PH P BPP Basic Definitions Language L belongs to class PP if - - PowerPoint PPT Presentation

Toda Theorem Part1

BPP PH ⊆

P ⊕

Basic Definitions

• Language L belongs to class PP if exists

polynomial NTM such that

5 . } 1 ) ( { > = ⇔ ∈ x M P L x

• Language L belongs to class BPP if

exists probabilistic polynomial algorithm such that x from L is accepted with probability more than 0.75 and x not form L is declined with the same probability.

• Language L belongs to class

P ⊕ (parity P) if exists polynomial-time non-deterministic Turing machine, such x lies in L is equivalent to that fact, that number of accepting computations is odd(or, exists polynomially checked relation R such that the number of y(R(x, y)= 1 ) is odd.

Class

P ⊕

• It can be quickly checked that

is complete for

SAT ⊕ P ⊕

Polynomial hierarchy

Ρ Σ +

= Σ = Σ

i

NP P P P

i 1

P PH

i i

Σ =

≥

U

• Polynomial Hierarchy

General Idea of proof

P

BPP NP

⊕

⊆

P

RP NP

⊕

⊆

A P A

BPP NP

⊕

⊆

P

BPP PH

⊕

⊆

N i ∈ ∀

P i

BPP P

⊕

⊆ Σ

• because because of

Valiant-Vazirani Lemma

• because Valiant-Vazirani Lemma can be

relativised

• We will prove, that by the

mathematical induction. It will prove, that

• Base i =1 is obvious

Idea of Proof

• Lemma 2:
• Lemma 3:
• Lemma 4:

So, considering this Lemmas being truth we’ll receive

A P A BPP

BPP P

⊕

⊆ ⊕

P P

P

⊕ ⊆ ⊕

⊕

A A BPP

BPP BPP ⊆

P P BPP P P BPP P BPP P P BPP P k

BPP BPP BPP BPP NP NP P

k

⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ Σ +

⊆ ⊆ ⊆ ⊆ ⊆ = Σ

1

and the first part of Toda’s Theorem will be proved.

Lemma 4

• In definition of BPP we can use, instead of (1-3/4),

where is a polynomially function of input length

• M uses L as oracle, M is wrong with pr=
• We can consider, that the lengths of all branches of NTM L

are equal and the number of its accesses to L is equal to l(n) in all branches

• We can replace every access to oracle L by the branch of

machine N(from the definition of BPP) with the probability

• f mistake
• The number of branches, which status will change is equal

to that can be made less than 0.25

A

BPP L ∈

i

p −

2

i

p

) (

2

n e −

) (

2

n e −

) ( ) ( ) (

) 2 1 ( 1 2

n l n i n e − −

− − +

) (

2

n i −

Lemma 2

• so exists NTM with oracle accepting

every for odd number of y. R is corresponding relationship

• So
• And we need:

A BPP

P L ⊕ ∈

A A

BPP B ∈

n

x } 1 , { ∈

) (

} 1 , {

n

y

γ

∈

A A BPP A BPP A B

BPP BPP P P R ∈ ∈ ∈ ∈

} } 2 ) 2 1 ( )} ( ) , , ( : { :# { |# {

) ( ) (

• dd

is L z y x z y x L

n n A ς π −

− ≥ Π ∈ = } 2 * 75 . } )} ( ) , , ( : { :# { |# {

) (n A

• dd

is L z y x y z x L

ς

≥ Π ∈ =

Lemma 2- end

• We can consider a table, (for fixed x, WLOG x is in

L), in (y, z) we’ll put result of П for given (y, z).The number of rows with more than

• nes is odd and the number of columns, which have

1 in intersection with this columns is more than

• In other rows is not many 1,so the number of ‘good’

rows is for

) ( ) ( ) ( ) (

2 * 2 * 2 2

n n n n γ ς π ς −

−

} 2 ) 2 1 (

) ( ) ( n n ς π −

−

) ( ) ( ) ( ) ( ) ( ) ( ) ( ) (

2 * 75 . 2 * 2 * 2 2 * 2 * 2 2

n n n n n n n n ς γ ς π γ ς π ς

≥ − −

− −

3 ) ( ) ( + ≥ n n γ π

Lemma 5

P co P ⊕ − = ⊕