exactly $L$ broken $V$-links. Similarly, if $\Delta_i=-3$ (as is the case between columns c and d in Fig. \[fig:nnnbonds\]), then there would be the same number of broken $H$-links, but there would now be $4$ broken $U$-links and $2$ broken $D$-links (that is, the number of broken $U$ and $D$ links switch from the previous case). From these examples we see that, in general, there are $\left| \Delta_i \right|$ broken $H$-links, $\left| \Delta_i - 1 \right|$ broken $U$-links, and $\left| \Delta_i +1 \right|$ broken $D$-links. It therefore follows that the step-edge energy is $$\begin{aligned}
  \label{eq:interfaceEnergy}
\frac{E(\{\Delta_i\})}{k_B T} &=&
\sum_{i=1}^L \Big(V + H\left| \Delta_i \right| + U\left| \Delta_i\! - \! 1\right| \nonumber + D\left| \Delta_i \! + \! 1\right| \Big)\\
&\equiv&
\sum_{i=1}^L K(\Delta_i).\end{aligned}$$

Because we seek the orientation dependence of $\beta$ and $\tilde{\beta}$, we constrain the step to have an overall offset $Y \equiv y_L - y_0 \equiv L \tan \theta = \sum_{i=1}^L \Delta_i$. (This constraint is represented in Fig. \[fig:nnnbonds\] by the shaded gray area. Equivalently, we specify that the overall slope of the step is $m \equiv \tan \theta$.) The constrained partition function is therefore $$\begin{aligned}
  \label{eq:partfunction}
  Z\left( Y\right) \equiv \sum_{\left\{ \Delta \right\} }\delta \left[
Y-\sum_{i=1}^L\Delta _i\right] e^{-E(\{\Delta_i\})/k_B T},\end{aligned}$$ 