Hamilton path and Euler circuits in Cycle graph
$begingroup$
Can $C_n$ has $2*n$ Euler circuits and $3*n$ Hamilton paths in the Cyclic graphs?
graph-theory hamiltonian-path eulerian-path
edited Dec 18 '18 at 17:46
Davide Giraudo
128k17154268
asked Dec 18 '18 at 17:10
JuneJune
122
$endgroup$
add a comment |
$begingroup$
Can $C_n$ has $2*n$ Euler circuits and $3*n$ Hamilton paths in the Cyclic graphs?
graph-theory hamiltonian-path eulerian-path
edited Dec 18 '18 at 17:46
Davide Giraudo
128k17154268
asked Dec 18 '18 at 17:10
JuneJune
122
$endgroup$
add a comment |
$begingroup$
Can $C_n$ has $2*n$ Euler circuits and $3*n$ Hamilton paths in the Cyclic graphs?
graph-theory hamiltonian-path eulerian-path
edited Dec 18 '18 at 17:46
Davide Giraudo
128k17154268
asked Dec 18 '18 at 17:10
JuneJune
122
$endgroup$
Can $C_n$ has $2*n$ Euler circuits and $3*n$ Hamilton paths in the Cyclic graphs?
graph-theory hamiltonian-path eulerian-path
graph-theory hamiltonian-path eulerian-path
edited Dec 18 '18 at 17:46
Davide Giraudo
128k17154268
asked Dec 18 '18 at 17:10
JuneJune
122
edited Dec 18 '18 at 17:46
Davide Giraudo
128k17154268
asked Dec 18 '18 at 17:10
JuneJune
122
edited Dec 18 '18 at 17:46
Davide Giraudo
128k17154268
edited Dec 18 '18 at 17:46
Davide Giraudo
128k17154268
edited Dec 18 '18 at 17:46
Davide Giraudo
128k17154268
128k17154268
asked Dec 18 '18 at 17:10
JuneJune
122
asked Dec 18 '18 at 17:10
JuneJune
122
asked Dec 18 '18 at 17:10
JuneJune
122
122
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Bt $C_n$ you mean the cycle with $n$ vertices, right?
A cycle on $n$ vertices has $2n$ Euler circuits: Each circuit is uniquely determined by the starting vertex ($n$ choices) and then the direction (2 choices).
Each Hamiltonian path is also uniquely determined by the starting vertex $n$ choices and then the direction 2 choices. So there are $2n$ Hamiltonian paths.
Each Hamiltonian cycle is also uniquely determined by the starting vertex $n$ choices and then the direction 2 choices. So there are $2n$ Hamiltonian cycles.
So to answer your question in short, yes there are $2n$ Eulerian circuits, no there are only $2n$ Hamiltonian paths.
answered Dec 18 '18 at 19:40
MikeMike
4,516512
$endgroup$
$begingroup$
Thanks @mike for the explanation :)
$endgroup$
– June
Dec 18 '18 at 20:18
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%2f3045419%2fhamilton-path-and-euler-circuits-in-cycle-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
$begingroup$
Bt $C_n$ you mean the cycle with $n$ vertices, right?
A cycle on $n$ vertices has $2n$ Euler circuits: Each circuit is uniquely determined by the starting vertex ($n$ choices) and then the direction (2 choices).
Each Hamiltonian path is also uniquely determined by the starting vertex $n$ choices and then the direction 2 choices. So there are $2n$ Hamiltonian paths.
Each Hamiltonian cycle is also uniquely determined by the starting vertex $n$ choices and then the direction 2 choices. So there are $2n$ Hamiltonian cycles.
So to answer your question in short, yes there are $2n$ Eulerian circuits, no there are only $2n$ Hamiltonian paths.
answered Dec 18 '18 at 19:40
MikeMike
4,516512
$endgroup$
$begingroup$
Thanks @mike for the explanation :)
$endgroup$
– June
Dec 18 '18 at 20:18
add a comment |
$begingroup$
Bt $C_n$ you mean the cycle with $n$ vertices, right?
A cycle on $n$ vertices has $2n$ Euler circuits: Each circuit is uniquely determined by the starting vertex ($n$ choices) and then the direction (2 choices).
Each Hamiltonian path is also uniquely determined by the starting vertex $n$ choices and then the direction 2 choices. So there are $2n$ Hamiltonian paths.
Each Hamiltonian cycle is also uniquely determined by the starting vertex $n$ choices and then the direction 2 choices. So there are $2n$ Hamiltonian cycles.
So to answer your question in short, yes there are $2n$ Eulerian circuits, no there are only $2n$ Hamiltonian paths.
answered Dec 18 '18 at 19:40
MikeMike
4,516512
$endgroup$
$begingroup$
Thanks @mike for the explanation :)
$endgroup$
– June
Dec 18 '18 at 20:18
add a comment |
$begingroup$
Bt $C_n$ you mean the cycle with $n$ vertices, right?
A cycle on $n$ vertices has $2n$ Euler circuits: Each circuit is uniquely determined by the starting vertex ($n$ choices) and then the direction (2 choices).
Each Hamiltonian path is also uniquely determined by the starting vertex $n$ choices and then the direction 2 choices. So there are $2n$ Hamiltonian paths.
Each Hamiltonian cycle is also uniquely determined by the starting vertex $n$ choices and then the direction 2 choices. So there are $2n$ Hamiltonian cycles.
So to answer your question in short, yes there are $2n$ Eulerian circuits, no there are only $2n$ Hamiltonian paths.
answered Dec 18 '18 at 19:40
MikeMike
4,516512
$endgroup$
Bt $C_n$ you mean the cycle with $n$ vertices, right?
A cycle on $n$ vertices has $2n$ Euler circuits: Each circuit is uniquely determined by the starting vertex ($n$ choices) and then the direction (2 choices).
Each Hamiltonian path is also uniquely determined by the starting vertex $n$ choices and then the direction 2 choices. So there are $2n$ Hamiltonian paths.
Each Hamiltonian cycle is also uniquely determined by the starting vertex $n$ choices and then the direction 2 choices. So there are $2n$ Hamiltonian cycles.
So to answer your question in short, yes there are $2n$ Eulerian circuits, no there are only $2n$ Hamiltonian paths.
answered Dec 18 '18 at 19:40
MikeMike
4,516512
answered Dec 18 '18 at 19:40
MikeMike
4,516512
answered Dec 18 '18 at 19:40
MikeMike
4,516512
answered Dec 18 '18 at 19:40
MikeMike
4,516512
4,516512
$begingroup$
Thanks @mike for the explanation :)
$endgroup$
– June
Dec 18 '18 at 20:18
add a comment |
$begingroup$
Thanks @mike for the explanation :)
$endgroup$
– June
Dec 18 '18 at 20:18
$begingroup$
Thanks @mike for the explanation :)
$endgroup$
– June
Dec 18 '18 at 20:18
$begingroup$
Thanks @mike for the explanation :)
$endgroup$
– June
Dec 18 '18 at 20:18
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%2f3045419%2fhamilton-path-and-euler-circuits-in-cycle-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
