CS计算机代考程序代写 algorithm Slide 1

Slide 1

Part III Theory of Parallel

Computation

Uniform Computation via

Nonuniform Computation

LECTURE 3-1

Polynomial-size Circuits

edges.-out its ofnumber theisfanout theand inputs)(or

edges-in its ofnumber theis its gate, logicaleach For

.log called are operationsBoolean by labeled nodes The

AND).(or n conjunctio and

OR),(or n disjunctio negation, assuch ,operationsBoolean by labeled are

nodesother and 1 and 0 or values esby variabl labeled are edge-in

no with nodes osedigraph wh acyclican is A

fanin

ical gates

rcuitBoolean ci

snsd0
便利貼
what is circuits? a circuit is a logic gate. Electronic circuit is a real magine. circuit can compute a boolean function.

. variables theaboveonly appear gatesnegation thesuch that

gates ofnumber most twiceat anddepth same with theone equivalent

an toed transformbecan circuit Boolean every law,Morgan de Using

Size = # of gates

THEOREM

circuits. size-polynomial has in set Every P

).()( such that

polynomial a exists thereif has languageA

).2 when )()(( )()(

is of The

.1)( where

functionsBoolean of sequence aby determined becan *}1,0{

).()( . computing and most at fanin of gatesth circuit wi of size

minimum theis of )( the, formulaBoolean aFor

,,,

,

,

2

npnCS

puts-size circpolynomialA

t==CnCS=CnCS

Aityze complexcircuit-si

Axx

A

fCfCft

ffComplexity circuit cf

A

AnAAntAt

An

An

t

=

=



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Simulate Transition function by circuits

PROOF

SIMULATE DTM COMPUTATION BY A CIRCUIT

P/poly

.by denoted is circuits size polynomial having languages of class The P/poly

. Thus, .)()2)((most at size hascircuit resultant The

2.fanin of gates 2)(most at by simulated becan gateeach So,

network. in the nodes 22)(most at are therebecause 12)(

most at isit ,restrictednot is gateeach offanin theAlthough .

function Boolean computing )(most at size ofcircuit aconsider

y,sufficienc see To obvious. isnecessity the,)()( Since

.)()(such that polynomial a exists there

,

P/polyAnpn+np

n+np

n ++npn ++np

χ

np

nCSnCS

npnCSpP/polyA

n,A

AA

,n



PROPOSITION

PROOF

Sparse and poly

sparse. is Then

}.,…,{for ,…, = (n) define , language aFor

.every for |)(||)(||such that polynomial a exists there, which,of

each for strings, tonumbers natural from functions ofset thedenote Let

.)(

such that polynomial a exists thereif is languageA

2121

Spolyh

wwwSwwwhS

xxp|xhph

poly

np||A

psparseA

s

knks

n



=

=

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Characterization of P/poly

.)(, ,|| with for

such that language a and function a is There (c)

.such that set sparse a exists There (b)

circuit. size-polynomial has Language (a)

.equivalent are statements following The

BnhxAxnxx

PBpolyh

PAS

A

S





(b)(c)(a)(b) 

PROOF

P/poly and NP

?/ and between iprelationshany say Can we

|)(|,

such that and exist there/

],|))[(|||,(

such that polynomial a and exist there

polyPNP

BxhxAx

polyhPBpolyPA

ByxxpyyAx

pPBNPA









(b)=>(a)

circuits. size-polynomial has then

circuits, size-polynomial has and If

A

BPA
B

LEMMA 1

circuits. size-polynomial hasset sparseEvery

LEMMA 2

(a)=>(c)

.}|)(|, {

and Clearly, . accepts )( decoding from obtainedcircuit the

and )(that condition thesatisfying s’, all ofset thebe Let

.function aobtain we

,)( string a into computingcircuit size-polynomial theEncoding
,

Bxhx|xA=

PBx|x|h

|x|y=hyxB

polyh

nh
An

(c)=>(b)

time.polynomialin accepts oracle with algorithm This

reject; else

accept then if

for-end

;0 then 10 if

;1 then 01 if

begin do )( to1for

;

;input

algorithm. following heConsider t .)()(such that polynomial a be Let

1. is )( of symbolth the01

0. is )( of symbolth the10

:set sparse a Define

AS

Bx#w

wS

wS

|x|pi

w

x

np|n|hp

nhiS

nhiS

S

in

in

in

in









P/poly and NP

?/ and between iprelationshany say Can we

|)(|,

such that and exist there/

],|))[(|||,(

such that polynomial a and exist there

polyPNP

BxhxAx

polyhPBpolyPA

ByxxpyyAx

pPBNPA









.,,,such that and different

havemay we,|||| and with ,For

],|))[(|||,(

such that polynomial a and exist there

221121

212121

ByxByxyy

nxxxxAxx

ByxxpyyAx

pPBNPA



==





/ Possibly, polyPNP 

BxhxAx

polyhPBpolyPA





|)(|,

such that and exist there/

.,,,such that same thehave

tohave we,|||| and with ,For

21

212121

ByxByxy

nxxxxAxx



==

2

2

1

1

|)(|,

such that and exist there/

],|))[(|||,(

such that polynomial a and exist there

BxhxAx

polyhPBpolyPA

ByxxpyyAx

pPBNPA









NPpolyP  /

. of role play thecannot Thus,

].,|))[(|||,exist may re the

].,|))[(||||),(|(

].,|))[(||||),(|(

./ Suppose

12

2

2

2

BB

Byxxpyy

ByxxpyxhyAx

ByxxpyxhyAx

polyPA



=

=

2

2

1

1

|)(|,

such that and exist there/

],|))[(|||,(

such that polynomial a and exist there

BxhxAx

polyhPBpolyPA

ByxxpyyAx

pPBNPA









NPpolyP  /

.sets recursive-non contains / Therefore,

.computable benot may Moreover,

polyP

h

./

.

.

2

2

polyPPHP/polyNP

P/poly

PHP/polyNP

pp

k

p

k

p



=

=

THEOREM

LEMMA 1

LEMMA 2

.,|’ and ,|’, then ,)( if)(

;,|’or ,|’, then ,)( if (b)

;,1 and ,0 (a)

,most at length of QBF-any For

:conditions following thesatisfies Then .)(Set

B].)(,SAT )[|| ,(

such that and exists thereThus,

./SAT So, .in -complete is which ,SATConsider

10

10

k

BwFBwFBwFF’xF= c

BwFBwFBwFF’xF=

BwBw

nF

wnw=h

nhFFnFF

polyh PB

polyP

xx

xx

k

k

k

p

k

p

m













==

==

PROOF OF LEMMA 1

. means This

)].,()()())[(||,))((||,(

Thus,

(c).apply can then we,’)( (b)apply can then we,’)( if

, variables with each For . variables1most at with

allfor trueisit that Assume true.isit (a),by then constant, a is if fact,In

.,SAT tobelongs length of QBF- athen

,conditions above thesatisfies ify that inductivel showcan we,Conversely

2

p

k

k

kk

SAT

BwFcbanFFnpwwSATF

FxFFxF

nFnF

F

BwFnF

w





==



PROOF OF LEMMA 2

trouble.a bringst replacemennotation theoflegality theidea, for this

However, proof. finish the would weone, into oracles sparse twoencoding

By oracles. sparse only two containing machine aobtain to and

)( g witnessinmachines twocombineThen .by replace

tois idea obtained-yimmediatelAn . that advantage by taking

set sparse somefor prove want to weand , have we

Now, .set sparse afor that so,/ ,hypothesis

induction By the .for Assume .consider ,1For

. since trivialisit ,1For .on induction by prove We

1

1

1

S

B

SSB

Sp

k

p

k

Bp

k

pp

k

PB

P/polyAP/polyNP

P/polyNP

S’PAPBNPA

SPBpolyP

BNPAAk>

NPk=kP/poly







=

. ))((,)(,)(,

and ,|| with any for )(|)(,| where

, )(,

.))((,)(,|)(|,

,|| with for Hence, .|)(|, ,|| with for

such that and exist there,/ since Moreover,

fact. apply thecan weThus,

.imply and that Note true.be wouldoracle

sparse afor t replacemen asuch shown that becan it fact, By this

.)(, ,|| with for

such that and exists thereset sparse afor

:fact following theusingby trouble thisaround go wefollowing, In the

npgnhxnx

nxxnpnhx

Rnx

RnpgnhxBxhxAx

nxxRxgxBxnxx

PRpolygpolyPNPB

NPAPBNPA

BnhxAxnxx

polyhNPBSNPA

SSB

S

=

















. togoes as zero togoes )8(2most at size

ofcircuit a hasfunction Boolean a ofy probabilit the, variables of functions

Boolean 2 are thereSince .2 << )O(2= ))22(2( ,)8(2For .))22(2(by bounded is most at size and variables of circuits ofnumber theSo, . size and variables of circuits ))22(2(most at have weTherefore, choices. 22most at hasit Thus, choices). (2constant aor choices), 2( literal a choices), most (at gate previous a becan nodes previousEach nodes. previous on two actshat operator t ORor ANDan with assigned is gateeach that Note . size and variables with circuits ofnumber count thefirst We ).2( size of circuits requires variables offunction Boolean every Almost nnn 224/22 2 2   + nn/ n n+s+sn/s= n+s+s sns nn+s+ n+s+n s sn /n n n nsn s s n THEOREM PROOF Open ).log()(such that language a Find nnnCSA A =