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
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A
xhaltxtime
MM
x
x
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otherwise 2||
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x
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)()(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
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and )( allfor 2/1)( with PTM a is Assume
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ntxtxMnt
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PROOF
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).()()(]0)(Pr[
and )(]1)(Pr[
Denote
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Complexity Class
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with timePTMby acceptedeach | languages{))((
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RPcoRPZPP
Characterization of ZPP
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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,
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each on simulates that ' PTM a define We).( and ,)(
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usly.simultaneo ' and simulate to'' PTM aConstruct
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such that ' and PTMs exist two there
MMM
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(=>)
Lecture 22
Power of Randomness
Does the randomness increases the power of computation?
PSPACEPPBPPRPZPPP
?ZPPP
?RPP
?BPPP
?PPP
ARE ALL OPEN!!!
polyPBPP /THEOREM
accepts. |))(|,(
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1)21(2222
least At .length of strings 2 are There
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such that PTM time-polynomial a exists There .Consider
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PROOF
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PROOF
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PTM time-polynomial a exists there, Since . aslength
same thehas | formula induced , ,..., luesboolean va and
formulaeach for that , techniquepadding simpleby assume, We
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error. no ,For
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halt; and reject else
1 then 1)|( if else
0 then 1)|( if
do to0for
;reject then 0)( if
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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