The class of problems for which there exists a BPP machine M such that, for all inputs x,
Alternatively defined as NPBPP.
Contains NP and BPP, and is contained in MA and SBP.
∃BPP seems obviously equal to MA, yet [FFK+93] constructed an oracle relative to which they're unequal! Here is the difference: if the answer is "yes," MA requires only that there exist a y such that for at least 2/3 of random strings r, M(x,y,r) accepts (where M is a P predicate). For all other y's, the proportion of r's such that M(x,y,r) accepts can be arbitrary (say, 1/2). For ∃BPP, by contrast, the probability that M(x,y) accepts must always be either at most 1/3 or at least 2/3, for all y's.
The class of decision problems solvable by a Merlin-Arthur protocol, which goes as follows. Merlin, who has unbounded computational resources, sends Arthur a polynomial-size purported proof that the answer to the problem is "yes." Arthur must verify the proof in BPP (i.e. probabilistic polynomial-time), so that
Defined in [Bab85].
An alternative definition requires that if the answer is "yes," then there exists a proof such that Arthur accepts with certainty. However, the definitions with one-sided and two-sided error can be shown to be equivalent (see [FGM+89]).
Contains NP and BPP (in fact also ∃BPP), and is contained in AM and in QMA.
There exists an oracle relative to which BQP is not in MA [Wat00].
Equals NP under a derandomization assumption: if E requires exponentially-sized circuits, then PromiseBPP = PromiseP, implying that MA = NP [IW97].
Shown in [San07] that MA/1 (that is, MA with 1 bit of advice) does not have circuits of size for any . In the same paper, the result was used to show that MA/1 cannot be solved on more than a fraction of inputs having length by any circuit of size . Finally, it was shown that MA does not have arithmetic circuits of size .
The class of decision problems for which the following holds. There exists a #P function f and an FP function g such that, for all inputs x,
Defined in [BGM02], where the following was also shown:
There exists an oracle relative to which SBP is not closed under intersection [GLM+15].
If SAT can be solved by an NP-machine with sub-exponential number of accepting paths, then SBP = AM [Vol20].