bpp contained in ph by michael sipser; clemens lautemann clemens lautemann presenter: jie meng

BPP Contained in PHBPP Contained in PH

By Michael Sipser;By Michael Sipser;

Clemens Lautemann Clemens Lautemann

Presenter: Jie Meng

Sipser-Lautemann TheoremSipser-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 14 215-217, 1983


OutlineOutline• 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.

3

1 ]1)([Pr L

3

2]1) ,([Pr L

r

r

x,rMx

rxMx


Trivially

NP ?BPP But,

BPPBPP

PSPACE PP RP P

PSPACE PP RP P

NP

BPP


Main Theorem:

pp22 BPPActually

PH BPP


Let C be a language, NPC be the class that L is in NPC 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 NPA, A is a language class.


NP: L is in NP if there exists a deterministic polynomial Turing Machine M, s.t. xL y M(x,y)=1


Acc Rej RejRejRejRejRejAccRej Rej Rej AccRejRej Rej Rej

L in NP, for any x in L


Rej Rej RejRejRejRejRejRejRej Rej Rej RejRejRej Rej Rej

L in NP, for any x not in L


ii

def

i

def

i

i

def

i

def

PNP

NP

11

1 co

1i

i

def

PH


Acc Rej RejRejRejRejRejAccRej Rej Rej AccRejRej Rej Rej

L in , x in L, L’ in NPp2

y in L’ ?

Yes/No


Acc Rej Rej RejRejRej Rej Rej

L in , x in L, p2


y in L’



L in , x in L, p2


y not in L’


Acc Acc Acc Acc

RejRej Rej Rej RejRejAccRejRejRej Rej Rej


Equivalent definition

NP: L is in NP, if and only if there exists a deterministic poly-time TM M, s.t. xL 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

p2 p

2


Technique

325 lbs

7’ 1’’

VS

fat

Thin


Technique


L in BPP:L in BPP:

By amplifying method and Chernoff Bound By amplifying method and Chernoff Bound

PROOF

3

1 ]1)([Pr L

3

2]1) ,([Pr L

r

r

x,rMx

rxMx

mrxM

mrxM

1]1),([Pr Lx

11]1),([Pr L x

m

m

{0,1}r

{0,1}r


PROOFPROOF

Wx={ y | M(x, y)=1}Wx={ y | M(x, y)=1}

x in L, |Wx|>2x in L, |Wx|>2m m (1-1/m), Wx is very fat;(1-1/m), Wx is very fat;

x not in L, |Wx|<2x not in L, |Wx|<2mm 1/m Wx is very thin; 1/m Wx is very thin;

{0,1}{0,1}mm is the whole space; is the whole space;


PROOFPROOF

Shifting:Shifting:

If Wx is fat, |Wx|>2If Wx is fat, |Wx|>2m m (1-1/m),(1-1/m), There exists a set of strings yThere exists a set of strings y11, y, y22, … y, … yrr, r=m/2 , r=m/2

s.t. s.t.

If Wx is thin, |Wx|<2If Wx is thin, |Wx|<2mm 1/m 1/m There is no such set of stringsThere is no such set of strings

z}y tWx, t|{zyWx

,{0,1} y m

mi 1} {0, y Wx

iy

mm rm

21

2


X in L, Wx is fat,

there exists a set of string y1, y2, … yr

Then for all z in {0, 1}m ,

That is, there exists i, s.t.

PROOFPROOF

mi 1} {0, y Wx

iy

y Wx z iyi

Wx y z i


PROOFPROOF

x in L, Wx is fat, x in L, Wx is fat, There exists yThere exists y11, y, y22, … y, … yrr, ,

For all z in {0,1}For all z in {0,1}mm, M(x, z y, M(x, z y ii)=1, for some i;)=1, for some i;

x in L, x in L,

1z)y M(x,or ...or 1z)y M(x,or 1z)y M(x,

z y ... y y

r21

r21


Question? Question?


HOMEWORKHOMEWORK

Finish the proof in case x is not in L, which Finish the proof in case x is not in L, which is to say, fill out the blank in the following sis to say, fill out the blank in the following statement:tatement:

L in BPP, xL in BPP, x L L […] […] […] […] M’(x,y,z)=[.]M’(x,y,z)=[.]

Give all necessary explanations about youGive all necessary explanations about your statement.r statement.

bpp contained in ph by michael sipser; clemens lautemann clemens lautemann presenter: jie meng

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