New answers tagged


co-$\mathsf{P} = \mathsf{P}$. To see this, pick your favorite problem $A$ in co-$\mathsf{P}$, let $T$ be a Turing machine for the complement of $A$ (in $\mathsf{P}$) and construct a new Turing Machine $T'$ that simulates $T$, accepts if $T$ rejects, and rejects if $T$ accepts.


Recall that a collection $\mathcal{F}$ of computable partial functions admits a computable numbering if there is a computable function $f : \mathbb{N} \to \mathbb{N}$ such that $\mathcal{F} = \{\varphi_{f(e)} \mid e \in \mathbb{N}\}$, where $(\varphi_e)_{e \in \mathbb{N}}$ is a standard enumeration of the computable partial functions. Let $\mathcal{I}$ be ...


You can compute the sumset (well, difference set) $X-Y$ using standard techniques (Computing Cardinality of Sumsets using Convolutions and FFT). As an optimization, I suggest first replacing $X,Y$ with $X'=X \bmod p$, $Y'=Y \bmod p$ where $p$ is at least $n+m$ or so. Then, enumerating through the elements of the multiset $X'-Y'$ by decreasing cardinality ...

Top 50 recent answers are included