**Introduction:**
Given a matrix A of a  $k$ regular graph G. The matrix A can be divided into 4 sub matrices  based on adjacency of vertex $x \in G$.
 $A_x$ is the symmetric matrix of the graph $(G-x)$, where $C$ is the symmetric matrix of the graph created by  vertices of $(G-x)$ which are  adjacent to $x$ and $D$ is the symmetric matrix of the graph created by  vertices of $(G-x)$ which are not adjacent to $x$. 

$$ A_x 
=
\left( \begin{array}{ccc}
C & E & 0 \\
E^{T} & D & 1\\
0 & 1 & 0\\
\end{array} \right) 
$$


It should be noted that

  1. Interchanging/swapping any two rows (or columns) of $C$ does not affect matrix $D$ (and vice versa).

  2. Any change in $C$ or $D$ or both $C$ and $D$ changes matrix $E$.


 
If some vertices of $G$ is rearranged (i.e., permuted), $A$  will be different, say, ***this new matrix is $B$. Again, matrix $B$ can be divided into 4 sub matrices  based on adjacency of vertex $x \in G$ and $ B_x$ can be obtained.***
$ B_x= \left( \begin{array}{cc}
S & R \\
R^{T} & Q\\
 \end{array} \right) $
<br>
**Given:** Both matrices $A,B$ are divided  based on same vertex $x$, and $E,R$ are zero matrices or matrices of all 1's. In order to get $A_x=B_x $, it must be shown , that,    $D=Q$ and $S=C$  (so that $A_x=B_x $ happen).
<br>
**Claim:**   ***To get $A=B$, reordering of $Q,S$(so that $A_x=B_x $ happens) can be done independently i.e. the total complexity of reordering$B_x$=complexity of reordering$S$+ complexity of reordering$Q$(both are independently added to the total complexity, not multiplied.  )   ***
<br>

<br>
**Question:**  Is this claim  correct?
This question is related to [this post][1].<br>
*** Please inform if something is not defined properly or unclear or miss-tagged. Also if you vote up/down it would be helpfull if you leave a comment.***  



  [1]:http://math.stackexchange.com/questions/1240637/counting-problem-of-combinations-of-symmetric-matrix