Linked Questions

Wikipedia only lists two problems under "unsolved problems in computer science": P = NP? The existence of one-way functions What are other major problems that should be added to this list? Rules: ...

This question is about problems for which there is a big open complexity gap between known lower bound and upper bound, but not because of open problems on complexity classes themselves. To be more ...

Many experts believe that the $\mathsf{P} \neq \mathsf{NP}$ conjecture is true and use it in their results. My concern is that the complexity strongly depends on the $\mathsf{P} \neq \mathsf{NP}$ ...

In my computer science education, I increasingly notice that most discrete problems are NP-complete (at least), whereas optimizing continuous problems is almost always easily achievable, usually ...

It is well known by Ladner's Theorem that if ${\mathsf P}\neq \mathsf {NP}$, then there exist infinitely many $\mathsf {NP}$-intermediate ($\mathsf{NPI}$) problems. There are also natural candidates ...

The sum of square roots problem asks, given two sequences $a_1, a_2, \dots, a_n$ and $b_1, b_2, \dots, b_n$ of positive integers, whether the sum $\sum_i \sqrt{a_i}$ less than, equal to, or greater ...

Assuming that P != NP, I believe it has been shown that there are problems which are not in P and not NP-Complete. Graph Isomorphism is conjectured to be such a problem. Is there any evidence of more ...

Are there any natural problems in $NP \cap coNP$ that are not (known to be/thought to be) in $UP \cap coUP$? Obviously the big one everyone knows about in $NP \cap coNP$ is the decision version of ...

How does one look at a problem and reason that it is likely NP-Intermediate as opposed to NP-Complete? It is often pretty simple to look at a problem and tell whether it is likely NP-Complete or not ...

The Graph Isomorphism problem (GI) is arguably the best known candidate for an NP-intermediate problem. The best known algorithm is sub-exponential algorithm with run-time $2^{O(\sqrt{n \log n})}$. ...

Two binary search trees are said to be linearly equivalent when they agree in their in-order traversals. The following theorem explains why tree rotations are so fundamental: Let A and B be binary ...

It is known that determining whether or not a given triangulated 3-manifold is a 3-sphere is in NP, via work by Saul Schleimer in 2004: "Sphere recognition lies in NP" arXiv:math/0407047v1 [math.GT]. ...

Here is a variant of the classic partition problem: Given a list of integers can it be partitioned into $S_1$, $S_2$, and $S_3$, with $S_1$ and $S_2$ nonempty, so the sum of elements in $S_1$ equals ...

Let's assume that $\mathsf{P} \neq \mathsf{NP}$. $\mathsf{NPI}$ is the class of problems in $\mathsf{NP}$ which are neither in $\mathsf{P}$ nor in $\mathsf{NP}$-hard. You can find a list of problems ...

Let distinct points $1 ... n$ sit in $\mathbb{R}^2$. We say points $i$ and $j$ are neighbors if $|i-j| < 3 \pmod{n-2}$, meaning each point is neighbors with points with indexes within $2$, ...