Suppose a problem A reduce to problem B and reduction is done in $O(n^2)$ time.

If problem B is solved in $O(n^3)$ time then what about the time complexity of problem A?


A is reduced to B . Here reduction is done at polynomial time. Here B is solved in polynomial time. So A should also be in polynomial time.

Now A can not be harder than B, So I think A can be $O(n^3)$ or $O(n^2)$. But logically if I reduced A to B and if B is $O(n^3)$ then it makes no sense for A to be $O(n^2)$, else why would I reduce it to higher complexity? So A is $O(n^3)$.

But my doubt is, we say while reduction that A can not be harder than B then how can we decide whether it is $O(n^2)$ or $O(n^3)$. Or is this argument only valid for P, NP classes of problem when I say A can not be harder than B or B must be at least as hard as A?

Answer given is "A is $O(n^3)$", but why can't it be $O(n^2)$ as "A can not be harder than B" but can be of equal complexity? Or is reducibility just an argument for complexity classes?

Explain if possible how reduction actually works, and why I couldn't apply it to my problem.


You have a reduction, but you don't know what size the new problem will be.

Say you have an instance of problem A of size n. And you can reduce it to an instance of problem B in time O (n^2). If you can reduce it to an instance of size $O(N^{1/2})$ then you are fine: You took O (N^2) for the reduction, and $O(N^{3/2})$ to solve the instance of B.

If you reduce the instance of problem A to an instance of size O (n log n) of problem B, then you will take O (n^3 log^3 n) to solve the instance of B.

So the time for the reduction and the time for solving B are not enough. You need to know the size of the new problem as well. (Of course you can't create an instance greater than O (n^2) in O (n^2) time, so the worst case is O (n^6), but it could be much better).

The answer that was given (O (n^3)) is definitely wrong. It doesn't follow from the information you were given.

  • $\begingroup$ can you answer my doubt in the comments below $\endgroup$ – CHETAN RAJPUT Dec 4 '18 at 10:33

Given an $O(n^2)$ reduction from $A$ to $B$ and an $O(n^3)$ algorithm for solving $B$, you can solve $A$ as follows:

  • Given an instance of $A$ of size $n$, reduce it to an equivalence instance $B$ of size $N = O(n^2)$.
  • Apply the algorithm for $B$, which runs in time $O(N^3) = O(n^6)$.

If we know more about the size of the $B$-instance produced by the $O(n^2)$ reduction, we might be able to tighten this analysis.

Also, all of this only gives an upper bound on the complexity of $A$. There might be other ways of solving $A$. Stated differently, the complexity of $A$ could be much smaller than $O(n^6)$.

  • $\begingroup$ If an algorithm used to covert the problem takes O(n^2) and reduced problem solved in O(n^3) then wont it be max(n^2,n^3 ) ? How n^2*n^3 as an upper bound for A ? $\endgroup$ – CHETAN RAJPUT Dec 4 '18 at 1:52
  • $\begingroup$ I am not exactly getting what you said, can you add some more points ? $\endgroup$ – CHETAN RAJPUT Dec 4 '18 at 1:54
  • $\begingroup$ For some reason you're assuming that the reduction maintains the size of the instance. This need not be the case. A reduction running in time $O(n^2)$ can produce an instance of that size. $\endgroup$ – Yuval Filmus Dec 4 '18 at 1:57
  • $\begingroup$ Then why answer given is O(n^3) i mean How they are so sure , it could be O(n^2) too i think ? $\endgroup$ – CHETAN RAJPUT Dec 4 '18 at 2:02
  • 1
    $\begingroup$ The output size cannot be exponential. Any transformation that runs in O (n^2) can only produce output of size O (n^2) at most. To produce output of size O (n^4), for example, would take at least O (n^4) time. $\endgroup$ – gnasher729 Dec 4 '18 at 9:39

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