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Results tagged with Search options answers only user 683
36 results
You are misinterpreting the statement quoted at point 1. We are unable to prove the existence of a single language in NP that is not in P, but we suspect that there are infinitely many, in fact, we su …
answered Jul 8 '14 by Yuval Filmus
Given an $O(n^2)$ reduction from $A$ to $B$ and an $O(n^3)$ algorithm for solving $B$, you can solve $A$ as follows: Given an instance of $A$ of size $n$, reduce it to an equivalence instance $B$ of …
answered Dec 4 '18 by Yuval Filmus
This is a rather strange question to ask, given that a large majority of complexity theorists believe that P is different from NP. All the evidence we have points at P being different from NP, which i …
answered Sep 18 '15 by Yuval Filmus
In order to break a one-way function, it suffices to be able to find a single preimage. Given $x$, $f(2x) = x$, so finding a preimage of an arbitrary input is easy. Hence it's not a one-way function a …
answered Jul 21 by Yuval Filmus
It is known that P$\subseteq$NP$\subset$R, where R is the set of recursive languages. Since R is countable and P is infinite (e.g. the languages $\{n\}$ for $n \in \mathbb{N}$ are in P), we get that P …
answered Dec 31 '12 by Yuval Filmus
Let me answer your concrete questions: An oracle is a language. One way to describe a language is to give a procedure for constructing it. The construction is not a counterexample to P=NP. It shows …
answered Sep 10 by Yuval Filmus
As Raphael explains, this question is ill-posed, since at most one of P=NP and P≠NP should be provable at all. However, a similar question arises in theoretical computer science in several guises, the …
answered Jan 28 '16 by Yuval Filmus
First of all, it is not clear what you mean by an algorithm which is able to solve any problem in NP; usually one algorithm solves one problem. However, given a polynomial time algorithm for solving s …
answered Apr 15 '15 by Yuval Filmus
Here is a more direct proof. Suppose that $P=NP$. In particular, SAT can be solved in polynomial time. Therefore $P^{SAT} \subseteq P^P = P \subseteq coNP$.
answered Sep 24 '17 by Yuval Filmus
We won't necessarily see any effects. Suppose that somebody finds an algorithm that solves 3SAT on $n$ variables in $2^{100} n$ basic operations. You won't be able to run this algorithm on any instanc …
answered Dec 30 '14 by Yuval Filmus
First, let me start by explaining what Boolean circuits are. You are probably familiar with Boolean formulas — these are formulas of the sort $(a \land b) \lor (\lnot a \land \lnot b)$. We can represe …
answered May 10 '18 by Yuval Filmus
You ask Doesn't polynomially bounding the length of possible solutions to a given instance mean that there are only polynomially many possible solution candidates? In fact, the number of binary …
answered Mar 18 '17 by Yuval Filmus
The time-hierarchy theorem shows that for each $k$ there are problems solvable in time $O(n^{k+1})$ but not in time $O(n^k)$; this problem is (roughly) the halting problem for Turing machines running …
answered Aug 24 '15 by Yuval Filmus
The polynomial hierarchy is not $\mathsf{DTIME}(n^k)$ for various $k$, but rather a polynomial time version of the arithmetical hierarchy.
answered Nov 27 '16 by Yuval Filmus
A deterministic polynomial time machine for a language $L$ can easily be converted to a non-deterministic polynomial time machine which has the same operational semantics (that is, it operates determi …
answered Dec 20 '15 by Yuval Filmus

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