0
$\begingroup$

What is the difference between $\text{SIZE}(n^k)$ and $\text{P}/\text{poly}$?

For reference:

  • $\text{SIZE}(n^k)$ is defined as the class of problems solvable with Boolean circuits (of fan-in two) with $O(n^k)$ gates.

  • $\text{P}/\text{poly}$ is defined as those problems over $\{0,1\}^*$ which can be solved by an infinite family of polynomial-size circuits ${C_n}$.

What is the difference between these classes?

$\endgroup$
1
$\begingroup$

SIZE($n^2$) consists only of problems that can be solved by circuit families of size at most $O(n^2)$. P/poly contains problems that can be solved by circuit families of size at most $O(n^3)$, and those solved by families of size at most $O(n^4)$, and so on.

In particular, P/poly = $\cup_k$ SIZE($n^k$).

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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