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# Lecture13 - Notes on Complexity Theory Last updated October 2011 Lecture 13 Jonathan Katz 1 Randomized Time Complexity 1.1 How Large is BPP We know

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Unformatted text preview: Notes on Complexity Theory Last updated: October, 2011 Lecture 13 Jonathan Katz 1 Randomized Time Complexity 1.1 How Large is BPP ? We know that P ⊆ ZPP = RP ∩ co RP ⊆ RP ∪ co RP ⊆ BPP ⊆ PSPACE . We currently do not have a very good unconditional bound on the power of BPP — in particular, it could be that BPP = NEXP . Perhaps surprisingly, especially in light of the many randomized algorithms known, the current conjecture is that BPP is not more powerful than P . We will return to this point later in the semester when we talk about derandomization. What (unconditional) upper bounds can we place on BPP ? Interestingly, we know that it is not more powerful than polynomial-size circuits; actually, the following theorem is also a good illustration of the power of non-uniformity. Theorem 1 BPP ⊂ P / poly . Proof Let L ∈ BPP . Using amplification, we know that there exists a polynomial-time Turing machine M such that Pr[ M ( x ) 6 = χ L ( x )] < 2-| x | 2 . Say M uses (at most) p ( | x | ) random coins for some polynomial p . (Note that p is upper-bounded by the running time of M .) An equivalent way of stating this is that for each n , and each x ∈ { , 1 } n , the set of “bad” coins for x (i.e., coins for which M ( x ) outputs the wrong answer) has size at most 2 p ( n ) · 2- n 2 . Taking the union of these “bad” sets over all x ∈ { , 1 } n , we find that the total number of random coins which are “bad” for some x is at most 2 p ( n ) · 2- n < 2 p ( n ) . In particular, there exists at least one set of random coins r * n ∈ { , 1 } p ( n ) that is “good” for every x ∈ { , 1 } n (in fact, there are many such random coins). If we let the sequence of “advice strings” be exactly { r * n } (using the alternate definition of P / poly ), we obtain the result of the theorem. We can also place BPP in the polynomial hierarchy: Theorem 2 BPP ⊆ Σ 2 ∩ Π 2 . Proof We show that BPP ⊆ Σ 2 ; since BPP is closed under complement, this proves the theorem. We begin by proving some probabilistic lemmas. Say S ⊆ { , 1 } m is large if | S | ≥ (1- 1 m )2 m , and is small if | S | < 2 m m . For a string z ∈ { , 1 } m define S ⊕ z def = { s ⊕ z | s ∈ S } ....
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## This note was uploaded on 01/13/2012 for the course CMSC 652 taught by Professor Staff during the Fall '08 term at Maryland.

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Lecture13 - Notes on Complexity Theory Last updated October 2011 Lecture 13 Jonathan Katz 1 Randomized Time Complexity 1.1 How Large is BPP We know

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