$NP$ is not in $P(n^k)$ for any fixed $k \geq 1$

I encountered this problem which asks to show that for any fixed $ k \geq 1$ , $ NP$ is not contained in $ P(n^k)$

As an attempt, I thought of using the time hierarchy theorem which says that there exists a language in $ P(n^{k+1})$ which is not decided in $ P(n^k)$ given that $ n^k \in o(n^{k+1})$ … Since the space of polynomial verifiers in $ NP$ is the union of all polynomials of $ n$ , using the time hierarchy theorem, this means that there exists a problem in $ NP$ that accepts instances if the accepting branch operates in time $ P(n^{k+1})$ , and so $ NP$ could not be contained in $ P(n^k)$

But is this correct ? I only assumed that such a problem exists in $ NP$ (i.e. which accepts in nondeterministic time $ P(n^{k+1})$ due to the time hierarchy theorem, but I have not really been able to construct a concrete problem…

If a decision problem’s solution is exponentially large, then is it impossible for it to be in $NP$?

Decision Problem: Is $ 2^k$ + $ M$ a prime?

The inputs for both $ K$ and $ M$ are integers only. The solution is the sum of $ 2^k$ +$ M$ . (Use AKS to decide prime)

The powers of 2 have approximately $ 2^n$ digits. Consider $ 2^k$ where $ K$ = 100000. Compare the amount of digits in $ K$ to the amount of digits in it’s solution!


Could the solution be verified in polynomial time even though the solution is exponentially large?

If no, can I say it is not in $ NP$ ?

Under what kind of oralces are $P$ and $NP$ equivalent?

How strong have the oracles needed to be for these two classes to be proven equivalent with respect to them?

For instance: is $ P^H$ = $ NP^H$ (ie. is $ P$ equipped with an oracle to solve the halting problem equivalent to $ NP$ equipped with an oracle to solve the halting problem)?

From Theodore Baker, John Gill, and Robert Solovay. Relativization of the P=?NP problem. Siam Journal of Computing, 4:432-442, 1975 [219] we know $ NP^A =P^A$ for their oracle A (which is a decision algorithm for a PSPACE complete problem).

If the oracle can perform an infinite amount of computation and return back the result in one step are these classes equal with respect to an oracle of this type? How about weaker ones? What is the weakest oracle we know of where $ P$ and $ NP$ are equal with respect to it?

An answer I’m looking for is something like: $ P^O$ =$ NP^O$ with respect to an oracle O and any oracle more powerful than it.

Must a decision problem in $NP$ have a complement in $Co-NP$, if I can verify the solutions to in polynomial-time?

Goldbach’s Conjecture says every even integer $ >$ $ 2$ can be expressed as the sum of two primes.

Let’s say $ N$ is our input and its $ 10$ . Which is an integer > 2 and is not odd.


1.Create list of numbers from $ 1,to~N$

2.Use prime-testing algorithm for creating a second list of prime numbers

3.Use my 2_sum solver that allows you to use primes twice that sum up to $ N$

for j in range(list-of-primes)):   if N-(list-of-primes[j]) in list-of-primes:    print('yes')    break 

4.Verify solution efficently

if AKS-primality(N-(list-of-primes[j])):     if AKS-primality(list-of-primes[j]):         print('Solution is correct') 


yes 7 + 3 Solution is correct 


If the conjecture is true, then the answer will always be Yes. Does that mean it can’t be in $ Co-NP$ because the answer is always Yes?

Is this in $NP$ or $coNP$?

Is ‘Given two codes with Generator matrices $ G_1$ and $ G_2$ do they have the same minimum distance?’ in $ NP$ or is it in $ coNP$ ?

If $ G_1$ is known to give minimum distance $ \geq d$ then the problem ‘is the minimum distance with $ G_2$ less than $ d$ ?’ is $ NP$ -complete.

$NP$, $P^{TFNP}$ and $P^{UP}$

Is it possible

  1. $ NP\in P^{TFNP}$ holds or

  2. $ NP\in P^{UP}$ holds

without the polynomial hierarchy collapsing?

Is there problems in

  1. one of each of the classes from $ NP$ , $ P^{UP}$ and $ P^{TFNP}$ but not in other two?

  2. two of each pair of the classes from $ NP$ , $ P^{UP}$ and $ P^{TFNP}$ but not in last?