Posted by: jonkatz | May 27, 2015

I will be participating in a Google hangout next week focusing on MIT alumni teaching MOOCs. Please join and ask questions, or just listen!

Registration can be done here

## Responses

1. Look this problem:

A succinct representation of a set of (distinct) b-bits positive integers is a Boolean circuit C with b input gates. The set represented by C, denoted S_{C}, is defined as follows: Every possible integer of S_{C} should be between 0 and (2^{b} – 1). And j is an element of S_{C} if and only if C accepts the binary representations of the b-bits integer j as input. The problem SUCCINCT MAXIMUM is now this: Given the succinct representation C of a set S_{C} and a b-bits integer x, where C is a Boolean circuit with b input gates, is x the maximum in S_{C}?

It is very easy to show this problem is not in P, because we should need n comparisons to know whether x is the maximum in a set of n (distinct) positive integers when the set is arbitrary. And this number of comparisons will be optimal. This would mean we cannot always accept every instance (C; x) of SUCCINCT MAXIMUM in polynomial-time, because we must use at least n = |S_{C}| comparisons for infinite amount of cases, where |S_{C}| is the cardinality of S_{C}. However, n could be exponentially more large than the size of (C; x).

But, at the same time, it is so easy to show this problem is in coNP. Certainly, given a b-bits integer y, we can check whether C accepts the binary representation of y (which means that y is an element of S_{C}) and x < y in polynomial-time, and thus, we could verify whether (C; x) is a "no" instance SUCCINCT MAXIMUM in polynomial-time.

However, the existence of a problem in coNP and not in P is sufficient to show that P is not equal to NP, because if P would be equal to NP, then P = coNP.

Basically is this, but you could see more in

https://hal.archives-ouvertes.fr/hal-01281254/document