The distance on a graph In page 14 of this article by Cushing et al. (2017), the Wassestein distance for the edge $(x,y)$ in the first case (when the edge lie on a triangle) was bounded $\leq \frac{2}{3}$ !!
It is known that the Wasserstein distance $$W_d(m_x,m_y)=min \sum \sum p(x,y)d(x,y)$$
Since the graph is $3-$regular then the degree of each vertex $d=3$ so $p(x,y)= \frac{1}{3}$. Then ,as I understood, we need to bound the minimum number of possible moves from both vertices $x \& y$ on the edge $(x,y)$ But I don't understand why this minimum number of moves is $\leq 2$ so that $$W_d(m_x,m_y)\leq 2\cdot\frac{1}{3}$$
So my problem how they counted the moves to $2$.
Many thanks for any kind of help.
 A: 
It is known that the Wasserstein distance $$W_d(m_x,m_y)=min \sum \sum p(x,y)d(x,y)$$
  Since the graph is $3-$regular then the degree of each vertex $d=3$ so
  $p(x,y)= \frac{1}{3}$.
  Then ,as I understood, we need to bound the
  minimum number of possible moves from both vertices $x \& y$ on the
  edge $(x,y)$

There don't seem to be enough variables in the version of the Wasserstein distance which you quote – or, more precisely, the $x$ on the left is not the same as the one on the right. In particular, you can't say $p(x,y) = \frac13$ because the $x,y$ of $p(x,y)$ are the bound variables of the sums.
Taking the notation of the paper, but renaming some variables to avoid confusion, $$W_1(\mu_x, \mu_y) = \inf_{\pi} \sum_{v\in V} \sum_{u\in V} d(u, v) \pi(u, v)$$
We have $\mu_x(u) = [x \sim u]\frac13$ (and similarly $\mu_y$), and the infimum is over $\pi : V \times V \to [0, 1]$ satisfying $\mu_x(u) = \sum_{v \in V} \pi(u, v)$ and $\mu_y(v) = \sum_{u \in V} \pi(u, v)$. Therefore we can easily conclude that $x \not\sim u \implies \forall v\in V:\pi(u, v) = 0$. (Note: this includes the case $x = u$, since there are no self-loops). Conversely, $\pi(u, v) > 0 \implies x \sim u$. Similarly, $\pi(u, v) > 0 \implies y \sim v$.
$x$ has degree $3$ and edges to $y$ and $z$. Call the third vertex to which it has an edge $s$. Similarly $y \sim t$. (Note: $s$ and $t$ are not necessarily distinct).
Then $\pi$ is only non-zero on $\{s,y,z\} \times \{t,x,z\}$. If we put all of the weight of $\pi$ on $\pi(s,x) = \pi(y,t) = \pi(z,z) = \frac13$ then since $d(s,x) = d(y,t) = 1$ and $d(z, z) = 0$ we get $W_1(\mu_x, \mu_y) \le 2 \cdot \frac13$
So if that $2$ counts anything then it isn't a number of moves but a number of distances of length $1$ to which we're able to assign all of the mass in the transport plan.

As an added bonus, we can say when the bound is tight. Consider the values of $d(u, v)$ for the possibly relevant pairs: $$\begin{array}{c|ccc}
u \setminus v & t & x & z \\
\hline
s & ? & 1 & ? \\
y & 1 & 1 & 1 \\
z & ? & 1 & 0 \\
\end{array}$$
We have to assign $\pi$ such that we put $\frac13$ in each row and column, then weight by those distances and sum. Clearly the lower bound on $W_1$ is $\frac13$, contributed by the middle row (or the middle column) however we assign its weight. If $s=t$ then that lower bound is achievable. Otherwise at most one of $s$ or $t$ can equal $z$, so we have either two rows or two columns with minimum distance $1$, and the bound of $\frac23$ is tight.
