fd4b89a39e
Approach: at each vertex of a simple closed cycle C in a 3-regular planar graph, define turn-sign(v) = +1 if the third edge (off-cycle) is in C's bounded region (interior), -1 if exterior. Compute Σ_v turn-sign(v). Empirical check on standard graphs (K_4, Q_3, dodecahedron, 3-prism): For a FACE boundary, Σ = -L_face (all third edges outside the face). For a NON-face cycle, Σ can range from -L to +L. Plan: under Lemma 5.2's alternation hypothesis (constancy on V(K_b) forces third edges to alternate sides along K_b), the signs alternate +,-,+,-,... yielding Σ = 0 for K_b of even length. This shows K_b is NOT a face boundary (= it bounds a region containing other vertices/edges), which is true but not a contradiction. A simple closed planar curve can have Σ = 0; that just means equal numbers of off-cycle edges are inside vs outside. So the winding-number approach (option 4) does not yield a direct contradiction under the chord-apex+Kempe + constancy hypothesis. Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>
167 lines
6.3 KiB
Python
167 lines
6.3 KiB
Python
"""Verify the combinatorial winding invariant for simple closed cycles
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in a 3-regular planar graph:
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Claim: For a simple closed cycle C in a 3-regular planar graph H
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(with C ⊆ E(H), |C|/2 edges of C at each vertex of C), walked CCW
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around its interior, define for each v ∈ V(C):
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turn-sign(v) = +1 if the third edge (= the one of v's 3 edges NOT
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on C) is on the OUTSIDE of C, locally at v.
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(= the walker turns LEFT into the interior).
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-1 if the third edge is on the INSIDE.
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Hypothesis: Σ_v turn-sign(v) = ±6 (= ±(L_C / |V|_C-cycle-curvature)).
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This is a combinatorial Gauss-Bonnet for cubic plane graphs:
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internal/external turn parity around any face / cycle.
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Sanity-check this on:
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- Triangle (length 3) in K_4.
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- Quadrilateral (length 4) in Q_3.
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- Pentagon (length 5) in dodecahedron.
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- Hexagon (length 6) in some triangulation dual.
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Also on Kempe cycles in chord-apex+Kempe colourings of reduced duals.
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If the invariant holds (always ±6), then combining with Lemma 5.2's
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alternation (= turn signs alternate +,-,+,- on any K_b under constancy),
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we'd have #(+) = #(-) = L_b / 2, so #(+) - #(-) = 0 ≠ ±6, giving a
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direct contradiction.
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Run with: sage experiments/check_combinatorial_winding.py
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"""
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import os
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import sys
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from sage.all import Graph
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from sage.graphs.graph_generators import graphs
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def cycle_winding(G, cycle_vertices):
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"""Compute Σ_v turn-sign(v) for a simple closed cycle in 3-reg planar G.
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cycle_vertices: list of vertices in walking order around the cycle.
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Walking direction: CCW (= traversal such that interior is on the LEFT).
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For each v in the cycle, the 3 edges at v are split as: 2 cycle
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edges (to v's neighbours in the cycle) and 1 third edge (off-cycle).
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The third edge is at some position in v's CW rotation system.
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turn-sign(v) = +1 if the third edge is OUTSIDE the cycle (= on the
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right of the CCW walk locally), -1 if INSIDE.
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"""
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G.is_planar(set_embedding=True)
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emb = G.get_embedding()
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L = len(cycle_vertices)
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total = 0
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for k in range(L):
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v = cycle_vertices[k]
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prev = cycle_vertices[(k - 1) % L]
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nxt = cycle_vertices[(k + 1) % L]
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# CW rotation at v: emb[v] is the list of neighbours in CW
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# order.
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nbrs = emb[v]
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# We need positions of prev, nxt, and third.
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if prev not in nbrs or nxt not in nbrs:
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return None
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# The third edge endpoint:
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third = [u for u in nbrs if u != prev and u != nxt]
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if len(third) != 1:
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return None
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third = third[0]
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# CW order of (prev, nxt, third) in emb[v]:
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idx_prev = nbrs.index(prev)
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idx_nxt = nbrs.index(nxt)
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idx_third = nbrs.index(third)
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# In CW order, what's the cyclic arrangement?
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# Walking CCW around the cycle means walker is on edge (prev, v),
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# at v, and exits via (v, nxt). The interior of the cycle is
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# on the LEFT of the walker.
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# The third edge is at idx_third in CW rotation.
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# If we go CW from idx_prev to idx_nxt:
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# - if idx_third is in this CW interval, third is on one side;
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# - else on the other side.
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# For 3-regular vertex with CW order (positions 0, 1, 2 cyclically),
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# idx_prev, idx_nxt, idx_third are some permutation of {0, 1, 2}.
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# Define: third is on the RIGHT of walker (= going CW from prev
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# to nxt without passing through third) ⟺ third is NOT between
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# prev and nxt in CW direction.
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# The CW order at v: nbrs[0], nbrs[1], nbrs[2].
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# CW direction at v: 0 → 1 → 2 → 0.
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# CW from prev to nxt: starting at idx_prev, going CW (= +1 mod 3),
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# we reach idx_nxt after some steps.
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# If we pass through idx_third in 1 step: third is between prev
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# and nxt in CW direction (i.e., on one specific side).
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# If we don't pass through idx_third (= go directly from prev to
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# nxt in 1 CW step): third is on the OTHER side.
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# 1 CW step from idx_prev: (idx_prev + 1) % 3.
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if (idx_prev + 1) % 3 == idx_nxt:
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# CW step from prev to nxt is direct (1 step).
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# Third is at the "other" position, NOT between in CW.
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# This means third is on the "left" of walker going CCW
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# along the cycle (= interior side).
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# turn-sign = -1 (third inside).
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sign = -1
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elif (idx_prev + 2) % 3 == idx_nxt:
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# CW step from prev to nxt is 2 steps (passing through
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# idx_third).
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# Third is BETWEEN prev and nxt in CW direction.
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# turn-sign = +1 (third outside, walker turns left into
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# interior).
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sign = +1
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else:
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return None
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total += sign
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return total
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def face_to_cycle(face):
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"""Convert face (= list of edges) to ordered vertex cycle."""
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if not face:
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return None
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verts = [face[0][0]]
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for e in face:
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verts.append(e[1])
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return verts[:-1] # last vertex = first vertex
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def main():
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tests = []
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# K_4 = tetrahedron
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K4 = graphs.CompleteGraph(4)
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K4.is_planar(set_embedding=True)
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tests.append(('K_4', K4))
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# Q_3 = cube
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Q3 = graphs.CubeGraph(3)
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Q3.is_planar(set_embedding=True)
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tests.append(('Q_3', Q3))
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# Dodecahedron
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Dod = graphs.DodecahedralGraph()
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Dod.is_planar(set_embedding=True)
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tests.append(('Dodecahedron', Dod))
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# Triangular prism
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Tprism = graphs.CompleteBipartiteGraph(2, 3) # not 3-regular
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# Use Q3 plus more
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tests.append(('Triangular prism (3-prism)', graphs.GeneralizedPetersenGraph(3, 1)))
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for name, G in tests:
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if not all(G.degree(v) == 3 for v in G.vertex_iterator()):
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print(f"\n{name}: NOT 3-regular (degree set = "
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f"{sorted(set(G.degree()))}); skipping")
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continue
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if not G.is_planar(set_embedding=True):
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print(f"\n{name}: NOT planar; skipping")
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continue
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print(f"\n{name}: |V|={G.order()}, |E|={G.size()}, faces:")
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faces = G.faces()
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for f in faces:
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cyc = face_to_cycle(f)
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if cyc is None:
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print(f" bad face")
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continue
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w = cycle_winding(G, cyc)
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print(f" face length {len(cyc)}, cycle={cyc}, winding={w}")
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if __name__ == '__main__':
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main()
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