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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
Sipser-Lautemann TheoremSipser-Lautemann Theorem
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
Sipser-Lautemann Theorem
Trivially
NP ?BPP But,
BPPBPP
PSPACE PP RP P
PSPACE PP RP P
NP
BPP
Sipser-Lautemann Theorem
Main Theorem:
pp22 BPPActually
PH BPP
Sipser-Lautemann Theorem
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.
Sipser-Lautemann Theorem
NP: L is in NP if there exists a deterministic polynomial Turing Machine M, s.t. xL y M(x,y)=1
Sipser-Lautemann TheoremSipser-Lautemann Theorem
Acc Rej RejRejRejRejRejAccRej Rej Rej AccRejRej Rej Rej
L in NP, for any x in L
Sipser-Lautemann TheoremSipser-Lautemann Theorem
Rej Rej RejRejRejRejRejRejRej Rej Rej RejRejRej Rej Rej
L in NP, for any x not in L
Sipser-Lautemann Theorem
ii
def
i
def
i
i
def
i
def
PNP
NP
11
1 co
1i
i
def
PH
Sipser-Lautemann Theorem
Acc Rej RejRejRejRejRejAccRej Rej Rej AccRejRej Rej Rej
L in , x in L, L’ in NPp2
y in L’ ?
Yes/No
Sipser-Lautemann Theorem
Acc Rej Rej RejRejRej Rej Rej
L in , x in L, p2
Acc Rej Rej RejRejRej Rej Rej
y in L’
Sipser-Lautemann Theorem
Acc Rej Rej RejRejRej Rej Rej
L in , x in L, p2
Acc Rej Rej RejRejRej Rej Rej
y not in L’
Sipser-Lautemann Theorem
Acc Acc Acc Acc
RejRej Rej Rej RejRejAccRejRejRej Rej Rej
Sipser-Lautemann Theorem
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
Sipser-Lautemann Theorem
Technique
325 lbs
7’ 1’’
VS
fat
Thin
Sipser-Lautemann Theorem
Technique
Sipser-Lautemann Theorem
Technique
Sipser-Lautemann TheoremSipser-Lautemann Theorem
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
Sipser-Lautemann TheoremSipser-Lautemann Theorem
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;
Sipser-Lautemann TheoremSipser-Lautemann Theorem
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
Sipser-Lautemann TheoremSipser-Lautemann Theorem
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
Sipser-Lautemann TheoremSipser-Lautemann Theorem
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
Sipser-Lautemann TheoremSipser-Lautemann Theorem
Question? Question?
Sipser-Lautemann TheoremSipser-Lautemann Theorem
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.