Take the 2-minute tour ×
Computer Science Stack Exchange is a question and answer site for students, researchers and practitioners of computer science. It's 100% free, no registration required.

Does anybody know a good definition of 2 decision / optimization problems being equivalent?

I am asking since for example allowing polynomial time computations any 2 problems in NP could be considered equivalent.

share|improve this question
You probably mean, every 2 ${\sf NP}$-complete problems can be considered equivalent. What you have said might not be true. –  A.Schulz Oct 31 '12 at 8:08
You should probably say "equivalent w.r.t. computational complexity". There are many other notions one could investigate. –  Raphael Oct 31 '12 at 9:53
add comment

1 Answer

up vote 5 down vote accepted

Suppose $L_1$ and $L_2$ are two decision problems, and $\mathcal{A}$ is a class of algorithms (like $P$). Say that $L_1$ is $\mathcal{A}$-reducible to $L_2$ if there is an $f \in \mathcal{A}$ such that $x \in L_1$ iff $f(x) \in L_2$. Say that $L_1$ and $L_2$ are $\mathcal{A}$-equivalent if each is $\mathcal{A}$-reducible to the other. For example, all NP-complete problems are $P$-equivalent. If you want a finer notion, you can consider more restricted classes, such as log-space or even AC0. Such notions of reducibility actually come up in refinements of Schaefer's dichotomy theorem.

As for optimization problems, let's consider maximization problems. With each maximization problem $L$ (which is a function mapping instances to optimal values), you can associate the decision problem $L'$ consisting of pairs $(x,v)$ such that $L(x) \geq v$. Now you can define reducibility and equivalence as before.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.