HC VIN CNG NGH BU CHNH VIN THNG

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KHOA CNG NGH THNG TIN 1

BI GING

TON RI RC 1

H Ni 2013

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LI GII THIU

Ton ri rc l lnh vc nghin cu v x l cc i tng ri rc. Ton ri rc dng m, quan st, v x l mi quan h gia cc i tng trong cc tp hp khc nhau. Bn cht tnh ton trn my tnh l ri rc. Chnh v vy, ton hc ri rc c xem l mn hc kinh in cho sinh vin cc ngnh Cng ngh thng tin v in t Vin thng. Ti liu hng dn mn hc ton hc ri rc c xy dng da trn c s kinh nghim ging dy mn hc v k tha nhng ni dung t gio trnh Ton hc ri rc ng dng trong tin hc ca Kenneth Rossen. Ti liu c trnh by thnh hai phn: L thuyt t hp (Ton ri rc 1) v L thuyt th (Ton ri rc 2).

Phn I trnh by nhng kin thc c bn v l thuyt t hp thng qua vic gii quyt bn bi ton c bn l: Bi ton m, Bi ton tn ti, Bi ton lit k v Bi ton ti u. Phn II trnh by nhng kin thc c bn v L thuyt th: khi nim, nh ngha, cc thut ton trn th, th Euler, th Hamilton. Mt s bi ton c ng dng thc tin quan trng khc ca l thuyt th cng c ch trng gii quyt l Bi ton t mu th, Bi ton tm ng i ngn nht v Bi ton lung cc i trong mng.

Trong mi phn ca ti liu, chng ti c gng trnh by ngn gn trc tip vo bn cht ca vn . Cc thut ton c trnh by v ci bng ngn ng lp trnh C++. Mc d rt cn trng trong qu trnh bin son, tuy nhin ti liu khng trnh khi nhng thiu st v hn ch. Chng ti rt mong c s gp qu bu ca tt c c gi v cc bn ng nghip.

H ni, thng 10 nm 2013 PT

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MC LC

CHNG 1. LOGIC, TP HP V NG DNG............................................ 5

1.1. Gii thiu chung ................................................................................................... 5 1.2. Nhng kin thc c bn v Logic mnh ........................................................... 6

1.2.1. nh ngha & php ton ................................................................................ 6 1.2.2. S tng ng gia cc mnh ................................................................. 7 1.2.3. Dng chun tc............................................................................................... 9

1.3. V t v lng t................................................................................................. 10 1.4. Mt s ng dng trn my tnh ........................................................................... 12 1.5. Nhng kin thc c bn v l thuyt tp hp ...................................................... 15

1.5.1. Khi nim & nh ngha ............................................................................... 15 1.5.2. Cc php ton trn tp hp ........................................................................... 16 1.5.3. Cc hng ng thc trn tp hp .................................................................. 17

1.6. Biu din tp hp trn my tnh .......................................................................... 18 1.7. Nhng ni dung cn ghi nh ............................................................................... 19

BI TP CHNG 1 ...................................................................................... 19

CHNG 2. BI TON M ........................................................................ 21

2.1. Nhng nguyn l m c bn.............................................................................. 21 2.1.1. Nguyn l cng ............................................................................................ 21 2.1.2. Nguyn l nhn ............................................................................................ 22

2.2. Nguyn l b tr ................................................................................................. 24 2.3. m cc hon v v t hp .................................................................................. 27

2.3.1. Chnh hp lp............................................................................................... 27 2.3.2. Chnh hp khng lp .................................................................................... 27 2.3.3. Hon v ........................................................................................................ 28 2.3.4. T hp.......................................................................................................... 28 2.3.5. T hp lp .................................................................................................... 30

2.4. H thc truy hi .................................................................................................. 31 2.4.1. nh ngha v v d ...................................................................................... 31 2.4.2. Gii cng thc truy hi tuyn tnh thun nht vi h s hng s ................... 34

2.5. Qui v cc bi ton n gin ............................................................................... 38 2.6. Phng php lit k ............................................................................................ 40

BI TP CHNG 2 ...................................................................................... 43

CHNG 3. BI TON LIT K.................................................................. 45

3.1- Gii thiu bi ton .............................................................................................. 45 3.2. Thut ton v phc tp tnh ton .................................................................... 46

3.2.1. V d v nh ngha ..................................................................................... 46

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3.2.2. Phng php biu din thut ton: ............................................................... 46 3.2.3. phc tp tnh ton ................................................................................... 48 3.2.4. Qui tc xc nh phc tp thut ton........................................................ 51

3.3. Phng php sinh ............................................................................................... 52 3.4. Thut ton quay lui (Back track) ........................................................................ 63 3.5. Nhng ni dung cn ghi nh ............................................................................... 69

BI TP CHNG 3 ...................................................................................... 70

CHNG 4. BI TON TI U ................................................................... 73

4.1. Gii thiu bi ton .............................................................................................. 73 4.2. Phng php duyt ton b ................................................................................. 76 4.3. Thut ton nhnh cn .......................................................................................... 79 4.4. K thut rt gn gii quyt bi ton ngi du lch............................................... 90

4.4.1.Th tc rt gn.............................................................................................. 91 4.4.2.Th tc chn cnh phn nhnh (r,c)............................................................... 94 4.4.3.Thut ton nhnh cn gii bi ton ngi du lch........................................... 99

4.5. Nhng im cn ghi nh ................................................................................... 100 BI TP CHNG 4 .................................................................................... 100

CHNG 5. BI TON TN TI ............................................................... 103

4.1. Gii thiu bi ton ............................................................................................ 103 5.2. Phng php phn chng .................................................................................. 106 5.3 Nguyn l Dirichlet............................................................................................ 107 5.4. Nhng ni dung cn ghi nh ............................................................................. 108

BI TP......................................................................................................... 109

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CHNG 1. LOGIC, TP HP V NG DNG

Ni dung chnh ca chng ny cp n nhng kin thc c bn v logic mnh , l thuyt tp hp v ng dng. Ni dung chnh ca chng bao gm: Logic mnh v ng dng. Logic v t v ng dng. L thuyt tp hp v ng dng. Mt s ng dng ca logic v tp hp trong tin hc. Bi tp chng 1. Bn c c th tm thy nhng kin thc su hn v chi tit hn trong cc ti liu [1]

v [2] ca ti liu tham kho.

1.1. Gii thiu chung

T hp l mt lnh vc quan trng ca ton hc ri rc cp ti nhiu vn khc nhau ca ton hc. L thuyt T hp nghin cu vic phn b cc phn t vo cc tp hp. Thng thng cc phn t ca tp hp l hu hn v vic phn b chng phi tho mn nhng iu kin nht nh no tu theo yu cu ca bi ton. Mi cch phn b c coi l mt cu hnh ca t hp. Cc cu hnh t hp c xem xt nh mt li gii ca bi ton m, bi ton lit k, bi ton tn ti hay bi ton ti u.

Bi ton m: y l dng bi ton nhm tr li cu hi c bao nhiu cu hnh tho mn iu kin nu?. Bi ton m c p dng c hiu qu vo nhng cng vic mang tnh cht nh gi nh xc sut xy ra ca mt s kin, thi gian tnh ton hay phc tp ca mt chng trnh my tnh.

Bi ton lit k: bi ton lit k quan tm n tt c cc cu hnh c th c c, v vy li gii ca n c biu din di dng thut ton vt cn tt c cc cu hnh. Bi ton lit k thng c lm nn cho nhiu bi ton khc. Hin nay, mt s bi ton tn ti, bi ton ti u, bi ton m vn cha c cch no gii quyt ngoi phng php lit k. Phng php lit k cng tr nn quan trng hn khi n c h tr bi cc h thng my tnh.

Bi ton ti u: khc vi bi ton lit k, bi ton ti u ch quan tm ti cu hnh tt nht theo mt ngha no . y l mt bi ton c nhiu ng dng thc tin c gii quyt bng l thuyt t hp.

Bi ton tn ti: nu nh bi ton m thc hin m bao nhiu cu hnh c th c, bi ton lit k xem xt tt c cc cu hnh c th c, bi ton ti u ch ra mt cu hnh tt nht. Bi ton tn ti hng n gii quyt nhng vn cn nghi vn. iu ny

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c ngha l ngay k c vn c hay khng mt cu hnh cng cha bit. Nhng bi ton ny thng l nhng bi ton kh. Do vy my tnh c xem l cng c hu hiu nht gii quyt bi ton tn ti.

1.2. Nhng kin thc c bn v Logic mnh

Cc qui tc c bn ca Logic cho ta ngha chnh xc ca cc mnh . Nhng qui tc ca logic chnh l cng c c s chng ta c th xy dng nn cc ngn ng lp trnh, cc bng mch my tnh, kim chng tnh ng n ca chng trnh v nhiu ng dng quan trng khc.

1.2.1. nh ngha & php ton

i tng nghin cu ca logic l cc mnh . Mt mnh c hiu l mt cu khng nh hoc ng hoc sai ch khng th va ng va sai. V d: Nhng cu khng nh sau y l mt mnh :

H ni l th ca Vit nam. 1 + 1 = 2 2 + 2 = 3

Cc mnh H ni l th ca vit nam, 1 +1 =2 l nhng mnh ng, mnh 2 +2 =3 l sai. Nhng nhng cu trong v d sau s khng phi l mt mnh v n nhng cu khng cho ta khng nh ng cng chng cho ta khng nh sai.

By gi l my gi ? Hy suy ngh iu ny cho k lng x +1 =2 x + y = z

Ta k hiu nhng ch ci A, B, C, D, p, q, r, s . . . l nhng mnh . Gi tr ca mt mnh ng c k hiu l T, gi tr mnh sai c k hiu l F. Tp gi tr T, F cn c gi l gi tr chn l ca mt mnh . nh ngha 1. Cho p l mt mnh . Php ph nh mnh p cng l mt mnh (k hiu l p hoc p). Mnh p c gi tr F khi v ch khi mnh p nhn gi tr T, nhn gi tr F khi v ch khi p nhn gi tr T. nh ngha 2. Cho p v q l hai mnh . Php hi gia mnh p vi mnh q l mt mnh (k hiu p q ). Mnh p q c gi tr T khi v ch khi p, q nhn gi tr T, c gi tr F khi v ch khi hoc p, q, hoc c hai nhn gi tr F.

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nh ngha 3. Cho p v q l hai mnh . Php tuyn gia mnh p vi mnh q l mt mnh (k hiu p p). Mnh p p c gi tr T khi v ch khi t nht mt trong hai mnh p, q nhn gi tr T, c gi tr F khi v ch khi c p, q u nhn gi tr F. nh ngha 4. Cho p v q l hai mnh . Php tuyn loi gia mnh p vi mnh q (c k hiu l pq) l mt mnh . Mnh pq ch ng khi mt trong p hoc q ng v sai trong cc trng hp khc cn li. nh ngha 5. Cho p v q l hai mnh . Php ko theo gia mnh p vi mnh q (k hiu p q) l mt mnh . Mnh p q nhn gi T khi v ch khi p v q nhn gi tr F hoc p v q cng nhn gi tr T. Mnh pq nhn gi tr F khi v ch khi p nhn gi tr T v q nhn gi tr F. nh ngha 6. Cho p v q l hai mnh . Php tng ng gia mnh p vi mnh q l mt mnh (k hiu p q). Mnh p q c gi tr ng khi p v q c cng gi tr chn l v sai trong cc trng hp khc cn li.

Cc php ton : , , , , , c th c nh ngha thng qua bng gi tr chn l sau:

Bng 1.1: Bng gi tr chn l ca cc php ton , , , , , p q p pq pq pq pq

p q

T T F T T F T T

T F F F T T F F

F T T F T T T F

F F T F F F T T

1.2.2. S tng ng gia cc mnh

Mt vn ht sc quan trng trong lp lun ton hc l vic thay th mt mnh bng mt mnh khc c cng gi tr chn l. Hai mnh c cng mt gi tr chn l chng ta c th hiu theo cch thng thng l chng tng ng nhau v ng ngha. Do vy, ta s tip cn v phn loi cc mnh phc hp thng qua cc gi tr chn l ca chng. nh ngha 7. Mt mnh phc hp lun lun ng vi bt k cc gi tr chn l ca cc mnh thnh phn c gi l hng ng (tautology). Mt mnh lun lun sai vi mi gi tr chn l ca cc mnh thnh phn c gi l mu thun.

V d: mnh phc hp p p l hng ng, p p l mu thun v gi tr chn l ca cc mnh trn lun lun ng, hoc lun lun sai nh c ch ra trong bng 1.2.

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Bng 1.2. V d v mnh hng ng & mnh mu thun p p p p p p

T F

F T

T T

F F

nh ngha 8. Hai mnh p, q c gi l tng ng logic vi nhau (k hiu : pq, hoc p q , hoc p=q) khi v ch khi cc ct cho gi tr chn l ca chng ging nhau. Hay mnh pq l hng ng.

V d 1. Hai mnh qp v qp l tng ng logic v cc ct gi tr chn l ca chng c th hin qua bng sau:

Bng 1.3. Bng gi tr chn l i vi qp v qp p q pq qp p q qp T T F F

T F T F

T T T F

F F F T

F F T T

F T F T

F F F T

Dng bng gi tr chn l chng minh tnh tng ng logic gia hai mnh phc hp cho ta mt phng php trc quan d hiu. Tuy nhin, vi nhng mnh logic phc hp c k mnh thnh phn th cn ti 2k t hp cc b gi tr chn l khc nhau. Do , dng bng chn l chng minh tnh tng ng logic gia hai mnh phc hp gp nhiu kh khn. Trong trng hp ny ta c th chng minh tnh tng logic bng vic thay th mt mnh phc hp bng nhng tng ng logic c trc.

Bng phng php bng chn l, d dng chng minh c s tng ng ca cc cng thc di y:

p q p q

pq (pq)(qp)

p p

Bng 1.4. Bng cc tng ng logic TNG NG TN GI

p T p Lut ng nht

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p F p

p T T

p F F

Lut nut

p p p

p p p

Lut lu ng

p p Lut ph nh kp

p q q p

p q q p

Lut giao hon

(p q) r p ( q r)

(p q) r p ( q r)

Lut kt hp

p ( q r) (p q ) (p r)

p ( q r) (p q) (p r)

Lut phn phi

qpqp

qpqp

Lut De Morgan

V d: Chng minh qpqpp ? Chng minh:

1.2.3. Dng chun tc

Cc cng thc (mnh ) tng ng c xem nh cc biu din khc nhau ca cng mt mnh . d dng vit cc chng trnh my tnh thao tc trn cc cng thc, chng ta cn chun ha cc cng thc, a chng v dng biu din chun c gi l dng chun hi. Mt cng thc c gi l dng chun hi nu n l hi ca cc

qp

FqpqpF

qppp

qppqpp

qppqpp

theo lut De Morgan th 2 theo lut De Morgan th 2 theo lut ph nh kp theo lut phn phi tng ng tin ch iu cn chng minh.

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mnh tuyn. Phng php bin i mt cng thc bt k v dng chun hi bng cch p dng cc th tc sau:

B cc php ko theo () bng cch thay (pq) bi qp .

Chuyn cc php ph nh ( ) vo st cc k hiu mnh bng cch p

dng lut De Morgan v thay p bi p.

p dng lut phn phi thay cc cng thc c dng (p(qr)) bi (pq)(pr).

V d. Ta chun ha cng thc srqp . Li gii.

sqprqpsrqpsrqp

srqpsrqp

Nh vy cng thc srqp c a v dng chun hi l sqprqp . 1.3. V t v lng t

Trong ton hc hay trong cc chng trnh my tnh chng ta rt hay gp nhng khng nh cha phi l mt mnh . Nhng khng nh u c lin quan n cc bin. Chng hn khng inh:

P(x) = x > 3 khng phi l mt mnh nhng ti nhng gi tr c th ca x=x0 no th P(x0) li l mt mnh . Hoc trong nhng on chng trnh gp cu lnh: if ( x > 3 ) then x:= x +1; th chng trnh s t gi tr c th ca bin x vo P(x), nu mnh P(x) cho gi tr ng x s c tng ln 1 bi cu lnh x:=x+1, P(x) c gi tr sai gi tr ca x c gi nguyn sau khi thc hin cu lnh if.

Chng ta c th phn tch mi khng nh thnh hai phn ch ng v v ng (hay v t), trong cu x ln hn 3 th x l ch ng, ln hn 3 l v ng. Hm P(x) c gi l hm mnh . Mt hm mnh c th c mt hoc nhiu bin. Gi tr chn l ca hm mnh ti nhng gi tr c th ca bin c xc nh nh nhng mnh thng thng.

V d. Cho Q(x, y, z) l hm mnh xc nh cu x2 = y2 +z2 hy xc nh gi tr chn l ca cc mnh Q (3, 2, 1), Q ( 5, 4, 3)?

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Li gii. t gi tr c th ca x , y , z vo Q(x,y,z) ta c : Q(3,2,1) l mnh 32 = 22 + 12 l sai do Q(3,2,1) l mnh sai. Trong ,

Q (5, 4, 3) l mnh 52 = 42 + 32 l mnh ng. Tng qut, gi s M l mt tp hp cc phn t no . M thng c gi l

trng hay min xc nh ca cc phn t thuc M. Khi , biu thc P(x) gi l v t xc nh trn trng M nu khi thay x bi mt phn t bt k ca trng M th P(x) s tr thnh mt mnh trn trng M.

Khi tt c cc bin ca hm mnh u c gn nhng gi tr c th, th mnh to ra s xc nh gi tr chn l. Tuy nhin, c mt phng php quan trng khc bin mt hm mnh thnh mt mnh m khng cn phi kim chng mi gi tr chn l ca hm mnh tng ng vi cc gi tr ca bin thuc trng ang xt. Phng php gi l s lng ho hay lng t. Chng ta xt hai lng t quan trng l lng t vi mi (k hiu :), lng t tn ti (k hiu : ).

nh ngha 1. Lng t vi mi ca P(x) k hiu l x P(x) l mt mnh P(x) ng vi mi phn t x thuc trng ang xt.

V d. Cho hm mnh P(x) = x2 + x + 41 l nguyn t. Xc nh gi tr chn l ca mnh P(x) vi x thuc khng gian bao gm cc s t nhin [0..39].

Li gii. V P(x) ng vi mi gi tr ca x [0..39] P(x) l ng. V d : Cho P(x) l hm mnh x + 1 > x. Xc nh gi tr chn l ca mnh

x P(x), trong khng gian cc s thc.

Li gii : V P(x) ng vi mi s thc x nn x P(x) l ng.

nh ngha 2. Lng t tn ti ca hm mnh P(x) (c k hiu l: x P(x) ) l mt mnh Tn ti mt phn t x trong khng gian sao cho P(x) l ng .

V d: Cho P(x) l hm mnh x > 3. Hy tm gi tr chn l ca mnh x P(x) trong khng gian cc s thc.

Li gii: V P(4) l 4 > 3 ng nn x P(x) l ng.

V d: Cho Q(x) l x + 1 > x. Hy tm gi tr chn l ca mnh x Q(x) trong khng gian cc s thc.

Li gii: v Q(x) sai vi mi x R nn mnh x Q(x) l sai.

Bng 1.5: Gi tr chn l ca lng t ,

x P(x) P(x) ng vi mi x C mt gi tr ca x P(x) sai

x P(x) C mt gi tr ca x P(x) ng P(x) sai vi mi x

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Dch nhng cu thng thng thnh biu thc logic: Dch mt cu c pht biu bng ngn ng t nhin (cu hi thng thng) thnh mt biu thc logic c vai tr ht sc quan trng trong xy dng cc ngn ng lp trnh, chng trnh dch v x l ngn ng t nhin. Qu trnh dch mt cu t ngn ng t nhin thnh mt biu thc s lm mt i tnh t nhin ca ngn ng v a s cc ngn ng u khng r rng, nhng mt biu thc logic li rt r rng cht ch t c php th hin n ng ngha ca cu. iu ny dn n phi c mt tp hp cc gi thit hp l da trn mt hm xc nh ng ngha cu cu . Mt khi cu c chuyn dch thnh biu thc logic, chng ta c th xc nh c gi tr chn l ca biu thc logic, thao tc trn biu thc logic, bin i tng ng trn biu thc logic. Chng ta s minh ho vic dch mt cu thng thng thnh biu thc logic thng qua nhng sau.

V d dch cu Bn khng c li xe my nu bn cao di 1.5 mt tr phi bn trn 18 tui thnh biu thc logic. Li gii.

Ta gi p l cu : Bn c li xe my. q l cu : Bn cao di 1.5m. r l cu : Bn trn 18 tui.

Khi : Cu hi trn c dch l: (q r ) p

V d: Dch cu Tt c cc sinh vin hc tin hc u hc mn ton hc ri rc Li gii: Gi P(x) l cu x cn hc mn ton hc ri rc v x c xc nh

trong khng gian ca cc sinh vin hc tin hc. Khi chng ta c th pht biu:x P(x). V d: Dch cu C mt sinh vin lp ny t nht tt c cc phng ca t

nht mt nh trong k tc x. Li gii : Gi tp sinh vin trong lp l khng gian xc nh sinh vin x, tp cc

nh trong k tc x l khng gian xc nh cn nh y, tp cc phng l khng gian xc nh phng z. Ta gi P(z,y) l z thuc y, Q(x,z) l x z. Khi ta c th pht biu :

x y z (P(z,y) Q(x,z));

1.4. Mt s ng dng trn my tnh

Cc php ton bt: Cc h thng my tnh thng dng cc bit (binary digit) biu din thng tin. Mt bt c hai gi tr chn l hoc 0 hoc 1. V gi tr chn l ca mt biu thc logic cng c hai gi tr hoc ng (T) hoc sai (F). Nu ta coi gi tr ng c gi tr 1 v gi tr sai l 0 th cc php ton vi cc bt trong my tnh c tng ng vi cc lin t logic.

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Mt xu bt (hoc xu nh phn) l dy khng hoc nhiu bt. Chiu di ca xu l s cc bt trong xu . V d xu nh 101010011 c di l 9. Mt s nguyn uc biu din nh mt xu nh phn c di 16 bt.

Cc php ton vi bt c xy dng trn cc xu bt c cng di, bao gm : AND bt (php v cp bt), OR (php hoc cp bt), XOR (php tuyn loi tr cp bt). V d: cho hai xu bt 01101 10110 v 11000 11101 hy tm xu AND bt, OR bt, XOR bt.

Bng 1.5. Cc php ton cp bt ng dng trong ngn ng LT

Gi tr ca A Gi tr ca B A and B A or B A xor B

A = 13 =1100 B = 8=1000 1000 1101 0101

Thut ton cc php tnh s nguyn: Cc thut ton thc hin cc php tnh vi cc s nguyn khi dng khai trin nh phn l ht sc quan trng trong b x l s hc ca my tnh. Nh chng ta bit, thc cht cc s nguyn c biu din trong my tnh l cc xu bt nh phn, do vy chng ta c th s dng biu din nh phn ca cc s thc hin cc php tnh. Gi s khai trin nh phn ca cc s nguyn a v b tng ng l: a = (an-1an-2 . . .a1a0)2 , b = (bn-1bn-2 . . .b1b0)2 . Khai trin ca a v b c ng n bt ( chp nhn nhng bt 0 u lm c n bt). Xt bi ton cng hai s nguyn vit dng nh phn. Th tc thc hin vic cng cng ging nh lm trn giy thng thng. Phng php ny tin hnh bng cch cng cc bt nh phn tng ng c nh tnh tng hai s nguyn. Sau y l m t chi tit cho qu trnh cng hai xu bt nh phn.

cng a vi b, trc ht ta cng hai bt phi nht, ngha l: a0 + b0 = c0*2 + s0; trong s0 l bt phi nht ca s nguyn tng a + b, c0 l s cn nh n c th bng 0 hoc 1. Sau ta cng hai bt tip theo v s nh: a1 + b1 + c0 = c1*2 + s1; s1 l bt tip theo ca s a + b, c1 l s nh. Tip tc qu trnh ny bng cch cng cc bt tng ng trong khai trin nh phn v s nh, giai on cui cng : an-1 + bn-1 + cn-2 = cn-1 * 2 + sn-1. Bt cui cng ca tng l cn-1. Khi khai trin nh phn ca tng a + b l (snan-1 . . .s1s0)2. V d: cng a =(1110)2, b = (1011)2 Li gii:

Trc ht ly:

a0 + b0 = 0 + 1 = 0 * 2 + 1 c0=0, s0 = 1 Tip tc:

a1 + b1 + c0 = 1 + 1 + 0 = 1 * 2 + 0 c1=1, s1 = 0

a2 + b2 + c1 = 1 + 0 + 1 = 1 * 2 + 0 c2=1, s2 = 0

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a3 + b3 + c2 = 1 + 1 + 1 = 1 * 2 + 1 c3=1, s3 = 1 Cui cng:

s4 = c3 = 1 a + b = (11001)2 Thut ton cng:

void Cong(a , b: positive integer) {

/*a = (an-1an-2 . . .a1a0)2 , b = (bn-1bn-2 . . .b1b0)2 */ c=0; for (j=0 ; j n-1; j++) { d= [( aj + bj + c)/ 2];

sj = aj + bj + c 2d; c = d;

} sn = c; khai trin nh phn ca tng l (snan-1 . . .s1s0)2;

} Thut ton nhn: nhn hai s nguyn n bt a, b ta bt u t vic phn tch: a = (an-1an-2. . .a1a0), b = (bn-1bn-2. . .b1b0)

Ta c th tnh a.b t phng trnh trn. Trc ht, ta nhn thy abj = a nu bj=1, abj=0 nu bj=0. Mi ln tnh ta nhn vi 2j hay dch chuyn sang tri j bt 0 bng cch thm j bt 0 vo bn tri kt qu nhn c. Cui cng, cng n s nguyn abj 2j (j=0..n-1) ta nhn c a.b. V d sau y s minh ho cho thut ton nhn:

V d: Tm tch ca a = (110)2, b= (101)2 Gii: Ta nhn thy

ab020 = (110)2*1*20 = (110)2 ab121 = (110)2*0*21 = (0000)2 ab222 = (110)2*1*22 = (11000)2 S dng thut ton tnh tng hai s nguyn a, b c biu din n bt ta nhn c(ta c th thm s 0 vo u mi ton hng): (0 110)2 + (0000)2 = (0110)2 ;

(0 0110)2 + (11000)2 = (11110)2 = ab. Thut ton nhn hai s nguyn n bt c th c m phng nh sau:

void Nhan( a, b : Positive integer){ /* khai trin nh phn tng ng ca a = (an-1an-2. . .a1a0),

b = (bn-1bn-2. . .b1b0) */ for (j=0; j n-1; j++) { if ( ( bj==1)

PTIT

15

cj = a * 2j; /* a c dch tri j bt 0 */ else cj =0; } /*c0, c1.., cn-1 l nhng tch ring ca abj 2j(j=0..n-1 */ p=0; for ( j=0 ; j n-1; j++) p= p + cj; /* p l gi tr ca tch ab */

}

1.5. Nhng kin thc c bn v l thuyt tp hp

1.5.1. Khi nim & nh ngha

Cc tp hp dng nhm cc i tng li vi nhau. Thng thng, cc i tng trong tp hp c cc tnh cht tng t nhau. V d, tt c sinh vin mi nhp trng to nn mt tp hp, tt c sinh vin thuc khoa Cng ngh thng tin l mt tp hp, cc s t nhin, cc s thc cng to nn cc tp hp. Ch rng, thut ng i tng c dng y khng ch r c th mt i tng no, s m t mt tp hp no hon ton mang tnh trc gic v cc i tng. nh ngha 1. Tp cc i tng trong mt tp hp c gi l cc phn t ca tp hp. Cc tp hp thng c k hiu bi nhng ch ci in hoa m nh A, B, X, Y . . ., cc phn t thuc tp hp hay c k hiu bi cc ch ci in thng nh a, b, c, u, v . . . ch a l phn t ca tp hp A ta vit a A, tri li nu a khng thuc A ta vit a A.

Tp hp khng cha bt k mt phn t no c gi l tp rng (k hiu l hoc { }). Tp hp A c gi l bng tp hp B khi v ch khi chng c cng chung cc phn t v c k hiu l A=B. V d tp A={ 1, 3, 5 } s bng tp B = { 3, 5, 1 }.

nh ngha 2. Tp A c gi l mt tp con ca tp hp B v k hiu l AB khi v ch khi mi phn t ca A l mt phn t ca B. Hay A B khi v ch khi lng t

x (x A x B) cho ta gi tr ng. T nh ngha trn chng ta rt ra mt s h qu sau:

Tp rng l tp con ca mi tp hp. Mi tp hp l tp con ca chnh n.

Nu A B v B A th A=B hay mnh :

x (x A xB ) x (xB x A) cho ta gi tr ng.

Nu A B v AB th ta ni A l tp con thc s ca B v k hiu l AB.

PTIT

16

nh ngha 3. Cho S l mt tp hp. Nu S c chnh xc n phn t phn bit trong S, vi n l s nguyn khng m th ta ni S l mt tp hu hn v n c gi l bn s ca S. Bn s ca S c k hiu l |S | hay N(S). nh ngha 4. Cho tp hp S. Tp lu tha ca S k hiu l P(S) l tp tt c cc tp con ca S.

V d S = { 0, 1, 2 } P(S) ={ , {0}, {1}, {2}, {0,1}, {0, 2}, {1, 2} {0, 1, 2}}. nh ngha 5. Dy sp th t (a1, a2,.., an) l mt tp hp sp th t c a1 l phn t th nht, a2 l phn t th 2, .., an l phn t th n. Chng ta ni hai dy sp th t l bng nhau khi v ch khi cc phn t tng ng ca chng l bng nhau. Ni cch khc (a1, a2,.., an) bng (b1, b2,.., bn) khi v ch khi ai = bi vi mi i =1, 2, ..n.

nh ngha 6. Cho A v B l hai tp hp. Tch cc ca A v B c k hiu l AB, l tp hp ca tt c cc cp (a,b) vi aA, b B. Hay c th biu din bng biu thc:

A B = { (a, b) | a A, b B }

nh ngha 7. Tch cc ca cc tp A1, A2, . ., An c k hiu l A1A2..An l tp hp ca dy sp th t (a1, a2,.., an) trong aiAi vi i = 1, 2,..n. Ni cch khc:

A1A2..An = { (a1, a2,.., an) | aiAi vi i = 1, 2,..n }

1.5.2. Cc php ton trn tp hp

Cc tp hp c th c t hp vi nhau theo nhiu cch khc nhau thng qua cc php ton trn tp hp. Cc php ton trn tp hp bao gm: Php hp (Union), php giao (Intersection), php tr (Minus).

nh ngha 1. Cho A v B l hai tp hp. Hp ca A v B c k hiu l AB, l tp cha tt c cc phn t hoc thuc tp hp A hoc thuc tp hp B. Ni cch khc:

AB = { x | x A x B }

nh ngha 2. Cho A v B l hai tp hp. Giao ca A v B c k hiu l AB, l tp cha tt c cc phn t thuc A v thuc B. Ni cch khc:

AB = { x | x A x B } nh ngha 3. Hai tp hp A v B c gi l ri nhau nu giao ca chng l tp rng (AB = ). nh ngha 4. Cho A v B l hai tp hp. Hiu ca A v B l tp hp uc k hiu l A-B, c cc phn t thuc tp hp A nhng khng thuc tp hp B. Hiu ca A v B cn c gi l phn b ca B i vi A. Ni cch khc:

A B = { x | x A x B }

nh ngha 5. Cho tp hp A. Ta gi A l phn b ca A l mt tp hp bao gm nhng phn t khng thuc A.

PTIT

17

AxxA | nh ngha 6. Cho cc tp hp A1, A2, . ., An. Hp ca cc tp hp l tp hp cha tt c cc phn t thuc t nht mt trong s cc tp hp Ai ( i=1, 2, . ., n). K hiu:

nn

i 211

nh ngha 7: Cho cc tp hp A1, A2, . ., An. Giao ca cc tp hp l tp hp cha cc phn t thuc tt c n tp hp Ai ( i=1, 2, . ., n).

n

iAnAAAi

1..21

1.5.3. Cc hng ng thc trn tp hp

Mi tp con ca tp hp tng ng vi mt tnh cht xc nh trn tp hp cho c gi l mnh . Vi tng ng ny, cc php ton trn tp hp c chuyn sang cc php ton ca logic mnh :

Ph nh ca A, k hiu A (hay NOT A) tng ng vi phn b A .

Tuyn ca A v B, k hiu A B (hay A or B) tng ng vi A B.

Hi ca A v B, k hiu A B (hay A and B) tng ng vi A B. Cc mnh cng vi cc php ton trn n lp thnh mt i s mnh (hay i s

logic). Nh th, i s tp hp v i s logic l hai i s ng cu vi nhau (nhng mnh pht biu trn i s logic tng ng vi mnh pht biu trn i s tp hp). Vi nhng trng hp c th, tu theo tnh hung, mt bi ton c th c pht biu bng ngn ng ca i s logic hay ngn ng ca i s tp hp. Bng 1.6 th hin mt s hng ng thc ca i s tp hp.

Bng 1.6. Mt s hng ng thc trn tp hp HNG NG THC TN GI

A = A

A U = A (U l tp v tr)

Lut ng nht

A U = U

A = A

Lut nut

AA = A

A A = A

Lut lu ng

PTIT

18

A = A Lut b

A B = B A

A B = B A

Lut giao hon

A (B C) = (A B)C

A (B C) = (AB) C

Lut kt hp

A (B C) = (A B) (A C )

A (B C) = (A B) (A C)

Lut phn phi

BABA

BABA

Lut De Morgan

1.6. Biu din tp hp trn my tnh

C nhiu cch khc nhau biu din tp hp trn my tnh, phng php ph bin l lu tr cc phn t ca tp hp khng sp th t. Vi vic lu tr bng phng php ny, ngoi nhng lng ph b nh khng cn thit, th qu trnh tnh hp, giao, hiu cc tp hp gp nhiu kh khn v mt nhiu thi gian v mi php tnh i hi nhiu thao tc tm kim trn cc phn t. Mt phng php lu tr cc phn t bng cch biu din c th t ca cc phn t ca mt tp v tr t ra hiu qu hn rt nhiu trong qu trnh tnh ton.

Gi s tp v tr U l hu hn gm n phn t(hu hn c hiu theo ngha cc phn t ca U lu tr c trong b nh my tnh). Gi s ta mun biu din tp hp A U. Trc ht ta chn mt th t tu no i vi cc phn t ca tp v tr U, gi s ta c b c th t a1,a2, . ., an. Sau xy dng mt xu bt nh phn c di n, sao cho nu bt th i c gi tr 1 th phn t aiA, nu ai =0 th aiA (i=1,2..,n). V d sau s minh ho k thut biu din tp hp bng xu bt nh phn.

V d. Gi s U = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }. Hy biu din tp hp A U l

1- Tp cc s nguyn l A U.

2- Tp cc s nguyn chn BU.

3- Tp cc s nguyn nh hn 5 C U.

4- Tm A B

5- Tm AC . . . Li gii. Trc ht ta coi th t cc phn t c sp xp theo th t tng dn tc

ai=i (i=1,2,..,10). Khi :

PTIT

19

1. Xu bt biu din cc s l trong U ( {1, 3, 5, 7, 9 } ) l xu c di n = 10 trong cc bt v tr th 1, 3, 5, 7, 9 c gi tr l 1, cc bt cn li c gi tr l 0. T ta c xu bt biu din tp hp A l: 1 0 1 0 1 0 1 0 1 0.

2. Xu bt biu din cc s chn trong U ( {2, 4, 6, 8, 10 } ) l xu c di n = 10 trong cc bt v tr th 2, 4, 6, 8, 10 c gi tr l 1, cc bt cn li c gi tr l 0. T ta c xu bt biu din tp hp B l: 0 1 0 1 0 1 0 1 0 1.

3. Xu bt biu din cc s nh hn 5 trong U ( {1, 2, 3, 4 } ) l xu c di n = 10 trong cc bt v tr th 1, 2, 3, 4 c gi tr l 1, cc bt cn li c gi tr l 0. T ta c xu bt biu din tp hp C l: 1 1 1 1 0 0 0 0 0 0.

4. Xu bt biu din tp hp A B l : (1 0 1 0 1 0 1 0 1 0 0 1 0 1 0 1 0 1 0 1) l xu 1 1 1 1 1 1 1 1 1 1. Nh vy, A B = U.

5. Tng t nh vy vi A C (1 0 1 0 1 0 1 0 1 0 1 1 1 1 0 0 0 0 0 0) l xu 1 0 1 0 0 0 0 0 0 0. Nh vy A C = { 1, 3 }

1.7. Nhng ni dung cn ghi nh

Cn hiu v nm vng c nhng ni dung sau: Cc php ton hi, tuyn, tuyn loi, suy ra, ko theo ca logic mnh . Cc phng php chng minh nh l dng bng chn l v cc tng ng

locgic. Phng php biu din cc cu hi thng thng bng logic v t. nh ngha v cc php ton trn tp hp. Phng php biu din tp hp trn my tnh

BI TP CHNG 1

1. Lp bng gi tr chn l cho cc mnh phc hp sau:

a) (p q) (qp) b) (p q) (q p)

c) (p q) (p q ) d) (p q) (p q )

e) (p q) (p q ) f) ( p q ) (pq)

g) ( p q) r h) (p q) r

i) (p q) ( q r) j) ( pq ) (qr)

PTIT

20

2- Dng bng chn l chng minh: a) Lut giao hon

p q q p

p q q p b) Lut kt hp

(p q) r p ( q r)

( p q) r p (q r) c) Lut phn phi

p (q r) (p q) (p r) 3- Chng minh cc cng thc sau y l ng nht ng bng cch lp bng gi tr chn

l:

a) ( X(YZ)) ((X Y)(XZ));

b) (XY)((XZ)(X(YZ)));

c) (XZ) ((YZ)((XY)Z)). 4. Chng minh cc cng thc sau y l tng ng logic

nn

nn

nn

nn

XXXXXXd

XXXXXXc

YXYXYXYYYXbYXYXYXYYYXa

2121

1121

2121

2121

)

()

)(...)()(...())(...)()(...()

5. Cho A, B, C l cc tp hp. Chng minh rng:

ABABAg

BABAf

ACBACABeBCCAd

CACBAcBACBAb

CBACBAa

()()

)

)()()())()()

)()())()()

)

PTIT

21

CHNG 2. BI TON M

m cc i tng c nhng tnh cht no l mt bi ton quan trng ca l thuyt t hp. Gii quyt tt bi ton m gip ta gii nhiu bi ton khc nhau trong nh gi phc tp tnh ton ca cc thut ton v tm xc sut ri rc cc bin c. Phng php chung gii bi ton m c da trn cc nguyn l m c bn (nguyn l cng, nguyn l nhn). Mt s bi ton m phc tp hn c gii bng phng php qui v cc bi ton con, xy dng cng thc truy hi hoc phng php hm sinh. Ni dung chnh c cp trong chng ny bao gm:

Cc nguyn l m c bn Nguyn l b tr Hon v v t hp H thc truy hi Qui v cc bi ton con Gii thiu bi ton tn ti

Bn c c th tm hiu nhiu k thut m cao cp hn trong ti liu [1], [2] trong phn tham kho ca ti liu ny.

2.1. Nhng nguyn l m c bn

2.1.1. Nguyn l cng

Gi s c hai cng vic. Vic th nht c th tin hnh bng n1 cch, vic th hai c th tin hnh bng n2 cch v nu hai vic ny khng th tin hnh ng thi. Khi s c n1 + n2 cch gii gii quyt mt trong hai vic trn.

Chng ta c th m rng qui tc cng cho trng hp nhiu hn hai cng vic. Gi s cc vic T1, T2,.., Tm c th lm tng ng bng n1, n2, .., nm cch v gi s khng c hai vic Ti, Tj no lm vic ng thi (i,j = 1, 2, .., m ; i j ). Khi , c n1 + n2 +.. +nm cch thc hin mt trong cc cng vic T1, T2, . ., Tm.

Qui tc cng c pht biu di dng ca ngn ng tp hp nh sau:

Nu A v B l hai tp ri nhau (A B = ) th : N(AB) = N(A) + N(B). Nu A1, A2, .., An l nhng tp hp ri nhau th:

N(A1 A2 . . An ) = N(A1) + N(A2) +..+ N(An). V d 1. Gi s cn chn hoc mt cn b hoc mt sinh vin tham gia mt hi

ng ca mt trng i hc. Hi c bao nhiu cch chn v i biu ny nu nh c 37 cn b v 63 sinh vin.

PTIT

22

Li gii: Gi vic th nht l chn mt cn b t tp cn b ta c 37 cch. Gi vic th hai l chn mt sinh vin t tp sinh vin ta c 63 cch. V tp cn b v tp sinh vin l ri nhau, theo nguyn l cng ta c tng s cch chn v i biu ny l 37 + 63 = 100 cch chn.

V d 2. mt on vn ng vin gm mn bn sng v bi c c i thi u nc ngoi. S vn ng vin nam l 10 ngi. S vn ng vin thi bn sng k c nam v n l 14 ngi. S n vn ng vin thi bi bng s vn ng vin nam thi bn sng. Hi on c bao nhiu ngi.

Li gii. Chia on thnh hai tp, tp cc vn ng vin nam v tp cc vn ng vin n. Ta nhn thy tp n li c chia thnh hai: thi bn sng v thi bi. Thay s n thi bi bng s nam thi bn sng , ta c s n bng tng s vn ng vin thi bn sng. T theo nguyn l cng ton on c 14 + 10 = 24 ngi.

V d 3. gi tr ca bin k s bng bao nhiu sau khi thc hin on chng trnh sau :

k := 0 for i1:= 1to n1 k:=k+1 for i2:= 1to n2 k:=k+1 . . . . . . . . . . for im:= 1 to nm k:=k+1

Li gii. coi mi vng for l mt cng vic, do ta c m cng vic T1, T2, . ., Tm. Trong Ti thc hin bi ni cch (i= 1, 2, .., m). V cc vng for khng lng nhau hay cc cng vic khng thc hin ng thi nn theo nguyn l cng tng tt c cc cch hon thnh T1, T2,.., Tm l k= n1 + n2 +.. + nm.

2.1.2. Nguyn l nhn

Gi s mt nhim v no c tch ra hai cng vic. Vic th nht c thc hin bng n1 cch, vic th hai c thc hin bng n2 cch sau khi vic th nht c lm, khi s c n1.n2 cch thc hin nhim v ny. Nguyn l nhn c th c pht biu tng qut bng ngn ng tp hp nh sau:

Nu A1, A2, .., Am l nhng tp hp hu hn, khi s phn t ca tch cc cc tp ny bng tch s cc phn t ca mi tp thnh phn. Hay ng thc:

N (A1 A2 .. Am ) = N (A1) N (A2) . . . N (Am). Nu A1 = A2 =. . Am th N(Ak) = N(A)k

PTIT

23

V d 1. gi tr ca k s bng bao nhiu sau khi ta thc hin on chng trnh sau:

k:=0 for i1 = 1 to n1 for i2 = 1 to n2 . . for in =1 to nm k:=k +1

Li gii. Gi tr khi to k=0. Mi vng lp kng nhau i qua gi tr ca k c tng ln 1 n v. Gi Ti l vic thi hnh vng lp th i. Khi , s ln vng lp l s cch thc hin cng vic. S cch thc hin cng vic Tj l nj (j=1,2, . ., n). Theo qui tc nhn ta vng lp kp c duyt qua n1 +n2 +..+nm ln v chnh l gi tr ca k.

V d 2. Ngi ta c th ghi nhn cho nhng chic gh ca mt ging ng bng mt ch ci v sau l mt s nguyn nh hn 100. Bng cch nh vy hi c nhiu nht bao nhiu chic gh c th ghi nhn khc nhau.

Li gii: c nhiu nht l 26 x 100 = 2600 gh c ghi nhn. V k t gn nhn u tin l mt ch ci vy c 26 cch chn cc ch ci khc nhau ghi k t u tin, tip theo sau l mt s nguyn dng nh hn 100 do vy c 100 cch chn cc s nguyn gn tip sau ca mt nhn. Theo qui tc nhn ta nhn c 26 x 100 = 2600 nhn khc nhau.

V d 3. C bao nhiu xu nh phn c di 7? Li gii. mt xu nh phn c di 7 gm 7 bt, mi bt c hai cch chn (hoc

gi tr 0 hoc gi tr 1), theo qui tc nhn ta c 2.2.2.2.2.2.2 = 27 = 128 xu bt n phn di 7.

V d 4. C bao nhiu hm n nh xc nh t mt tp A c m phn t nhn gi tr trn tp B c n phn t?

Li gii. Trc tin ta nhn thy, nu m >n th tn ti t nht hai phn t khc nhau ca A cng nhn mt gi tr trn B, nh vy vi m>n th s cc hm n nh t AB l 0. Nu m

24

gm 3 ch s v nhm m my gm 4 ch s. V nhng nguyn nhn k thut nn c mt s hn ch i vi mt s con s. Ta gi s, X biu th mt s c th nhn cc gi tr t 0..9, N l s c th nhn cc ch s t 2..9, Y l cc s c th nhn cc ch s 0 hoc 1. Hi theo hai d n nh s NYX NNX XXXX v NXX NXX XXXX c bao nhiu s in thoi c nh s khc nhau Bc M?

Li gii: nh s theo d n NYX NNX XXXX c nhiu nht l : 8 x 2 x 10 x 8 x 8 x10 x10 x10 x 10 x 10 x10 = 2 x 83 x 106 = 1 024. 106

nh s theo d n NXX NXX XXXX c nhiu nht l : 8 x 10 x 10 x 8 x 10 x10 x10 x10 x 10 x 10 x10 = 82 x 108 = 64. 108 V d 6. Dng qui tc nhn hy ch ra rng s tp con ca mt tp S hu hn l

2N(S). Li gii. ta lit k cc phn t ca tp S l s1, s2, .., sN(S). Xy dng mt xu bt nh

phn di N(S) bt, trong nu bt th i c gi tr 0 th phn t si S, nu bt th i c gi tr 1 th phn t siS (i=1, 2, .., N(S) ). Nh vy, theo nguyn l nhn, s tp con ca tp hp S chnh l s xu bt nh phn c di N(S). Theo v d 3, chng ta c 2N(S) xu bt nh phn di N(S).

2.2. Nguyn l b tr

Trong mt s bi ton m phc tp hn. Nu khng c gi thit g v s ri nhau gia hai tp A v B th N(AB) = N(A) + N(B) N(AB).

V d 1. lp ton hc ri rc c 25 sinh vin gii tin hc, 13 sinh vin gii ton v 8 sinh vin gii c ton v tin hc. Hi lp c bao nhiu sinh vin nu mi sinh vin hoc gii ton hoc hc gii tin hc hoc gii c hai mn?

Li gii. Gi A tp l tp cc sinh vin gii Tin hc, B l tp cc sinh vin gii ton. Khi A B l tp sinh vin gii c ton hc v tin hc. V mi sinh vin trong lp hoc gii ton, hoc gii tin hc hoc gii c hai nn ta c tng s sinh vin trong lp l N(AB). Do vy ta c:

N(AB) = N(A) + N(B) N(AB) = 25 + 13 8 = 30. V d 2. C bao nhiu s nguyn khng ln hn 1000 chia ht cho 7 hoc 11. Li gii. Gi A l tp cc s nguyn khng ln hn 1000 chia ht cho 7, B l tp

cc s nguyn khng ln hn 1000 chia ht cho 11. Khi tp s nguyn khng ln hn 1000 hoc chia ht cho 7 hoc chia ht cho 11 l N(AB). Theo cng thc 1 ta c:

N(AB) = N(A) + N(B) N(AB) = 1000/7+ 1000/11 - 1000/7.11 = 142 + 90 12 = 220.

Trc khi a ra cng thc tng qut cho n tp hp hu hn. Chng ta a ra cng thc tnh s phn t ca hp 3 tp A, B, C.

PTIT

25

Ta nhn thy N(A) + N(B) + N(C) m mt ln nhng phn t ch thuc mt trong ba tp hp. Nh vy, s phn t ca A B, AC, BC c m hai ln v bng N(AB), N(AC), N(BC), c m ba ln l nhng phn t thuc ABC. Nh vy, biu thc:

N(ABC) N(AB)- N(AC) N(BC) ch m cc phn t ch thuc mt trong ba tp hp v loi b i nhng phn t

c m hai ln. Nh vy, s phn t c m ba ln cha c m, nn ta phi cng thm vi giao ca c ba tp hp. T ta c cng thc i vi 3 tp khng ri nhau:

N(ABC)= N(A)+N(B)+N(C)N(AB)N(AC)N(BC)+N(ABC) nh l. Nguyn l b tr. Gi s A1, A2, . ., Am l nhng tp hu hn. Khi

N(A1A2 . . .Am) = N1- N2 + . . +(-1)m-1Nm, trong Nk l tng phn t ca tt c cc giao ca k tp ly t m tp cho. (ni

ring N1=N(A1) + N(A2) + . .+ N(Am), Nm = N(A1 A2 . . .Am ) . Ni cch khc:

)..()1(...

()()()...(

211

1 ,1 121

nn

kni nji nkji

jijiin

AAAN

AAANAANANAAAN

nh l c chng minh bng cch ch ra mi phn t ca hp n tp hp c m ng mt ln. Bn c c th tham kho cch chng minh trong ti liu [1].

V d 3. Tm cng thc tnh s phn t ca 4 tp hp. Li gii. T nguyn l b tr ta c

N(A1A2A3A4) = N(A1) + N(A2) + N(A3) + N(A4) N(A1A2)

N(A1A3) N(A1A4) N(A2A3) N(A2A4) N(A3A4) + N(A1A2A3) + N(A1A2A4) + N(A1A3A4 + N(A2A3A4) N(A1A2A3A4).

V d 4. Hi trong tp X = { 1, 2, . ., 10000} c bao nhiu s khng chia ht cho bt c s no trong cc s 3, 4, 7?

Li gii. Gi A l tp cc s nh hn 10000 chia ht cho 3, B l tp cc s nh hn 10000 chia ht cho 4, C l tp cc s nh hn 10000 chia ht cho 7. Theo nguyn l b tr ta c:

N(A BC)= N(A)+N(B) + N(C) N(AB N(AC) N(BC) + N(ABC) trong :

N(A) + N(B) + N (C) = [10 000/3] + [10 000/4] + [10 000/7] = 3333 + 2500 + 1428 = 7261

N(AB) = N(A) + N(B) N(AB) = 3333 + 2500 [10000/3x4] = 833

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26

N(AC) = N(A) + N(C) N(AC) = 3333 + 1428 [10000/3x7] = 476

N(BC) = N(B) + N(C) N(BC) = 2500 + 1428 [10000/4x7] = 357

N(AB) + N(AC) + N(BC) = 833 + 476 + 357 = 1666

N(ABC) = [10000/3x4x7] = 119. =>S cc s nh hn 10000 cn m l :

1000 - N(ABC) = 7261 1666 + 119 = 4286. V d 5. C bao nhiu xu nh phn di 10 bt u bi 00 hoc kt thc bi 11. Li gii. Gi A l s xu nh phn di 10 bt u bi 00 , B l s xu nh phn

di 10 kt thc bi 11. D rng nhn thy, N(A) = N(B) = 256, N(AB) = 26 = 64. Theo nguyn l b tr ta c:

N(AB) = N(A) + N(B) N(AB) = 256 + 256 64 = 448.

V d 6. Bi ton b th. C n l th v n phong b ghi sn a ch. B ngu nhin cc l th vo cc phong b. Hi xc sut xy ra khng mt l th no b ng a ch l bao nhiu?

Li gii: C tt c n! cch b th. Vn t ra l m s cch b th sao cho khng l th no ng a ch. Gi X l tp hp tt c cc cch b th v Ak l tnh cht l th k b ng a ch. Khi theo nguyn l b tr ta c:

nn NNNNN )1(...21

Trong N l s cn tm, N = n!, Nk l s tt c cc cch b th sao cho c k l th ng a ch. Nhn xt rng, Nk l mi cch ly k l th t n l, vi mi cch ly k l th, c (n-k )! cch b k l th ny ng a ch, t ta nhn c.

!!)!)(,(

knknknCN k v !

)1(!2

1!1

11(!n

nNn

T ta c xc xut cn tm l:

1!)1(

!21

!111 e

n

n

S c tnh nh trn c gi l s mt th t v c k hiu l Dn. Di y l mt vi gi tr ca Dn, s tng nhanh ca Dn mt ln na cho ta thy r s bng n t hp.

N 2 3 4 5 6 7 8 9 10 11

Dn 1 2 9 44 265 1845 14833 133496 1334961 4890741

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27

2.3. m cc hon v v t hp

2.3.1. Chnh hp lp

nh ngha 1. Mt chnh hp lp chp k ca n phn t l b c th t gm k thnh phn ly t n phn t ca tp cho.

Nh vy, mt chnh hp lp chp k ca n phn t c th xem l phn t ca tch cc Ak vi A l tp cho. Theo nguyn l nhn, s cc tt c cc chnh hp lp chp k ca n s l nk.

V d 1. Tnh s hm t tp c k phn t vo tp c n phn t? Li gii: Biu din mi hm bng mt b k thnh phn, trong thnh phn th i

l nh ca phn t th i (1

28

Li gii: S hm n nh t tp k phn t sang tp n phn t chnh l P(n,k). V d 2. Gi s c tm vn ng vin chy thi. Ngi v nht s c nhn hun

chng vng, ngi v nh nhn hun chng bc, ngi v ba nhn huy chng ng. Hi c bao nhiu cch trao huy chng nu tt c cc kt cc u c th xy ra?

Li gii. S cch trao huy chng chnh l s chnh hp chp 3 ca tp hp 8 phn t. V th c P(8,3) = 8.7.6 = 336 cch trao huy chng.

V d 3. C bao nhiu cch chn 4 cu th khc nhau trong i bng gm 10 cu th tham gia cc trn u n.

Li gii. C P(10,4) = 10.9.8.7 = 5040 cch chn.

2.3.3. Hon v

nh ngha 3. Ta gi cc hon v ca n phn t l mt cch xp c th t cc phn t . S cc hon v ca tp n phn t c th coi l trng hp ring ca chnh hp khng lp vi k = n.

Ta cng c th ng nht mt hon v vi mt song nh t tp n phn t ln chnh n. Nh vy, s hon v ca tp gm n phn t l P(n, n) = n!.

V d 1. C 6 ngi xp thnh hng chp nh. Hi c th b tr chp c bao nhiu kiu khc nhau?

Li gii. Mi kiu nh l mt hon v ca 6 ngi. Do c 6! = 720 kiu nh khc nhau c th chp.

V d 2. Cn b tr thc hin n chng trnh trn mt my tnh. Hi c bao nhiu cch b tr khc nhau?

Li gii. S chng trnh c nh s t 1, 2, . ., n. Nh vy, s chng trnh cn thc hin trn mt my tnh l s hon v ca 1, 2, .., n.

V d 3. Mt thng nhn i bn hng ti tm thnh ph. Ch ta c th bt u hnh trnh ca mnh ti mt thnh ph no nhng phi qua 7 thnh ph kia theo bt k th t no m ch ta mun. Hi c bao nhiu l trnh khc nhau m ch ta c th i?

Li gii. V thnh ph xut pht c xc nh. Do vy thng nhn c th chn tu 7 thnh ph cn li hnh trnh. Nh vy, tt c s hnh trnh ca thng nhn c th i qua l 7! = 5040 cch.

2.3.4. T hp

nh ngha 4. Mt t hp chp k ca n phn t l mt b khng k th t gm k thnh phn khc nhau ly t n phn t cho. Ni cch khc, ta c th coi mt t hp chp k ca n phn t l mt tp con k phn t ly trong n phn t. S t hp chp k ca n phn t k hiu l C(n,k).

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29

Ta c th tnh c trc tip s cc t hp chp k ca tp n phn t thng qua chnh hp khng lp ca k phn t. Xt tp hp tt c cc chnh hp khng lp chp k ca n phn t. Sp xp chng thnh nhng lp sao cho hai chnh hp thuc cng mt lp ch khc nhau v th t. R rng mi lp nh vy l mt t hp chp k ca n phn t(P(n,k)). S chnh hp trong mi lp u bng nhau v bng k! (s hon v k phn t: P(k,k) ). S cc lp bng s t hp chp k ca n (P(n,k)). T ta c:

)!(!!

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

knkn

kknPknCkkPknCknP

V d 1. Cho S = { a, b, c, d } tm C(4,2)? Li gii. R rng C(4,2) = 6 tng ng vi 6 tp con {a, b}, {a, c}, {a, d}, {b,c},

{b, d} {c,d}. V d 2. C n i bng thi u vng trn. Hi phi t chc bao nhiu trn u. Li gii: C hai i bng th c mt trn. T suy ra s trn u s bng s cch

chn 2 trong n i, ngha l bng C(n, 2) = n! / 2!(n-2)! = n(n-1)/2 trn u. V d 3. Chng minh a) C(n,k) = C(n, n-k) b) C(n, 0) = C(n,n)= 1 c) C(n,k) = C(n-1,k-1) + C(n-1,k) Li gii a. C(n,n-k) = n!/(n-k)! (n-n+k)! = n!/k!(n-k)! = C(n,k).

Hoc C(n, k) = n!/k!(n-k)! = n!/ (n-k)! (n-(n-k))! = C(n, n-k); b) Ch 0!=1 => b hin nhin ng c) C(n,k) = C(n-1,k-1) + C(n-1,k)

).()!(!

!)()!1()!1(

)!.1(11)!1()!1(

)!1()!1(!

)!1()!11()!1(

)!1(),1()1,1(

knCknk

nknknkk

nnkknknk

nknk

nknk

nknCknC

T nhng tnh cht trn, ta c th tnh tt c cc h s t hp ch bng php cng. Cc h s ny c tnh v vit ln lt theo dng, trn mi dng ta tnh v thc hin theo ct. Bng c dng tam gic chnh l tam gic Pascal.

Cc h s t hp c lin quan cht ch ti vic khai trin lu tha ca mt nh thc. Thc vy, trong tch

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30

(x+y)n = (x+y)(x+y). . .(x+y) h s ca xkyk-n s l s cch chn k phn t (x+y) m t ly ra x v ng thi (n-k) nhn t cn li ly ra y, ngha l:

kn

kknnnnnn yxknCynCxynnCyxnCxnCyx 011 ),()0,()1,(...)1,()0,()(

Cng thc trn cn c gi l khai trin nh thc Newton, cc h s t hp cn c gi l h s nh thc. Chng hn lu tha bc 8 ca nh thc (x+y)8 c khai trin nh sau:

876253443526788 828567056288)( yxyyxyxyxyxyxyxxyx

Trong trng hp y=1, tc khai trin (x+1)n ta c:

),()1,(...)1,()0,()1( 1 nnCxnnCxnCxnCx nnn

Hoc ng thc sau s c rt ra t khai trin nh thc Newton:

),()1,(...)1,()0,()11(2 nnCnnCnCnCnn

C th ni rt nhiu ng thc v h s t hp s c suy ra. Nh tnh cc tp l, o hm.

2.3.5. T hp lp

T nh ngha t hp, ta c th tnh ton c s t hp lp chp k t tp n phn t bng C( n + k -1, k). V d 1. Phng trnh x1 + x2 + x3 = 11 c bao nhiu nghim nguyn khng m? Li gii. Mi nghim nguyn khng m ca phng trnh ng vi mt cch chn 11 phn t t mt tp c 3 loi, sao cho c x1 phn t loi 1 c chn, x2 phn t loi 2 c chn, x3 phn t loi 3 c chn. S ny chnh bng s t hp lp chp 11 t tp c 3 phn t. V vy, s nghim nguyn khng m ca phng trnh l: C( 3 + 11 -1, 11 ) = C (13, 11) = (13.12) / 2 = 78. V d 2. Phng trnh x1 + x2 + x3 = 11 c bao nhiu nghim nguyn khng m tha mn x11, x22, x33? Li gii. Mi nghim nguyn khng m ca phng trnh ng vi mt cch chn 11 phn t t mt tp c 3 loi, sao cho c x1 phn t loi 1 c chn, x2 phn t loi 2 c chn, x3 phn t loi 3 c chn. Trong , c t nht mt phn t loi 1, hai phn t loi 2 v ba phn t loi 3. V th ta chn mt phn t loi 1, hai phn t loi 2, ba phn t loi 3 sau chn thm 5 phn t na. S ny chnh bng s t hp lp chp 5 t tp c 3 phn t. V vy, s nghim nguyn khng m ca phng trnh l: C( 3 + 5 -1, 5 ) = C (7, 5) = (7.6) / 2 = 21. V d 3. Phng trnh x1 + x2 + x3 = 11 c bao nhiu nghim nguyn khng m tha mn 1 x1 3?

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31

Li gii. Ta nhn thy, s nghim nguyn khng m ca phng trnh tha mn 1 x1 3 chnh bng s nghim nguyn khng m ca phng trnh tha mn x11 sau tr bt i nhng nghim x1 4. T V d 1 ta tnh c s nghim nguyn khng m ca phng trnh tha mn 1 x1 3 l C( 3 + 10 -1, 10 ) - C(3 +7-1, 7) = C (12, 10) - C(9, 7).

2.4. H thc truy hi

Thng thng ngi ta thng quan tm ti nhng bi ton m trong kt qu m ph thuc vo mt tham s u vo (m ta k hiu l n), chng hn nh cc s mt th t Dn . Vic biu din kt qu ny nh mt hm ca n bng mt s hu hn cc php ton khng phi l n gin. Trong nhiu trung hp, vic tm ra mt cng thc trc tip gia kt qu m v n l ht sc kh khn v nhiu khi khng gii quyt c, trong khi cng thc lin h gia kt qu m ng vi gi tr n vi cc kt qu b hn n li n gin v d tm. Thng qua cng thc ny v mt vi gi tr ban u, ta c th tnh mi gi tr cn li khc. Cng thc gi l cng thc truy hi hay cng thc qui. c bit, cng thc truy hi rt thch hp vi lp trnh trn my tnh. N cng cho php gim ng k phc tp cng nh gia tng n nh ca qu trnh tnh ton.

2.4.1. nh ngha v v d

nh ngha 1. H thc truy hi i vi dy s {an} l cng thc biu din an qua mt hay nhiu s hng i trc ca dy, c th l a1, a2, .., an-1 vi mi nn0 nguyn dng. Dy s c gi l li gii hay nghim ca h thc truy hi nu cc s hng ca n tho mn h thc truy hi.

V d 1. Li kp. Gi s mt ngi gi 10000 la vo ti khon ca mnh ti mt ngn hng vi li xut kp 11% mi nm. Hi sau 30 nm anh ta c bao nhiu tin trong ti khon ca mnh?

Li gii. Gi Pn l tng s tin c trong ti khon sau n nm. V s tin c trong ti khon sau n nm bng s tin c c trong n-1 nm cng vi li xut nm th n. Nn dy {Pn} tho mn h thc truy hi :

Pn = Pn-1 + 0.11Pn-1 = 1.11Pn-1 Chng ta c th dng phng php lp tm cng thc trn cho Pn. D nhn thy rng: P0 = 10000 P1 = 1.11P0 P2 = 1.11P1 = (1.11)2P0 . . . . . . . . . . . . . . . . . . . . . Pn = 1.11Pn-1 = (1.11)n-1P0

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32

Ta c th chng minh tnh ng n ca cng thc truy hi bng qui np. Thay P0= 10000, v n = 30 ta c: P30 = (1.11)3010000 = 228922,97 $

V d 2. H nh th v s Fibonaci. Mt cp th sinh i (mt con c v mt con ci) c th ln mt hn o. Gi s rng cp th s cha sinh sn c trc khi y hai thng tui. T khi chng y hai thng tui, mi thng chng sinh thm c mt cp th. Tm cng thc truy hi tnh s cp th trn o sau n thng vi gi s cc cp th l trng th.

S thng S cp sinh sn S cp th con Tng s cp th

1 0 1 1

2 0 1 1

3 1 1 2

4 1 2 3

5 2 3 5

6 3 5 8

.. .. .. ..

Li gii: Gi s fn l s cp th sau n thng. Ta s ch ra rng f1, f2, . ., fn (n=1, 2, . ., n) l cc s ca dy fibonaci. Cui thng th nht s cp th trn o l f1 = 1. V thng th hai cp th vn cha n tui sinh sn c nn trong thng th hai f2 =1. V mi cp th ch c sinh sn sau t nht hai thng tui, nn ta tm s cp th sau thng th n bng cch cng s cp th sau thng n-2 v thng n-1 hay fn = fn-1 + fn-2. Do vy, dy { fn} tho mn h thc truy hi fn = fn-1 + fn- 2 vi n>=3 v f1 = 1, f2 = 1. V d 3. Tnh s mt th t Dn.

Li gii: nh s th v phong b th t 1 n n (th i gi ng a ch nu b vo phong b i). Mt cch b th uc ng nht vi hon v (a1, a2, . ., an) ca { 1, 2, . ., n }. Mt mt th t c nh ngha l l mt hon v (a1, a2, . ., an) sao cho aii vi mi i. Thnh phn a1 c th chp nhn mi gi tr ngoi 1. Vi mi gi tr k (k1) ca a1, xt hai trng hp:

1. ak =1, khi cc thnh phn cn li c xc nh nh mt mt th t ca n-2 phn t, tc l s mt th t loi ny bng Dn-2.

2. ak1, khi cc thnh phn t 2 n n c xc nh nh mt mt th t ca n-1 phn t cn li, tc l s mt th t ny thuc loi Dn-1.

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33

T ta nhn c cng thc Dn = (n-1) (Dn-1 + Dn-2), n>=3 vi D1 = 0, D2 =1.

Mi gi tr cn li c tnh n gin nh lut k tha. D3 = (3 1) (0 +1) =2 D4 = (4 1 )( 1 + 2) = 9 D5 = (5 1 )( 9 + 2) = 44 D6 = (6 1 )(9 + 44) = 265 D7 = (7 1 )( 44 + 265) = 1854 D8 = (8 1 )( 265 + 1854) = 14833 . . . . . . . . . . . .. . . . . . . . . . . . . . . . . cng thc ng vi n = 2, ta coi D0 = 1 C th nhn c s mt th t thng qua cng thc truy hi trn v: ))1(())(1( 21121 nnnnnnn DnDnDDDDnD

t 1 nnn nDDV ta c:

nnnnnn VVVnDD )1()1( 1111 . Hay ta c th vit:

!)1(

)!1(!1

nnD

nD nnn

. Cng cc h thc trn vi n = 1, 2, .., n ta c:

!)1(

!21

!111

! nnD nn . T thu li c cng thc c:

)!)1(

!21

!111(!

nnD

n

n

V d 3. Tnh h s t hp C(n,k). Li gii. Chn phn t c nh a trong n phn t ang xt. Chia s cch chn tp

con k phn t ny thnh hai lp (lp cha a v lp khng cha a). Nu a c chn th ta cn b xung k-1 phn t t n-1 phn t cn li, t lp cha a gm C(n-1, k-1) cch. Nu a khng c chn, th ta phi chn k phn t t n-1 phn t cn li, t lp khng cha a gm C(n-1, k) cch. Theo nguyn l cng ta c cng thc truy hi: C(n, k) = C(n-1, k-1) + C(n-1,k) vi cc gi tr bin c suy ra trc tip: C(n,0) = C(n,n) = 1. Phng php ny c gi l phng php kh. Khng phi lc no cng d dng kh c cng thc truy hi a v cng thc trc tip. Tuy nhin, trong mt s trng hp c bit ta c th a ra phng php tng qut gii cng thc truy hi.

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2.4.2. Gii cng thc truy hi tuyn tnh thun nht vi h s hng s

nh ngha 1. Mt h thc truy hi tuyn tnh thun nht bc k vi h s hng s l h thc truy hi c dng:

knknnn acacaca 2111 (1)

trong c1,c2, . ., ck l cc s thc v ck 0 Ta cn tm cng thc trc tip cho s hng an ca dy s {an} tho mn cng thc (1). Theo nguyn l th hai ca qui np ton hc th dy s tho mn nh ngha trn c xc nh duy nht nu nh n tho mn k iu kin u: a0 = C0, a1 = C1, . ., ak-1 = Ck-1, trong C1, C2, . ., Ck-1 l cc hng s.

V d 1. H thc truy hi Pn=(1.11)Pn-1 l h thc truy hi tuyn tnh thun nht bc 1. H thc truy hi fn = fn-1 + fn-2 l h thc truy hi tuyn tnh thun nht bc 2. H thc truy hi an = an-5 l h thc truy hi tuyn tnh thun nht bc 5. H thc truy hi Bn=nBn-1 khng phi l h thc truy hi tuyn tnh thun nht v n khng c h s hng s. Phng php c bn gii h thc truy hi tuyn tnh thun nht l tm nghim di dng an = rn, trong r l hng s. Cng cn ch rng an = rn l nghim ca h thc truy hi an = c1an-1 + c2 an-2 + . .+ ckan-k nu v ch nu

an = c1rn-1 + c2rn-2 + . .+ ckrn-k . Chia c hai v cho rn-k ta nhn c

rk c1rk-1 c2rk-2 - . . ck-1r ck =0 (2) Vy dy {an} vi an=rn l nghim nu v ch nu r l nghim ca (2). Phng trnh

2 cn c gi l phng trnh c trng ca h thc truy hi, nghim ca n l nghim c trng ca h thc truy hi. Nghim ca phng trnh c trng dng biu din cng thc tt c cc nghim ca h thc truy hi.

Chng ta s trnh by cc kt qu vi h thc truy hi tuyn tnh thun nht bc hai. Sau ta s nu ra nhng kt qu tng t cho trng hp tng qut khi bc ln hn hai.

nh l 1. Cho c1, c2 l cc hng s thc. Gi s r2 c1r + c2 =0 c hai nghim phn bit r1, r2. Khi dy {an} l nghim ca h thc truy hi an = c1an-1 + c2an-2 khi v ch khi an = 1r1n + 2rn2 vi n =1, 2, . .., 1, 2 l cc hng s.

Chng minh: chng minh nh l ny ta cn thc hin hai vic. u tin ta cn ch ra rng nu r1, r2 l hai nghim ca phng trnh c trng v 1, 2 l hai hng s th dy {an} vi nnn rra 2211 l nghim ca h thc truy hi. Ngc li, cn phi chng minh rng nu {an} l nghim th nnn rra 2211 vi 1, 2 l cc hng s no .

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35

(): Gi s r1 v r2 l hai nghim phn bit ca r2 c1r + c2=0 , khi 221

2221

21

21 ; crcrcrcr ng thi ta thc hin dy cc php bin i sau:

nnn

nn

nn

nnnnnn

arrrrrr

crcrcrcrrrcrrcacac

2211

22

222

21

211

2212

222111

11

222

2112

222

11112211

)()()()(

iu ny chng t dy {an} vi nnn rra 2211 l nghim ca h thc truy hi cho.

(): chng minh ngc li, ta gi s dy {an} l mt nghim bt k ca h thc truy hi. Ta chn 1, 2 sao cho dy {an} vi nnn rra 2211 tho mn cc iu kin u a0 =C0, a1 = C1. Thc vy,

221111

2100

rrCaCa

T phng trnh u ta c 2 = C0 - 1 th vo phng trnh th hai ta c: 20211210111 )()( rCrrrCrC ; T y suy ra:

21

110

21

2010102

21

2011

)()(;)(rrCrC

rrrCCCC

rrrCC

.

Nh vy, khi chn nhng gi tr trn cho 1, 2 dy {an} vi nnn rra 2211 tho mn cc iu kin u. V h thc truy hi v cc iu kin u c xc nh duy nht nn nnn rra 2211 . nh l c chng minh.

V d 1. Tm nghim ca h thc truy hi an = an-1 +2an-2 vi a0 = 2, a1 = 7. Li gii. Phng trnh c trng ca h thc truy hi c dng r2 - r - 2 =0. Nghim

ca n l r=2 v r = -1. Theo nh l 1, dy {an } l nghim ca h thc truy hi nu v ch nu :

an = 12n +2(-1)n vi 1, 2 l cc hng s no . T cc iu kin u suy ra:

a0 = 2 = 1 +2

a1 = 7 = 12 +2(-1)

Gii ra ta c 1=3, 2=-1. Vy nghim ca biu thc truy hi vi iu kin u l dy {an} vi an = 3.2n (-1)n.

V d 2. Tm cng thc hin ca cc s fibonaci. Gii: Cc s fibonaci tho mn h thc fn = fn-1 + fn-2 v cc iu kin u f0 = 0,

f1=1. Cc nghim ca phng trnh c trng l:

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36

251;

251

21 rr theo nh l 1 ta suy ra s fibonaci c cho bi cng

thc sau:

nn

nf

251

251

21 vi 1, 2 l hai hng s. Cc iu kin u f0=0,

f1=1 c dng xc nh cc hng s 1, 2.

12

512

51

0

211

210

f

f

T hai phng trnh ny ta suy ra 5

1;5

121 do cc s fibonaci c cho

bng cng thc dng hin nh sau: nn

nf

251

51

251

51

nh l 1 khng dng c trong trng hp nghim ca phng trnh c trng l nghim bi. Khi phng trnh c trng c nghim bi ta s dng nh l sau.

nh l 2. Cho c1, c2 l cc hng s thc, c20. Gi s r2 c1r c2 = 0 ch c mt nghim r0. Dy {an} l nghim ca h thc truy hi an = c1an-1 + c2an-2 khi v ch khi

nnn nrra 0201 vi n = 1, 2, . . trong 1, 2 l nhng hng s.

V d 3. Tm nghim ca cng thc truy hi an = 6an-1 9an-2 vi cc iu kin u a0=1, a1 = 6. Gii: Phng trnh c trng r2 6r 9 =0 c nghim kp r=3. Do nghim ca h thc truy hi c dng:

nnn na 33 21 vi 1, 2 l cc hng s no . T cc iu kin u ta suy ra:

a0 = 1=1

a1 = 6 = 13+23 1 =1, 2=1 vy nghim ca h thc truy hi v cc iu kin u cho l:

an = 3n+n3n nh l 1, nh l 2 khng dng c trong trng hp nghim ca phng trnh c trng l nghim phc. Khi phng trnh c trng c nghim phc ta s dng nh l sau.

nh l 3. Cho c1, c2 l cc hng s thc, c20. Gi s r2 c1r c2 = 0 c hai nghim phc lin hp

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37

sin.cossin.cos

2

1

irrirr

Khi , dy {an} l nghim ca h thc truy hi an = c1an-1 + c2an-2 khi v ch khi nnra nn sincos 21 vi n = 1, 2, . . trong 1, 2 l nhng hng s.

V d. Tm nghim ca h thc truy hi sau an = 2an-1 - 4an-2, vi a0 = 4, a1 =4? Li gii. Bc 1. Tm nghim ca phng trnh c trng:

3/sin3/cos23/sin3/cos2

0422

12

irir

rr

Bc 2. Xy dng cng thc tng qut cho an:

)3/sin()3/cos(2)sin()cos( 2121 ncncncncra nnn Bc 3. Tm cc hng s thng qua iu kin u:

3;1

423

214

423

212

21

21

21

cc

cc

Bc 4. Hon chnh cng thc cho an.

3/sin33/cos2 nna nn By gi ta pht biu kt qu tng qut v nghim cc h thc truy hi tuyn tnh

thun nht vi cc h s hng s. nh l 4. Cho c1 , c2 , . . , ck l cc s thc. Gi s phng trnh c trng

rk c1rk-1-. .-ck = 0 c k nghim phn bit r1, r2, . . , rk. Khi dy {an} l nghim ca h thc truy hi knknnn acacaca 2211 khi v ch khi

nkk

nnn rrra 2211 vi n=0,1,2, . ., trong 1, 2, . ., k l cc hng s.

V d 4. Tm nghim ca h thc truy hi an = 6an-1 11an-2+6an-3 vi iu kin u a0=2, a1 =5, a2=15.

Gii: a thc c trng ca h thc truy hi l : r3 6r2 + 11r 6 c cc nghim l r1=1, r2 = 2, r3 = 3. Do vy nghim ca h thc truy hi c dng: nnnna 321 321 .

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38

tm cc hng s 1, 2, 3 ta da vo nhng iu kin ban u:

a0 = 2 = 1 + 2 + 3

a1 = 5 = 1 + 22 +33

a2 = 15=1 + 24 +39

Gii: h phng trnh ny ta nhn c 1 = 1, 2 =- 1, 3=2. V vy nghim duy nht ca h thc truy hi ny v cc iu u cho l dy {an} vi:

nnna 3.221

2.5. Qui v cc bi ton n gin

Mt trong nhng phng php gii quyt bi ton m phc tp l qui bi ton ang xt v nhng bi ton nh hn. S phn chia ny c thc hin mt cch lin tip cho ti khi nhn c li gii ca bi ton nh mt cch d dng. Tuy nhin iu ny khng phi lc no cng thc hin c v n i hi mt s phn tch su sc cu hnh cn m.

Gi s rng c mt thut ton phn chia bi ton c n thnh a bi ton nh, trong mi bi ton nh c c n/b( n gin ta gi s n chia ht cho b); trong thc t cc bi ton nh thng c c l s nguyn gn nht vi n/b. Gi s tng cc php ton thm vo khi thc hin phn chia bi ton c n thnh cc bi ton c nh hn l g(n). Khi nu f(n) l s cc php ton cn thit gii bi ton cho th f tho mn h thc truy hi sau:

)()( ngbnafnf

; h thc ny c tn l h thc chia tr.

V d 1. Xt thut ton nhn hai s nguyn kch c 2n bt. K thut ny gi l thut ton nhn nhanh c dng k thut chia tr.

Gii: Gi s a v b l cc s nguyn c biu din nh phn l 2n bt (c th thm cc bt 0 vo u chng c th di bng nhau) . 2012212 )( aaaaa nn v 2012212 )( bbbbb nn

Gi s a = 2nA1 +A0, b = 2nB1 +B0 trong

2122121 )( nnnn aaaaA ; 201210 )( aaaaA nn

2122121 )( nnnn bbbbB ; 201210 )( bbbbB nn

Thut ton nhn nhanh c da trn ng thc:

001001112 )12())((2)22( BABBAABAab nnnn

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39

iu ny ch ra rng php nhn hai s nguyn 2n bt c th thc hin bng cch dng 3 php nhn cc s nguyn n bt v cc php cng, tr dch chuyn. Nh vy, nu f(n) l tng cc php ton nh phn cn thit nhn hai s n bt th

Cnnfnf )(3)2(

Ba php nhn cc s nh phn n bt cn 3f(n) php ton nh phn. Mi mt php ton cng, tr, dch chuyn dng mt hng s nhn vi n ln chnh l Cn. V d 2. Bi ton xp khch ca Lucas. C mt bn trn, xung quanh c 2n gh. Cn sp ch cho n cp v chng sao cho cc ng ngi sen k cc b v khng c hai cp v chng no ngi cnh nhau. Hi c tt c bao nhiu cch xp ? Li gii. Gi s phi tm l Mn. Xp cho cc b trc(c xp mt gh th mt gh trng dnh cho cc ng), s cch xp cho cc b l 2n! cch. Gi s cch xp cho cc ng ng vi mt cch xp cc b l Un ta c s cch xp l

Mn = 2n! x Un .Vn cn li l tnh s Un. nh s cc b ( xp) t 1 n n, nh s cc ng tng ng vi cc b (ng i l

chng b i), sau nh s cc gh trng theo nguyn tc: gh s i nm gia b i v b i+1 (cc php cng c hiu ly modul n ngha l n +1 = 1). Mi cch xp cc ng c biu din bng mt php th trn tp {1, 2, . ., n } vi qui c (i) = j c ngha l gh i c xp cho ng j. Theo gi thit phi tho mn:

(i) i v (i)i+1 (*)

Nh vy, Un l s tt c cc php th tho mn iu kin (*). Trong ton hc gi Un l s phn b.

Xt tp hp tt c cc php th ca { 1, 2, . ., n }. Trn tp ny ta gi Pi l tnh cht (i) = i, Qi l tnh cht (i) = i+1. t Pn+i = Qi, theo nguyn l b tr tng ng vi 2n tnh cht Pi ta c:

...! 21 NNnNU n trong Nk l tng s tt c cc php th tho mn k tnh cht ly t 2n tnh cht ang xt. Cn ch rng, khng th xy ra ng thi tho mn Pi v Qi. Do trong cc php ly ra k tnh cht t 2n tnh cht ang xt cn thm vo iu kin: Pi v Qi hoc Pi+1 v Qi khng c ng thi c mt. Gi s cc cch ny l g(2n, k) ( ni ring g(2n,k)=0 khi k>n). Vi mi cch ly ra k tnh cht nh vy (k

40

Bi ton 1. C bao nhiu cch ly ra k phn t trong n phn t xp trn ng thng sao cho khng c hai phn t no k nhau cng c ly ra.

Li Gii. Khi ly k phn t, ta cn n-k phn t. Gia n-k phn t cn li c n-k+1 khong trng (k c hai u). Mi cch ly ra k khong t cc khong ny s tng ng vi mt cch chn k phn t tho mn yu cu nu. Vy s cch chn cn tm l C(n-k+1, k).

Bi ton 2. Ging nh bi ton 1 nhng n phn t xp trn vng trn. Li gii. C nh phn t a c chn chia cc cch ly thnh 2 lp 1- Cc cch m a c chn khi 2 phn t k a s khng c chn v phi

ly k-1 phn t t n-3 phn t cn li. Cc phn t ny xem nh kt qu ca bi ton 1. Theo bi ton 1, s cch thuc lp kiu ny l C(n-k-1, k-1).

2- Cc cch m a khng c chn, khi b a i v bi ton tr v bi ton 1 chn k phn t t n-1 phn t xp trn ng thng. Theo bi ton 1 s cch xp kiu ny l C(n-k,k).

Vy theo nguyn l cng s cch cn tm l :

),(),()1,1( kknCkn

nkknCkknC

T kt qu ca hai bi ton trn ta nhn c:

),2(2

2),2( kknCkn

nkng

v s phn b Un c tnh bng

),2(2)1()!2)(2,22(22

2)!1)(1,12(12

2! nnCnnnnC

nnnnC

nnnU nn

Di y l mt s gi tr ca Un, mt ln na chng ta li c quan st hin

tng bng n t hp.

n 2 3 4 5 6 7 8 9 10

Un 0 1 2 13 80 579 4783 43387 439792

2.6. Phng php lit k

Vic tm mt cng thc cho kt qu m ngay c trong trng hp cng thc truy hi khng phi d dng v lc no cng thc hin c. Cho n nay cn nhiu bi ton m cha c li gii di dng mt cng thc. i vi nhng bi ton nh vy, ngi ta ch cn cch ch ra mt phng php lit k, theo c th i qua c tt c cc cu hnh cn m. R rng bn thn phng php lit k khng ch ra c mt kt qu c th no nhng qua ngi ta c th lp trnh cho my tnh in t m h.

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minh ho cho phng php lit k, ta xt mt cu hnh t hp ni ting l cc hnh ch nht la tinh. Gi s S l tp gm n phn t. Khng mt tnh tng qut ta gi s S = {1, 2, . ., n} Mt hnh ch nht la tinh trn S l mt bng gm p dng, q ct sao cho mi dng ca n l mt chnh hp khng lp chp q ca S v mi ct ca n l mt chnh hp khng lp chp p ca S.

Theo nh ngha ta c pn, qn. c bit trong trng hp q = n, mi dng ca hnh ch nht la tinh l mt hon v ca S, sao cho khng c ct no cha hai phn t lp li. Hnh ch nht la tinh dng ny c gi l chun nu dng u ca n l hon v 1, 2,. ., n.

V d. Hnh la tinh chun trn tp S = {1, 2, 3, 4, 5, 6, 7 } 1 2 3 4 5 6 7 2 3 4 5 6 7 1 3 4 5 6 7 1 2

Gi L(p,n) l s hnh ch nht la tinh p x n, cn K(p,n) l s hnh ch nht la tinh chun p x n ta c: L(p,n) = n! K(p,n)

D dng nhn thy rng, s mt Dn l s hnh la tinh chun 2 x n, s phn b Un l s hnh ch nht la tinh chun 3 x n vi hai dng u l: 1 2 . . . n-1 n 2 3 . . . n 1

Riodan J(1946) chng minh cng thc:

knm

k kkn UDDknCnK 20 ),(),3( trong m= [n/2], U0 = 1. Bi ton m vi s dng nhiu hn n nay vn cha c gii quyt. Ngi ta mi ch a ra c mt vi dng tim cn ca L(p,n). Nu p=q=n, th hnh ch nht la tinh c gi l hnh vung la tinh. Mt hnh vung la tinh cp n c gi l chun nu c dng u v ct u l hon v 1, 2,. .n. Th d mt hnh vung la tinh chun cp 7. 1 2 3 4 5 6 7 2 3 4 5 6 7 1 3 4 5 6 7 1 2 4 5 6 7 1 2 3 5 6 7 1 2 3 4 6 7 1 2 3 4 5

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7 1 2 3 4 5 6 Gi ln l s cc hnh vung nh th ta c L(n,n) = n!(n-1)!ln Vic tm mt cng thc cho ln n nay vn b ng. Tuy nhin ta c th nh my

tnh lit k tt c cc hnh vung chun cp n. Di y l mt vi gi tr tnh c.

N 1 2 3 4 5 6 7

ln 1 1 1 4 56 9408 16942080

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43

BI TP CHNG 2

1. Xu thun nghch c l mt xu khi vit theo th t ngc li cng bng chnh n.

Hy m s xu nh phn c di n l thun nghch c. 2. C du v ch r mi bn bn ng thnh mt hng chp nh. Hi c bao nhiu

cch xp hng nu: a) C du ng cnh ch r b) C du khng ng cnh ch r c) C du ng pha bn phi ch r

3. C bao nhiu xu nh phn di 10 c nm s 0 lin nhau hoc nm s 1 lin nhau. 4. C bao nhiu xu nh phn di bng 8 c 3 s 0 lin nhau hoc 4 s 1 lin nhau. 5. Mi sinh vin lp ton hc ri rc hoc gii ton hoc gii tin hc hoc gii c hai

mn ny. Trong lp c bao nhiu sinh vin nu 38 ngi gii tin (k c ngi gii c hai mn), 23 ngi gii ton (k c ngi gii c hai mn), v 7 ngi gii c hai mn.

6. Chng t rng, trong n+1 s nguyn dng khng vt qu 2n tn ti t nht mt s chia ht cho mt s khc.

7. Chng minh rng, trong dy gm n2 + 1 s thc phn bit u c mt dy con di n+1 hoc thc s tng, hoc thc s gim.

8. Gi s trong mt nhm 6 ngi mi cp hai hoc l bn, hoc l th. Chng t rng trong nhm c ba ngi l bn ca nhau hoc l k th ca nhau.

9. Hy ch ra rng, trong 102 ngi c chiu cao khc nhau ng thnh mt hng c th tm c 11 ngi c chiu cao tng dn hoc gim dn m khng cn thay i th t ca h trong hng.

10. Mt vt tay tham gia thi u ginh chc v ch trong 75 gi. Mi gi anh ta thi u t nht mt trn, nhng ton b anh ta khng thi u qu 125 trn. Chng t rng, c nhng gi lin tip anh ta thi u 24 trn.

11. Mt nhn vin bt u lm vic ti cng ty t nm 1987 vi mc lng khi im l 50000 la. Hng nm anh ta c nhn thm 1000 la v 5% lng ca nm trc. a) Hy thit lp h thc truy hi tnh lng ca nhn vin n nm sau nm 1987. b) Lng vo nm 1995 ca anh ta l bao nhiu? c) Hy tm cng thc tng minh tnh lng ca nhn vin ny n nm sau nm 1987.

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44

12. Tm h thc truy hi cho s hon v ca tp n phn t. Dng h thc truy hi tnh hon v ca tp n phn t.

13. Mt my bn tem t ng ch nhn cc ng xu mt la v cc loi t tin 1 la v 5 la. a) Hy tm h thc truy hi tnh s cch t n la vo trong my bn hng, trong

th t cc ng xu, cc t tin l quan trng. b) Tm cc iu kin u. c) C bao nhiu cch t 10 la vo my mua mt b tem.

14. Gii cc h thc truy hi vi cc iu u sau:

a) an = an-1 + 6an-2 vi n 2, a0 = 3, a1 = 6.

b) an = 7an-1 - 6an-2 vi n 2, a0 = 2, a1 = 1.

c) an = 6an-1 - 8an-2 vi n 2, a0 = 4, a1 = 10.

d) an = 2an-1 - an-2 vi n 2, a0 = 4, a1 = 1.

e) an = an-2 vi n 2, a0 = 5, a1 = -1.

f) an = -6an-1 - 9an-2 vi n 2, a0 = 3, a1 = -3.

g) an+2 = -4an+1 + 5an vi n 0, a0 = 2, a1 = 8. 15. Tm cc nghim c trng ca h thc truy hi tuyn tnh thun nht

a) an = 2an-1 - 2an-2 b) Tm nghim tho mn h thc truy hi trn v cc iu kin u a0 =1, a1 =2.

16. a) Tm nghim c trng ca h thc truy hi tuyn tnh thun nht an = an-4 b) Tm nghim tho mn h thc truy hi trn v cc iu kin u a0=1, a1=0,

a2=-1, a3=1. 17. Mt bo co v th trng my tnh c nhn cho bit c 65000 ngi s mua modem

cho my tnh ca h trong nm ti, 1 250 000 ngi s mua t nht mt sn phm phn mm. Nu bo co ny ni rng 1.450.000 ngi s mua hoc l modem hoc l t nht mt sn phm phn mm th s c bao nhiu ngi s mua c modem v mua t nht mt sn phm phn mm.

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CHNG 3. BI TON LIT K

Ni dung bi ton m l m xem c bao nhiu cu hnh t hp tha mn mt s tnh cht no . Bi ton lit khng ch m c cc cu hnh t hp tha mn cc tnh cht t ra m cn xem xt tng cu hnh t hp l g. i vi mi bi ton, khi cha tm c thut gii th lit k c xem l bin php cui cng thc hin vi s h tr ca my tnh. C th ni, lit k c xem l phng php gii vn nng mt bi ton bng my tnh. Ni dung chnh ca chng ny tp chung gii quyt nhng vn c bn sau: Gii thiu bi ton lit k. Thut ton v phc tp tnh ton. Gii quyt bi ton lit k bng phng php sinh. Gii quyt bi ton lit k bng phng php quay lui. Bn c c th tm thy cch gii nhiu bi ton lit k trong cc ti liu [1] v [2]

trong ti liu tham kho.

3.1- Gii thiu bi ton

Bi ton a ra danh sch tt c cc cu hnh t hp c th c c gi l bi ton lit k t hp. Khc vi bi ton m l tm kim mt cng thc cho li gii, bi ton lit k li cn xc nh mt thut ton theo c th xy dng c ln lt tt c cc cu hnh cn quan tm. Mt thut ton lit k phi m bo hai nguyn tc:

Khng c lp li bt k mt cu hnh no. Khng c b st bt k mt cu hnh no.

V d 1. Cho tp hp cc s a1, a2, .., an v s M. Hy tm tt c cc tp con k phn t ca dy s {an} sao cho tng s cc phn t trong tp con ng bng M. Li gii: Nh chng ta bit, s cc tp con k phn t ca tp gm n phn t l C(n,k). Nh vy chng ta cn phi duyt trong s C(n,k) tp k phn t ly ra nhng tp c tng cc phn t ng bng M. V khng th xc nh c c bao nhiu tp k phn t t tp n phn t c tng cc phn t ng bng M nn chng ta ch cn cch lit k cc cu hnh tho mn iu kin cho. V d 2. Mt thng nhn i bn hng ti tm thnh ph. Ch ta c th bt u hnh trnh ca mnh ti mt thnh ph no nhng phi qua 7 thnh ph kia theo bt k th t no m ch ta mun. Hy ch ra l trnh ngn nht m ch ta c th i. Li gii: V thnh ph xut pht c xc nh. Do vy thng nhn c th chn tu 7 thnh ph cn li hnh trnh. Nh vy, tt c s hnh trnh ca thng nhn c th

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46

i qua l 7! = 5040 cch. Tuy nhin trong 5040 cch chng ta phi duyt ton b ch ra mt hnh trnh l ngn nht.

C th ni phng php lit k l bin php cui cng nhng cng l bin php ph dng nht gii quyt cc bi ton t hp. Kh khn chnh ca phng php ny l s bng n t hp. xy dng chng 1 t cu hnh (con s ny khng phi l ln i vi cc bi ton t hp nh s mt th t Dn, s phn b Un, s hnh vung la tinh ln), ta gi s cn 1 giy lit k mt cu hnh th chng ta cng cn 31 nm mi gii quyt xong. Tuy nhin vi s pht trin nhanh chng ca my tnh, bng phng php lit k, nhiu bi ton kh ca l thuyt t hp c gii quyt v gp phn thc y s pht trin ca nhiu ngnh ton hc.

3.2. Thut ton v phc tp tnh ton

3.2.1. V d v nh ngha

nh ngha. Dy h hn cc thao tc s cp F=F1F2..Fn(Input)Output c gi l mt thut ton trn tp thng tin vo Input c c kt qua ra Output. Dy cc thao tc s cp y c hiu l cc php ton s hc, cc php ton logic, cc php ton so snh. Mt thut ton cn tha mn cc tnh cht di y:

Tnh n nh. mi bc ca thut ton, cc thao tc s cp phi ht sc r rng, khng gy nn s ln xn, nhp nhng, a ngha. Thc hin ng cc bc ca thut ton trn tp d liu vo, ch cho duy nht mt kt qu ra.

Tnh dng. Thut ton khng c ri vo qu trnh v hn. Phi dng li v cho kt qu sau mt s hu hn cc bc.

Tnh ng. Sau khi thc hin tt c cc bc ca thut ton theo ng qui trnh nh, ta phi nhn c kt qu mong mun vi mi b d liu u vo. Kt qu c kim chng bng yu cu ca bi ton.

Tnh ph dng. Thut ton phi d sa i thch ng c vi bt k bi ton no trong lp cc bi ton cng loi v c th lm vic trn nhiu loi d liu khc nhau.

Tnh kh thi. Thut ton phi d hiu, d ci t, thc hin c trn my tnh vi thi gian cho php.

3.2.2. Phng php biu din thut ton:

Thng thng, biu din mt thut ton ta c th s dng cc phng php sau: Biu din bng ngn ng t nhin. Ngn ng t nhin l phng tin

giao tip gia con ngi vi con ngi. Ta c th s dng chnh ngn ng ny vo vic biu din thut ton.

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47

Ngn ng hnh thc. Ngn ng hnh thc l phng tin giao tip trung gian gia con ngi v h thng my tnh. V d ngn ng s khi, ngn ng ta t nhin, ngn ng c t. c im chung ca cc loi ngn ng ny l vic s dng n rt gn vi ngn ng t nhin v ngn ng my tnh.

Ngn ng my tnh. L phng tin giao tip gia my tnh v my tnh. Trong trng hp ny ta c th s dng bt k nn ng lp trnh no m t thut ton.

Ghi ch. Trong cc phng php biu din thut ton, phng php biu din bng ngn ng hnh thc c s dng rng di v n gn vi ngn ng t nhin v khng ph thuc vo ngn ng my tnh.

V d 1. Biu din thut ton tm USCLN (a, b) bng ngn ng t nhin. u vo (Input). Hai s t nhin a, b. u ra (Output). S nguyn u ln nht a v b u chia ht cho u. Thut ton (Euclide Algorithm): Bc 1. a vo hai s t nhin a v b.

Bc 2. Nu b 0 th chuyn n bc 3, nu b=0 th thc hin bc 4. Bc 3. t r = a mod b; a = b; b = r ; Sau quay tr li bc 2. Bc 4 (Output). Kt lun u=a l s nguyn cn tm.

V d 2. Biu din Biu din thut ton tm USCLN (a, b)bng ngn ng hnh thc. Thut ton Euclide:

u vo (Input): aN, aN.

u ra (Output): s = max { u N : a mod u =0 and b mod u =0}. Format : s = Euclide (a, b). Actions :

while (b 0 ) do r = a mod b; a = b; b = r; endwhile; return(a); Endactions.

V d 3. Biu din thut ton tm USCLN (a, b) bng ngn ng my tnh (C++). Int USCLN( int a, int b) { while ( b != 0 ) { r = a % b; a = b; b = r;

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48

} return(a); }

3.2.3. phc tp tnh ton

Mt bi ton c th thc hin bng nhiu thut ton khc nhau. Chn gii thut nhanh nht gii bi ton l mt nhu cu ca thc t. V vy ta cn phi c s c lng c th minh chng bng ton hc mc nhanh chm ca mi gii thut. Khi nim phc tp thut ton: Thi gian thc hin mt gii thut bng chng trnh my tnh ph thuc vo cc yu t:

Kch thc d liu vo: D liu cng ln th thi gian x l cng chm. Phn cng my tnh: my c tc cao thc hin nhanh hn trn my c

tc thp. Tuy vy, yu t ny khng nh hng n qu trnh xc nh thi gian thc hin ca thut ton nu xem xt thi gian thc hin thut ton nh mt hm ca di d liu T(n).

Tng qut, cho hai hm f(x), g(x) xc nh trn tp cc s nguyn dng hoc tp cc s thc vo tp cc s thc. Hm f(x) c gi l O(g(x)) nu tn ti mt hng s C>0 v n0 sao cho: |f(x)| C.|g(x)| v mi xn0.

iu ny c ngha vi cc gi tr x n0 hm f(x) b chn trn bi hng s C nhn vi g(x). Nu f(x) l thi gian thc hin ca mt thut ton th ta ni gii thut c cp g(x) hay phc tp thut ton l O(g(x)). Ghi ch. Cc hng s C, n0 tha mn iu kin trn l khng duy nht. Nu c ng thi f(x) l O(g(x)) v h(x) tha mn g(x) < h(x) vi x>n0 th ta cng c f(x) l O(h(n)).

V d 1. Cho 0111 ... axaxaxaxf nnnn . Trong , ai l cc s thc (i =0,1, 2, ..,n). Khi f(x) = O(xn). Chng minh. Thc vy, vi mi x>1:

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49

011

011

011

011

1

011

1

...).(.

...

...

...

...

aaaaCxOxC

aaaax

xaxaxaxa

axaxaxa

axaxaxaxf

nn

nn

nnn

nnnn

nn

nn

nn

nn

nn

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50

V d 2. Tm phc tp thut ton sp xp kiu Bubble-Sort? Void Bubble-Sort ( int A[], int n ) { for ( i=1; i

51

Cc dng hm nh gi phc tp thut ton:

Dng nh gi Tn gi

O(1) Hng s

O(lg lg n) Log log

O(lg n) Logarithm

O(n) Tuyn tnh

O(n2) Bc hai

O(n3) Bc 3

O(nm) a thc

O(mn) Hm m

O(n!) Giai tha

3.2.4. Qui tc xc nh phc tp thut ton

Qui tc tng: Nu f1(x) c phc tp l O(g1(x)) v f2(x) c phc tp l O(g2(x)) th phc tp ca (f1(x) + f2(x) l O( Max(g1(x), g2(x)). Chng minh.

V f1(x) c phc tp l O(g1(x) nn tn ti hng s C1 v k1 sao cho |f1(x)||g1(x)| vi mi x k1;

V f2(x) c phc tp l O(g2(x)) nn tn ti hng s C2 v k2 sao cho |f2(x)||g2(x)| vi mi x k2;

Ta li c :

|f1(x)+ f2(x)| |f1(x)| + |f2(x)|

C1|g1(x)| + C2|g2(x)|

C|g(x)| vi mi x >k; Trong , C = C1 + C2; g(x) = max( g1(x), g2(x)); k = max (k1, k2). Tng qut. Nu phc tp ca f1(x), f2(x),.., fm(x) ln lt l O(g1(x)), O(g2(x)),.., O(gn(x)) th phc tp ca f1(x) + f2(x) + ..+fm(x) l O(max(g1(x), g2(x),..,gm(x)).

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Qui tc nhn: Nu f(x) c phc tp l O(g(x) th phc tp ca fn(x) l O(gn(x). Trong : fn(x) = f(x).f(x).f(x). //n ln f(x). gn(x) = g(x).g(x)g(x).//n ln g(x) Ni cch khc, on chng trnh P c thi gian thc hin T(n)= O(f(n)). Khi , nu thc hin k(n) ln on chng trnh P vi k(n) l O(g(n)) th phc tp tnh ton l O(f(n). g(n)). Chng minh. Tht vy theo gi thit f(x) l O(g(x)) nn tn ti hng s C v k sao cho vi mi x>k th |f(x)| C.|g(x). Ta c:

xgOxgC

xgCxgCxgC

xfxfxfxf

nnn

n

nn

.......

....21

21

3.3. Phng php sinh

Phng php sinh c th p dng gii cc bi ton lit k t hp t ra nu nh hai iu kin sau c thc hin:

(i) C th xc nh c mt th t trn tp cc cu hnh t hp cn lit k. T c th xc nh c cu hnh t hp u tin v cui cng trong th t c xc nh.

(ii) Xy dng c thut ton t cu hnh cha phi l cui cng ang c a ra cu hnh k tip sau n.

Ta gi thut ton trong iu kin (ii) l thut ton sinh k tip. R rng thut ton ny ch thc hin c khi c mt cu hnh c xc nh theo iu kin (i). Gi s mt bi ton u tho mn cc iu kin trn, khi phng php sinh k tip c th c m t bng th tc nh sau:

procedure Generation ( ){ ; stop =false while (! stop) { ; ; } }

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V d 1. Lit k tt c cc dy nh phn di n.

Li gii. Vit dy nh phn di dng b1b2..bn, trong bi{0, 1 }. Xem mi dy nh phn b=b1b2..bn l biu din nh phn ca mt s nguyn p(b). Khi th t hin nhin nht c th xc nh trn tp cc dy nh phn l th t t in c xc nh nh sau:

Ta ni dy nh phn b = b1b2..bn i trc dy nh phn b = b1b2..bn theo th t t in v k hiu b 0 && bi !=0 ) { //Nu bi=1 th gn thnh 0

bi = 0; i = i-1; }

if ( i > 0 ) bi = 1; //Nu cha l cu hnh cui cng th i>0 else OK = False; ///Nu l cu hnh cui cng th i=0 }

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Di y l chng trnh lit k cc xu nh phn c di n. #include #include #define MAX 100 #define TRUE 1 #define FALSE 0 int n, X[MAX], OK=TRUE, dem=0; void Init (void ){ coutn; for (int i=1; i

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Ket qua buoc 12:1 0 1 1 Ket qua buoc 13:1 1 0 0 Ket qua buoc 14:1 1 0 1 Ket qua buoc 15:1 1 1 0 Ket qua buoc 16:1 1 1 1

V d 2. Lit k tp con m phn t ca tp n phn t. Cho X = { 1, 2, . ., n }. Hy lit k tt c cc tp con k phn t ca X (k n). Li gii: Mi tp con ca tp hp X c th biu din bng b c th t gm k thnh phn a =(a1a2..ak) tho mn 1 a1 a2 . . ak n. Trn tp cc tp con k phn t ca X c th xc nh nhiu th t khc nhau. Th t d nhn thy nht l th t t in c nh ngha nh sau:

Ta ni tp con a = a1a2. . . ak i trc tp con a = a1a2. . .ak trong th t t in v k hiu l a

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Thut ton lit k tp con k tip m phn t ca tp n phn t: void Next_Combination(void){ i = k; //Xut pht t v tr th k

while ( i> 0 && ai == n-k+i) //Xc nh i ai n-k+i i = i -1; if (i>0) { //Nu cha phi l t hp cui cng th i>0 ai = ai + 1; for ( j = i+1; j

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void Next_Combination(void) { int i= k; while (i>0 && X[i]==n-k+i)i--; if (i > 0 ) { X[i] = X[i] +1; for (int j = i+1; j

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a = (a1, a2, .., an) tho mn ai X, i = 1, 2, .., n, ap aq, p q. Trn tp cc hon v t n phn t ca X c th xc nh nhiu th t khc nhau. Tuy nhin, th t d thy nht l th t t in c nh ngha nh sau: Ta ni hon v a = a1a2. . . an i trc hon v a = a1a2. . .an trong th t t in v k hiu l a

59

Thut ton sinh hon v k tip: void Next_Permutation( void ){ j = n-1; //Duyt t v tr j=n-1 while (j> 0 && aj > aj +1 ) //Tm v tr j aj > aj +1 j = j -1; if (j>0) { // Nu j>0 th hon v cha phi cui cng k = n; //Xut pht t v tr k=n while (aj > ak ) // Tm k aj < ak k= k - 1; temp =aj; aj = ak; ak = temp;//i ch aj cho ak r = j + 1; s = n; while ( r < s) {//Lt ngc li on t j+1 n n temp = ar; ar = as; as = temp; r = r +1; s = s - 1; }

} else OK = False;//Nu l hon v cui cng th i=0

} Chng trnh lit k hon v c th hin nh sau:

#include #define MAX 100 #define TRUE 1 #define FALSE 0 int n,X[MAX],OK=TRUE,dem=0; void Init (void ){ coutn; for (int i=1; i

60

void Next_Permutation(void) { int j= n-1; while (j>0 && X[j]>X[j+1])j--; if (j > 0 ) { int k =n; while(X[j]>X[k]) k--; int t = X[j]; X[j]=X[k]; X[k]=t; int r = j +1, s =n; while (rb2>...> bk, v duyt theo trnh t t in ngc. Chng hn vi n = 5, chng ta c th t t in ngc ca cc cch phn chia nh sau:

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5 4 1 3 2 3 1 1 2 2 1 2 1 1 1 1 1 1 1 1 1

Nh vy, cch chia u tin chnh l n. Cch chia cui cng l dy n s 1. By gi chng ta ch cn xy dng thut ton sinh k tip cho mi cch phn chia cha phi l cui cng. Thut ton sinh cch phn chia k tip:

void Next_Division(void){ int i, j, R, S, D; i = k; //Xut pht t cui cch chia trc while(i>0 && C[i]==1) //Tm i sao cho C[i]1 i--; if(i>0){ // Nu cha phi l cch chia cui cng th i>0 C[i] = C[i]-1; //Gim C[i] i mt n v. D = k - i +1; R = D / C[i]; S = D % C[i]; k = i; if(R>0){ for(j=i+1; j0){ k=k+1; C[k] = S; } } else Stop=TRUE; }

PTIT

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Chng trnh lit k cc cch chia s n thnh tng cc s nh hn: #include #include #define MAX 100 #define TRUE 1 #define FALSE 0 int n, k, X[MAX], dem =0, OK =TRUE; void Init(void ){ coutn; k = 1; X[k] = n; } void Result(void) { cout

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3.4. Thut ton quay lui (Back track)

Phng php sinh k tip c th gii quyt c cc bi ton lit k khi ta nhn bit c cu hnh u tin & cu hnh cui cng ca bi ton. Tuy nhin, khng phi cu hnh sinh k tip no cng c sinh mt cch n gin t cu hnh hin ti, ngay k c vic pht hin cu hnh ban u cng khng phi d tm v nhiu khi chng ta phi chng minh s tn ti ca cu hnh. Do vy, thut ton sinh k tip ch gii quyt c nhng bi ton lit k n gin. gii quyt nhng bi ton t hp phc tp, ngi ta thng dng thut ton quay lui (Back Track) s c trnh by di y.

Ni dung chnh ca thut ton ny l xy dng dn cc thnh phn ca cu hnh bng cch th tt c cc kh nng. Gi s cn phi tm mt cu hnh ca bi ton x = (x1, x2, . ., xn) m i-1 thnh phn x1, x2, . ., xi-1 c xc nh, by gi ta xc nh thnh phn th i ca cu hnh bng cch duyt tt c cc kh nng c th c v nh s cc kh nng t 1 . .ni. Vi mi kh nng j, kim tra xem j c chp nhn c hay khng. Khi c th xy ra hai trng hp:

Nu chp nhn j th xc nh xi theo j, nu i=n th ta c mt cu hnh cn tm, ngc li xc nh tip thnh phn xi+1.

Nu th tt c cc kh nng m khng c kh nng no c chp nhn th quay li bc trc xc nh li xi-1.

im quan trng nht ca thut ton l phi ghi nh li mi bc i qua, nhng kh nng no c th trnh s trng lp. nh li nhng bc duyt trc , chng trnh cn phi c t chc theo c ch ngn xp (Last in first out). V vy, thut ton quay lui rt ph hp vi nhng php gi qui. Thut ton quay lui xc nh thnh phn th i c th c m t bng th tc Try(i) nh sau:

void Try( int i ) { for ( j = 1; j < ni; j ++) {

if ( ) { if (i==n) ; else Try(i+1); }

} }

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C th m t qu trnh tm kim li gii theo thut ton quay lui bng cy tm kim li gii sau: Gc Kh nng chn x1 Kh nng chn x2 vi x1 chn Kh nng chn x3 vi x1, x2 chn Hnh 3.1. Cy lit k li gii theo thut ton quay lui. Di y l mt s v d in hnh s dng thut ton quay lui. V d 1. Lit k cc xu nh phn di n.

Biu din cc xu nh phn di dng b1, b2,..., bn, trong bi{0, 1 }. Th tc qui Try(i) xc nh bi vi cc gi tr c cho bi l 0 v 1. Cc gi tr ny mc nhin c chp nhn m khng cn phi tho mn iu kin g (do bi ton khng cn n bin trng thi). Th tc Init khi to gi tr n v bin m count. Th tc kt qu in ra dy nh phn tm c. Chng hn vi n =3 , cy tm kim li gii c th hin nh hnh 3.2. Gc 0 1 0 1 0 1 0 1 0 1 0 1 0 1 000 001 010 011 100 101 110 111 Hnh 3.2. Cy tm kim li gii lit k dy nh phn di 3 Vn bn chng trnh lit k cc xu nh phn c di n s dng thut ton quay lui c thc hin nh sau:

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#include #include #define MAX 100 #define TRUE 1 #define FALSE 0 int n, X[MAX], dem=0; void Init (void ){

coutn; } void Result(void){ cout

66

Chng trnh lit k cc tp con k phn t trong tp n phn t c th hin nh sau: #include #include #define MAX 100 #define TRUE 1 #define FALSE 0 int n,k,X[MAX],dem=0; void Init (void ){ coutn; coutk; X[0] = 0; } void Result(void){ cout

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Gc 1 2 3 2 3 1 3 1 2 3 2 3 1 2 1 1,2,3 1,3, 2 2,1,3 2,3,1 3,1,2 3,2,1 Hnh 3.4. Cy tm kim li gii bi ton lit k hon v ca {1,2,3} Sau y l chng trnh gii quyt bi ton lit k cc hon v ca 1, 2, . ., n.

#include #include #define MAX 100 #define TRUE 1 #define FALSE 0 int n,X[MAX],chuaxet[MAX],dem=0; void Init (void ){

coutn; for (int i=1; i

68

V d 4. Bi ton Xp Hu. Lit k tt c cc cch xp n qun hu trn bn c n x n sao cho chng khng n c nhau. Li gii. Bn c c n hng c nh s t 1 n n, n ct c nh s t 1 n n; Bn c c n*2 -1 ng cho xui c nh s t 1 n 2*n -1, 2 *n -1 ng cho ngc c nh s t 1n 2*n -1. V d: vi bn c 8 x 8, chng ta c 8 hng c nh s t 1 n 8, 8 ct c nh s t 1 n 8, 15 ng cho xui, 15 ng cho ngc c nh s t 1 . .15.

V trn mi hng ch xp c ng mt qun hu, nn chng ta ch cn quan tm n qun hu c xp ct no. T dn n vic xc nh b n thnh phn x1, x2, . ., xn, trong xi = j c hiu l qun hu ti dng i xp vo ct th j. Gi tr ca i c nhn t 1 n n; gi tr ca j cng c nhn t 1 n n, nhng tho mn iu kin (i,j) cha b qun hu khc chiu n theo ct, ng cho xui, ng cho ngc.

Vic kim sot theo hng ngang l khng cn thit v trn mi hng ch xp ng mt qun hu. Vic kim sot theo ct c ghi nhn nh dy bin logic chuaxetj vi qui c chuaxetj=True nu ct j cn trng, ct chuaxetj= False nu ct j khng cn trng. ghi nhn ng cho xui v ng cho ngc c chiu ti (i,j) hay khng, ta s dng phng trnh i + j = const v i - j = const, ng cho th nht c ghi nhn bi dy bin XUOIj, ng cho th 2 c ghi nhn bi dy bin NGUOCj vi qui c nu ng cho no cn trng th gi tr tng ng ca n l True ngc li l False. Nh vy, ct j c chp nhn khi c 3 bin chuaxetj, XUOIi+j, NGUOCi+j u c gi tr 1. Cc bin ny phi c khi u gi tr True trc , gn li gi tr False khi xp xong qun hu th i v tr li gi tr 1 khi a ra kt qu.

Di y l chng trnh bng thut ton quay lui. #include #include #define MAX 100 #define TRUE 1 #define FALSE 0 int X[MAX], XUOI[MAX], NGUOC[MAX], chuaxet[MAX]; int n, dem =0; void Init (void ) {

coutn; for (int i=1; i

69

void Try(int i){

for (int j =1; j

70

BI TP CHNG 3

1. Lit k tt c cc xu nh phn di 5 khn cha hai s 0 lin tip. 2. Lit k tt c cc phn t ca tp

njZxbxaxxxxD jn

jjjn ,,2,1,,:,,,(

121

Trong a1,a2,..,an, b l cc s nguyn dng. 3. Hnh vung thn b ma phng bc n l ma trn vung cp n vi cc phn t l

cc s t nhin t 1 n n2 tha mn cc tnh cht: Tng cc phn t trn mi dng, mi ct v mi mt trong hai ng cho c cng mt gi tr. Hy lit k tt c cc ma phng bc 3, 4 khng sai khc nhau bi cc php bin hnh n gin (quay, i xng).

V d di y l mt ma phng bc 3 tha mn tnh cht tng hng, ct, ng cho u l 15.

294753618

4. Tam gic thn b. Cho mt li vung gm n x n v s nguyn dng k. Tm

cch in cc s t nhin t 1 n 3n-3 vo cc ct u tin, dng cui cng v ng cho chnh sao cho tng cc s in trong ct u tin, dng cui cng v ng cho chnh ca li u bng k. V d n=5, k = 35 ta c cch in sau:

11

10 3

9 2

1 7

4 5 6 8 12

Pht trin thut ton da trn thut ton quay lui ch ra vi gi tr ca n, k cho trc bi ton c li gii hay khng. Nu c cu tr li ch cn a ra mt li gii.

PTIT

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5. Tm tp con di nht c th t tng dn, gim dn. Ch