Mixing
Clarification of Edge-connectivity Menger's theorem
Edge-connectivity Menger's Theorem: A graph is k-edge-connected if and only if every pair of vertices has k pairwise edge-disjoint (p.e.d.) paths in between.
I can see that the later "k" is actually a lower bound, i.e. there might be more than k p.e.d. paths in between some pairs of vertices. But then, if between all pairs of vertices there are more than k p.e.d. paths, wouldn't the graph be not k-edge-connected? Or is it the case that this construction is simply not possible?
graph-theory graph-connectivity
asked Nov 20 '18 at 21:34
ensbana
282113
add a comment |
Edge-connectivity Menger's Theorem: A graph is k-edge-connected if and only if every pair of vertices has k pairwise edge-disjoint (p.e.d.) paths in between.
I can see that the later "k" is actually a lower bound, i.e. there might be more than k p.e.d. paths in between some pairs of vertices. But then, if between all pairs of vertices there are more than k p.e.d. paths, wouldn't the graph be not k-edge-connected? Or is it the case that this construction is simply not possible?
graph-theory graph-connectivity
asked Nov 20 '18 at 21:34
ensbana
282113
The construction is not possible. What you are saying is merely that if G is k edge connected then G has at most k edge disjoint paths. So you are “proving by contradiction” one direction of the theorem. In fact you are stating that it is obvious and I totally agree. However, the difficulty and beauty of this problem is to show that the converse is also true.
– Sorin Tirc
Nov 20 '18 at 21:53
add a comment |
Edge-connectivity Menger's Theorem: A graph is k-edge-connected if and only if every pair of vertices has k pairwise edge-disjoint (p.e.d.) paths in between.
I can see that the later "k" is actually a lower bound, i.e. there might be more than k p.e.d. paths in between some pairs of vertices. But then, if between all pairs of vertices there are more than k p.e.d. paths, wouldn't the graph be not k-edge-connected? Or is it the case that this construction is simply not possible?
graph-theory graph-connectivity
asked Nov 20 '18 at 21:34
ensbana
282113
Edge-connectivity Menger's Theorem: A graph is k-edge-connected if and only if every pair of vertices has k pairwise edge-disjoint (p.e.d.) paths in between.
I can see that the later "k" is actually a lower bound, i.e. there might be more than k p.e.d. paths in between some pairs of vertices. But then, if between all pairs of vertices there are more than k p.e.d. paths, wouldn't the graph be not k-edge-connected? Or is it the case that this construction is simply not possible?
graph-theory graph-connectivity
graph-theory graph-connectivity
asked Nov 20 '18 at 21:34
ensbana
282113
asked Nov 20 '18 at 21:34
ensbana
282113
asked Nov 20 '18 at 21:34
ensbana
282113
asked Nov 20 '18 at 21:34
ensbana
282113
asked Nov 20 '18 at 21:34
ensbana
282113
282113
The construction is not possible. What you are saying is merely that if G is k edge connected then G has at most k edge disjoint paths. So you are “proving by contradiction” one direction of the theorem. In fact you are stating that it is obvious and I totally agree. However, the difficulty and beauty of this problem is to show that the converse is also true.
– Sorin Tirc
Nov 20 '18 at 21:53
add a comment |
The construction is not possible. What you are saying is merely that if G is k edge connected then G has at most k edge disjoint paths. So you are “proving by contradiction” one direction of the theorem. In fact you are stating that it is obvious and I totally agree. However, the difficulty and beauty of this problem is to show that the converse is also true.
– Sorin Tirc
Nov 20 '18 at 21:53
The construction is not possible. What you are saying is merely that if G is k edge connected then G has at most k edge disjoint paths. So you are “proving by contradiction” one direction of the theorem. In fact you are stating that it is obvious and I totally agree. However, the difficulty and beauty of this problem is to show that the converse is also true.
– Sorin Tirc
Nov 20 '18 at 21:53
The construction is not possible. What you are saying is merely that if G is k edge connected then G has at most k edge disjoint paths. So you are “proving by contradiction” one direction of the theorem. In fact you are stating that it is obvious and I totally agree. However, the difficulty and beauty of this problem is to show that the converse is also true.
– Sorin Tirc
Nov 20 '18 at 21:53
add a comment |
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%2f3006933%2fclarification-of-edge-connectivity-mengers-theorem%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.
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%2f3006933%2fclarification-of-edge-connectivity-mengers-theorem%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
The construction is not possible. What you are saying is merely that if G is k edge connected then G has at most k edge disjoint paths. So you are “proving by contradiction” one direction of the theorem. In fact you are stating that it is obvious and I totally agree. However, the difficulty and beauty of this problem is to show that the converse is also true.
– Sorin Tirc
Nov 20 '18 at 21:53