# Understanding the transition rule for the Markov chain in the JSV algorithm for approximating the permanent

I was making my way through the paper by Jerrum, Sinclair, and Vigoda on developing a randomized polynomial time procedure (FRPAS) for approximating the permanent of a matrix $$A$$ with non-negative entries. The paper mainly argues with the case when the entries of $$A$$ are in $$\{0,1 \}$$ and aims to approximate the number of perfect matchings of the bipartite graph $$G = (V, E)$$ with adjacency matrix $$\begin{bmatrix}0&A \\ A^{T}&0 \end{bmatrix}.$$

The paper does this by introducing a Markov chain with sample space $$\Omega = \mathcal{M} \cup_{u,v \in V} \mathcal{M}(u,v)$$ where $$\mathcal{M}$$ is the set of perfect matchings and $$\mathcal{M}(u,v)$$ is the set of "near perfect" matchings, these are matchings that cover all the vertices except the vertices $$u$$ and $$v$$. The Markov chain proceeds from a state $$M$$ either to state $$M'$$ or remains at state $$M$$

The transition rule and state $$M'$$ is given on pg 678 of the work. I'm having trouble understanding the case where if you are at a state $$M \in \mathcal{M}(u,v)$$, in particular, the cases 2ii and 2iii.

It says that if you are at a state $$M \in \mathcal{M}(u,v)$$ then you choose $$z$$ uniformly from $$V = V_{1} \cup V_{2}$$. If $$z \notin \{u, v \}$$ then $$z \in V_{1}$$ or $$z \in V_{2}$$. For example, if $$z \in V_{1}$$ and the edge $$(z,v) \in E$$ then $$M' = M \cup (z,v) \setminus (z,y).$$

I just don't understand what $$y$$ is here?