lecture 21 probabilistic tm
DESCRIPTION
LECTURE 21 Probabilistic TM. Work tape. Random-bit tape. Random-bit Generator Ø. Finite control. 0. 1. B. B. Random-bit tape. Random-bit Generator Ø. Finite control. Work tape. Random-bit tape. Finite control. Random-bit Generator Ø. Halt and Accept. PROBABILITY. - PowerPoint PPT PresentationTRANSCRIPT
![Page 1: LECTURE 21 Probabilistic TM](https://reader036.vdocument.in/reader036/viewer/2022062517/56812dc1550346895d930218/html5/thumbnails/1.jpg)
LECTURE 21
Probabilistic TM
![Page 2: LECTURE 21 Probabilistic TM](https://reader036.vdocument.in/reader036/viewer/2022062517/56812dc1550346895d930218/html5/thumbnails/2.jpg)
Random-bit tape
Random-bit Generator Ø
Finite control
Work tape
![Page 3: LECTURE 21 Probabilistic TM](https://reader036.vdocument.in/reader036/viewer/2022062517/56812dc1550346895d930218/html5/thumbnails/3.jpg)
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
![Page 4: LECTURE 21 Probabilistic TM](https://reader036.vdocument.in/reader036/viewer/2022062517/56812dc1550346895d930218/html5/thumbnails/4.jpg)
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
![Page 5: LECTURE 21 Probabilistic TM](https://reader036.vdocument.in/reader036/viewer/2022062517/56812dc1550346895d930218/html5/thumbnails/5.jpg)
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
![Page 6: LECTURE 21 Probabilistic TM](https://reader036.vdocument.in/reader036/viewer/2022062517/56812dc1550346895d930218/html5/thumbnails/6.jpg)
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
![Page 7: LECTURE 21 Probabilistic TM](https://reader036.vdocument.in/reader036/viewer/2022062517/56812dc1550346895d930218/html5/thumbnails/7.jpg)
.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
![Page 8: LECTURE 21 Probabilistic TM](https://reader036.vdocument.in/reader036/viewer/2022062517/56812dc1550346895d930218/html5/thumbnails/8.jpg)
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
![Page 9: LECTURE 21 Probabilistic TM](https://reader036.vdocument.in/reader036/viewer/2022062517/56812dc1550346895d930218/html5/thumbnails/9.jpg)
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
![Page 10: LECTURE 21 Probabilistic TM](https://reader036.vdocument.in/reader036/viewer/2022062517/56812dc1550346895d930218/html5/thumbnails/10.jpg)
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
![Page 11: LECTURE 21 Probabilistic TM](https://reader036.vdocument.in/reader036/viewer/2022062517/56812dc1550346895d930218/html5/thumbnails/11.jpg)
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
(=>)
![Page 12: LECTURE 21 Probabilistic TM](https://reader036.vdocument.in/reader036/viewer/2022062517/56812dc1550346895d930218/html5/thumbnails/12.jpg)
Lecture 22
Power of Randomness
Does the randomness increases the power of computation?
![Page 13: LECTURE 21 Probabilistic TM](https://reader036.vdocument.in/reader036/viewer/2022062517/56812dc1550346895d930218/html5/thumbnails/13.jpg)
PSPACEPPBPPRPZPPP
?ZPPP
?RPP
?BPPP
?PPP
ARE ALL OPEN!!!
![Page 14: LECTURE 21 Probabilistic TM](https://reader036.vdocument.in/reader036/viewer/2022062517/56812dc1550346895d930218/html5/thumbnails/14.jpg)
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
![Page 15: LECTURE 21 Probabilistic TM](https://reader036.vdocument.in/reader036/viewer/2022062517/56812dc1550346895d930218/html5/thumbnails/15.jpg)
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
![Page 16: LECTURE 21 Probabilistic TM](https://reader036.vdocument.in/reader036/viewer/2022062517/56812dc1550346895d930218/html5/thumbnails/16.jpg)
.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
![Page 17: LECTURE 21 Probabilistic TM](https://reader036.vdocument.in/reader036/viewer/2022062517/56812dc1550346895d930218/html5/thumbnails/17.jpg)
ppBPP 22 THEOREM
![Page 18: LECTURE 21 Probabilistic TM](https://reader036.vdocument.in/reader036/viewer/2022062517/56812dc1550346895d930218/html5/thumbnails/18.jpg)
Relationship
ZPPP
RP
RPco -
BPP
NP
NPco -
pp22
PP
PSPACE
![Page 19: LECTURE 21 Probabilistic TM](https://reader036.vdocument.in/reader036/viewer/2022062517/56812dc1550346895d930218/html5/thumbnails/19.jpg)
Puzzle 1
on?distributicertain under
timeaverage polynomialin solved be problem hard-NP aCan
![Page 20: LECTURE 21 Probabilistic TM](https://reader036.vdocument.in/reader036/viewer/2022062517/56812dc1550346895d930218/html5/thumbnails/20.jpg)
Puzzle 2
happen? d what woul, tobelongs problem complete-NP a If ZPP
![Page 21: LECTURE 21 Probabilistic TM](https://reader036.vdocument.in/reader036/viewer/2022062517/56812dc1550346895d930218/html5/thumbnails/21.jpg)
Thanks, End