Assuming a graph $G$ with $N$ nodes distributed in a $\mathcal{L}\times\mathcal{L}$ area randomly. There is an edge between two nodes if and only if the Euler distance between them is equal or less than $\mathcal{R}$. The adjacent matrix is $A$. We define an operation on adjacent matrix "$\circ$". For adjacent matrix $A$ and $B$ $$C=A\circ B$$ $$ c_{ij}=\left\{ \begin{array}{rcl} 0 & & {\sum_{k=1}^{N}a_{ik}b_{kj}=0}\\ \\ 1 & & {\sum_{k=1}^{N}a_{ik}b_{kj}>0} \end{array} \right. $$ $A^{[1]}=A$, $A^{[i+1]}=A^{[i]}\circ A$. Define the intial adjacent matrix $A_0=A$. We select a pair of elements randomly from $A_i$ :$A_i(m,n)$ and $A_i(n,m)$. $A_{i+1}(m,n)=A_{i+1}(n,m)=1A_{i}(m,n)=1A_{i}(n,m)$. Other element of $A_{i+1}$ is equal to the corresponding element of $A_i$. Then we can get a series of matrix $A_0, A_1, \cdots, A_s$. There comes the question: $A_0^{[\infty]} \circ A_1^{[\infty]} \circ A_2^{[\infty]} \circ \cdots \circ A_s^{[\infty]}$=$?$ I have waited for more than a week? Could anyone help?

First, observe a few facts:
All of the above are easily proven. Here we claim that if the added/removed edge is a bridge, then $A_i^{[\infty]} \circ A_{i+1}^{[\infty]}$ equals to the operand that corresponds to the graph with the extra edge. Now we prove the stronger proposition that if $G_B$ is obtained from $G_A$ by addition of some number of edges, then $A^{[\infty]} \circ B^{[\infty]} = B^{[\infty]}$. Let $v_i$ and $v_j$ be two arbitrary vertices. There are three cases for $v_i$ and $v_j$:
That completes the proof that $A^{[\infty]} \circ B^{[\infty]}=B^{[\infty]}$. By this proposition, it can be proven by induction that $A_0^{[\infty]} \circ A_1^{[\infty]} \circ A_2^{[\infty]} \circ \cdots \circ A_s^{[\infty]}$ equal to the transitive closure of the graph that has every edge that any of the $G_{A_i}$ had. The exact value, of course, can not be evaluated without knowledge of $s$ and the way we choose the edges to be added/removed. 

