We have a procedure $P(n)$ that makes multiple calls to a procedure $Q(m)$, and runs in polynomial time in n. Unfortunately, a significant flaw was discovered in $Q(m)$, and it had to be replaced by $R(m)$, which runs in exponential time in m. Thankfully, P is still correct when we replace each call to $Q(m)$ with a call to $R(m)$ instead. Which of the following can we definitely say about the modified version of P?

(a) $P(n)$ still runs in polynomial time in $n$.

(b) $P(n)$ requires exponential time in $n$.

(c) $P(n)$ runs in polynomial time in $n$ if the number of calls made to Q is proportional to $log n$

(d) $P(n)$ runs in polynomial time in $n$ if, for each call $Q(m)$, $m ≤ log n$.

The answer here is: (d)

Their reasoning: If $m ≤ log n$ for each call to $Q$ with argument $m$, then the running time of $R(m)$ is exponential in $m$ but polynomial in $n$. Since the number of calls to $Q$ originally (and $R$ after the modification) is a polynomial in $n$, the overall running time of $P$ is polynomial. None of the other options definitively hold

I don't understand how they reached that conclusion: Since $R(m)$ is exponential, calling it even once would result into exponential time and since $R(m)$ is a sub-routine to $P(n)$, how can they say that "running time of $R(m)$ is exponential in m but polynomial in $n$"

I am totally confused by this, i can understand reductions ( Theory of computation ), my assumption was according to programming if we call an exponential function inside a polynomial function. The overall complexity would be exponential since that is the bottle-neck

Can someone shed a light on how to approach these problems?


2 Answers 2


If $R(n)$ is exponential in $n$, $R(\log n)$ is polynomial in $n$.

E.g. $$T(n)=4^n\to T(\log_2n)=4^{\log_2n}=n^2.$$


If $R(m)$ runs in exponential time in $m$, that means that the running time of $R(m)$ is, informally, something like $O(2^m)$. More formally, its running time is $O(c^m)$ for some constant $c$.

Now plug in $m=\log n$, simplify, and see what conclusions you can draw about the running time as a function of $n$.


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