The meaning of relativization

Question

I don't understand the notion of relativization. I expose with an example. Consider a class $A$ that contains $P$, e.g. $NP$. Why $P^A$ is not necessarily equal to $A$? I can naively think that if one can decide any problem in $A$, one can also decide any problem that uses polynomially many steps in a Turing machine plus calls to $A$. One can have at most polynomially many calls to $A$, thus $P^A$ seems there is only a polynomial overhead compare to $A$. If $A$ is $NP$, I could give a certification for all the (polinomially many) oracle calls, and also easily give a certification of the polynomial rest of the computation. I understand that if it is true, then $NP$ is the same as $coNP$, both being contained in $P^{NP}$ from the polynomial hierarchy. So, where is the error in my naive argument?

Answer 1

Take $A=NP$, as you requested. $P^{NP}$ is not necessarily equal to $NP$.

Let me give an example why not. Consider TAUTOLOGY (given a boolean formula $\varphi$, is it true for all possible assignments to the variables?). TAUTOLOGY is known to be co-NP-complete. Therefore, TAUTOLOGY most likely is not in NP, since if TAUTOLOGY were in NP, it would follow that NP = co-NP, which is widely suspected not to be the case.

However, TAUTOLOGY is certainly in $P^{NP}$. Here's why. Given an oracle for SAT, we can solve TAUTOLOGY as follows: Given $\varphi$, query whether $\neg\varphi$ is in SAT; then flip the answer (if $\neg\varphi$ is in SAT, answer "no, $\varphi$ is not in TAUTOLOGY"; if $\neg\varphi$ is not in SAT, answer "yes, $\varphi$ is in TAUTOLOGY"). This shows how to solve TAUTOLOGY in polynomial time given an oracle for SAT, and SAT is in NP, so this shows that TAUTOLOGY is in $P^{NP}$.

So TAUTOLOGY is an example of a problem that is in $P^{NP}$, but probably isn't in NP (unless NP = co-NP). Hopefully this helps illustrate why NP is not the same as $P^{NP}$.

If not, here's a little more intuition. NP is a particular class. $P^{NP}$ is the class of problems that can be solved by querying an oracle for NP, possibly many times. The "many times" part is the key difference between NP and $P^{NP}$. A (many-one) reduction between two NP problems can only query a NP-oracle once, while $P^{NP}$ allows you to query the oracle as many times as you like. That seems to be a lot more powerful, and it's the reason why $P^{NP}$ seems to be a bigger class than NP.

Answer 2

Why might $P^A$ not equal to $A$?

$A$ might be weaker than $P$

$P^A$ can't be weaker than $P$, but $A$ might be. For example, if $A$ is the complexity class of linear-time problems $O(n)$ then $P^A = P^{O(n)}=P$.

$A$ might have different strengths than $P$

For example, consider $B=O(1)^{3SAT}$ and $A=3Coloring$. The only problem $A$ can solve is 3-coloring. It can't solve 3-Sat. $B$ can solve 3-sat, and various constant time problems, but it can't solve 3-coloring because it doesn't have enough time to do so directly or to translate the problem into 3-sat.

On the other hand, $B^A = {O(1)^{3SAT}}^{3Coloring}$ can solve both 3-sat and 3-coloring, so it's stronger than either $A$ or $B$ individually.

In the case of $P$, a class with different strengths could be the class of problems that takes $O(n)$ space. Some problems in $P$ require more-than-linear space, and some problems that require $O(n)$ space use exponential time. So $P^{O(n) space}$ gains both the powers of $P$ and $O(n) space$, making it larger than either of them individually.

$A$ and $P$ might synergize into something better

For example, $P^{NP}$ can trivially solve $coNP$ problems, but it's suspected that $P \subset coNP$ and $NP \neq coNP$ and $(P \cup coNP) \not \subset coNP$.

This is even more striking if you think of it as $NP^{NP}$. Complexity classes don't have to be low for themselves.