Why is the average-case running time of quicksort $O(n log_2 n)$?
$begingroup$
What is a mathematically rigorous explanation of why quicksort's average-case running time is $O(n log_2 n)$?
algorithms
asked Dec 6 '18 at 7:22
Austin ConlonAustin Conlon
2621412
$endgroup$
add a comment |
$begingroup$
What is a mathematically rigorous explanation of why quicksort's average-case running time is $O(n log_2 n)$?
algorithms
asked Dec 6 '18 at 7:22
Austin ConlonAustin Conlon
2621412
$endgroup$
add a comment |
$begingroup$
What is a mathematically rigorous explanation of why quicksort's average-case running time is $O(n log_2 n)$?
algorithms
asked Dec 6 '18 at 7:22
Austin ConlonAustin Conlon
2621412
$endgroup$
What is a mathematically rigorous explanation of why quicksort's average-case running time is $O(n log_2 n)$?
algorithms
algorithms
asked Dec 6 '18 at 7:22
Austin ConlonAustin Conlon
2621412
asked Dec 6 '18 at 7:22
Austin ConlonAustin Conlon
2621412
asked Dec 6 '18 at 7:22
Austin ConlonAustin Conlon
2621412
asked Dec 6 '18 at 7:22
Austin ConlonAustin Conlon
2621412
asked Dec 6 '18 at 7:22
Austin ConlonAustin Conlon
2621412
2621412
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
To do average case analysis, we need to consider all possible permutation of the input array and calculate the time taken by every permutation which is not easy. We can get an idea of average case by considering the case when partition cuts $O(frac{n}{9})$ elements in one set and $O(frac{9n}{10})$ elements in other set. Following this we get
begin{equation}
T(n) = T(frac{n}{9}) + T(frac{9n}{10}) + O(n)
end{equation}
which is shown to be $O(n log n)$. If you average over all such permutations, you will also get $O(n log n)$.
answered Dec 6 '18 at 7:37
Ahmad BazziAhmad Bazzi
8,3622824
$endgroup$
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%2f3028171%2fwhy-is-the-average-case-running-time-of-quicksort-on-log-2-n%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
$begingroup$
To do average case analysis, we need to consider all possible permutation of the input array and calculate the time taken by every permutation which is not easy. We can get an idea of average case by considering the case when partition cuts $O(frac{n}{9})$ elements in one set and $O(frac{9n}{10})$ elements in other set. Following this we get
begin{equation}
T(n) = T(frac{n}{9}) + T(frac{9n}{10}) + O(n)
end{equation}
which is shown to be $O(n log n)$. If you average over all such permutations, you will also get $O(n log n)$.
answered Dec 6 '18 at 7:37
Ahmad BazziAhmad Bazzi
8,3622824
$endgroup$
add a comment |
$begingroup$
To do average case analysis, we need to consider all possible permutation of the input array and calculate the time taken by every permutation which is not easy. We can get an idea of average case by considering the case when partition cuts $O(frac{n}{9})$ elements in one set and $O(frac{9n}{10})$ elements in other set. Following this we get
begin{equation}
T(n) = T(frac{n}{9}) + T(frac{9n}{10}) + O(n)
end{equation}
which is shown to be $O(n log n)$. If you average over all such permutations, you will also get $O(n log n)$.
answered Dec 6 '18 at 7:37
Ahmad BazziAhmad Bazzi
8,3622824
$endgroup$
add a comment |
$begingroup$
To do average case analysis, we need to consider all possible permutation of the input array and calculate the time taken by every permutation which is not easy. We can get an idea of average case by considering the case when partition cuts $O(frac{n}{9})$ elements in one set and $O(frac{9n}{10})$ elements in other set. Following this we get
begin{equation}
T(n) = T(frac{n}{9}) + T(frac{9n}{10}) + O(n)
end{equation}
which is shown to be $O(n log n)$. If you average over all such permutations, you will also get $O(n log n)$.
answered Dec 6 '18 at 7:37
Ahmad BazziAhmad Bazzi
8,3622824
$endgroup$
To do average case analysis, we need to consider all possible permutation of the input array and calculate the time taken by every permutation which is not easy. We can get an idea of average case by considering the case when partition cuts $O(frac{n}{9})$ elements in one set and $O(frac{9n}{10})$ elements in other set. Following this we get
begin{equation}
T(n) = T(frac{n}{9}) + T(frac{9n}{10}) + O(n)
end{equation}
which is shown to be $O(n log n)$. If you average over all such permutations, you will also get $O(n log n)$.
answered Dec 6 '18 at 7:37
Ahmad BazziAhmad Bazzi
8,3622824
answered Dec 6 '18 at 7:37
Ahmad BazziAhmad Bazzi
8,3622824
answered Dec 6 '18 at 7:37
Ahmad BazziAhmad Bazzi
8,3622824
answered Dec 6 '18 at 7:37
Ahmad BazziAhmad Bazzi
8,3622824
8,3622824
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%2f3028171%2fwhy-is-the-average-case-running-time-of-quicksort-on-log-2-n%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
