61 NP Complete

NP - Complete

2013-11-25 2

Mt s bi ton ti u ri rc

Bi ton ngi du lch:

Cho n im trn mt phng (thnh ph), gia hai thnh ph bt k c xc nh mt thng s l chi ph i li. Mt hnh trnh l mt cch i xut pht t mt thnh ph no , qua n thnh ph v quay v ni xut pht.

OP (Optimization Problem): Tm hnh trnh * c tng chi ph b nht.

DP (Decision Problem): C tn ti mt hnh trnh vi chi ph D?

2013-11-25 3


Bi ton t mu th:

Cho th G ={V,E}.

OP: S mu t nht t th G?DP: Cho s nguyn K. C tn ti hay khng cch t mu th

G vi s mu khng qu K?

Mt cch t mu th l mt phng n gn cho mi nh mt mu, sao cho hai nh lin k c hai mu khc nhau.

2013-11-25 4


Bi ton ci ti:

Cho n vt vi kch thc l cc s nguyn s1, s2, ..., sn v cc ti

vi kch thc l s nguyn T.

OP: Tm s ti t nht xp cc vt.

DP: Cho s nguyn K. C tn ti cch xp cc vt vo khng qu K ti vi sc cha T?

2013-11-25 5


Bi ton tp con:

Cho s nguyn dng T v tp X gm n s nguyn dng a1, a2, ..., an.

OP: Xc nh tp con ca X sao cho tng ca chng gn nht v khng qu T.

DP: C tn ti tp con sao cho tng kch thc ng bng T.

2013-11-25 6


Bi ton phn cng cng vic:Gi thit c n cng vic:

Mi thi im ch thc hin mt cng vic,

Thi gian thc hin t1, t2, ..., tn,

Thi hn hon thnh d1, d2, ..., dn (tnh t khi bt u cng vic u tin),

Mc pht i vi mi cng vic b chm l p1, p2, ..., pn.

Phn cng cng vic l mt hon v ca tp J={1, 2,..., n}: J(1), J(2), ..., J(n).

Tng gi tr pht ca phn cng :

OP: Tm lch sp xp cng vic c gi tr hm pht thp nht P() min.

DP: Cho trc k, xc nh lch c mc pht khng qu k: P() k.

n

jjjj pdttP

1)()()()1( 0elsethen...if

2013-11-25 7

Lp P

Thut ton c phc tp O(f(n)) nu vi mi b s liu c di n, s

php tnh phi thc hin khng qu C*f(n), vi C >0.

Thut ton c phc tp O(p(n)), vi p(n) l mt a thc, gi l c

phc tp a thc.

nh ngha:

P l lp cc bi ton c gii vi thi gian a thc.

Ch :

Khng phi mi bi ton thuc lp P c thut ton hiu qu.

Nu bi ton khng thuc lp P th u phi tr gi rt t v thi gian

hoc thm ch khng gii c n trong thc t.

2013-11-25 8

Lp NP (Nondeterministic Polynomial)

NP l lp cc bi ton quyt nh m vic kim tra li gii i

vi d liu vo c thc hin vi thi gian a thc.

2013-11-25 9

Lp NP (Nondeterministic Polynomial)

Thut ton bt nh (nondeterministic algorithm):

Pha bt nh:

Mt xu k t S bt k c sinh ra trong b nh, c th coi nh

li gii ngh.

Pha tin nh:

c d liu vo (S c th b b qua). Thut ton c th kt thc vi

khng nh Yes, No, hoc ri vo vng lp khng dng.

C th coi l pha kim tra li gii ngh S.

NP l lp bi ton gii c bng thut ton bt nh vi thi gian a thc (nondeterministic polynomial bouded).

Thut ton bt nh l a thc nu tn ti a thc p sao cho vi mi d liu vo c kch thc n v c tr li yes vi tnh ton ca thut ton l a thc.

2013-11-25 10

Lp NP

V d:

th c 5 inh, 8 cnh v k =4:

V ={1, 2, 3, 4, 5},

E={(1,2), (1,4), (2,4), (2,3), (3,5), (2,5), (3,4), (4,5)}

K hiu: R (Red), B (Blue), G (Green), O (Orange), Y (Yellow).

1

2

4

3

5

Dng 4 mu, cc cp nh k khc muYesRGRBY

Dng ti 5 muNoRBYGO

nh 5 khng c t muNoRGRB

nh 2 v 5 cng muNoRGRBG

ReasonOutputS

2013-11-25 11

Lp P vs. NP

Hin nhin: P NP

NP \ P = ?

(NP P?)

P

Sp xpCy khung b nhtNhn ma trnTm kim tun tng i ngn nht...

NP

Bi ton ci ti

Bi ton ba l

Bi ton ngi du lch

2013-11-25 12

NP-y (NP-complete)

Hin nhin: P NP.

Nhng: NP \ P = ?

Chng minh c:

Trong NP c nhng bi ton kh khng

km bt c bi ton no khc trong NP.

A NP: nu c mt thut ton a thc

no gii c A th vi mi bi ton B

NP u c thut ton a thc gii B.

Bi ton A c gi l NP-y .

Ni cch khc, A NP c gi l NP-

y nu A P th suy ra P=NP.

P


NP

Bi ton ci ti

Bi ton ba l

Bi ton ngi

du lch

2013-11-25 13

NP-y (NP-complete)

Hin nhin: P NP.

Nhng: NP \ P = ?

Chng minh c:

1. Nu NP P th c bi

ton thuc NP nhng

khng thuc P v cng

khng phi NP-y .

2. Nhiu bi ton l NP-y

.

P


NP-

y

NP

2013-11-25 14

NP-y (NP-complete)

Nh vy, ch ra mt bi ton no l NP-y cn ch ra rng:

Tn ti mt thut ton a thc bt nh gii n (tc l ch ra nthuc lp NP);

C mt bi ton NP-y dn v n.

P

Sp xpCy khung bnhtNhn ma trnTm kim tun tng i ngn nht...

NP-

y

NP

2013-11-25 15

NP-y (NP-complete)

Gi thit cn gii bi ton A v c thut ton gii bi ton B.

Gi s c nh x T chuyn mi d liu vo x ca bi ton A thnh d liu vo T(x) ca bi ton B

T: x T(x)sao cho:

Li gii y ca bi ton A vi d liu vo x

tng ng vi

Li gii z ca bi ton B vi d liu vo T(x)

(nu y l yes th z cng l yes).

Thut ton gii bi ton A = {nh x T + Thut ton gii B}

2013-11-25 16

Bi ton CNF-SAT

M t:

1) Bin logic nhn mt trong hai gi tr: true hoc false. K hiu a l bin logic, l ph nh ca a. Nu a nhn gi tr true th nhn gi tr false v ngc li, nu a nhn gi tr false th nhn gi tr true. Mt tn bin lmt bin logic hoc ph nh ca bin logic (cng l mt bin logic).

2) Mt mnh l mt dy cc tn bin c xen k bi php ton logic OR ().

3) Mt biu thc logic trong dng lin kt chun (conjunction nomal form -CNF) l mt dy cc mnh c kt ni bi php ton AND ().

Bi ton quyt nh:

Khi cc bin nhn gi tr true hoc false, biu thc c nhn gi tr true hay khng?

2013-11-25 17

Bi ton CNF-SAT

nh l Cook:

Bi ton CNF-SAT l NP-y

Cn ch ra rng bt k bi ton NP no (A) cng dn chuyn v CNF-SAT:

Cn xy dng nh x T dn chuyn input x ca bi ton A v mt biu thc logic.

Li gii ca bi ton quyt nh A trng vi gi trtrue hoc false ca CNF-SAT.

2013-11-25 18

Bi ton CNF-SAT

nh l Cook:

Bi ton CNF-SAT l NP-y

xd liu voca A

yeshoc no

T(x)T

CNF-SAT

Algorithm for A

2013-11-25 19

Bi ton CNF-SAT

V d: Xt bi ton 3-CG:

th G ={V,E} nh hnh vc t c bng 3 mu?

a

db

c

Chuyn dn bi ton 3-CG v bi ton CNF-SAT?

2013-11-25 20

Bi ton CNF-SAT

u1

u4u3

u2K hiu: xkj nh k t bi mu j

Xt 32 mnh :

C(k) = {xk1, xk2, xk3}, nh k c t bi t nht 1 mu

A(k) = {xk1, xk2,}, nh k khng cng c t mu 1 v 2

B(k) = {xk2, xk3}, nh k khng cng c t mu 2 v 3

C(k) = {xk1, xk3}, nh k khng cng c t mu 1 v 3

k = a, b, c, d

M = C(a) & ..&C(d)&A(a)&..&A(d)&B(a)&..&B(d)&C(a)&..&C(d)

M = true nu mi nh c t bi ng 1 trong ba mu {1, 2, 3}

2013-11-25 21

Bi ton CNF-SAT

u1

u4u3

u2

K hiu: e=(u,v) E, u,v V

D(e,j) ={ xuj, xvj } - nh u v v khng cng t bi mu j

N = D((u1,u2),1)&D((u1,u3),1)&D((u1,u4),1)& D((u2,u3),1)&D((u2,u4),1)&D((u3,u4),1)

&...&

D((u1,u2),3)&D((u1,u3),3)&D((u1,u4),3)& D((u2,u3),3)&D((u2,u4),3)&D((u3,u4),3)

N = true nu hai nh khng c t bi cng 1 trong ba mu {1, 2, 3}

G l th 3 mu chkhi M&N c gi trtrue.

2013-11-25 22

NP-y V d 2: Bi ton tng tp con chuyn dn c v bi ton lp lch

Cho s nguyn dng T v tp n s nguyn dng

X={ a1, a2, ..., an}

Tm tp con ca X sao cho tng ca chng gn nht vi T.

n cng vic: J1, J2, ..., Jn;

thi gian thc hin: t1, t2, ..., tn;

thi hn hon thnh (t khi bt u d n): d1, d2, ..., dnmc pht chm qu thi hn: p1, p2, ..., pn. Mi thi im ch thc hin mt cng vic.

Phn cng cng vic : {1, 2,..., n} {(1), (2),..., (n)}

Th t thc hin: J(1), J(2),...,J(n).

Tng gi tr pht ca phn cng trn l

n

jjjj pdttP

1)()()()1( 0elsethen...if

2013-11-25 23


Nu

i=1,..,nsi < T

bi ton tp con khng c li gii.

C th chuyn dn phng n thnh bt k d liu vo cho bi ton phn cng cng vic m khng c li gii.

V d:

ti= 2, di = pi = 1, vi i = 1, 2, ..., n.

and k=0.

(Bi ton quyt nh:

C tn ti mt tp con sao cho tng ng bng T?

C tn ti mt lch cng tc m tng gi tr pht P k ?.)

2013-11-25 24


Xt trng hp i=1,..,nsi T

t ti = si,

di = T,

pi = si, vi i = 1, ..., n v

k = i=1,..,nsi T.

Gi s J NS = {1, 2,...,n}

iJsi = T

t: : J { 1, 2, ..., |J|}, : NS\J { |J|+1,...,n},

TtssJ

ii

J

ii

Jii

||

1)(

||

1)( ||...,,2,1,

1)( JjTt

j

ii

Cc cng vic t |J|+1 cho n n u qu hn, gi tr pht:

kTsssspn

ii

Jii

n

ii

n

Jii

n

Jii

111||)(

1||)(

2013-11-25 25


Ngc li: gi s l mt phn cng cng vic vi mc pht khng qu k. K hiu m l s nguyn ln nht tho mn:

Ttm

ii

1)(

Tskpn

mi

n

iii

1 1)(Gi tr pht:

TssTssn

mii

m

ii

n

ii

n

mii

1)(

1)(

11)(

m

iisT

1)(

Tsm

ii

1)(

Vy

s(1), s(2), ..., s(m)

l li gii bi ton tng tp con.

2013-11-25 26

NP-y

S chng minh A l bi ton NP-y :

Ch ra rng A thuc lp NP;

Chn bi ton NP-y no , v d l B v chuyn dn B v A

Tht vy:A l NP-y nu

A thuc lp NP vmi bi ton X thuc lp NP c XA.

B l NP-y nn mi bi ton C trong NP:

CBDo BA, suy ra CA. Ngha l mi bi ton C trong NP u chuyn dn v A c.

2013-11-25 27

NP-y

1) Thng thng, vic ch ra rng mt bi ton A no thuc lp NP l kh (phi ch ra c thut ton bt nh gii n trong thi gian a thc).

2) Vic tm c bi ton NP-y chuyn dn v A xem ra c th n gin hn.

Vic chuyn dn c chng t n c kh tng ng vi bi ton NP-y . Nhng bi ton nh vy gi l NP-kh.

2013-11-25 28

Remark

Bi ton tm ng ng i ngn nht ni hai nh ca th lin thng c trng s c thut ton hiu qu gii. Tuy nhin, bi ton tm ng i di nht ni hai nh l bi ton NP-y .

Bi ton xc nh th G-2 mu kh n gin. Tuy nhin, bi ton kim tra G-3 mu li l NP-y .

Cho th c hng G={V,E}, s nguyn k, tp con E E tho mn |E|k. E c cha cnh ca cc chu trnh trong G? Bi ton ny c chng mnh l NP-y , tuy nhin nu xt bi ton ny vi th v hng, n li thuc lp P.

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61 NP Complete