Sum of the degrees of second neighbours of a vertex in a graph
$begingroup$
The question is: "Using only matrix formalism find the vector $pmb{v}$ whose element $i$ is the sum of the degrees of vertex's $i$ second neighbours".
My attempt:
Let $A$ be the adjacent matrix of the graph. The vector $pmb{k}$ whose elements are the degree of the vertices is $pmb{k}=Acdotpmb{1}$ where $pmb{1}$ is a column vector off all ones. Now, to find the sum of the degrees of the second neighbours I need a proper matrix to rearrange and sum the elements of the vector $pmb{k}$. For instance if I want instead the sum of the degrees of the first neighbours the proper matrix is the adjacent matrix. Indeed $Acdotpmb{k}$ is the vector whose element $i$ is the sum of the degrees of vertex $i$-th.
I tried to use a matrix similar to: $A^2-diag(pmb{k})$, but, of course, it will not work due to repetitions and loops.
Thanks
graph-theory network adjacency-matrix
$endgroup$
add a comment |
$begingroup$
The question is: "Using only matrix formalism find the vector $pmb{v}$ whose element $i$ is the sum of the degrees of vertex's $i$ second neighbours".
My attempt:
Let $A$ be the adjacent matrix of the graph. The vector $pmb{k}$ whose elements are the degree of the vertices is $pmb{k}=Acdotpmb{1}$ where $pmb{1}$ is a column vector off all ones. Now, to find the sum of the degrees of the second neighbours I need a proper matrix to rearrange and sum the elements of the vector $pmb{k}$. For instance if I want instead the sum of the degrees of the first neighbours the proper matrix is the adjacent matrix. Indeed $Acdotpmb{k}$ is the vector whose element $i$ is the sum of the degrees of vertex $i$-th.
I tried to use a matrix similar to: $A^2-diag(pmb{k})$, but, of course, it will not work due to repetitions and loops.
Thanks
graph-theory network adjacency-matrix
$endgroup$
$begingroup$
Well, $Apmb{k}$ counts the degree of each vertex, $A^2pmb{k}$ counts the sum of the degrees of the neighbours of each vertex, and something like $A^3pmb{k} - diag^2(A pmb{k})$ counts the sum of the degrees of the second neighbours of each vertex, the problem being caused by multiple counting due to cycles of length 3 or 4. What was the exact wording of the question?
$endgroup$
– Mike
Oct 4 '18 at 20:22
add a comment |
$begingroup$
The question is: "Using only matrix formalism find the vector $pmb{v}$ whose element $i$ is the sum of the degrees of vertex's $i$ second neighbours".
My attempt:
Let $A$ be the adjacent matrix of the graph. The vector $pmb{k}$ whose elements are the degree of the vertices is $pmb{k}=Acdotpmb{1}$ where $pmb{1}$ is a column vector off all ones. Now, to find the sum of the degrees of the second neighbours I need a proper matrix to rearrange and sum the elements of the vector $pmb{k}$. For instance if I want instead the sum of the degrees of the first neighbours the proper matrix is the adjacent matrix. Indeed $Acdotpmb{k}$ is the vector whose element $i$ is the sum of the degrees of vertex $i$-th.
I tried to use a matrix similar to: $A^2-diag(pmb{k})$, but, of course, it will not work due to repetitions and loops.
Thanks
graph-theory network adjacency-matrix
$endgroup$
The question is: "Using only matrix formalism find the vector $pmb{v}$ whose element $i$ is the sum of the degrees of vertex's $i$ second neighbours".
My attempt:
Let $A$ be the adjacent matrix of the graph. The vector $pmb{k}$ whose elements are the degree of the vertices is $pmb{k}=Acdotpmb{1}$ where $pmb{1}$ is a column vector off all ones. Now, to find the sum of the degrees of the second neighbours I need a proper matrix to rearrange and sum the elements of the vector $pmb{k}$. For instance if I want instead the sum of the degrees of the first neighbours the proper matrix is the adjacent matrix. Indeed $Acdotpmb{k}$ is the vector whose element $i$ is the sum of the degrees of vertex $i$-th.
I tried to use a matrix similar to: $A^2-diag(pmb{k})$, but, of course, it will not work due to repetitions and loops.
Thanks
graph-theory network adjacency-matrix
graph-theory network adjacency-matrix
edited Jan 15 at 14:55
amWhy
1
1
asked Oct 4 '18 at 11:06
AlexAlex
32319
32319
$begingroup$
Well, $Apmb{k}$ counts the degree of each vertex, $A^2pmb{k}$ counts the sum of the degrees of the neighbours of each vertex, and something like $A^3pmb{k} - diag^2(A pmb{k})$ counts the sum of the degrees of the second neighbours of each vertex, the problem being caused by multiple counting due to cycles of length 3 or 4. What was the exact wording of the question?
$endgroup$
– Mike
Oct 4 '18 at 20:22
add a comment |
$begingroup$
Well, $Apmb{k}$ counts the degree of each vertex, $A^2pmb{k}$ counts the sum of the degrees of the neighbours of each vertex, and something like $A^3pmb{k} - diag^2(A pmb{k})$ counts the sum of the degrees of the second neighbours of each vertex, the problem being caused by multiple counting due to cycles of length 3 or 4. What was the exact wording of the question?
$endgroup$
– Mike
Oct 4 '18 at 20:22
$begingroup$
Well, $Apmb{k}$ counts the degree of each vertex, $A^2pmb{k}$ counts the sum of the degrees of the neighbours of each vertex, and something like $A^3pmb{k} - diag^2(A pmb{k})$ counts the sum of the degrees of the second neighbours of each vertex, the problem being caused by multiple counting due to cycles of length 3 or 4. What was the exact wording of the question?
$endgroup$
– Mike
Oct 4 '18 at 20:22
$begingroup$
Well, $Apmb{k}$ counts the degree of each vertex, $A^2pmb{k}$ counts the sum of the degrees of the neighbours of each vertex, and something like $A^3pmb{k} - diag^2(A pmb{k})$ counts the sum of the degrees of the second neighbours of each vertex, the problem being caused by multiple counting due to cycles of length 3 or 4. What was the exact wording of the question?
$endgroup$
– Mike
Oct 4 '18 at 20:22
add a comment |
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$begingroup$
Well, $Apmb{k}$ counts the degree of each vertex, $A^2pmb{k}$ counts the sum of the degrees of the neighbours of each vertex, and something like $A^3pmb{k} - diag^2(A pmb{k})$ counts the sum of the degrees of the second neighbours of each vertex, the problem being caused by multiple counting due to cycles of length 3 or 4. What was the exact wording of the question?
$endgroup$
– Mike
Oct 4 '18 at 20:22