Adjacency matrix of subdivision graph.
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
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
add a comment |
$begingroup$
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
$endgroup$
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
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $E$.
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3021590%2fadjacency-matrix-of-subdivision-graph%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $E$.
$endgroup$
add a comment |
$begingroup$
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 $E$.
$endgroup$
add a comment |
$begingroup$
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 $E$.
$endgroup$
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 $E$.
answered Dec 3 '18 at 4:12
Alex RavskyAlex Ravsky
39.9k32282
39.9k32282
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3021590%2fadjacency-matrix-of-subdivision-graph%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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