Complexity of many constant time steps with occasional logarithmic steps Planned maintenance...
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Complexity of many constant time steps with occasional logarithmic steps
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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
asked yesterday
rtheunissenrtheunissen
1324
$endgroup$
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 $O(log{n})$, 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
asked yesterday
rtheunissenrtheunissen
1324
$endgroup$
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
yesterday
$begingroup$
@ryan k is constant. (I have edited the question to specify this)
$endgroup$
– rtheunissen
yesterday
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 $O(log{n})$, 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
asked yesterday
rtheunissenrtheunissen
1324
$endgroup$
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
asked yesterday
rtheunissenrtheunissen
1324
asked yesterday
rtheunissenrtheunissen
1324
edited yesterday
rtheunissen
asked yesterday
rtheunissenrtheunissen
1324
asked yesterday
rtheunissenrtheunissen
1324
asked yesterday
rtheunissenrtheunissen
1324
1324
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
yesterday
$begingroup$
@ryan k is constant. (I have edited the question to specify this)
$endgroup$
– rtheunissen
yesterday
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
yesterday
$begingroup$
@ryan k is constant. (I have edited the question to specify this)
$endgroup$
– rtheunissen
yesterday
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
yesterday
$begingroup$
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
yesterday
$begingroup$
@ryan k is constant. (I have edited the question to specify this)
$endgroup$
– rtheunissen
yesterday
$begingroup$
@ryan k is constant. (I have edited the question to specify this)
$endgroup$
– rtheunissen
yesterday
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 $O(1 + frac{log n}{k})$. This follows from the definition of amortized complexity.
answered yesterday
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
yesterday
8
$begingroup$
If $k$ is constant, the amortized complexity is $O(log n)$.
$endgroup$
– Yuval Filmus
yesterday
$begingroup$
@YuvalFilmus might want to note that this is per step, the overall complexity cannot be better than O(n)
$endgroup$
– Frank Hopkins
yesterday
$begingroup$
Of course, I just mean the cost of the task itself.
$endgroup$
– rtheunissen
yesterday
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
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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 $O(1 + frac{log n}{k})$. This follows from the definition of amortized complexity.
answered yesterday
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
yesterday
8
$begingroup$
If $k$ is constant, the amortized complexity is $O(log n)$.
$endgroup$
– Yuval Filmus
yesterday
$begingroup$
@YuvalFilmus might want to note that this is per step, the overall complexity cannot be better than O(n)
$endgroup$
– Frank Hopkins
yesterday
$begingroup$
Of course, I just mean the cost of the task itself.
$endgroup$
– rtheunissen
yesterday
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 $O(1 + frac{log n}{k})$. This follows from the definition of amortized complexity.
answered yesterday
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
yesterday
8
$begingroup$
If $k$ is constant, the amortized complexity is $O(log n)$.
$endgroup$
– Yuval Filmus
yesterday
$begingroup$
@YuvalFilmus might want to note that this is per step, the overall complexity cannot be better than O(n)
$endgroup$
– Frank Hopkins
yesterday
$begingroup$
Of course, I just mean the cost of the task itself.
$endgroup$
– rtheunissen
yesterday
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 $O(1 + frac{log n}{k})$. This follows from the definition of amortized complexity.
answered yesterday
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 $O(1 + frac{log n}{k})$. This follows from the definition of amortized complexity.
answered yesterday
Yuval FilmusYuval Filmus
197k15185349
answered yesterday
Yuval FilmusYuval Filmus
197k15185349
answered yesterday
Yuval FilmusYuval Filmus
197k15185349
answered yesterday
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
yesterday
8
$begingroup$
If $k$ is constant, the amortized complexity is $O(log n)$.
$endgroup$
– Yuval Filmus
yesterday
$begingroup$
@YuvalFilmus might want to note that this is per step, the overall complexity cannot be better than O(n)
$endgroup$
– Frank Hopkins
yesterday
$begingroup$
Of course, I just mean the cost of the task itself.
$endgroup$
– rtheunissen
yesterday
add a comment |
$begingroup$
Does that mean that if k is constant, the amortized complexity is O(1 + (log n / k))?
$endgroup$
– rtheunissen
yesterday
8
$begingroup$
If $k$ is constant, the amortized complexity is $O(log n)$.
$endgroup$
– Yuval Filmus
yesterday
$begingroup$
@YuvalFilmus might want to note that this is per step, the overall complexity cannot be better than O(n)
$endgroup$
– Frank Hopkins
yesterday
$begingroup$
Of course, I just mean the cost of the task itself.
$endgroup$
– rtheunissen
yesterday
$begingroup$
Does that mean that if k is constant, the amortized complexity is O(1 + (log n / k))?
$endgroup$
– rtheunissen
yesterday
$begingroup$
Does that mean that if k is constant, the amortized complexity is O(1 + (log n / k))?
$endgroup$
– rtheunissen
yesterday
8
8
$begingroup$
If $k$ is constant, the amortized complexity is $O(log n)$.
$endgroup$
– Yuval Filmus
yesterday
$begingroup$
If $k$ is constant, the amortized complexity is $O(log n)$.
$endgroup$
– Yuval Filmus
yesterday
$begingroup$
@YuvalFilmus might want to note that this is per step, the overall complexity cannot be better than O(n)
$endgroup$
– Frank Hopkins
yesterday
$begingroup$
@YuvalFilmus might want to note that this is per step, the overall complexity cannot be better than O(n)
$endgroup$
– Frank Hopkins
yesterday
$begingroup$
Of course, I just mean the cost of the task itself.
$endgroup$
– rtheunissen
yesterday
$begingroup$
Of course, I just mean the cost of the task itself.
$endgroup$
– rtheunissen
yesterday
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
yesterday
$begingroup$
@ryan k is constant. (I have edited the question to specify this)
$endgroup$
– rtheunissen
yesterday