Mixing
Could a non simple graph have at least two vertices of the same degree?
$begingroup$
I tried to prove that a simple graph $G$ with at least two vertices has two vertices of the same degree. It's easy to prove that this is true by the pigeonhole principle. If we remove the hypothesis that $G$ is simple can we affirm that $G$ necessarily has two vertices of the same degree? For example, if a $G$ has 3 vertices and between 1,2 there are two edges and between 2,3 only one we have one vertex of degree 2, one of degree 3 and the last of degree 1. Is it correct?
discrete-mathematics graph-theory
asked Jan 27 at 15:41
JackJack
756
$endgroup$
add a comment |
$begingroup$
I tried to prove that a simple graph $G$ with at least two vertices has two vertices of the same degree. It's easy to prove that this is true by the pigeonhole principle. If we remove the hypothesis that $G$ is simple can we affirm that $G$ necessarily has two vertices of the same degree? For example, if a $G$ has 3 vertices and between 1,2 there are two edges and between 2,3 only one we have one vertex of degree 2, one of degree 3 and the last of degree 1. Is it correct?
discrete-mathematics graph-theory
asked Jan 27 at 15:41
JackJack
756
$endgroup$
add a comment |
$begingroup$
I tried to prove that a simple graph $G$ with at least two vertices has two vertices of the same degree. It's easy to prove that this is true by the pigeonhole principle. If we remove the hypothesis that $G$ is simple can we affirm that $G$ necessarily has two vertices of the same degree? For example, if a $G$ has 3 vertices and between 1,2 there are two edges and between 2,3 only one we have one vertex of degree 2, one of degree 3 and the last of degree 1. Is it correct?
discrete-mathematics graph-theory
asked Jan 27 at 15:41
JackJack
756
$endgroup$
I tried to prove that a simple graph $G$ with at least two vertices has two vertices of the same degree. It's easy to prove that this is true by the pigeonhole principle. If we remove the hypothesis that $G$ is simple can we affirm that $G$ necessarily has two vertices of the same degree? For example, if a $G$ has 3 vertices and between 1,2 there are two edges and between 2,3 only one we have one vertex of degree 2, one of degree 3 and the last of degree 1. Is it correct?
discrete-mathematics graph-theory
discrete-mathematics graph-theory
asked Jan 27 at 15:41
JackJack
756
asked Jan 27 at 15:41
JackJack
756
asked Jan 27 at 15:41
JackJack
756
asked Jan 27 at 15:41
JackJack
756
asked Jan 27 at 15:41
JackJack
756
756
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Not necessarily. Consider the graph on 3 vertices $x,y,z$ where there are
2 edges between $x$ and $y$, 1 edge between $x$ and $z$ (and no edges between $y$ and $z$). So $d(x) = 2+1=3$, $d(y) = 2$ and $d(z)=1$.
answered Jan 27 at 17:54
MikeMike
4,461512
$endgroup$
add a comment |
$begingroup$
Your example is correct. If multiple edges between two vertices are allowed, it may not be the case that there are two vertices of the same degree.
answered Jan 27 at 17:56
Matt SamuelMatt Samuel
39k63769
$endgroup$
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%2f3089720%2fcould-a-non-simple-graph-have-at-least-two-vertices-of-the-same-degree%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
$begingroup$
Not necessarily. Consider the graph on 3 vertices $x,y,z$ where there are
2 edges between $x$ and $y$, 1 edge between $x$ and $z$ (and no edges between $y$ and $z$). So $d(x) = 2+1=3$, $d(y) = 2$ and $d(z)=1$.
answered Jan 27 at 17:54
MikeMike
4,461512
$endgroup$
add a comment |
$begingroup$
Not necessarily. Consider the graph on 3 vertices $x,y,z$ where there are
2 edges between $x$ and $y$, 1 edge between $x$ and $z$ (and no edges between $y$ and $z$). So $d(x) = 2+1=3$, $d(y) = 2$ and $d(z)=1$.
answered Jan 27 at 17:54
MikeMike
4,461512
$endgroup$
add a comment |
$begingroup$
Not necessarily. Consider the graph on 3 vertices $x,y,z$ where there are
2 edges between $x$ and $y$, 1 edge between $x$ and $z$ (and no edges between $y$ and $z$). So $d(x) = 2+1=3$, $d(y) = 2$ and $d(z)=1$.
answered Jan 27 at 17:54
MikeMike
4,461512
$endgroup$
Not necessarily. Consider the graph on 3 vertices $x,y,z$ where there are
2 edges between $x$ and $y$, 1 edge between $x$ and $z$ (and no edges between $y$ and $z$). So $d(x) = 2+1=3$, $d(y) = 2$ and $d(z)=1$.
answered Jan 27 at 17:54
MikeMike
4,461512
answered Jan 27 at 17:54
MikeMike
4,461512
answered Jan 27 at 17:54
MikeMike
4,461512
answered Jan 27 at 17:54
MikeMike
4,461512
4,461512
add a comment |
add a comment |
$begingroup$
Your example is correct. If multiple edges between two vertices are allowed, it may not be the case that there are two vertices of the same degree.
answered Jan 27 at 17:56
Matt SamuelMatt Samuel
39k63769
$endgroup$
add a comment |
$begingroup$
Your example is correct. If multiple edges between two vertices are allowed, it may not be the case that there are two vertices of the same degree.
answered Jan 27 at 17:56
Matt SamuelMatt Samuel
39k63769
$endgroup$
add a comment |
$begingroup$
Your example is correct. If multiple edges between two vertices are allowed, it may not be the case that there are two vertices of the same degree.
answered Jan 27 at 17:56
Matt SamuelMatt Samuel
39k63769
$endgroup$
Your example is correct. If multiple edges between two vertices are allowed, it may not be the case that there are two vertices of the same degree.
answered Jan 27 at 17:56
Matt SamuelMatt Samuel
39k63769
answered Jan 27 at 17:56
Matt SamuelMatt Samuel
39k63769
answered Jan 27 at 17:56
Matt SamuelMatt Samuel
39k63769
answered Jan 27 at 17:56
Matt SamuelMatt Samuel
39k63769
39k63769
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%2f3089720%2fcould-a-non-simple-graph-have-at-least-two-vertices-of-the-same-degree%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
