Complexity of many constant time steps with occasional logarithmic steps












1












$begingroup$


I have a data structure that can perform a task $T$ in constant time, $O(1)$. However, every $k$th invocation requires $O(log{n})$, where $k$ is constant.



Is it possible for this task to ever take amortized constant time, or is it impossible because the logarithm will eventually become greater than $k$?



If an upper bound for $n$ is known as $N$, can $k$ be chosen to be less than $log{N}$?






algorithm-analysis runtime-analysis amortized-analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    It depends on how $k$ relates to $n$. For instance $k=2$ then this will not matter and $n$ operations will take $O(n log n)$. If $k = n$ then after $n$ operations we have time $O(n + log n)$. How does $k$ relate to $n$?
    $endgroup$
    – ryan
    1 hour ago










  • $begingroup$
    @ryan k is constant. (I have edited the question to specify this)
    $endgroup$
    – rtheunissen
    7 mins ago


















1












$begingroup$


I have a data structure that can perform a task $T$ in constant time, $O(1)$. However, every $k$th invocation requires $O(log{n})$, where $k$ is constant.



Is it possible for this task to ever take amortized constant time, or is it impossible because the logarithm will eventually become greater than $k$?



If an upper bound for $n$ is known as $N$, can $k$ be chosen to be less than $log{N}$?






algorithm-analysis runtime-analysis amortized-analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    It depends on how $k$ relates to $n$. For instance $k=2$ then this will not matter and $n$ operations will take $O(n log n)$. If $k = n$ then after $n$ operations we have time $O(n + log n)$. How does $k$ relate to $n$?
    $endgroup$
    – ryan
    1 hour ago










  • $begingroup$
    @ryan k is constant. (I have edited the question to specify this)
    $endgroup$
    – rtheunissen
    7 mins ago
















1












1








1





$begingroup$


I have a data structure that can perform a task $T$ in constant time, $O(1)$. However, every $k$th invocation requires $O(log{n})$, where $k$ is constant.



Is it possible for this task to ever take amortized constant time, or is it impossible because the logarithm will eventually become greater than $k$?



If an upper bound for $n$ is known as $N$, can $k$ be chosen to be less than $log{N}$?






algorithm-analysis runtime-analysis amortized-analysis







share|cite|improve this question











$endgroup$




I have a data structure that can perform a task $T$ in constant time, $O(1)$. However, every $k$th invocation requires $O(log{n})$, where $k$ is constant.



Is it possible for this task to ever take amortized constant time, or is it impossible because the logarithm will eventually become greater than $k$?



If an upper bound for $n$ is known as $N$, can $k$ be chosen to be less than $log{N}$?






algorithm-analysis runtime-analysis amortized-analysis




algorithm-analysis runtime-analysis amortized-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 5 mins ago







rtheunissen

















asked 1 hour ago









rtheunissenrtheunissen

1204




1204








  • 1




    $begingroup$
    It depends on how $k$ relates to $n$. For instance $k=2$ then this will not matter and $n$ operations will take $O(n log n)$. If $k = n$ then after $n$ operations we have time $O(n + log n)$. How does $k$ relate to $n$?
    $endgroup$
    – ryan
    1 hour ago










  • $begingroup$
    @ryan k is constant. (I have edited the question to specify this)
    $endgroup$
    – rtheunissen
    7 mins ago
















  • 1




    $begingroup$
    It depends on how $k$ relates to $n$. For instance $k=2$ then this will not matter and $n$ operations will take $O(n log n)$. If $k = n$ then after $n$ operations we have time $O(n + log n)$. How does $k$ relate to $n$?
    $endgroup$
    – ryan
    1 hour ago










  • $begingroup$
    @ryan k is constant. (I have edited the question to specify this)
    $endgroup$
    – rtheunissen
    7 mins ago










1




1




$begingroup$
It depends on how $k$ relates to $n$. For instance $k=2$ then this will not matter and $n$ operations will take $O(n log n)$. If $k = n$ then after $n$ operations we have time $O(n + log n)$. How does $k$ relate to $n$?
$endgroup$
– ryan
1 hour ago




$begingroup$
It depends on how $k$ relates to $n$. For instance $k=2$ then this will not matter and $n$ operations will take $O(n log n)$. If $k = n$ then after $n$ operations we have time $O(n + log n)$. How does $k$ relate to $n$?
$endgroup$
– ryan
1 hour ago












$begingroup$
@ryan k is constant. (I have edited the question to specify this)
$endgroup$
– rtheunissen
7 mins ago






$begingroup$
@ryan k is constant. (I have edited the question to specify this)
$endgroup$
– rtheunissen
7 mins ago












1 Answer
1






active

oldest

votes


















2












$begingroup$

If every $k$th operation takes $O(log n)$ time, then the best bound you can get on the amortized complexity is $O(1 + frac{log n}{k})$. This follows from the definition of amortized complexity.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Does that mean that if k is constant, the amortized complexity is O(1 + (log n / k))?
    $endgroup$
    – rtheunissen
    6 mins ago






  • 1




    $begingroup$
    If $k$ is constant, the amortized complexity is $O(log n)$.
    $endgroup$
    – Yuval Filmus
    4 mins ago












Your Answer








StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "419"
};
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: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
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
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcs.stackexchange.com%2fquestions%2f106957%2fcomplexity-of-many-constant-time-steps-with-occasional-logarithmic-steps%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









2












$begingroup$

If every $k$th operation takes $O(log n)$ time, then the best bound you can get on the amortized complexity is $O(1 + frac{log n}{k})$. This follows from the definition of amortized complexity.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Does that mean that if k is constant, the amortized complexity is O(1 + (log n / k))?
    $endgroup$
    – rtheunissen
    6 mins ago






  • 1




    $begingroup$
    If $k$ is constant, the amortized complexity is $O(log n)$.
    $endgroup$
    – Yuval Filmus
    4 mins ago
















2












$begingroup$

If every $k$th operation takes $O(log n)$ time, then the best bound you can get on the amortized complexity is $O(1 + frac{log n}{k})$. This follows from the definition of amortized complexity.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Does that mean that if k is constant, the amortized complexity is O(1 + (log n / k))?
    $endgroup$
    – rtheunissen
    6 mins ago






  • 1




    $begingroup$
    If $k$ is constant, the amortized complexity is $O(log n)$.
    $endgroup$
    – Yuval Filmus
    4 mins ago














2












2








2





$begingroup$

If every $k$th operation takes $O(log n)$ time, then the best bound you can get on the amortized complexity is $O(1 + frac{log n}{k})$. This follows from the definition of amortized complexity.






share|cite|improve this answer









$endgroup$



If every $k$th operation takes $O(log n)$ time, then the best bound you can get on the amortized complexity is $O(1 + frac{log n}{k})$. This follows from the definition of amortized complexity.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 42 mins ago









Yuval FilmusYuval Filmus

197k15185349




197k15185349












  • $begingroup$
    Does that mean that if k is constant, the amortized complexity is O(1 + (log n / k))?
    $endgroup$
    – rtheunissen
    6 mins ago






  • 1




    $begingroup$
    If $k$ is constant, the amortized complexity is $O(log n)$.
    $endgroup$
    – Yuval Filmus
    4 mins ago


















  • $begingroup$
    Does that mean that if k is constant, the amortized complexity is O(1 + (log n / k))?
    $endgroup$
    – rtheunissen
    6 mins ago






  • 1




    $begingroup$
    If $k$ is constant, the amortized complexity is $O(log n)$.
    $endgroup$
    – Yuval Filmus
    4 mins ago
















$begingroup$
Does that mean that if k is constant, the amortized complexity is O(1 + (log n / k))?
$endgroup$
– rtheunissen
6 mins ago




$begingroup$
Does that mean that if k is constant, the amortized complexity is O(1 + (log n / k))?
$endgroup$
– rtheunissen
6 mins ago




1




1




$begingroup$
If $k$ is constant, the amortized complexity is $O(log n)$.
$endgroup$
– Yuval Filmus
4 mins ago




$begingroup$
If $k$ is constant, the amortized complexity is $O(log n)$.
$endgroup$
– Yuval Filmus
4 mins ago


















draft saved

draft discarded




















































Thanks for contributing an answer to Computer Science 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














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcs.stackexchange.com%2fquestions%2f106957%2fcomplexity-of-many-constant-time-steps-with-occasional-logarithmic-steps%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Why is a white electrical wire connected to 2 black wires?

Image
1 I just opened my ceiling light to replace it with another mount. I see a white wire connected to 2 black wires, is that right? The middle switch controls this ceiling light. electrical ceiling light share | improve this question asked 4 hours ago konyak konyak 108 4 New contributor konyak is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. ...
Read more

Waikiki

Image
Denne artikkelen kan ha godt av ein språkvask , som reinskar opp målføringa og/eller innfører same språkstilen overalt. Denne artikkelen kan ha godt av ei opprydding for å nå ein høgare standard og/eller for å verta i tråd med standardoppsettet. Sjå korleis du redigerer ei side og stilmanualen for hjelp. Waikiki Land   USA Stat Hawaii Stad Honolulu Wikimedia Commons: Waikiki Utsyn over Waikiki Waikīkī er ein bydel i Honolulu, sør på øya Oahu, som er ei av Hawaii-øyane. Waikīkīstranda er ein av dei meste berømte strendene i verda. Waikīkī strekkjer seg langs stillehavskysten og har ei rekkje velrenommerte hotell. Bydelen er Hawaii-turismens høyborg. På Waikīkī Beach står ei statue av den lokale sportsstjerna Duka Kahanamoku. Kjelder | Denne artikkelen bygger på «Waikiki» frå Wikipedia på bokmål, den 4. november 2018. Denne artikkelen manglar kjelder eller referansar . Hjelp Wikipedia med å finna truverdige kjelder...
Read more

What are all the squawk codes?

Image
2 $begingroup$ I love aviation and I love learning new things about it. I was wondering if anyone knew all of the squawk codes and in particular what squawk 7500 means. squawk-codes share | improve this question edited 15 mins ago Jamiec ♦ 14.8k 4 60 91 asked 5 hours ago Zac Callaghan Zac Callaghan 11 2 New contributor Zac Callaghan is a new co...
Read more