LECTURE 12-13

Complete Problems in PH

.for -complete is Clique-Exact 2pp

T

Clique.-ExactClique Therefore,

][ ))((

.Clique-Exact

Clique is Clique-Exact

2

2

pT

pm

pT

pT

p

pNP

pT

BA

BANPBA

P

THEOREM

PROOF

moves? within accept can ,0 string a

and input an , NTMan of code Given the

tyM

yMxt

BOUNDED HALTING PROBLEM (BHP)

PROOF

THEOREM

complete.-NP is BHP

NP.)(BHPThen moves. within input on

simulate to NTM universalan Construct

u

u

MLty

MM

.0,,construct , of input For

. bound timepolynomial

with NTMan for )(NP

|)(|ypyMMy

p

MMLAA

BHPA

moves. within accept can

,0 string a and input an , oracle with NTMan Given t

tyM

yAMA

LEMMA

.NPfor -complete is BHP Apm

A

.NP)(BHPThen moves. in input on

simulate to oracle with NTM universal aConstruct A

uA

u

MLtyM

AM

.0,,construct , of input For

. bound timepolynomial

and oracle with NTMan for )(NP

|)(|yp

A

yMMy

p

AMMLBB

PROOF

LEMMA

.NP then ,for -complete is If 1pk

Apk

pmA

PROOF

.NPP

.NP,NP

]NP)[)((

.NP have we, Since

1

1

AApm

Apm

C

Cpk

pk

pk

Apk

CCAC

BACB

BCB

A

THEOREM

DEFINITION

.BHP, 10kA

kAA

.for -complete is pk

pmkA

PROOF by induction on k

.NPfor -complete is

lemma,by Then .for -complete is Suppose

.NPfor -complete is BHPBHP

11

11

pk

Apmk

pk

pmk

ppm

kA

A

A

Natural Problem

Machine Related Problem

(Machine unrelated)

problems. natural containsit ift significan is class complexityA

SAT k

.even for and , oddfor denotes where

1|})1,0{,(

})1,0{,})(1,0{,(

: trueis following he whether terminedet

,over formulaBoolean a

and ,,...,, variablesof sets Given

,...,,

2211

21

21

21

kkQ

FXQ

XX

XXXXF

XXXk

k

kkk

k

k

k

THEOREM

.for -complete is SAT pk

pmk

PROOF (odd k)

Byyxnqyy

nqyynqyyAx

nxx

qBA

kkk

pk

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

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

,|| with *any for such that

polynomial and Pexist there

1

2211

1|})1,0{,(

})1,0{,})(1,0{,(

such that ,..., and construct we*,each for

,SAT show To

,...,,

2211

1

21

kFX

XXAx

XXFx

A

kk

k

kpm

; (c)

state; final thecontains (b)

ion;configurat initial theis (a)

1

0

ii

m

such that

|),,,(|||

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

, ofpath n computatio a is there))[(||,(

))(||,))((||,( iff

. bound time-polynomial has where)(Let

1

110

2211

ki

km

kk

yyxp

yyxpm

Mnqyy

nqyynqyyAx

pMMLB

such that ,,, formulasBoolean Construct

otherwise 0

symbol holds ),( cell if 1

).()1)((),,,(Then

.###,,,Let

4321

,,

1

11

ffff

ajiz

nrnqknyyxp

yyxyyx

aji

k

kk

; i.e., ,holds (c)1

state; final thecontains i.e., holds, (b)1

ion;configurat initial theis i.e., holds, (a)1

symbol. oneexactly holds cellEach 1

14

3

02

1

ii

m

f

f

f

f

BByyxxsx kn ###]#[ 1210

Bnr

nqnynqnynnxnxsx

z

zzzzzzzfnqn

),(,1

#),(2,1),(1,1,2,1#,1,1,1,12#,112

)(,11121

).)((

witheach replacingby 'Obtain

for variablegintroducinby

aconstruct want to we,each From

0,,1,1,,1,

,,12

,,

)(,11,1

,

jhijhi

yj

hihi

inqi

zxzx

zf

yx

Xyyy

hi

1|})1,0{,(

})1,0{,})(1,0{,(

Then .'Set

,...,,

2211

4321

21

kFX

XXAx

ffffF

kk

3-CNF SAT k

.even for and , oddfor denotes where

1|})1,0{,(

})1,0{,})(1,0{,(

: trueis following he whether terminedet

,over CNF-3 a

and ,,...,, variablesof sets Given

,...,,11

111

21

21

21

kkQ

FX

XQXQ

XXXXF

XXXk

k

kkkkkk

k

k

k

THEOREM

.even is when for -complete is SAT 3CNF

.odd is when for -complete is SAT 3CNF

k

kpk

pmk

pk

pmk

Generalized Ramsey Number

. size of

clique blue aor size of clique red aeither exists there

, from extended coloring-2any for that it true is ,0

integeran given and blue,*})red,{:( colored-2

partially edges with ),(graph complete aGiven

K

K

cK

Ec

EVG

THEOREM

.for -complete is GRN 2pp

m PROOF

GRN SAT CNF-3 2pm

KcGYXF ,,),( formulaBoolean CNF-3

Such that

Construction

]color same in the size of clique a has )[','(

1]|})[1,0{,})(1,0{,( ,

KGccc

FYX

2ix1ix

2ix1ix 1'ix

2'ix1'ix

2'ix

Variables ix

2ix1ix

2ix1ix 1'ix

2'ix1'ix

2'ix

Variables

Clauses 1kC jk yC :1'

jk yC :3

2kC

2ix1ix

2ix1ix 1'ix

2'ix1'ix

2'ix

Variables

Clauses 1kC ik xC :1'

ik xC :3

2kC

2ix1ix

2ix1ix 1'ix

2'ix1'ix

2'ix

Variables

Clausesik xC :1 '1' : ik xC

3kC2kC

2ix1ix

2ix1ix 1'ix

2'ix1'ix

2'ix

Variables

Clausesii xC :1 '1' : ii xC

3iC2iC

R

4|| KR

12x11x

12x11x 21x

22x21x

22x

Variables

Clauses

111 : xC121 : xC

213 : xC112 : yC

R

))(( 211211 yyxxyx

223 : yC122 : yC

12x11x

12x11x 21x

22x21x

22x

Variables

Clauses

111 : xC121 : xC

213 : xC112 : yC

))(( 211211 yyxxyx

223 : yC122 : yC

12x11x

12x11x 21x

22x21x

22x

Variables

Clauses

111 : xC121 : xC

213 : xC112 : yC

R

))(( 211211 yyxxyx

223 : yC122 : yC

12x11x

12x11x 21x

22x21x

22x

111 : xC121 : xC

213 : xC112 : yC

R

223 : yC122 : yC

color. samein , size of clique a , ofextention as ' coloring

]1|))(})[(1,0{:})(1,0{:( ,211211

Kcc

yyxxyxYX

12x11x

12x11x 21x

22x21x

22x

111 : xC121 : xC

213 : xC112 : yC

R

. somefor red),('),(' 2121 ixxcxxc iiii

223 : yC122 : yC

Case 1

clique. red required a is ThereAnswer:

4|| KR

12x11x

12x11x 21x

22x21x

22x

111 : xC121 : xC

213 : xC112 : yC

R

occur.not does 1 Case

223 : yC122 : yC

Case 2

clique. blue required a is ThereAnswer:

4|| KR

12x11x

12x11x 21x

22x21x

22x

111 : xC121 : xC

213 : xC112 : yC

occur.not does 1 Case

223 : yC122 : yC

Case 2

clique. blue required a is ThereAnswer:

1

1

21

21

yy

xx

clauses of #

of #

2

k

xn

knK

i

12x11x

12x11x 21x

22x21x

22x

111 : xC121 : xC

213 : xC112 : yC

R

223 : yC122 : yC

color. samein , size of clique a , ofextention as ' coloring

]1|))(})[(1,0{:})(1,0{:( ,211211

Kcc

yyxxyxYX

color. samein , size of clique a , ofextention as ' coloring

]1|))(})[(1,0{:})(1,0{:( ,211211

Kcc

yyxxyxYX

]1|))(})[(1,0{:( ,211211 yyxxyxY

.),(',),(' then ,1)( If

.),(',),(' then ,1)( If

2121

2121

bluexxcredxxcx

redxxcbluexxcx

iiiii

iiiii

Consider

Define a coloring c’:

No red clique of size K exists.

}1 ,0{: XTo prove

12x11x

12x11x 21x

22x21x

22x

111 : xC121 : xC

213 : xC112 : yC

R

223 : yC122 : yC

121 xx

color. samein , size of clique a , ofextention as ' coloring

]1|))(})[(1,0{:})(1,0{:( ,211211

Kcc

yyxxyxYX

)])()[(( 211211 yyxxyxY

.),(',),(' then ,1)( If

.),(',),(' then ,1)( If

2121

2121

bluexxcredxxcx

redxxcbluexxcx

iiiii

iiiii

Consider

Define a coloring c’:

No red clique of size K exists.

}1,0{: XTo prove

There must exist a blue clique Q of size K.

color. samein , size of clique a , ofextention as ' coloring

]1|))(})[(1,0{:})(1,0{:( ,211211

Kcc

yyxxyxYX

12x11x

12x11x 21x

22x21x

22x

111 : xC121 : xC

213 : xC112 : yC

R

223 : yC122 : yC

121 xx

color. samein , size of clique a , ofextention as ' coloring

]1|))(})[(1,0{:})(1,0{:( ,211211

Kcc

yyxxyxYX

otherwise 0

, size of clique blue in the appearsit if 1 Kyi

PPROPERTY OF BLUE CLIQUE Q

.in appears and of onemost At .3

. clauseeach for oneexactly contains .2

.0)( if },{

1)( if },{ contains .1

21

21

Qyy

CCQ

xxx

xxxQ

ii

iij

iii

iii

Assignment for y

Then F(X,Y) is satisfied

1. equals at literal then , If ijij CQC

12x11x

12x11x 21x

22x21x

22x

111 : xC121 : xC

213 : xC112 : yC

R

223 : yC122 : yC

121 xx

color. samein , size of clique a , ofextention as ' coloring

]1|))(})[(1,0{:})(1,0{:( ,211211

Kcc

yyxxyxYX