Complexity of many constant time steps with occasional logarithmic steps












3












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
    4 hours ago










  • $begingroup$
    @ryan k is constant. (I have edited the question to specify this)
    $endgroup$
    – rtheunissen
    3 hours ago


















3












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
    4 hours ago










  • $begingroup$
    @ryan k is constant. (I have edited the question to specify this)
    $endgroup$
    – rtheunissen
    3 hours ago
















3












3








3


1



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 3 hours ago







rtheunissen

















asked 5 hours ago









rtheunissenrtheunissen

1324




1324








  • 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
    4 hours ago










  • $begingroup$
    @ryan k is constant. (I have edited the question to specify this)
    $endgroup$
    – rtheunissen
    3 hours ago
















  • 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
    4 hours ago










  • $begingroup$
    @ryan k is constant. (I have edited the question to specify this)
    $endgroup$
    – rtheunissen
    3 hours 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
4 hours 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
4 hours ago












$begingroup$
@ryan k is constant. (I have edited the question to specify this)
$endgroup$
– rtheunissen
3 hours ago






$begingroup$
@ryan k is constant. (I have edited the question to specify this)
$endgroup$
– rtheunissen
3 hours ago












1 Answer
1






active

oldest

votes


















6












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Does that mean that if k is constant, the amortized complexity is O(1 + (log n / k))?
    $endgroup$
    – rtheunissen
    3 hours ago






  • 6




    $begingroup$
    If $k$ is constant, the amortized complexity is $O(log n)$.
    $endgroup$
    – Yuval Filmus
    3 hours ago










  • $begingroup$
    @YuvalFilmus might want to note that this is per step, the overall complexity cannot be better than O(n)
    $endgroup$
    – Frank Hopkins
    51 mins ago










  • $begingroup$
    Of course, I just mean the cost of the task itself.
    $endgroup$
    – rtheunissen
    23 mins ago












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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









6












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Does that mean that if k is constant, the amortized complexity is O(1 + (log n / k))?
    $endgroup$
    – rtheunissen
    3 hours ago






  • 6




    $begingroup$
    If $k$ is constant, the amortized complexity is $O(log n)$.
    $endgroup$
    – Yuval Filmus
    3 hours ago










  • $begingroup$
    @YuvalFilmus might want to note that this is per step, the overall complexity cannot be better than O(n)
    $endgroup$
    – Frank Hopkins
    51 mins ago










  • $begingroup$
    Of course, I just mean the cost of the task itself.
    $endgroup$
    – rtheunissen
    23 mins ago
















6












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Does that mean that if k is constant, the amortized complexity is O(1 + (log n / k))?
    $endgroup$
    – rtheunissen
    3 hours ago






  • 6




    $begingroup$
    If $k$ is constant, the amortized complexity is $O(log n)$.
    $endgroup$
    – Yuval Filmus
    3 hours ago










  • $begingroup$
    @YuvalFilmus might want to note that this is per step, the overall complexity cannot be better than O(n)
    $endgroup$
    – Frank Hopkins
    51 mins ago










  • $begingroup$
    Of course, I just mean the cost of the task itself.
    $endgroup$
    – rtheunissen
    23 mins ago














6












6








6





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 4 hours 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
    3 hours ago






  • 6




    $begingroup$
    If $k$ is constant, the amortized complexity is $O(log n)$.
    $endgroup$
    – Yuval Filmus
    3 hours ago










  • $begingroup$
    @YuvalFilmus might want to note that this is per step, the overall complexity cannot be better than O(n)
    $endgroup$
    – Frank Hopkins
    51 mins ago










  • $begingroup$
    Of course, I just mean the cost of the task itself.
    $endgroup$
    – rtheunissen
    23 mins ago


















  • $begingroup$
    Does that mean that if k is constant, the amortized complexity is O(1 + (log n / k))?
    $endgroup$
    – rtheunissen
    3 hours ago






  • 6




    $begingroup$
    If $k$ is constant, the amortized complexity is $O(log n)$.
    $endgroup$
    – Yuval Filmus
    3 hours ago










  • $begingroup$
    @YuvalFilmus might want to note that this is per step, the overall complexity cannot be better than O(n)
    $endgroup$
    – Frank Hopkins
    51 mins ago










  • $begingroup$
    Of course, I just mean the cost of the task itself.
    $endgroup$
    – rtheunissen
    23 mins ago
















$begingroup$
Does that mean that if k is constant, the amortized complexity is O(1 + (log n / k))?
$endgroup$
– rtheunissen
3 hours ago




$begingroup$
Does that mean that if k is constant, the amortized complexity is O(1 + (log n / k))?
$endgroup$
– rtheunissen
3 hours ago




6




6




$begingroup$
If $k$ is constant, the amortized complexity is $O(log n)$.
$endgroup$
– Yuval Filmus
3 hours ago




$begingroup$
If $k$ is constant, the amortized complexity is $O(log n)$.
$endgroup$
– Yuval Filmus
3 hours ago












$begingroup$
@YuvalFilmus might want to note that this is per step, the overall complexity cannot be better than O(n)
$endgroup$
– Frank Hopkins
51 mins ago




$begingroup$
@YuvalFilmus might want to note that this is per step, the overall complexity cannot be better than O(n)
$endgroup$
– Frank Hopkins
51 mins ago












$begingroup$
Of course, I just mean the cost of the task itself.
$endgroup$
– rtheunissen
23 mins ago




$begingroup$
Of course, I just mean the cost of the task itself.
$endgroup$
– rtheunissen
23 mins ago


















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