Let $e$ be an edge of minimum weight in the connected weighted graph $G$. Every minimum spanning tree of $G$...
$begingroup$
Let $e$ be an edge of minimum weight in the connected weighted graph
$G$. Every minimum spanning tree of $G$ contains $e$.
I have been told that this is not true. But I also know this property :
Must a minimum weight spanning tree for a graph contain the least weight edge of every vertex of the graph?
Is it not true because that $e$ is not necessary the only minimum weight edge?
graph-theory
asked Jan 21 at 17:51
Marine GalantinMarine Galantin
860316
$endgroup$
add a comment |
$begingroup$
Let $e$ be an edge of minimum weight in the connected weighted graph
$G$. Every minimum spanning tree of $G$ contains $e$.
I have been told that this is not true. But I also know this property :
Must a minimum weight spanning tree for a graph contain the least weight edge of every vertex of the graph?
Is it not true because that $e$ is not necessary the only minimum weight edge?
graph-theory
asked Jan 21 at 17:51
Marine GalantinMarine Galantin
860316
$endgroup$
add a comment |
$begingroup$
Let $e$ be an edge of minimum weight in the connected weighted graph
$G$. Every minimum spanning tree of $G$ contains $e$.
I have been told that this is not true. But I also know this property :
Must a minimum weight spanning tree for a graph contain the least weight edge of every vertex of the graph?
Is it not true because that $e$ is not necessary the only minimum weight edge?
graph-theory
asked Jan 21 at 17:51
Marine GalantinMarine Galantin
860316
$endgroup$
Let $e$ be an edge of minimum weight in the connected weighted graph
$G$. Every minimum spanning tree of $G$ contains $e$.
I have been told that this is not true. But I also know this property :
Must a minimum weight spanning tree for a graph contain the least weight edge of every vertex of the graph?
Is it not true because that $e$ is not necessary the only minimum weight edge?
graph-theory
graph-theory
asked Jan 21 at 17:51
Marine GalantinMarine Galantin
860316
asked Jan 21 at 17:51
Marine GalantinMarine Galantin
860316
asked Jan 21 at 17:51
Marine GalantinMarine Galantin
860316
asked Jan 21 at 17:51
Marine GalantinMarine Galantin
860316
asked Jan 21 at 17:51
Marine GalantinMarine Galantin
860316
860316
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $G$ contains a cycle of edges each of the same minimum weight, then it is not possible for $G$ to have a spanning tree without omitting at least one of those edges.
Otherwise the answer is yes: Let $T$ be a spanning tree of $G$ NOT containing $e$. Then $T+e$ contains one cycle. By assumption, there is an edge in the cycle of non-minimum weight. Remove that edge, and we have a new spanning tree $T'$ of lower weight.
answered Jan 21 at 17:59
PMarPMar
461
$endgroup$
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%2f3082172%2flet-e-be-an-edge-of-minimum-weight-in-the-connected-weighted-graph-g-every%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
$begingroup$
If $G$ contains a cycle of edges each of the same minimum weight, then it is not possible for $G$ to have a spanning tree without omitting at least one of those edges.
Otherwise the answer is yes: Let $T$ be a spanning tree of $G$ NOT containing $e$. Then $T+e$ contains one cycle. By assumption, there is an edge in the cycle of non-minimum weight. Remove that edge, and we have a new spanning tree $T'$ of lower weight.
answered Jan 21 at 17:59
PMarPMar
461
$endgroup$
add a comment |
$begingroup$
If $G$ contains a cycle of edges each of the same minimum weight, then it is not possible for $G$ to have a spanning tree without omitting at least one of those edges.
Otherwise the answer is yes: Let $T$ be a spanning tree of $G$ NOT containing $e$. Then $T+e$ contains one cycle. By assumption, there is an edge in the cycle of non-minimum weight. Remove that edge, and we have a new spanning tree $T'$ of lower weight.
answered Jan 21 at 17:59
PMarPMar
461
$endgroup$
add a comment |
$begingroup$
If $G$ contains a cycle of edges each of the same minimum weight, then it is not possible for $G$ to have a spanning tree without omitting at least one of those edges.
Otherwise the answer is yes: Let $T$ be a spanning tree of $G$ NOT containing $e$. Then $T+e$ contains one cycle. By assumption, there is an edge in the cycle of non-minimum weight. Remove that edge, and we have a new spanning tree $T'$ of lower weight.
answered Jan 21 at 17:59
PMarPMar
461
$endgroup$
If $G$ contains a cycle of edges each of the same minimum weight, then it is not possible for $G$ to have a spanning tree without omitting at least one of those edges.
Otherwise the answer is yes: Let $T$ be a spanning tree of $G$ NOT containing $e$. Then $T+e$ contains one cycle. By assumption, there is an edge in the cycle of non-minimum weight. Remove that edge, and we have a new spanning tree $T'$ of lower weight.
answered Jan 21 at 17:59
PMarPMar
461
answered Jan 21 at 17:59
PMarPMar
461
answered Jan 21 at 17:59
PMarPMar
461
answered Jan 21 at 17:59
PMarPMar
461
461
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%2f3082172%2flet-e-be-an-edge-of-minimum-weight-in-the-connected-weighted-graph-g-every%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
