- 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>
Adds the concrete construction (40 vertices, 60 edges, cubic + planar
+ proper 3-edge-coloured) on which h_φ is simultaneously constant on
two Kempe cycles sharing an edge:
- K_{red, blue} = 8-cycle (the outer frame): all h_φ = -1
- K_{red, green} = 12-cycle (outer frame + upper-left ladder side):
all h_φ = -1
- They share the colour-red edge (0, 7) (and others).
The graph is drawn in TikZiT and stored as
papers/face_monochromatic_pairs/constant_heawood_counterexample.tikz
The Sage transcription + Heawood/Kempe verification + PNG renderer is
papers/face_monochromatic_pairs/experiments/counterexample_conj_5_5.py
Rendered PNG (with the four bent outer-face / trapezoid arcs matching
the tikz drawing) is at
papers/face_monochromatic_pairs/figures/no-two-constant-kempe-counterexample.png
Globally h_φ has 16 vertices at +1 and 24 at -1; the +1 vertices are
concentrated in the inner "tilted ladder" region, leaving the outer
and the K_{red,green}-extension all at -1. This is the structural
reason both Kempe cycles can be constant.
Also includes the TikZiT styles file default.tikzstyles defining the
red/blue/green edge styles used by the .tikz file.
Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>
didericis