Question regarding the connectedness of spanning 2-regular subgraphs
$begingroup$
If a 4-regular connected graph does not have one or more cut vertices then we can say it has a 2-regular spanning sub graph which is connected, isn't it? (A spanning sub graph with one component)
There might be several 2-regular spanning subgraphs, some which are a union of disjoint cycles, but if there is no cut vertex above type of a spanning sub graph will also be present, right?
Can someone please guide me to understand this.
Thanks a lot in advance.
graph-theory
$endgroup$
add a comment |
$begingroup$
If a 4-regular connected graph does not have one or more cut vertices then we can say it has a 2-regular spanning sub graph which is connected, isn't it? (A spanning sub graph with one component)
There might be several 2-regular spanning subgraphs, some which are a union of disjoint cycles, but if there is no cut vertex above type of a spanning sub graph will also be present, right?
Can someone please guide me to understand this.
Thanks a lot in advance.
graph-theory
$endgroup$
add a comment |
$begingroup$
If a 4-regular connected graph does not have one or more cut vertices then we can say it has a 2-regular spanning sub graph which is connected, isn't it? (A spanning sub graph with one component)
There might be several 2-regular spanning subgraphs, some which are a union of disjoint cycles, but if there is no cut vertex above type of a spanning sub graph will also be present, right?
Can someone please guide me to understand this.
Thanks a lot in advance.
graph-theory
$endgroup$
If a 4-regular connected graph does not have one or more cut vertices then we can say it has a 2-regular spanning sub graph which is connected, isn't it? (A spanning sub graph with one component)
There might be several 2-regular spanning subgraphs, some which are a union of disjoint cycles, but if there is no cut vertex above type of a spanning sub graph will also be present, right?
Can someone please guide me to understand this.
Thanks a lot in advance.
graph-theory
graph-theory
edited Dec 2 '18 at 10:46
Buddhini Angelika
asked Nov 30 '18 at 18:03
Buddhini AngelikaBuddhini Angelika
15710
15710
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
A "2-regular spanning subgraph which is connected" is simply a Hamiltonian cycle, and in general when you have a reasonable-sounding condition for a Hamiltonian cycle to exist, it's probably not good enough.
This MathOverflow answer gives one counter-example. (Here, a Hamiltonian cycle does not exist, because to visit every vertex, we would have to use the left and right vertices multiple times.)
The Meredith graph is an even stronger counter-example: it is not just $2$-connected but $4$-connected (the best we can hope for a $4$-regular graph) yet is not Hamiltonian.
$endgroup$
$begingroup$
Thanks a lot @MishaLavrov in the graph in that link if we remove 2 vertices then it will be disconnected. Can't we get a good condition which will enable to decide that a connected 4 regular graph has a connected 2 regular spanning sub graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 3:59
$begingroup$
In general, determining if a 4-regular graph is Hamiltonian is NP-complete, so we won't have a good sufficient and necessary condition. Many sufficient conditions for Hamiltonian cycles don't work for sparse graphs, but it is a theorem of Tutte that, for example, a 4-connected planar graph is always Hamiltonian.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 4:05
$begingroup$
Thank you. Can we take a 4 regular graph as a 4 connected graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 4:32
$begingroup$
I'm not sure what you're asking. Not all 4-regular graphs are 4-connected, but some are.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 4:33
$begingroup$
Thanks @MishaLavrov. I have edited the question by adding a 4 regular graph. In that graph there exist 2 edge disjoint spanning subgraphs with one component. I.e. a Hamiltonian cycle. But I was trying to give a reason for the existence of a connected 2 regular spanning sub graph.
$endgroup$
– Buddhini Angelika
Dec 2 '18 at 10:50
|
show 6 more comments
$begingroup$
Is this a good hint? If the graph $G $ is connected then because all degree of the vertices are even, $G $ has an eulérienne cycle. If $G $ is not connected, then each component of $G $ has an eulérienne cycle.
$endgroup$
$begingroup$
This seems irrelevant to the question. Eulerian cycles are not $2$-regular spanning subgraphs.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 1:20
$begingroup$
Thanks a lot all! Is there any good condition which will enable to decide that a 4 regular connected graph has a 2 regular spanning sub graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 3:55
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%2f3020419%2fquestion-regarding-the-connectedness-of-spanning-2-regular-subgraphs%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 "2-regular spanning subgraph which is connected" is simply a Hamiltonian cycle, and in general when you have a reasonable-sounding condition for a Hamiltonian cycle to exist, it's probably not good enough.
This MathOverflow answer gives one counter-example. (Here, a Hamiltonian cycle does not exist, because to visit every vertex, we would have to use the left and right vertices multiple times.)
The Meredith graph is an even stronger counter-example: it is not just $2$-connected but $4$-connected (the best we can hope for a $4$-regular graph) yet is not Hamiltonian.
$endgroup$
$begingroup$
Thanks a lot @MishaLavrov in the graph in that link if we remove 2 vertices then it will be disconnected. Can't we get a good condition which will enable to decide that a connected 4 regular graph has a connected 2 regular spanning sub graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 3:59
$begingroup$
In general, determining if a 4-regular graph is Hamiltonian is NP-complete, so we won't have a good sufficient and necessary condition. Many sufficient conditions for Hamiltonian cycles don't work for sparse graphs, but it is a theorem of Tutte that, for example, a 4-connected planar graph is always Hamiltonian.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 4:05
$begingroup$
Thank you. Can we take a 4 regular graph as a 4 connected graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 4:32
$begingroup$
I'm not sure what you're asking. Not all 4-regular graphs are 4-connected, but some are.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 4:33
$begingroup$
Thanks @MishaLavrov. I have edited the question by adding a 4 regular graph. In that graph there exist 2 edge disjoint spanning subgraphs with one component. I.e. a Hamiltonian cycle. But I was trying to give a reason for the existence of a connected 2 regular spanning sub graph.
$endgroup$
– Buddhini Angelika
Dec 2 '18 at 10:50
|
show 6 more comments
$begingroup$
A "2-regular spanning subgraph which is connected" is simply a Hamiltonian cycle, and in general when you have a reasonable-sounding condition for a Hamiltonian cycle to exist, it's probably not good enough.
This MathOverflow answer gives one counter-example. (Here, a Hamiltonian cycle does not exist, because to visit every vertex, we would have to use the left and right vertices multiple times.)
The Meredith graph is an even stronger counter-example: it is not just $2$-connected but $4$-connected (the best we can hope for a $4$-regular graph) yet is not Hamiltonian.
$endgroup$
$begingroup$
Thanks a lot @MishaLavrov in the graph in that link if we remove 2 vertices then it will be disconnected. Can't we get a good condition which will enable to decide that a connected 4 regular graph has a connected 2 regular spanning sub graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 3:59
$begingroup$
In general, determining if a 4-regular graph is Hamiltonian is NP-complete, so we won't have a good sufficient and necessary condition. Many sufficient conditions for Hamiltonian cycles don't work for sparse graphs, but it is a theorem of Tutte that, for example, a 4-connected planar graph is always Hamiltonian.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 4:05
$begingroup$
Thank you. Can we take a 4 regular graph as a 4 connected graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 4:32
$begingroup$
I'm not sure what you're asking. Not all 4-regular graphs are 4-connected, but some are.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 4:33
$begingroup$
Thanks @MishaLavrov. I have edited the question by adding a 4 regular graph. In that graph there exist 2 edge disjoint spanning subgraphs with one component. I.e. a Hamiltonian cycle. But I was trying to give a reason for the existence of a connected 2 regular spanning sub graph.
$endgroup$
– Buddhini Angelika
Dec 2 '18 at 10:50
|
show 6 more comments
$begingroup$
A "2-regular spanning subgraph which is connected" is simply a Hamiltonian cycle, and in general when you have a reasonable-sounding condition for a Hamiltonian cycle to exist, it's probably not good enough.
This MathOverflow answer gives one counter-example. (Here, a Hamiltonian cycle does not exist, because to visit every vertex, we would have to use the left and right vertices multiple times.)
The Meredith graph is an even stronger counter-example: it is not just $2$-connected but $4$-connected (the best we can hope for a $4$-regular graph) yet is not Hamiltonian.
$endgroup$
A "2-regular spanning subgraph which is connected" is simply a Hamiltonian cycle, and in general when you have a reasonable-sounding condition for a Hamiltonian cycle to exist, it's probably not good enough.
This MathOverflow answer gives one counter-example. (Here, a Hamiltonian cycle does not exist, because to visit every vertex, we would have to use the left and right vertices multiple times.)
The Meredith graph is an even stronger counter-example: it is not just $2$-connected but $4$-connected (the best we can hope for a $4$-regular graph) yet is not Hamiltonian.
edited Dec 1 '18 at 4:07
answered Dec 1 '18 at 1:13
Misha LavrovMisha Lavrov
44.9k556107
44.9k556107
$begingroup$
Thanks a lot @MishaLavrov in the graph in that link if we remove 2 vertices then it will be disconnected. Can't we get a good condition which will enable to decide that a connected 4 regular graph has a connected 2 regular spanning sub graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 3:59
$begingroup$
In general, determining if a 4-regular graph is Hamiltonian is NP-complete, so we won't have a good sufficient and necessary condition. Many sufficient conditions for Hamiltonian cycles don't work for sparse graphs, but it is a theorem of Tutte that, for example, a 4-connected planar graph is always Hamiltonian.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 4:05
$begingroup$
Thank you. Can we take a 4 regular graph as a 4 connected graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 4:32
$begingroup$
I'm not sure what you're asking. Not all 4-regular graphs are 4-connected, but some are.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 4:33
$begingroup$
Thanks @MishaLavrov. I have edited the question by adding a 4 regular graph. In that graph there exist 2 edge disjoint spanning subgraphs with one component. I.e. a Hamiltonian cycle. But I was trying to give a reason for the existence of a connected 2 regular spanning sub graph.
$endgroup$
– Buddhini Angelika
Dec 2 '18 at 10:50
|
show 6 more comments
$begingroup$
Thanks a lot @MishaLavrov in the graph in that link if we remove 2 vertices then it will be disconnected. Can't we get a good condition which will enable to decide that a connected 4 regular graph has a connected 2 regular spanning sub graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 3:59
$begingroup$
In general, determining if a 4-regular graph is Hamiltonian is NP-complete, so we won't have a good sufficient and necessary condition. Many sufficient conditions for Hamiltonian cycles don't work for sparse graphs, but it is a theorem of Tutte that, for example, a 4-connected planar graph is always Hamiltonian.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 4:05
$begingroup$
Thank you. Can we take a 4 regular graph as a 4 connected graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 4:32
$begingroup$
I'm not sure what you're asking. Not all 4-regular graphs are 4-connected, but some are.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 4:33
$begingroup$
Thanks @MishaLavrov. I have edited the question by adding a 4 regular graph. In that graph there exist 2 edge disjoint spanning subgraphs with one component. I.e. a Hamiltonian cycle. But I was trying to give a reason for the existence of a connected 2 regular spanning sub graph.
$endgroup$
– Buddhini Angelika
Dec 2 '18 at 10:50
$begingroup$
Thanks a lot @MishaLavrov in the graph in that link if we remove 2 vertices then it will be disconnected. Can't we get a good condition which will enable to decide that a connected 4 regular graph has a connected 2 regular spanning sub graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 3:59
$begingroup$
Thanks a lot @MishaLavrov in the graph in that link if we remove 2 vertices then it will be disconnected. Can't we get a good condition which will enable to decide that a connected 4 regular graph has a connected 2 regular spanning sub graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 3:59
$begingroup$
In general, determining if a 4-regular graph is Hamiltonian is NP-complete, so we won't have a good sufficient and necessary condition. Many sufficient conditions for Hamiltonian cycles don't work for sparse graphs, but it is a theorem of Tutte that, for example, a 4-connected planar graph is always Hamiltonian.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 4:05
$begingroup$
In general, determining if a 4-regular graph is Hamiltonian is NP-complete, so we won't have a good sufficient and necessary condition. Many sufficient conditions for Hamiltonian cycles don't work for sparse graphs, but it is a theorem of Tutte that, for example, a 4-connected planar graph is always Hamiltonian.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 4:05
$begingroup$
Thank you. Can we take a 4 regular graph as a 4 connected graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 4:32
$begingroup$
Thank you. Can we take a 4 regular graph as a 4 connected graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 4:32
$begingroup$
I'm not sure what you're asking. Not all 4-regular graphs are 4-connected, but some are.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 4:33
$begingroup$
I'm not sure what you're asking. Not all 4-regular graphs are 4-connected, but some are.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 4:33
$begingroup$
Thanks @MishaLavrov. I have edited the question by adding a 4 regular graph. In that graph there exist 2 edge disjoint spanning subgraphs with one component. I.e. a Hamiltonian cycle. But I was trying to give a reason for the existence of a connected 2 regular spanning sub graph.
$endgroup$
– Buddhini Angelika
Dec 2 '18 at 10:50
$begingroup$
Thanks @MishaLavrov. I have edited the question by adding a 4 regular graph. In that graph there exist 2 edge disjoint spanning subgraphs with one component. I.e. a Hamiltonian cycle. But I was trying to give a reason for the existence of a connected 2 regular spanning sub graph.
$endgroup$
– Buddhini Angelika
Dec 2 '18 at 10:50
|
show 6 more comments
$begingroup$
Is this a good hint? If the graph $G $ is connected then because all degree of the vertices are even, $G $ has an eulérienne cycle. If $G $ is not connected, then each component of $G $ has an eulérienne cycle.
$endgroup$
$begingroup$
This seems irrelevant to the question. Eulerian cycles are not $2$-regular spanning subgraphs.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 1:20
$begingroup$
Thanks a lot all! Is there any good condition which will enable to decide that a 4 regular connected graph has a 2 regular spanning sub graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 3:55
add a comment |
$begingroup$
Is this a good hint? If the graph $G $ is connected then because all degree of the vertices are even, $G $ has an eulérienne cycle. If $G $ is not connected, then each component of $G $ has an eulérienne cycle.
$endgroup$
$begingroup$
This seems irrelevant to the question. Eulerian cycles are not $2$-regular spanning subgraphs.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 1:20
$begingroup$
Thanks a lot all! Is there any good condition which will enable to decide that a 4 regular connected graph has a 2 regular spanning sub graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 3:55
add a comment |
$begingroup$
Is this a good hint? If the graph $G $ is connected then because all degree of the vertices are even, $G $ has an eulérienne cycle. If $G $ is not connected, then each component of $G $ has an eulérienne cycle.
$endgroup$
Is this a good hint? If the graph $G $ is connected then because all degree of the vertices are even, $G $ has an eulérienne cycle. If $G $ is not connected, then each component of $G $ has an eulérienne cycle.
answered Nov 30 '18 at 23:17
nafhgoodnafhgood
1,801422
1,801422
$begingroup$
This seems irrelevant to the question. Eulerian cycles are not $2$-regular spanning subgraphs.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 1:20
$begingroup$
Thanks a lot all! Is there any good condition which will enable to decide that a 4 regular connected graph has a 2 regular spanning sub graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 3:55
add a comment |
$begingroup$
This seems irrelevant to the question. Eulerian cycles are not $2$-regular spanning subgraphs.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 1:20
$begingroup$
Thanks a lot all! Is there any good condition which will enable to decide that a 4 regular connected graph has a 2 regular spanning sub graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 3:55
$begingroup$
This seems irrelevant to the question. Eulerian cycles are not $2$-regular spanning subgraphs.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 1:20
$begingroup$
This seems irrelevant to the question. Eulerian cycles are not $2$-regular spanning subgraphs.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 1:20
$begingroup$
Thanks a lot all! Is there any good condition which will enable to decide that a 4 regular connected graph has a 2 regular spanning sub graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 3:55
$begingroup$
Thanks a lot all! Is there any good condition which will enable to decide that a 4 regular connected graph has a 2 regular spanning sub graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 3:55
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%2f3020419%2fquestion-regarding-the-connectedness-of-spanning-2-regular-subgraphs%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