User:Wim van Dam

About this page
This is my personal page, which I will use mostly as a sandbox. To get some proper information about me, go to my home page at UC Santa Barbara. This is very much my sandbox.

Sandbox on Reductions in Complexity Theory
In complexity theory reductions between problems play an important role, as they allow us talk about the relative hardness of problems. When there is a reduction from $$A$$ to $$B$$ such that $$B$$ is at least as hard as $$A$$, we denote this by a no greater than symbol between $$A$$ and $$B$$ with a subscript indicating the specific reduction that is used. Here are the three most relevant ones.


 * $$A\leq_T B$$ Turing or Cook reduction, if there exists a polytime deterministic algorithm that decides for each possible $$x$$ whether or not $$x\in A$$ using queries to an oracle for the problem $$B$$.
 * $$A \leq_{tt} B$$ truth table reduction if there exist a polytime deterministic algorithm that decides membership of $$A$$ using non-adaptive queries to the oracle for $$B$$.
 * $$A \leq B$$ or $$A \leq_m B$$ Karp or many one reduction, if there exists a polytime computable function $$F$$ such that for all $$x$$ it holds that $$x \in A$$ if and only if $$F(x)\in B$$.

Some examples of the difference between these reductions are: In general $$\mathsf{P}^B := \{A : A \leq_T B\}$$.
 * Although the traveling salesman problem sits in $$\mathsf{NP}$$, finding the exact length has to be done adaptively (most likely), hence random decision problems about the shortest length does not sit in $$\mathsf{NP}$$, although it Turing reduces to it, hence such problems sit in in the complexity class $$\Delta_2^p = \mathsf{P}^{\mathsf{NP}} := \{ L : L \leq_T \mathrm{SAT}\}$$.


 * Consider the decision problem "Given two Boolean formulas, determine whether or not exactly one of them is satisfiable." This requires two non-adaptive queries, hence this problem sits in $$\Theta_2^p = \mathsf{P}^{\mathsf{NP}}_{\|} := \{ L : L \leq_{tt} \mathrm{SAT}\}$$, but it does not sit in $$\mathsf{NP}$$ or in $$\mathsf{co}$$-$$\mathsf{NP}$$ as far as we know (the $$\|$$ indicates the pararallelness of the queries).
 * The fact that $$\mathrm{SAT}$$ is $$\mathsf{NP}$$-complete means that for all $$A\in \mathsf{NP}$$ we have $$A\leq \mathrm{SAT}$$.