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## Hot answers tagged oracles

26

You can just take the oracle A s.t. NP$^A$=EXP$^A$ since EXP is not in i.o.-subexp. For SAT$^A$ it depends on the encoding, for example if the only valid SAT instances have even length then it is easy to solve SAT on odd-length strings. But if you use a language like $L=\{\phi 01^*\ |\ \phi\in SAT^A\}$ then you should be fine.

23

The oracle is basically just an implementation of the predicate you want to search for a satisfying solution to. For example, suppose you have a 3-sat problem: (¬x1 ∨ ¬x3 ∨ ¬x4) ∧ (x2 ∨ x3 ∨ ¬x4) ∧ (x1 ∨ ¬x2 ∨ x4) ∧ (x1 ∨ x3 ∨ x4) ∧ (¬x1 ∨ x2 ∨ ¬x3) Or, in table form with each row being a 3-clause, x meaning "this variable false", o ...

17

Yes. First, since it took me a minute to figure this out myself, let me formalize the difference between your question and $\mathsf{AlmostP}$; it's the order of quantifiers. $\mathsf{AlmostP} := \{L : Pr_R(L \in \mathsf{P}^R) = 1\}$, and the result you allude to is $\forall L\, L \in \mathsf{BPP} \iff Pr_R(L \in \mathsf{P}^R) = 1$. If I've understood ...

16

You don't have to go to the lengths Lance was suggesting. For example, relative to a random oracle, using the oracle as a one-way function (say, evaluated on consecutive bit postions) is exponentially hard to invert on all but finitely many lengths. This problem directly reduces to SAT on the same length input, so it does follow that SAT^A is not in ...

14

You are exactly right. The relativization operation $B\mapsto B^A$ is not well defined. P and PA are independently defined objects. The names are suggestive, but you cannot formally define PA from the set P. (You can define P from PA by setting A to be the empty set.) Think of PA as being some kind of generalization of P, which equals P when A is empty, ...

13

Yes. Indeed, an oracle $A$ satisfies $\mathsf{NP}^A=\mathsf{NP}$ if and only if $A \in \mathsf{NP} \cap \mathsf{coNP}$. This class is called $\mathsf{Low(NP)}$ or sometimes $\mathsf{L_1 P}$ (see the link and the paper cited there for more of an explanation of the low hierarchy in general). Your intuition about "determinism" is actually somewhat correct (...

13

Because one of the principal applications of Type Theory in formalizations has been to study programing languages and computation in general, a lot of thought has gone into ways of representing possibly non-terminating programs. I won't make a complete survey here, but I'll try and give pointers to the main thrusts of different directions. The "...

12

For question 1, the BGS construction can be performed in exponential time, so you can construct such $B \in \mathsf{EXP}$. (For question 2: Sasho Nikolov's answer was originally only for $\mathsf{\Sigma_k P}$-complete languages, and I pointed out that one can also take any $B' \in \mathsf{NP} \cap \mathsf{coNP}$, since $\mathsf{NP}^{\mathsf{NP} \cap \mathsf{... 11 First, this result is listed in the complexity zoo: https://complexityzoo.uwaterloo.ca/Complexity_Zoo:N#npiconp. Alternatively, it's possible to prove without much trouble (which I do below). We want to show that$P^{NP \cap coNP} = NP \cap coNP$. Clearly, one direction is obviously true:$NP \cap coNP \subseteq P^{NP \cap coNP}$. To prove the other ... 11 This version of the answer incorporates feedback from Emil Jeřábek. As far as I can see, the main twist is that there is a language in$\mathsf{EXP}^{\Sigma^\mathsf{P}_2}$of exponential circuit complexity. In particular, fix a binary encoding of boolean circuits and define$L$as the language defined by$L_n$is not decided by any circuit of size$2^{n/2}$,... 11 For question 2, you can take any$B' \in \mathsf{PH}$(this means you cannot bring down the$B$in the BGS result down from$\mathsf{EXP}$to$\mathsf{PH}$without resolving the big question). Clearly for any$B'$,$P \subseteq \mathsf{P}^{B'} \subseteq \mathsf{NP}^{B'}$. Let$B' \in \Sigma_i^{\mathsf{P}}$. Recall that, by the definition of the Polynomial ... 10 Elaborating on Joe's earlier answer: note that$\textrm{FACTORING} \in \mathsf{NP \cap coNP}$. The latter is the second lowest class in the "low" hierarchy: which is to say that$\mathsf{NP^{NP \cap coNP} = NP}$. This implies in particular that $$\mathsf{P^{\textrm{FACTORING}} \subseteq NP^{\textrm{FACTORING}}} \subseteq \mathsf{NP}.$$ We may make similar ... 9 I believe if you trace through the argument given, e.g., in Section 4.1 of Ker-I Ko's survey, you get an upper bound of$\mathsf{DTIME}(2^{2^{O(n^2)}})$. In fact, we can replace$n^2$here with any function$nf(n)$where$f(n) \to \infty$as$n \to \infty$. This isn't quite what was asked for, but it's close. In particular, using the translation between ... 8 Yes. Beigel CCC '89 showed$\mathsf{P} \neq \mathsf{UP} \neq \mathsf{NP}$with probability 1. Combined with Rossman-Servedio-Tan, this gives the result you want. You should always try the Complexity Zoo for questions like this... 8 This is equivalent to$LOGSPACE≠NP$(which is obviously open). The proof of that equivalence relativizes (at least under the usual oracle models). And there are oracles making$LOGSPACE = NP$(the PSPACE-complete TQBF works) and making them not equal (the oracle separating P from NP works). 7 While the order of quantifiers between what you are asking and almost P differ, it is not too hard to show that they are equivalent. First, for any fixed L, the question of whether L \in P^O does not depend on any finite initial segment of O. it follows that the probability that L \in P^R is either 0 or 1. From the almost -P result, for each computable ... 7 I think you may be misunderstanding the sentence "Note that the general notion of a reduction (i.e., Cook-reduction) seems inherent." This is not about reductions being inherent to self-reducibility (in the sense Goldreich uses it), but rather about Cook reductions being inherent to this notion, in the sense that they cannot be replaced by Karp reductions. ... 7 The quoted Beigel, Buhrman, and Fortnow paper gives a solution to 2 in Theorem 1.8: there is an oracle relative to which$\mathrm{P=Mod_3P}$(which implies$\mathrm{P=UP}$), and$\mathrm{\oplus P=NP=EXP}$(which, together with the first equality, actually implies$\mathrm{EXP=ZPP}$). 7$\mathsf{UP} \neq \mathsf{EXP}$is open. A UP-generic oracle* should make$\mathsf{P} \neq \mathsf{UP} = \mathsf{EXP}$, and since$\mathsf{UP} \subseteq \mathsf{\oplus P} \subseteq \mathsf{EXP}$relative to any oracle, this should resolve 1. (I say "should" because I haven't checked all the details...) *UP-generic oracles are discussed, for example, by ... 7$\mathsf{MA_{EXP}} \not\subseteq \mathsf{P/poly}$but there is an oracle relative to which this is false; both were proved in H. Buhrman, L. Fortnow, T. Thierauf. Nonrelativizing separations. CCC '98. (freely available author's version) 6 On popular request, here is my comment as an answer: There is an oracle separating$\mathrm{PP}$from$\mathrm{PSPACE}$: Jacobo Toran, A combinatorial technique for separating counting complexity classes, ICALP 1989. The best result for$\mathrm{P}^\mathrm{PP}$that I know is a conditional result by Heribert Vollmer: Relating polynomial time to constant ... 6 Yes, yes, and yes. The basic idea is to consider the characteristic function of a language$L$(the oracle you're constructing) at length$n$as a string of length$2^n$that will be an input to a ("big") circuit. An OR gate in that circuit will correspond to a polynomially-bounded existential quantifier, and an AND gate to a polynomially-bounded ... 6 The oracle you ask for has$P=NP\ne BQP=NEXP$, and therefore it has$BQP\ne PH$. Finding any oracle relative to which$BQP\ne PH$was an open problem for twenty years until Raz and Tal [1] found such an oracle last year. In summary, the oracle you ask for currently is not known to exist, but people are looking. There are oracles relative to which$P\ne BPP=...

5

Anindya Sen and I have a paper in ALT '13 where we give an $\tilde O(n \sqrt{n})$ algorithm for this problem. We don't know if a better algorithm is possible.

5

(I assume this question will eventually get migrated to CS.SE, but I am posting my answer to it here on cstheory for now.) Technically, one doesn't usually think of relativization as an "operator" or "function"; however, I don't see a reason why you couldn't take a statement and map the statement to a relativized version of it. The trick is that, as others ...

5

Yes on Question 1 (assuming ZFC is consistent). You don't need $f$ to be random exactly, any $f$ will do. And for the proof you need to also use the fact that there is an oracle $h$ with NP$^h=$P$^h$.

5

If you just want oracle separations with $\#P$, you don't need to use the new result of Raz and Tal. You can use the classic Parity/Majority not in $AC^0$ results from the 1980s. For example, the strongest quantitative version of Parity not in $AC^0$ says that the $n$-bit Parity function cannot be computed by a quasi-polynomial size $AC^0$ circuit of depth ...

4

This is quite unlikely to hold, because $\mathrm{EXP_{poly}^{NEXP}}$ actually coincides with $\Theta^{\exp}_2$, the exponential analogue of the class $\Theta^P_2$, which is presumably a strict subclass of $\mathrm{EXP^{NP}}$ (which is the exponential analogue of $\Delta^P_2$). $\Theta^{\exp}_2$ can be variously defined as \Theta^{\exp}_2=\mathrm{EXP^{\|NP}=...

3

Using $n$ calls to the halting oracle and time $O(n^2)$, you can compute the first $n$ bits of the Chaitin's constant. Using the $n$ bits of the Chaitin's constant and unbounded time, all queries to the halting oracle of length approximately $≤n$ (depending on the representation) can be eliminated. Finally, the result can be obtained in bounded time by ...

3

Given an instance of 3SAT with $m$ clauses, you can find the set of clauses that are not satisfied in some optimal assignment with $O(m)$ calls to the oracle. The algorithm: Call the oracle on the original input to find $k^*$, the maximum number of clauses that can be satisfied. For each clause, try removing the clause and then call the oracle. If the ...

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