Giao Trinh Toan Roi Rac - Chuong 2

8/6/2019 Giao Trinh Toan Roi Rac - Chuong 2

1/15

CHNG II

BI TON ML thuyt t hp l mt phn quan trng ca ton hc ri rc chuyn nghi

s phn b cc phn t vo cc tp hp. Thng thng cc phn t ny l hu hvic phn b chng phi tho mn nhng iu kin nht nh no , ty theo yca bi ton cn nghin cu. Mi cch phn b nh vy gi l mt cu hnh t h ny c nghin cu t th k 17, khi nhng cu hi v t hp c nu ranhng cng trnh nghin cu cc tr chi may ri. Lit k, m cc i tng c ntnh cht no l mt phn quan trng ca l thuyt t hp. Chng ta cn phi i tng gii nhiu bi ton khc nhau. Hn na cc k thut m c dnhiu khi tnh xc sut ca cc bin c.2.1. C S CA PHP M.2.1.1. Nhng nguyn l m c bn:1) Quy tc cng:Gi s c k cng vic T1, T2, ..., Tk . Cc vic ny c th lm tngng bng n1, n2, ..., nk cch v gi s khng c hai vic no c th lm ng thi. Khis cch lm mt trong k vic l n1+n2+ ... + nk .Th d 1: 1) Mt sinh vin c th chn bi thc hnh my tnh t mt trong ba dasch tng ng c 23, 15 v 19 bi. V vy, theo quy tc cng c 23 + 15 + 19 cch chn bi thc hnh.2)Gi tr ca bin m bng bao nhiu sau khi on chng trnh sau c thc hin

m := 0for i1 := 1to n1m := m+1

for i2 :=1 to n2m := m+1

.......................for ik := 1to nk

m := m+1Gi tr khi to ca m bng 0. Khi lnh ny gm k vng lp khc nhau. Sau

bc lp ca tng vng lp gi tr ca k c tng ln mt n v. Gi Ti l vic thihnh vng lp th i. C th lm Ti bng ni cch v vng lp th i c ni bc lp. Do ccvng lp khng th thc hin ng thi nn theo quy tc cng, gi tr cui cng bng s cch thc hin mt trong s cc nhim v Ti, tc l m = n1+n2+ ... + nk .

Quy tc cng c th pht biu di dng ca ngn ng tp hp nh sau: N1,A2, ..., Ak l cc tp hp i mt ri nhau, khi s phn t ca hp cc tp hp bng tng s cc phn t ca cc tp thnh phn. Gi s Ti l vic chn mt phn t t

22


2/15

tp Ai vi i=1,2, ..., k. C |Ai| cch lm Ti v khng c hai vic no c th c lmcng mt lc. S cch chn mt phn t ca hp cc tp hp ny, mt mt bng st ca n, mt khc theo quy tc cng n bng |A1|+|A2|+ ... +|Ak |. Do ta c:

|A1 A2 ... Ak | = |A1| + |A2| + ... + |Ak |.2) Quy tc nhn:Gi s mt nhim v no c tch ra thnh k vic T1, T2, ..., Tk . Nu vic Ti c th lm bng ni cch sau khi cc vic T1, T2, ... Ti-1 c lm, khi c n1.n2....nk cch thi hnh nhim v cho.Th d 2:1) Ngi ta c th ghi nhn cho nhng chic gh trong mt ging bng mt ch ci v mt s nguyn dng khng vt qu 100. Bng cch nhnhiu nht c bao nhiu chic gh c th c ghi nhn khc nhau?

Th tc ghi nhn cho mt chic gh gm hai vic, gn mt trong 26 ch csau gn mt trong 100 s nguyn dng. Quy tc nhn ch ra rng c 26.100=cch khc nhau gn nhn cho mt chic gh. Nh vy nhiu nht ta c th g

cho 2600 chic gh.2)C bao nhiu xu nh phn c di n.Mi mt trong n bit ca xu nh phn c th chn bng hai cch v mi bit h

bng 0 hoc bng 1. Bi vy theo quy tc nhn c tng cng 2n xu nh phn khc nhauc di bng n.3) C th to c bao nhiu nh x t tp A c m phn t vo tp B c n phn t

Theo nh ngha, mt nh x xc nh trn A c gi tr trn B l mt php tngng mi phn t ca A vi mt phn t no ca B. R rng sau khi chn ca i - 1 phn t u, chn nh ca phn t th i ca A ta c n cch. V vy thtc nhn, ta c n.n...n=nm nh x xc nh trn A nhn gi tr trn B.4)C bao nhiu n nh xc nh trn tp A c m phn t v nhn gi tr trn tp B phn t?

Nu m > n th vi mi nh x, t nht c hai phn t ca A c cng mt nh c ngha l khng c n nh t A n B. By gi gi s m n v gi cc phn tca A l a1,a2,...,am. R rng c n cch chn nh cho phn t a1. V nh x l n nh nnnh ca phn t a2 phi khc nh ca a1 nn ch c n - 1 cch chn nh cho phn t a2. Ni chung, chn nh ca ak ta c n - k + 1 cch. Theo quy tc nhn, ta c

n(n 1)(n 2)...(n m + 1) =n

n m

!( )!

n nh t tp A n tp B.5)Gi tr ca bin k bng bao nhiu sau khi chng trnh sau c thc hin?

m := 0for i1 := 1to n1

for i2 := 1to n2.......................

23


3/15

for ik := 1to nk k := k+1

Gi tr khi to ca k bng 0. Ta c k vng lp c lng nhau. Gi Ti l vic thihnh vng lp th i. Khi s ln i qua vng lp bng s cch lm cc vic T1, T2, ...,Tk . S cch thc hin vic T j l n j (j=1, 2,..., k), v vng lp th j c duyt vi mi gtr nguyn i j nm gia 1 v n j. Theo quy tc nhn vng lp lng nhau ny c duyqua n1.n2....nk ln. V vy gi tr cui cng ca k l n1.n2....nk .

Nguyn l nhn thng c pht biu bng ngn ng tp hp nh sau. N1,A2,..., Ak l cc tp hu hn, khi s phn t ca tch Descartes ca cc tp ny tch ca s cc phn t ca mi tp thnh phn. Ta bit rng vic chn mt phtch Descartes A1 x A2 x...x Ak c tin hnh bng cch chn ln lt mt phn t A1, mt phn t ca A2, ..., mt phn t ca Ak . Theo quy tc nhn ta c:

|A1 x A2 x ... x Ak | = |A1|.|A2|...|Ak |.

2.1.2. Nguyn l b tr:Khi hai cng vic c th c lm ng thi, ta khng th dng quy tc cntnh s cch thc hin nhim v gm c hai vic. tnh ng s cch thc hinv ny ta cng s cch lm mi mt trong hai vic ri tr i s cch lm ng thhai vic. Ta c th pht biu nguyn l m ny bng ngn ng tp hp. Cho A1, A2 lhai tp hu hn, khi

|A1 A2| = |A1| + |A2| |A1 A2|.T vi ba tp hp hu hn A1, A2, A3, ta c:

|A1 A2 A3| = |A1| + |A2| + |A3| |A1 A2| |A2 A3| |A3 A1| + |A1 A2 A3|,v bng quy np, vi k tp hu hn A1, A2, ..., Ak ta c:

| A1 A2 ... Ak | = N1 N2 + N3 ... + (1)k-1 Nk ,trong Nm (1 m k) l tng phn t ca tt c cc giao m tp ly t k tp ngha l

Nm = |...|...1 21 21 mmi

k iiiii A A A


4/15


5/15

Theo nguyn l Dirichlet, s hc sinh cn tm l 102, v ta c 101 kt qu thi khc nhau.

3) Trong s nhng ngi c mt trn tri t, phi tm c hai ngi c hmrng ging nhau. Nu xem mi hm rng gm 32 ci nh l mt xu nh phn c

chiu di 32, trong rng cn ng vi bit 1 v rng mt ng vi bit 0, th c tt 232 = 4.294.967.296 hm rng khc nhau. Trong khi s ngi trn hnh tinh ny

l vt qu 5 t, nn theo nguyn l Dirichlet ta c iu cn tm.2.2.2. Nguyn l Dirichlet tng qut:Mnh :Nu c N vt c t vo trong k hp th s tn ti mt hp cha ] N/k [ vt.( y,] x[ l gi tr ca hm trn ti s thc x, l s nguyn nh nht c gi trhn hoc bng x. Khi nim ny i ngu vi [x] gi tr ca hm sn hay hm nguyn ti x l s nguyn ln nht c gi tr nh hn hoc bng x.)

Chng minh: Gi s mi hp u cha t hn] N/k [ vt. Khi tng s vt l

k (] N

k [ 1) < k

N

k = N .

iu ny mu thun vi gi thit l c N vt cn xp.Th d 5: 1) Trong 100 ngi, c t nht 9 ngi sinh cng mt thng.

Xp nhng ngi sinh cng thng vo mt nhm. C 12 thng tt c. Vynguyn l Dirichlet, tn ti mt nhm c t nht] 100/12[ = 9 ngi.2) C nm loi hc bng khc nhau. Hi rng phi c t nht bao nhiu sinh vichc chn rng c t ra l 6 ngi cng nhn hc bng nh nhau.

Gi N l s sinh vin, khi ] N/5[ = 6 khi v ch khi 5 < N/5 6 hay 25 < N 30. Vy s N cn tm l 26.3) S m vng cn thit nh nht phi l bao nhiu m bo 25 triu my itrong nc c s in thoi khc nhau, mi s c 9 ch s (gi s s in thoi c0XX - 8XXXXX vi X nhn cc gi tr t 0 n 9).

C 107 = 10.000.000 s in thoi khc nhau c dng 0XX - 8XXXXX. V theo nguyn l Dirichlet tng qut, trong s 25 triu my in thoi t nht c]25.000.000/10.000.000[ = 3 c cng mt s. m bo mi my c mt s cn cnht 3 m vng.2.2.3. Mt s ng dng ca nguyn l Dirichlet.

Trong nhiu ng dng th v ca nguyn l Dirichlet, khi nim vt vcn phi c la chn mt cch khn kho. Trong phn ny c vi th d nh vTh d 6: 1) Trong mt phng hp c n ngi, bao gi cng tm c 2 ngi cngi quen trong s nhng ngi d hp l nh nhau.

S ngi quen ca mi ngi trong phng hp nhn cc gi tr t 0 n n 1. Rrng trong phng khng th ng thi c ngi c s ngi quen l 0 (tc l k

26


6/15

quen ai) v c ngi c s ngi quen l n 1 (tc l quen tt c). V vy theo s lnngi quen, ta ch c th phn n ngi ra thnh n1 nhm. Vy theo nguyn l Dirichlettn tai mt nhm c t nht 2 ngi, tc l lun tm c t nht 2 ngi c s quen l nh nhau.2) Trong mt thng gm 30 ngy, mt i bng chuyn thi u mi ngy t nht 1nhng chi khng qu 45 trn. Chng minh rng tm c mt giai on gm mngy lin tc no trong thng sao cho trong giai on i chi ng 14 trn.

Gi a j l s trn m i chi t ngy u thng n ht ngy j. Khi 1 a1< a2 < ... < a30 < 45

15 a1+14 < a2+14 < ... < a30+14 < 59.Su mi s nguyn a1, a2, ..., a30, a1+ 14, a2 + 14, ..., a30+14 nm gia 1 v 59. Do theo nguyn l Dirichlet c t nht 2 trong 60 s ny bng nhau. V vy tn ti i vcho ai = aj + 14 (j < i). iu ny c ngha l t ngy j + 1 n ht ngy i i c

ng 14 trn.3) Chng t rng trong n + 1 s nguyn dng khng vt qu 2n, tn ti t nht mchia ht cho s khc.

Ta vit mi s nguyn a1, a2,..., an+1 di dng a j = jk 2 q j trong k j l s nguynkhng m cn q j l s dng l nh hn 2n. V ch c n s nguyn dng l nh hnn theo nguyn l Dirichlet tn ti i v j sao cho qi = q j = q. Khi ai= ik 2 q v aj =

jk 2 q. V vy, nu k i k j th a j chia ht cho ai cn trong trng hp ngc li ta c aichia ht cho a j.

Th d cui cng trnh by cch p dng nguyn l Dirichlet vo l thuyt t m vn quen gi ll thuyt Ramsey, tn ca nh ton hc ngi Anh. Ni chung, lthuyt Ramsey gii quyt nhng bi ton phn chia cc tp con ca mt tp cc phTh d 7.Gi s trong mt nhm 6 ngi mi cp hai hoc l bn hoc l th. Chrng trong nhm c ba ngi l bn ln nhau hoc c ba ngi l k th ln nhau.

Gi A l mt trong 6 ngi. Trong s 5 ngi ca nhm hoc l c t nhngi l bn ca A hoc c t nht ba ngi l k th ca A, iu ny suy ra t nl Dirichlet tng qut, v] 5/2[ = 3. Trong trng hp u ta gi B, C, D l bn ca Anu trong ba ngi ny c hai ngi l bn th h cng vi A lp thnh mt b ba

bn ln nhau, ngc li, tc l nu trong ba ngi B, C, D khng c ai l bn ai chng t h l b ba ngi th ln nhau. Tng t c th chng minh trong trnc t nht ba ngi l k th ca A.2.3. CHNH HP V T HP SUY RNG.2.3.1. Chnh hp c lp.

Mt cch sp xp c th t k phn t c th lp li ca mt tp n phn tgi l mt chnh hp lp chp k t tp n phn t. Nu A l tp gm n phn t

27


7/15

chnh hp nh th l mt phn t ca tp Ak . Ngoi ra, mi chnh hp lp chp k t tn phn t l mt hm t tp k phn t vo tp n phn t. V vy s chnh hp lpt tp n phn t l nk .2.3.2. T hp lp.

Mt t hp lp chp k ca mt tp hp l mt cch chn khng c th t kt c th lp li ca tp cho. Nh vy mt t hp lp kiu ny l mt dy khth t gm k thnh phn ly t tp n phn t. Do c th l k > n.Mnh 1: S t hp lp chp k t tp n phn t bngk k nC 1+ .Chng minh.Mi t hp lp chp k t tp n phn t c th biu din bng mt d1thanh ng v k ngi sao. Ta dng n 1 thanh ng phn cch cc ngn. Ngn th icha thm mt ngi sao mi ln khi phn t th i ca tp xut hin trong t hp.hn, t hp lp chp 6 ca 4 phn t c biu th bi:

* * | * | | * * *

m t t hp cha ng 2 phn t th nht, 1 phn t th hai, khng c phn t t3 phn t th t ca tp hp.Mi dy n 1 thanh v k ngi sao ng vi mt xu nh phn di n + k 1 vi k

s 1. Do s cc dy n 1 thanh ng v k ngi sao chnh l s t hp chp k t tp+ k 1 phn t. l iu cn chng minh.Thi d 8: 1) C bao nhiu cch chn 5 t giy bc t mt kt ng tin gm nhn1000, 2000, 5000, 10.000, 20.000, 50.000, 100.000. Gi s th t m ctin c chn l khng quan trng, cc t tin cng loi l khng phn bit v mc t nht 5 t.

V ta khng k ti th t chn t tin v v ta chn ng 5 ln, mi ln ly 1 trong 7 loi tin nn mi cch chn 5 t giy bc ny chnh l mt t hp lp ch7 phn t. Do s cn tm l5 157 +C = 462.2)Phng trnh x1 + x2 + x3 = 15 c bao nhiu nghim nguyn khng m?

Chng ta nhn thy mi nghim ca phng trnh ng vi mt cch ch phn t t mt tp c 3 loi, sao cho c x1 phn t loi 1, x2 phn t loi 2 v x3 phn tloi 3 c chn. V vy s nghim bng s t hp lp chp 15 t tp c 3 ph bng 15 1153 +C = 136.2.3.3. Hon v ca tp hp c cc phn t ging nhau.

Trong bi ton m, mt s phn t c th ging nhau. Khi cn phi cntrnh m chng hn mt ln. Ta xt th d sau.Th d 9:C th nhn c bao nhiu xu khc nhau bng cch sp xp li cc chca t SUCCESS?

V mt s ch ci ca t SUCCESS l nh nhau nn cu tr li khng phi hon v ca 7 ch ci c. T ny cha 3 ch S, 2 ch C, 1 ch U v 1 ch E. nh s xu khc nhau c th to ra c ta nhn thy c C(3,7) cch chn 3 ch

28


8/15

ch S, cn li 4 ch trng. C C(2,4) cch chn 2 ch cho 2 ch C, cn li 2 ch C th t ch U bng C(1,2) cch v C(1,1) cch t ch E vo xu. Theo nguynhn, s cc xu khc nhau c th to c l:

37C . 24C . 12C . 11C =

7 4 2 1

3 4 2 2 1 1 1 0

! ! ! !

!. !. !. !. !. !. !. !=

7

3 2 1 1

!

!. !. !. != 420.

Mnh 2:S hon v ca n phn t trong c n1 phn t nh nhau thuc loi 1, n2 phn t nh nhau thuc loi 2, ..., v nk phn t nh nhau thuc loi k, bng

!!....!.!

21 k nnn

n.

Chng minh. xc nh s hon v trc tin chng ta nhn thy c1nnC cch gi n1ch cho n1 phn t loi 1, cn li n - n1 ch trng. Sau c 2

1

n

nnC

cch t n2 phn tloi 2 vo hon v, cn li n - n1 - n2 ch trng. Tip tc t cc phn t loi 3, loi 4,loi k - 1vo ch trng trong hon v. Cui cng ck

k

n

nnnC

11 ... cch t nk phn tloi k vo hon v. Theo quy tc nhn tt c cc hon v c th l:

1nnC . 2 1

n

nnC

.... k k n

nnnC

11 ... = !!....!.!

21 k nnnn .

2.3.4. S phn b cc vt vo trong hp.Th d 10:C bao nhiu cch chia nhng xp bi 5 qun cho mi mt trong 4 ngchi t mt c bi chun 52 qun?

Ngi u tin c th nhn c 5 qun bi bng552C cch. Ngi th hai c thc chia 5 qun bi bng547C cch, v ch cn 47 qun bi. Ngi th ba c th nhc 5 qun bi bng542C cch. Cui cng, ngi th t nhn c 5 qun bi b

537C cch. V vy, theo nguyn l nhn tng cng c

552C . 547C . 542C . 537C =

52!

5 5 5 5 32!!. !. !. !.

cch chia cho 4 ngi mi ngi mt xp 5 qun bi.Th d trn l mt bi ton in hnh v vic phn b cc vt khc nhau

cc hp khc nhau. Cc vt l 52 qun bi, cn 4 hp l 4 ngi chi v s cn trn bn. S cch sp xp cc vt vo trong hp c cho bi mnh sauMnh 3: S cch phn chia n vt khc nhau vo trong k hp khc nhau sao cc ni vt c t vo trong hp th i, vi i = 1, 2, ..., k bng

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

121 k k nnnnnnn

.2.4. SINH CC HON V V T HP.2.4.1. Sinh cc hon v:

C nhiu thut ton c pht trin sinh ra n! hon v ca tp {1,2,...,ns m t mt trong cc phng php , phng php lit k cc hon v c{1,2,...,n} theo th t t in. Khi , hon v a1a2...an c gi l i trc hon v b1 b2...bn nu tn ti k (1 k n), a1 = b1, a2 = b2,..., ak -1 = bk -1 v ak < bk .

29


9/15

Thut ton sinh cc hon v ca tp {1,2,...,n} da trn th tc xy dng hok tip, theo th t t in, t hon v cho trc a1 a2 ...an. u tin nu an-1 < an th rrng i ch an-1 v an cho nhau th s nhn c hon v mi i lin sau hon v c Nu tn ti cc s nguyn a j v a j+1 sao cho a j < a j+1 v a j+1 > a j+2> ... > an, tc l tm cps nguyn lin k u tin tnh t bn phi sang bn tri ca hon v m s u nhs sau. Sau , nhn c hon v lin sau ta t vo v tr th j s nguyn nhtrong cc s ln hn a j ca tp a j+1, a j+2, ..., an, ri lit k theo th t tng dn ca cc scn li ca a j, a j+1, a j+2, ..., an vo cc v tr j+1, ..., n. D thy khng c hon v no i sahon v xut pht v i trc hon v va to ra.Th d 11:Tm hon v lin sau theo th t t in ca hon v 4736521.

Cp s nguyn u tin tnh t phi qua tri c s trc nh hn s sau l a3 = 3v a4 = 6. S nh nht trong cc s bn phi ca s 3 m li ln hn 3 l s 5. t vo v tr th 3. Sau t cc s 3, 6, 1, 2 theo th t tng dn vo bn v tr cn

Hon v lin sau hon v cho l 4751236.procedureHon v lin sau (a1, a2, ..., an) (hon v ca {1,2,...,n} khc (n, n1, ..., 2, 1)) j := n 1whilea j> a j+1

j := j 1 {j l ch s ln nht m a j< a j+1}k := nwhilea j > ak

k := k - 1 {ak l s nguyn nh nht trong cc s ln hn a j v bn phi a j}i ch (a j, ak )

r := ns := j + 1whiler > s

i ch (ar , as)r := r - 1 ; s := s + 1

{iu ny s xp phn ui ca hon v sau v tr th j theo th t tng dn.}

2.4.2. Sinh cc t hp:

Lm th no to ra tt c cc t hp cc phn t ca mt tp hu hnhp chnh l mt tp con, nn ta c th dng php tng ng 1-1 gia cc tp co{a1,a2,...,an} v xu nh phn di n.

Ta thy mt xu nh phn di n cng l khai trin nh phn ca mt s ngunm gia 0 v 2n 1. Khi 2n xu nh phn c th lit k theo th t tng dn ca nguyn trong biu din nh phn ca chng. Chng ta s bt u t xu nh phnht 00...00 (n s 0). Mi bc tm xu lin sau ta tm v tr u tin tnh t ph

30


10/15

tri m l s 0, sau thay tt c s 1 bn phi s ny bng 0 v t schnh v tr ny.procedureXu nh phn lin sau (bn-1 bn-2...b1 b0): xu nh phn khc (11...11)

i := 0while bi = 1begin

bi := 0i := i + 1

end bi := 1Tip theo chng ta s trnh by thut ton to cc t hp chp k t n ph

{1,2,...,n}. Mi t hp chp k c th biu din bng mt xu tng. Khi c th cc t hp theo th t t in. C th xy dng t hp lin sau t hp a1a2...ak bng cch

sau. Trc ht, tm phn t u tin ai trong dy cho k t phi qua tri sao cho ai n k + i. Sau thay ai bng ai + 1 v a j bng ai + j i + 1 vi j = i + 1, i + 2, ..., k.Th d 12:Tm t hp chp 4 t tp {1, 2, 3, 4, 5, 6} i lin sau t hp {1, 2, 5, 6}.

Ta thy t phi qua tri a2 = 2 l s hng u tin ca t hp cho tha miu kin ai 6 4 + i. nhn c t hp tip sau ta tng ai ln mt n v, tc a2 =3, sau t a3 = 3 + 1 = 4 v a4 = 3 + 2 = 5. Vy t hp lin sau t hp cho l{1,3,4,5}. Th tc ny c cho di dng thut ton nh sau.procedure T hp lin sau ({a1, a2, ..., ak }: tp con thc s ca tp {1, 2, ..., n} khn bng {n k + 1, ..., n} vi a1 < a2 < ... < ak )

i := k whileai = n k + i

i := i 1ai := ai + 1

for j := i + 1to k a j := ai + j i

2.5. H THC TRUY HI.2.5.1. Khi nim m u v m hnh ha bng h thc truy hi:

i khi ta rt kh nh ngha mt i tng mt cch tng minh. Nhng cd dng nh ngha i tng ny qua chnh n. K thut ny c gi l quy.ngha quy ca mt dy s nh r gi tr ca mt hay nhiu hn cc s hng v quy tc xc nh cc s hng tip theo t cc s hng i trc. nh ngha th dng gii cc bi ton m. Khi quy tc tm cc s hng t cc s htrc c gi l cc h thc truy hi.

31


11/15

nh ngha 1: H thc truy hi (hay cng thc truy hi) i vi dy s {an} l cngthc biu din an qua mt hay nhiu s hng i trc ca dy. Dy s c gi lgii hay nghim ca h thc truy hi nu cc s hng ca n tha mn h thc tny.Th d 13(Li kp): 1) Gi s mt ngi gi 10.000 la vo ti khon ca mnhmt ngn hng vi li sut kp 11% mi nm. Sau 30 nm anh ta c bao nhiu tin tti khon ca mnh?

Gi Pn l tng s tin c trong ti khon sau n nm. V s tin c trong ti khsau n nm bng s c sau n 1 nm cng li sut ca nm th n, nn ta thy dy {Pn}tho mn h thc truy hi sau:

Pn = Pn-1 + 0,11Pn-1 = (1,11)Pn-1vi iu kin u P0 = 10.000 la. T suy ra Pn = (1,11)n.10.000. Thay n = 30 chota P30 = 228922,97 la.

2)Tm h thc truy hi v cho iu kin u tnh s cc xu nh phn v khng c hai s 0 lin tip. C bao nhiu xu nh phn nh th c di bng 5?Gi an l s cc xu nh phn di n v khng c hai s 0 lin tip. nh

c h thc truy hi cho {an}, ta thy rng theo quy tc cng, s cc xu nh phn di n v khng c hai s 0 lin tip bng s cc xu nh phn nh th kt thc bncng vi s cc xu nh th kt thc bng s 0. Gi s n 3.

Cc xu nh phn di n, khng c hai s 0 lin tip kt thc bng s 1 chnxu nh phn nh th, di n 1 v thm s 1 vo cui ca chng. Vy chng c tt l an-1. Cc xu nh phn di n, khng c hai s 0 lin tip v kt thc bng s 0 phi c bit th n 1 bng 1, nu khng th chng c hai s 0 hai bit cui cng. Trotrng hp ny chng c tt c l an-2. Cui cng ta c c:

an = an-1 + an-2 vi n 3.iu kin u l a1 = 2 v a2 = 3. Khi a5 = a4 + a3 = a3 + a2 + a3 = 2(a2 + a1) + a2 = 13.2.5.2. Gii cc h thc truy hi.nh ngha 2:Mt h thc truy hi tuyn tnh thun nht bc k vi h s hng s l hthc truy hi c dng:

an = c1an-1 + c2an-2 + ... + ck an-k ,

trong c1, c2, ..., ck l cc s thc v ck 0.Theo nguyn l ca quy np ton hc th dy s tha mn h thc truy htrong nh ngha c xc nh duy nht bng h thc truy hi ny v k iu kina0 = C0, a1 = C1, ..., ak-1 = Ck-1.

Phng php c bn gii h thc truy hi tuyn tnh thun nht l tm ndi dng an = r n, trong r l hng s. Ch rng an = r n l nghim ca h thc truy hian = c1an-1 + c2an-2 + ... + ck an-k nu v ch nu

32


12/15

r n = c1r n-1 + c2r n-2 + ... + ck r n-k hay r k c1r k -1 c2r k -2 ... ck -1r ck = 0.Phng trnh ny c gi l phng trnh c trng ca h thc truy hi, nghin gi l nghim c trng ca h thc truy hi.Mnh : Cho c1, c2, ..., ck l cc s thc. Gi s rng phng trnh c trng

r k c1r k -1 c2r k -2 ... ck -1r ck = 0c k nghim phn bit r 1, r 2, ..., r k . Khi dy {an} l nghim ca h thc truy hi an =c1an-1 + c2an-2 + ... + ck an-k nu v ch nu an = 1r 1n + 2r 2n + ... + k r k n, vi n = 1, 2, ...trong 1, 2, ..., k l cc hng s.Th d 14: 1)Tm cng thc hin ca cc s Fibonacci.

Dy cc s Fibonacci tha mn h thc f n = f n-1 + f n-2 v cc iu kin u f 0 = 0v f 1 = 1. Cc nghim c trng l r 1 = 1 5

2

+ v r 2 = 1 52

. Do cc s Fibonacci

c cho bi cng thc f n = 1( 1 52+ )n + 2( 1 52

)n. Cc iu kin ban u f 0 = 0 =

1 + 2 v f 1 = 1 = 1( 1 52+

) + 2( 1 52

). T hai phng trnh ny cho ta 1 =1

5, 2 = -

1

5. Do cc s Fibonacci c cho bi cng thc hin sau:

f n=1

5( 1 5

2

+ )n - 15

( 1 52

)n.

2)Hy tm nghim ca h thc truy hi an = 6an-1 - 11an-2 + 6an-3 vi iu kin banu a0 = 2, a1 = 5 v a2 = 15.

a thc c trng ca h thc truy hi ny l r 3 - 6r 2 + 11r - 6. Cc nghim ctrng l r = 1, r = 2, r = 3. Do vy nghim ca h thc truy hi c dng

an = 11n + 22n + 33n.Cc iu kin ban u a0 = 2 = 1 + 2 + 3

a1 = 5 = 1 + 22 + 33a2 = 15 = 1 + 24 + 39.

Gii h cc phng trnh ny ta nhn c 1= 1, 2 = 1, 3 = 2. V th, nghim duynht ca h thc truy hi ny v cc iu kin ban u cho l dy {an} vi

an = 1 2n + 2.3n.2.6.QUAN H CHIA TR.

2.6.1. M u: Nhiu thut ton quy chia bi ton vi cc thng tin vo cho thnh mtnhiu bi ton nh hn. S phn chia ny c p dng lin tip cho ti khi c thc li gii ca bi ton nh mt cch d dng. Chng hn, ta tin hnh vic tmnh phn bng cch rt gn vic tm kim mt phn t trong mt danh sch ti vi phn t trong mt danh sch c di gim i mt na. Ta rt gn lin tip nhcho ti khi cn li mt phn t. Mt v d khc l th tc nhn cc s nguyn. Th

33


13/15

ny rt gn bi ton nhn hai s nguyn ti ba php nhn hai s nguyn vi s bit gi mt na. Php rt gn ny c dng lin tip cho ti khi nhn c cc s ngc mt bit. Cc th tc ny gi l cc thut ton chia tr.2.6.2. H thc chia tr:

Gi s rng mt thut ton phn chia mt bi ton c n thnh a bi ton

trong mi bi ton nh c c nb ( n gin gi s rng n chia ht cho b; trong th

t cc bi ton nh thng c c [n

b] hoc]

n

b[ ). Gi s rng tng cc php ton thm

vo khi thc hin phn chia bi ton c n thnh cc bi ton c c nh hn l g(n)., nu f(n) l s cc php ton cn thit gii bi ton cho th f tha mn htruy hi sau:

f(n) = af(n

b) + g(n)

H thc ny c tn l h thc truy hi chia tr.

Th d 15:1) Thut ton tm kim nh phn a bi ton tm kim c n v bi ton tm k phn t ny trong dy tm kim c n/2, khi n chn. Khi thc hin vic rt gn cn hai snh. V th, nu f(n) l s php so snh cn phi lm khi tm kim mt phn t trong datm kim c n ta c f(n) = f(n/2) + 2, nu n l s chn.

2)C cc thut ton hiu qu hn thut ton thng thng nhn hai s ng y ta s c mt trong cc thut ton nh vy. l thut ton phn nhanh, c k thut chia tr. Trc tin ta phn chia mi mt trong hai s nguyn 2n bit thai khi mi khi n bit. Sau php nhn hai s nguyn 2n bit ban u c thu v php nhn cc s nguyn n bit cng vi cc php dch chuyn v cc php cng.

Gi s a v b l cc s nguyn c cc biu din nh phn di 2n la = (a2n-1a2n-2 ... a1 a0)2 v b = (b2n-1 b2n-2 ... b1 b0)2.

Gi s a = 2nA1 + A0 , b = 2nB1 + B0 , trong A1 = (a2n-1 a2n-2 ... an+1 an)2 , A0 = (an-1 ... a1 a0)2B1 = (b2n-1 b2n-2 ... bn+1 bn)2 , B0 = (bn-1 ... b1 b0)2.

Thut ton nhn nhanh cc s nguyn da trn ng thc:ab = (22n + 2n)A1B1 + 2n(A1 - A0)(B0 - B1) + (2n + 1)A0B0.

ng thc ny ch ra rng php nhn hai s nguyn 2n bit c th thc hin bng

dng ba php nhn cc s nguyn n bit v cc php cng, tr v php dch chuyiu c ngha l nu f(n) l tng cc php ton nh phn cn thit nhn hnguyn n bit th

f(2n) = 3f(n) + Cn.Ba php nhn cc s nguyn n bit cn 3f(n) php ton nh phn. Mi mt trong cc pcng, tr hay dch chuyn dng mt hng s nhn vi n ln cc php ton nh phCn l tng cc php ton nh phn c dng khi lm cc php ton ny.

34


14/15

Mnh 1: Gi s f l mt hm tng tho mn h thc truy hi f(n) = af(n

b ) + c vimi n chia ht cho b, a 1, b l s nguyn ln hn 1, cn c l s thc dng. Khi

f(n) =

O nlogb a , a 1

O logn , a = 1

{

.

Mnh 2: Gi s f l hm tng tho mn h thc truy hi f(n) = af(n

b ) + cnd vi mi

n = bk , trong k l s nguyn dng, a 1, b l s nguyn ln hn 1, cn c v d l ccs thc dng. Khi

f(n) =

d d

d d

d a

banO

bannO

banO b

,)(,)log(

,)( log

.

Th d 16:Hy c lng s php ton nh phn cn dng khi nhn hai s nguyn n bng thut ton nhn nhanh.

Th d 15.2 ch ra rng f(n) = 3f(n/2) + Cn, khi n chn. V th, t Mnh suy ra f(n) = O( 3log 2n ). Ch l log23 1,6. V thut ton nhn thng thng dngO(n2) php ton nh phn, thut ton nhn nhanh s thc s tt hn thut ton n

thng thng khi cc s nguyn l ln.

BI TP CHNG II:

1. Trong tng s 2504 sinh vin ca mt khoa cng ngh thng tin, c 1876 theomn ngn ng lp trnh Pascal, 999 hc mn ngn ng Fortran v 345 hc ngn ng Ngoi ra cn bit 876 sinh vin hc c Pascal v Fortran, 232 hc c Fortran v Chc c Pascal v C. Nu 189 sinh vin hc c 3 mn Pascal, Fortran v C th trtrng hp c bao nhiu sinh vin khng hc mn no trong 3 mn ngn ngtrnh k trn.2. Mt cuc hp gm 12 ngi tham d bn v 3 vn . C 8 ngi pht bvn I, 5 ngi pht biu v vn II v 7 ngi pht biu v vn III. Ngong 1 ngi khng pht biu vn no. Hi nhiu lm l c bao nhiu ng biu c 3 vn .3. Ch ra rng c t nht 4 ngi trong s 25 triu ngi c cng tn h vit tt bch ci sinh cng ngy trong nm (khng nht thit trong cng mt nm).

35


15/15

4. Mt tay vt tham gia thi u ginh chc v ch trong 75 gi. Mi gi anh tanht mt trn u, nhng ton b anh ta c khng qu 125 trn. Chng t rng c gi lin tip anh ta u ng 24 trn.5. Cho n l s nguyn dng bt k. Chng minh rng lun ly ra c t n s mt s s hng thch hp sao cho tng ca chng chia ht cho n.6. Trong mt cuc ly kin v 7 vn , ngi c hi ghi vo mt phiu tr bng cch nguyn hoc ph nh cc cu tr li tng ng vi 7 vn nu

Chng minh rng vi 1153 ngi c hi lun tm c 10 ngi tr li ht nhau.7. C 17 nh bc hc vit th cho nhau trao i 3 vn . Chng minh rng luc 3 ngi cng trao i mt vn .8. Trong k thi kt thc hc phn ton hc ri rc c 10 cu hi. C bao nhiu cchim cho cc cu hi nu tng s im bng 100 v mi cu t nht c 5 im.

9.Phng trnh x1 + x2 + x3 + x4 + x5 = 21 c bao nhiu nghim nguyn khng m?10. C bao nhiu xu khc nhau c th lp c t cc ch ci trong tMISSISSIPI ,yu cu phi dng tt c cc ch?11. Mt gio s ct b su tp gm 40 s bo ton hc vo 4 chic ngn t, mng 10 s. C bao nhiu cch c th ct cc t bo vo cc ngn nu:

1)Mi ngn c nh s sao cho c th phn bit c;2)Cc ngn l ging ht nhau?

12.Tm h thc truy hi cho s mt th t Dn.13. Tm h thc truy hi cho s cc xu nh phn cha xu 01.14. Tm h thc truy hi cho s cch i ln n bc thang nu mt ngi c th bhai hoc ba bc mt ln.15. 1) Tm h thc truy hi m R n tho mn, trong R n l s min ca mt phng b phn chia bi n ng thng nu khng c hai ng no song song v khng ng no cng i qua mt im.b) Tnh R n bng phng php lp.16.Tm nghim ca h thc truy hi an = 2an-1 + 5an-2 - 6an-3 vi a0 = 7, a1 = -4, a2 = 8.

36

Documents

Giao Trinh Toan Roi Rac - Chuong 2