Complexity of many constant time steps with occasional logarithmic stepsUpper-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
Numerical second order differentiation
How can a flywheel makes engine runs smoothly?
What is the context for Napoleon's quote "[the Austrians] did not know the value of five minutes"?
How do I run a script as sudo at boot time on Ubuntu 18.04 Server?
Redirecting output only on a successful command call
Having some issue with notation in a Hilbert space
The instant an accelerating object has zero speed, is it speeding up, slowing down, or neither?
How can the US president give an order to a civilian?
Converting 3x7 to a 1x7. Is it possible with only existing parts?
New Site Design!
Co-worker is now managing my team. Does this mean that I'm being demoted?
Is swap gate equivalent to just exchanging the wire of the two qubits?
Is it a bad idea to have a pen name with only an initial for a surname?
What is this plant I saw for sale at a Romanian farmer's market?
Background for black and white chart
How to sort human readable size
How could I create a situation in which a PC has to make a saving throw or be forced to pet a dog?
Time at 1G acceleration to travel 100000 light years
How can I detect if I'm in a subshell?
Digital signature that is only verifiable by one specific person
Is this set open or closed (or both?)
SQL Server has encountered occurences of I/O requests taking longer than 15 seconds
What do I put on my resume to make the company i'm applying to think i'm mature enough to handle a job?
Do my partner and son need an SSN to be dependents on my taxes?
Complexity of many constant time steps with occasional logarithmic steps
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
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
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 Apr 14 at 19:13
rtheunissenrtheunissen
1416
$endgroup$
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 Apr 14 at 19:13
rtheunissenrtheunissen
1416
$endgroup$
2
$begingroup$
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
Apr 14 at 20:09
$begingroup$
@ryan k is constant. (I have edited the question to specify this)
$endgroup$
– rtheunissen
Apr 14 at 21:04
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 Apr 14 at 19:13
rtheunissenrtheunissen
1416
$endgroup$
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 Apr 14 at 19:13
rtheunissenrtheunissen
1416
asked Apr 14 at 19:13
rtheunissenrtheunissen
1416
edited Apr 14 at 21:06
rtheunissen
asked Apr 14 at 19:13
rtheunissenrtheunissen
1416
asked Apr 14 at 19:13
rtheunissenrtheunissen
1416
asked Apr 14 at 19:13
rtheunissenrtheunissen
1416
1416
2
$begingroup$
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
Apr 14 at 20:09
$begingroup$
@ryan k is constant. (I have edited the question to specify this)
$endgroup$
– rtheunissen
Apr 14 at 21:04
add a comment |
2
$begingroup$
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
Apr 14 at 20:09
$begingroup$
@ryan k is constant. (I have edited the question to specify this)
$endgroup$
– rtheunissen
Apr 14 at 21:04
2
2
$begingroup$
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
Apr 14 at 20:09
$begingroup$
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
Apr 14 at 20:09
$begingroup$
@ryan k is constant. (I have edited the question to specify this)
$endgroup$
– rtheunissen
Apr 14 at 21:04
$begingroup$
@ryan k is constant. (I have edited the question to specify this)
$endgroup$
– rtheunissen
Apr 14 at 21:04
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 Apr 14 at 20:29
Yuval FilmusYuval Filmus
202k15197359
$endgroup$
$begingroup$
Does that mean that if k is constant, the amortized complexity is O(1 + (log n / k))?
$endgroup$
– rtheunissen
Apr 14 at 21:05
8
$begingroup$
If $k$ is constant, the amortized complexity is $O(log n)$.
$endgroup$
– Yuval Filmus
Apr 14 at 21:07
$begingroup$
@YuvalFilmus might want to note that this is per step, the overall complexity cannot be better than O(n)
$endgroup$
– Frank Hopkins
Apr 14 at 23:46
$begingroup$
Of course, I just mean the cost of the task itself.
$endgroup$
– rtheunissen
Apr 15 at 0:14
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 Apr 14 at 20:29
Yuval FilmusYuval Filmus
202k15197359
$endgroup$
$begingroup$
Does that mean that if k is constant, the amortized complexity is O(1 + (log n / k))?
$endgroup$
– rtheunissen
Apr 14 at 21:05
8
$begingroup$
If $k$ is constant, the amortized complexity is $O(log n)$.
$endgroup$
– Yuval Filmus
Apr 14 at 21:07
$begingroup$
@YuvalFilmus might want to note that this is per step, the overall complexity cannot be better than O(n)
$endgroup$
– Frank Hopkins
Apr 14 at 23:46
$begingroup$
Of course, I just mean the cost of the task itself.
$endgroup$
– rtheunissen
Apr 15 at 0:14
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 Apr 14 at 20:29
Yuval FilmusYuval Filmus
202k15197359
$endgroup$
$begingroup$
Does that mean that if k is constant, the amortized complexity is O(1 + (log n / k))?
$endgroup$
– rtheunissen
Apr 14 at 21:05
8
$begingroup$
If $k$ is constant, the amortized complexity is $O(log n)$.
$endgroup$
– Yuval Filmus
Apr 14 at 21:07
$begingroup$
@YuvalFilmus might want to note that this is per step, the overall complexity cannot be better than O(n)
$endgroup$
– Frank Hopkins
Apr 14 at 23:46
$begingroup$
Of course, I just mean the cost of the task itself.
$endgroup$
– rtheunissen
Apr 15 at 0:14
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 Apr 14 at 20:29
Yuval FilmusYuval Filmus
202k15197359
$endgroup$
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 Apr 14 at 20:29
Yuval FilmusYuval Filmus
202k15197359
answered Apr 14 at 20:29
Yuval FilmusYuval Filmus
202k15197359
answered Apr 14 at 20:29
Yuval FilmusYuval Filmus
202k15197359
answered Apr 14 at 20:29
Yuval FilmusYuval Filmus
202k15197359
202k15197359
$begingroup$
Does that mean that if k is constant, the amortized complexity is O(1 + (log n / k))?
$endgroup$
– rtheunissen
Apr 14 at 21:05
8
$begingroup$
If $k$ is constant, the amortized complexity is $O(log n)$.
$endgroup$
– Yuval Filmus
Apr 14 at 21:07
$begingroup$
@YuvalFilmus might want to note that this is per step, the overall complexity cannot be better than O(n)
$endgroup$
– Frank Hopkins
Apr 14 at 23:46
$begingroup$
Of course, I just mean the cost of the task itself.
$endgroup$
– rtheunissen
Apr 15 at 0:14
add a comment |
$begingroup$
Does that mean that if k is constant, the amortized complexity is O(1 + (log n / k))?
$endgroup$
– rtheunissen
Apr 14 at 21:05
8
$begingroup$
If $k$ is constant, the amortized complexity is $O(log n)$.
$endgroup$
– Yuval Filmus
Apr 14 at 21:07
$begingroup$
@YuvalFilmus might want to note that this is per step, the overall complexity cannot be better than O(n)
$endgroup$
– Frank Hopkins
Apr 14 at 23:46
$begingroup$
Of course, I just mean the cost of the task itself.
$endgroup$
– rtheunissen
Apr 15 at 0:14
$begingroup$
Does that mean that if k is constant, the amortized complexity is O(1 + (log n / k))?
$endgroup$
– rtheunissen
Apr 14 at 21:05
$begingroup$
Does that mean that if k is constant, the amortized complexity is O(1 + (log n / k))?
$endgroup$
– rtheunissen
Apr 14 at 21:05
8
8
$begingroup$
If $k$ is constant, the amortized complexity is $O(log n)$.
$endgroup$
– Yuval Filmus
Apr 14 at 21:07
$begingroup$
If $k$ is constant, the amortized complexity is $O(log n)$.
$endgroup$
– Yuval Filmus
Apr 14 at 21:07
$begingroup$
@YuvalFilmus might want to note that this is per step, the overall complexity cannot be better than O(n)
$endgroup$
– Frank Hopkins
Apr 14 at 23:46
$begingroup$
@YuvalFilmus might want to note that this is per step, the overall complexity cannot be better than O(n)
$endgroup$
– Frank Hopkins
Apr 14 at 23:46
$begingroup$
Of course, I just mean the cost of the task itself.
$endgroup$
– rtheunissen
Apr 15 at 0:14
$begingroup$
Of course, I just mean the cost of the task itself.
$endgroup$
– rtheunissen
Apr 15 at 0:14
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
2
$begingroup$
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
Apr 14 at 20:09
$begingroup$
@ryan k is constant. (I have edited the question to specify this)
$endgroup$
– rtheunissen
Apr 14 at 21:04