Eccc Tr 97 - 045
نویسنده
چکیده
We provide another proof of the Sipser{Lautemann Theorem by which BPP MA (PH). The current proof is based on strong results regarding the ampliication of BPP, due to Zuck-erman. Given these results, the current proof is even simpler than previous ones. Furthermore, extending the proof leads to two results regarding MA: MA ZPP NP (which seems to be new), and that two-sided error MA equals MA. Finally, we survey the known facts regarding the fragment of the polynomial-time hierarchy which contains MA.
منابع مشابه
Another Proof that BPP ⊆ PH ( and more )
We provide another proof of the Sipser–Lautemann Theorem by which BPP ⊆ MA (⊆ PH). The current proof is based on strong results regarding the amplification of BPP, due to Zuckerman (1996). Given these results, the current proof is even simpler than previous ones. Furthermore, extending the proof leads to two results regarding MA: MA ⊆ ZPP (which seems to be new), and that two-sided error MA equ...
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