# Tag Info

### Is there a task that is solvable in polynomial time but not verifiable in polynomial time?

This is only possible if there are many admissible outputs for a given input. I.e., when the relation $R$ is not a function because it violates uniqueness. For instance, consider this problem: ...
Accepted

### Assuming P = NP, how would one solve the graph coloring problem in polynomial time?

There are two cases: $P = NP$ non-constructively: this means we have derived a contradiction from the assumption that $P \neq NP$, and thus can conclude that $P = NP$ by the law of the excluded ...
Accepted

### How is the traveling salesman problem verifiable in polynomial time?

NP is the class of problems where you can verify "yes" instances. No guarantee is given that you can verify "no" instances. The class of problems where you can verify "no" instances in polynomial ...
Accepted

### Can a subset of an NP-complete problem be in P?

Your question doesn't make sense: The problem is NP-complete (proven) for all input data (without exception). This is not a thing. NP-completeness is a property of entire sets, not of specific ...
Accepted

### Evolving artificial neural networks for solving NP problems

No. This direction is unlikely to be useful, for two reasons: Most computer scientists believe that P $\ne$ NP. Assuming P $\ne$ NP, this means there does not exist any polynomial-time algorithm to ...
Accepted

### Is determining if there is a prime in an interval known to be in P or NP-complete?

So your problem is as follows: Input: integers $\ell,u$ Question: does there exist a prime in $[\ell,u]$? As far as I know, it is not known whether that problem is in P or not. Here's what I do know: ...
Accepted

### NP-complete problems not "obviously" in NP

There are at least four such $NP$-complete problems listed in the appendix of Garey and Johnson's COMPUTERS AND INTRACTABILITY: A Guide to the Theory of NP-Completeness. [AN6] NON-DIVISIBILITY OF A ...
Accepted

### Is Post Correspondence Problem in NP?

The Post correspondence problem is undecidable, and in particular it is not in NP. The reason that your idea doesn't work is that the witness is not necessarily of polynomial size (in fact, you just ...

### Is the class NP closed under complement?

First of all, the question you are asking is open, since an affirmative answer shows that $\sf NP = coNP$. In fact it is one of the most prominent open problems in computer science. If $\sf P= NP$, ...
Accepted

### Is detecting easy instances of NP-hard problems easy?

The problem isn't really well-posed. For any particular instance, there is a single solution, say $S$. Consequently, we can imagine an algorithm that has the answer $S$ hardcoded in: no matter what ...
Accepted

### Why rectangle packing is NP-hard but maybe not in NP?

In order for a language $L$ to be in NP, there needs to be a way to certify that instance $x$ belongs to $L$. This "way" is a polynomial size witness which can be verified in polynomial time....
Accepted

### is FIND WORDS in P?

Your language is in P. Suppose that the matrix is $n\times n$ and that the words have total length $\ell$. Each word can start at at most $n^2$ positions and be written in $O(1)$ many orientations, ...

### What does the 2 in a 2-approximation algorithm mean?

Typically, we use $\alpha < 1$ for maximization problems, and $\alpha > 1$ for minimization problems, where $\alpha$ is the approximation guarantee. So, a $2$-approximation algorithm returns a ...
Accepted

### Why are NP-complete problems so different in terms of their approximation?

One reason that we see different approximation complexities for NP-complete problems is that the necessary conditions for NP-complete constitute a very coarse grained measure of a problem's complexity....
Accepted

### Does every problem in NP have an exponential time algorithm?

Yes, every NP problem has an exponential-time algorithm. One definition of NP is the "succinct certificates" definition: a language $L$ is in NP if, and only if, there is a relation $R$ on ...
Accepted

### Can any finite problem be in NP-Complete?

If a finite problem is NP-complete then P=NP, since every finite problem has a polynomial time algorithm (even a constant time algorithm). When we say that Sudoku is NP-complete, we mean that a ...

### There are pretty much no algorithms in P

Actually you're wrong. In the example you give, you take a list of integers, but even with binary representation for the numbers in the list, the input length is still linear in the number of elements ...

### Why are NP-complete problems so different in terms of their approximation?

One way to consider the difference between decision version and optimization version is by considering different optimization versions of the same decision version. Take for example the MAX-CLIQUE ...
Accepted

### Proof Complexity of a Proof or Disproof of P = NP

It is known that any proof of super-polynomial circuit lower bounds (which are slightly stronger statements than $P\neq NP$) require super-polynomial, even exponential size proofs in weak proof ...
Accepted

### Why does the solution of an NP problem have to be polynomial size?

First, let's refresh the proof that the definition given is indeed equivalent to the standard definition of NP, i.e., the class of languages accepted by polynomial-time nondeterministic Turing ...

### How do we know any problem is in NP-complete if we don't know all problems in NP?

We do know all problems in NP. Each problem in NP is given by a non-deterministic Turing machine running in polynomial time. Steve Cook (and, independently, Leonid Levin) proved that SAT is NP-...
Accepted

### Problem A is polynomially reducible to problem B... what can we say about A and B?

Your intuition about "relative hardness" is correct, the underlying mathematics is why III. is true. However your justification about I. is a little off (not wrong, but the reasoning is possibly not ...
Accepted

### What do we know about NP ∩ co-NP and its relation to NPI?

The fact that P ≠ NP does not preclude the possibility that NP = co-NP, in which case NP ∩ co-NP = NP. So to further the discussion, let us assume that NP ≠ co-NP. In that case, Corollary 9 in ...