Van Tru Hoc Chuong

8/2/2019 Van Tru Hoc Chuong

1/35

Vn Tr Hc

H Tn: Bi B Nguyn

MSSV: 0711151

2010


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CHNG 13: M HNH CY VI THI GIAN THAY I

1. Gii thiuMinimum spanning tree (MST)

Vi mt mng N, vn l tm mt mng kt ni thay Tm mch h, kt ni mng con ko di

ni tt ccc nh ca N, nh vy tng chi ph (hay di) ca cc vng cung thnh phn ca T l

ti thiu. Vn l tip theo gi l ti thiu bt ngun t cy bao trm T phi c gc r ca n ti s.

tm cy bao trm ti thiu ca mt mng no l mt trong nhng vn ni ting l lnh vc

ti u ha mng,xem Ahuja et al (1993); Graham et al (1985); Gabow et al(1986); Recski(1988).

Mc d, vn MST v bin th ca n c nghin cu trong vn hc, hu ht cc cng trnh

cng bcho n nay u gii quyt vn nh l mt phn tnh hc , ni ngi ta cho rng thi gian

khng cn thit i t mt nh n nh khc, v rng tt c cc thuc tnh ca mng l thi gian -

bt bin. R rng, khi chng ti cho thy trc , cc ginh ny chc xp x ca cc vn

th gii thc. trong hu ht cc tnh hung thc t, cc mng c xem xt chc chn c ththay i

theo thi gian.

Chng ti s nghin cu vn MST thi gian khc nhau trong chng ny. m hnh ca chng ti

xem xt mt mng li hp dng tnh thi gian i b (x, y, t) l cn thit i qua mt vng cung

(x, y), ti mt chi ph c (x, y, t) Hn th na, c thi gian vn chuyn b (x, y, t) v cc chi ph c (x, y,

t) l thi gian khc nhau, l nhng hm ca thi gian khi hnh ti nh ca x. chi ti mt

nh x c thc php, nm bt nhng thi gian tt nht khi hnh t x. Cho mt k thi hn

v mt nt s , vn l tm mt cy ko di bt ngun trang tri tt ccc nh trong mng, do

tng chi ph ca cc vng cung thnh phn ca cy bao trm l ti thiu, trong khi bt knh z

ca mng c tht c, trc khi thi hn k, cng mt con ng trong cy bao trm ca kt ni

s v z

Mc d phin bn tnh ca vn MST l chui a thc gii c (xem, v d, Ahuja et al (1993)),

chng ti s cho rng phin bn thi gian khc nhau ni chung l NP-y . Sau khi t xung mt

s khi nim c bn v thut ng trong phn 2 di y, chng ti s tp trung vo cc vn MST

v mt loi ca cc mng cung chui-song song (phn 3). Chng ti s cho thy rng vn ny l


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NP-y theo ngha thng thng (mc 3.1), v trnh by mt thut ton c th tm mt gii php

ti u trong thi gian O ((m + n) MK^2) , vi m v n l slng ca cc vng cung v cc nh ,

tng ng (mc 3.2). Sau chng ti s xem xt mt mng tng qut hn, mt mng li v hng

khng cha thcon ng cu K4 (phn 4). Chng ti s ly c mt thut ton c th tm mt

gii php chnh xc ti u trong thi gianO ((m + n) MK^2) . Cc trng hp chung sc nghin

cu trong phn 5 phc tp vtnh y mnh-NP sc kim tra.. Thut ton Heuristic scpht trin v v thi gian phc tp v sai st sc phn tch. Cui cng, mt s tham chiu b sung

v pht biu sc a ra trong phn 6.

2. Nhng khi nim v nhng kin v vn ta xtCho N = (V, A, b, c) l mt mng thi gian khc nhau Mt nh trong N c bit n nh l ngun

gc (root), k hiu l s. Gi s b (x, y, t) l mt snguyn dng, t = 0,1, ...., k vi k l mt s

nguyn dng, i din cho mt thi hn nht nh. Bng mt cch tip cn tng tnh mc 5,

chng 1, cc kt qu pht trin trong chng ny cng c thc tng qut cho cc tnh hung m

cc thi gian vn chuyn b l s nguyn khng m.

Nhli rng mt con ng nng ng trong mng thi gian khc nhau N l mt con ng P m tt

c cc ln khi hnh, thi gian n, v thi gian chi ti mi nh trn P c quy nh. hin nay

chng ti gii thiu khi nim ca mng con ng gy ra, nh di y.

nh ngha 2.1

i vi xj V, k = 1,2, ..., J, vi 1 J n, gi s Pj (s, xj) l mt ng dn ng ti xj vi hu ht t

. Cho V (Pj) v A (Pj) l tp nh v tp cc vng cung ca Pj, tng ng, v cho V '= Uj V (Pj), A' =

Uj A (Pj). Hn na, (Pj) l tp ca b ba

(x, y, (x)) trn Pj, vi (x) l thi gian khi hnh ti nh x trn Pj, v

= Uj (Pj).

xc nh Ij (x) = {[ (x), (x)]} l tp hp cc khong thi gian (trong thi gian chi ti cc nh

trn Pj), v I (x) = Uj Ij(x). Sau , N '= (V', A ', b, c), cng vi v I (x) (x V'), c cho l mt

ng dn mng con ca N. Mng con N ' l tng cc ng dn Pj, 1 j J, k hiu l N' = [P1,P2, ..., PJ].

R rng, mt con ng nng ng duy nht P (s, x) trong N cng l mt con ng sinh ra mng

con ca N 'ca N. Mt khc, a ra mt con ng-mng con c sinh t N 'ca N, th phi ra khi


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con ng P1, P2, ..., Pk, nh vy N' = [P1, P2, ... Pk]. nh ngha sau y cung cp cho cc khi

nim vcy bao trm nng ng.

nh ngha 2.2

Cho N '= [P1, P2, ... PJ] l mt con ng gy ra mng con ca Nvaf t n . N 'c gi l mt ng

m rng cy ca hu ht thi gian t , k hiu l T (t), nu c cc iu kin sau y:(i) i vi mi x V, c mt li ra con ng nng ng P (s, x) ca hu ht thi gian t trong N ';

(ii) Nu x l nh cui ca ng dn Pi N ', sau x phi khng c Pi nh l mt nh trung gian,

cng khng phi trong bt kcon ng khc Pj N', ni m i j v 1 i, j J.

nh ngha 2.3

t T (t) l mt cy bao trm nng ng cahu ht thi gian ti t , v cho

(T(t)) = +

Mt mrng nng ng cy T* (t) c cho l mt cy bao trm ti thiu thi gian t trn mng thigian khc nhau N (V, A, b, c), nu

(T*(t))=

Vi (t) l tp hp tt ccc cy bao trm nng ng ca thi gian t .

Vn cc cy bao trm ti thiu vi thi gian khc nhau (TMST) l xc nh cy bao trm ti thiu

T * (k) cho thi gian khc nhau cho mng N.

3. mng cung v dng song song

Chng ti s tp trung, trong phn, v vn TMST v thi gian chi ti mt nh l ty cho

php, v mng c cu trc song song vi dng h quang. loi mng trong cc ng dng (xem ng

dng ca i sBoolean chuyn i cy thi gian khc nhau ko di ti thiu (TMST) vn l

xc nh cy bao trm ti thiu T * (k) vi thi gian khc nhau cho mng N.

Ngoi ra, kt quthu c v kiu ca cc mng c th cung cp nhng hiu bit c gi tr cho vic

nghin cu nhiu vn chung.

Mt mng li p ng cc c tnh sau y l mt mng li cung dng -song song (ASP mng):

(i) Mt vng cung (s, ) l mt mng li ASP;

(ii) Nu G1 v G2 l hai mng ASP v s1, s2 v 1, 2 l ngun v ht ca G1, G2, tng ng, sau

Gp to ra t G1 v G2 bng cch kt hp s1 vi s2 v 1 vi 2 l mt mng li ASP . Ngoi ra,

Gs hnh thnh t G1 v G2 bng cch kt hp 1 vi s2 l mt mng li ASP.


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Trong thc t, bt k mng ASP c thc gy ra bi hai c tnh trn. Tnh ton, chng ti c th

cho bit mt mng cho l h quang dng song song, bng cch s dng cc thut ton ca Valdes

v cng s (1982), chy trong thi gian O (m + n).

Di y, chng ti s nghin cu s phc tp ca vn TMST theo thi gian, thay i ASP mng.

sau chng ti s trnh by mt thut ton chnh xc m c th tm thy mt cy bao trm ti thiu

trong thi gian ngn . Lu rng khi chng ta ni rng vn chng ti xem xt c mt c cu songsong vi dng hquang, chng ti lun lun c ngha l cc nh ngun (root) ca vn l nh bt

u trong mng ASP.

3.1 mrng

By gichng ta thy rng, vn TMST trn mt mng ASP l NP-y .

TMS-SPN. Ta t l dy cung thi gian khc nhau, song song ca mng

N (V, A, b, c), Mt thi gian gii hn k, v mt gi trngng K, liu c tn ti mt cy bao trm T(k) ca thi gian hu ht k nht, m (T (k)) 1; hoc nh mt hi t nu d-(x)

1>, trong d +(x) v d- (x) l bc ngoi v bc trong ca x, tng ng, pht trin cc thut ton

ca chng ti, chng ti gii thiu cc khi nim ca kim cng nh sau.

nh ngha 2.4:

cho f thuc V , g thuc V ,f l mt nh lan rng, v g l nh hi t. nu c hai ng dn tf n g

khng cha nh ly lan v hi tno khc, sau th con gy ra bi nhng con ng ko ny

c gi l "kim cng" (diamond).

K hiu D (f, g)

nh ngha 2.5

Gi s rng P (f, g)L mt con ng trongN, trong c khng c cc nh ly lan v hi t no

khc f, g. i vi mi nh x th con ng ny, xc nh df, g (x, ts.t) nh l chi ph ca nhng


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ng tf n x m ng ny c c gii hn trong thi gian [ ts,t ]. Nu nh ng ny khng

tn ti thdf,g(x ,ts.t)= .

B 2.1:

cho P = (f, g) l mt con ng m khng c nh lan rng v hi tkhc hn so vi f, g. sau ,

chng ti c df, g (x, ts.t) = 0 cho tt c cc0


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c) Ngc li, chn mt diamond D trong N. Tnh tonI

vE

v c mt php ton thu hp

nh sau:

a) Xa1P v 2P trong N ngoi trnh f v g.

b) To mt nh x v 2 i s(f,x) v (x,g). Chng ta c ( , , ) ( , )s E s

x t t D t v

, , ( , , ) ( ', )s I s E sg t t D t t D t vi 0 st t . Khi mt diamond c

chuyn i thnh mt ng i ( , , )P f x g .c) Nu f va g khng cn lan truyn v v mt nh hi t trong mng mi. Gi s rn

ng ( , )P f g cha ( , )P f g nh l mt ng i con. Tnh gi tr , ( , , )f g sd x t t

vi mi nh x trong ng i ( , )P f g bng cng thc 2.3.

d) Lp li bc (c) cho ti khi N trthnh mt ng i hoc mt diamond.Ch rng sau mi ln thu hp mt mng miNc hnh thnh, cha ng i ( , )P f g .

Hn th na, mt cy bao trm T trong mng N s bao gm mt i s(f,x) hoc (f,x) v

(f,g), khi ( ') ( ') 1d x d x khi ( ')d x v ( ')d x l bc trong v bc ngoi ca x

trong mng N. Trc o, chng ta c thtnh ton cc ph ca T nh sau:

( , ) ( )\( , )

( , ) ( )\ , , ,

( , , ( )) ( , ( ), ),( , ) ( )

( ), , , , , ,

x y A T f x

x y A T f x x g

c x y x x f x g A T

Tc x y x x f x g f g

B 2.3

Chng ta c ,P f g l mt ng i, ,0 , , , 0s f g st d f t t vi tt c st t v

, , ,f g sd y t t vi tt cy f. chng ta c:

i. Nu (x,y) khong l mt i s th

, , ,| , ,, , min , , 1 , 1 , min , , , ,f g s f g s f g su u b x y u t d y t t d y t t c y t d x t u c x y u

ii. Nu (x,y) l mt i s th

, , ,, , min , , 1 , 1 ,min , , , ,s

f g s f g s f g st u t

d y t t d y t t c y t d x t u y u t

Khi x l nh lin trc ca y trong P

Chng ta c 1 2 1, , ,..., , , , ,...r rP f x x x x g x g l mt ng i , v , ,P f x g l mt ng i c

c t mt diamond D. tin li chng ta gi s rng 1 1,

, ,i js p s

d x t t

c biu hin cho cc ph

nh nht ca cy bao trm trong h mng con c to bi D.v ng i 1, , ..., ,rP f x x f , vi


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sf t khi , , ,f g sd g t t c biu th cho gi tr nh nht ca cy bao trm trong h mng

con c sinh ra tD v ng i 1, , ..., ,rP f x x f vi sf t v g t

Chng ta sn sng trnh by thut ton sau:

Algorithm TMST-SP

Begin

While N khng phi l ng i hay diamond doChn mt cch ty mt diamond D trong N;

Tnh , , ,f g sd x t t vi tt c x V D v vi tt c t: 0 st t ;

Tnh , ,I sD t t v ,E sD t vi 0 st t ;

Xa D, loi nh f ,g t N;To mt ng iXa D, loi nh f ,g t N;

To mt ng i , ,P f x g trong N v chng ta c , , ,s E sx t t D t v

, , , , ,s I s E sg t t D t t D t vi 0 st t ;End While;

IfN l mt ng i then , ,0,s pT d ;

IfN l mt diamond then ,0,IT N ;

End

nh l 2.2 TMST-SP c th tm mt cch gii quyt ti u cho thi gian thay i cy bao trm nhnht vi mt cong tnh ton mng song song trong chng ny

nh l 2.3 TMST-SP c thc ci t vi phc tp 2O m n m

V d 2.2

Hnh 2.3 Mt v d m t


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s

x x x x

p p

p

1

1 2 3 4

2

Arc t = 0b,c

t = 1b,c

t = 2b,c

t = 3b,c

t = 4b,c

1,s x 2,5 1,3 2,6

2,s x 1,1 2,1 2,2

3,s x 1,4 1,3 2,5

4,s x 2,5 2,3 1,4

1 1,x 1,2 1,1 1,3

2 1,x 2,3 1,4

3 2,x 2,1 1,4

4 2,x 1,1 2,1 1,2

1, 1,1 1,1 2,1

2 , 1,2 1,2 2,1

Lin quan ca mt ASP mng N c m t trong hnh 2.3(a). Vi mi nt c 2 thng s thi gian dichuyn b(x,y,t) v cc ph c(x,y,t), c biu din trong hnh 2.3(b) (mt trng trong bn ng vib=c=). Tt ccc ph chc(x,t) = 0 Cho 4 , mc ch ca chng ta l tm thi gian c cyti tiu T ca N,

u tin, ly ra mt diamond 1 1 2 1, , ,D s x x . Vi mi 1x V D v 1 20 t t , tnh

1, 1 2

, ,s

d x t t . V d 1 1, , 2,0,0 0, ,0,0s sd s d x v


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1 1, 2 , 2 , 2,0,1 min ,0,0 , ,0,0 , ,0

min ,0 1 1

s s sd x d x d s c s x

Khi tn ti u=0 tha mn 2, , 1u b s x u . Gi tr ca 1, 1 2, ,sd x t t c biu din trong bn 2.1,

khi 1 1 2, ,I D t t v 1 1,E D t c ly trong bn 2.2.

Mng N by gi c chuyn thnh mt mng c m t trong hnh 2.4(a). Khi ta ly

2 3 4 2, , ,D s x x . Vi mi 1( )x V D v 1 20 t t , tnh 2, 1 2, ,sd x t t , xem bn 2.3. Gi tr ca

2 1 2, ,I D t t v 2 1,E D t c ghi trong bn 2.4

Bng 2.1. Gi tr ca1,s

d

cho diamond1

D

1, 1 1 2

, ,sd x t t 2t =1 2 3 4 1, 2 1 2, ,sd x t t 2 1t 2 3 4

10t

3 3 31

0t 1 1 1 1

13 3 3 1 1 1

2 6 2 2

3

3

11, 1 1 2

, ,x

sd t t 2t =1 2 3 4 2 1, 2 1 2, ,x

sd t t 2 1t 2 3 4

10t

5 41

0t 7 7 4

1 5 4 1 5

2 2

3

3

Bng 2.2. Gi tr ca1

v E cho diamond 1D


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1 1 1 2, ,D t t 2 1t 2 3 4 1t 1 1,E D t

10t

10 6 5 0 4

1 6 5 1 4

2 2 8

3 3

Mt mng mi c nhn c nh trong hnh 2.4(b).

Cui cng , mng hin thi nh mt diamond, chng ta c th tnh ton 1 2, ,I D t t vi

1 20 t t . Kt quc biu din trong bn 2.5

Khi , 0,4 14I D , chng ta c 14T . Cy vo trm ti tiu c biu din trong hnh

2.5(a). S trong vi arc l thi gian lch v s nm ngoi l cc ph di chuyn. Hnh 2.5(b) biu

din mt cy bao trm ti tiu ca mng N vi 3 .

Bng 2.3 Gi tr ca2,s

d cho diamond 2D

2, 3 1 2

, ,sd x t t 2t =1 2 3 4 2, 4 1 2, ,sd x t t 2 1t 2 3 4

10t

3 3 31

0t 5 3 3

13 3 3 1 3 3

2 5 2 4 4

3 3

2

3

, 2 1 2, ,

x

sd t t 2t =1 2 3 4 4 2, 2 1 2, ,

x

sd t t 2 1t 2 3 4

10t

6 6 41

0t 6 6

1 4 1

2 2

3 3

Bng 2.4 Gi tr caI

vE cho diamond D


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1 2 1 2, ,D t t 2 1t 2 3 4 1t 2 1,E D t

10t

9 9 7 0 6

1 7 1 6

2 2 9

3 3 10

Bng 2.5 Gi tr caI

cho diamond D

1 2, ,I D t t 2 1t 2 3 4

10t 16 14

1 14

2 3

Hnh 2.4 Mt v d m phng thut ton TMST-SP

s

x x x

p p

p

1

1 3 4

2

s

x x

p p

p

1

1 2

2

(a) (b)

Hnh 2.5 Mt gii php ti u


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s

x x x

p p

p

1

1 34

2

1

2

1

1

2x

0

3

1

1

1

3

1

3

3

2

s

x x x

p p

p

1

1 34

2

1

2

1

0

2x

1

2

1

2

2

3

2

4

3

1

(a) (b)

4. Mng khng cha th con lin kt vi K4

Trong phn ny, chng ta s khi qut cc m hnh trong mc 3,3 n nhng mng c th khngcha th con lin kt vi K4( th 4 cnh)

4.1 Tnh cht v phc tp

Liu v Geldmacher (1976) ch ra rng, bt k th no khng cha th con lin kt vi K4 uc th chuyn i 1 cch quy, bng cch p dng 4 nguyn tc chuyn i ( nh ngha 2.7 phadi), n 1 nh n. H tip tc ngh ra mt thut ton thi gian tuyn tnh m c th quyt nhkhi m 1 thc 1 th con lin kt vi K4


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nh ngha 2.7 'G l th kt qu sau khi p dng quy tc chuyn i T1, T2, T3 v T4 vi thG cho n khi khng quy tc no c p dng c na, vi

T1: Thay th 1 vng lp vv vi 1 nh v

T2: Thay th 1 cnh l lng vi nh u.

T3: Thay th 1 cp cnh uv v vw vi 1 cnh uw.T4: Thay th 1 cp cnh song song uv v uv vi 1 cnh uv.

Nu G ch c duy nht 1 nh, th chng ta gi G l kh quy.Ngc li, chng ta gi G l khng khquy.

Lu rng trong nh ngha trn, chng ta theo Liu v Geldmacher (1976)

s dng thut ngkhquy, m khc nhau t cc thut ng "khquy" c s dng trong tch phnNP-y (xem phn 3.2).

Hai thuc tnh sau c thit lp bi Liu v Geldmacher (1976).

Tnh cht 2.1: Nu T1, T2, T3 v T4 c p dng cho 1 thcho n khi khng th p dng cna, kt qul c c 1 th duy nht, c lp vi trnh t ca ng dng ca T1, T2, T3 v T4.

Tnh cht 2.2: 1 th l khng kh quy nu v ch nu n c 1 th con lin kt vi K4

Tng ng vi khi nim ca th khquy, chng ta nh ngha thut ng mng khquy nh sau:

nh ngha 2.8 A l 1 mng kh quy nu n nhng thc sca n khng cha th con linkt vi K4

Lu rng 1 th c hng lot cnh song song ( thc sca 1 mng c hng lot cnh songsong ) l kh quy l kh quy v n khng cha th con lin kt vi K4. Mt khc, chng ta c ccmng khc m thc sca n khng c hng lot cnh song song, nhng l kh quy, hnh 2.6bn di l 1 v d.

V d 2.3 1 mng ASP l 1 trng hp c bit ca mng khquy. Do , tnh l 2.1 , 2.5 v 2.6(bn di) , chng ta c nh l sau:

nh l 2.4Bi ton cy khung thay i theo thi gian trn 1 mng kh quy l NP-y theo ngha thng thng.

Hnh 2.6: 1 th kh quy m khng c hng lot cnh song song

Hnh 2.7: 2 trng hp c bit ca mng kh quy


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4.2 Mt thut ton chnh xc

Hy nhrng chng ta xem 1 mng thay i theo thi gian khng c cnh song song no, lkhng tn ti 2 cnh cng 1 hng gia 2 nh. N c th c, tuy nhin 2 cnh c hng tri ngcnhau gia 2 nh. n gin ha vic trnh by bng hnh, trong phn ny chng ta s dng lin kt ch ra 1 cnh n hoc 1 cp cnh i din gia 2 nh.

Chng ta cng xem xt 2 trng hp c bit sau:

Trng hp 1 Mng N c th hin trong hnh 2.7a, ni s l nh ngun. Bng nh ngha 2.5 ,ds,x(x, ts,t) m cc ph ca ng i ngn nht t s ti x trong khon thi gian [ts,t] ( Nu ng ikhng tn ti th ds,x(x, ts,t)= ). Sau , chng ta d nhn ra l

(T()) = ds,y(y, 0,)+ ds,z(z, 0,)

vi (T()) l gi tr ti thiu ca cy khung ca N vi thi gian gii hn .

Trng hp 2 Mng N c th hin trong hnh 2.7b. Trong trng hp ny chng ta c

(T()) = min{ds,xi(xi, 0,)+ ds,xi+1 (xi+1, 0,)}

vi ds,xi(xi, 0,) c tnh ton da theo ng i P(x1,x2, ..., xi) trong khi ds,xi+1 (xi+1, 0,) c tnhton da theo ng P(xr,xr1, ..., xi+1). Mt ln na, lu rng mc nh chng ta nh ngha ds,x(x,ts,t)= nu khng c ng i no tn ti t s ti x.

Mt nh v thuc V c gi l 1 nh lin hp, nu n c nhiu hn 2 nh lin k trong N. Mtng i P(x, y) ( hoc 1 chu trnh C(x = x1, ..., xr = x)) cng vi 1 nh lin hp x v s khng thucV (P)\{x} (hoc or s khng thuc V (C)\{x}) c gi l ng i treo ( hoc chu trnh treo)( Hnh2.8). Mt chu trnh khng c 2 nh lin hp th c gi l c dng hnh thoi. Lu rng nh ngha

ny l mt s tng qut ca nh ngha 2.4

Hnh 2.8 : l v e l 2 nh lin hp, trong khi P(a, e) l 1 ng i treo, chu trnh C(e, g, h) l 1 chutrnh treo, v D(P(s,w, l),P(s, f, l)) l mt dng hnh thoi.

C 2 cu trc, ng i v dng hnh thoi, gi vai tr chnh trong thut ton m chng ta xem xtbn di. Lu rng nu 1 ng i khng c 2 nh lin hp nh 2 im kt thc v s khng nmtrong n, dng chy c th chy trong hoc ngoi mi nh lin hp ca ng i. Hnh 2.9 chothy iu ny.

Hnh 2.9 Chra 1 dng chy vo hoc ra t1 ng vi y v z l 2 nh lin hp.

T2 nh x v y, c(x, y, t) c th bng vi c(y,x, t), chng ta cn tnh ton cc ph ca ng i dnv3 ng khung trn. iu ny dn ti nh ngha sau.

nh ngha 2.9: Gi s rng P(y = x1,x2, ..., xr= z) l 1 ng i trong N,vi y v z l 2 nh lin hpv s khng thuc V (P)\{y, z}. nh ngha


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(1)dyI ,zO(ty,tz) nh l gi tr ca cy khung ti tiu P, vi y l nh dng chy trong, z l nh dngchy ngoi, (y) ty v (z) tz;

(2)dyO,zI (ty,tz) l cc ph ca cy khung ti tiu P, vi y l nh dng chy ngoi, z l nh dngchy trong, (y) tyv (z) tz;

(3)dyI ,zI (ty,tz) l rng nh nht ca P, vi c2 nh y v z l nh dng chy trong, (y) tyand (z) tz.

Nu khng c cy no tn ti th hy cc ph l .

Nhc li nh ngha 2.5 ch df,g(x, ts,t). Bsau a ra phng php tnh dyI ,zO(ty,tz), dyO,zI(ty,tz), v dyI ,zI (ty,tz).

B 2.4: Gi s rng P(y = x1,x2, ..., xr= z) l 1 ng i trong N, vi y v z l 2 nh lin hp, vs khng thuc V (P)\{y, z}. Sau chng ta c

dyI ,zO(ty,tz)= dy,z(z, ty,tz), 0 ty tz

dyO,zI (ty,tz)= dz,y(y, tz,ty), 0 tz ty

dyI ,zI (ty,tz)= min {dy,z(xi,ty,)+dz,y(xi+1,tz,)}, 0 ty,tz .

nh ngha 2.10 Gi s rng D l 1 dng hnh thoi trong N, vi y v z l 2 nh lin hp, v s khngthuc V (D)\{y, z}. nh ngha yI ,zO(ty,tz) l cc ph ca cy khung ti tiu ca D y l nh dngchy trong, z l nh dng chy ngoi, (y) tyand (z) tz. nh ngha yO,zI (ty,tz) v yI ,zI (ty,tz)trong 1 ng i ging nhau. Nu khng c cy khung no tn ti th cc ph l .

Hnh 2.10: Ch ra 1 dng chy vo hoc ra t 1 dng hnh thoi, vi y v z l 2 nh lin hp.

B 2.5 Gi s rng D =(P1(y, z),P2(y, z)) v 1 hnh hp ch nht m tn ti 2 ng i P1 v P2 viy v z l 2 nh treo, v s khng thuc V (D)\{y, z}. Sau , chng ta c

yI ,zO(ty,tz)=min{d1yI ,zO(ty,tz)+ d2yI ,zI (ty,tz),d2yI ,zO(ty,tz) +d1yI ,zI (ty,tz)}, 0 ty tz

yO,zI (ty,tz)=min{d1

yO,zI (ty,tz)+ d2

yI ,zI (ty,tz),d2

yO,zI (ty,tz) +d1

yI ,zI (ty,tz)}, 0 tz ty

yI ,zI (ty,tz)= d1

yI ,zI (ty,tz)+ d2

yI ,zI (ty,tz), 0 ty,tz

vi dk biu thcc ph cho Pk (k =1, 2).

Svn hnh IXem xt 1 ng i treo ( hay chu trnh treo) ca N. Trc khi a ra 1 s vn hnhchng ta gii thiu nh ngha sau trc:

nh ngha 2.11 Gi sP(x, ..., y) l 1 ng treo trong N v x l 1 nh lin hp. nh ngha:(x, t)= dxI ,yO(t, )

vi 0 t . Trong trng hp y = x, P trthnh 1 chu trnh treo C(x, ..., x). Sau , nh ngha

(x, t)= dxI ,xI (t, t).


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Tht ra, (x, t) l cc ph ca ng i ngn nht ca P hoc cc ph ca cy khung ti tiu ca Cvi thi gian bt u t ti nh x. S vn hnh l loi bng i t P ti N (trnh x), v thm(x, t)(0 t ) ti nh x trong N. hon thnh, cho (z, t)=0, (0 t ) m i z thuc V(N) nukhng tn ti 1 ng i hay 1 chu trnh treo no ti z. Hn na nu c nhiu hn 1 ng i hay chutrnh treo ti nh x, cho (x, t) l tng cc ph.

Svn hnh II: Cho 1 dng hnh thoi D(P1(e, ..., g),P2(e, ..., g)), vi e v g l 2 nh lin hp. By

gi, b D trong N trnh e v g, v thm ng i bt kP(e, x,g) vo N c 1 mng mi N,vi x l 1 nh bt k, ex v xg l 2 cnh bt k(hnh 2.11). Tuy nhin, thm eI ,gO(te,tg), eI,gO(te,tg), v eI ,gO(te,tg) vo nh x.

Hnh 2.11: Mt dng hnh thoi c thay i vo 1 ng i

Lu rng, mng N l khquy, do sau khi thc hin 2 s vn hnh ny cho n khi khng thc na, mng cui cng l cng i hoc chu trnh c m ttrong trng hp I v II.

Sau khi hon thnh s vn hnh, N cha nhng nh bt k, nhng cnh bt k v nhng bin thmvo bt k(x, t). Do chng ta c thnh ngha cc ph ca 1 cy khung N nh sau:

nh ngha 2.12: Gi sN l 1 mng t c sau s vn hnh, v T l 1 cy khung ca N. nhngha

(x)= eI ,gI ((e),(g)), nu (e, x) thuc E(T),(x,g) khng thuc E(T),

hoc (x,g) thuc E(T), (e,x) khng thuc E(T)

eI ,gO((e), (g)), nu (e,x), (x,g) thuc E(T), (e) (g)

eO,gI ( (e),(g)), nu (e,x), (x,g) thuc E(T), (e) >(g)

e v g l 2 nh lin hp ca D, v A(N) l tp nh gi ca N.

Trong trng hp c nhiu nh gi, chng ta c thnh ngha cc ph ca cy khung bng cchtng t. nh ngha 2.12 l s tng qut ha ca nh ngha 2.3. Khi chng ta c th thy rngnu N khng c ng i gi v tt c = 0, th nh ngha 2.12 trthnh nh ngha 2.3.

Hn th, cng thc tnh ton de,g(x, te,tg)and dg,e(x, tg,te) th c xt li nh sau:

B 2.6: Vi P(e, ..., x, y, z, ..., g) l 1 ng i ca N v e, g l 2 nh lin hp. Sau , de,g(e,te,t)=0 cho tt c0 te t , de,g(y, te,te)= cho tt cy = e, v, cho 0 te


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de,g(y, te,t)=min{de,g(y, te,t1), min {de,g(x, te,u)+xI ,zO(u, t)}},

(b)Mt khc, nu y khng l 1 nh gi thde,g(y, te,t)= (y, t)+min{de,g(y, te,t 1),de,g(x, te,t)},

Tng t chng ta c th tnh ton dg,e(x, tg,te).

Cng vy, b2.4 nn c vit li nh sau:

B 2.7: Gi s rng P(y = x1,x2, ..., xr= z) l 1 ng i trong N vi y v z l 2 nh treo, v skhng thuc V (P)\{y, z}. Sau chng ta c

dyI ,zO(ty,tz)= dy,z(z, ty,tz), 0 ty tz

dyO,zI (ty,tz)= dz,y(y, tz,ty), 0 tz ty

dyI ,zI (ty,tz)= min i, 0 ty,tz

vi

i = dy,z(xi,ty,)+ dz,y(xi+1,tz,), c 2 xi v xi+1u khng l nh gi

min {dy,z(xi1,ty,u1) + (u1,) + dz,y(xi+1,tz,)},xil 1 nh gi

min {dy,z(xi,ty,)+(, u2) +dz,y(xi+2,tz,u2)} xi+1 l 1 nh gi

Bng b 2.6 v b 2.7, cng thc (2.3) v (2.4) gic vit li nh l (2.5) v (2.6) 1 cch quy.

(T*)= dsI ,yO(0,)+ dsI ,zO(0,)+ (s, 0),

(T*)= dsI ,sI (0, 0) + (s, 0).

Nhng bc c bn ca thut ton l nh sau:

(a)Nu N l 1 ng i P(y, ..., s, ..., z), sau (T()) c tnh ton bi cng thc (2.5) (Gi trban u ca (x, t) l 0 vi mi x thuc V v t =0, 1, ..., ).

(b)Nu N l 1 chu trnh C (s = x1,x2, ..., xr= s), sau (T()) th c tnh bi cng thc (2.6).(c)Nu N bao gm 1 ng i treo P(x, y), vi x l 1 nh lin hp, sau (x, t)= dxI ,yO(t,

)+ (x, t)for t =0, 1, ..., , v thc hin vn hnh I.(d)Nu N cha 1 chu trnh treo C(x = x1,x2, ..., xr = x), vi x l 1 nh lin hp, sau cho t =0,

1, ..., , let (x, t)= dsI ,sI (t, t)+ (s, t)+ (x, t),v thc hin vn hnh I.

(e)Mt khc, chn 1 dng hnh thoi D(P1(y, z),P2(y, z)) trong N ( nu c nhiu hn 1 dng hnhthoi, chn i 1 ci), vi y v z l 2 nh treo ca D. Theo b2.5, tnh yI ,zO(ty,tz) cho r0 ty tz , yO,zI (ty,tz) cho 0 tz ty , v yI ,zI (ty,tz) cho 0 ty,tz .

(f) biu th mng mi nh N. Lp li cc bc t (c) ti (e) cho ti khi N trthnh 1 ng ihay 1 chu trnh.

By gita chun b vo thut ton

Thut ton TMST-RN


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Bt u

Cho (x, t)=0 vi mi x thuc V(N) v mi 0 t ;

WhileN khng l 1 ng i hay chu trnh vi s lm

Lp li s vn hnh I xa ng i hay chu trnh treo;

Chn i 1 dng hnh thoi D(y, z) trong N;

Tnh yI ,zO(ty,tz), yO,zI (ty,tz), and yI ,zI (ty,tz) cho D v cho tt c0 ty,tz ;

Thc hin s vn hnh II;

End While;

NuN l 1 ng i P(x, ..., s, ..., y) th (T)= dsI ,yO(0,)+ dsI ,zO(0,)+ (s, 0);

Nu N l 1 chu trnh C(s = x1,x2, ..., xr= s) th (T)= dsI ,sI (0, 0)+ (s, 0);

Kt thc.

nh l 2.5: TMST-RN c th gii quyt mt cch ti u bi ton thi gian thay i ca cy mrngti tiu trn mt kh mng.

CM: Chng ta cm rng, khi thut ton kt thc, (T) l cc ph ca cy ti tiu N. S dng qui npvi m=|E(N)|. Xt TH m=1. Khi ch c mt cung, N l mt ng i v iu yu cu l hin nhin.

Gi s rng khi m


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, . Bng nh ngha ca chng ta bit rng c mtcy mrng T ca P(x,y) vi cc ph . Nh vy chng ta c th to ra T bng cchthm vo T thu c mt cy mrng xut pht ca N. By gichng ta thy rng (T) (T). Ch rng P l mt ng i treo trong N, ch l gi thuyt, tt ccc nh y trong P cthc thm t x. Gi s rng t0 l thi gian xut pht ti x trong T. Chng ta to T bng cchct bng i P trong T trim x, v

l cy mrng ti tiu ca ng i P vi thi

gian xut phat t0 ti im x.

Sau , chng ta c . Bng nh ngha chng ta c(T) = +

= + +

+ + +

= (T).V vy yu cu c CM.

(ii) N c mt chu trnh treo. Phn tch tng t (i).(iii) N khng c ng i treo hay chu trnh treo, khi N gim, phi tn ti 1 sn vi 2 nh lin

kt. Chn 1 sn D trong N vi 2 nh lin kt gi l e v g. Chng ta tnh , v . Chuyn D thnh P(e,x,g) thu c mt mng mi N.Mt cch tng t. u tin chng ta cm rng vi bt k cy mrng T ca N, n cthc to ra t cy mrng T ca N. Ch rng trong N, bc ca x l 2, v vy cnh

ex( hay xg) phi nm trong T. Gi s rng ex nm trong T. v xg T. Chng ta tocnh ex v im g trong mt rng F ca sn D, v thm F vo T thu c mt cymrng T ca N. Mt phn tch tng t c thc p dng vo TH vi ex,xg E(T)hay chc xgE(T). By gichng ta thy rng vi bt k cy mrng T ca N, c mtcy mrng T ca N, do (T) (T). Gi s rng F l mt rng mrng ca Dtrong T vi = , v = .Chng ta to T bng cch thay th F bng cnh ex. th hin cc ph carng mrng ti tiu ca D.V vy, chng ta c

(T)= + (F) + = (T)Chng minh vi TH: T bao gm 1 cy mrng ca D c thc thit lp 1 cch tngt. Ch rng N vn cn l mng gim thi gian thay i vi |E(N)|


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nh l 2.6. TMST-RN c thc tin hnh trong khong thi gian O((m+n)m).CM: Vic t vi mi im x V v vi mi khong thi gian t cn khong thi gianO(n). Trong vng lp, tm 1 ng i treo hoc chu trnh treo, chng ta c thi theo chiu suth nht (xem Gilberg (2001)). Bc c thc lm trong khoang thi gian O(m). khong thigian cn cho vic la chn 1 sn trong N l O(m+n) (xem Valdes(1982)). tnh v

vi tt c0 t1 t2

, chng ta cn khong thi gian O((m+n)m

). Vic thay th 1

sn bi 1 ng i nhn to cn khong thi gian O(m). V vy, 1 vng lp cn khong thi gianO((m+n)m). Khi mi vng lp gim t nht 1 cnh, chng ta cn t nht m vng lp. Nh vy, thigian i tng cng c xc nh trn l O((m+n)m).5. Mng tng qut.

By gichng ta tm hiu bi ton thi gian thay i ca cy mrng ti tiu trn mng tng qut.u tin chng ta s kim tra s phc tp ca n trong thi hn ca cy NP-bo trm mnh, v sau pht trin thut ton c th tm thy, trong nhiu khong thgian o, cc nghim xp x.

5.1. NP-xp x mnh.

N c hiu tt rng bi ton cy mrng ti tiu cin c gii vi nhiu gin tr. Chng ta c bit trong phn 3 v 4 trn, bi ton thi gian thay i ca cy mrng ti tiu trn mngnhm cung song song, hoc mt mng gim, d nh th no, NP-y nm trong s nhn bitthng thng. Trong phn ny chng ta s thy su hn rng bi ton TMST tng qut l NP-y trong s nhn bit mnh bng phng nu th c s ca N l 1 cy vi b(x,y,t)=b(x,y), hocc(x,y,t)=c(x,y) vi bt k cung (x,y) A. Chng ta s thy rng bi ton tp ph ti tiu (MSC)gim ti TMST.

nh ngha 2.13.MSC: 1 tp hu hn C={C1,C2,,Cm} cho trc v 1 s Ks, c tn ti 1 tp phCsao cho |C| Ks hay khng ?

tm hiu s phc tp ca n trong k hn ca NP-bao trm, chng ta kim tra cch thc gii quytca bi ton TMST mc thp hn.

nh ngha 2.14.TMST: cho trc 1 mng thi gian thay i N v 2 s nguyn k v , c tn ti 1cy mrng vi thi gian gii hn do cc ph tng cng khng ln hn k khng ?u tin chng ta kim tra bi ton vi s min cng rng vic chi bt knh no l khngtha. Kt qu ca chng ta thu c trong nh l 2.7 v nh l 2.8 sau.

nh l 2.7

Nu vic chi ti bt k nh no l khng tha, TMST l NP-y trong s nhn bit mnh,nu thc sca N l 1 cy, v

(i) c(x,y,t) l thi gian thay i v b(x,y,t)=b(x,y), (x,y) A;hoc


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(ii) b(x,y,t) l thi gian thay i v c(x,y,t)=c(x,y), (x,y) A.Chng minh: 1 cch d hiu, TMST nm trong NP. By gi, chng ta s cm rng MSC gim tiTMST v khng chi tha bt k nh no. Vi mi u , chng ta to 1 nh x i dincho n trong N. Hn na, thm 1 gc s v 1 mc xch v0 trong N. Dng cung (s,v0) v (v0,u) nicc nh v c

b(s,v0, )=, b(s,v0,)=1, t=0,1,2,, -1.c(v0,u,0)=, c(s,v0,t)=1,t=1,2,, .

b(v0,u,t)= t=1,2,,

c(v0,u,t)=, t=1,2,, .C k=Ks v =m. Cui cng, gi s rng ti bt k xV, khng chi c tha.

Mng N c nh ngha trn c gi tr b l thi gian thay i v c ch ph thuc cung (x,y).Mt cch d hiu, s ngn gn c thc tin hnh trong nhiu khong thi gian (xem hinh2.12).

By gichng ta chng minh rng ng vi MSC th tng ng cng ng vi TMST.Nu MSC c 1 tp phC vi |C|=l Ks, cy mrng vi thi gian chi 0 ti bt k 1 nh

gia c th bt u c xy dng vi gc s. khng mt tnh tng qut, gi s CiC, 1 I l (ch rng l Ks m ). C=C1 v Ci= Ci\Cj, 2 i l. Mt cch ngn gn, C=Ci=Ci. Vi miCi, nu |Ci|0, chng ta chn ng i Pi(s,v0) bt u ti s ti thi gian i-1, v dn ti ti v0 vi thigian i. nu u l yu t xy ra trong Ci

Chng ta thm cung (v0,u) vo Pi(s,v0) vi thi gian bt u i v i+1 vo cy con, th hin biTi. Vic ph hp vi s ngn gn ca chng ta, mi im u Ci c thc kt ni vi thi gian


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. Hin nhin l cy c xy dng bng cch kt hp tt c Ti l 1 cy mrng T vi thi gian =mv tng cc ph t nht km hn k=Ks (xem hinh 2.13).

Nu TMST c 1 cy mrng T, cc ph khng vt qu k= Ks, cc ph hn ch bo mrng tn ti t hn k cy con. Ti bao gm cc cnh t s ti cc yu t u . Chn tt c cc yut u trong Ti c dng tp Ci. Mt cch hin nhin, C=Ci bao gm tt c cc yu t u , Cl 1 tp ph vi |C| k Ks.

Vi TH b(x,y,t)=b(x,y) v c l thi gian thay i, chng ta sa i s ngn gn nh sau:

b(x,y,t)=1, (x,y) .c(s,v0, )=.

c(s,v0, )= t=0,1,,.

Tt c cc phn tch cn li khng thay i. iu hon tt cm.

Mt bi ton c t ra trong lp ca APX ,nu c 1 li khng i sp x thut ton,,mtthut ton c th tm trong nhiu khong thi gian, 1 nghim xp x vi mt li c th, lmt chc chn khng i (ch rng li c thc tm thy nh sau: v l nghim ti u vnghim xp x. c gi l c 1 li c th l nu . Lund v Yannakakis(1993)ch rarng MSC khng nm trong lp ca APX ,, tm mt nhn tkhng i xp x thut ton choMSC t nht kh nh l chng minh P=NP. Ngn gn chng ta xy dng cc trn rng bi tonTMST cng khng nm trong lp ca APX. iu cho chng ta kt qu sau.

nh l 2.8. Xt bi ton TMST khng chi tha bt cnh no. Khng c li khng ia thi gian xp x cho bi ton TMST nu PNP, nu thc sca N l 1 cy, thi gian chiti bt k nh no phi l 0, v 1 trong 2 thng s, b hoc c, l khong thi gian c lp.

By gichng ta xt vng xung quanh ni m vic chi bt k nh no tha mt cchc lp.(). Chng ta s thit lp NP-bao trm ca n, hn na s dng s ngn gn t MSC. Ch rng trong TH chi ti bt k nh no l b cm, chng ta c th thy cy bao trm bng phngca n khi thc sca N l 1 cy. iu d thno chng na cng khng mrng ti TH vivic chi ti bt k nh no ang c tha mt cch c lp. S phn tch ngm gn by gic xy dng trn 1 mng cy khng c gc; xem bn di.

nh l 2.9. Xt bi ton TMST ni m vic chi ti bt k nh no c tha mt

cch c lp l NP-y trong s nhn bit mnh nu c(x,y,t)=c(x,y) v b(x,y,t)=b(x,y), (x,y) A.CM: Vi bt k th d vMSC cho trc, chng ta xy dng 1 th d cho TMST nh sau: vi miCiC, to 1 cp nh xi v x i (1 i m). Vi mi yu t ujCi, to 1 nh xm+j. Nhng nh cng nhau vi 1 im b sung ,gc s, tp c sp xp V. To cung (s,xi), (s,xi) v (xi,xi), 1 i m,


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nh trong hnh 2.14 vi mi nh xm+i, to 1 cung (xj,xj+1) nu ujCi(xem hinh 2.14). Nhng cung l tp cc cung c sp xp.

b(s,xi,t)= b(s,xi,t)= b(xi,xi,t)= b(xj,xm+j,t)=1, 1 i m, 1 j n, 0 t c(s,xi,t)=1, c(s,xi,t)= c(xi,xi,t)=0, 1 i m, 0 t

c(xj,xm+j,t)=0, 1 i m, 1 j n, 0 t Cui cng, cho k=Ks, =2. Mng N(V,A,b,c) nh tm thy trn c gi tr l c 2 c v b. l cckhong thi gian c lp(ch rng chng ta khng c bt k bin php hn ch no trong thi gianchi ti bt k nh no ).Trong nhng g tha, chng ta cm rng ng vi SMC tng ngng vi TMST. Nu MSC c 1 tp phC vi |C|= l Ks, 1 cy mrng vi nhng khong thigian chi ti cc nh ca n c thc xy dng nh sau : khng mt tnh tng qut, gi s

CiC, 1 I l (ch rng l Ks m), v CC1 v Ci=Ci\Cj, 2 I l. vi mi Ci, nu |Ci|0, chng ta chn ng i Pi(s,xi) vi (s)=0 v (xi)=1. Nu uj l yu t trong Ci, chng tathm cung (xi,xm+j) vo ng i Pi(s,xi) vi (s)=1 v (xi)=2.

Hnh 2.4

Khi c nhiu hn 1 cung c thm vo Pi(s,xi), chng ta c 1 cy con, c th hin bi Ti. Khi

Ci= Ci= Ci, mi nh xm+j c thc lien kt vi khong thi gian la 2. Vi nhng Ci


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C, chng ta chn ng i P(s,x i)= (s,xi,xi) vi (s)=0, (xi)=(xi)=1 v (xi)=2. Hin nhin l ttccc nh trong V c thc lin kt vi khong thi gian l 2. Vic kt hp tt c cc Ti v Pichng ta thu c 1 cy mrng T vi khong thi gian =2. Khi cc ph ca Ti gn bng 1, vcc ph ca Pi l 0, chng ta c (T) k Ks.

By gi, nu TMST c 1 cy mrng T(2) vi (T(2)) k Ks,cc ph v thi gian tri quatrong bi ton c xy dng trn c bo m rng tn ti l k cy con, bao gm ng i (s) t

s ti nh xm+j vi (xm+j)=2. Chn cc tp C i nu xm+j xut hin trong Ti. C= Ci. Khi T(2) baogm tt cnh xm+j, chng ta c tt c cc yu t uj Ci. Nh vy, C l 1 tp ph vi |C| k =Ks. Trong bng tm tt,chng ta hon thnh cm.

Lp li, tnh ngha ca lp ca APX, ta c:

nh l 2.10. Xt bi ton TMST ni m vic chi ti bt k nh no tha hon ton c lp.Khng c li khng i a thi gian xp x thut ton cho bi ton nu PNP, nu vi bt k cung(x,y) , b(x,y,t)=b(x,y) v c(x,y,t)=c(x,y).

Cui cng, chng ta xt bi ton TMST tng qut ni m thi gian chi ti bt k im no clc rng buc bi 1 hm c lp. Khi chng bao gm vng quanh ni m thi gian chi ti 1im phi la 0. Nh l 1 TH c bit, nh l 2.11 v 2.12 theo sau tha st vi inh l 2.7 v 2.8, 1cch lien quan.

nh l 2.11. Bi ton TMST tng qut l NP-y trong nhn thc mnh, nu thc sca Nl 1 cy v

(i) c(x,y,t) l thi gian thay i v b(x,y,t)=b(x,y), (x,y)A; hoc(ii) b(x,y,t) l thi gian thay i v c(x,y,t)=c(x,y), (x,y)A.

nh l 2.12. Khng c sai s hng a thi gian xp x thut ton cho bi ton TMST tng qut nuP NP, nu thc sca N l 1 cy v 1 trong 2 thng s b hoc c l khong thi gian c lp.

5.2. Thut ton Heuristic:

Ta s miu t thut ton ny bng cch tm nghim xp x trong gia thc thi gian. Thut ton nybao gm 2 bc chnh: , ta nh ngha ng i ngn nht tsnx ti hu ht k. Vitt c nhng ng i ny, ta thu c mt mng con ca N. Sau , ta b nhng cung tha trong

mng con ny v thu c mt cy bao trm caN. Ta chng minh rng cy ny, k hiu l ( )AT k , l

mt nghip xp x ca bi ton TMST, vi tng thigian cn thit i ht ( )AT k l mt gia thc.

p dng thut ton chng 1 tm ng i ngn nht, trong phn sau ta s gii hn bi ton, vi

thi gian ch ti vecto x (x V ), b chn trn bi xu . Ch y rng khi 0xu vi mi x V , ta c

trng hp khng c thi gian i. Khi , nu ,xu x V , ta c trng hp thi gian i tix l

ty .

5.2.1 Tm ng i ngn nht:


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nh ngha di y cp n n hm tr v ca phn gii thiu bn di.

nh ngha 2.17:

Cho ( , )b

d y t l cc ph ca ng i ngn nht t s n y ca thi gian tuyt i x . Nu ng i

n khng tn ti, t ( , )bd y t .

Ta vit li thut ton TSP-BWnh sau:

Quy trnh DP

Begin

Khi to: ( ,0) : 0bd s , v ( ,0)bd x , ;x s ( , ) :bd x t , x v 0t ; : { ( , 0)x bHeap d x v

( ,0) : ( ,0), ;m

b bd x d x x

Sp xp tt c cc gi tr ( , , )u b x y u cho tt c cc 0,1,...,u k v cho tt c cc cung ( , )x y A ;

For 1,...,t k do

For vi mi cung ( , )x y A do ( , , ) :b x y t ;

For tt c cc cung ( , )x y A v tt c Du sao cho ( , , )D Du b x y u t

Do

( , , ) : min{ ( , , ), ( , ) ( , , )}mb b b D D

x y t x y t d x u c x y u ;

For vi mi vectoydo { |( , ) }( , ) : min ( , , )b x x y A bd y t x y t ;

For vi mi vectoryupdate

Insert-heap( )y

( , )bd y t ;

If yt u then delete-heap ( )y ( ,1 1)b yd y u ;

For vi mi vectorydo

( ):A yu Minimum heap ;

( , ) : ( , );m

b b Ad y t d y u

For miydo * 0( ) : min ( , )b t k bd y d y t ;

End.


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5.2.2 Loi b nhng cung tha:

Sau khi p dng thut ton trn tm nhng ng i ngn nht gia nt gc s v mi nt x V ,ta c thtm c mt mng con caNbng cch kt hp nhng ng i ny vi nhau. Bc tiptheo l loi b nhng cung tha trong mng con. Ta s biu din cc bc xa vecto/cung nh sau:

(i) Xa nhng vecto trung gian: Nu mt vectorx xut hin trong 2 ng ii

P vj

P , v nu

[ ( ), ( )] [ ( ), ( ) ]j j i i xx x x x u , th cung ( , )y x b xa v yc thay th bi x trong

jP , v tt c cung ni cax trong jP u c thn c s thong qua ( , )iP s x trong ng

iP v thi gian i bt buc khng b vi phm.Do , ta c th xa cung ( , )y x . Sau ,

ngj

P kt thc ti vector y, v mt ng i mi c to ra bao gm 2 phn: phn

( , )iP s x trong ng iP , v phn bt u tx trong ng jP vi thi gian xut pht l

( )j x . Lp li qu trnh ny cho n khi khng cn vector trung gian. Sau , chuyn qua

bc (ii).

(ii)

Xa nhng vactor cui tha: Cho

0

jx l vector cui ca ng jP . Nu

0

jx xut hin trongng i khc, hoc xut hin trong

jP nh mt vector trung gian, khi ta xa vector

ny v nhng vector ni vi n. Khi , vector lin trc 0jx trthnh vector cui. Lp li

qu trnh ny cho n khi cng i b xa ht hoc vector cui cajP khng xut hin

trong nhng ng i khc hoc trongj

P nh mt vector trung gian.

Theo nh ngha 2.2, mng con - ng cm ng 'N to ra sau biu din 2 qu trnh bn trn lmt cy bao trm. Trong phn sau, ta sdng 2 phng php DSIV (Deleting Ahsre IntermediateVertices) v DREV (Deleting Redundant End Vertices) thy r th t ca 2 qu trnh ny.

(1)Quy trnh DSIV:DSIV c dng xa nhng vector trung gian trong cc ng i. hn ch thi gian, qutrnh ny dng mt mng 3-th nguyn ( , , )e x t i . Nu vectorx xut hin trong ng jP vi thi

gian n l ( )j x v thi gian xut pht l ( )j x , khi ( , ,1)e x t j v ( , ,2) ( )je x t x , vi

( )jt x . Ban u, n c gn bng 0. Qu trnh bao gm 2 bc c bn sau:

(i) Nu c hn mt ng i cha x vi cng mt ti gian n ( )x , khi 0jP c thigian xut pht l 0 ( )j x . Xa tt c cc cung ( , )iy x trong iP , vi iy l nt lin trc

ca x trongi

P v 0i j .

(ii) Kin tra mng e. Vi mi vectorx, nu tn ti u v tsao cho t u v ( , ,2) xe x u t u (c ngha l c 2 ng i

iP v

jP c ( , ,1)e x t i , ( , ,1)e x u j , ( )it x , ( )ju x


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v ( , , 2) ( )je x u x , v tha [ ( ), ( )] [ ( ), ( ) ])j j i i xx x x x u , khi cung ( , )jy x c

thc xa trongj

P .

Sau khi hon thnh 2 bc trn, tt c cc vector tha s b xa.

Quy trnh DSIV:

Begin

For \{ }x V s v 0,1,...,t k do ( , ,1) : ( , ,2) : 0e x t e x t ;

For vi mi ng i ( 1,..., 1)jP j n v mi ( )jx V P do

If ( , ( ),1) 0e x x then ( , ( ),1):e x x j , ( , ( ), 2) : ( )e x x x ;

Else If ( ) ( , ( ), 2)x e x x then xa cung ( , )iy x trong iP , ( , ( ),1):e x x j ,

( , ( ), 2) : ( )e x x x ;

For mi \{ }x V s do

t : 0 ;

For 0,1,...,t k do

If ( , ,1) 0e x t then : t ;

Else If ( , ,2) xe x t u thent : ( , ( ),1)i e x x , xa cung ( , )iy x trong iP ,

( ) :i

x , ( , ,1) : ( , ,2) : 0e x t e x t ;

End.

phc tp ca quy trnh ny c thc phn tch nh sau. Nhnh Fordo u tin v th3 c phc tp l ( )O nk . Nhnh th2 c phc tp l ( )O nk v mi ng i jP gm hu ht vector k.

Do , tng thi gian chy quy trnh ny b chn bi ( )O nk .

(2)Quy trnh DREV:Quy trnh ny c dng xa nhng vector cui tha ca ng i. tng c bn l to bin

m ( )num x cho mi vectorxlu li slng ca chng trong ng i. Sau , khi ta kim travector x ca mt ng i c nm v tr no khc khng, ta ch cn kin tra ( )num x . Nu

( ) 1num x , r rang rngx phi xut hin mt vtr khc v do ta c th xa n ra khi ng i,

v sau tng ( )num x thm 1.

Quy trnh DREV


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Begin

(0) : 0;num

For mi \{ }x V s do (0) : 0;num For mi \{ }x V s v 0,1,...,t k do

If ( , ,1) 0e x t then (0) : ( ) 1num num x ;

For mi ng ijP do

nh ngha 0jx ca jP ;

While 0( ) 1jnum x do

t 0 0( ) : ( ) 1j jnum x num x ;

Xa0

jx trong

jP ;Endwhile;

End.

Quy trnh DREV kin tra mi vecto trong tt ccc ng i.. V c 1n ng i (v c 1n vectocui ) trong ( )AT x v mi ng i gm kvecto (v thi gian vn chuyn b l mt snguyn dng

v thi gian n mt vecto phi ln hn thi gian n nt lin ktrc n), nn quy trnh DREV c phc tp l ( )O nk .

5.2.3 Thut ton A-TMST:

Thut ton ny c trnh by nh sau:

Thut ton A-TMST

Begin

Call quy trnh DP tm ng i ngn nht ( , )jP s x t s n mi vecto \{ }x V s , vi

1 1j n

Iftn ti ng i ( , )P s x vi *( )d x then t ( ( )) :AT k v dng li;

Call quy trnh DSIV xa nhng vector trung gian.

Call quy trnh DREV xa nhng vecto cui tha.

Combine tt ccc ng i cn li to ra mt mng con ( )AT k .

End.

Mt nghim xp x b chn bifc gi l mtf-nghim xp x. t L l tt c cc nhnh trong

( )AT k c c trong A-TMST v | |l L , ta c nh l sau:


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nh l 2.13:

Thut ton A-TMST c th tm ra , vi phc tp ( ( log ))O k m n k , mt l-nghim xp x ca

TMST. Hn na, l l gii hn chp nhn c tt nht ca thut ton A-TMST.

Chng minh:

R rang, ( )AT k thu c bi thut ton ny l mt cy bao trm ca Nvi thi gian k, theo nhngha 2.2.

Cho trc thi gian cn thit ca A-TMST. Thi gian chy quy trnh DP l ( ( log ))O k m n k . V quy

trnh DSIV v DREV u c phc tp l ( )O nk , nn tng thi gian chy ca A-TMST b chn bi

( ( log ))O k m n k .

Ta phn tch sai s gii hn ca nghim xp x ( )AT k . t li l l s nhnh trong ( )AT k . D thy

rng:

*( ( )) ( ( , )) ( ( , )) ( ( )),Ax L x L

T k P s x P s x l T k

Vi ( )T k l cy bao trm ti u v ( , )P s x l ng i trong ( )T k tsnx, vi *( , )P s x l ng

i ngn nht tsnx thu c bng quy trnh DP.

Hnh 2.15:Phn tch mt mng sai sgii hn

chng t gii hn l ny l kt qu tt nht ca thut ton A-TMST, ta xem cc lp c bit sauy ca mng thi gian thay iN, v minh ha cho nghim xp xthu c khi p dng thut ton

A-TMST cho mng ny l s t c gii hn l. t ( , , , )N V A b c (xem hnh 2.15), vi

1 2{ ( ), ,..., }nV x s x x , 2{( , ),2 ,( , ),3 }i jA s x i n x x j n , v

( , , ) 1, ( , , ) 1,2 ,0i ib s x t c s x t i n t k


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2 2( , , ) , ( , , ) 1,3 ,0j jb x x t c x x t j n t k .

Ch rng khi 2k , cy bao trm ti u ( )T k hnh 2.16 vi ( ( )) 1 ( 2)T k n . Nghim ca

A-TMST, ( )AT k , hnh 2.17 vi ( ( )) 1AT k n . Cho tin v 0, ta c:

0( ( )) 11

( ( )) 1 ( 2)

AT k n n l

T k n

Vi l l s nhn ca ( )AT k . Chng minh xong.

Nhn xt: Ch rng phc tp ca thut ton A-TMST c gii hn bi cc quy trinh tmng i ngn nht.

tm sai s gii hn ca thut ton A-TMST, ta c th ci tin hn na khng? nh l 2.13 s chra rng gii hn l chp nhn c tt nht. Tuy nhin, trong mt strng hp c bit, khi nhngiu kin hin nhin c tha mn, ta c th chn mt gii hn tt hn. Phn sau y s tho lun vnhng trng hp ny.

5.3 Cc sai s gii hn ca thut ton heuristic trong mt strng hp c bit:

chng ti trnh by cc li rng buc ca thut ton A-TMST c thc ci thin cho mt loi

mng nhiu giai on. nh mt mng li shu mt cu trc a giai on, mi giai on bt u t

mt nh ngun chung:

t Ni biu th cho tiu mng ca mt thi ktrong N, 1 i k; . Hn na, t L(Ni) biu th

cc l caTA (Ni), ni m TA(Ni) l cy tiu ca TA (k) bao gm Ni, v cho Li = | L(Ni)| (1 ik).

H qu 2.1: cc nghim xp xthu c bng thut ton A-TMST c mt sai s gii hn f = max

{L1, L2 ,...., Lk}, nu n l mt lch trnh cng vic mng v:


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(i) chi ti nh bt kc cho php m khng c bt k hn ch.(ii) cb(x,y,t) and c(x,y,t) th khng gia tng theo thi gian T.

chng minh : u tin ta s trnh by cc bt ng thc sau y.

vi 2 i k.

chng ta ci tin bt ng thc (1) nh sau :

(T(k)) +....+

+....+

L1(k)) + L2(k)) +........+ Lk(k))

max{L1,..........,Lk} (T(k)).ni (k) l cy bao trm ca T (k) trn mng ph Ni, 1 i k, l:

max{L1,.......,Lk}

Do , chng ti hon thnh bng chng

5.4 mt chng trnh gn ng cho vn ny vi hn ch chi ty :cho chui thi gian vn mrng cy ti thiu trn mngban u, nhng theo thi gian vi hn ch

chi ty , chng ta c mt chng trnh gn ng, c th gii quyt hiu qu vn hn. n

bao gm 2 bc chnh:

th nht, i vi 1 mng thi gian ban u, to ra mt biu bao trm ca N, k hiu l N ', mkhng cha th con ng hnh ti K4.

sau , ta p dng thut ton TMST-RN trn N '. cho T 'l gii php ti u thu c bng thutton. r rng, T 'l mt gii php gn ng ca mng ban u N. chai 2 bc c thc thc

hin trong thi gian a thc thi gian gi.

5.4.1 To ra mt mng rng ln kh quy:A (N) l tp hp cc cung ca mng ban u N v s l gc ca N.

(i)cho Q l mt tp nh.Thit lp Q = {s}. biu tht (x) l thi gian khi hnh sm nht c th tix. cho t (s) = 0.


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(ii) Ly mt nh x t Q. cho adj (x) = {(x, y) | (x, y) A (N), (x, y) khng c nh du}, tt ccc vng cung lin kkhng c kim sot ca x, v sp xp theo th ttrong khng gia tng v

gi tr ca c (x, y, t (x)). lm nhiu ln cho n khi adj (x) trnn trng rng:

a)chn cung u tin (x, y) trong adj (x) v thm n vo N '. kim tra xem N 'cha mt th conng hnh ti K4 hay khng. nu cu tr li l "c", loi b (x, y) t N ', nu khng li (x, y)

trong N' (vn c gi l mng li mi thu c l N '). nu y Q sau cho Q = Q + y.b)xo (x, y) tadj (x) v (x, y) c kim tra.(iii) lm (ii) nhiu ln cho n khi Q trnn trng rng.

r rng, N 'l mt mng li tri rng kh quy ca N vi cc cnh cng nhiu cng tt.

by gichng ti cung cp cho cc thut ton nh di y.

Thut ton TMST-A

Begin

SetE(N)=0 and V(N)=0 . let Q={s}, t(s)=0;

While Q 0 do;

chn 1 vecto x t Q;

sp xp adj(x) trong trt t khng tng trn gi trc(x,y,t(x));

Whileadj(x)0 do ;

Chn cnh u tin (x,y) adj (x);

IfN 'cha mt thcon ng hnh n K4, sau loi b (x,y);

Else let A(N)=A(N)+ (x,y) and V(N)=V(N)+y if y V(N);

let adj(x)=adj(x)\(x,y);

Endwhile;

Let Q=Q\x;

Endwhile;

End


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r rng, N 'thu c bng thut ton l mt mng li kh quy ca N.

nh l 2.14: thut ton TMST-A c th thc hin trong O(m.max{m,n}).

5.4.2 S th nghim.Bng 2,6 minh ha kt qu bng s. phc tp ca bi ton c lit k trn ct u tin vi

snh n, s cnh m, v thi gian T tng ng. tt c cc thi gian vn chuyn b chi ph c c tora ngu nhin. ct bn phi ca bng 2.6, m ' l s mng mrng kh quy to ra trn mng bi

thut gi TMST-A.

V d , gii hn 1 l 30 nh , 80 cnh vi thi gian T = 30. cc cnh ca cc mng kh quy m

rng N 'l 38. chi ph ca cy bao trm ti thiu ca N 'l 31, v thi gian s dng 2s CPU . nh so

snh, chng ti p dng phng php Monte Carlo cng mt vn . to ra 1000 cy bao trm ti

thiu cho cc N mng ban u v la chn tt nht, trong c chi ph 70 v chi ph 170 s thi gian

CPU. tt c kt qu th nghim cho thy s thut ton gn ng ca chng ti l nhiu hn so vi

phng php bt Monte Carlo.

Bng 2.6 : kt qu ca sth nghip:

Kch thc phng php Monte Carlo thut ton gn ng

n m T Thi gian lp gi CPU(s) m gi CPU(s)

30 80 30

50 120 50

80 260 60

100 300 80

150 400 100

200 500 100

1000 70 170

5000 113 454

5000 202 515

5000 265 631

5000 394 948

5000 527 1282

38 31 2

62 49 5

98 80 36

127 99 69

176 149 204

228 202 366

6. B sung ti liu tham kho v ly kin.

NG DNG ca MST c nghin cu trong cc ti liu. bao gm :

Nhng thit k h thng vt l (Prim (1957); loberman v cng s (1957) al; Dijkstra (1959)). Thit k mng (magnanti v cng s (1984)).


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Ti u thng qua (Abdel-wahad v cng s (1997); Prim( 1957)). M hnh phn loi (dude v cng s (1973)). Hnh nh ch bin (osteen v cng s (1974); Abdel-wahab v cng s (1997)). V phn tch tin cy mng (Van Slyke v cc cng s (1972)). Cc ng dng ca vn graham v cng s (1985). Thut ton hiu qu cho cc vn kruskal (1956); Prim (1957), v Dijkstra (1959). vn MST o c thc gii quyt bi mt thut ton Edmonds (1965).

Van Tru Hoc Chuong

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