BPP Contained in PH
By Michael Sipser;
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 14
215-217, 1983
Sipser-Lautemann Theorem
Outline
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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.
2
x  L  Prr [ M ( x, r )  1] 
3
1
x  L  Prr [ M ( x,r)  1] 
3
Sipser-Lautemann Theorem

Trivially
P  RP  BPP  PP  PSPACE
P  RP  NP  PP  PSPACE
BPP  BPP
But, BPP  ? NP
Sipser-Lautemann Theorem

Main Theorem:
BPP  PH
Actually BPP   2p   2p
Sipser-Lautemann Theorem
Let C be a language, NPC be the class
that L is in NPC if there is a nondeterministic 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. xL  y M(x,y)=1
Sipser-Lautemann Theorem
L in NP, for any x in L
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Rej Rej Rej Rej Acc Rej Rej Rej Acc Rej Rej Acc Rej Rej Rej Rej
Sipser-Lautemann Theorem
L in NP, for any x not in L
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Rej Rej Rej Rej Rej Rej Rej Rej Rej Rej Rej Rej Rej Rej Rej Rej
Sipser-Lautemann Theorem
def
def
1  NP
def
 i 1  NP
def 
 i  co   i
i
PH    i
i 1
def
 i 1  P
i
Sipser-Lautemann Theorem
L in  2p , x in L, L’ in NP
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y in L’ ?
Yes/No
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Rej Rej Rej Rej Acc Rej Rej Rej Acc Rej Rej Acc Rej Rej Rej Rej
Sipser-Lautemann Theorem
L in  2p , x in L,
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y in L’
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Rej Rej Rej Rej Acc Rej Rej Rej Rej Rej Rej Rej Acc Rej Rej Rej
Sipser-Lautemann Theorem
L in  2p , x in L,
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y not in L’
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Rej Rej Rej Rej Acc Rej Rej Rej Rej Rej Rej Rej Acc Rej Rej Rej
Sipser-Lautemann Theorem
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Rej Rej Rej Rej Acc Acc Acc Acc Rej Rej Rej Rej Rej Acc Rej Rej
Sipser-Lautemann Theorem
Equivalent definition
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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
 2p: L is in  2p , 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
fat
325 lbs
7’ 1’’
VS
Thin
Sipser-Lautemann Theorem
Technique
Sipser-Lautemann Theorem
Technique
Sipser-Lautemann Theorem
PROOF

L in BPP:
2
x  L  Prr [ M ( x, r )  1] 
3
1
x  L  Prr [ M ( x,r)  1] 
3

By amplifying method and Chernoff Bound
1
x  L  Prr{0,1}m [ M ( x , r )  1]  1 
m
1
x  L  Prr{0,1}m [ M ( x , r )  1] 
m
Sipser-Lautemann Theorem
PROOF
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Wx={ y | M(x, y)=1}
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x in L, |Wx|>2m (1-1/m),
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Wx is very fat;
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x not in L, |Wx|<2m 1/m
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Wx is very thin;
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{0,1}m is the whole space;
Sipser-Lautemann Theorem
PROOF
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Shifting:
y {0,1}m ,
Wx  y  {z | t  Wx, t  y  z}
If Wx is fat, |Wx|>2m (1-1/m),
 There exists a set of strings y1, y2, … yr, r=m/2
s.t.
 yi Wx  yi  {0, 1}m

If Wx is thin, |Wx|<2m 1/m
 There is no such set of strings

2m
1
 r  2m
m
Sipser-Lautemann Theorem
PROOF
X in L, Wx is fat,
there exists a set of string y1, y2, … yr
 yi Wx  yi  {0, 1}
m
Then for all z in {0, 1}m ,
z  yi Wx  y i
That is, there exists i, s.t.
z  yi  Wx
Sipser-Lautemann Theorem
PROOF
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x in L, Wx is fat,
 There exists y1, y2, … yr,
 For all z in {0,1}m, M(x, z  yi)=1, for some i;
 x in L,
 y1 y 2 ... y r  z
M(x, y1  z)  1 or M(x, y 2  z)  1 or ... or M(x, y r  z)  1
Sipser-Lautemann Theorem
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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.