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 )

Definition (PH):

• 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

Type of formula complete for i?

# of alternation independent of size

Polynomial Time Hierarchy:1=NP •‘’ on all SAT variables

1=coNP •Complement of NP

Hierarchy: • ii+1 and ii+1

Limit •PHPSPACE

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

4

PSPACE

1=NP

1=coNP

P

Accept Prob.

• Prr[M(x,r)]

LBPP if:• probabilistic poly-time TM M,

x Prr[M(x,r) = `x L’] > 2/3

Probabilistic TM

• Special random tape

Probabilistic TMs

5

a a

b a

b -

_ _

_ _

_ -

q8

For any input x

6

Random Divide

NoteTMs that are right on most x’s (e.g for PRIMES: always say ‘NO’) are something else (average case complexity)

All random strings

Strings M errs on

Claim:

• 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))

Proof:

• M’ return the majority ofm2 independent runs of ML;m = #random bits M uses -Apply Chernoff bound 7

Amplification

A function of the number of random bitsOne can get stronger amplification – this suffices here

Maybe

• BPPNP

Theorem [Sipser,Lautemann]:

• BPP2

Proof:

• InsightLBPP poly-time probabilistic TM M, s.t. n and x{0,1}n: xLs1,…,sm{0,1}mr{0,1}m1imM(x,rsi)(where m=p(n))

8

BPP in PH

Does this suffice?

Not known!

9

Yes-instance

{0, 1}m

10

No-instance

{0, 1}m

Assume

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

Utilize

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

Completeness:

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

Soundness:

• Union Bound11

Proof

m1 m R

mm

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

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

12

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

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

It follows that:

• 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

Hence

• L2 BPP2

14

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

15

WWindex

16

Polynomial Time Hierarchy

BPP TQBF

SAT Probablistic Turing Machine