[Solved]: Anti-symmetry of polynomial time reductions

Problem Detail: 

I read somewhere that, if $A\leq_p B$ and $B\leq_p A$, then it is said that $A\equiv_p B$. What exactly does this mean? Is it saying that both $A$ and $B$ are the exact same level of complexity?

Asked By : agent154

Answered By : Shaull

$A\equiv_p B$ is by definition $A\le_p B$ and $B\le_p A$.

It means that they are in the "exact same level of complexity", but only with respect to polynomial time reductions! For example, if $A\in LOGSPACE$ and $B\in NLOGSPACE$, then trivially $A\equiv_p B$. But clearly $A$ and $B$ are not in the "same level of complexity".

This is part of a general observation about reductions: A reduction that is computable by a machine that works in $t(n)$ time (resp. $s(n)$ space) is only useful if you use it in classes that are provable at least as strong as $TIME[t(n)]$ (resp. $SPACE[s(n)]$), and are conjectured to be stronger.

So, polynomial time reduction are used in $NP$ and $PSPACE$, $LOGSPACE$ reductions are used in $NLOGSPACE$, etc.

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Question Source : http://cs.stackexchange.com/questions/9985