Exact sampling by CFTP needs c1c2 ≥ a1a2 and c1c2 ≥ b1b2. Outside that region no monotone coupling of the update exists, so no CFTP scheme is valid there. Raise the c weights, or lower a and b, to re-enable it.
a1=a2=1 fixed (WLOG by gauge freedom -- the model only depends on 2 independent ratios). Δ>1: ferroelectric. −1≤Δ<1: disordered. Δ<−1: antiferroelectric. Exact sample is a mathematically exact draw: the sequential transfer-matrix sampler for n ≤ 14 at any weights, and Coupling From The Past (n ≤ 250) wherever c1c2 dominates a1a2 and b1b2. That region contains the whole antiferroelectric phase Δ<−1, so Δ=−3 is exactly samplable.
How long CFTP runs. The stopping time T* = min{T : the composed update map from −T to 0 is constant} measured exactly, against Δ, one curve per lattice size: view the plot. It is non-monotone, with a minimum near Δ = −0.5 to −1, and the exponent in T* ~ nα is flat at 2.30 for Δ ≥ −1.5 and rises to 2.86 at Δ = −4.
Does the island exist? A yes/no test on a single configuration: call a c-vertex solid when all eight of its neighbours are c-vertices, and say the island exists when the solid c-vertices span the central N/5 window. P(island) against Δ, for N = 32, 48, 64. It is essentially zero for Δ ≥ −1 at every size (at most 0.025) and climbs below, reaching 0.74 at Δ = −3 and rising with N. Solidity is what matters: the largest c-region is macroscopic on both sides of −1, so a test on its size alone answers yes everywhere.
Where does it become macroscopic? The crossing test above needs a window, and any fixed window biases the threshold deep, since near the transition the island is smaller than any fixed fraction of N. The window-free quantity is A = (largest solid core)/N²: A and its decay exponent. Fitting A ~ N−θ, the core covers N2−θ faces, so θ = 1 is a filament and θ = 0 a macroscopic island. Measured, θ falls from about 1 near Δ = −1 to 0.16 at Δ = −3 — filaments above, a thickening blob below — but with no sharp jump at −1 at these sizes.