Mixing
Proving that the complement graph is not euler graph.
$begingroup$
I came across with the following question from a book in Graphs (not in English):
Let $T(V,E)$ be a tree with $ngeq 5$ vertices and exactly $3$ leafs.
A. Prove that $T$ has exactly one vertex with degree $3$.
B. Prove that the complement graph of $T$ (lets call it $T'$), is not euler graph.
I proved the first theorem but I think I found an example which disproves B:
Consider graph T with 6 vertices:
The complement graph of $T$ is:
I checked for a few time (I hope it's correct). There are exactly two vertices $v_2,v_5$ that has an odd degree so it's an euler graph.
Where is my mistake?
Leaving that example, I tried to prove it by saying the following:
T has 3 lefts so those vertices in graph T' will have n-1 degree. if n is even then we have 3 vertices that has an odd degree meaning that T' is not euler. But what can I say about the n even-case?
EDIT:
Euler graph is a connectivity finite graph which follows one of those conditions:
- Has exactly two vertices of odd degree. In that case its not a circle.
- All of the vertices with even degree. In that case its a circle.
combinatorics graph-theory
asked Jan 18 at 16:35
vesiivesii
3108
$endgroup$
|
show 3 more comments
$begingroup$
I came across with the following question from a book in Graphs (not in English):
Let $T(V,E)$ be a tree with $ngeq 5$ vertices and exactly $3$ leafs.
A. Prove that $T$ has exactly one vertex with degree $3$.
B. Prove that the complement graph of $T$ (lets call it $T'$), is not euler graph.
I proved the first theorem but I think I found an example which disproves B:
Consider graph T with 6 vertices:
The complement graph of $T$ is:
I checked for a few time (I hope it's correct). There are exactly two vertices $v_2,v_5$ that has an odd degree so it's an euler graph.
Where is my mistake?
Leaving that example, I tried to prove it by saying the following:
T has 3 lefts so those vertices in graph T' will have n-1 degree. if n is even then we have 3 vertices that has an odd degree meaning that T' is not euler. But what can I say about the n even-case?
EDIT:
Euler graph is a connectivity finite graph which follows one of those conditions:
- Has exactly two vertices of odd degree. In that case its not a circle.
- All of the vertices with even degree. In that case its a circle.
combinatorics graph-theory
asked Jan 18 at 16:35
vesiivesii
3108
$endgroup$
$begingroup$
The term "Euler graph" is sometimes used in the weaker sense for a graph in which every vertex has even degree. Have you checked the definition in your book?
$endgroup$
– Servaes
Jan 18 at 16:49
$begingroup$
@Servaes Hey, thanks for the reply. Please see my definition.
$endgroup$
– vesii
Jan 18 at 16:56
$begingroup$
I think you are right, and the exercise in this form is incorrect.
$endgroup$
– Leen Droogendijk
Jan 18 at 17:12
$begingroup$
Btw, you remark in the edit session is incorrect. A graph in which every vertex has degree $4$ is Eulerian, but not a circle.
$endgroup$
– Leen Droogendijk
Jan 18 at 17:18
1
$begingroup$
@LeenDroogendijk I suspect this is a translation problem, and that the intended word is 'cycle'.
$endgroup$
– Servaes
Jan 18 at 18:24
|
show 3 more comments
$begingroup$
I came across with the following question from a book in Graphs (not in English):
Let $T(V,E)$ be a tree with $ngeq 5$ vertices and exactly $3$ leafs.
A. Prove that $T$ has exactly one vertex with degree $3$.
B. Prove that the complement graph of $T$ (lets call it $T'$), is not euler graph.
I proved the first theorem but I think I found an example which disproves B:
Consider graph T with 6 vertices:
The complement graph of $T$ is:
I checked for a few time (I hope it's correct). There are exactly two vertices $v_2,v_5$ that has an odd degree so it's an euler graph.
Where is my mistake?
Leaving that example, I tried to prove it by saying the following:
T has 3 lefts so those vertices in graph T' will have n-1 degree. if n is even then we have 3 vertices that has an odd degree meaning that T' is not euler. But what can I say about the n even-case?
EDIT:
Euler graph is a connectivity finite graph which follows one of those conditions:
- Has exactly two vertices of odd degree. In that case its not a circle.
- All of the vertices with even degree. In that case its a circle.
combinatorics graph-theory
asked Jan 18 at 16:35
vesiivesii
3108
$endgroup$
I came across with the following question from a book in Graphs (not in English):
Let $T(V,E)$ be a tree with $ngeq 5$ vertices and exactly $3$ leafs.
A. Prove that $T$ has exactly one vertex with degree $3$.
B. Prove that the complement graph of $T$ (lets call it $T'$), is not euler graph.
I proved the first theorem but I think I found an example which disproves B:
Consider graph T with 6 vertices:
The complement graph of $T$ is:
I checked for a few time (I hope it's correct). There are exactly two vertices $v_2,v_5$ that has an odd degree so it's an euler graph.
Where is my mistake?
Leaving that example, I tried to prove it by saying the following:
T has 3 lefts so those vertices in graph T' will have n-1 degree. if n is even then we have 3 vertices that has an odd degree meaning that T' is not euler. But what can I say about the n even-case?
EDIT:
Euler graph is a connectivity finite graph which follows one of those conditions:
- Has exactly two vertices of odd degree. In that case its not a circle.
- All of the vertices with even degree. In that case its a circle.
combinatorics graph-theory
combinatorics graph-theory
asked Jan 18 at 16:35
vesiivesii
3108
asked Jan 18 at 16:35
vesiivesii
3108
edited Jan 18 at 16:55
vesii
asked Jan 18 at 16:35
vesiivesii
3108
asked Jan 18 at 16:35
vesiivesii
3108
asked Jan 18 at 16:35
vesiivesii
3108
3108
$begingroup$
The term "Euler graph" is sometimes used in the weaker sense for a graph in which every vertex has even degree. Have you checked the definition in your book?
$endgroup$
– Servaes
Jan 18 at 16:49
$begingroup$
@Servaes Hey, thanks for the reply. Please see my definition.
$endgroup$
– vesii
Jan 18 at 16:56
$begingroup$
I think you are right, and the exercise in this form is incorrect.
$endgroup$
– Leen Droogendijk
Jan 18 at 17:12
$begingroup$
Btw, you remark in the edit session is incorrect. A graph in which every vertex has degree $4$ is Eulerian, but not a circle.
$endgroup$
– Leen Droogendijk
Jan 18 at 17:18
1
$begingroup$
@LeenDroogendijk I suspect this is a translation problem, and that the intended word is 'cycle'.
$endgroup$
– Servaes
Jan 18 at 18:24
|
show 3 more comments
$begingroup$
The term "Euler graph" is sometimes used in the weaker sense for a graph in which every vertex has even degree. Have you checked the definition in your book?
$endgroup$
– Servaes
Jan 18 at 16:49
$begingroup$
@Servaes Hey, thanks for the reply. Please see my definition.
$endgroup$
– vesii
Jan 18 at 16:56
$begingroup$
I think you are right, and the exercise in this form is incorrect.
$endgroup$
– Leen Droogendijk
Jan 18 at 17:12
$begingroup$
Btw, you remark in the edit session is incorrect. A graph in which every vertex has degree $4$ is Eulerian, but not a circle.
$endgroup$
– Leen Droogendijk
Jan 18 at 17:18
1
$begingroup$
@LeenDroogendijk I suspect this is a translation problem, and that the intended word is 'cycle'.
$endgroup$
– Servaes
Jan 18 at 18:24
$begingroup$
The term "Euler graph" is sometimes used in the weaker sense for a graph in which every vertex has even degree. Have you checked the definition in your book?
$endgroup$
– Servaes
Jan 18 at 16:49
$begingroup$
The term "Euler graph" is sometimes used in the weaker sense for a graph in which every vertex has even degree. Have you checked the definition in your book?
$endgroup$
– Servaes
Jan 18 at 16:49
$begingroup$
@Servaes Hey, thanks for the reply. Please see my definition.
$endgroup$
– vesii
Jan 18 at 16:56
$begingroup$
@Servaes Hey, thanks for the reply. Please see my definition.
$endgroup$
– vesii
Jan 18 at 16:56
$begingroup$
I think you are right, and the exercise in this form is incorrect.
$endgroup$
– Leen Droogendijk
Jan 18 at 17:12
$begingroup$
I think you are right, and the exercise in this form is incorrect.
$endgroup$
– Leen Droogendijk
Jan 18 at 17:12
$begingroup$
Btw, you remark in the edit session is incorrect. A graph in which every vertex has degree $4$ is Eulerian, but not a circle.
$endgroup$
– Leen Droogendijk
Jan 18 at 17:18
$begingroup$
Btw, you remark in the edit session is incorrect. A graph in which every vertex has degree $4$ is Eulerian, but not a circle.
$endgroup$
– Leen Droogendijk
Jan 18 at 17:18
1
1
$begingroup$
@LeenDroogendijk I suspect this is a translation problem, and that the intended word is 'cycle'.
$endgroup$
– Servaes
Jan 18 at 18:24
$begingroup$
@LeenDroogendijk I suspect this is a translation problem, and that the intended word is 'cycle'.
$endgroup$
– Servaes
Jan 18 at 18:24
|
show 3 more comments
0
active
oldest
votes
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%2f3078471%2fproving-that-the-complement-graph-is-not-euler-graph%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3078471%2fproving-that-the-complement-graph-is-not-euler-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
$begingroup$
The term "Euler graph" is sometimes used in the weaker sense for a graph in which every vertex has even degree. Have you checked the definition in your book?
$endgroup$
– Servaes
Jan 18 at 16:49
$begingroup$
@Servaes Hey, thanks for the reply. Please see my definition.
$endgroup$
– vesii
Jan 18 at 16:56
$begingroup$
I think you are right, and the exercise in this form is incorrect.
$endgroup$
– Leen Droogendijk
Jan 18 at 17:12
$begingroup$
Btw, you remark in the edit session is incorrect. A graph in which every vertex has degree $4$ is Eulerian, but not a circle.
$endgroup$
– Leen Droogendijk
Jan 18 at 17:18
1
$begingroup$
@LeenDroogendijk I suspect this is a translation problem, and that the intended word is 'cycle'.
$endgroup$
– Servaes
Jan 18 at 18:24