Polynomial-Time

Hierarchy

And Random Computations

Goal:

• Introduce the Polynomial-time Hierarchy (PH)

• Introduce BPP• Show the relationship between the

two

Plan:

• PH as an extension of SAT• Define probabilistic TMs and BPP• BPP2

Alternating QuantifiersSAT: NP-Complete

x1 …xn (x1 …xn )

TQBF: PSPACE-Complete

x1x2…xn (x1 …xn )

• TQBF with first quantifier and i-1 alternation is i-complete: x1x2… xn (x1,…xn)

• close under Karp-reduction• i = co-i

• PH = i i

Definition (PH):

Type of formula

complete for i?

# of alternation independent of

size

Polynomial Time Hierarchy:

• ‘’ on all SAT variables 1=NP

• Complement of NP1=coNP

• ii+1 and ii+1Hierarch

y:

• PHPSPACELimit

• NP=coNP PH=NP• By induction on i, i=NPCollapse

4

PSPACE

2

2

1=NP

1 =co

NP

P

• Prr[M(x,r)]

Accept Prob.

• for given x, the probability M (over a random r) accepts

which is:

• Special random tape

Probabilistic TM

Probabilistic TMs

5

a a b a b - -

_ _ _ _ _ - -

q8

• Prr[M(x,r)]

Accept Prob.

• probabilistic poly-time TM M, x, x L Prr[M(x,r) ]> 0

LNP if:

• probabilistic poly-time TM M, x, x L Prr[M(x,r) ] > ½

LPP if:

• probabilistic poly-time TM M, x Prr[M(x,r) = ‘x L’] > 2/3

LBPP if:

BPP

6

PPNP :-!

For any input x

7

Random DivideN

oteTMs that are

right on most

x’s (e.g for

PRIMES: always

say ‘NO’)

are something

completely

different (average case

complexity)

All random strings

Strings M errs

on M(x,r)

L(x)]

• LBPP probabilistic poly-time TM M’ and a polynomial p(n) s.t. x{0,1}n Prr{0,1}

p(n)[M’(x,r)L(x)] < 1/(3p(n))

Claim:

• M’ return the majority ofm2 independent runs of ML;m = #random bits M uses -Apply Chernoff bound

Proof:

8

Amplification

A function of the

number of random

bits p(n)=m3

With proper use of

Chernoff, one can

get stronger amplification – this

suffices here

• BPPNP

Maybe

• BPP2

Theorem [Sipser,Lautemann]:

• InsightLBPP poly-time probabilistic TM M (uses m=p(n) random bits), s.t. n and x{0,1}n:xLs1,…,sm{0,1}m r{0,1}m 1imM(x,rsi)

Proof:

9

BPP in PH

Why does this suffice?

Not known!

10

Yes-instance

{0, 1}m

11

No-instance

{0, 1}m

• M uses m random bits and errs w.p. <1/3m

Assume

• The Probabilistic Method:Pra[a has property P] > 0 a with property P

Utilize

• Probability that s1,…,sm{0,1}m dissatisfyr{0,1}m 1imM(x,rsi) small

Completeness:

• Union Bound

Soundness:

12

Proof

mR

m

ir {0,1}i 1

Pr M x,r s 1

13

For xL

union-bound m

m

ir {0,1}i 1

Pr M x,r s 1

1

m 13m

xL

m1 m R

mm

is ,...,s {0,1}i 1

Pr r 0,1 , M x,r s 0

14

Probability Random si’s is Bad

union-bound m

1 m Rm

m

is ,...,s {0,1}i 1r {0,1}

Pr M x,r s 0

m1 m Rm

m

is ,...,s {0,1}i 1r {0,1}

Pr M(x,r s) 0

si’s independent

mR

mm

s {0,1}i 1

2 Pr M x,s 0

r:s random rs random

mm 1

2 13m

xL

• LBPP there’s a poly. prob. TM M, s.t for any x there is m=poly(|x|) s.txL s1,…,sm

r 1imM(x,rsi)=1

It follows that:

• L2 BPP2

Hence

15

Q.E.D!

• the polynomial-time hierarchy • Saw NP PH PSPACE• NP=coNP PH=NP (“the hierarchy

collapses”)PH

• probabilistic TMs• Defined the complexity class BPP• How to amplify randomized

computations• We proved P BPP 2

BPP

16

WWindex

17

Polynomial Time Hierarchy

BPP

TQBF SAT

Probablistic Turing Machine