BPP Contained in PH By Michael Sipser; Clemens Lautemann Clemens Lautemann Presenter: Jie Meng

Sipser-Lautemann Theorem M. Sipser. A complexity theoretic approach to randomness, In Proceedings of the 15th ACM STOC, 1983 C. Lautemann, BPP and the polynomial hierarchy, Information Process Letter , 1983

Sipser-Lautemann Theorem Outline Definition and Background Techniques Proof Questions Homework

Sipser-Lautemann Theorem BPP: A language L is in BPP if and only if there exists a randomized Turing Machine M, s.t.

Sipser-Lautemann Theorem Trivially

Sipser-Lautemann Theorem Main Theorem:

Sipser-Lautemann Theorem Let C be a language, NP C be the class that L is in NP C if there is a non- deterministic Turing Machine M, which can accept L, with the power that M can query an oracle such questions like “if y is in C” and get the correct answer in one step. This can be generalized to NP A, A is a language class.

Sipser-Lautemann Theorem NP: L is in NP if there exists a deterministic polynomial Turing Machine M, s.t. x  L   y M(x,y)=1

Sipser-Lautemann Theorem     Acc Rej Acc Rej Acc Rej L in NP, for any x in L

Sipser-Lautemann Theorem     Rej L in NP, for any x not in L

Sipser-Lautemann Theorem

     Acc Rej Acc Rej Acc Rej L in, x in L, L’ in NP y in L’ ? Yes/No

Sipser-Lautemann Theorem     Acc Rej L in, x in L,     Acc Rej y in L’

Sipser-Lautemann Theorem     Acc Rej L in, x in L,     Acc Rej y not in L’

Sipser-Lautemann Theorem   Acc   Rej AccRej

Sipser-Lautemann Theorem Equivalent definition NP: L is in NP, if and only if there exists a deterministic poly-time TM M, s.t. x  L   y M(x,y)=1 : L is in, if and only if there exists a deterministic poly-time TM M’, s.t. x  L  y  z M’(x,y,z)=1 x L  y  z M’(x,y,z)=0

Sipser-Lautemann Theorem Technique 325 lbs 7’ 1’’ VS fat Thin

Sipser-Lautemann Theorem Technique

Sipser-Lautemann Theorem Technique

Sipser-Lautemann Theorem  L in BPP:  By amplifying method and Chernoff Bound PROOF

Sipser-Lautemann Theorem PROOF  Wx={ y | M(x, y)=1}  x in L, |Wx|>2 m (1-1/m), Wx is very fat;  x not in L, |Wx|<2 m 1/m Wx is very thin;  {0,1} m is the whole space;

Sipser-Lautemann Theorem PROOF  Shifting:  If Wx is fat, |Wx|>2 m (1-1/m),  There exists a set of strings y 1, y 2, … y r, r=m/2 s.t.  If Wx is thin, |Wx|<2 m 1/m  There is no such set of strings

Sipser-Lautemann Theorem X in L, Wx is fat, there exists a set of string y 1, y 2, … y r Then for all z in {0, 1} m, That is, there exists i, s.t. PROOF

Sipser-Lautemann Theorem PROOF  x in L, Wx is fat,  There exists y 1, y 2, … y r,  For all z in {0,1} m, M(x, z y i )=1, for some i;  x in L,

Sipser-Lautemann Theorem  Question?

Sipser-Lautemann Theorem HOMEWORK  Finish the proof in case x is not in L, which is to say, fill out the blank in the following statement:  L in BPP, x L  […]  […] M’(x,y,z)=[.]  Give all necessary explanations about your statement.