Complexity of many constant time steps with occasional logarithmic steps Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Upper-bounding the number of comparisons for Sorting to $Theta(n)$ using a physically big number like Number of Particles in the UniverseWhy does this mergesort variant not do Θ(n) comparisons on average?Why does the total credit associated with a data structure must be nonnegative at all times for the accounting method?How does the token method of amortized analysis work in this example?In Amortized Analysis, can we chose how big $n$ is?Incremental strongly connected componentsThe validity of the potential function for splay treeClock page replacement policy vs LRU page replacement policy, is Clock more efficient?if binary heap potential function is c*size(binary heap)) then insert will not take O(logn)and extract min will not take O(1) amortized timeMergable heap with no key knowledge cannot EXTRACT-MIN in $o(log n)$ amortized time
Why is "Captain Marvel" translated as male in Portugal?
Estimate capacitor parameters
What do you call a plan that's an alternative plan in case your initial plan fails?
Array/tabular for long multiplication
Need a suitable toxic chemical for a murder plot in my novel
Why don't the Weasley twins use magic outside of school if the Trace can only find the location of spells cast?
How do you clear the ApexPages.getMessages() collection in a test?
A constraint that implies convexity
What LEGO pieces have "real-world" functionality?
Interesting examples of non-locally compact topological groups
Would an alien lifeform be able to achieve space travel if lacking in vision?
Statistical model of ligand substitution
How to colour the US map with Yellow, Green, Red and Blue to minimize the number of states with the colour of Green
Is above average number of years spent on PhD considered a red flag in future academia or industry positions?
What is the electric potential inside a point charge?
Slither Like a Snake
Can the prologue be the backstory of your main character?
How many spell slots should a Fighter 11/Ranger 9 have?
Can't figure this one out.. What is the missing box?
If A makes B more likely then B makes A more likely"
Single author papers against my advisor's will?
How to dynamically generate the hash value of a file while it gets downloaded from any website?
Fishing simulator
I'm having difficulty getting my players to do stuff in a sandbox campaign
Complexity of many constant time steps with occasional logarithmic steps
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Upper-bounding the number of comparisons for Sorting to $Theta(n)$ using a physically big number like Number of Particles in the UniverseWhy does this mergesort variant not do Θ(n) comparisons on average?Why does the total credit associated with a data structure must be nonnegative at all times for the accounting method?How does the token method of amortized analysis work in this example?In Amortized Analysis, can we chose how big $n$ is?Incremental strongly connected componentsThe validity of the potential function for splay treeClock page replacement policy vs LRU page replacement policy, is Clock more efficient?if binary heap potential function is c*size(binary heap)) then insert will not take O(logn)and extract min will not take O(1) amortized timeMergable heap with no key knowledge cannot EXTRACT-MIN in $o(log n)$ amortized time
$begingroup$
I have a data structure that can perform a task $T$ in constant time, $O(1)$. However, every $k$th invocation requires $O(logn)$, 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 $logN$?
algorithm-analysis runtime-analysis amortized-analysis
asked 2 hours ago
rtheunissenrtheunissen
1254
$endgroup$
add a comment |
$begingroup$
I have a data structure that can perform a task $T$ in constant time, $O(1)$. However, every $k$th invocation requires $O(logn)$, 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 $logN$?
algorithm-analysis runtime-analysis amortized-analysis
asked 2 hours ago
rtheunissenrtheunissen
1254
$endgroup$
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
1 hour ago
add a comment |
$begingroup$
I have a data structure that can perform a task $T$ in constant time, $O(1)$. However, every $k$th invocation requires $O(logn)$, 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 $logN$?
algorithm-analysis runtime-analysis amortized-analysis
asked 2 hours ago
rtheunissenrtheunissen
1254
$endgroup$
I have a data structure that can perform a task $T$ in constant time, $O(1)$. However, every $k$th invocation requires $O(logn)$, 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 $logN$?
algorithm-analysis runtime-analysis amortized-analysis
algorithm-analysis runtime-analysis amortized-analysis
asked 2 hours ago
rtheunissenrtheunissen
1254
asked 2 hours ago
rtheunissenrtheunissen
1254
edited 1 hour ago
rtheunissen
asked 2 hours ago
rtheunissenrtheunissen
1254
asked 2 hours ago
rtheunissenrtheunissen
1254
asked 2 hours ago
rtheunissenrtheunissen
1254
1254
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
1 hour ago
add a comment |
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
1 hour 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
1 hour ago
$begingroup$
@ryan k is constant. (I have edited the question to specify this)
$endgroup$
– rtheunissen
1 hour ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If every $k$th operation takes $O(log n)$ time, then the best bound you can get on the amortized complexity is $O(1 + fraclog nk)$. This follows from the definition of amortized complexity.
answered 1 hour ago
Yuval FilmusYuval Filmus
197k15185349
$endgroup$
$begingroup$
Does that mean that if k is constant, the amortized complexity is O(1 + (log n / k))?
$endgroup$
– rtheunissen
1 hour ago
3
$begingroup$
If $k$ is constant, the amortized complexity is $O(log n)$.
$endgroup$
– Yuval Filmus
1 hour ago
add a comment |
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
);
);
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%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
$begingroup$
If every $k$th operation takes $O(log n)$ time, then the best bound you can get on the amortized complexity is $O(1 + fraclog nk)$. This follows from the definition of amortized complexity.
answered 1 hour ago
Yuval FilmusYuval Filmus
197k15185349
$endgroup$
$begingroup$
Does that mean that if k is constant, the amortized complexity is O(1 + (log n / k))?
$endgroup$
– rtheunissen
1 hour ago
3
$begingroup$
If $k$ is constant, the amortized complexity is $O(log n)$.
$endgroup$
– Yuval Filmus
1 hour ago
add a comment |
$begingroup$
If every $k$th operation takes $O(log n)$ time, then the best bound you can get on the amortized complexity is $O(1 + fraclog nk)$. This follows from the definition of amortized complexity.
answered 1 hour ago
Yuval FilmusYuval Filmus
197k15185349
$endgroup$
$begingroup$
Does that mean that if k is constant, the amortized complexity is O(1 + (log n / k))?
$endgroup$
– rtheunissen
1 hour ago
3
$begingroup$
If $k$ is constant, the amortized complexity is $O(log n)$.
$endgroup$
– Yuval Filmus
1 hour ago
add a comment |
$begingroup$
If every $k$th operation takes $O(log n)$ time, then the best bound you can get on the amortized complexity is $O(1 + fraclog nk)$. This follows from the definition of amortized complexity.
answered 1 hour ago
Yuval FilmusYuval Filmus
197k15185349
$endgroup$
If every $k$th operation takes $O(log n)$ time, then the best bound you can get on the amortized complexity is $O(1 + fraclog nk)$. This follows from the definition of amortized complexity.
answered 1 hour ago
Yuval FilmusYuval Filmus
197k15185349
answered 1 hour ago
Yuval FilmusYuval Filmus
197k15185349
answered 1 hour ago
Yuval FilmusYuval Filmus
197k15185349
answered 1 hour 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
1 hour ago
3
$begingroup$
If $k$ is constant, the amortized complexity is $O(log n)$.
$endgroup$
– Yuval Filmus
1 hour ago
add a comment |
$begingroup$
Does that mean that if k is constant, the amortized complexity is O(1 + (log n / k))?
$endgroup$
– rtheunissen
1 hour ago
3
$begingroup$
If $k$ is constant, the amortized complexity is $O(log n)$.
$endgroup$
– Yuval Filmus
1 hour ago
$begingroup$
Does that mean that if k is constant, the amortized complexity is O(1 + (log n / k))?
$endgroup$
– rtheunissen
1 hour ago
$begingroup$
Does that mean that if k is constant, the amortized complexity is O(1 + (log n / k))?
$endgroup$
– rtheunissen
1 hour ago
3
3
$begingroup$
If $k$ is constant, the amortized complexity is $O(log n)$.
$endgroup$
– Yuval Filmus
1 hour ago
$begingroup$
If $k$ is constant, the amortized complexity is $O(log n)$.
$endgroup$
– Yuval Filmus
1 hour ago
add a comment |
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.
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%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
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
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
1 hour ago