant on $V(K_1)$.
\end{conjecture}
\begin{remark}[Disproof of
Conjecture~\ref{conj:no-two-constant-kempe-cycles}]
\label{rem:no-two-constant-kempe-cycles-counterexample}
Conjecture~\ref{conj:no-two-constant-kempe-cycles} is \emph{false}.
Figure~\ref{fig:no-two-constant-kempe-counterexample} exhibits a
concrete counterexample: a cubic plane graph $H$ on $40$ vertices with
a proper $3$-edge-colouring $\varphi$ (colours red/blue/green) in
which both
\begin{align*}
K_{\mathrm{red},\mathrm{blue}}
&= \text{the outer $8$-cycle, and} \\
K_{\mathrm{red},\mathrm{green}}
&= \text{the $12$-cycle (outer frame $+$ upper-left ``ladder'' side)}
\end{align*}
share the colour-red edge $(0, 7)$ and satisfy
$h_\varphi \equiv -1$ on the vertex set of each. Globally $h_\varphi$
takes value $+1$ on $16$ vertices and $-1$ on $24$ vertices, with all
of the $+1$-vertices concentrated in the inner ``tilted ladder''
region, so that both Kempe cycles miss them entirely.
\smallskip
The underlying graph appears to be a fresh ad-hoc construction: it has
$60$ edges, vertex- and edge-connectivity $3$, girth $3$ (two
triangles), is Hamiltonian, is not bipartite, has trivial
automorphism group, and its $22$ faces have lengths distributed
as $\{3{:}2,\,4{:}4,\,5{:}10,\,6{:}1,\,7{:}1,\,8{:}2,\,9{:}1,\,10{:}1\}$.
Its canonical \texttt{graph6} string (via
\texttt{G.canonical\_label().graph6\_string()}) is
\begin{center}
\small\ttfamily
ggE\_?C?O\_B?B?@C?P???B??\_?@???\_??Sa???W????@a??\_B????G\_???@?????O??@\_???@?\_???K?????\_????K?????o???B?????AG?????B?????G?????AG?????F
\end{center}
The construction, the proper $3$-edge-colouring, Kempe-cycle tracing,
and the Heawood-number computation (via the CW rotation at each
vertex) are all in \texttt{experiments/counterexample\_conj\_5\_5.py};
the source drawing is in \texttt{constant\_heawood\_counterexample.tikz}.
\begin{remark}[On the face-length hypothesis]
\label{rem:no-two-constant-kempe-face-length-hypothesis}
Without the hypothesis that every face has length $\ge 5$, the
conjecture is \emph{false}, with very small counterexamples:
\begin{itemize}
\item \textbf{$K_4$ (the tetrahedron, $n = 4$):} The standard
proper $3$-edge-colouring has $h_\varphi$ constant (all $-1$ or
all $+1$, depending on planar embedding orientation) on every vertex,
and every pair of distinct Kempe cycles shares colour-edges. (All
four faces of $K_4$ are triangles.)
\item \textbf{An $n = 8$ cubic plane graph with girth $3$
(graph6 \texttt{G\}GOW[}):} Found by the brute-force enumeration in
\texttt{experiments/search\_smaller\_counterexample.py}; both Kempe
cycles are $8$-cycles visiting every vertex, and $h_\varphi$ is
constant on each.
\item \textbf{An ad-hoc $n = 40$ cubic plane graph with girth $3$:}
The graph in \texttt{constant\_heawood\_counterexample.tikz} (face
lengths $\{3{:}2,\,4{:}4,\,5{:}10,\,6{:}1,\,7{:}1,\,8{:}2,\,9{:}1,\,
10{:}1\}$, $|V| = 40$) has an $8$-cycle and a $12$-cycle that share a
colour-red edge and both have $h_\varphi \equiv -1$, with $24$ of the
$40$ vertices ($60\%$) lying outside $V(K_0) \cup V(K_1)$ --- so this
is a structurally non-trivial counterexample, unlike $K_4$ and the
$n = 8$ example where the two Kempe cycles already cover all
vertices.
\end{itemize}
All three counterexamples have at least one face of length
$\le 4$. The face-length-$\ge 5$ hypothesis above is the smallest
restriction that simultaneously kills all currently known
counterexamples; whether it suffices to make the conjecture true is
open. The smallest cubic plane graphs satisfying this hypothesis have
$|V(H)| \ge 20$ (with $|V| = 20$ forced to be the dodecahedron, all
faces pentagonal); brute-force search up to some moderate $|V|$ is in
\texttt{experiments/search\_smaller\_counterexample.py}.
\end{remark}
\begin{figure}[h]
\centering
\includegraphics[width=0.85\textwidth]{figures/no-two-constant-kempe-counterexample.png}
\caption{Counterexample to
Conjecture~\ref{conj:no-two-constant-kempe-cycles}: a cubic plane
graph on $40$ vertices with a proper $3$-edge-colouring on which
$h_\varphi$ is simultaneously constant ($\equiv -1$) on the outer
red/blue $8$-cycle and on the red/green $12$-cycle (outer frame plus
the upper-left ladder side), which share the colour-red edge
$(0, 7)$.}
\includegraphics[width=0.7\textwidth]{figures/no-two-constant-kempe-counterexample.png}
\caption{The $n = 40$ counterexample to the unrestricted version of
Conjecture~\ref{conj:no-two-constant-kempe-cycles} (cf.\
Remark~\ref{rem:no-two-constant-kempe-face-length-hypothesis}). Both
the outer red/blue $8$-cycle and the red/green $12$-cycle (outer
frame plus the upper-left ladder side) have $h_\varphi \equiv -1$ and
share the colour-red edge $(0, 7)$. The face lengths of this drawing
are $\{3, 4, 5, 6, 7, 8, 9, 10\}$ with two triangles, so it does not
satisfy the face-length-$\ge 5$ hypothesis above.}
\label{fig:no-two-constant-kempe-counterexample}
\end{figure}