41 votes

Would the P vs. NP problem become trivial as a result of the development of universal quantum computers?

No, there will be absolutely no implication, for several reasons: The P vs. NP problem is about classical computation rather than quantum computation. Even if quantum computers could solve NP-hard ...
Yuval Filmus's user avatar
22 votes
Accepted

Why is NP in EXPTIME?

Any problem in NP is in EXPTIME because you can either use exponential time to try all possible certificates or to enumerate all possible computation paths of a nondeterministic machine. More ...
David Richerby's user avatar
22 votes

Would the P vs. NP problem become trivial as a result of the development of universal quantum computers?

No implications are known either way: classical simulation of quantum computers tells us nothing about how hard NP search problems are; fast solutions to NP search problems tell us nothing about how ...
Lieuwe Vinkhuijzen's user avatar
14 votes
Accepted

Do any decision problems exist outside NP and NP-Hard?

If $P = NP$, then any non-trivial language is NP-hard, and any trivial language belongs to NP. Hence, we do not get anything which is neither NP or NP-hard in this case. If, however, $P \neq NP$, ...
Arno's user avatar
  • 3,075
14 votes
Accepted

Notation: SPACE(n) vs SPACE(O(n))

It depends on what definitions you use. Sipser [1] defines $\mathrm{SPACE}(f(n))$ to be the class of languages decided by Turing machines using $O(f(n))$ cells on their work tapes for inputs of ...
David Richerby's user avatar
14 votes
Accepted

Can any problem in P be converted to any other problem in P in polynomial time?

If by convert you mean reduce (through a Karp-reduction), then it is possible to reduce any problem $A$ in $P$ to any non-trivial problem $B$ in $P$. Here "non trivial" means that $B$ has at least ...
Steven's user avatar
  • 29.5k
12 votes

What is the definition of P, NP, NP-complete and NP-hard?

From the P vs. NP and the Computational Complexity Zoo video. For a computer with a really big version of a problem... P problems easy to solve (rubix cube) NP problems hard to solve - but ...
icc97's user avatar
  • 233
12 votes
Accepted

If NP is the class of problems that cannot be solved in polynomial time, what is co-NP?

Your prof was absolutely not rigorous (i.e. completely wrong), that's why the distinction between NP and co-NP doesn't make sense with his definition. Better definition: Def.: A decision problem (...
gnasher729's user avatar
11 votes

Relationship of algorithm complexity and automata class

Here are some known results: $\mathsf{REG} = \mathsf{DSPACE}(1) = \mathsf{NSPACE}(1) = \mathsf{DSPACE}(o(\log\log n)) = \mathsf{NSPACE}(o(\log\log n))$, where $\mathsf{REG}$ is the set of regular ...
Yuval Filmus's user avatar
11 votes
Accepted

Complexity classes pertaining to listing all solutions?

The concept you are looking for is called enumeration complexity, which is the study of the computational complexity of enumerating (listing) all the solutions to a problem (or the members of a ...
mdxn's user avatar
  • 1,301
10 votes
Accepted

Is DTIME(n) = DTIME(2n) true? (unlike Rosenberg's results)

$\mathrm{DTIME}(O(n))$ is the set of problems that can be solved in deterministic $O(n)$ time for some constant implicit in $O$, in other words, it is the union of the $\mathrm{DTIME}(cn)$ for all $c&...
Gro-Tsen's user avatar
  • 393
10 votes
Accepted

Why is PH in PSPACE?

No, it is not necessary to remember all $y$'s tried before. In order to remember that I've tried the numbers $1,2,\ldots,200$, I do not need to remember $3,4,5,6,\ldots,199$. If you try them in order, ...
Tom van der Zanden's user avatar
9 votes
Accepted

Why does P/Poly can also receive bad advice?

Lets review the definition of the class $P/poly$ by Turing machines which take advice. The class $T(n)/a(n)$ is the set of languages decidable by a Turing machine which runs in time $T(n)$ with ...
Ariel's user avatar
  • 13.4k
9 votes
Accepted

Known problems in BQP \ NP?

If there was a problem known to be in $\text{BQP}$ but not $\text{NP}$, that would prove $\text{BQP} \not\subset \text{P}$. But $\text{BQP}$ vs $\text P$ is also still an open problem. It is ...
Craig Gidney's user avatar
  • 5,852
9 votes

What is the definition of P, NP, NP-complete and NP-hard?

P, NP, NP-complete and NP-hard are complexity classes, classifying problems according to the algorithmic complexity for solving them. In short, they're based on three properties: Solvable in ...
Thomas C. G. de Vilhena's user avatar
9 votes

What is the relation between EXPTIME and NP HARD complexity classes?

There are NP-hard problems that are not in EXPTIME and vice versa. This is to be expected as NP-hard is defined by a lower bound and EXPTIME mainly by an upper bound. NP is contained in EXPTIME, ...
Pontus's user avatar
  • 687
8 votes
Accepted

Relationship of algorithm complexity and automata class

Yes, there are relationships in many cases! For example, it is known that any language which is accepted by reversal-bounded counter machines are in $P$ (see here). Similarly, we know that all ...
Joey Eremondi's user avatar
8 votes
Accepted

Is NEXP = co-NEXP?

It is known that $\mathsf{NP} = \mathsf{coNP}$ implies $\mathsf{NEXP} = \mathsf{coNEXP}$, using a padding argument. However, both are considered unlikely. The difference between classes like $\mathsf{...
Yuval Filmus's user avatar
8 votes

What is the relation between EXPTIME and NP HARD complexity classes?

The two classes are incomparable: neither is a subset of the other. There are problems in EXPTIME that are not NP-hard. The languages $\emptyset$ and $\Sigma^*$ are both in EXPTIME but are definitely ...
David Richerby's user avatar
8 votes
Accepted

Oracle Turing Machine EXP^EXP

No, $\mathsf{EXP^{EXP}=2EXP}$, a set of languages decidable in $O\left(2^{2^{\mathrm{poly}(n)}}\right)$ time. This is just because you can give exponentially long input to an oracle which can solve ...
rus9384's user avatar
  • 1,632
8 votes
Accepted

Is every PSPACE-complete problem complete with respect to logspace reductions?

If every PSPACE complete problem is also complete under logspace reduction, then $\mathsf{P\neq PSPACE}$. To see why, suppose for the purpose of contradiction that the definition of completeness ...
Ariel's user avatar
  • 13.4k
8 votes
Accepted

Is DISCRETE LOG a NP hard problem?

No one knows, but: It is suspected that neither factoring nor discrete logarithm are NP-complete, but we have no proof. (Evidence for the suspicion: they are in NP $\cap$ coNP. See, e.g., https://...
D.W.'s user avatar
  • 159k
7 votes
Accepted

If NP is not a proper subset of coNP, why does NP not equal coNP?

$\{2,3\}$ is not a proper subset of $\{3,4\}$, yet the two clearly are not equal. Comparing sets is not like comparing numbers: two sets might not be comparable. Additionally, NP is not a subset of ...
Joey Eremondi's user avatar
7 votes
Accepted

$\mathsf{PP=RP}$ consequences

$\newcommand{\co}{\mathsf{co}\text{-}}$ $\newcommand{\class}[1]{\mathsf{#1}}$ If $\class{PP}=\class{RP}$ then you have a collapse to the first level, namely $\class{PH}=\class{NP}$. Assume $\class{...
Ariel's user avatar
  • 13.4k
7 votes
Accepted

Is complexity class $\Sigma^1_1$ from polynomial hierarchy decidable?

The confusion lies in using the same notation for the polynomial hierarchy and the arithmetical hierarchy. The similarities are obvious, both notations talk about formulas with increasing quantifier ...
Ariel's user avatar
  • 13.4k
7 votes
Accepted

Is exponentiation in P?

Your problem is not in P, for two different reasons: P is a class of decision problems, but your problem is a function problem. Instead of P, you should consider its functional equivalent FP. The ...
Yuval Filmus's user avatar
7 votes
Accepted

Are any "standard" complexity classes uncountably infinite?

Nonuniform complexity class P/poly is uncountable. We can just choose for each input length its own circuit, so for any subset $S \subset \mathbb{N}$ the language $L_S = \{w \colon |w| \in S \}$ is in ...
Vladimir Lysikov's user avatar
6 votes

Proofs of $P^{\#P}\subseteq P^{PP}$ and $\#P\subseteq FP^{PP}$

This is not the full proof but the complete idea: Let Turing machine $M$, define $L_M$ be the language defined as $L_M = \{ ((x, y) \in \Sigma^*\times \mathbb{N} ~|~ \#\text{accepting path of } {M}...
Thinking's user avatar
  • 151
6 votes
Accepted

Select a subset of the columns in $2\times n$ matrix, is it easy?

Hint: Show that it is NP-hard by reduction from EQUAL-PARTITION, the variant of PARTITION in which we require both parts to have equal sizes (see a question on math.se). Given an instance $\{x_1,\...
Yuval Filmus's user avatar
6 votes
Accepted

Maximal class for which function equivalence is decidable

There is no such maximal class. Proof: suppose that there is a class $C$ of representations of total functions for which function equivalence is decidable. If $C$ is finite, then $C$ augmented with ...
Gilles 'SO- stop being evil''s user avatar

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