LECTURE 21

Probabilistic TM

Random-bit tape

Random-bit Generator Ø

Finite control

Work tape

Random-bit tape

Random-bit Generator Ø

Finite control

10 B B

right. tomove always and

only read is bit tape-random on the heads The 1.or 0

bit a generatesgenerator random themove,each In

Random-bit tape

Random-bit Generator Ø

Finite control

Work tape

),,,(ion Configurat tsq

q state

tsB B

BB

t of symbol1st at the head

of symbollast theheads

Halt and Accept

state. final a

in isit ifion configurat an ision configuratA

it. from done becan

move no ifion configurat a ision configuratA

accept.halt PTM,In

accepting

halting

accepts} ),( {0,1}*, | 2{)(

halts} ),( {0,1}*, | 2{)(||

||

xMxaccept

xMxhalt

M

M

PROBABILITY

Time Complexity

}.|| |)({ max)(

moves}. ||exactly in halts ),( | {

where

otherwise } |{| max

1)( if )(

nxxtiment

xMA

A

xhaltxtime

MM

x

x

MM

}.|| |)({expe max)(ˆ

otherwise 2||

1)( if )(exp ||

nxxttiment

xhaltxecttime

MM

xA

M

M

x

.moves] ||in haltingnot ),(Pr[ So,

)()(ˆmoves] ||in haltingnot ),(Pr[)/)((

.input on get might 'y that probabiliterror

extra thebounds which ,most at is moves ||in haltingnot ),(

ofy probabilit Then the ./)(|| with stringbinary a be Let .moves

/)(most at for input each on simulates that PTM thebe 'Let

.)()( , allfor such that

/)()( with ' PTM equivalentan exists Then there ).()(ˆ

and )( allfor 2/1)( with PTM a is Assume

'

'

xM

ntxtxMnt

xM

xM

nt

ntxMM

xerrxerrx

ntxtMntnt

MLxxacceptM

M

M

MM

M

LEMMA

PROOF

Accepting Input

. )]()(Pr[)(

}.2/1]1)(Pr[|{)(

.2/1]1)(Pr[ if accepted is input An

).()()(]0)(Pr[

and )(]1)(Pr[

Denote

)( xxMxerr

xMxML

xMx

xacceptxhaltxrejectxM

xacceptxM

MLM

MMM

M

Complexity Class

)}()( complexity

with timePTMby acceptedeach | languages{))((

ntnt

ntRTIME

M

)(0k

k nRTIMEPP

)}( allfor 0)(th wi

PTM time-polynomial aby acceptedeach | languages{

MLxxerr

MRP

M

}allfor 1/2)(

with PTM timepolynomial aby acceptedeach | languages{

xxerr

BPP

M

RPcoRPZPP

Characterization of ZPP

.polynomial somefor )()(ˆ and input allfor 0)(with

computes such that PTM a exists thereset A

npntxxerr

AMMZPPA

PROOF (<=)

. Thus,

).(for 0)( (b)

,4/1)( (a)

Then . rejects ' otherwise accepts; accepts ' then moves,

)(4after halts ofn computatio a If moves. )(4most at for input

each on simulates that ' PTM a define We).( and ,)(

timeexpected has andy probabiliterror zero has that PTM a be Let

'

'

RPA

MLxxerr

xerr

xMMM

npMnpx

MMMLAnp

M

M

M

usly.simultaneo ' and simulate to'' PTM aConstruct

.1]1)('Pr[ and 2/1]0)(Pr[

2/1]1)('Pr[ and 1]1)(Pr[

such that ' and PTMs exist two there

MMM

xMxMAx

xMxMAx

MMRPcoRPA

(=>)

Lecture 22

Power of Randomness

Does the randomness increases the power of computation?

PSPACEPPBPPRPZPPP

?ZPPP

?RPP

?BPPP

?PPP

ARE ALL OPEN!!!

polyPBPP /THEOREM

accepts. |))(|,(

Then ).( called and such fixed a Choose

.||

with allfor )(),(for which ),(||}*,1,0{ string

1)21(2222

least At .length of strings 2 are There

.2|)}(),( ),(|| |{|

is, that ,2)( ,|| with each for

such that PTM time-polynomial a exists There .Consider

)(2)()(

2)(

2

xhxMAx

nh

nx

xxxMnp

n

xxMnp

xerrnxx

MBPPA

A

nnpnnpnnp

n

nnpA

nM

PROOF

pPHpolyPNPBPPNP 2/ THEOREM

.RPNPBPPNP THEOREM

PROOF

.2by boumdedy probabiliterror with accepting

PTM time-polynomial a exists there, Since . aslength

same thehas | formula induced , ,..., luesboolean va and

formulaeach for that , techniquepadding simpleby assume, We

. |or | that Note

. show tosufficesIt

||

,...,1

10

11

F

bxbxi

xx

Sat

MBPPSatF

Fbb

F

SatFSatFSatF

RPSatBPPSat

ii

.1||for

2/1)21(y probabilit with accepts algorithm the,For

error. no ,For

.reject else accept then 1| if

halt; and reject else

1 then 1)|( if else

0 then 1)|( if

do to0for

;reject then 0)( if

.Input

:follows as assignment truth a producecan weNow,

||

,...,

1,,...,

0,,...,

11

1111

1111

nF

FSatF

SatF

FFF

F

aFM

aFM

ni

FFM

F

nF

axax

ixaxax

ixaxax

nn

iii

iii

ppBPP 22 THEOREM

Relationship

ZPPP

RP

RPco -

BPP

NP

NPco -

pp22

PP

PSPACE

Puzzle 1

on?distributicertain under

timeaverage polynomialin solved be problem hard-NP aCan

Puzzle 2

happen? d what woul, tobelongs problem complete-NP a If ZPP

Thanks, End