Complexity of many constant time steps with occasional logarithmic steps Planned maintenance scheduled April 23, 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 23, 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 Apr 14 at 19:13
rtheunissenrtheunissen
1394
$endgroup$
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
1394
$endgroup$
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
1394
$endgroup$
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
1394
asked Apr 14 at 19:13
rtheunissenrtheunissen
1394
edited Apr 14 at 21:06
rtheunissen
asked Apr 14 at 19:13
rtheunissenrtheunissen
1394
asked Apr 14 at 19:13
rtheunissenrtheunissen
1394
asked Apr 14 at 19:13
rtheunissenrtheunissen
1394
1394
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
197k15185349
$endgroup$
$begingroup$
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
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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 Apr 14 at 20:29
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
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
197k15185349
$endgroup$
$begingroup$
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
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 Apr 14 at 20:29
Yuval FilmusYuval Filmus
197k15185349
answered Apr 14 at 20:29
Yuval FilmusYuval Filmus
197k15185349
answered Apr 14 at 20:29
Yuval FilmusYuval Filmus
197k15185349
answered Apr 14 at 20:29
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
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 |
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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
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