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
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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
1204
$endgroup$
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
1204
$endgroup$
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
20 mins 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
1204
$endgroup$
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
1204
asked 2 hours ago
rtheunissenrtheunissen
1204
edited 18 mins ago
rtheunissen
asked 2 hours ago
rtheunissenrtheunissen
1204
asked 2 hours ago
rtheunissenrtheunissen
1204
asked 2 hours 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
20 mins 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
20 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
20 mins ago
$begingroup$
@ryan k is constant. (I have edited the question to specify this)
$endgroup$
– rtheunissen
20 mins 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 55 mins 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
19 mins ago
1
$begingroup$
If $k$ is constant, the amortized complexity is $O(log n)$.
$endgroup$
– Yuval Filmus
17 mins ago
add a comment |
Your Answer
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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 55 mins 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
19 mins ago
1
$begingroup$
If $k$ is constant, the amortized complexity is $O(log n)$.
$endgroup$
– Yuval Filmus
17 mins 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 55 mins 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
19 mins ago
1
$begingroup$
If $k$ is constant, the amortized complexity is $O(log n)$.
$endgroup$
– Yuval Filmus
17 mins 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 55 mins 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 55 mins ago
Yuval FilmusYuval Filmus
197k15185349
answered 55 mins ago
Yuval FilmusYuval Filmus
197k15185349
answered 55 mins ago
Yuval FilmusYuval Filmus
197k15185349
answered 55 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
19 mins ago
1
$begingroup$
If $k$ is constant, the amortized complexity is $O(log n)$.
$endgroup$
– Yuval Filmus
17 mins ago
add a comment |
$begingroup$
Does that mean that if k is constant, the amortized complexity is O(1 + (log n / k))?
$endgroup$
– rtheunissen
19 mins ago
1
$begingroup$
If $k$ is constant, the amortized complexity is $O(log n)$.
$endgroup$
– Yuval Filmus
17 mins ago
$begingroup$
Does that mean that if k is constant, the amortized complexity is O(1 + (log n / k))?
$endgroup$
– rtheunissen
19 mins ago
$begingroup$
Does that mean that if k is constant, the amortized complexity is O(1 + (log n / k))?
$endgroup$
– rtheunissen
19 mins ago
1
1
$begingroup$
If $k$ is constant, the amortized complexity is $O(log n)$.
$endgroup$
– Yuval Filmus
17 mins ago
$begingroup$
If $k$ is constant, the amortized complexity is $O(log n)$.
$endgroup$
– Yuval Filmus
17 mins ago
add a comment |
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$begingroup$
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
20 mins ago