Mixing
Definition of 2-factorable Graph Theory
$begingroup$
I would appreciate a simpler explanation of the following info
I am confused on what the definition of 2-factor is. It sounds like a perfect matching to me, but that's not what this paper is saying
graph-theory
$endgroup$
add a comment |
$begingroup$
I would appreciate a simpler explanation of the following info
I am confused on what the definition of 2-factor is. It sounds like a perfect matching to me, but that's not what this paper is saying
graph-theory
$endgroup$
1
$begingroup$
This seems like a non-standard use of "2-factorable" as applied to the coloring; there's no stated requirement in the contracted graph to use every edge across a set of subgraphs. Perhaps that also explains the non-standard use of "2-cycle" which would not usually be allowed in a 2-factor.
$endgroup$
– Joffan
Apr 19 '17 at 16:15
add a comment |
$begingroup$
I would appreciate a simpler explanation of the following info
I am confused on what the definition of 2-factor is. It sounds like a perfect matching to me, but that's not what this paper is saying
graph-theory
$endgroup$
I would appreciate a simpler explanation of the following info
I am confused on what the definition of 2-factor is. It sounds like a perfect matching to me, but that's not what this paper is saying
graph-theory
graph-theory
edited Apr 20 '17 at 20:56
asked Apr 19 '17 at 15:51
user424929
1
$begingroup$
This seems like a non-standard use of "2-factorable" as applied to the coloring; there's no stated requirement in the contracted graph to use every edge across a set of subgraphs. Perhaps that also explains the non-standard use of "2-cycle" which would not usually be allowed in a 2-factor.
$endgroup$
– Joffan
Apr 19 '17 at 16:15
add a comment |
1
$begingroup$
This seems like a non-standard use of "2-factorable" as applied to the coloring; there's no stated requirement in the contracted graph to use every edge across a set of subgraphs. Perhaps that also explains the non-standard use of "2-cycle" which would not usually be allowed in a 2-factor.
$endgroup$
– Joffan
Apr 19 '17 at 16:15
1
1
$begingroup$
This seems like a non-standard use of "2-factorable" as applied to the coloring; there's no stated requirement in the contracted graph to use every edge across a set of subgraphs. Perhaps that also explains the non-standard use of "2-cycle" which would not usually be allowed in a 2-factor.
$endgroup$
– Joffan
Apr 19 '17 at 16:15
$begingroup$
This seems like a non-standard use of "2-factorable" as applied to the coloring; there's no stated requirement in the contracted graph to use every edge across a set of subgraphs. Perhaps that also explains the non-standard use of "2-cycle" which would not usually be allowed in a 2-factor.
$endgroup$
– Joffan
Apr 19 '17 at 16:15
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
A graph $G$ is 2-factorable if and only if $G$ is $2k-regular$ for some positive integer $k$.
A perfect matching is 1-factorable. If $M$ is a perfect matching in a graph $G$, then $G[M]$ is a 1-regular spanning subgraph of $G$.
If you want a simple way of determining if a graph is 2-factorable, it must be able to be factored into Hamiltonian cycles.
answered Apr 19 '17 at 17:48
AnonymousAnonymous
74
$endgroup$
$begingroup$
Your last sentence is false: a $2$-factor is any $2$-regular spanning subgraph which is a union of disjoint cycles, not necessarily a Hamiltonian cycle. This post has an example of a $4$-regular (and therefore $2$-factorable) graph which is not Hamiltonian.
$endgroup$
– Misha Lavrov
Apr 20 '17 at 21:13
add a comment |
$begingroup$
Ordinarily... a $k$-factor is a $k$-regular spanning subgraph. Thus
A $1$-factor is a perfect matching.
A $2$-factor will be a spanning subgraph which is a union of disjoint cycles.
This is the ordinary definition used in, say, Petersen's $2$-factor Theorem.
So here's an example of a graph with a $2$-factor highlighted as colored cycles:
A $k$-factorization is a decomposition into $k$-factors. A graph is $k$-factorable if it admits a $k$-factorization.
The above graph isn't $2$-factorable because it has vertices of odd degree.
Judging from the (now deleted) definition, the author:
includes isolated edges as $2$-cycles, and
defines a graph as "$2$-factorable" if there exists a $2$-factor for which after contracting the cycles, we obtain another graph with a $2$-factor.
According to this definition, a $1$-factor (perfect matching) would also be a $2$-factor. And the above graph would be $2$-factorable.
I've never seen this definition before.
answered Apr 25 '17 at 4:15
Rebecca J. StonesRebecca J. Stones
21k22781
$endgroup$
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%2f2242030%2fdefinition-of-2-factorable-graph-theory%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A graph $G$ is 2-factorable if and only if $G$ is $2k-regular$ for some positive integer $k$.
A perfect matching is 1-factorable. If $M$ is a perfect matching in a graph $G$, then $G[M]$ is a 1-regular spanning subgraph of $G$.
If you want a simple way of determining if a graph is 2-factorable, it must be able to be factored into Hamiltonian cycles.
answered Apr 19 '17 at 17:48
AnonymousAnonymous
74
$endgroup$
$begingroup$
Your last sentence is false: a $2$-factor is any $2$-regular spanning subgraph which is a union of disjoint cycles, not necessarily a Hamiltonian cycle. This post has an example of a $4$-regular (and therefore $2$-factorable) graph which is not Hamiltonian.
$endgroup$
– Misha Lavrov
Apr 20 '17 at 21:13
add a comment |
$begingroup$
A graph $G$ is 2-factorable if and only if $G$ is $2k-regular$ for some positive integer $k$.
A perfect matching is 1-factorable. If $M$ is a perfect matching in a graph $G$, then $G[M]$ is a 1-regular spanning subgraph of $G$.
If you want a simple way of determining if a graph is 2-factorable, it must be able to be factored into Hamiltonian cycles.
answered Apr 19 '17 at 17:48
AnonymousAnonymous
74
$endgroup$
$begingroup$
Your last sentence is false: a $2$-factor is any $2$-regular spanning subgraph which is a union of disjoint cycles, not necessarily a Hamiltonian cycle. This post has an example of a $4$-regular (and therefore $2$-factorable) graph which is not Hamiltonian.
$endgroup$
– Misha Lavrov
Apr 20 '17 at 21:13
add a comment |
$begingroup$
A graph $G$ is 2-factorable if and only if $G$ is $2k-regular$ for some positive integer $k$.
A perfect matching is 1-factorable. If $M$ is a perfect matching in a graph $G$, then $G[M]$ is a 1-regular spanning subgraph of $G$.
If you want a simple way of determining if a graph is 2-factorable, it must be able to be factored into Hamiltonian cycles.
answered Apr 19 '17 at 17:48
AnonymousAnonymous
74
$endgroup$
A graph $G$ is 2-factorable if and only if $G$ is $2k-regular$ for some positive integer $k$.
A perfect matching is 1-factorable. If $M$ is a perfect matching in a graph $G$, then $G[M]$ is a 1-regular spanning subgraph of $G$.
If you want a simple way of determining if a graph is 2-factorable, it must be able to be factored into Hamiltonian cycles.
answered Apr 19 '17 at 17:48
AnonymousAnonymous
74
answered Apr 19 '17 at 17:48
AnonymousAnonymous
74
answered Apr 19 '17 at 17:48
AnonymousAnonymous
74
answered Apr 19 '17 at 17:48
AnonymousAnonymous
74
74
$begingroup$
Your last sentence is false: a $2$-factor is any $2$-regular spanning subgraph which is a union of disjoint cycles, not necessarily a Hamiltonian cycle. This post has an example of a $4$-regular (and therefore $2$-factorable) graph which is not Hamiltonian.
$endgroup$
– Misha Lavrov
Apr 20 '17 at 21:13
add a comment |
$begingroup$
Your last sentence is false: a $2$-factor is any $2$-regular spanning subgraph which is a union of disjoint cycles, not necessarily a Hamiltonian cycle. This post has an example of a $4$-regular (and therefore $2$-factorable) graph which is not Hamiltonian.
$endgroup$
– Misha Lavrov
Apr 20 '17 at 21:13
$begingroup$
Your last sentence is false: a $2$-factor is any $2$-regular spanning subgraph which is a union of disjoint cycles, not necessarily a Hamiltonian cycle. This post has an example of a $4$-regular (and therefore $2$-factorable) graph which is not Hamiltonian.
$endgroup$
– Misha Lavrov
Apr 20 '17 at 21:13
$begingroup$
Your last sentence is false: a $2$-factor is any $2$-regular spanning subgraph which is a union of disjoint cycles, not necessarily a Hamiltonian cycle. This post has an example of a $4$-regular (and therefore $2$-factorable) graph which is not Hamiltonian.
$endgroup$
– Misha Lavrov
Apr 20 '17 at 21:13
add a comment |
$begingroup$
Ordinarily... a $k$-factor is a $k$-regular spanning subgraph. Thus
A $1$-factor is a perfect matching.
A $2$-factor will be a spanning subgraph which is a union of disjoint cycles.
This is the ordinary definition used in, say, Petersen's $2$-factor Theorem.
So here's an example of a graph with a $2$-factor highlighted as colored cycles:
A $k$-factorization is a decomposition into $k$-factors. A graph is $k$-factorable if it admits a $k$-factorization.
The above graph isn't $2$-factorable because it has vertices of odd degree.
Judging from the (now deleted) definition, the author:
includes isolated edges as $2$-cycles, and
defines a graph as "$2$-factorable" if there exists a $2$-factor for which after contracting the cycles, we obtain another graph with a $2$-factor.
According to this definition, a $1$-factor (perfect matching) would also be a $2$-factor. And the above graph would be $2$-factorable.
I've never seen this definition before.
answered Apr 25 '17 at 4:15
Rebecca J. StonesRebecca J. Stones
21k22781
$endgroup$
add a comment |
$begingroup$
Ordinarily... a $k$-factor is a $k$-regular spanning subgraph. Thus
A $1$-factor is a perfect matching.
A $2$-factor will be a spanning subgraph which is a union of disjoint cycles.
This is the ordinary definition used in, say, Petersen's $2$-factor Theorem.
So here's an example of a graph with a $2$-factor highlighted as colored cycles:
A $k$-factorization is a decomposition into $k$-factors. A graph is $k$-factorable if it admits a $k$-factorization.
The above graph isn't $2$-factorable because it has vertices of odd degree.
Judging from the (now deleted) definition, the author:
includes isolated edges as $2$-cycles, and
defines a graph as "$2$-factorable" if there exists a $2$-factor for which after contracting the cycles, we obtain another graph with a $2$-factor.
According to this definition, a $1$-factor (perfect matching) would also be a $2$-factor. And the above graph would be $2$-factorable.
I've never seen this definition before.
answered Apr 25 '17 at 4:15
Rebecca J. StonesRebecca J. Stones
21k22781
$endgroup$
add a comment |
$begingroup$
Ordinarily... a $k$-factor is a $k$-regular spanning subgraph. Thus
A $1$-factor is a perfect matching.
A $2$-factor will be a spanning subgraph which is a union of disjoint cycles.
This is the ordinary definition used in, say, Petersen's $2$-factor Theorem.
So here's an example of a graph with a $2$-factor highlighted as colored cycles:
A $k$-factorization is a decomposition into $k$-factors. A graph is $k$-factorable if it admits a $k$-factorization.
The above graph isn't $2$-factorable because it has vertices of odd degree.
Judging from the (now deleted) definition, the author:
includes isolated edges as $2$-cycles, and
defines a graph as "$2$-factorable" if there exists a $2$-factor for which after contracting the cycles, we obtain another graph with a $2$-factor.
According to this definition, a $1$-factor (perfect matching) would also be a $2$-factor. And the above graph would be $2$-factorable.
I've never seen this definition before.
answered Apr 25 '17 at 4:15
Rebecca J. StonesRebecca J. Stones
21k22781
$endgroup$
Ordinarily... a $k$-factor is a $k$-regular spanning subgraph. Thus
A $1$-factor is a perfect matching.
A $2$-factor will be a spanning subgraph which is a union of disjoint cycles.
This is the ordinary definition used in, say, Petersen's $2$-factor Theorem.
So here's an example of a graph with a $2$-factor highlighted as colored cycles:
A $k$-factorization is a decomposition into $k$-factors. A graph is $k$-factorable if it admits a $k$-factorization.
The above graph isn't $2$-factorable because it has vertices of odd degree.
Judging from the (now deleted) definition, the author:
includes isolated edges as $2$-cycles, and
defines a graph as "$2$-factorable" if there exists a $2$-factor for which after contracting the cycles, we obtain another graph with a $2$-factor.
According to this definition, a $1$-factor (perfect matching) would also be a $2$-factor. And the above graph would be $2$-factorable.
I've never seen this definition before.
answered Apr 25 '17 at 4:15
Rebecca J. StonesRebecca J. Stones
21k22781
answered Apr 25 '17 at 4:15
Rebecca J. StonesRebecca J. Stones
21k22781
answered Apr 25 '17 at 4:15
Rebecca J. StonesRebecca J. Stones
21k22781
answered Apr 25 '17 at 4:15
Rebecca J. StonesRebecca J. Stones
21k22781
21k22781
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%2f2242030%2fdefinition-of-2-factorable-graph-theory%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
$begingroup$
This seems like a non-standard use of "2-factorable" as applied to the coloring; there's no stated requirement in the contracted graph to use every edge across a set of subgraphs. Perhaps that also explains the non-standard use of "2-cycle" which would not usually be allowed in a 2-factor.
$endgroup$
– Joffan
Apr 19 '17 at 16:15