Mixing
Revolving door graph
$begingroup$
How can I prove that revolving door graph $RD^d$ is $d$-regular graph?
$RD^d$ is bipartite graph in which the vertices are $(d-1)$-element subsets and $d$-element subsets of a $(2d-1)$-element ground set. Two vertices are adjacent if one of the corresponding sets is a subset of the other.
graph-theory
asked Jan 17 at 10:04
KateKate
104
$endgroup$
add a comment |
$begingroup$
How can I prove that revolving door graph $RD^d$ is $d$-regular graph?
$RD^d$ is bipartite graph in which the vertices are $(d-1)$-element subsets and $d$-element subsets of a $(2d-1)$-element ground set. Two vertices are adjacent if one of the corresponding sets is a subset of the other.
graph-theory
asked Jan 17 at 10:04
KateKate
104
$endgroup$
add a comment |
$begingroup$
How can I prove that revolving door graph $RD^d$ is $d$-regular graph?
$RD^d$ is bipartite graph in which the vertices are $(d-1)$-element subsets and $d$-element subsets of a $(2d-1)$-element ground set. Two vertices are adjacent if one of the corresponding sets is a subset of the other.
graph-theory
asked Jan 17 at 10:04
KateKate
104
$endgroup$
How can I prove that revolving door graph $RD^d$ is $d$-regular graph?
$RD^d$ is bipartite graph in which the vertices are $(d-1)$-element subsets and $d$-element subsets of a $(2d-1)$-element ground set. Two vertices are adjacent if one of the corresponding sets is a subset of the other.
graph-theory
graph-theory
asked Jan 17 at 10:04
KateKate
104
asked Jan 17 at 10:04
KateKate
104
asked Jan 17 at 10:04
KateKate
104
asked Jan 17 at 10:04
KateKate
104
asked Jan 17 at 10:04
KateKate
104
104
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First focus on a $(d-1)$-element subset $X$. In how many ways can you choose another element from the $2d-1$ elements to enlarge $X$ to size $d$?
Then focus on a $d$-element subset $Y$. In how many ways can you remove an element from $Y$ to shrink it to size $d-1$.
Conclusion?
answered Jan 17 at 10:10
Leen DroogendijkLeen Droogendijk
6,1351716
$endgroup$
$begingroup$
$2d-1-(d-1)=d$, so $d$ ways. In second case also $d$ ways. So is that enough to prove it?
$endgroup$
– Kate
Jan 17 at 10:27
$begingroup$
Yes, we have shown that every element has $d$ edges to other elements. But the fact that you ask this question shows that you have not yet properly mastered the concepts involved. Do some more study until you thoroughly understand this, then write your own proof.
$endgroup$
– Leen Droogendijk
Jan 17 at 10:54
$begingroup$
Is that standard terminology? How is that graph like a revolving door?
$endgroup$
– bof
Jan 17 at 11:24
$begingroup$
Try to change one subset into another one (using a path on the graph). You have to change one element at a time.
$endgroup$
– Leen Droogendijk
Jan 17 at 12:01
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%2f3076801%2frevolving-door-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
$begingroup$
First focus on a $(d-1)$-element subset $X$. In how many ways can you choose another element from the $2d-1$ elements to enlarge $X$ to size $d$?
Then focus on a $d$-element subset $Y$. In how many ways can you remove an element from $Y$ to shrink it to size $d-1$.
Conclusion?
answered Jan 17 at 10:10
Leen DroogendijkLeen Droogendijk
6,1351716
$endgroup$
$begingroup$
$2d-1-(d-1)=d$, so $d$ ways. In second case also $d$ ways. So is that enough to prove it?
$endgroup$
– Kate
Jan 17 at 10:27
$begingroup$
Yes, we have shown that every element has $d$ edges to other elements. But the fact that you ask this question shows that you have not yet properly mastered the concepts involved. Do some more study until you thoroughly understand this, then write your own proof.
$endgroup$
– Leen Droogendijk
Jan 17 at 10:54
$begingroup$
Is that standard terminology? How is that graph like a revolving door?
$endgroup$
– bof
Jan 17 at 11:24
$begingroup$
Try to change one subset into another one (using a path on the graph). You have to change one element at a time.
$endgroup$
– Leen Droogendijk
Jan 17 at 12:01
add a comment |
$begingroup$
First focus on a $(d-1)$-element subset $X$. In how many ways can you choose another element from the $2d-1$ elements to enlarge $X$ to size $d$?
Then focus on a $d$-element subset $Y$. In how many ways can you remove an element from $Y$ to shrink it to size $d-1$.
Conclusion?
answered Jan 17 at 10:10
Leen DroogendijkLeen Droogendijk
6,1351716
$endgroup$
$begingroup$
$2d-1-(d-1)=d$, so $d$ ways. In second case also $d$ ways. So is that enough to prove it?
$endgroup$
– Kate
Jan 17 at 10:27
$begingroup$
Yes, we have shown that every element has $d$ edges to other elements. But the fact that you ask this question shows that you have not yet properly mastered the concepts involved. Do some more study until you thoroughly understand this, then write your own proof.
$endgroup$
– Leen Droogendijk
Jan 17 at 10:54
$begingroup$
Is that standard terminology? How is that graph like a revolving door?
$endgroup$
– bof
Jan 17 at 11:24
$begingroup$
Try to change one subset into another one (using a path on the graph). You have to change one element at a time.
$endgroup$
– Leen Droogendijk
Jan 17 at 12:01
add a comment |
$begingroup$
First focus on a $(d-1)$-element subset $X$. In how many ways can you choose another element from the $2d-1$ elements to enlarge $X$ to size $d$?
Then focus on a $d$-element subset $Y$. In how many ways can you remove an element from $Y$ to shrink it to size $d-1$.
Conclusion?
answered Jan 17 at 10:10
Leen DroogendijkLeen Droogendijk
6,1351716
$endgroup$
First focus on a $(d-1)$-element subset $X$. In how many ways can you choose another element from the $2d-1$ elements to enlarge $X$ to size $d$?
Then focus on a $d$-element subset $Y$. In how many ways can you remove an element from $Y$ to shrink it to size $d-1$.
Conclusion?
answered Jan 17 at 10:10
Leen DroogendijkLeen Droogendijk
6,1351716
answered Jan 17 at 10:10
Leen DroogendijkLeen Droogendijk
6,1351716
answered Jan 17 at 10:10
Leen DroogendijkLeen Droogendijk
6,1351716
answered Jan 17 at 10:10
Leen DroogendijkLeen Droogendijk
6,1351716
6,1351716
$begingroup$
$2d-1-(d-1)=d$, so $d$ ways. In second case also $d$ ways. So is that enough to prove it?
$endgroup$
– Kate
Jan 17 at 10:27
$begingroup$
Yes, we have shown that every element has $d$ edges to other elements. But the fact that you ask this question shows that you have not yet properly mastered the concepts involved. Do some more study until you thoroughly understand this, then write your own proof.
$endgroup$
– Leen Droogendijk
Jan 17 at 10:54
$begingroup$
Is that standard terminology? How is that graph like a revolving door?
$endgroup$
– bof
Jan 17 at 11:24
$begingroup$
Try to change one subset into another one (using a path on the graph). You have to change one element at a time.
$endgroup$
– Leen Droogendijk
Jan 17 at 12:01
add a comment |
$begingroup$
$2d-1-(d-1)=d$, so $d$ ways. In second case also $d$ ways. So is that enough to prove it?
$endgroup$
– Kate
Jan 17 at 10:27
$begingroup$
Yes, we have shown that every element has $d$ edges to other elements. But the fact that you ask this question shows that you have not yet properly mastered the concepts involved. Do some more study until you thoroughly understand this, then write your own proof.
$endgroup$
– Leen Droogendijk
Jan 17 at 10:54
$begingroup$
Is that standard terminology? How is that graph like a revolving door?
$endgroup$
– bof
Jan 17 at 11:24
$begingroup$
Try to change one subset into another one (using a path on the graph). You have to change one element at a time.
$endgroup$
– Leen Droogendijk
Jan 17 at 12:01
$begingroup$
$2d-1-(d-1)=d$, so $d$ ways. In second case also $d$ ways. So is that enough to prove it?
$endgroup$
– Kate
Jan 17 at 10:27
$begingroup$
$2d-1-(d-1)=d$, so $d$ ways. In second case also $d$ ways. So is that enough to prove it?
$endgroup$
– Kate
Jan 17 at 10:27
$begingroup$
Yes, we have shown that every element has $d$ edges to other elements. But the fact that you ask this question shows that you have not yet properly mastered the concepts involved. Do some more study until you thoroughly understand this, then write your own proof.
$endgroup$
– Leen Droogendijk
Jan 17 at 10:54
$begingroup$
Yes, we have shown that every element has $d$ edges to other elements. But the fact that you ask this question shows that you have not yet properly mastered the concepts involved. Do some more study until you thoroughly understand this, then write your own proof.
$endgroup$
– Leen Droogendijk
Jan 17 at 10:54
$begingroup$
Is that standard terminology? How is that graph like a revolving door?
$endgroup$
– bof
Jan 17 at 11:24
$begingroup$
Is that standard terminology? How is that graph like a revolving door?
$endgroup$
– bof
Jan 17 at 11:24
$begingroup$
Try to change one subset into another one (using a path on the graph). You have to change one element at a time.
$endgroup$
– Leen Droogendijk
Jan 17 at 12:01
$begingroup$
Try to change one subset into another one (using a path on the graph). You have to change one element at a time.
$endgroup$
– Leen Droogendijk
Jan 17 at 12:01
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%2f3076801%2frevolving-door-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
