perpendicular case $\Delta_{i_0} \! =\! +2$ and $\Delta_{i_0+1} \! =\! -2$, implying an added energy of $4H + 4(U\! +\!D)$. Counting bonds we see that the parallel configuration has one more $\epsilon_1$ bond and two more $\epsilon_2$ bonds than the perpendicular configuration. Invoking $H \! =\! -\epsilon_1/2k_BT$, we see that $U\! +\!D \! =\! -\epsilon_2/k_BT$; if $U\! =\!D$, then $D \! =\! -\epsilon_2/2k_BT$. The factor-of-2 difference between broken links in Eq. (\[eq:interfaceEnergy\]) and broken bonds was noted (for H links) already in the classic exposition by Leamy et al.[@Leamy] An alternate argument, presented over a decade ago,[@ZELD] for this factor of 2 is that the ragged edge is created by severing bonds along the selected path through an infinite square. This leads to the formation of two complementary irregular boundary layers (with opposite values of $\{\Delta_{i}\}$, so that the associated energy of each is half that of the broken bonds (at least when $U\! =\!D$).

Evaluation of the Free Energy
-----------------------------

As detailed in the first part of the Appendix, the sum in the Fourier transform of $Z(Y)$, which we denote by $W(\mu )$, factorizes. Thus, it can be written as $$W(\mu )=\exp \left[ -Lg(i\mu )/k_BT\right] ,$$ where $g(i\mu )$ 