c947ce75ff
- New paper papers/even_level_graph_generators/: defines Even Level Graph (every level cycle even), derived level graphs, intertwining trees, and the disjunction conjecture (every maximal planar graph is a derived level graph or intertwining tree). Empirically tested through n=11: every iso class is at least an intertwining tree, so the disjunction holds trivially in this range. The intertwining tree disjunct fails at the Tutte graph dual (n=25), so the disjunction becomes non-trivial past some unknown threshold. - Level Switching paper: adds Section 4 (Reachability via edge switches) with the two-step argument (Sleator-Tarjan-Thurston for Case 1; face-merges for Case 2) and Theorem 4.1 (O(n) edge switches suffice to reach all-depth-0). Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>
152 lines
5.3 KiB
Python
152 lines
5.3 KiB
Python
"""User-proposed simpler algorithm:
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1) Assign each face x = min dual-tree distance to any leaf face (ear).
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2) When face F at depth d has no balanced surface switch, edge-switch
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on the edge between F and F's neighbour with the smallest x.
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3) Repeat until the number of depth-d faces has strictly decreased.
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"""
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import os
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import sys
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sys.path.insert(0, os.path.dirname(os.path.abspath(__file__)))
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import math
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import networkx as nx
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from stress_test_termination import (
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compute_depths, apply_switch, check_balanced, face_edges
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)
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def compute_x(faces, outer_set):
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"""x(F) = min dual-tree distance from F to any leaf face."""
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D = nx.Graph()
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D.add_nodes_from(range(len(faces)))
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for i, fi in enumerate(faces):
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for j, fj in enumerate(faces):
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if i < j and face_edges(fi) & face_edges(fj):
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D.add_edge(i, j)
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leaves = [i for i in D.nodes if D.degree(i) == 1]
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x = {}
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for i in range(len(faces)):
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if not leaves:
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x[i] = float('inf')
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continue
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x[i] = min(nx.shortest_path_length(D, i, lf) for lf in leaves)
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return x, D
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def neighbour_at_edge(faces, F_idx, e):
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for j, fj in enumerate(faces):
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if j != F_idx and e in face_edges(fj):
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return j
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return None
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def run_leaf_distance_algorithm(faces, outer_set, max_steps=500,
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verbose=True):
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"""Run the algorithm. If F admits a balanced switch, take it; else
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pick the neighbour with smallest x and switch on that edge."""
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total = 0
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log = []
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for step in range(max_steps):
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depth = compute_depths(faces, outer_set)
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d_max = max(depth.values())
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if d_max == 0:
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log.append(f'terminated at step {step}')
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return faces, total, log
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max_d_faces = [i for i, d in depth.items() if d == d_max]
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F_idx = max_d_faces[0]
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F = faces[F_idx]
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ok, _, fp_idx, e = check_balanced(F_idx, faces, depth, outer_set)
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if ok:
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Fp = faces[fp_idx]
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u, v = tuple(e)
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w = [vert for vert in F if vert not in (u, v)][0]
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x = [vert for vert in Fp if vert not in (u, v)][0]
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log.append(f'step {step}: BALANCED on edge ({u},{v}) '
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f'with F\' = {Fp}')
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faces = apply_switch(faces, (u, v), (w, x))
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total += 1
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continue
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# Find x values
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x_vals, D = compute_x(faces, outer_set)
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# Among neighbours of F (interior edges, hence in dual graph),
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# pick the one with smallest x.
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nb_choices = []
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for e_test in face_edges(F):
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if e_test in outer_set:
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continue
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nb_idx = neighbour_at_edge(faces, F_idx, e_test)
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if nb_idx is None:
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continue
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nb_choices.append((x_vals[nb_idx], e_test, nb_idx))
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if not nb_choices:
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log.append(f'step {step}: no interior neighbour; stuck')
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break
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nb_choices.sort()
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x_min, e_chosen, nb_idx = nb_choices[0]
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Fnb = faces[nb_idx]
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u, v = tuple(e_chosen)
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w = [vert for vert in F if vert not in (u, v)][0]
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xv = [vert for vert in Fnb if vert not in (u, v)][0]
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log.append(f'step {step}: x-preprocess on edge ({u},{v}) '
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f'(F\'={Fnb}, x={x_min}; other choices: '
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f'{[(c[0]) for c in nb_choices[1:]]})')
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faces = apply_switch(faces, (u, v), (w, xv))
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total += 1
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log.append(f'hit max_steps; final max depth = '
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f'{max(compute_depths(faces, outer_set).values())}')
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return faces, total, log
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# ----- TEST CASE 1: 9-vertex example -----
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n9 = 9
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outer9 = [(i, (i + 1) % n9) for i in range(n9)]
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outer_set9 = {frozenset(e) for e in outer9}
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faces9 = [
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(0, 1, 2), (0, 2, 3), (3, 4, 5), (3, 5, 6),
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(6, 7, 8), (6, 8, 0), (0, 3, 6),
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]
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print('=== 9-vertex example ===')
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_, total, log = run_leaf_distance_algorithm(faces9, outer_set9)
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print('\n'.join(log))
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print(f'total switches: {total}\n')
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# ----- TEST CASE 2: 24-vertex doubly-lopsided example -----
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n24 = 24
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outer24 = [(i, (i + 1) % n24) for i in range(n24)]
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outer_set24 = {frozenset(e) for e in outer24}
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def arm(a, b):
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return [
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(a, a + 1, a + 2), (a, a + 2, b), (a + 2, a + 3, a + 4),
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(a + 2, a + 4, b), (a + 4, a + 5, a + 6), (a + 4, a + 6, b),
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(a + 6, a + 7, b),
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]
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faces24 = [(0, 8, 16)]
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for (a, b) in [(0, 8), (8, 16), (16, 24)]:
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fs = arm(a, b)
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fs = [tuple(0 if v == 24 else v for v in vt) for vt in fs]
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faces24.extend(fs)
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print('=== 24-vertex doubly-lopsided example ===')
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_, total, log = run_leaf_distance_algorithm(faces24, outer_set24)
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print('\n'.join(log[:25]))
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if len(log) > 25:
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print(f'... ({len(log) - 25} more lines)')
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print(f'total switches: {total}\n')
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# ----- TEST CASE 3: random outerplanar n=40 (one of the previously-slow seeds) -----
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from stress_test_termination import random_triangulation
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seed = 94476710
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outer, _, faces = random_triangulation(40, seed=seed)
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outer_set = {frozenset(e) for e in outer}
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print(f'=== Random n=40 (seed={seed}) with the new algorithm ===')
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_, total, log = run_leaf_distance_algorithm(faces, outer_set, max_steps=1000)
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print(f'total switches: {total} (random algorithm with cap=10000 needed 863)')
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print(f'first 5 lines: {log[:5]}')
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print(f'last 2 lines: {log[-2:]}')
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