# Possibility of Disconnected Subgraphs of a $k$ Connected $r$ regular Graph under a given condition

Context: Given a adjacency matrix A of a $r$-regular graph $G$ (not complete graph $K_{r+1}$) . $G$ is $k$ connected.

The matrix A can be divided into 4 sub-matrices based on adjacency of vertex $x \in G$. $A_x$ is the adjacency matrix of the graph $(G-x)$, where $C$ is the adjacency matrix of the graph created by vertices which are not adjacent to $x$, and $D$ is the adjacency matrix of the graph created by vertices which are adjacent to $x$. $C,D$ are sub-graphs(regular) of graph $G$, $|V(C)|>|V(D)|$ where $|V(C)|,|V(D)|$ are total vertices number of graph $C,D$ respectively. $$A_x = \left( \begin{array}{ccc} C & E \\ E^{T} & D \\ \end{array} \right)$$

again, this process can be done recursively, where $A_{y+1}=D_y$, $y=$ iteration number of the recursive process. $$A_yx = \left( \begin{array}{ccc} C_y & E_y \\ E_y^{T} & D_{y} \\ \end{array} \right)$$

$A_x$(=$A_1x$) is the matrix of 1st iteration, for 2nd iteration, $A$ matrix would be $A_2x$. $x$ is always a vertex of $A_yx$.

This recursive process can be done a maximum of $\log_2(|V(G)|)$ times.

Question: In this recursive process (Given that, $C_y , D_y$ are always regular in this recursive process, for each iteration $y$),

1) Is it possible to have an $E_y$ matrix as a zero matrix, i.e., is it possible to have disconnected sub-graphs $C_y,D_y$ under the given condition that $G$ is $k$ connected and $C_y , D_y$ are always regular in this recursive process?

2) If possible, then what is the maximum number of happening this incident (e.g $|V(G)|/2$ times maximum)?

This question is connected to this post.

Request: Please, inform if anything is unclear/undefined . Also, inform, if the question is miss-tagged.Thanks.

• Jim, please limit the number of edits you make to a post. It might be better to write an offline draft first, and think about it for a while, rather than require >10 edits. – Scott Morrison Jul 26 '15 at 7:17
• @ScottMorrison , duly noted. – Jim Jul 26 '15 at 7:19