Isomorphism in regular graphs
up vote
0
down vote
favorite
Let $G$ be a connected regular graph. Consider two different vertices $u,v$ of $G$. Let $H_1$ be the graph obtained from $G$ by deleting vertex $u$, and $H_2$ be the graph obtained from $G$ by deleting vertex $v$. Is it necessary that $H_1$ and $H_2$ are isomorphic?
graph-theory graph-isomorphism
asked Nov 27 at 13:42
Ranveer Singh
199110
add a comment |
up vote
0
down vote
favorite
Let $G$ be a connected regular graph. Consider two different vertices $u,v$ of $G$. Let $H_1$ be the graph obtained from $G$ by deleting vertex $u$, and $H_2$ be the graph obtained from $G$ by deleting vertex $v$. Is it necessary that $H_1$ and $H_2$ are isomorphic?
graph-theory graph-isomorphism
asked Nov 27 at 13:42
Ranveer Singh
199110
1
First you can ask yourself: is a connected regular graph necessarily vertex-transitive? The answer is no and if you start from those counterexamples then you will work out a counterexample to your question.
– Michal Adamaszek
Nov 27 at 14:07
the en.wikipedia.org/wiki/Frucht_graph is a counterexample.
– gilleain
Nov 27 at 16:42
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $G$ be a connected regular graph. Consider two different vertices $u,v$ of $G$. Let $H_1$ be the graph obtained from $G$ by deleting vertex $u$, and $H_2$ be the graph obtained from $G$ by deleting vertex $v$. Is it necessary that $H_1$ and $H_2$ are isomorphic?
graph-theory graph-isomorphism
asked Nov 27 at 13:42
Ranveer Singh
199110
Let $G$ be a connected regular graph. Consider two different vertices $u,v$ of $G$. Let $H_1$ be the graph obtained from $G$ by deleting vertex $u$, and $H_2$ be the graph obtained from $G$ by deleting vertex $v$. Is it necessary that $H_1$ and $H_2$ are isomorphic?
graph-theory graph-isomorphism
graph-theory graph-isomorphism
asked Nov 27 at 13:42
Ranveer Singh
199110
asked Nov 27 at 13:42
Ranveer Singh
199110
edited Nov 27 at 13:56
asked Nov 27 at 13:42
Ranveer Singh
199110
asked Nov 27 at 13:42
Ranveer Singh
199110
asked Nov 27 at 13:42
Ranveer Singh
199110
199110
1
First you can ask yourself: is a connected regular graph necessarily vertex-transitive? The answer is no and if you start from those counterexamples then you will work out a counterexample to your question.
– Michal Adamaszek
Nov 27 at 14:07
the en.wikipedia.org/wiki/Frucht_graph is a counterexample.
– gilleain
Nov 27 at 16:42
add a comment |
1
First you can ask yourself: is a connected regular graph necessarily vertex-transitive? The answer is no and if you start from those counterexamples then you will work out a counterexample to your question.
– Michal Adamaszek
Nov 27 at 14:07
the en.wikipedia.org/wiki/Frucht_graph is a counterexample.
– gilleain
Nov 27 at 16:42
1
1
First you can ask yourself: is a connected regular graph necessarily vertex-transitive? The answer is no and if you start from those counterexamples then you will work out a counterexample to your question.
– Michal Adamaszek
Nov 27 at 14:07
First you can ask yourself: is a connected regular graph necessarily vertex-transitive? The answer is no and if you start from those counterexamples then you will work out a counterexample to your question.
– Michal Adamaszek
Nov 27 at 14:07
the en.wikipedia.org/wiki/Frucht_graph is a counterexample.
– gilleain
Nov 27 at 16:42
the en.wikipedia.org/wiki/Frucht_graph is a counterexample.
– gilleain
Nov 27 at 16:42
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
No, they're not necessarily isomorphic. For easy counterexamples look at the disjoint union of two different regular graphs of the same degree (e.g. different cyclic graphs), then delete a point from each of the graphs.
answered Nov 27 at 13:53
Gnampfissimo
18011
Sorry. I meant a connected regular graph. I have edited the question.
– Ranveer Singh
Nov 27 at 13:56
@Ranveer Singh: if you want connected, take the complement of the above examples.
– Chris Godsil
Nov 27 at 19:02
@ChrisGodsil I understood that the complement of the above graphs (let's say disjoint union of two cycles of a different order) is a regular graph, call it $G$. But I don't think it will give a counterexample that the graphs $H_1$, $H_2$ obtained from $G$ are isomorphic, atleast degree sequence of $H_1, H_2$ is the same no matter what vertex we remove.
– Ranveer Singh
Nov 28 at 14:05
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',
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%2f3015780%2fisomorphism-in-regular-graphs%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
up vote
1
down vote
No, they're not necessarily isomorphic. For easy counterexamples look at the disjoint union of two different regular graphs of the same degree (e.g. different cyclic graphs), then delete a point from each of the graphs.
answered Nov 27 at 13:53
Gnampfissimo
18011
Sorry. I meant a connected regular graph. I have edited the question.
– Ranveer Singh
Nov 27 at 13:56
@Ranveer Singh: if you want connected, take the complement of the above examples.
– Chris Godsil
Nov 27 at 19:02
@ChrisGodsil I understood that the complement of the above graphs (let's say disjoint union of two cycles of a different order) is a regular graph, call it $G$. But I don't think it will give a counterexample that the graphs $H_1$, $H_2$ obtained from $G$ are isomorphic, atleast degree sequence of $H_1, H_2$ is the same no matter what vertex we remove.
– Ranveer Singh
Nov 28 at 14:05
add a comment |
up vote
1
down vote
No, they're not necessarily isomorphic. For easy counterexamples look at the disjoint union of two different regular graphs of the same degree (e.g. different cyclic graphs), then delete a point from each of the graphs.
answered Nov 27 at 13:53
Gnampfissimo
18011
Sorry. I meant a connected regular graph. I have edited the question.
– Ranveer Singh
Nov 27 at 13:56
@Ranveer Singh: if you want connected, take the complement of the above examples.
– Chris Godsil
Nov 27 at 19:02
@ChrisGodsil I understood that the complement of the above graphs (let's say disjoint union of two cycles of a different order) is a regular graph, call it $G$. But I don't think it will give a counterexample that the graphs $H_1$, $H_2$ obtained from $G$ are isomorphic, atleast degree sequence of $H_1, H_2$ is the same no matter what vertex we remove.
– Ranveer Singh
Nov 28 at 14:05
add a comment |
up vote
1
down vote
up vote
1
down vote
No, they're not necessarily isomorphic. For easy counterexamples look at the disjoint union of two different regular graphs of the same degree (e.g. different cyclic graphs), then delete a point from each of the graphs.
answered Nov 27 at 13:53
Gnampfissimo
18011
No, they're not necessarily isomorphic. For easy counterexamples look at the disjoint union of two different regular graphs of the same degree (e.g. different cyclic graphs), then delete a point from each of the graphs.
answered Nov 27 at 13:53
Gnampfissimo
18011
answered Nov 27 at 13:53
Gnampfissimo
18011
answered Nov 27 at 13:53
Gnampfissimo
18011
answered Nov 27 at 13:53
Gnampfissimo
18011
18011
Sorry. I meant a connected regular graph. I have edited the question.
– Ranveer Singh
Nov 27 at 13:56
@Ranveer Singh: if you want connected, take the complement of the above examples.
– Chris Godsil
Nov 27 at 19:02
@ChrisGodsil I understood that the complement of the above graphs (let's say disjoint union of two cycles of a different order) is a regular graph, call it $G$. But I don't think it will give a counterexample that the graphs $H_1$, $H_2$ obtained from $G$ are isomorphic, atleast degree sequence of $H_1, H_2$ is the same no matter what vertex we remove.
– Ranveer Singh
Nov 28 at 14:05
add a comment |
Sorry. I meant a connected regular graph. I have edited the question.
– Ranveer Singh
Nov 27 at 13:56
@Ranveer Singh: if you want connected, take the complement of the above examples.
– Chris Godsil
Nov 27 at 19:02
@ChrisGodsil I understood that the complement of the above graphs (let's say disjoint union of two cycles of a different order) is a regular graph, call it $G$. But I don't think it will give a counterexample that the graphs $H_1$, $H_2$ obtained from $G$ are isomorphic, atleast degree sequence of $H_1, H_2$ is the same no matter what vertex we remove.
– Ranveer Singh
Nov 28 at 14:05
Sorry. I meant a connected regular graph. I have edited the question.
– Ranveer Singh
Nov 27 at 13:56
Sorry. I meant a connected regular graph. I have edited the question.
– Ranveer Singh
Nov 27 at 13:56
@Ranveer Singh: if you want connected, take the complement of the above examples.
– Chris Godsil
Nov 27 at 19:02
@Ranveer Singh: if you want connected, take the complement of the above examples.
– Chris Godsil
Nov 27 at 19:02
@ChrisGodsil I understood that the complement of the above graphs (let's say disjoint union of two cycles of a different order) is a regular graph, call it $G$. But I don't think it will give a counterexample that the graphs $H_1$, $H_2$ obtained from $G$ are isomorphic, atleast degree sequence of $H_1, H_2$ is the same no matter what vertex we remove.
– Ranveer Singh
Nov 28 at 14:05
@ChrisGodsil I understood that the complement of the above graphs (let's say disjoint union of two cycles of a different order) is a regular graph, call it $G$. But I don't think it will give a counterexample that the graphs $H_1$, $H_2$ obtained from $G$ are isomorphic, atleast degree sequence of $H_1, H_2$ is the same no matter what vertex we remove.
– Ranveer Singh
Nov 28 at 14:05
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3015780%2fisomorphism-in-regular-graphs%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
First you can ask yourself: is a connected regular graph necessarily vertex-transitive? The answer is no and if you start from those counterexamples then you will work out a counterexample to your question.
– Michal Adamaszek
Nov 27 at 14:07
the en.wikipedia.org/wiki/Frucht_graph is a counterexample.
– gilleain
Nov 27 at 16:42