From c482bc5633b18f8051867a9b501acb369188efec Mon Sep 17 00:00:00 2001
From: didericis <eric@dideric.is>
Date: Wed, 17 Jun 2026 21:13:35 -0400
Subject: [PATCH] Option 2 (direct transversal) clean form is dead too

transversal.py: a STRONG transversal (n-2 faces whose Boolean assignment
single-valuedly and injectively determines the boundary) would give a
constructive proof. It exists in 0/2948 disks with k>=1 -- once there is
any interior vertex, fixing n-2 faces leaves boundary-visible completion
freedom, so the boundary is never single-valued in them. Works only at
k=0 (base case). Both elementary routes (reduction localization, direct
transversal) now closed.

Co-Authored-By: Claude Opus 4.8 <noreply@anthropic.com>
---
 .../experiments/transversal.py                | 123 ++++++++++++++++++
 1 file changed, 123 insertions(+)
 create mode 100644 papers/heawood_restrictions_on_nested_tire_graph_duals/experiments/transversal.py

diff --git a/papers/heawood_restrictions_on_nested_tire_graph_duals/experiments/transversal.py b/papers/heawood_restrictions_on_nested_tire_graph_duals/experiments/transversal.py
new file mode 100644
index 0000000..c53b548
--- /dev/null
+++ b/papers/heawood_restrictions_on_nested_tire_graph_duals/experiments/transversal.py
@@ -0,0 +1,123 @@
+"""
+Option 2: a direct "strong transversal" for |Phi(D)| >= 2^(n-2).
+
+In the Boolean reformulation, feasible x in {0,1}^F satisfy |x|_w ≡ -deg(w) at
+interior w, and the boundary sequence is (deg(v)+|x|_v mod 3)_{v in C}. A STRONG
+TRANSVERSAL is a set A of n-2 faces such that
+  (i)  every assignment x_A in {0,1}^A extends to a feasible x,
+  (ii) the boundary sequence is single-valued in x_A (all feasible completions of
+       a given x_A give the same boundary), and
+  (iii) the map x_A -> boundary is injective.
+If such A exists, the 2^(n-2) assignments give 2^(n-2) distinct boundary sequences,
+proving the bound CONSTRUCTIVELY. We test whether a strong transversal exists.
+
+Heuristic worry: over GF(3) the internal completion freedom (dim k) exceeds the
+boundary-invisible freedom (dim k-1), so (ii) may fail. Test settles it.
+"""
+
+import sys
+from collections import defaultdict
+from itertools import combinations, product
+
+import numpy as np
+from scipy.spatial import Delaunay
+
+np.seterr(all="ignore")
+
+
+def disk(n, k, rng):
+    ang = 2 * np.pi * np.arange(n) / n
+    rad = 1.0 + 0.18 * rng.random(n)
+    bpts = np.c_[rad * np.cos(ang), rad * np.sin(ang)]
+    if k:
+        r = 0.8 * np.sqrt(rng.random(k)); t = 2 * np.pi * rng.random(k)
+        pts = np.vstack([bpts, np.c_[r * np.cos(t), r * np.sin(t)]])
+    else:
+        pts = bpts
+    return [tuple(int(x) for x in s) for s in Delaunay(pts).simplices]
+
+
+def valid(faces, n, k):
+    if len(faces) != 2 * k + n - 2:
+        return False
+    from collections import Counter
+    ec = Counter()
+    for a, b, c in faces:
+        for e in ((a, b), (b, c), (a, c)):
+            ec[frozenset(e)] += 1
+    return all(ec[frozenset((i, (i + 1) % n))] == 1 for i in range(n))
+
+
+def feasible_table(faces, n):
+    """Return list of (x tuple, boundary tuple) over feasible x in {0,1}^F."""
+    interior = sorted(set(v for f in faces for v in f if v >= n))
+    F = len(faces)
+    if F > 14:
+        return None
+    Bint = np.zeros((len(interior), F), dtype=np.int64)
+    Cinc = np.zeros((n, F), dtype=np.int64)
+    deg = np.zeros(n, dtype=np.int64)
+    iidx = {w: r for r, w in enumerate(interior)}
+    degw = np.zeros(len(interior), dtype=np.int64)
+    for j, (a, b, c) in enumerate(faces):
+        for v in (a, b, c):
+            if v >= n:
+                Bint[iidx[v], j] = 1; degw[iidx[v]] += 1
+            else:
+                Cinc[v, j] = 1; deg[v] += 1
+    xs = np.array(list(product((0, 1), repeat=F)), dtype=np.int64)
+    if len(interior):
+        ok = np.all((xs @ Bint.T - (-degw)) % 3 == 0, axis=1)
+        xs = xs[ok]
+    if xs.shape[0] == 0:
+        return []
+    bnd = (deg + xs @ Cinc.T) % 3
+    return list(zip(map(tuple, xs.tolist()), map(tuple, bnd.tolist())))
+
+
+def has_strong_transversal(table, F, n):
+    target = 2 ** (n - 2)
+    for A in combinations(range(F), n - 2):
+        groups = defaultdict(set)
+        for x, b in table:
+            groups[tuple(x[i] for i in A)].add(b)
+        if len(groups) != target:
+            continue                              # not every x_A feasible
+        if any(len(bs) != 1 for bs in groups.values()):
+            continue                              # not single-valued (cond ii)
+        reps = [next(iter(bs)) for bs in groups.values()]
+        if len(set(reps)) == target:              # injective (cond iii)
+            return True
+    return False
+
+
+def main():
+    seed = int(sys.argv[1]) if len(sys.argv) > 1 else 0
+    rng = np.random.default_rng(seed)
+    tested = 0; have = 0; fails = []
+    for _ in range(3000):
+        n = int(rng.integers(4, 6)); k = int(rng.integers(1, 3))
+        faces = disk(n, k, rng)
+        if not valid(faces, n, k) or len(faces) > 12:
+            continue
+        tbl = feasible_table(faces, n)
+        if not tbl:
+            continue
+        tested += 1
+        if has_strong_transversal(tbl, len(faces), n):
+            have += 1
+        elif len(fails) < 5:
+            fails.append((n, k, len(faces)))
+    print(f"disks tested (k>=1): {tested}")
+    print(f"  have a STRONG transversal (size n-2, single-valued, injective): "
+          f"{have}/{tested} ({100*have/max(tested,1):.1f}%)")
+    if fails:
+        print(f"  failures (n,k,F): {fails}")
+    if have == tested and tested:
+        print("  => strong transversals always exist: constructive proof viable.")
+    elif have == 0:
+        print("  => strong transversals NEVER exist: this clean form of option 2 is dead.")
+
+
+if __name__ == "__main__":
+    main()