Diameter of Cayley graphs of symmetric groups
2
$begingroup$
Let $S_n$ denote the symmetric group on $n$ letters. If we consider the Cayley graph $Gamma(S_n,C)$, where $C={(12),(12cdots n)}$, is there any formula for calculating the diameter of $Gamma(S_n,C)$?
In fact, for an example, I need to know the diameter of this Cayley graph for big numbers, e.g. $n=100$ or $n=1000$.
finite-groups cayley-graphs
asked Nov 26 '18 at 7:51
kherskhers
778
$endgroup$
|
show 1 more comment
2
$begingroup$
Let $S_n$ denote the symmetric group on $n$ letters. If we consider the Cayley graph $Gamma(S_n,C)$, where $C={(12),(12cdots n)}$, is there any formula for calculating the diameter of $Gamma(S_n,C)$?
In fact, for an example, I need to know the diameter of this Cayley graph for big numbers, e.g. $n=100$ or $n=1000$.
finite-groups cayley-graphs
asked Nov 26 '18 at 7:51
kherskhers
778
$endgroup$
$begingroup$
We can get an upper bound on the number of elements distance $k$ away from a point, which gives us lower bound on the diameter. Fix a starting point (such at $e$). Every vertex in the graph is of degree $3$, and so there is $1$ vertex distance $0$ away, $3$ distance $1$, and at most $3cdot 2^{k-1}$ distance $k$ away for $kgeq 1$. Summing, there are at most $1+3cdot 2^k$ vertices distance at most $k$ away. Since there are $n!$ vertices, if $k< log_2((n!-1)/3)$, then we will not have exhausted all the vertices. Therefore, this is a lower bound for the diameter.
$endgroup$
– Aaron
Nov 26 '18 at 9:06
$begingroup$
@Aaron Thanks so much for your help, Is there any upper bound for this diameter which is tight? Thanks again
$endgroup$
– khers
Nov 26 '18 at 10:45
$begingroup$
I don't know myself (I've never thought about this before). But google allowed me to find the following, which says the answer is $O(n^2)$. math.uchicago.edu/~may/VIGRE/VIGRE2011/REUPapers/Tan.pdf
$endgroup$
– Aaron
Nov 26 '18 at 11:09
$begingroup$
@Aaron Thanks so much again. I found the upper bound $(n−1).5n$ in the paper, However, I do not know that it is tight or not. Thanks again.
$endgroup$
– khers
Nov 26 '18 at 11:28
$begingroup$
I don't know how tight it is, but from the lower bound and stirlings approximation, we know that the true order is somewhere between $nlog n$ and $n^2$. This isn't enough to pin things down up to a constant, but it's not too wide a range.
$endgroup$
– Aaron
Nov 26 '18 at 11:44
|
show 1 more comment
2
2
2
1
$begingroup$
Let $S_n$ denote the symmetric group on $n$ letters. If we consider the Cayley graph $Gamma(S_n,C)$, where $C={(12),(12cdots n)}$, is there any formula for calculating the diameter of $Gamma(S_n,C)$?
In fact, for an example, I need to know the diameter of this Cayley graph for big numbers, e.g. $n=100$ or $n=1000$.
finite-groups cayley-graphs
asked Nov 26 '18 at 7:51
kherskhers
778
$endgroup$
Let $S_n$ denote the symmetric group on $n$ letters. If we consider the Cayley graph $Gamma(S_n,C)$, where $C={(12),(12cdots n)}$, is there any formula for calculating the diameter of $Gamma(S_n,C)$?
In fact, for an example, I need to know the diameter of this Cayley graph for big numbers, e.g. $n=100$ or $n=1000$.
finite-groups cayley-graphs
finite-groups cayley-graphs
asked Nov 26 '18 at 7:51
kherskhers
778
asked Nov 26 '18 at 7:51
kherskhers
778
edited Nov 26 '18 at 8:39
khers
asked Nov 26 '18 at 7:51
kherskhers
778
asked Nov 26 '18 at 7:51
kherskhers
778
asked Nov 26 '18 at 7:51
kherskhers
778
778
$begingroup$
We can get an upper bound on the number of elements distance $k$ away from a point, which gives us lower bound on the diameter. Fix a starting point (such at $e$). Every vertex in the graph is of degree $3$, and so there is $1$ vertex distance $0$ away, $3$ distance $1$, and at most $3cdot 2^{k-1}$ distance $k$ away for $kgeq 1$. Summing, there are at most $1+3cdot 2^k$ vertices distance at most $k$ away. Since there are $n!$ vertices, if $k< log_2((n!-1)/3)$, then we will not have exhausted all the vertices. Therefore, this is a lower bound for the diameter.
$endgroup$
– Aaron
Nov 26 '18 at 9:06
$begingroup$
@Aaron Thanks so much for your help, Is there any upper bound for this diameter which is tight? Thanks again
$endgroup$
– khers
Nov 26 '18 at 10:45
$begingroup$
I don't know myself (I've never thought about this before). But google allowed me to find the following, which says the answer is $O(n^2)$. math.uchicago.edu/~may/VIGRE/VIGRE2011/REUPapers/Tan.pdf
$endgroup$
– Aaron
Nov 26 '18 at 11:09
$begingroup$
@Aaron Thanks so much again. I found the upper bound $(n−1).5n$ in the paper, However, I do not know that it is tight or not. Thanks again.
$endgroup$
– khers
Nov 26 '18 at 11:28
$begingroup$
I don't know how tight it is, but from the lower bound and stirlings approximation, we know that the true order is somewhere between $nlog n$ and $n^2$. This isn't enough to pin things down up to a constant, but it's not too wide a range.
$endgroup$
– Aaron
Nov 26 '18 at 11:44
|
show 1 more comment
$begingroup$
We can get an upper bound on the number of elements distance $k$ away from a point, which gives us lower bound on the diameter. Fix a starting point (such at $e$). Every vertex in the graph is of degree $3$, and so there is $1$ vertex distance $0$ away, $3$ distance $1$, and at most $3cdot 2^{k-1}$ distance $k$ away for $kgeq 1$. Summing, there are at most $1+3cdot 2^k$ vertices distance at most $k$ away. Since there are $n!$ vertices, if $k< log_2((n!-1)/3)$, then we will not have exhausted all the vertices. Therefore, this is a lower bound for the diameter.
$endgroup$
– Aaron
Nov 26 '18 at 9:06
$begingroup$
@Aaron Thanks so much for your help, Is there any upper bound for this diameter which is tight? Thanks again
$endgroup$
– khers
Nov 26 '18 at 10:45
$begingroup$
I don't know myself (I've never thought about this before). But google allowed me to find the following, which says the answer is $O(n^2)$. math.uchicago.edu/~may/VIGRE/VIGRE2011/REUPapers/Tan.pdf
$endgroup$
– Aaron
Nov 26 '18 at 11:09
$begingroup$
@Aaron Thanks so much again. I found the upper bound $(n−1).5n$ in the paper, However, I do not know that it is tight or not. Thanks again.
$endgroup$
– khers
Nov 26 '18 at 11:28
$begingroup$
I don't know how tight it is, but from the lower bound and stirlings approximation, we know that the true order is somewhere between $nlog n$ and $n^2$. This isn't enough to pin things down up to a constant, but it's not too wide a range.
$endgroup$
– Aaron
Nov 26 '18 at 11:44
$begingroup$
We can get an upper bound on the number of elements distance $k$ away from a point, which gives us lower bound on the diameter. Fix a starting point (such at $e$). Every vertex in the graph is of degree $3$, and so there is $1$ vertex distance $0$ away, $3$ distance $1$, and at most $3cdot 2^{k-1}$ distance $k$ away for $kgeq 1$. Summing, there are at most $1+3cdot 2^k$ vertices distance at most $k$ away. Since there are $n!$ vertices, if $k< log_2((n!-1)/3)$, then we will not have exhausted all the vertices. Therefore, this is a lower bound for the diameter.
$endgroup$
– Aaron
Nov 26 '18 at 9:06
$begingroup$
We can get an upper bound on the number of elements distance $k$ away from a point, which gives us lower bound on the diameter. Fix a starting point (such at $e$). Every vertex in the graph is of degree $3$, and so there is $1$ vertex distance $0$ away, $3$ distance $1$, and at most $3cdot 2^{k-1}$ distance $k$ away for $kgeq 1$. Summing, there are at most $1+3cdot 2^k$ vertices distance at most $k$ away. Since there are $n!$ vertices, if $k< log_2((n!-1)/3)$, then we will not have exhausted all the vertices. Therefore, this is a lower bound for the diameter.
$endgroup$
– Aaron
Nov 26 '18 at 9:06
$begingroup$
@Aaron Thanks so much for your help, Is there any upper bound for this diameter which is tight? Thanks again
$endgroup$
– khers
Nov 26 '18 at 10:45
$begingroup$
@Aaron Thanks so much for your help, Is there any upper bound for this diameter which is tight? Thanks again
$endgroup$
– khers
Nov 26 '18 at 10:45
$begingroup$
I don't know myself (I've never thought about this before). But google allowed me to find the following, which says the answer is $O(n^2)$. math.uchicago.edu/~may/VIGRE/VIGRE2011/REUPapers/Tan.pdf
$endgroup$
– Aaron
Nov 26 '18 at 11:09
$begingroup$
I don't know myself (I've never thought about this before). But google allowed me to find the following, which says the answer is $O(n^2)$. math.uchicago.edu/~may/VIGRE/VIGRE2011/REUPapers/Tan.pdf
$endgroup$
– Aaron
Nov 26 '18 at 11:09
$begingroup$
@Aaron Thanks so much again. I found the upper bound $(n−1).5n$ in the paper, However, I do not know that it is tight or not. Thanks again.
$endgroup$
– khers
Nov 26 '18 at 11:28
$begingroup$
@Aaron Thanks so much again. I found the upper bound $(n−1).5n$ in the paper, However, I do not know that it is tight or not. Thanks again.
$endgroup$
– khers
Nov 26 '18 at 11:28
$begingroup$
I don't know how tight it is, but from the lower bound and stirlings approximation, we know that the true order is somewhere between $nlog n$ and $n^2$. This isn't enough to pin things down up to a constant, but it's not too wide a range.
$endgroup$
– Aaron
Nov 26 '18 at 11:44
$begingroup$
I don't know how tight it is, but from the lower bound and stirlings approximation, we know that the true order is somewhere between $nlog n$ and $n^2$. This isn't enough to pin things down up to a constant, but it's not too wide a range.
$endgroup$
– Aaron
Nov 26 '18 at 11:44
|
show 1 more 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
});
}
});
draft saved
draft discarded
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%2f3014007%2fdiameter-of-cayley-graphs-of-symmetric-groups%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
draft saved
draft discarded
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.
draft saved
draft discarded
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%2f3014007%2fdiameter-of-cayley-graphs-of-symmetric-groups%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
$begingroup$
We can get an upper bound on the number of elements distance $k$ away from a point, which gives us lower bound on the diameter. Fix a starting point (such at $e$). Every vertex in the graph is of degree $3$, and so there is $1$ vertex distance $0$ away, $3$ distance $1$, and at most $3cdot 2^{k-1}$ distance $k$ away for $kgeq 1$. Summing, there are at most $1+3cdot 2^k$ vertices distance at most $k$ away. Since there are $n!$ vertices, if $k< log_2((n!-1)/3)$, then we will not have exhausted all the vertices. Therefore, this is a lower bound for the diameter.
$endgroup$
– Aaron
Nov 26 '18 at 9:06
$begingroup$
@Aaron Thanks so much for your help, Is there any upper bound for this diameter which is tight? Thanks again
$endgroup$
– khers
Nov 26 '18 at 10:45
$begingroup$
I don't know myself (I've never thought about this before). But google allowed me to find the following, which says the answer is $O(n^2)$. math.uchicago.edu/~may/VIGRE/VIGRE2011/REUPapers/Tan.pdf
$endgroup$
– Aaron
Nov 26 '18 at 11:09
$begingroup$
@Aaron Thanks so much again. I found the upper bound $(n−1).5n$ in the paper, However, I do not know that it is tight or not. Thanks again.
$endgroup$
– khers
Nov 26 '18 at 11:28
$begingroup$
I don't know how tight it is, but from the lower bound and stirlings approximation, we know that the true order is somewhere between $nlog n$ and $n^2$. This isn't enough to pin things down up to a constant, but it's not too wide a range.
$endgroup$
– Aaron
Nov 26 '18 at 11:44