Adjacency matrix of subdivision graph.












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Is there a way to get from an adjacency matrix of a simple graph to an arbitrary subdivision of the graph? I believe I found a way to get from $G$ to $G^{1/2}$ using the incidence matrix, but in particular I want a way to get from $G$ to $G^{1/3}$ without needing to manually type out 1600+ entries.






graph-theory algorithms







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$endgroup$








  • 1




    $begingroup$
    In the context of a graph, what do you mean by $G^{1/2}$ and $G^{1/3}$?
    $endgroup$
    – Misha Lavrov
    Dec 1 '18 at 17:56










  • $begingroup$
    @MishaLavrov, sorry I've been using that notation so long I thought it was standard. $G^{1/2}$ is the first subdivision graph (one vertex has been put on every edge), while $G^{1/3}$ is the second (two vertices have been put on every edge).
    $endgroup$
    – GraphicalTests
    Dec 2 '18 at 0:35
















1












$begingroup$


Is there a way to get from an adjacency matrix of a simple graph to an arbitrary subdivision of the graph? I believe I found a way to get from $G$ to $G^{1/2}$ using the incidence matrix, but in particular I want a way to get from $G$ to $G^{1/3}$ without needing to manually type out 1600+ entries.






graph-theory algorithms







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    In the context of a graph, what do you mean by $G^{1/2}$ and $G^{1/3}$?
    $endgroup$
    – Misha Lavrov
    Dec 1 '18 at 17:56










  • $begingroup$
    @MishaLavrov, sorry I've been using that notation so long I thought it was standard. $G^{1/2}$ is the first subdivision graph (one vertex has been put on every edge), while $G^{1/3}$ is the second (two vertices have been put on every edge).
    $endgroup$
    – GraphicalTests
    Dec 2 '18 at 0:35














1












1








1





$begingroup$


Is there a way to get from an adjacency matrix of a simple graph to an arbitrary subdivision of the graph? I believe I found a way to get from $G$ to $G^{1/2}$ using the incidence matrix, but in particular I want a way to get from $G$ to $G^{1/3}$ without needing to manually type out 1600+ entries.






graph-theory algorithms







share|cite|improve this question









$endgroup$




Is there a way to get from an adjacency matrix of a simple graph to an arbitrary subdivision of the graph? I believe I found a way to get from $G$ to $G^{1/2}$ using the incidence matrix, but in particular I want a way to get from $G$ to $G^{1/3}$ without needing to manually type out 1600+ entries.






graph-theory algorithms




graph-theory algorithms






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 1 '18 at 17:24









GraphicalTestsGraphicalTests

10515




10515








  • 1




    $begingroup$
    In the context of a graph, what do you mean by $G^{1/2}$ and $G^{1/3}$?
    $endgroup$
    – Misha Lavrov
    Dec 1 '18 at 17:56










  • $begingroup$
    @MishaLavrov, sorry I've been using that notation so long I thought it was standard. $G^{1/2}$ is the first subdivision graph (one vertex has been put on every edge), while $G^{1/3}$ is the second (two vertices have been put on every edge).
    $endgroup$
    – GraphicalTests
    Dec 2 '18 at 0:35














  • 1




    $begingroup$
    In the context of a graph, what do you mean by $G^{1/2}$ and $G^{1/3}$?
    $endgroup$
    – Misha Lavrov
    Dec 1 '18 at 17:56










  • $begingroup$
    @MishaLavrov, sorry I've been using that notation so long I thought it was standard. $G^{1/2}$ is the first subdivision graph (one vertex has been put on every edge), while $G^{1/3}$ is the second (two vertices have been put on every edge).
    $endgroup$
    – GraphicalTests
    Dec 2 '18 at 0:35








1




1




$begingroup$
In the context of a graph, what do you mean by $G^{1/2}$ and $G^{1/3}$?
$endgroup$
– Misha Lavrov
Dec 1 '18 at 17:56




$begingroup$
In the context of a graph, what do you mean by $G^{1/2}$ and $G^{1/3}$?
$endgroup$
– Misha Lavrov
Dec 1 '18 at 17:56












$begingroup$
@MishaLavrov, sorry I've been using that notation so long I thought it was standard. $G^{1/2}$ is the first subdivision graph (one vertex has been put on every edge), while $G^{1/3}$ is the second (two vertices have been put on every edge).
$endgroup$
– GraphicalTests
Dec 2 '18 at 0:35




$begingroup$
@MishaLavrov, sorry I've been using that notation so long I thought it was standard. $G^{1/2}$ is the first subdivision graph (one vertex has been put on every edge), while $G^{1/3}$ is the second (two vertices have been put on every edge).
$endgroup$
– GraphicalTests
Dec 2 '18 at 0:35










1 Answer
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I assume that the graphs $G$ and $G^{1/3}$ are undirected. Let $A=|a_{ij}|$ be an adjacency $ntimes n$-matrix of the graph $G$. Let $E={(i,j): i<j$ and $a_{ij}=1}$. To obtain an adjacency matrix for the graph $G^{1/3}$ we add to the matrix $A$ $2|E|$ rows and $2|E|$ columns. Each of these rows and columns is indexed by a $(i,j,t)$, where $(i,j)in E$ and $tin {1,2}$.
The only non-zero elements of the extended matrix $A$ are $a_{i, (i,j,1)}$, $a_{(i,j,1),i}$, $a_{j, (i,j,2)}$, $a_{(i,j,2),j}$, $a_{(i,j,1), (i,j,2)}$, and $a_{(i,j,2), (i,j,1)}$, where $i$ ranges over $1,dots, n$ and $(i,j)$ ranges over $E$.






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    $begingroup$

    I assume that the graphs $G$ and $G^{1/3}$ are undirected. Let $A=|a_{ij}|$ be an adjacency $ntimes n$-matrix of the graph $G$. Let $E={(i,j): i<j$ and $a_{ij}=1}$. To obtain an adjacency matrix for the graph $G^{1/3}$ we add to the matrix $A$ $2|E|$ rows and $2|E|$ columns. Each of these rows and columns is indexed by a $(i,j,t)$, where $(i,j)in E$ and $tin {1,2}$.
    The only non-zero elements of the extended matrix $A$ are $a_{i, (i,j,1)}$, $a_{(i,j,1),i}$, $a_{j, (i,j,2)}$, $a_{(i,j,2),j}$, $a_{(i,j,1), (i,j,2)}$, and $a_{(i,j,2), (i,j,1)}$, where $i$ ranges over $1,dots, n$ and $(i,j)$ ranges over $E$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      I assume that the graphs $G$ and $G^{1/3}$ are undirected. Let $A=|a_{ij}|$ be an adjacency $ntimes n$-matrix of the graph $G$. Let $E={(i,j): i<j$ and $a_{ij}=1}$. To obtain an adjacency matrix for the graph $G^{1/3}$ we add to the matrix $A$ $2|E|$ rows and $2|E|$ columns. Each of these rows and columns is indexed by a $(i,j,t)$, where $(i,j)in E$ and $tin {1,2}$.
      The only non-zero elements of the extended matrix $A$ are $a_{i, (i,j,1)}$, $a_{(i,j,1),i}$, $a_{j, (i,j,2)}$, $a_{(i,j,2),j}$, $a_{(i,j,1), (i,j,2)}$, and $a_{(i,j,2), (i,j,1)}$, where $i$ ranges over $1,dots, n$ and $(i,j)$ ranges over $E$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        I assume that the graphs $G$ and $G^{1/3}$ are undirected. Let $A=|a_{ij}|$ be an adjacency $ntimes n$-matrix of the graph $G$. Let $E={(i,j): i<j$ and $a_{ij}=1}$. To obtain an adjacency matrix for the graph $G^{1/3}$ we add to the matrix $A$ $2|E|$ rows and $2|E|$ columns. Each of these rows and columns is indexed by a $(i,j,t)$, where $(i,j)in E$ and $tin {1,2}$.
        The only non-zero elements of the extended matrix $A$ are $a_{i, (i,j,1)}$, $a_{(i,j,1),i}$, $a_{j, (i,j,2)}$, $a_{(i,j,2),j}$, $a_{(i,j,1), (i,j,2)}$, and $a_{(i,j,2), (i,j,1)}$, where $i$ ranges over $1,dots, n$ and $(i,j)$ ranges over $E$.






        share|cite|improve this answer









        $endgroup$



        I assume that the graphs $G$ and $G^{1/3}$ are undirected. Let $A=|a_{ij}|$ be an adjacency $ntimes n$-matrix of the graph $G$. Let $E={(i,j): i<j$ and $a_{ij}=1}$. To obtain an adjacency matrix for the graph $G^{1/3}$ we add to the matrix $A$ $2|E|$ rows and $2|E|$ columns. Each of these rows and columns is indexed by a $(i,j,t)$, where $(i,j)in E$ and $tin {1,2}$.
        The only non-zero elements of the extended matrix $A$ are $a_{i, (i,j,1)}$, $a_{(i,j,1),i}$, $a_{j, (i,j,2)}$, $a_{(i,j,2),j}$, $a_{(i,j,1), (i,j,2)}$, and $a_{(i,j,2), (i,j,1)}$, where $i$ ranges over $1,dots, n$ and $(i,j)$ ranges over $E$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 3 '18 at 4:12









        Alex RavskyAlex Ravsky

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