hexagon model is $$\begin{picture}(20,40)(-10,0)
\put(0,20){\line(0,1){20}}
\put(0,10){\circle{20}}
\put(0,20){\circle*{4}}
\put(0,40){\circle*{4}}
\end{picture}$$ This diagram is to be understood as follows. If the system is in the bottom state at position $i$, that is $\sigma_i=0$, then at position $i+1$ it can either be in the bottom state again so that $\sigma_{i+1}=0$ as indicated by the loop or it can be in the top state which means $\sigma_{i+1}=1$ as indicated by the line. On the other hand, if the system is in the top state at position $i$ then at the next position it has to be in the bottom state, meaning that $\sigma_i=1$ implies $\sigma_{i+1}=0$. This corresponds to condition on paths that $\sigma_i\sigma_{i+1}=0$ for all $i$. Up to a ${\mathbb{Z}}_2$ symmetry this state diagram is isomorphic to $$\label{equiv}
\raisebox{-.3cm}{
\begin{picture}(20,40)(-10,0)
\put(0,20){\line(0,1){20}}
\put(0,10){\circle{20}}
\put(0,20){\circle*{4}}
\put(0,40){\circle*{4}}
\end{picture}
}
\quad\cong\quad
\raisebox{-1cm}{
\begin{picture}(4,60)(-2,0)
\put(0,0){\line(0,1){60}}
\put(0,0){\circle*{4}}
\put(0,20){\circle*{4}}
\put(0,40){\circle*{4}}
\put(0,60){\circle*{4}}
\end{picture}}
\;\;\Big{/}\;\; {\mathbb{Z}}_2.$$ Under this correspondence the paths of the hard hexagon model become paths in a strip of height 4 with steps going up or down at each position. For example, with the convention that the paths start at height three, the path in Figure \[fig path\] becomes the path in Figure \[fig:newpath\].

(210,80)(-10,0) (10,60)[(1,1)[20]{}]{} (30,80)[(1,-1)[20]{}]{} (50,60)[(1,-1)[20]{}]{} (70,40)[(1,1)[20]{}]{} (90,60)[(1,1)[20]{}]{} 