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