Let $v$ be a vertex of a 2-connected graph $G$. Prove that $v$ has a neighbor $u$ such that $G − u − v$...
$begingroup$
Let $v$ be a vertex of a 2-connected graph $G$. Prove that $v$ has a neighbor
$u$ such that $G − u − v$ is connected.
I'm not sure I understood that prove. Please anyone can explain me that ?
Prove that in a 2-connected graph like G which has the vertex $v$, $v$ has a neighbor $u$ such that $G-v-u$ is connected
Thanks!
graph-theory graph-connectivity
asked Jun 8 '18 at 8:04
hemihemi
1127
$endgroup$
add a comment |
$begingroup$
Let $v$ be a vertex of a 2-connected graph $G$. Prove that $v$ has a neighbor
$u$ such that $G − u − v$ is connected.
I'm not sure I understood that prove. Please anyone can explain me that ?
Prove that in a 2-connected graph like G which has the vertex $v$, $v$ has a neighbor $u$ such that $G-v-u$ is connected
Thanks!
graph-theory graph-connectivity
asked Jun 8 '18 at 8:04
hemihemi
1127
$endgroup$
add a comment |
$begingroup$
Let $v$ be a vertex of a 2-connected graph $G$. Prove that $v$ has a neighbor
$u$ such that $G − u − v$ is connected.
I'm not sure I understood that prove. Please anyone can explain me that ?
Prove that in a 2-connected graph like G which has the vertex $v$, $v$ has a neighbor $u$ such that $G-v-u$ is connected
Thanks!
graph-theory graph-connectivity
asked Jun 8 '18 at 8:04
hemihemi
1127
$endgroup$
Let $v$ be a vertex of a 2-connected graph $G$. Prove that $v$ has a neighbor
$u$ such that $G − u − v$ is connected.
I'm not sure I understood that prove. Please anyone can explain me that ?
Prove that in a 2-connected graph like G which has the vertex $v$, $v$ has a neighbor $u$ such that $G-v-u$ is connected
Thanks!
graph-theory graph-connectivity
graph-theory graph-connectivity
asked Jun 8 '18 at 8:04
hemihemi
1127
asked Jun 8 '18 at 8:04
hemihemi
1127
asked Jun 8 '18 at 8:04
hemihemi
1127
asked Jun 8 '18 at 8:04
hemihemi
1127
asked Jun 8 '18 at 8:04
hemihemi
1127
1127
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $u_1$ be any neighbor of $v$. If $G - v - u_1$ is connected then we have found our vertex. If $G - v - u_1$ is disconnected, look at the set of components. If any component contains no neighbors of $v$, then $G-u_1$ must actually be disconnected, which is a contradiction. So we may assume all components contain neighbors of $v$. Choose some such component, and let $u_2$ be one of the neighbors of $v$ lying in it.
Now consider $G - v - u_2$. If this is connected then again we have found our vertex. If it is disconnected, look at the set of components, which again must each contain neighbors of $v$. Choose some component that does NOT contain $u_1$, choose a neighbor of $v$ in it, and call that neighbor $u_3$.
The idea is that we repeat this process, and when we look at the components of $G - v - u_k$ we can observe that $u_1 cdots u_{k-1}$ must all actually lie in the same component. Thus, we never end up in a circle, and the process must eventually terminate. When it terminates, we have found the desired $u$.
answered Jun 8 '18 at 8:42
CarlCarl
2,22311127
$endgroup$
$begingroup$
This is, of course, assuming the graph is finite or at least that the vertices have finite degree. If the vertices don't have finite degree, then I don't know if the result is even still true.
$endgroup$
– Carl
Jun 8 '18 at 8:45
add a comment |
$begingroup$
Take a vertex $v$ of $G$, and let $U:=N(v)$ be the set of neighbors of $v$, then let $T$ be the smallest connected subgraph of $G-v$ containing all vertices in $U$. Then $T$ is a tree as you can remove edges on the any cycle to make it a tree. Also, all leafs of $T$ are vertices in $U$ as otherwise you can just remove the leaf to make a smaller connected subgraph containing all vertices in $U$. Take $u$ be any leaf of $T$, and suppose $G-v-u$ is not connected. Then let $C$ be a component of $G-v-u$ not containing $T-u$. But then $C$ does not contain any neighbors of $v$, so $G-v$ is disconnected contradicting 2-connectivity of $G$.
answered Dec 18 '18 at 16:06
nafhgoodnafhgood
1,803422
$endgroup$
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%2f2812268%2flet-v-be-a-vertex-of-a-2-connected-graph-g-prove-that-v-has-a-neighbor-u%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
$begingroup$
Let $u_1$ be any neighbor of $v$. If $G - v - u_1$ is connected then we have found our vertex. If $G - v - u_1$ is disconnected, look at the set of components. If any component contains no neighbors of $v$, then $G-u_1$ must actually be disconnected, which is a contradiction. So we may assume all components contain neighbors of $v$. Choose some such component, and let $u_2$ be one of the neighbors of $v$ lying in it.
Now consider $G - v - u_2$. If this is connected then again we have found our vertex. If it is disconnected, look at the set of components, which again must each contain neighbors of $v$. Choose some component that does NOT contain $u_1$, choose a neighbor of $v$ in it, and call that neighbor $u_3$.
The idea is that we repeat this process, and when we look at the components of $G - v - u_k$ we can observe that $u_1 cdots u_{k-1}$ must all actually lie in the same component. Thus, we never end up in a circle, and the process must eventually terminate. When it terminates, we have found the desired $u$.
answered Jun 8 '18 at 8:42
CarlCarl
2,22311127
$endgroup$
$begingroup$
This is, of course, assuming the graph is finite or at least that the vertices have finite degree. If the vertices don't have finite degree, then I don't know if the result is even still true.
$endgroup$
– Carl
Jun 8 '18 at 8:45
add a comment |
$begingroup$
Let $u_1$ be any neighbor of $v$. If $G - v - u_1$ is connected then we have found our vertex. If $G - v - u_1$ is disconnected, look at the set of components. If any component contains no neighbors of $v$, then $G-u_1$ must actually be disconnected, which is a contradiction. So we may assume all components contain neighbors of $v$. Choose some such component, and let $u_2$ be one of the neighbors of $v$ lying in it.
Now consider $G - v - u_2$. If this is connected then again we have found our vertex. If it is disconnected, look at the set of components, which again must each contain neighbors of $v$. Choose some component that does NOT contain $u_1$, choose a neighbor of $v$ in it, and call that neighbor $u_3$.
The idea is that we repeat this process, and when we look at the components of $G - v - u_k$ we can observe that $u_1 cdots u_{k-1}$ must all actually lie in the same component. Thus, we never end up in a circle, and the process must eventually terminate. When it terminates, we have found the desired $u$.
answered Jun 8 '18 at 8:42
CarlCarl
2,22311127
$endgroup$
$begingroup$
This is, of course, assuming the graph is finite or at least that the vertices have finite degree. If the vertices don't have finite degree, then I don't know if the result is even still true.
$endgroup$
– Carl
Jun 8 '18 at 8:45
add a comment |
$begingroup$
Let $u_1$ be any neighbor of $v$. If $G - v - u_1$ is connected then we have found our vertex. If $G - v - u_1$ is disconnected, look at the set of components. If any component contains no neighbors of $v$, then $G-u_1$ must actually be disconnected, which is a contradiction. So we may assume all components contain neighbors of $v$. Choose some such component, and let $u_2$ be one of the neighbors of $v$ lying in it.
Now consider $G - v - u_2$. If this is connected then again we have found our vertex. If it is disconnected, look at the set of components, which again must each contain neighbors of $v$. Choose some component that does NOT contain $u_1$, choose a neighbor of $v$ in it, and call that neighbor $u_3$.
The idea is that we repeat this process, and when we look at the components of $G - v - u_k$ we can observe that $u_1 cdots u_{k-1}$ must all actually lie in the same component. Thus, we never end up in a circle, and the process must eventually terminate. When it terminates, we have found the desired $u$.
answered Jun 8 '18 at 8:42
CarlCarl
2,22311127
$endgroup$
Let $u_1$ be any neighbor of $v$. If $G - v - u_1$ is connected then we have found our vertex. If $G - v - u_1$ is disconnected, look at the set of components. If any component contains no neighbors of $v$, then $G-u_1$ must actually be disconnected, which is a contradiction. So we may assume all components contain neighbors of $v$. Choose some such component, and let $u_2$ be one of the neighbors of $v$ lying in it.
Now consider $G - v - u_2$. If this is connected then again we have found our vertex. If it is disconnected, look at the set of components, which again must each contain neighbors of $v$. Choose some component that does NOT contain $u_1$, choose a neighbor of $v$ in it, and call that neighbor $u_3$.
The idea is that we repeat this process, and when we look at the components of $G - v - u_k$ we can observe that $u_1 cdots u_{k-1}$ must all actually lie in the same component. Thus, we never end up in a circle, and the process must eventually terminate. When it terminates, we have found the desired $u$.
answered Jun 8 '18 at 8:42
CarlCarl
2,22311127
answered Jun 8 '18 at 8:42
CarlCarl
2,22311127
answered Jun 8 '18 at 8:42
CarlCarl
2,22311127
answered Jun 8 '18 at 8:42
CarlCarl
2,22311127
2,22311127
$begingroup$
This is, of course, assuming the graph is finite or at least that the vertices have finite degree. If the vertices don't have finite degree, then I don't know if the result is even still true.
$endgroup$
– Carl
Jun 8 '18 at 8:45
add a comment |
$begingroup$
This is, of course, assuming the graph is finite or at least that the vertices have finite degree. If the vertices don't have finite degree, then I don't know if the result is even still true.
$endgroup$
– Carl
Jun 8 '18 at 8:45
$begingroup$
This is, of course, assuming the graph is finite or at least that the vertices have finite degree. If the vertices don't have finite degree, then I don't know if the result is even still true.
$endgroup$
– Carl
Jun 8 '18 at 8:45
$begingroup$
This is, of course, assuming the graph is finite or at least that the vertices have finite degree. If the vertices don't have finite degree, then I don't know if the result is even still true.
$endgroup$
– Carl
Jun 8 '18 at 8:45
add a comment |
$begingroup$
Take a vertex $v$ of $G$, and let $U:=N(v)$ be the set of neighbors of $v$, then let $T$ be the smallest connected subgraph of $G-v$ containing all vertices in $U$. Then $T$ is a tree as you can remove edges on the any cycle to make it a tree. Also, all leafs of $T$ are vertices in $U$ as otherwise you can just remove the leaf to make a smaller connected subgraph containing all vertices in $U$. Take $u$ be any leaf of $T$, and suppose $G-v-u$ is not connected. Then let $C$ be a component of $G-v-u$ not containing $T-u$. But then $C$ does not contain any neighbors of $v$, so $G-v$ is disconnected contradicting 2-connectivity of $G$.
answered Dec 18 '18 at 16:06
nafhgoodnafhgood
1,803422
$endgroup$
add a comment |
$begingroup$
Take a vertex $v$ of $G$, and let $U:=N(v)$ be the set of neighbors of $v$, then let $T$ be the smallest connected subgraph of $G-v$ containing all vertices in $U$. Then $T$ is a tree as you can remove edges on the any cycle to make it a tree. Also, all leafs of $T$ are vertices in $U$ as otherwise you can just remove the leaf to make a smaller connected subgraph containing all vertices in $U$. Take $u$ be any leaf of $T$, and suppose $G-v-u$ is not connected. Then let $C$ be a component of $G-v-u$ not containing $T-u$. But then $C$ does not contain any neighbors of $v$, so $G-v$ is disconnected contradicting 2-connectivity of $G$.
answered Dec 18 '18 at 16:06
nafhgoodnafhgood
1,803422
$endgroup$
add a comment |
$begingroup$
Take a vertex $v$ of $G$, and let $U:=N(v)$ be the set of neighbors of $v$, then let $T$ be the smallest connected subgraph of $G-v$ containing all vertices in $U$. Then $T$ is a tree as you can remove edges on the any cycle to make it a tree. Also, all leafs of $T$ are vertices in $U$ as otherwise you can just remove the leaf to make a smaller connected subgraph containing all vertices in $U$. Take $u$ be any leaf of $T$, and suppose $G-v-u$ is not connected. Then let $C$ be a component of $G-v-u$ not containing $T-u$. But then $C$ does not contain any neighbors of $v$, so $G-v$ is disconnected contradicting 2-connectivity of $G$.
answered Dec 18 '18 at 16:06
nafhgoodnafhgood
1,803422
$endgroup$
Take a vertex $v$ of $G$, and let $U:=N(v)$ be the set of neighbors of $v$, then let $T$ be the smallest connected subgraph of $G-v$ containing all vertices in $U$. Then $T$ is a tree as you can remove edges on the any cycle to make it a tree. Also, all leafs of $T$ are vertices in $U$ as otherwise you can just remove the leaf to make a smaller connected subgraph containing all vertices in $U$. Take $u$ be any leaf of $T$, and suppose $G-v-u$ is not connected. Then let $C$ be a component of $G-v-u$ not containing $T-u$. But then $C$ does not contain any neighbors of $v$, so $G-v$ is disconnected contradicting 2-connectivity of $G$.
answered Dec 18 '18 at 16:06
nafhgoodnafhgood
1,803422
answered Dec 18 '18 at 16:06
nafhgoodnafhgood
1,803422
answered Dec 18 '18 at 16:06
nafhgoodnafhgood
1,803422
answered Dec 18 '18 at 16:06
nafhgoodnafhgood
1,803422
1,803422
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%2f2812268%2flet-v-be-a-vertex-of-a-2-connected-graph-g-prove-that-v-has-a-neighbor-u%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
