Polygon recutting singularity position conjecture

Let $m\in\mathbb{Z}$ and let $\tilde z$ be an $m$-closed polygon recutting lattice map, meaning $\tilde z_{i,j}=\tilde z_{i+m,j-m}$ for all $i,j$. Assume $\tilde z_{i,0}=0$ for all $i\in 2\mathbb{Z}$, define $y_i=\tilde z_{2i+1,1}$ and $\varrho_i=\ell_{2i}=\ell_{2i+1}$, and let $\mathcal I_k$ be the set of $k$-subsets of $\{0,1,\dots,m-1\}$. Assume polygon recutting $T$ can be applied to $\tilde z$ at least $m-1$ times. Polygon recutting singularity position conjecture. If $m\in 2\mathbb{Z}+1$ and $k=(m-1)/2$, then, for every $i\in 2\mathbb{Z}+1$, $\tilde z_{i,m}$ equals the stated ratio of sums over $\mathcal I_k$; if $m\in 2\mathbb{Z}$ and $k=m/2$, then, for every $i\in 2\mathbb{Z}$, $\tilde z_{i,m}$ equals the stated ratio of sums over $\mathcal I_{k-1}$ and $\mathcal I_k$. Explicitly, the odd case is

$$\tilde z_{i,m}=(-1)^k\frac{\sum\limits_{I\in\mathcal I_k}\left[\prod\limits_{r\in I}(-1)^r\prod\limits_{r\notin I}y_r\prod\limits_{r,s\in I,r<s}(\varrho_r^2-\varrho_s^2)\prod\limits_{r,s\notin I,r<s}(\varrho_r^2-\varrho_s^2)\right]}{\sum\limits_{I\in\mathcal I_k}\left[\prod\limits_{r\in I}(-1)^r\prod\limits_{r\in I}y_r\prod\limits_{r,s\in I,r<s}(\varrho_r^2-\varrho_s^2)\prod\limits_{r,s\notin I,r<s}(\varrho_r^2-\varrho_s^2)\right]}.$$

The even case is

$$\tilde z_{i,m}=(-1)^{k+1}\frac{\sum\limits_{I\in\mathcal I_{k-1}}\left[\prod\limits_{r\in I}(-1)^r\varrho_r^2\prod\limits_{r\notin I}y_r\prod\limits_{r,s\in I,r<s}(\varrho_r^2-\varrho_s^2)\prod\limits_{r,s\notin I,r<s}(\varrho_r^2-\varrho_s^2)\right]}{\sum\limits_{I\in\mathcal I_k}\left[\prod\limits_{r\in I}(-1)^r\varrho_r^2\prod\limits_{r\in I}y_r\prod\limits_{r,s\in I,r<s}(\varrho_r^2-\varrho_s^2)\prod\limits_{r,s\notin I,r<s}(\varrho_r^2-\varrho_s^2)\right]}.$$

The conjecture gives an explicit formula for the position of the singularity following the known polygon recutting Devron phenomenon; the source does not state whether these formulas have been proved.