Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...0 B GIO DC V O TO TRNG I HC NNG NGHIP H NI V KIM THNH TON RI RC (Gio trnh dnh cho sinh vin ngnh cng ngh thng tin) H ni2008 Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...1 MC LC Li ni u 5 Chng 1. THUT TON1. nh ngha 2. M t thut ton bng lu 3. M t thut ton bng ngn ng phng Pascal 4. phc tp ca thut ton 5. Thut ton tm kim 6. Thut ton quy7. Mt s thut ton v s nguyn BI TP CHNG 1 7 7 8 9 14 18 19 23 28 Chng 2. BI TON M 1. Nguyn l cng v nguyn l nhn 2. Chnh hp. Hon v. T hp. 3. Nguyn l b tr 4. Gii cc h thc truy hi 5. Bi ton lit k. 6. Bi ton tn ti BI TP CHNG 2 32 32 35 42 44 51 61 64 Chng 3. CC KHI NIM C BN V TH 1. Cc nh ngha v th v biu din hnh hc ca th 2. Biu din th bng i s 3. S ng cu ca cc th 4. Tnh lin thng trong th 5. S n nh trong, s n nh ngoi v nhn ca th 6. Sc s ca th BI TP CHNG 3 69 69 79 82 84 88 91 93 Chng 4. TH EULER, TH HAMILTON, TH PHNG 1. th Euler 2. th Hamilton 3. thi phng BI TP CHNG 4 98 98 103 108 113 Chng 5. CY V MT S NG DNG CA CY 1. Cy v cc tnh cht c bn ca cy 2. Cy nh phn v php duyt cy 3. Mt vi ng dng ca cy 117 118 122 126 Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...2 4. Cy khung (cy bao trm) ca th5. H chu trnh c lp 6. Cy khung nh nhtBI TP CHNG 5 131 134 136 142 Chng 6. MT S BI TON TI U TRN TH 1. Bi ton ng i ngn nht trong th 2. Tm, Bn knh, ng knh ca th 3. Mngv Lung4. Bi ton du lch BI TP CHNG 6 147 147 152 153 160 166 Chng 7. I S BOOLE 1. Hm Boole 2. Biu thc Boole 3. nh ngha i s Boole theo tin 4. Biu din cc hm Boole 5. Cc cng logic 6Ti thiu ho hm Boole BI TP CHNG 7 172 172 174 176 177 183 185 193 Ph chng. I CNG V TON LOGIC 1. Lgic mnh 2. Cng thc ng nht ng v cng thc ng nht bng nhau tronglgic mnh 3. iu kin ng nht ng trong lgic mnh 4. Lgic v t BI TP PH CHNG197 197 201 205 208 213 Mt s bi tp lm trn my tnh Mt s thut ng dng trong gio trnh Ti liu tham kho 216 218 221 Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...3 LI NI U Ton Ri rc (Discrete mathematics) l mn ton hc nghin cu cc i tng ri rc. N c ng dng trong nhiu ngnh khoa hc khc nhau, c bit l trong tin hc bi qu trnh x l thng tin trn my tnh thc cht l mt qu trnh ri rc.Phm vi nghin cu ca Ton Ri rc rt rng, c th chia thnh cc mn hc khc nhau. Theo quy nh ca chng trnh mn hc, gio trnh ny cp n cc lnh vc: Thuttonvbitonm;Lthuytth;isLogicvcchiathnh8 chng: -Chng1cpnmttrongccvncbnnhtcaThuttonl phc tp v thi gian ca thut ton. -Chng 2 ni v cc nguyn l c bn ca Bi ton m.-Cc chng 3, 4, 5 v 6 trnh by v L thuyt th v cc ng dng. y l phn chim t trng nhiu nht ca gio trnh. Trong c cc chng v cc khi nim c bn ca th, cc th c bit nh th Euler, th Hamilton, th phng, Cy cng cc ng dng cacc thi c bit ny. Ringchng 6 dnhchomt vn trng l mt s bi ton ti u trn th hoc bi ton ti u c gii bng cch ng dng l thuyt th. -Chng7lcckinthccbnvisBoole,mtcngchuhiutrong vic thit k cc mch in, in t. Cuigiotrnhlphchng:NhngkhinimcbnvtonLogicngi hc c th t nghin cu thm v Ton Logic. Trongmichngchngticgngtrnhbycckinthccbnnhtca chngcngccthdminhhacth.Vkhunkhstithcnnchngti lc b mt s chng minh phc tp. Cui mi chng u c cc bi tp ngi hc ng dng, kim chng cc l thuyt hc, ng thi cng cung cp mt s p s ca cc bi tp cho. CngcnnithmrngtonRirckhngchcngdngtrongtinhcm cncngdngtrongnhiungnhkhoahckhc.Bivygiotrnhcngcch cho nhng ai cn quan tm n cc ng dng khc ca mn hc ny. Tcgixinchnthnhcmnccbnngnghipngvinvgpcho vicbinsongiotrnhny.cbitchngtixincmnNhgioutNguyn nh Hin hiu nh v cho nhiu kin ng gp b ch v thit thc. V trnh c hn v gio trnh c bin son ln u nn khng trnh khi cc thiu st. Tc gi rt mong nhn c cc kin ng gp ca cc ng nghip v bn c v cc khim khuyt ca cun sch. TC GI Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...4 CHNG1. THUT TON 1. nh ngha. 2. M t thut ton bng lu . 3. M t thut ton bng ngn ng phng Pascal. 3.1. Cu lnh Procedure (th tc) hoc Function (hm). 3.2. Cu lnh gn. 3.3. Khi cu lnh tun t. 3.4. Cu lnh diu kin. 3.5. Cc cu lnh lp. 4. phc tp ca thut ton. 4.1. Khi nim tng ca hm. 4.2. tng ca t hp cc hm. 4.3. phc tp ca thut ton. 5. Thut ton tm kim 5.1.Thut tontm kim tuyn tnh (cn gi l thut ton tm kim tun t). 5.2.Thut ton tm kim nh phn. 6. Thut ton quy.6.1. Cng thc truy hi. 6.2. Thut ton quy. 6.3. quy v lp 7.Mt s thut ton v s nguyn. 7.1. Biu din cc s nguyn. 7.2. Cng v nhn trong h nh phn. 1.nh ngha Thutton(algorithm)lmtdyccquytcnhmxcnhmtdyccthaotc trn cc i tng sao cho saumtshuhn bc thchin s tcmctiu t ra.

T nh ngha ca thut ton cho thy cc c trng (tnh cht) c bn ca thut ton l: a.Yu t vo, ra: u vo (Input): Mi thut ton c mt gi tr hoc mt b gi tr u vo t mt tp xc nh c ch r. u ra(Output): T cc gi tr u vo, thut ton cho ra ccgi tr cn tm gi l kt qu ca bi ton. Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...5 b.Tnh dng: Sau mt s hu hn bc thao tc thut ton phi kt thc v cho kt qu . c.Tnh xc nh: Cc thao tc phi r rng, cho cng mt kt qu d c x l trn cc b x l khc nhau (hai b x l trong cng mt iu kin khng th cho hai kt qu khc nhau). d.Tnh hiu qu Sau khi a d liu vo cho thut ton hot ng phi a ra kt qu nh mun. e.Tnh tng qut Thuttonphicpdngchomibitoncngdngchkhngphichcho mt tp c bit cc gi tr u vo. Cnhiucchmtthuttonnh:Dngngnngtnhin;dnglu(s khi); dng mt ngn ng lp trnh no (trong gio trnh ny dng loi ngn ng m t gn nh ngn ng lp trnh Pascal v gi l phng Pascal); 2.M t thut ton bng lu Sau khi c thut ton gii bi ton, trc khi chuyn sang ngn ng lp trnh, ngi ta thng phi th hin thut ton di dng s . S gi l lu ca thut ton. Cc k hiu quy c dng trong lu c trnh by trong bng 1. Bng 1. Cc k hiu quy c dung trong lu thut ton Tn k hiuK hiuVai tr ca k hiu Khi m u hoc kt thc Dngmuhocktthc thut ton Khi vo ra a d liu vo v in kt qu Khi tnh ton Biudincccngthctnhton v thay i gi tr cc i tng Khi iu kin Kim tra cc iu kin phn nhnh Chng trnh con Gi cc chng trnh con Hng i cathut ton Hngchuynthngtin,linh gia cc khi Th d: Thut ton gii phng trnh bc haiax2 + bx + c = 0 gm cc bc sau: 1) Xc nh cc h s a, b, c (thng tin u vo) 2) Kim tra h s a: -Nu a = 0:Phng trnh cho l phng trnh bc nht, nghim l: bcx = . -Nu a 0:Chuyn sang bc 3. Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...6 3) Tnh bit thc = b2 4ac. 4) Kim tra du ca bit thc - Nu 0:Phng trnh c nghim thc - Nu < 0:Phng trnh c nghim phc 5)In kt qu Lu ca thut ton c trnh by trong hnh 1 3.M t thut ton bng ngn ng phng Pascal giibitontrnmytnhintphivitchngtrnhtheomtngnnglp trnh no (Pascal, C, Basic, ...). Mi ngn ng lp trnh c mt quy tc cu trc ring. thayvicmtthuttonbngli,cthmtthuttonbngcccutrclnh tng t nh ngn ng lp trnh Pascal v gi l ngn ng phng Pascal. Ccculnhchnhdngmtthuttongmc:ProcedurehocFunction;cu lnhgn;ccculnhiukin;ccvnglp.Ngoirakhicngiithchccculnh bng li, c th cc li gii thch trong du (* ... *) hoc {}.Ngha l ngn ng phng Pascal hon ton tng t ngn ng lp trnh Pascal, nhng khngcphnkhaibo.Tuynhin,phinuruvo(Input)vura(output)ca thut ton. Bt u Nhp a, b, c Saia = 0ng = b2 = 4ac bcx = Sai 0 ng Phn thc = a 2b 2abx1 + =Phn o=2a 2abx2 + = Thng bo kt qu Kt thc Hnh 1.Lu gii phng trnh bc hai Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...7 3.1.Cu lnh Procedure (th tc) hoc Function (hm) ng ngay sau cu lnh ny l tn ca th tc hoc tn hm. Cc bc thc hin ca thut ton c m t trong th tc (hm) c bt u bi t kha begin v kt thc bi t kha end. Th d 1 Function Max(a, b, c)(* Hm tm s ln nht trong 3 s a, b, c *) Begin (* thn hm*) End; Th d 2 Procedure Giai_phuong_trnh_bac_hai (* Th tc gii phng trnh bc hai *) Begin (* thn th tc *) End; Chrng,khimtthuttonbngfunction,trckhiktthc(end)thutton phi tr v (ghi nhn) gi tr ca function . 3.2.Cu lnh gn Dng gn gi tr cho cc bin. Cch vit: Tn bin := gi tr gn Th d:x := a;(*bin x c gn gi tr a*) max := b; (*bin max c gn gi tr b*) 3.3.Khi cu lnh tun t c m u bng t kha begin v kt thc bng end nh sau: begin Cu lnh 1; Cu lnh 2; ... ..... Cu lnh n; end; Cc lnh c thc hin tun t t cu lnh th nht n cu lnh cui cng. 3.4.Cu lnh iu kin C hai dng: dng n gin v dng la chn. a. Dng n gin: Cch vit: ifthencu lnh hoc khi cu lnh; Khi thc hin, iu kin c kim tra, nu iu kin tha mn th cu lnh (khi cu lnh) c thc hin, nu iu kin khng tha mn th lnh b b qua. Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...8 b.Dng la chn: Cch vit: ifthencu lnh hoc khi cu lnh 1elsecu lnh hoc khi cu lnh 2; Khi thc hin, iu kin c kim tra, nu iu kin tha mn th cu lnh (khi cu lnh) 1 c thc hin, nu iu kin khng c tha mn th cu lnh (khi cu lnh) 2 c thc hin. Th d 1. Thut ton tm s ln nht trong 3 s thc a, b, c. -u tin cho max = a; -So snh max vi b, nub > max th max = b; -So snh max vi c, nu c > max th max = c. Functionmax(a,b,c) Input:3 s thc a,b,c; Output:S ln nht trong 3 s nhp; Begin x := a; ifb > xthenx:= b; ifc > xthenx:= c; max := x; End; Th d 2. Thut ton gii phng trnh bc haiax2 + bx + c = 0ProcedureGiai_phuong_trinh_bac2; Input:Cc h s a, b, c; Output:Nghim ca phng trnh; begin ifa = 0thenx := -c/b; ifa 0 then begin := b2 4ac; if 0then begin x1 = ( b )/2a ; x2 = ( b + )/2a; end else begin Phn thc := -b/2a; Phn o := ( )/2a; end; end; end; Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...9 3.5.Cc cu lnh lp C hai loi: Loi c bc lp xc nh v loi c bc lp khng xc nh. a.Loi c bc lp xc nh:Cch vit nh sau: forbin iu khin := gi tr utogi tr cuidocu lnh hoc khi cu lnh; Khi thc hin, bin iu khin c kim tra, nu bin iu khin nh hn hoc bng gitrcuithculnh(khiculnh)cthchin.Tipbiniukhinstng thm 1 n v v qu trnh c lp li cho n khi bin iu khin ln hn gi tr cui th vnglpdngvchoktqu.Nhvyhtvnglpforsbclplgitrcui(ca bin iu khin) tr gi tr u cng mt. Th d:Tm gi tr ln nht ca dy s a1, a2, ,an. Thut ton:u tin cho gi tr ln nht (max) bng a1, sau ln lt so snh max vi cc s ai (i = 2, 3, , n), nu max < ai th max bng ai, nu max > ai th max khng i. Functionmax_day_so; Input: Dy s a1, a2, ,an; Output:Gi tr ln nht (max) ca dy s nhp; begin max := a1 ; fori:= 2tondoif ai > maxthenmax := ai ; max_day_so := max; end; Ch thch:Vng lp for cn cch vit li bin iu khin nh sau: for bin iu khin := gi tr cui downto gi tr u do cu lnh hoc khi cu lnh; Vic thc hin cu lnh ny tng t nh khi vit bin iu khin tng dn. b.Loi c bc lp khng xc nh:C hai cch vit Cch th nht:whileiu kindocu lnh hoc khi cu lnh; Khithchin,iukinckimtra,nuiukincthomnthculnh (khiculnh)cthchin.Nuiukinkhngthomnthvnglpdngvcho kt qu. Th d:Kim tra xem s nguyn dng m cho c phi l s nguyn t khng? Thut ton nh sau: S m l s nguyn t nu n khng chia ht cho bt c s nguyn dng khc 1 no nh hn hoc bngm.Thtvy,numlmthps(khngphilsnguynt),nghaltnticcs nguyn dng a, b sao cho: m = a.ba m hoc b m Vy, nu m l s nguyn t th m khng chia ht cho mi s nguyn dng i, 2 im Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...10 Procedure nguyento(m); Input:S nguyn dng m; Output:True, nu m l s nguyn t; False, nu m khng phi l s nguyn t; begin i := 2; whileim do begin ifm mod i= 0thennguyento := false elsenguyento := true; i := i+1; end; end; Cch th hai:repeatcu lnh hoc khi cu lnhuntiliu kin; Khithchin,culnh(khiculnh)cthchin,sauiukinckim tra, nu iu kin sai th vng lp c thc hin, nu iu kin ng th vng lp dng v cho kt qu. Thd:Thutton-clittmcschunglnnhtcahaisnguyndnga,b nh sau: Gi s a > b v a chia cho b c thng l q v s d l r, trong a, b, q, r l cc s nguyn dng: a = bq + r suy ra:CLN(a, b) = CLN(b, r) v s d cui cng khc khng l c s chung ln nht ca a v b. Tht vy: Gi s d l c s chung ca hai s nguyn dng a v b, khi :r = a bq chia ht cho d. Vy d l c chung ca b v r. Ngc li, nu d l c s chung ca b v r, khi do bq + r = a cng chia ht cho d. Vy d l c s chung ca a v b. Chng hn, mun tm c s chung ln nht ca 111 v 201 ta lm nh sau: 201 = 1. 111 + 90 111 =1. 90 + 21 90 =4. 21 + 6 21 =3.6 + 3 6 =2. 3 Vy CLN(111, 201) = 3 (3 l s d cui cng khc 0). Function UCLN(a, b) Input:a, b l 2 s nguyn dng; Output:UCLN(a, b); begin Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...11 x := a; y : = b; repeat begin r := x mod y;(* r l phn d khi chia x cho y *) x : = y;y := r ; ify 0 then uc := y; end; untily =0; UCLN := uc; end; Ch : Khi gii cc bi ton phc tp th thng phi phn chia bi ton thnh cc bi ton nh hn gi l cc bi ton con. Khi phi xy dng cc th tc hoc hm gii cc bi ton con , sau tp hp cc bi ton con ny gii bi ton ban u t ra. Thut ng tin hc gi cc chng trnh gii bi ton con l cc chng trnh con. Th d:Tm s nguyn t nh nht ln hn s nguyn dng m cho. ProcedureSo_nguyen_to_lon_hon(m); Input: S nguyn dng m; Output:n l s nguyn t nh nht ln hn m; begin n := m + 1; whilenguyento(n) = falsedon := n + 1; end; 4. phc tp ca thut ton C hai l do lm cho mt thut ton ng c th khng thc hin c trn my tnh. l do my tnh khng b nh thc hin hoc thi gian tnh ton qu di. Tng ng vi hai l do trn ngi ta a ra hai khi nim: - phc tp khng gian ca thut ton, phc tp ny gn lin vi cc cu trc d liu c s dng. Vn ny khng thuc phm vi ca mn hc ny. - phc tp thi gian ca thut ton, phc tp ny c th hin qua s lng cc cu lnh v cc php gn, cc php tnh s hc, php so snh, c s dng trong thut ton khi cc gi tr u vo c kch thc cho. 4.1.Khi nim tng ca hm Trc ht xt th d: Gi s thi gian tnh ton ca mt thut ton ph thuc vo kch thc n ca u vo theo cng thc: t(n) = 60n2 + 9n + 9 Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...12 Bng sau cho thy khi n ln, t(n) xp x s hng 60n2 : nt(n) = 60n2 + 9n + 960n2 10 100 1 000 10 000 9099 600 909 60 009 009 6 000 090 009 6000 600 000 60 000 000 6 000 000 000 nh ngha:Cho f(x) vg(x) l hai hm t tps nguyn hoc tp s thc vo tp cc s thc. Ngi ta ni f(x) l O(g(x)) hay f(x) c quan h big-O vi g(x), k hiu f(x) = O(g(x)), nu tn ti hai hng s C v k sao cho: | f(x) |C. | g(x) |, x k. Th d 1.t(n) = 60n2 + 9n + 9 = O(n2) Tht vy: n 1, ta c: | 60n2 + 9n +9| =60n2 + 9n +9 =||

\|+ +22n9n960 n n2 (60 + 9 + 9)=78n2. VyC = 78; k = 1. Tng t ta c th chng minh: Pn(x) = a0xn + a1xn-1 + ... + an = O(xn) , x R. Th d 2. f(n) = 1 + 2 + 3 + ... + n k2 Chn k = max(k1; k2) th c hai bt ng thc u tho mn. Do : 1)|(f1 + f2)(x)| = |f1(x) + f2(x)| |f1(x)| + |f2(x)| C1|g1(x)| + C2|g2(x)| (C1 + C2)g(x) y g(x) = max{|g1(x)|, |g2(x)|}. Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...13 2)|(f1.f2)(x)| = |f1(x)|.|f2(x)| C1C2 |g1(x)|.|g2(x)|=C1C2|g1(x)g2(x)| H qu: Nuf1(x) = O(g(x)),f2(x) = O(g(x))th(f1+f2)(x) = O(g(x)) Th d.Cho nh gi O ca cc hm: 1/f(n) = 2nlg(n!) + (n3 + 3)lgn, n N. 2/f(x) = (x + 1)lg(x2 + 1) + 3x2 ,x R. Gii:1)Ta c:lg(n!) = O(nlgn) 2nlg(n!) = O(n2lgn) (n3 + 3)lgn = O(n3lgn) Vyf(n) = O(n3lgn) 2)Ta c:lg(x2 + 1) lg2x2 = lg2 + 2lgx 3lgx ,x > 1 lg(x2 + 1) = O(lgx)(x + 1)lg(x2 + 1) = O(xlgx) Mt khc: 3x2= O(x2) vmax{xlgx; x2}= x2. Vy f(x) = O(x2).4.3. phc tp ca thut ton Nh ni phn u ca mc 4, chng ta ch cp n phc tp v thi gian ca thut ton. phc tp v thi gian ca thut ton c nh gi qua s lngcc php ton m thut ton s dng. V vy phi m cc php ton c trong thut ton. Cc php ton phi m l: - Cc php so snh cc s nguyn hoc s thc; - Cc php tnh s hc: cng, tr, nhn, chia; - Cc php gn; - V bt k mt php tnh s cp no khc xut hin trong qu trnh tnh ton. Gi s s cc php ton ca thut ton l f(n), trong n l kch thc u vo, khi ngi ta thng quy phc tp v thi gian ca thut ton v cc mc: phc tp O(1), gi l phc tp hng s, nu f(n) = O(1). phc tp O(logan), gi l phc tp logarit, nu f(n) = O(logan). (iu kin a > 1) phc tp O(n), gi l phc tp tuyn tnh, nu f(n) = O(n). phc tp O(nlogan), gi l phc tp nlogarit nu f(n) = O(logan). (iu kin a > 1) phc tp O(nk), gi l phc tp a thc, nu f(n) = O(nk). phc tp O(an), gi l phc tp m, nu f(n) = O(an). (iu kin a > 1) phc tp O(n!), gi l phc tp giai tha, nu f(n) = O(n!). Th d 1.Tm phc tp ca thut ton gii bi ton: Tm s ln nht trong dy n s nguyn a1, a2, , an cho: Proceduremax(a1, a2, , an); Input:Dy sa1, a2, ... , an; Output:S ln nht (max) ca dy s cho; begin max := a1; for i := 2 tondo Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...14 ifai > max thenmax := ai; end;Mi bc ca vng lp for phi thc hin nhiu nht 3 php ton: php gn bin iu khin i, php so snh ai vi max v c th l php gn ai vo max; vng lp c (n 1) bc (i = 2, 3, , n) do nhiu nht c c thy 3(n - 1) php ton phi thc hin. Ngoi ra thut ton cn phi thc hin php gn u tin max := a1.Vy s php ton nhiu nht m thut ton phi thc hin l: 3(n 1) + 1 = 3n 2 = O(n) phc tp v thi gian ca thut ton l phc tp tuyn tnh. Th d 2. phc tp ca thut tonnhn ma trn. Procedure nhn_matran; Input:2 ma trnA = (aij)m x p vB = (bij)p x n; Output: ma trn tchAB = (cij)m x n ; Begin fori:=1 tomdo forj:=1 tondo begin cij := 0; fork:=1topdo cij := cij + aikbkj ; end; End. S php cng v s php nhn trong thut ton trn l: Vi mi phn t cij phi thc hin p php nhn v p php cng. C tt c m.n phn t cij, vy phi thc hin 2mnp php cng v php nhn.xcnhphctpcathutton,tagisA,Blhaimatrnvungcpn, nghalm=n=p.nhvyphicn2n3phpcngvphpnhn.Vyphctpca thut ton l O(n3) phc tp a thc. Mtiuthvl,khinhnt3matrntrlnthsphptnhcngvnhnph thucvothtnhnccmatrny.ChnghnA,B,Clccmatrnckchthc tng ng l 3020, 2040, 4010. Khi : Nu thc hin theo th t ABC=A(BC) th tchBC l ma trn kch thc 2010 v cnthchin20.40.10=8000phptnhcngvnhn.MatrnA(BC)ckchthc 3010vcnthchin30.20.10=6000phpcngvnhn.Tsuyracnthchin 8000+6000 = 14000 php tnh cng v nhn hon thnh tch ABC. Tngt,nuthchintheothtABC=(AB)Cthcnthchin30.20.40php tnh cng v nhn thc hin tch AB v 30.40.10 php cng v nhn thc hin tch (AB)C.DosccphptnhcngvnhnphithchinhonthnhtchABCl 24000+12000 = 36000 php tnh. R rng hai cch nhn cho kt qu v s lngcc php tnh phi thchin l khc nhau. Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...15 5.Thut ton tm kim Bi ton tm kim c pht biu nh sau: Tm trong dy s a1, a2, , an mt phn t c gi tr bng s a cho trc v ghi li v tr ca phn t tm c.Bitonnycnhiungdngtrongthct.Chnghnvictmkimttrongt in, vic kim tra li chnh t ca mt on vn bn, C hai thut ton c bn gii bi ton ny: Thut ton tm kim tuyn tnh v thut ton tm kim nh phn. Chng ta ln lt xt cc thut ton ny. 5.1.Thut ton tm kim tuyn tnh (cn gi l thut ton tm kim tun t). em so snh a ln lt vi ai (i = 1, 2, , n) nu gp mt gi tr ai = a th ghi li v tr ca ai, nu khng gp gi tr ai = a no (ai a. i) th trong dy khng c s no bng a. ProcedureTim_tuyen_tinh_phan_tu_bang_a; Input: a v dy s a1, a2, ... , an; Output:V tr phn t ca dy c gi tr bng a, hoc l s 0 nu khng tm thy a trong dy; begin i := 1; while(i nand ai a)doi := i + 1; ifi nthen vitri := ielse vitri := 0; end; Nh vy nu a c tm thy v tr th i ca dy (ai = a) th cu lnh i := i + 1 trong vng lp while c thc hin i ln (i = 1, 2, , n). Nu a khng c tm thy, cu lnh phi thc hin n ln. Vy s php ton trung bnh m thut ton phi thc hin l:O(n)21 n2n1) n(nnn 2 1=+=+=+ + + . . . Vy, phc tp ca thut ton tm kim tuyn tnh l phc tp tuyn tnh. 5.2.Thut ton tm kim nh phn. Githitrngccphntcadycxptheothttngdn.Khisosnha vi s gia dy, nu a < amvi ((

+=2n 1m(cn nhc li rng phn nguyn ca x: [x] l s nguyn nh nht c trong x) th tm a trong dy a1, ,am , nu a > am th tm a trong dy am+1, , an. i vi mi dy con (mt na ca dy cho) c lm tng t ch phi tmphntcgitrbngamtnadycon.Qutrnhtmkimktthckhitm thy v tr ca phn t c gi tr bng a hoc khi dy con ch cn 1 phn t. Chng hn vic tm s 8 trong dy s5, 6, 8, 9, 11, 12, 13, 15, 16, 17, 18, 19, 20, 22 c tin hnh nh sau: Dy cho gm 14 s hng, chia dy thnh 2 dy con: 5, 6, 8, 9, 11, 12, 13v 15, 16, 17, 18, 19, 20, 22 Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...16 V8 < 13 nn tip theo ch cn tm dy u tin. Tip tc chia i thnh 2 dy: 5, 6, 8, 9 v 11, 12, 13; v 8 < 9 nn li ch phi tm dy 5, 6, 8, 9. Li chia i dy ny thnh 2 dy con 5, 6 v 8, 9 thy ngay 8 thuc v dy con 8, 9 v qu trnh tm kim kt thc, v tr ca s 8 trong dy cho l th ba ProcedureTim_nhi_phan_phan_tu_bang_a; Input: a v dy s a1, a2, ... , an xp theo th t tng; Output:V tr phn t ca dy c gi tr bng a, hoc l s 0 nu khng tm thytrong dy; Begin i := 1;(* i l im mt tri ca khong tm kim*) j := n;(* j l im mt phi ca khong tm kim*) while i < j do begin ((

+=2j 1: m ; if a > am then i := m+1 else j := m; end; ifa = aithen vitri := ielse vitri := 0; end; phctpcathuttontmkimnhphncnhginhsau:Khnggim tngqutcthgisdicadya1,a2,,anln=2kviklsnguyndng. (Nu n khng phi l ly tha ca 2, lun tm c s k sao cho 2k 1 < n < 2k do c th xem dy cho l mt phn ca dy c 2k phn t). Nh vy phi thc hin nhiu nht k ln chia i cc dy s (mi na dy ca ln chia i th nht c 2k 1 phn t, ca ln chia i th hai c 2k2 phn t, , v ca ln chia i th k l 2kk = 20 = 1 phn t). Ni cch khc l nhiu nht c k vng lp while c thc hin trong thut ton tm kim nh phn. Trong mi vng lp while phi thc hin hai php so snh, v vng lp cui cng khi ch cn 1 phn t phi thc hin 1 php so snh bit khng cn 1 phn t no thm na v1phpsosnhbitacphilphnthaykhng.Tthyrngthutton phi thc hin nhiu nht 2k + 2 = 2[log2n] + 2 = O(logn) php so snh. Vy, phc tp ca thut ton tm kim nh phn l phc tp logarit. 6.Thut ton quy 6.1.Cng thc truy hi i khi rt kh nh ngha mt i tng no mt cch tng minh, nhng c th nh ngha i tng qua chnh n vi u vo nh hn. Cch nh ngha nh vy gi l cch nh ngha bng truy hi hoc nh ngha bng quy v n cho mt cng thc gi l cng thc truy hi. Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...17 nhngha:nhnghabngtruyhibaogmccquytcxcnhcci tng, trong c mt s quy tc dng xc nh cc i tng ban u gi l cc iu kin ban u;cn cc quy tc khc dng xc nh cc i tng tiptheogi l cng thc truy hi. Th d 1.Dy s an c nh ngha bng quy nh sau: a0 = 3;an = an 1 + 3.Trong a0 = 3 l iu kin ban u, cn an = an 1 + 3l cng thc truy hi. Th d 2.nh ngha bng quy giai tha ca s t nhin n l: GT(0) = 1; GT(n) = n.GT(n 1). V GT(n) = n! = n(n-1)(n-2)1 = n.GT(n-1). Trong GT(0) = 1 l iu kin ban u, cn GT(n) = n.GT(n 1) l cng thc truy hi. Th d 3.Dy s F0, F1, F2, , Fn c nh ngha: F0 = 0; F1 = 1; Fn = Fn 1 + Fn 2 . chnh l nh ngha bng quyca dy s c tn l dyFibonacci. Trong F0 = 0, F1 = 1 l cc iu kin ban u, cn Fn = Fn 1 + Fn 2 l cng thc quy. D thy mt s s hng u tin ca dy l: 0; 1; 1; 2; 3; 5; 8; 13; 21; 6.2.Thut ton quy. Nhiu khi vic gii bi ton vi u vo xc nh c th a v vic gii bi ton vi gi tr u vo nh hn. Chng hn: n! = n . (n-1)! hay UCLN(a, b) = UCLN(a mod b, b) ,a > b nh ngha:Mt thut ton gi l quy nu thut ton gii bi ton bng cch rt gn lin tip bi ton ban u ti bi ton cng nh vy nhng vi d liuu vo nh hn. D thy c s ca thut ton l cng thc truy hi. Th d 1.Tnh giai tha ca s t nhin n bng quy. Function GT(n); Input: S t nhin n; Output: Gi tr ca n!; Begin ifn = 0thenGT(0) := 1 else GT(n) := n*GT(n 1); End; Th d 2. Tnh s hng ca dy Fibonacci bng quy. Function Fibonacci(n); Input:V tr th n ca dy Fibonacci; Output:Gi trFn ca dy Fibonaci; Begin Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...18 ifn = 0thenFibonacci(0) := 0 elseifn = 1thenFibonacci(1) := 1 else Fibonacci(n) := Fibonacci(n-1) + Fibonacci(n-2); End; Th d 3. Thut ton quy tm UCLN(a, b). FunctionUCLN(a, b); Input:Hai s nguyn dng a v b; Output: c s chung ln nht ca a v b; Begin ifb = 0thenUCLN(a, b) := a elseif a > bthenUCLN(a, b) := UCLN(a mod b, b) elseUCLN(a, b) := UCLN(b mod a, a); End; By gi chng ta th tm phc tp v thi gian ca mt vi thut ton vit bng quy.ChnghnxtthuttonquytnhshngcadyFibonacci,tnhFntabiu din Fn = Fn1 + Fn2 , sau thay th c hai s ny bng tng ca hai s Fibonacci bc thp hn. Qu trnh tip tc nh vy cho n khi F0 v F1 xut hin th c thay bng cc gi tr ca chng trong nh ngha. MibcquychotikhiF0vF1xuthin,ccsFibonaccictnhhailn. Chng hn gin cy hnh 2 cho ta hnh dung cch tnh F5 theo thut ton quy. T c th thy rng tnh Fn cn thc hin Fn + 1 1 php cng.

phctpcathuttonquytmcschunglnnhtcahaisnguyn dng a, b (th d 3): UCLN(a,b) = UCLN(a mod b,b), nu a b (a mod b l phn d khi chia a cho b) c nh gi bng cch ng dng dy Fibonacci. Trchtbngquynptonhcchngtachngminhshngtngqutcady Fibonacci tha mn: F5 F4F3 F3 F2F2 F1 F2F1 F1F0 F1F0 F1F0 Hnh 2. Lc tnh F5 bng quy Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...19 Fn > n 2, n 3,trong 25 1+= . (1) Tht vy:Ta c: < 2 = F3,ngha l (1) ng vi n = 3. Gi s Fn > n 2 ng vi n, xt vi n+1. D thy l mt nghim ca phng trnhx2 x 1 = 0nn suy ra2 = + 1.T : n 1 = 2n 3 = ( + 1)n 3 = n 2 + n 3 . Theo gi thit quy np, nu n 4, ta cFn 1 > n3 vFn > n2. Thay vo nh nghaca dy Fibonacci: Fn + 1 = Fn + Fn 1 > n 2 + n 3 = n 1 VyFn > n 2,n 3. Cng thc (1) c chng minh. Tr li thut ton quy tm c s chung ln nht ca hai s nguyn dng a, b (a b).phctpcathuttoncnhgiquaslngccphpchiadngtrong thut ton ny. t r0 = a,r1 = b, ta c: r0 = r1q1 + r2; 0 r2 < r1, r1 = r2q2 + r3; 0 r3 < r2, ..... rn 2 = rn 1 qn 1 + rn ; 0 rn < rn 1, rn 1 = rn qn ; Nh vy phi dng n php chia tm rn = UCLN(a,b). Cc thng q1, q2, , qn 1 lun ln hn hoc bng 1, cn qn 2. T suy ra: rn1=F2, rn 12rn = 2F2 =F3, rn 2 rn 1 + rn F3 + F2 = F4, ...r2 r3 + r4Fn 1 + Fn 2 = Fn, b = r1 r2 + r3 Fn + Fn 1 = Fn + 1. trong Fn l s hng th n trong dy Fibonacci. Vy nu n l s cc php chia trong thut ton -clit tm c s chung ln nht ca hai s nguyn dng a, b thb Fn + 1, trong Fn l s Fibonacci th n. Do Fn + 1 > n 1vin > 2 v 25 1+= nnb > n 1. T :lgb>(n 1) lg > 51 n ( vlg 0,208 > 51). Vy: n 1 < 5lgb n < 5lgb + 1 = O(lgb). Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...20 Ngha l phc tp ca thut ton -clit vit theo quy l O(lgb). Qua y c th thy trong nhiu trng hp, vic nh gi phc tp ca mt thut ton qu l khng d dng. 6.3. quy v lp Hunhccbitongiicbnglpthcnggiicbngquy.Thng thng, nu dng phng php lp th s php tnh s t hn; chng ta minh ha iu ny bng th tc tnh s Fibonacci bng thut ton quy v thut ton lp. Trong th d 2, mc 6.2. chng ta tnh s php cng theo thut ton quy tnh s Fibonacci th n l Fn + 1 1. By gi ta xt th tc lp tnh s Fibonacci:Function Lap_Fibonacci(n); Input:V tr thn + 1 ca dy Fibonacci; Output:Gi trfn ca dy; Begin ifn = 0theny := 0else begin x := 0;y:=1 for i := 1to n 1begin z := x + y; x :=y; y := z end; end; Lap_Fibonacci := z; End. Thut ton ny khi to x nh l F0 = 0 v y nh l F1 = 1. Qua mi bc lp tng ca x v y c gn cho bin ph z. Sau x c gn gi tr ca y v y c gn gi tr ca z. Vy qua vng lp th nht, ta c x = F1 v y = F2. Khi qua vng lp th n 1 th x = Fn 1. Vy ch c n 1 php cng c thc hin tnh Fn. R rng s lng cc php tnh ny nh hn khi tnh bng quy. Tuy s lng cc php tnh khi dng quy nhiu hn khi dng lp, nhng nhiu khi ngi ta vn thch s dng quy hn. C l ldo l ch chng trnh quy thng gn hn, mt khc c nhng bi ton ch gii c bng quy m khng gii c bng lp. 7.Mt vi thut ton v s nguyn 7.1.Biu din cc s nguyn nh l: Cho b l mt s nguyn dng ln hn 1, khi nu n l s nguyn dng ty th n c th biu din duy nht di dng:n = ak 1 bk 1 + ak 2 bk 2 + + a1 b + a0. (1) Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...21 trong k l s nguyn khng m, a0, a1, , ak 1 l cc s nguyn khng m v nh hn b ng thi ak 1 0. Chng ta khng chng minh nh l ny. Biu din ca s n theo cng thc (1) gi l khai trin c s b ca n v c k hiu l (ak 1 ak 2a0)b. c bit, nu l c s 10 (h thp phn, b = 10), ngi ta quy c khng cn vit c s km theo. Ngoi h thp phn cn c h nh phn (c s 2), h bt phn (c s 8) v h thp lc phn (c s 16) l hay c dng trong tin hc. Th d 1: Xc nh khai trin thp phn ca s nguyn t cc h c s khc: (245)8 = 2.82 + 4.8 + 5 = 165(10101111)2 = 1.27 + 0.26 + 1.25 + 0.24 + 1.23 +1.22 + 1.2 + 1 = 175 H thp lc phn c 16 ch s 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F, trong A, B, C, D, E, F tng ng vi cc s 10, 11, 12, 13, 14, 15 trong h thp phn. chng hn: (2AE0B)16 = 2.164 + 10.163 + 14.162 + 0.16 + 11 = 175 627 Bygixtbitonngc,xcnhkhaitrintheocsbcastnhinncho trong h thp phn. Trc ht chia n cho b c s d l a0: n = bq0 + a0,0 a0 b ( a0 = n mod b) khi a0 l chnh l ch s cui cng trong h c s b. Tip theo chia q0 cho b c s d a1 chnh l s ng trc a0 trong h m c s b: q0 = bq1 + a1,0 a1 b ( a1 = q0 mod b) Qu trnh tip tc cho n khi nhn c thng bng 0. Th d 2: Tm khai trin c s 8 ca 12345. Ta c: 12345 =8. 1543 + 1 1543 =8. 192 + 7 192 =8. 24 + 0 24 =8 . 3 + 0 3 =8 . 0 + 3 Vy: 12345 = (30071)8.Thut ton tm khai trin c s b ca s t nhin n nh sau: Procedure Khai_trien_co_so_b; Input:S t nhin n v c s b; Output: Khai trin n theo c s b (akak 1a0)b; Begin q := n; k := 0; while q 0 do beginak :=q mod b; Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...22

((

=bq: q; k := k + 1; end; End;7.2.Cng v nhn trong h nh phn Ccthuttonthchinccphptnhviccsnguynhnhphncvaitr quantrngtrongtinhc.Bivyphnnymthaithuttonthchinphpcngv php nhn hai php ton s hc c bn nht cc s nguyn trong h nh phn. Gi s c hai s nguyn a v b di dng nh phn: a = (an 1 an 2 a1 a0)2v b = (bn 1 bn 2 b1 b0)2 . (Nu cn thit c th t thm cc bt 0 ln phn u ca cc khai trin nh phn ca a hoc b) a.Php cng cng a v b, trc ht cng hai bit cui: a0 + b0 = c0.2 + s0. y s0 l bit cui trong khai trin nh phn ca tng a + b v c0 l s nh. Sau l cng hai bit tip theo v s nh: a1 + b1 + c0 = c1.2 + s1. y s1 l bit tip theo trong khai trin nh phn ca tng a + b v c1 l s nh. Tip tc nh vy cho n: an 1 + bn 1 + cn 2 = cn 1.2 + sn 1 . t sn = cn 1 ta c khai trin nh phn ca tng a + b: a + b = (sn sn 1 s1 s0)2. Thut ton c m t bng ngn ng phng Pascal nh sau: ProcedureCong_hai_so_nguyen_duong_dang_nhi_phan; Input:Hai s nguyn dng a v b di dng nh phn(an 1 a1 a0)2v(bn 1 b1 b0)2 ; Output: Tng a + b di dng nh phn (sn sn 1 s1 s0)2; Begin c := 0 for j:=1 to n do begin ((

+ +=2c b a: dj j ; Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...23 =0aabjsj := aj + bj + c 2d; c := d; end; sn := c; End; C th hiu l php cng trong h nh phn cng c tin hnh nh h thp phn, nhng lu rng: 0 + 0 = 0;0 + 1 = 1 + 0 = 1; 1 + 1 = 10 Chng hn tnh (1110)2 + (10111)2 l: + 0 1110 1 0111 10 0101 phc tp ca thut ton c nh gi nh sau: Mi bit trong khai trin nh phn ca tng a + b c tnh bng cch cng mt cp bit v c th phi cng thm bit nh. Nh vy c mi bit ca tng cn nhiu nht 3 php cng. Mi khai trin nh phn c n bit. Vy phi thc hin ti a 3n php cng hai s nguyn n bit, ngha l thut ton ton c phc tp tuyn tnh O(n). b. Php nhn Theo lut phn phi, ta c: a.b = a( ) ===1 n0 jjj1 n0 jjj2 ab 2 bD thy:,nu bj = 1 ,nu bj = 0 v mi ln nhn mt khai trin nh phn vi 2 l dch chuyn khai trin mt v tr v bn tribngcchthmmts0vocuikhaitrin.Docthnhnctch(abj)2j bng cch dch khai trin nh phn ca abj v bn tri j v tr, ni cch khc l thm j bit 0 vo cui khai trin nh phn abj. Cui cng nhn c tch ab bng cch cng n s nguyn dng abj2j vi j = 0, 1, , n 1. Th d:Tm tch caa = (110)2 vb = (101)2.ab0.20 = (110)2.1.20 = (110)2 ab1.21 = (110)2.0.21 = (0000)2 ab2.22 = (110)2.1.22 = (11000)2 v, cui cng ta cng (110)2, (0000)2 v (11000)2 vi nhauc ab = (11110)2. Dthycchthchinphpnhntronghnhphncnggingnhtronghthp phn. Thut ton c m t bng ngn ng phng Pascal mh sau: 110 101 + 110 000 110 11110 Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...24 ProcedureNhan_hai_so_nguyen_dang_nhi_phan; Input:Hai s nguyn dng a v b di dng nh phn(an 1 a1 a0)2v(bn 1 b1 b0)2 ; Output: Tch ab di dng nh phn; Begin for j := 0ton do ifbj = 1 thencj := ac dch sang tri j v tr else cj := 0; (* c0, c1, , cn 1 l cc tch ring phn *) p := 0; forj := 0 to n 1 do p := p + cj ;(* p l gi tr ca tch ab *) End; phctpcathuttonnhn2snguyndngnhphnbngcchtnhscc php cng bit v dch bt ca thut ton c nh gi nh sau: Cc tch ringcj = abj2j khng cn dchchuyn khi bj = 0, v khi cj = 0 v n phi dch chuyn j v tr khi bj = 1. V th vi tt c n tch ring cj cn thc hin ti a: 2) 1 n ( n) 1 n ( . . . 2 1 0= + + + +php dch ch. Vy s php dch ch ti a l O(n2). Sau khi dich ch snguyn cn1 = abn12n1 c 2 n bit, theo thut ton cng hai s nguyn s c ti a O(n) s php cng bit cng tt c n cc tch ring cj. Vy phc tp ca thut ton nhn 2 s nguyn dng nh phn l O(n2). Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...25 BI TP CHNG 1 1.1.Chng minh rng: a) ) x ( O1 x1 x2=++;b)2n + 17 = O(3n). 1.2. Tm mt s nguyn n nh nht sao cho f(x) = O(xn), nu: a) f(x) = 2x3 + x2lgx;b) f(x) = 3x5 + (lgx)4; c) 1 x1 x x) x ( f32 4++ +=; d) 1 xx lg 5 x) x ( f43++= . 1.3.Chng minh rngx2 + 4x + 17 l O(x3), nhngx3 khng l O(x2 + 4x + 17). 1.4.Hy cho mt nh gi big-O tt nht c th c i vi cc hm sau: a)(n2 + 8)(n + 1); b) (nlgn + n2)(n3 + 2); c)(n! + 2n)(n3 + lg(n2 + 1)); d) (2n + n2)(n3 + 3n); e)(x2 + x(lgx)3)(2x + x3). 1.5.Lp v m t thut ton bng ngn ng phng Pascal cho cc bi ton sau: a) Tnh xn vi x l s thc v n l s nguyn. (Gi :trc ht lp th tc tnh xn vi n l s nguyn dng, sau m rng th tc cho n l s nguyn m nnx1x =, nN+) b)Ch dng cc lnh gn bin b ba s (x, y, z) thnh (y, z, x). S ti thiu cc lnh gn l bao nhiu? c)Chn1snguynxvovtrthchhptrongdyccsnguyna1,a2,...,an c xp theo th t tng dn. d)Tm s nh nht v s ln nht trong dy gm n s nguyn cho. 1.6. Xc nh s cc php nhn c dng tnhk2xbt u vi x ri lin tip bnh phng ( tm x2, x4, ). Cch ny c hiu qu hn cch nhn x vi chnh n mt s ln thch hp khng? 1.7.Lp v m t thut ton bng ngn ng phng Pascal cho bi ton sau: m cc bit 1 trongmtxubitbngcchkimtratngbitcaxuxemcphilbit1khng.Cho nh gi big-O i vi cc php so snh c dngtrong thut ton. 1.8.Thut ton tnh gi tr ca a thcP(x) = anxn + an 1 xn 1 + ... + a1x + a0 ti x = c nh sau: function Dathuc(c, a0, a1, ..., an); u vo:c, a0, a1, ..., an ; u ra:y = P(c); begin power := 1; y : = a0 ; Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...26 = 01b aij ij= 01b aij ij for i := 1 to ndo begin power := power *c ; y := y + ai*power ; end; Dathuc := y; end; a)Tm P(2) nuP(x) = 3x2 + x + 1 bng cch thc hin tng bc thut ton trn. b)C bao nhiu php nhn v php cng dng tng bin trong vng lp? 1.9. Thut ton Horner tm gi tr P(c) ca a thc P(x) = anxn + an 1xn 1 + + a1x + a0 nh sau: function Horner(c, a0, a1, ..., an); u vo:c, a0, a1, ..., an ; u ra:y = P(c); beginy := an ; for i := 1 tondo y := y*c + an i ; Horner :=y; end; Cc cu hi nh trong bi tp 1.8. 1.10. Lp thut ton tm s hng u tin trong mt dy cc s nguyn cho trc sao cho s hng tm thy bng mt s hng no ng trc n. Tm phc tp v thi gian ca thut ton lp. 1.11.Lp thut ton tm trong dy s nguyn dng cho trc s hng u tin nh hn s hng ng ngay trc n. Xc nh phc tp ca thut ton lp. 1.12.Ma trn lgic Ma trn m cc phn t l 0 hoc 1 c gi l ma trn lgic. 1)Cho hai ma trn lgic kch thc mn: A = [aij]m, n v B = [bij]m, n , ta c: A B l ma trn lgic hp gia A v B c cc phn t hng i, ct j l: nuaij = 1hocbij = 1 nuaij = bij = 0 A B l ma trn lgic giao gia A v B c cc phn t hng i ct j l nu aij = bij = 1 nu aij = 0hocbij = 0 2)ChohaimatrnlgicA=[aij]m,pkchthcmpvB=[bij]p,nkch thcpn.TchboolecaAviB,khiuAB,lmatrnlgickchthc mn m phn t cij (phn t hng i, ct j) c xc nh nh sau: cij = (ai1b1j) (ai2b2j) (aipbpj) Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...27 a)Vit thut ton dng phng Pascal cho cc php ton ca cc ma trn lgic nh cc nh ngha nu trn. b)C bao nhiu php ton bit c dng tnh AB vi A, B l cc ma trn lgic vung cp n. c)Chngminhrng:A(BC)=(AB)Cvigithitccphptontrongbiu thc cho l thc hin c, 1.13.Tm v m t thut ton quy cho cc bi ton sau: a)Tm tng ca n s nguyn dng l u tin. b)Tm s cc tiu ca tp hu hn cc s nguyn. 1.14.Hy a ra thut ton quy tm n! mod m, tromg m, n, l cc s nguyn dng. 1.15.Tmthuttonquytnh n2a ,trongalmtsthcvnlmtsnguyn dng (Gi : dng ng thc ( )22 2n 1 na a =+ ). 1.16.Tm thut ton qui tm c s chung ln nht ca 2 s nguyn khng m a, b(a < b) bit rng UCLN(a, b) = UCLN(a,b a). P S 1.2.a)3;b) 5; c)1; d) 0 1.4.a)O(n3);b)O(n5); c) O(n3n!);d) O(6n); e) O(x22x). 1.5. c)Procedure insert (x, a1, a2, ,an : integer); {dy xp theo th t tng:a1 a2 an} an := x + 1; i := 1; while x > ai do i := i + 1; for j := 0 to n-i do an j + 1 := an j ; aj := x; {x c chn vo v tr ng} 1.7.Procedureones (a: xu bit,a = a1a2 ... an); begin ones := 0; fori:= 1 to n do if ai = 1 thenones := ones + 1; end;{ones l s cc bt 1 trong xu a} Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...28 1.10.Procedurefind duplicate (a1a2 ... an : integer); location := 0; i := 2; whilei n and location = 0 do begin j := 1; while j < i andlocation := 0 do if ai = ajthen location := ielsej := j +1 i := i +1; end;{location l ch s ca gi tr u tin lp li gi tr trc trong dy} O(n2) 1.11. Procedurefind decrease (a1 a2 ... annguyn dng); location := 0; i := 2; whilei nandlocation = 0 do ifai < ai 1 thenlocation :=ielse i : = i + 1; end;{location l ch s ca gi tr u tin nh hn gi tr ngay trc n} phc tp ca thut ton l O(n). 1.13. a) Proceduresum of odds (n : nguyn dng); if n = 1 then sum of odds := 1 elsesum of odds : = sum of odds(n 1) + 2n 1; b) Proceduresmalless (a1, a2, , an : nguyn dng); if n = 1 thensmalless (a1, a2, , an) := a1 else smalless (a1, a2, , an) := min(smalless (a1, a2, , an 1), an); 1.14. Proceduremodfactorial (n, m : nguyn dng); ifn = 0 then modfactorial (n, m) : = 1 elsemodfactorial (n, m) :=(n * modfactorial (n-1, m))mod m;

Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...29 CU HI N TP CHNG 1 1.Phtbiunhnghavccctrngcathutton.Ccphngphpmtthut ton. 2.Trnhbyquanhbig-Ocacchmsvkhinimphctpvthigianca thut ton. nh gi phc tp ca thut ton nhn ma trn 3.M t cc thut ton tm kim v nh gi phc tp ca chng. 4.nh ngha bng quy l g? Th no l cng thc truy hi? Trnh by thut ton quy. Mi lin h gia quy v lp. 5.M t cc thut ton lp v quy : Tm tng ca mt dy s; Tm c s chung nh nht ca hai s nguyn dng; Tm s Fibonacci 6.Nuthuttonchuynicscscastnhin.Cchcngvnhnccs nguyn trong h nh phn? Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...32 CHNG 2 BI TON M 1.Nguyn l cng v nguyn l nhn 1.1.Nguyn l cng 1.2.Nguyn l nhn 2. Chnh hp. Hon v. T hp. 2.1.Chnh hp 2.2. Hon v 2.3.T hp 2.4.Chnh hp v T hp suy rng 3. Nguyn l b tr 4. Gii cc h thc truy hi 4.1. Phng php kh tm cng thc trc tip t cng thc truy hi 4.2. Gii cc h thc truy hi tuyn tnh. 5. Bi ton lit k. 5.1. Phng php sinh 5.2. Thut ton quay lui 6. Bi ton tn ti 6.1. Nguyn l Di-ric-le (Dirichlet) 6.2. Mt vi bi ton ng dng nguyn l Di-ric-l. 6.3. Phng php phn chng L thuyt t hp l mt phn rt quan trng ca ton ri rc, n chuyn nghin cu s sp xp cc i tng (phn t ca tp hp) theo mt quy tc no . Mi kt qu ca mt cchspxpccphntcamttphptheomtquytcxcnhchocgil mtcuhnhthp.mccphntcamttphp,hoccccuhnhthplni dung ca bi ton m. Chng ta cng cn c mt quan nim rng ri v khi nim m, chng hn nu ni c 100 th l mt cch m, nhng c th thay cch ni bng cch ni c 10 t 5 v 25 t 2. Ni cch khc, lit k cc cu hnh t hp l mt cch m. Cctphpnintrongchngnylcctphpchuhnccphnt.S lng cc phn t ca tp hp A, k hiu l N(A), c gi l lc lng (hay bn s) ca tp hp A. 1. Nguyn l cng v nguyn l nhn. 1.1.Nguyn l cng Nu A v B l 2 tp hp ri nhau (AB = ) th N(AB) = N(A) + N(B). Nguyn l ny c th m rng nh sau: Gi s A1, A2, ... , An l cc tp con ca mt tp X no tho mn: Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...33 -Cc Ai i mt ri nhau (AiAj = , khi i j) -X = A1A2 ... An . (Cc tp A1, A2, , An tho mn 2 tnh cht trn c gi l mt phn hoch ca tp X) Khi : N(X) = N(A1) + N(A2) + ... + N(An) c bit, ta cN(A) = N(X) N( A) trong A l tp b ca tp A trong X. Thd1:Mtkhohcc3danhschlachnccbithchnh:Danhschth nht c 10 bi; danh sch th hai c 15 bi; danh sch th ba c 25 bi. Mi sinh vin c chn 1 bi trong 3 danh sch trn lm thc hnh. Hi mi sinh vin c bao nhiu cch la chn bi thc hnh. Gii:C 10 cch la chn trong danh sch th nht, 15 cch la chn trong danh sch thhaiv25cchlachntrongdanhschthba.Vyc10+15+25=50cchla chn cho mi sinh vin. Th d2:Mtonvnngvincamt aphngccithiu2mn: bnsngv bn cung. Trong on c10 nam v s vn ng vin bn sngl14 (k c nam v n). S n vn ng vin bn cung bng s nam vn ng vin bn sng. Hi on c bao nhiu ngi. Gii:GiA,Btngnglcctpvnngvinnam,vnngvinn,tac N(A) = 10 A1, A2 tng ng l cc tp vn ng vin nam bn sng, vn ng vin nam bn cung, ta cN(A1) + N(A2) = 10 B1,B2tngnglcctpvnngvinnbnsng,vnngvinnbn cung, ta cN(B2) = N(A1) v N(B1) + N(A1) = 14 T N(B1) + N(B2) = 14, ngha l on c 14 vn ng vin n. Vy ton on c 10 + 14 = 24 vn ng vin. Th du 3:Xt on chng trnh phng Pascal sau k := 0;for i1:= 1 to n1 dok := k + 1; for i2:= 1 to n2 dok := k + 1; for i3:= 1 to n3 dok := k + 1; Hi sau khi thc hin xong on chng trnh trn, gi tr ca k bng bao nhiu? Gii:Gi tr ban u gn cho k l 0. Khi lnh gm 3 vng lp tun t, sau mi bc lp ca tng vng lp gi tr ca k c tng thm 1, vng lp th i (i = 1, 2, 3) c ni bc nn sau vng lp th i gi tr ca k c tng thm ni. Do cc vng lp l tun t nn sau c 3 vng lp th gi tr ca k l:k = n1 + n2 + n3. 1.2.Nguyn l nhn Nu mi thnh phn th i (tc l ai) ca b c th t gm k thnh phn (a1, a2, ... , ak) c ni cch chn (i = 1, 2, , k) th s lngcc b k thnh phn s l n1n2 nk (tch ca s cc cch chn ca mi thnh phn). Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...34 C th pht biu nguyn l nhn di dng s phn t ca tch -cc ca cc tp hp: N(A1 A2 Ak) = N(A1). N(A2) N(Ak) Thd1:C4phngtinitHnivothnhphHChMinhl:t,Tu ho, Tu thu v Mybay.C 2 phng tin i t thnh ph H ch Minh ra Cn o l My bay v Tu thu. Hi c my cch i t H ni n Cn o nu bt buc phi i qua thnh ph H Ch Minh? Gii: Mi cch i t H ni vo thnh ph H Ch Minh c 2 cch i ra Cn o. Vy vi4cchitHnivothnhphHChMinhc4.2=8cchitHniqua thnh ph H Ch Minh ri n Cn o. Th d 2:Mt xu nh phn l mt dy lin tip cc ch s 0 hoc 1 (gi l bit 0 v bit 1). di ca xu l s cc ch s 0, 1 c mt trong xu. Hai xu cng di c gi l khc nhau nu c t nht mt v tr ti cc bit l khc nhau. Hi c bao nhiu xu nh phn c di 8 ? Gii: Ti mi mt trong 8 v tr c 2 cch la chn l bit 0 hoc bit 1. Do theo quy tc nhn cho thy c 28 = 256 xu nh phn khc nhau c di 8. Thd3:Mtbinsxemycamtaphngccutogmhainhm. Nhm u gm 2 k t: u tin l mt ch ci sau l mt ch s. Nhm th 2 gm mt dy 4 ch s lin tip (k c cc s trng nhau). Hi c th pht hnh bao nhiu bin s xe ti a phng .Gii:C tt c 24 cch chn cho v tr ch ci u tin. Nm v tr tip theo, mi v tr c 10 cch chn. Theo quy tc nhn, s bin s ng k xe my c th l: 24.10.10.10.10.10 = 24.105 = 2 400 000 Th d 4: Gi tr ca bin k bng bao nhiu sau khi on chng trnh sau c thc hin? k := 0; for i1 := 1 to n1 do for i2 := 1 to n2 do for i3:= 1 to n3 do k := k + 1; Gii:u tin k c gn gi tr bng 0. C 3 vng lp lng nhau. Sau mi ln lp ca vng for trong cng, gi tr ca k tng thm 1. Vi mi gi tr ca i1, vng for th hai thc hinn2lnvvimigitrcai2vngforth3thchinn3ln.Vytheonguynl nhn,saukhic3vnglpktthcthvnglptrongcng(i3)thchinn1.n2.n3ln, ngha lk = n1.n2.n3. Chrngnhiubitonmphigiibngcchkthpcnguynlcngv nguyn l nhn. Th d:Tn bin trong ngn ng lp trnh Pascal l mt xu gm ch ci tingAnh (khng phn bit ch ci thng v ch ci hoa) v ch s, trong khng c bt u bng ch s. Hi c th t c bao nhiu tn bin khc nhau c di khng qu 2 k t, bit rng c 10 t kho c di 2 (chng hn nh to, do, go, if, or, in, ) Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...35 Gii:Gi A l tp tn bin c 1 k t, th N(A) = 26 (c 26 tn bin ng vi 26 ch ci tin Anh) Gi B l tp tn bin c 2 k t. K t u c 26 cch la chn, cn k t th hai c 36cchlachn(26chciv10chs).VyN(B)=26.36=936.Doc10xub cm nn tng s tn bin c di khng qu 2 k t l: 26 + 936 10 = 952 2. Chnh hp. Hon v. T hp. 2.1.Chnh hpnhngha:Mtchnhhpchpkcanphntlmtbcthtgmk phn t ly t n phn t cho (k n). S cc chnh hp chp k ca n phn t, k hiu l knA , c tnh theo cng thc: k)! (nn!= + = ) 1 k n ( . . . ) 2 n )( 1 n ( n Akn

Tht vy,v c n phn t cho nn c n cch chn phn t ng u,tip theo c n 1 cch chn phn t ng th hai, n 2phn t ng th ba, ..., n k + 1 cch chn phn t thk. Theo nguyn l nhn c cng thc cn chng minh. Th d 1.C 10 vn ng vin thi chy. Hi c bao nhiu cch d on cc vn ng vin v nht, nh, ba? Bit rng cc vn ng vin u c cng kh nng nh nhau. Gii:S cch d on l s cch chn c th t 3 trong 10 vn ng vin, tc l: 310A = 10. 9. 8 = 720 cch d on. Th d 2.C th lp c 39A 9= 9.9.8.7 = 4536 s nguyn c 4 ch s khc nhau. Th d 3.C bao nhiu n nh t tp hp A c k phn t vo tp hp B c n phn t (k n)? Gii :Gi s cc phn t ca tp hp A tng ng vi cc s 1, 2, ..., k;khi mi n nh chnh l mt b c th t cc nh ca cc phn t ca tp hp A. Vy c knAcc n nh t A vo B. 2.2.Hon v: nh ngha: Mt hon v ca n phn t cho l mt cch sp xp c th t ca n phn t . D nhn ra rng, mt hon v ca n phn t chnh l mt chnh hp chp n ca n. Do : S cc hon v ca n phn t , k hiu Pn , l: ! n 1 . . . ) 1 n ( n A Pnn n= = =Th d 1.Mi cch xp hng (ngang hoc dc) ca 5 ngi l mt hon v ca 5 phn t. Vy c P5 = 5! = 120 cch xp hng ca 5 ngi.Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...36 Th d 2.Mt ngiphi i kim tra 8 da im khc nhau theo bt kth t no. Hi ngi c bao nhiu cch i nu bt u t mt trong 8 a im sao cho mi a im u c kim tra ng mt ln. Gii:Ngi c P7 = 7! = 5040 cch i. 2.3.T hp nh ngha:Mt t hp chp k ca n phn t l mt cch chn khng k th t k phn t t n phn t cho (n k). S cc t hp chp k ca n phn t , k hiu l knC , c tnh nh sau: Vimi b k phn t khng sp th t tng ng vi k! cch sp xp c th t. Ni cch khc mi t hp chp k ca n tng ng vik! chnh hp chp k ca n. T suy ra: knknC ! k A = ( )! k n k!n!k!ACkn kn= =Th d 1. C 6 i bng thi u vng trn mt lt. Hi phi t chc bao nhiu trn u? Gii:C 2 igp nhau mt trn suy ra s trn u l26C= 15 trn Th d 2.Tnh s giao im ca cc ng cho nm pha trong mt a gic li n cnh (n 4). Bit rng khng c 3 ng cho no ng quy ti 1 im pha trong a gic . Gii:C 4 nh ca a gic cho ta mt giao im tho mn bi ton. Vy s im cn m l: 243) 2)(n 1)(n n(nC4n =Thd3.Chomtlicm.nchnht.Ccntcnhst0nntheo chiu t tri sang phi v t 0 n m theo chiu t di ln trn (xem hnh v). Hi c bao nhiungikhcnhautnt(0,0)nnt(n,m),bitrngmibcilsdch chuyn hoc sang phi hoc ln trn ca mt hnh ch nht (khng dch chuyn sang tri hoc xung di) (0,m)(n,m) (0,0)(n,0) Gii: Mt ng i nh vyc xem l gmn+m bc, mi bc ch cchn mttronghaigitrl0nuilnv1nuisangphi.Nhvymingitng ng vi mt dy nh phn c di m+n trong c ng m gi tr 1 (v tt nhin c ng n gi tr 0). Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...37 Bi ton dn n vic tm xem c bao nhiu dy nh phn c di m+n trong c ng m thnh phn 1. Ni cch khc l s cch chn m v tr trong m+n v tr xp s 1. Vy s ng i s l mn mC+ (hoc nn mC+) Th d 4.Xt on chng trnh phng Pascal sau: fori := 1ton 1 do forj := i + 1to ndo ifai > aj then begin {i ch ai, aj } t : = ai ; ai := aj ; aj := t ; end; (y l mt trong cc thut ton xp dy s a1, a2, ..., an theo th t tng dn) Hy m s php so snh cc ai c thc hin trong on chng trnh trn. Gii:Ti mi vng lp i = k th vng lp j phi thc hin n k + 1 bc. Ngha l vi mi i = k phi thc hinn k +1 php so snh (k = 2, 3, ..., n). T theo nguyn l cng, s cc php so snh l: 21) n(n1 2) (n 1) (n= + + + . . .Cngcthgiinhsau:Vmicpai,aj(ij)uphisosnhvinhau,do tng cc php so snh l 2nC . Mt vi tnh cht T cng thc tnh knC , d dng chng minh c: 1)1 C Cnn0n= =; 2) k nnknC C= ; 3) 1 k1 nk1 nknC C C + = ; Tnh cht 3 chnh l cng thc truy hi tnh knC . Ta c: n\ k012345 01 111 2121 31331 414641 515101051 Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...38 Hay: 00C 1101C C 221202C C C . .. nn2n1n0nC . . . C C CMi s hngca hng di, tr s hng u vs hngcui bng 1, bng tng 2 s hng hng trn cng ct v ct trc ca s hng cn tnh, ngi ta thng gi chng l cc tam gic Pascal. Bng l lun t hp c th chng minh c: ( )n nn2 2n1n0 0nnx C . . . x C x C x C 1 x + + + + = + . Tht vy, h s ca xk v phi ca cng thc c c bngcch nhn k s hng x trong k nhn t (x + 1) vi n ks 1 trong n knhn t (x + 1) cn li v tri. iu tngngviscchchnknhnt(x+1)trongnnhnt vtri, scchchn chnh l knC . Vy cng thc c chng minh. T cng thc trn, ta c cng thc nh thc Niu-tn: ( )k n kn0 kknn0 kkkn knn0 kkknnnn ny x Cyxy CyxC y 1yxy y x = = == =|||

\|=|||

\|+ = + Ch 1. Nu chn x = y = 1, ta c: 2 nn1n0n2 C . . . C C = + + + . Tuy nhin, knCchnh l s cc tp con c k phn t ca tp hp c n phn t . Vy, tng s tp con (k c tp rng v chnh tp hp ) ca mt tp hp c n phn t l 2n. Ch 2. Nu chnx = 1,y= 1, ta c:0 C ) 1 ( . . . C C Cnnn 2n1n0n= + + + . 2.4.Chnh hp v T hp suy rng a.Chnh hp lp nh ngha:Mt b c th t gm k phn t ly t n phn t cho, trong mi phntcthclynhiulncgilmtchnhhplpchpktnphnt cho. Ch rng do mi phn t c th c ly nhiu ln nn rt c th k > n. S cc chnh hp lp chp k ca n c k kiu l knAv c tnh theo cng thc: k knn A = Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...39 Tht vy, vc n cch chn cho mi phn t th i( i = 1, 2, ..., k); nn theo quy tc nhn c cng thc cn chng minh. Th d 1.Tnh s cc dy nh phn c di n. Mi dy nh phn c di n l mt b c th t gm n thnh phn c chn trong tp {0, 1}. Do s dy nh phn c di n l 2n. Th d 2.C th lp c9. 310A= 9. 103 = 9000 s nguyn dng c 4 ch s. Th d 3. C bao nhiu nh x t tp hp A c k phn t vo tp hp B c n phn t ? Lluntngtnhthdtnhsnnhtrongmcchnhhp;domiphnt ca tp hp B c th l nh ca nhiu phn t thuc A nn s nh x c th c l nk. Th d 4.Trong ngn ng Pascal chun quy nh tn bin khng qu 8 k t (mi k t l mt trong 26 ch ci ting Anh hoc mt trong 10 ch s) v phi bt u bng mt ch ci, k t l ch khng phn bit ch hoa v ch thng. Hi c th nh ngha c bao nhiu bin khc nhau? Gii:Tt c c 8 loi bin: 1 k t, 2 k t, ..., 8 k t. Theo quy tc nhn th s bin c k k t l 26.36 k 1, (k = 1, 2, ..., 8). T p dng quy tc cng c s cc bin khc nhau trong ngn ng Pascal l: 26(1 + 36 + 362 + ... + 367) = 2 095 681 645 538 2. 1012. y l mt con s rt ln, khng th no duyt ht c chng. Hin tng cc s t hp tng nh vy gi l hin tng bng n t hp. S k t1234567 S bin2696234 6581 247 71444 917 7301 617 038 30658 213 379 042 b.T hp lp nh ngha:Mt t hp lp chp k ca n phn t l mt cch chn khng k th t k phn ttn phn t cho, trong mi phn t c th c chn nhiu ln. Ch rng do mi phn t c th c ly nhiu ln nn rt c th k > n. C rt nhiu bi ton m phi s dng khi nim t hp lp. Chng hn, trong hp c 3 loi bi: bi , bi xanh v bi trng; mi loi c t nht 4 vin. Hi c bao nhiu cch ly ra 4 vin bi t hp , nu khng phn bit th t ly v cc vin bi ly ra c th cng 1 loi. Gii:C th lit k cc cch ly nh sau: 4 bi 4 bi xanh4 bi trng3 bi , 1 bi xanh 3 bi , 1 bi xanh 3 bi xanh, 1 bi 3 bi xanh, 1 bi trng 3 bi trng, 1 bi 3 bi trng, 1 bi xanh 2 bi , 2 bi xanh 2 bi , 2 bi trng 2 bi xanh,2 bi trng 2 bi , 1 bi xanh, 2 bi xanh, 1 bi , 2 bi trng, 1 bi , 1 bi trng1 bi trng 1 bi xanh C tt c 15 cch ly 4 bi t hp cho. Mi cch ly nh vy l mt t hp lp chp 4 t 3 phn t cho. Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...40 Chng ta khng th tnh s t hp lp chp k t n phn t mt cch th cng nh vy, bi khng phi lc no cng lit k ht cc t hp c. Gi s mt hp c 3 ngn ng 3 loi bi khc nhau: ****...**** bi bi xanh bi trng bi bi xanhbi trng Vic chn 4 bi tng ng vi vic t 4 bi vo trong 3 ngn cho. Loi tr 2 vch ngnngoicnglcnh,cn2vchngnphatrongv4vinbicthichcho nhau. Nh vys cch chn 4 bi tng ng vi s cch chn 4 trong 6 v tr t 4 bi. Ni cch khc s cch chn 4 bi l 46C= 15 Khiqutcthlplunnhsau:Mithplpchpkcanphntcthxem nh cch xp k du * vo mt hp c n ngn, trong cc vch ngn hp v cc du * c th dch chuyn c tr hai vch ngn hai u hp l c nh. Nh vy mi t hp lp chp k ca n phn t chnh l s cch chn k v tr trong n + k 1 (gm k v tr t du * v n 1 vch ngn) t k du *. Vy, k hiu s cc t hp lp chp k ca n phn t l knC , ta c: k1 k nknC C +=Th d 1. Mt ca hng c 4 loi bnh hp khc nhau. Hi c bao nhiu cch chn 6 hp bnh? Gii:Theo khi nim t hp lp, ta c6961 6 464C C C = = + = 84 cch chn. Th d 2. C bao nhiu cch chn 5 t tin t 7 loi tin giy 1000, 2000, 5000, 10 000, 20 000, 50 000 v 100 000? Gii:S cch chn l511C C57= = 462. Th d 3. Phng trnhx + y + z = 11 c bao nhiu nghim nguyn khng m? Gii:Mi nghim ca phng trnh ng vi mt cch phn chia 11 phn t (11 s 1) cho thnh ba loi, sao cho c x phn t loi 1, y phn t loi 2 v z phn t loi 3. V vy s nghim ca phng trnh bng s t hp lp chp 11 t tp c 3 loi phn t , ngha l c: 111 11 3C C113 += = 78nghim. Cthtmcnghimcaphngtrnhchovicciukinrngbucca cc n s. Chng hn tm s nghim nguyn khng m ca phng trnh cho trn tho mn iu kin: x 1; y 2; z 3. Khi mt nghim ca phng trnh ng vi mt cch phn chia 11 phn t cho thnh 3 loi, sao cho c x phn t loi 1, y phn t loi 2 v z phn t loi 3 trong c t nht 1 phn t loi 1 (x c sn mt s 1), c t nht 2 phn t loi 2 v c t nht 3 phn t loi 3. V th trc ht chn 1 phn t loi 1, 2 phn t loi 2 v3 phn t loi 3; cn li 5 phn t c chia tip cho ba loi. Do snghim ca phng trnh l s t hp lp chp 5 t 3 loi phn t cho: 5753C C == 21(nghim) Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...41 c.Hon v ca cc tp hp c cc phn t ging nhau Trong cc bi ton m, mt s phn t ca tp hp cho c th ging nhau. Khi phi cn thn ko ta m chng nhiu hn mt ln. Th d: C th lp c bao nhiu xu khc nhau bng cch sp xp li cc ch ci trong cm tNHANDANANHHUNG (nhn dn anh hng)? Gii: Trong t cho c 14 ch ci gm 5 ch N, 3 ch H, 3 ch A v cc ch D, U, G. L ra c 14! cch sp xp, nhng trong mi cch sp xp nu i ch 5 ch N (c 5! cchich),ich3chH(c3!cchich)vich3chA(c3!cchi ch), th xu thu c l khng i. Vy nu gi s xu thu c l m th theo nguyn l nhn, ta c: m.5!.3!.3! = 14! T s cc xu khc nhau c c l: 160 180 28! 3 !. 3 !. 5! 14m = = Bng l lun tng t ta c:S hon v ca n phn t trong c n1 phn t ging nhau thuc loi mt, n2 phn t ging nhau thuc loi hai, , nk phn t ging nhau thuc loi k (n1 + n2 + + nk n) l: ! n ... ! n ! n! nk 2 1 Ch :C nhiu cch suy ra cng thc tnh s hon v lp. Chng hn, trong n v tr ta chn n1 v tr xp cc phn t loi mt, nh vy c 1nnCcch chn; sau chn n2 v tr trong n n1 v tr cn li xp cc phn t loi hai, tc l c 21nn nC; v.v T dng nguyn l nhn v cng thc tnh s t hp c kt qu nh bit. d.S phn chia cc tp hp thnh cc tp con khc nhau Trc ht xt th d: C bao nhiu cch chia mt c bi 52 qun cho 4 ngi sao cho mt ngi c 10 qun, nhng ngi khc ln lt c 8, 8, 7 qun? Gii:Ngi u tin nhn 10 qun bi c 1052Ccch nhn khc nhau, cn li 42 qun nn ngi tip theo c 842Ccch nhn 8 qun bi, cn li 34 qun bi nn ngi th ba c 834Ccch nhn 8 qun bi, v ngi cui cng c 726Ccch nhn 7 qun bi. Theo nguyn l nhn th s cch chia cc qun bi tha mn u bi l: ! 19 ! 7 ! 8 ! 8 ! 10! 52! 19 ! 7! 26! 26 ! 8! 34! 34 ! 8! 42! 42 ! 10! 52C C C C7268348421052= = 2,23.1031. Tng qut, tac: S cch phn chia mt tp hp gm n phn t thnh ktp sao cho tp con th i c ni phn t ( n1 + n2 + + nk = n) l: ! n ... ! n ! n! nk 2 1 Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...42 Ch l cng thc va lp c cho thy th t cc tp con khng nh hng n kt qu phn chia tp hp cho. 3.Nguyn l b tr. Nu khng c gi thit A, B l 2 tp hp ri nhau, ta c nguyn l cng tng qut: N(A B) = N(A) + N(B) N(A B) (1) R rng, nu A, B ri nhau th A B = nn N(A B) =0 v c nguyn l cng n gin (mc 1.1): N(A B) = N(A) + N(B) Cng thc (1) cn gi l nguyn l b tr cho 2 tp hp. Tng qut hn, ta c: nh l:Gi s A1, A2, ..., An l n tp hp hu hn. Khi : ) A ... A N(A 1) ( ... ) A A N(A ) A N(A ) N(A) A . . . A N(An 2 11 nk j inj i in 2 1k j i 1 i j i + + + == = (2) Chngminh:ChngtabitrngscctpgiaoAiAjl2nC ,scctpgiao AiAjAk l 3nC, ... . Tng qut s cc tp giao ca r tp l rnC , r = 1, 2, ...,n. Cng thc (2) c chng minh bng cch ch ra rng, mi phn t ca tp hpA1 A2 ... Anc m ng mt ln v phi ca (2). Xt phn t tu x ca tp hp A1 A2 ... An . Gi s x c mt trong r tp ca A1, A2, ..., An trong r = 1, 2, ..., n. Phn t x ny c m 1rCln trong tng N(Ai), m 2rCtrong tng N(AiAj), ... Bi vy tng cc ln m ca x v phi ca (2) l: ] C ) 1 ( ... C C C 1 [ 1 C ) 1 ( ... C C Crrr 3r2r1rrrr 3r2r1r + + + = + + = 1 (1 1)r = 1 Vy x c m ng mt ln trong v phi ca (2). nh l c chng minh. C th vit cng thc (2) di dng khc: ng nht tp Ai vi tnh cht Ai cho trn mt tp hp X no v phi m xem c bao nhiu phn t ca X khng tho mn bt c tnh cht Ai no (i = 1, 2, ..., n) Gi N* l s cn m;N = N(X) (l s phn t ca X), ta c: N* = N N(A1 A2 ... An) = N N1 + N2 ... + (1)nNn . (3) trong Nk l tng cc phn t ca X tho mnk tnh cht ly t n tnh cht cho. Th d 1:Trong mt lp hc c 50 sinh vin th c 30 ngi bit ting Anh, 20 ngi bittingPhp,trongssinhvinbit ngoingc10ngibitctingAnhvting Php. Hi trong lp c bao nhiu sinh vin khng bit c ting Anh v ting Php? Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...43 Gii:Theo cng thc (1), ta c s sinh vin bit t nht 1 trong 2 th ting Anh, Php l:30 + 20 10 = 40 s sinh vin khng bit c ting Anh v ting Php l 10 ngi. Th d 2.C bao nhiu xu nh phn c di 8 hoc bt u bng 1 hoc kt thc bng 00? Gii: Do v tr u tin l c nh, ch c 1 cch chn duy nht, mi mt trong 7 v tr tip theo c 2 cch chn (0 hoc 1), suy ra s xu nh phn c di 8 bt u bng 1 l 27 = 128 xu. Tng t s xu nh phn c di 8 kt thc bng 00 l 26 = 64 v s xu nh phn c di 8 bt u bng 1 v kt thc 00 l 25 = 32. Vy, s xu nh phn c di 8 hoc bt u bng 1 hoc kt thc bng 00 l128 + 64 32 = 160(xu) Th d 3.C bao nhiu s nguyn dng khngln hn 100 hoc chia ht cho 4 hoc chia ht cho 6 nhng khng ng thi chia ht cho c 4 v 6? Gii:S cc s chia ht cho 4 l:254100=((

,S cc s chia ht cho 6 l:166100=((

, S cc s ng thi chia ht cho c 4 v 6 l:812100=((

Vy s cn tm l:25 + 16 8 = 33. Th d 4.C bao nhiu s nguyn nh hn hoc bng 1000 khng chia ht cho bt c s no trong cc s 3, 4, 5? Gii:Gi X l cc s nguyn dng nh hn hoc bng 1000 th N(X) = 1000. Ai = {x X| xchia ht cho i },i = 3, 4, 5. Khi A3 A4 A5 l tp cc s trong X chia ht cho t nht mt trong cc s 3, 4, 5 v:N(A3 A4 A5) = N1 N2 + N3 . Ta c: N1 = N(A3) + N(A4) + N(A5) == ((

+((

+((

510004100031000 = 333 + 250 + 200 = 783 N2 = N(A3 A4) + N(A3 A5) + N(A4 A5) = = ((

+((

+((

4.510003.510003.41000 = 83 + 66 + 50 = 199 N3 = N(A3 A4 A5) =163.4.51000=((

T c p s ca bi ton l: 1000 N1 + N2 N3 = 400. Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...44 Th d 5. Bi ton b th C n phong b ghi sn a ch v n l th vit cho n a ch . B ngu nhin cc l th vo cc phong b (mi l th cho vo 1 phong b). Hy tm xem c bao nhiu cch b th sao cho khng c l th no c b ng a ch. Gii:S cch b th vo phong b l n!. Ni cch khc, nu gi X l tt c cc cch b th vo phong b th N(X) = n! . GiAk l tnh cht c k l th b ng a ch. Gi N*l s cch b th sao cho khng c l th no ng a ch. Theo cng thc (3), ta c: N* = n! N1 + N2 ... + (1)nNn , trong Nk l s tt c cc cch b th sao cho c k l th b ng a ch. tnh s Nk ta l lun nh sau: C knCcch ly k l th, mi cch ly k l th c (n k)! cch b k l th ny ng a ch (v khi n k l th cn li c b ty ), do : k!n!k)! (n C Nkn k= =T :|||

\|+ + =n!1) (2!11!11 n!n*NS N* trong bi ton b th c gi l s mt th t, k hiu l Dn. Di y l mt vi gi tr ca Dn: n234567891011 Dn 12944265185414833133 4961 334 9614 890 741 S mt th t Dn tng rt nhanh, ngi ta gi l hin tng bng n t hp. 4.Gii cc h thc truy hi Thngccbitonmphthcvothamsuvolstnhinn,chng hn nh s t hp, s chnh hp, s mt th t, ... Vic biu din tng minh cc kt qu ny nh mt hm ca s t nhin n gi l cng thc trc tip. Trong nhiu trng hp vic tm cng thc trc tip nh th l kh khn. C mt phng php biu din kt qu m c ng vi gi tr u vo thng qua cc gi tr u vo nh hn v cng thc tnh thu c gi l cng thc truy hi. Cc gi tr u tin m k t cng thc truy hi c hiu lc gi l gi tr ban u ca cng thc truy hi (xem li nh ngha trong 6.1. chng 1).Thd:Cthtnhsthp knC nhsau:Chnmtphntacnhtrongn phn t ang xt. Khi , nu a c chn vo t hp th phi chn thmk 1phn tna t n 1 phn t cn li, s cch chn trong trng hp ny l 1 k1 nC; cn nu a khng c chn vo t hp th phi chn c k phn t trong n 1 phn t cn li, s cch chn trong trng hp ny l knC . Theo quy tc cng ta c: k1 n1 k1 nknC C C+ =Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...45 l cng thc truy hi tnh s knC . Cc gi tr ban u ca cng thc ny l1 C C1101= = . Diychngtaxtmtsphngphpacngthctruyhivcngthc trc tip. 4.1. Phng php kh tm cng thc trc tip t cng thc truy hiPhng php ny c xt qua mt s th d sau: Th d 1: Bi ton chia mt phng. Trn mt phng, k n ng thng sao cho khng c 2 ng no song song v khng cqu2ngthngngquytimtim.Himtphngcchiathnhbaonhiu phn? Gii:Gi s phn ca mt phng c chia bi n ng thng nh vy l Sn. Gi s k c n 1 ng thng, k thm ng thng th n th s phn c thm s bng s giao im c thm cng 1. S giao im c thm l s giao im m ng thng va k ct n 1 ng thng k trc , ngha l bng n 1. T nhn c cng thc truy hi: Sn = Sn 1 + n ;n 1; S0 = 1. (S0l gi tr ban u) T cng thc trn, d dng tnh c: S1 = 1 + 1 = 2,S2 = 2 + 2 = 4,S3 = 4 + 3 = 7 S4 = 7 + 4 = 11, S5 = 11+ 5 = 16,S6 = 16 + 6 = 22, ... tm cng thc trc tip, ta c: S0 = 1 S1 = S0 + 1 S2 = S1 + 2 S3 = S2 + 3 ...... Sn = Sn 1+ n Cng v vi v ca cc ng thc trn c: Sn = S0 + 1 + 2 + 3 + ... + n = 22 n n21) n(n12+ +=+Th d2: Bi ton li kp. Mt ngi gi tit kim 10 triu ng vi li sut hng nm l 10%. Ht mt nm gi tin, nu khng rt tin ra ngi c cng s li vo gc tnh li cho nm tip theo (li kp). Hi sau 20 nm gi khng rt ra ln no th s tin ca ngi l bao nhiu? Gii:Gi Pn l s tin ca ngi sau n nm gi, khi s tin ca ngi bng s tin ca nm thn 1cng vi li sut ca nm th n. Nh vy: Pn = Pn 1 + 0,1 Pn 1 = 1,1 Pn 1. l cng thc truy hi tnh s tin sau n nm gi ca ngi . Gi tr ban u ca bi ton l P0 = 10 (triu). Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...46 Th ln ltPn 1 = 1,1 Pn 2 vo Pn ,Pn 2 = 1,1 Pn 3 vo Pn 2 ; ... Cui cng ta c: Pn = 1,1n P0. Sau 20 nm s tin ca ngi l:P20 = 1,120.10 6,7275. 10 = 67,275 triu ng. Th d 3: Bi ton thp H ni. Tng truyn rng, ti mt ngi cha H ni c mt tm bng ng, trn c 3 ci cc bng kim cng. Lc khai thin lp a, trn mt trong 3 ci cc c Pht 64 chic a bng vng vi ng knh nh dn (hnh 1). Ngy m cc nh s phi dch chuyn a sang mt cc khc theo nguyn tc: mi ln ch c chuyn 1 a t cc ny sangcc khc bt knhng khng c t mt chic a to ln trn mt chic a khc nh hn. Hi phi mt bao lu cc nh s mi chuyn ht s a sang mt cc khc, bit rng mi ln chuyn 1 a ht 1 giy? Hnh 1.Trng thi ban u Hnh 2. Trng thi sau Hn1 ln chuyn Gii:GiHn l s ln dch chuyn a nu cc ban u c n a. Gi s lc u n a trn cc 1 (hnh 1). Sau Hn 1 ln dch chuyn cc nh s chuyn c n 1 a sang cc 3 (hnh 2). T y phi c 1 ln chuyn chic a ln nht t cc 1 sang cc 2 v Hn 1 ln dch chuynn 1 chic a t cc 3 sang cc 2. Nh vy, ta c: Hn=2 Hn 1 + 1 l cng thc truy hi tnh s ln dch chuyn n a t cc ny sang cc khc. D thy gi tr ban u l H1 = 1 (cc ch c 1 a). Chng ta tm cng thc trc tip tnh Hn. Ta c: Hn =2 Hn 1 + 1=2 (2 Hn 2 + 1) + 1 =22 Hn 2 + 2 + 1 = =22 (2 Hn 3 + 1) + 2 + 1=23 Hn 3 + 22 + 2 + 1 = . . . . . =2n 1H1 + 2n 2 + ... + 2 + 1 = =2n 1. Vi64chic a cho, cc nh s phi dch chuyn: H64 = 264 1 = 18,4467.1018(hn 18 t t ln) Numilndchchuynht1giy,ccnhscnkhong580tnmmihonthnh vic chuyn 64 a t cc ny sang cc khc. Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...47 Th d 4:Lp cng thc truy hi tnh s mt th t Dn. Gii:nh s th v phong b t 1 n n, th th i ng vi a ch phong b i . Mt cch b th ng nht vi mt hon v (a1, a2, ..., an) ca (1, 2, ..., n). Mt cch mt th t l mt hon v (a1, a2, ..., an) sao cho ai i, i. Thnh phn a1 ca hon v c th nhnn 1 gi tr ngoi s 1. Vi mi gi tr k ca a1 (k 1) c th xy ra 2 trng hp: 1)ak = 1,khiccthnhphncn lic xc nh nh s cch mt th t can 2 phn t , s ny chnh l Dn 2 . 2)ak 1,khiccthnhphncn lic xc nh nh s cch mt th t can 1 phn t (xem gi tr 1 nh l gi tr k), s ny chnh l Dn 1. T c cng thc truy hi: Dn=(n 1)(Dn 2 + Dn 1) , n 3 . Cc gi tr ban u c th tnh trc tip t nh ngha v c D1 = 0, D2 = 1.Nu coi D0 = 1 th cng thc truy hi trn ng n 2. Ta c:D3 = (3 1)(0 + 1) = 2 ,D4 = (4 1)(1 + 2) = 9, D5 = (5 1)(2 + 9) = 44 ,D6 = (6 1)(9 + 44) = 265,... C th a cng thc truy hi trn v cng thc trc tip nh sau: Dn = (n 1)(Dn 2 + Dn 1 ) Dn nDn 1 = [Dn 1 (n 1)Dn 2] t In =Dn nDn 1 . Ta c: Dn n Dn 1 =In= In 1 = In 2 = ... = (1)n 1 I1 = (1)n. Chia c 2 v chon! , c: 2 n ,n!1) (1)! (nDn!Dn1 n n = Cng v vi v ca h thc trn c : |||

\|+ + + = + + + =n!1) (2!11!11 n! Dn!1) (2!21!11n!Dnnnn... ...Cng thc thu c l cng thc bit trong th d 5, mc 3 (Nguyn l b tr). 4.2. Gii cc h thc truy hi tuyn tnh. Trong phn ny xt cch tm cng thc trc tip t cng thc truy hi c dng c bit gi l dng tuyn tnh. a.Cc khi nim nh ngha:Cng thc truy hi tuyn tnh thun nht bc k c h s hng s l cng thc truy hi c dng: an = c1 an 1 + c2 an 2 + ... + ck an k (1) trong c1, c2, ..., ck l cc s thc (hng s) v ck 0. Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...48 S d c thut ngcng thc truy hi bc k v s hng an c tnh qua k s hng trc n. Cc gi tr ban u ca cng thc truy hi trn gm k gi tr: a0, a1, ..., ak 1 . C th tm cng thc trc tip tnh an di dng: an = r n . (2) Trong r l mt s thc cn xc nh. Dy { an } xc nh theo (2) s tho mn (1) nu r l nghim ca phng trnh: rn = c1 rn 1 + c2 rn 2 + ... + ck rn k . hay:rn c1 rn 1 c2 rn 2 ... ck rn k = 0. (3) Phngtrnh (3)gi lphngtrnh c trngca cng thc (1),v nghim ca (3) gi l nghim c trng. Trc ht xt vik = 2, tc l cng thc truy hi tuyn tnh bc hai: an =c1 an 1 + c2 an 2 . Phng trnh c trng ca cng thc truy hi tuyn tnh bc hai l: r2 c1 r c2=0 b.Giicngthctruyhituyntnhbchaikhiphngtrnhctrngc2 nghim phn bit. nh l 1:Cho c1, c2 l hai s thc. Nu phng trnh c trng r2 c1 r c2 = 0 c hai nghim thc phn bit r1, r2 th iu kin cn v dy s{an } l nghim ca cng thc truy hi: an =c1 an 1 + c2 an 2

l:an=1 r1n + 2 r2n , n = 1, 2, ... trong1,2lccsthccthxcnhcnhcciukinbanucacng thc truy hi. Chng minh:iu kin cn:Nu r1, r2 l 2 nghim ca phng trnh c trng, ta phi chng minh an=1 r1n + 2 r2n . vi1, 2 l cc hng s tho mn cng thc truy hi an =c1 an 1 + c2 an 2. Tht vy:Ta c:r12 c1 r1 c2=0v r22 c1 r2 c2=0 hay: r12 =c1 r1 + c2 v r22 = c1 r2 + c2 T :an =c1 an 1 + c2 an 2 =c1(1 r1n 1 + 2 r2n 1 ) + c2(1 r1n 2 + 2 r2n 2 ) = = 1 r1n 2(c1 r1 + c2) + 2 r2n 2(c1 r2 + c2) = = 1 r1n 2 r12+2 r2n 2 r22= = 1 r1n + 2 r2n . iu kin : Gi s { an } l mt dy bt k tho mn h thc truy hian = c1 an 1 + c2 an 2. Ta phi chng minh rng s chn c cc s1 , 2 sao cho an=1 r1n + 2 r2n . Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...49 Tht vy, gi s a0 v a1 l 2 gi tr ban u ca h thc truy hi. Ta c: + =+ =2 2 1 1 12 1 0r r aa y l h phng trnh tuyn tnh 2 n Cramer, n c nghim duy nht: 2 11 1 022 10 2 11r ra r a;r ra r a== Vyan=1 r 1n + 2 r2nc xc nh. Th d:DyFibonacci, nh bit c xc nh bi cng thc truy hi: Fn = Fn 1 + Fn 2 ;F0 = 0 ; F1 = 1. lcngthctruyhituyntnhbchaichskhngi.Taitmcngthc trc tip tnh Fn. Gii: Phng trnh c trng ca cng thc l: 25 1r 0 1 r r1,22= = T , theo nh l 1, ta c: n2n1 n25 125 1F|||

\|+|||

\|+= Thay cc iu kin ban uF0 = 0,F1 = 1 vo c h phng trnh: = ++ += + 125 125 102 12 1 Gii h phng trnh ny c nghim:51- ;512 1= =Vy:.25 15125 151n n|||

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\|+=nFc.Giicngthctruyhituyntnhbchaikhiphngtrnhctrngc nghim kp nh l 2: Cho c1, c2 l hai s thc. Nu phng trnh c trng r2 c1 r c2 = 0 c nghim kp r0 th iu kin cn v dy s {an } l nghim ca cng thc truy hi:an =c1 an 1 + c2 an 2

l:an=1 r0n + 2 n r0n , n = 1, 2, ... trong1,2lccsthccthxcnhcnhcciukinbanucacng thc truy hi. Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...50 Vic chng minh nh l 2 tng t chng minh nh l 1. Th d: Tm nghim ca cng thc truy hi: an = 6an 1 9an 2;a0 = 1,a1 = 6. Gii:Phng trnh c trngr2 6r + 9 = 0c nghim kpr = 3. Vyan c dng: an =1 3n + 2 n 3n. Thayn = 0 , n = 1, ta c: 1 6 a 3 31 a 2 11 2 10 1= = = = += = Vy: an=3n + n3n=(n + 1)3n. d.Ch thch C th m rng nh l 1 cho h thc truy hi tuyn tnh bc k nh sau:Cho c1, c2, , ck l cc s thc. Gi s phng trnh c trng: rk c1 rk 1 ck 1 r ck=0 c k nghim phn bit r1, r2, , rk . Khi iu kin cn v dy {an} l nghim ca cng thc truy hi: an = c1 an -1 + c2 an - 2 + + ck an - k l:an = 1 r1n + 2 r2n + + k rkn trong 1, 2, , k l cc s thc c th xc nh c nh cc iu kin ban u ca cng thc truy hi cho. Th d:Tm nghim ca cng thc truy hi: an = 6an 1 1an 2 + 6an 3 ; a0 = 2,a1 = 5,a2 = 15. Gii:Phng trnh c trng ca cng thc truy hi cho l: r3 6r2 + 11r 6 = 0 Cc nghim ca phng trnh c trng l: r = 1, r = 2, r = 3. Vy cng thc trc tip tnh an c dng: an = 1 1n + 22n + 3 3n. Thay n = 0, n = 1, n = 2, n = 3 vo cng thc tnh an, c h phng trnh tuyn tnh theo 1, 2, 3: = + + = + + = + + 15 9 45 3 223 2 13 2 13 2 1 Gii h phng trnh c nghim: 1 = 1, 2 = 1, 3 = 2. Vy dng hin ca cng thc truy hi cho l: an = 1 2n + 2. 3n. Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...51 5.Bi ton lit k. C nhng bi ton m cc cu hnh ca t hp, ngoi vic m s lng cc cu hnh tho mn cn phi lit k tt c cc cu hnh cn m. l cc bi ton phi tm cc cu hnh t hp tho mn thm mt hoc mt s iu kin no . Th d nh phi tm cc tp con ca mt tp hp c 5 s sao cho tng ca cc phn t ca tp con bng 50, khi phi kim tra tng cc s trong mi tp ca 25 = 32 tp con ca tp cho tm cc tp tho mn iu kin nu. Mun vy, phi lit k cc phn t ca mi tp trong 32 tp con .Vic lit k cc cu hnh phi tho mn cc nguyn tc sau: -Khng c lp li mt cu hnh m. -Khng c st mt cu hnh. Sau y chng ta tm hiu mt s thut ton lit k thng gp. 5.1.Phng php sinh thc hin phng php ny, bi ton cn tho mn 2 iu kin: 1- C th xc nh c mt th t trn tp cc cu hnh cn m. T c th xc nh c cu hnh u tin v cu hnh cui cng theo th t xc nh. 2- Xy dng c thut ton t mt cu hnh cha phi l cui cng ang c c th a ra mt cu hnh k tip theo th t xc nh. t tn thut ton theo iu kin 2 l thut ton Sinh_k_tip. C th m phng thut ton sinh nh sau: Agorithm:Tng qut v thut ton sinh. Procedue Phng php sinh; Begin < Xy dng cu hnh ban u > stop := false; whilenot stopdo begin < a ra cu hnh ang c > Sinh_k_tip ; end; End; Bin stop l mt bin Boole dng dng chng trnh, khi n cu hnh cui cng th stop nhn gi tr true v chng trnh dng. Sinh_k_tipl mt th tc con nhm xy dng cu hnh k tip ca cu hnh ang c. Ch rng th t cc cu hnh xc nh trong iu kin 1 cn la chn sao cho c th xy dng c thut ton sinh_k_tip. Xt 3 bi ton c th s dng phng php sinh. Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...52 a.Lit k cc xu nh phn c di n Mi xu nh phnb = bn 1 bn 2 b0, trong bi {0, 1} c th xem l biu din nhphnca1snguynP(b)no.Vycthlythttngcasnguynxc nh th t cc xu nh phn. Xu nh phn b = bn 1 bn 2 b0 c gi l i trc xu a = an 1 an 2 ... a0, (hay a k tip b) nu P(b) < P(a), k hiu l b < a. Cch xc nh th t nh vy gi l th t t nhin hay cn gi l th t t in. Chng hn vi n = 3, c th t t in ca cc xu nh sau: Th t ca xu01234567 Xu c th000001010011100101110111 Nh vy xu u tin l gm ton s 0 v xu cui cng gm ton s 1. T suy ra quy tc sinh xu nh phn k tip nh sau: Xu u tin gm ton s 0:00 ... 0 Gi s ang c xu bn 1 bn 2 ... b0, khi : - Tm t phi sang, tc l t b0, b1,... n khi gp bt 0 u tin, chng hn l bi, - Thay bi = 1 v bj = 0 j < ixu mi thu c l xu k tip ca xu ang c. Th d:Tm xu k tip ca xu 1 000 100 111 (n =10). i t bn phi sang, bit u tin bng 0 l bit th 4, tc l b3 = 0. Vy thay b3 = 1, b2 = b1 = b0 = 0 c xu k tip ca xu cho l 1 000 101 000. (R rng l:1 000 100 111 = 29+ 25+ 22+21+20 = 541 v 1 000 101 000 = 29+ 25+ 23 = 542)Thut ton va trnh by c th m t nh sau: Algorithm: Sinh xu nh phn k tip. ProcedureXu nh phn k tip; Input: Xu nh phn bn 1 bn 2 ... b0,bi = 0; Output:Xu nh phn k tipan 1 an 2 ... a0; Begin i:= 0; whilebi = 1do begin bi := 0; i:=i + 1 ; end; bi := 1; End; b.Lit k cc t hp chp k ca mt tp hp c n phn tXt bi ton lit k tt c cc t hp chp k ca tp hpX = {1, 2, ..., n} Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...53 V mi t hp chp k ca n phn t cho l mt tp con gm k phn t ly t n phn t nn phi xp th t cc tp con . Trc ht biu din mi tp con k phn t thnh mt b c th t: a = (a1, a2, ...,ak) ;1 a1 < a2 < ... < ak n vnhngha: Tp cona =(a1, a2, ...,ak)gi l i trc tp con a =(a1, a2, ...,ak) theo th t t in, k hiua a, nu tm c mt ch s i(1 i k) sao cho: a1 = a1 , a2 = a2 , ... ,ai 1 = ai 1 ,ai < ai . Th d:Cho X = {1, 2, 3, 4, 5}. Cc t hp chp 3 ca X c lit k theo th t t in l: {1, 2, 3}{1, 2, 4}{1, 2, 5} {1, 3, 4}{1, 3, 5} {1, 4, 5} {2, 3, 4} {2, 3, 5}{2, 4, 5} {3, 4, 5} Nh vy tp con u tin l {1, 2, ..., k} v tp con cui cng l {n k+1, nk+2, ..., n}.T suy ra quy tc sinh tp con k tip nh sau: Tm t phi qua tri ca dy phn t u tin tho mn ai n k + i , Thay ai bng ai + 1 , Thay aj bng ai + j i , vij = i + 1, i + 2, ..., k. Th d: Tm t hp chp 4 t tp{1, 2, 3, 4, 5, 6} k tip t hp {1, 2, 5, 6} Gii:Li t phi sang tri thy a2 = 2 l s hng u tin tho mn: a2 = 2 6 4 + 2= 4. Nh vy t hp k tip l: a1 = 1, a2 = 2 + 1 = 3, a3 = 3 + 3 2= 4, a4 = 3 + 4 2= 5, tc l tp {1, 3, 4, 5} Algorithm:Sinh t hp k tip. ProcedureT hp k tip; Input:T hp{a1, a2, ...,ak} khng trng vi {n k + 1, ... , n}; Output: T hp k tip{a1, a2, ...,ak} ; Begin i := k ; while ai = n k + ido i := i 1; ai := ai + 1 ; forj := i + 1 to k doaj = ai + j i ; End; c. Lit k cc hon v ca tp hp c n phn tBi ton t ra l: Lit k tt c cc hon v ca tp hp X = {1, 2, ..., n} Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...54 Mi hon v ca X c th biu din bi b c th t n thnh phn a = (a1, a2, ...,an) ; ai ajkhii j nh ngha th t t in ca cc hon v nh sau: Hon v a = (a1, a2, ...,an) gi l i trc hon v b = (b1, b2, ...,bn), k hiu a b, nu tm c mt ch s k (1 kn) sao cho: a1 = b1 , a2 = b2 , ... ,ak 1 = bk 1 ,ak < bk . Th d:Cc hon v ca{1, 2, 3} c lit k theo th t t in l: (123)(132)(213)(231)(312)(321) Nh vy hon v u tin l(1, 2, ..., n) v hon v cui cng l(n, n 1, ...,1). T c th ch ra quy tc sinh hon v k tip hon v a = (a1, a2, ...,an) (n, n1,..., 1) nh sau: Tm t phi qua tri ca hon v ang xt c ch s i u tin tho mn ai < ai + 1. Tm ak bn phi ai (tc lk > i) theo iu kin ak l s nh nht trong cc s c chn sao choak > ai , i ch ai, ak. Lt ngc on t ai + 1 nan (xp li cc s t ai + 1 n an theo th t tng dn) Th d:Tm hon v k tip ca hon v (3, 6, 2, 5, 4, 1) S u tin tnh t cui ln m nh hn s sau n l a3 = 2 < a4 = 5. S nh nht trong cc s ng sau a3 = 2 m ln hn a3 l a5 = 4. i ch a3 = 2 v a5 = 4 ta c (3, 6, 4, 5, 2, 1). Lt ngc(a4, a5, a6) = (5, 2, 1) c hon v k tip l (3, 6, 4, 1, 2, 5). Thut ton c m phng nh sau: AlgorithmSinh hon v k tip. ProcedureHon v k tip; Input: Hon v(a1, a2, ...,an) (n, n 1, ..., 1); Output:Hon v k tip(b1, b2, ...,bn); Begin (* Tm ch s j ln nht tho mn aj < aj + 1*) j := n 1; whileaj > aj + 1do j:= j 1; (* Tm ak l s nh nht cn ln hn aj bn phi aj*) k := n; while aj > akdok := k 1; i ch (aj ,ak ) (* i ch aj viak *) (*Lt ngc on t aj + 1 nan *) r := n; s := j + 1;while r > s do begin Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...55 i ch (ar, as); r := r 1; s := s + 1; end; End; (* ich (x, y) l th tc con nhm i ch x vi y: z := x; x := y; y := z *) Ni chung, phng php sinh c nhiu hn ch v khng phi cu hnh no cng sinh ccuhnhktipmtcchngin.Mtkhc,ngayccuhnhutincngnh cu hnh cui cng cng khng d xc nh c. gii bi ton lit k cc cu hnh t hp, ngi ta thng dng thut ton quay lui di y. 5.2.Thut ton quay lui Ni dung ca thut ton l tm dn cc thnh phn ca cu hnh bngcch th tt c cc kh nng c th c. C th nh sau: Gi s cn tm cu hnh c biu din bng mt b n thnh phn (x1, x2, ..., xn) v tm c k 1 thnh phn ca cu hnh l x1, x2, ..., xk 1. Khi thnh phn xk ca cu hnh c xc nh theo cch sau: Gi Tk l tp hp tt c cc thnh phn m xk c th nhn c. Tp Tk l hu hn. Gi s N(Tk) = nk, do ln lt duyt tt c cc phn t (thnh phn) ca Tk. vic duyt c thun li, cn nh s cho cc phn t ca Tk. Vi mi kh nng j, tc l chn phn t xj (1 j nk) phi th xem xj c chp nhn c khng. Khi xy ra 2 tnh hung: Nu xj chp nhn c th a xj vo cu hnh ang xt, tc l thnh phn xk ca cu hnh l xj. Nu k = n th cu hnh cn lit k c xc nh, nu k < n th tin hnh xc nh thnh phn xk + 1. Nu xj khng chp nhn c th quay li bc trc xc nh li xk. Th tc quy cho thut ton quay lui nh sau: AlgorithmThut ton quay lui. ProcedureTh (k: integer); varj: integer; Begin forj := 1tondo ifthen begin < Xc nh xk theo j >; if k = nthen< ghi nhn mt cu hnh > elseTh (k + 1); end; End; iu cn quan tm trong th tc trn l: Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...56 1)Lm th no xc nh c tp Tk l tp tt c cc thnh phn m xk c th nhn. 2)Xcnhgitrbiuthclogic.Gitrnyngoivicphthuc vo j cn ph thuc vo cc thnh phn chn tik 1 bc trc. Nh vy cn ghinhtrngthicsauligiTh(k+1).Cctrngthinycghinhnnh mt bin logic (bin Boole) v gi l bin trng thi. Thut ton quay lui c th m t bng cy tm kim nh sau:

Ngoi th tc quy Th(k), chng trnh cn c thm cc th tc Khi to v In kt qu. Chng trnh chnh gii bi ton lit k c dng: Begin Khoi_tao; Thu; End. Th d 1: Lit k tt c cc hon v ca n s t nhin dng u tin. Mi hon v c th c biu din di dngx[1], x[2], ..., x[n] trong x[i] nhn gi tr t 1 n n v x[i] x[j]. Cc gi tr t 1 n n ln lt c cho x[k], trong gi tr j c chp nhn nu n cha c dng. Ni cch khc, ta c T1 = N = {1, 2, ..., n} vnu xc nhc x[1], x[2], ..., x[k 1] thTk = N \{x[1], x[2], ..., x[k1]}. V th cn phi ghi nh xem gi tr j c dng hay cha. iu nyc thc hin nh mt dy bin logic b[j], trong b[j] l true nu j cha c dng. Do chng c gn gi tr true cho mi b[j] trong th tc khi to v sau khi gn gi tr j cho xk th b[j] nhn gi tr false, cui cng li phi gn true cho b[j] sau khi thc hin xong bc in kt qu hay thc hin xong th tc thu(k+1) s dng cho bc sau. Program Liet_ke_hoan_vi; uses crt; varn: integer; Gc

Kh nng chn x1 Kh nng chn x2 vi x1 chn Kh nng chn x3 vi x1, x2 chn Hnh 3.Cy lit k li gii theo thut ton quay lui Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...57 x: array[1..20] of integer; b: array[1..20] of boolean; dem: word; Procedure khoitao; var i: integer; begin write (' So phan tu n = '); readln(n); for i:=1 to n do b[i] := true; dem := 0; end; Procedure in_kq; var i: integer; begin dem := dem+1; write (' Hoan vi thu',dem:3,':'); for i:=1 to n do write(x[i]:4); writeln; end; Procedure thu(k: integer); var j: integer; begin for j:=1 to n do if b[j] then begin x[k] := j; b[j] := false;{chuyn trng thi mi} if k=n then in_kq else thu(k+1); b[j] := true;{tr li trng thi c} end; end; BEGIN {main} clrscr; khoitao; thu(1); readln; END. Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...58 Th d 2:Lit k tt c cc chui nh phn c di n Mi chui nh phn di n c dng x[1] x[2] ... x[n] vi x[i] B; B = {0, 1}. Gi s xc nh c x[1] x[2] ... x[k1], khi x[k] c chn mt trong 2 phn t ca B. Vy c chng trnh nh sau: Program Lietke_Day_Nhiphan; uses crt; varn: integer; b: array[1..20] of 0..1; dem: word; ProcedureKhoi_tao; begin write (' Do dai cua chuoi la: n = '); readln(n); dem := 0; end; Procedure In_kq; vari: integer; begin dem := dem+1; if dem mod 20 = 0 then readln; write ( 'Chuoi thu',dem:5,':'); for i:=1 to n do write(b[i]:2); writeln; end; Procedure Thu(k:integer); var j: integer; begin for j:=0to 1 do begin b[k] := j; if k = n then In_kqelse Thu(k+1); end; end; BEGIN clrscr; Khoi_tao; Thu(1); Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...59 write(' Tong so co', dem:4,'chuoi nhi phan'); readln; END. Th d 3:Bi ton xp hu. Lit k tt c cc cch xp n con hu trn bn c nn sao cho chng khng n c nhau.Tclkhngc2conhunocnghng,cngct,cngngchochnhhoc ng cho ph. Gii:Gi s n hng, n ct ca bn c c nh s t 1 n n. Chng ta s dng ch s k ch hng v ch s j ch ct (1 k, j n). Quy c x[k] = j ch v tr con hu hngk,ctj.Mihngcxpngmtconhu,vncnlilxemmiconhu c xp vo ct no.D thy T1 = N = {1, 2, ..., n}.Gi s xc nh c x[1], x[2], ..., x[k1], cn phi xp tip x[k] tho mn yu cu bi ton.Vickimsottheochiunganglkhngcnthitvmihngcxpngmt con hu. Vic kim sot theo chiu dc c thc hin nh dy bin logic a[j] vi quy c a[j] bng true nu ct j cn trng Bncc2ngcho:ngchochnhvccngsongsongngcho chnh c k j v tr (1n kj n1) c kim sot nh dy bin logic c[j]. ng cho ph ck + j v tr (2 k+j 2n) c kimsot nh dybin logic b[j]. Cng nh bin a[j], cc bin b[j] v c[j] s bng true nu tng ng cn trng. Cc bin trng thi a[j], b[j], c[j] c khi gn gi tr true trong th tc khi to. Nh vy gi tr j c chp nhn khi v ch khi c 3 bin a[j], b[k+j] v c[kj] c gi tr true, cc bin ny cn c gn li false khi xp xong qun hu th j v tr li true sau khi gi thu(i+1) hay sau khi in kt qu. Program Xep_hau; usescrt; var n: integer; x: array[1..20] of integer; a: array[1..20] of boolean; b: array[2..40] of boolean; c: array[19..19] of boolean; dem: word; Procedure Khoi_tao; var i: integer; begin write(' Kich thuoc ban co, n = '); readln(n); for i:=1 to n do a[i] := true; for i:=2 to 2*n do b[i] := true; for i:=1n to n1 do c[i] := true; Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...60 dem := 0; end; Procedure In_kq; var i: integer; begin dem := dem+1; if dem mod 20 = 0 then readln; write(' Phuong an', dem:3,' :'); for i:=1 to n dowrite(x[i]:4); writeln; end; Procedure Thu(k: integer); var j: integer; begin for j:=1 to n do if a[j] and b[k+j] and c[kj] then begin x[k] := j; a[j] := false; b[k+j] := false; c[kj] := false; if k=n then in_kq else thu(k+1); a[j] := true; b[k+j] := true; c[kj] := true; end; end; BEGIN clrscr; Khoi_tao; Thu(1); readln; write(' Tong so co:',dem:3,' phuong an'); readln; END. Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...61 6.Bi ton tn ti Trong cc bi ton trn chng ta a ra cc thut ton m cc cu hnh t hp. Trongccbiton,stnticacccuhnhlcthtvchngtachvicm chng. Tuy nhin, c nhng bi ton ngay c vic ch ra s tn ti ca cu hnh tho mn iukinchocnglmtkhkhn.Nhvyxuthinbiton:Khngnhchay khng c t ra l mt cu hnh tho mn cc iu kin cho. Cc bi ton nh vy c gi l bi ton tn ti. 6.1. Nguyn l Di-ric-le (Dirichlet) a.nh l 1.Nguyn l lng chim b cu Nucnhiuhnnitngcnxpvoncihpthctnhtmthpccha nhiu hn 1 i tng. Nguyn l ny xut pht t bi ton: C mt n chim b cu bay v t ca chng, nu s ngn chung t hn s chim th t nht c mt ngn c nhiu hn 1 con chim. nhlcchngminhbngphnchng.Giskhngcmthpnocnhiu hnmtitng,khitngsitngcxptrongcchpkhngvtqus hp, ngha l s i tng l t hn s hp. iu bytri vi gi thit s i tng nhiu hn s hp. Th d:1.Trong s 367 ngi khc nhau lun lun tm c 2 ngi c cng ngy sinh. (V 1 nm ch c nhiu nht l 366 ngy) 2.Trong k thi hc sinh gii, thang im cho l t 0 n 20, v im thi c lm trn n 0,5 im. Hi phi c t nht bao nhiu hc sinh d thi chc chn c t nht 2 hc sinh c cng im s. Gii:C 41 loi im s khc nhau. Vy cn phi c t nht 42 hc sinh d thi. b.nh l 2. Nguyn l Di-ric-le tng qut Nu xpnitngvokcihp,th lunlun tmcmthpckhngthn 1kn+((

i tng.Chng minh:Gi s khng c hp no trong k hp c cha nhiu hn ((

kn i tng. Khi tng s i tng c cha trong k ci hp nhiu nht l: nknkknk = 10, a2 > 10a3 > 20 a4 > 30a5 > 50a6 > 80a7 > 130. iu ny tri vi gi thit ca bi ton. Vy bi ton c chng minh. Thd2.Ccnhcamtthpgiclicgnccsnguyn0,1,2,...,9mt cchngunhin;minhlmtskhcnhau.Chngminhrnglunluntmc3 nh lin tip sao cho tng cc s c gn ln hn 13. Gii:Gi x1, x2, ..., x10 l cc s c gn cho cc nh th nht, th hai, ..., th mi ca thp gic. Gi s khng tm c 3 nh lin tip no c tng cc s gn ln hn 13, ngha l: k1 = x1 + x2 + x3 13 ; k2 = x2 + x3 + x4 13 ; ............ k9 = x9 + x10 + x1 13 ; k10 = x10 + x1 + x2 13 ; Suy ra: k1 + k2 + ... + k10 130. Mt khc:k1 + k2 + ... + k10 = 3(x1 + x2 + ... + x10) = 3(0 + 1+ ... + 9) = 135 T c:135 = k1 + k2 + ... + k10 130. V l. Vy phi tn ti 3 nh lin tip tng cc s c gn l ln hn 13. Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...64 BI TP CHNG 2 T hp, Chnh hp, Hon v 2.1.Mtnhmsinhvingmnnamvnn.Cbaonhiucchxpthnhmthng ngang sao cho nam n ng xen k nhau? 2.2. Mt tp hp c 10 phn t c bao nhiu tp con vi s phn t l? 2.3. Mt t c 13 ngi. a) C bao nhiu cch chn 10 ngi i lm mt cng vic nh sn? Bit rng mi ngi u lm c cng vic . b)Cbaonhiucchchn10ngilm10vickhcnhau?Bitrngmi ngi u lm c mi vic nh nhau.c) C bao nhiu cch chn 10 ngi sao cho c t nht 1 n? Bit trong t c 3 n. 2.4. C bao nhiu xu nh phn c di 10 nu trong xu c: a) ng 3 s 0. b) t nht 3 s 1. c) S cc s 0 bng s cc s 1. d) C t nht 3 s 0 v t nht 3 s 1. 2.5. C bao nhiu bin ng k xe. Bit mi bin gm 3 ch ci ting Anh khc nhau v tip n l 3 ch s khc nhau? 2.6. Chng minh rng (ng thc Van-dec-non) a) knr0 kk rmrn mC . C C=+= ;b)r1 r nr0 kkk nC C+ +=+= Gi : a) Xt 2 tp hp c tngng m, nphn t.Lyt 2 tp hp ra r phn t theo hai cch: Cch I l trn ln 2 tp hp vi nhau; Cch II l ly tp hp ny k phn t, cn tp hp kiar kphn t (k = 0, 1, 2, ..., r) b)Chng minh bng quy np theo r , xut pht tC0n = C0n + 1 = 1. 2.7. C bao nhiu cch phn cng 3 cng vic cho 5 ngi lm, nu 1 ngi c th lm c c 3 vic. 2.8.Mtconlntit kimc50ngxu. Cthcbaonhiut hpkhcnhaucc ng xu: 200, 500, 1000, 2000, 5000 trong ? 2.9. C 100 vin bi hon ton ging nhau. C bao nhiu cch xp cc vin bi vo 3 ci hp? bit mi ci hp u c th cha n 100 vin bi. 2.10.Phng trnh:x1 + x2 + x3 + x4 = 17c bao nhiu nghim nguyn khng m? 2.11. Phng trnh:x1 + x2 + x3 + x4 + x5 = 21c bao nhiu nghim nguyn khng m sao cho: a) x1 1;b) xi2vii = 1, 2, 3, 4, 5 c) 0 x1 10 ; d) 0 x1 3 v1 x2 4vx3 15 Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...65 2.12. C bao nhiu xu c cc ch ci trong t MISSISSIPPI ? 2.13. C bao nhiu xu khc nhau c 7 hoc nhiu hn cc k t c th lp c t cc ch ci ca t NONGNGHIEP ? 2.14. C bao nhiu xu nh phn khc nhau gm 6 ch s 0 v 8 ch s 1? 2.15. Mt thi c 10 cu hi. C bao nhiu cch gn im cho cc cu hi nu tng s im l 100 v mi cu c t nht 5 im? 2.16. Bt phng trnh:x1 + x2 + x3 11 c bao nhiu nghim nguyn khng m? Gi :B sung thm bin ph x4 sao cho: x1 + x2 + x3 + x4 = 11. 2.17. Trong khng gian Oxyz mt con b di chuyn bng cch nhy tng bc, mi bc 1 n v theo hng hoc truc Ox, hoc trc Oy, hoc trc Oz v khng c nhy git li. Tnh s cch con b c th di chuyn t gc ta (0, 0, 0) n im (4, 3, 5). Nguyn l b tr 2.18.Hy tm s phn t ca ABC nu mi tp A, B, C u c 100 phn t v nu cc tp hp A, B, C tha mn: a)i mt ri nhau. b) C 50 phn t chung ca mi cp tp v khng c phn t no chung ca c 3 tp hp. c)C 50 phn t chung ca mi cp tp v 25 c phn t chung ca c ba tp hp. 2.19. Kim tra 270 sinh vin mi tt nghip ca mt trng i hc. Kt qu cho thy c 64 sinh vin thnh tho ting Anh,94 sinh vin thnh tho ting Php, 108 sinh vin thnh tho vi tnh vn phng, 26 sinh vin va thnh tho ting Php va thnh tho vi tnh, 28 sinhvinvathnhthotingAnhvathnhthovitnh,22sinhvinvathnhtho ting Php va thnh tho ting Anh v 11 sinh vin thnh tho c 3 th (ting Anh, ting Php v vi tnh). Hi trong s 270 sinh vin c bao nhiu ngi khng bit c 3 th k trn? 2.20. C bao nhiu xu nh phn c di 8 v khng cha 6 s 0 lin tip? 2.21.C bao nhiu s nguyn dng khng vt qu 100 v hoc l s l hoc l s chnh phng. 2.22.C bao nhiu hon v ca 10 ch s nguyn dng u tin hoc l bt u bng ba s 987, hoc l cha cc s 45 v tr th 5 v th 6, hoc l kt thc bng 3 s 123. Gii cc h thc truy hi. 2.23.S lng mt loi sinh vt s tng gp ba sau mi gi. a) Lp cng thc truy hi tnh s lng vi sinh vt sau n gi. b) Nu lc u c 100 con th sau 10 gi s c bao nhiu con? 2.24.a)Tmhthctruyhitnhscchilnnbccuthangcamtngi.Bit ngi c th bc mi ln mt hoc hai bc. b) C bao nhiu cch ngi i ln mt cu thang c 8 bc? Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...66 2.25. Chng minh rng cc s Fibonacci tho mn h thc truy hi: fn = 5fn 4 + 3fn 5 , n 5; f0 = 0,f1 = 1,f2 = 1,f3 = 2,f4 = 3. Dng h thc ny chng minh rngf5n chia ht cho 5 vin = 1, 2, 3, ... 2.26. Tm h thc truy hi tnh s xu nh phn c di n v: a)Khng c 2 s 0 lin tip. b)C 2 s 0 lin tip. Tnh c th s xu nh phn tho mn iu kin a hoc b nu n = 5 2.27.Mt ngi u t 100 triu ng vo mt c s sn xut. Sau 1 nm ngi c hai khon li: Khon th nht l 20% tng s tin trong nm cui; Khon th hai l 13% ca tng s tin c trong nm trc . a) Tm cng thc truy hi tnh {Pn}, trong Pn l tng s tin c vo cui nm th n. b) Tm cng thc trc tip tnh Pn. Gi thit rng ngi khng rt bt c khon tin no trong c n nm . 2.28.Vi cc tm lt kch thc 12 v 22 c th lt mt bng 2n bng bao nhiu cch? 2.29. Gii cc h thc truy hi sau: a) an = 5an 1 6an 2 , n 2; a0 = 1,a1 = 0. b) an = 4an 1 4an 2 ,n 2; a0 = 6,a1 = 8. c) 4aa2 nn= ,n 2; a0 = 1,a1 = 0. d) an =7an 2 + 6an 3 , a0 = 9,a1 = 10,a2 = 32. e) an = 2an 1 + 5a n 2 6an 3 , a0 = 7,a1 = - 4,a2 = 8. 2.30.Xt ma trn |||

\|=1 11 0Aa)Chngminhrng: |||

\|=+1 n nn 1 n nF FF FA ,trongFnlshngthncady Fibonacci. b)Tnhdet(An). t suy ra cng thc: Fn + 1 Fn 1 (Fn)2 = (1)n. Bi ton tn ti. 2.31.Cho n l mt s nguyn dng. Chng minh rng trong mi tp hp c n s nguyn dng lin tip c ng mt s chia ht cho n. 2.32. Cho (xi, yi, zi), i = 1, 2, ..., 9 l mt tp hp gm 9 im khc nhau c to nguyn trong khng gian Oxyz. Chng minh rng cc im gia ca ng ni ca 2 trong 9 im c t nht mt im c to nguyn. 2.33.Trongmthpknc10vinbimuv10bimuxanh.Mtngilyngu nhinccvinbitrongbngti.Hingicnlytnhtbaonhiuvinbichc chn ly c t nht 2 vin bi cng mu. Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...67 2.34.Chng t rng trong 5 s chn t 8 s nguyn dng u tin chc chn c mt cp c tng bng 9. 2.35.a) Trong phng c 6 my tnh. Mi my c ni trc tip hoc khng ni vi cc my khc. Hy chng t rng c t nht hai my c cng s kt ni vi cc my khc hoc c t nht hai my khng c ni vi my khc. b)Mt mng my tnh c 6 my. Mi myc ni trc tip vi t nht mt my khc. Hy chng t rng c t nht hai my c cng s kt ni vi cc my khc. 2.36. C 12 cu th bng mc o mang s t 1 n 12 dng thnh mt vng trn gia sn. Chng minh rng lun lun tm c 3 ngi ng lin tip nhau c tng cc s trn o t nht bng 20. P S 2.1. 2(n!)2. 2.2. 29. 2.3. a)286; b)1 716; c)1 037 836 800. 2.4. a)120; b)968; c)252;d)912. 2.5. 12 232 000. 2.7. 60. 2.8. 316 251. 2.9. 5 151. 2.10. 1 140. 2.11. a)10 626; b)1 365;c)11 649; d)106. 2.12. 34 650. 2.13. 838 320. 2.14. 3 003. 2.15. 12 565 671 261. 2.16. 364. 2.17. 27 720. 2.18. a)300;b)150;c)175. 2.19. 69. 2.20. 248. 2.21. 55. 2.22. 50 138. Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...68 2.23. a)an = 3an 1 ,a0 = 100; b)5 904 900. 2.24. a)an = an 1 + an 2 , a0 = 1, a1 = 1; b)34. 2.26. a)an = an 1 + an 2 ,a1 = 2,a2 = 3; a5 = 13; b)an = an 1 + a n 2 + 2n 2,a0 = 0,a2 = 1; a5 = 19. 2.27. a)Pn = 1,2Pn 1 + 0,13Pn 2 ,P0 = 100;P1 = 120. b)( )n n n 3n) 1 , 0 ( 143 , 7 ) 3 , 1 ( 857 , 92 ) 1 , 0 ( ) 3 , 1 .( 1314100P + + =2.28. an = an 1 + 2an 2 ,a1 = 1,a2 = 3; ( )n 1 nn) 1 ( 231a + =+.2.29.a)an = 3.2n 2 . 3n; b)an = (6 2n)2n; c)1 n 1 nn2121a+ +||

\| ||

\|= d)an = 8(1)n 3(2)n + 4. 3n ; d)an = 5 + 3(2)n 3n. 2.33.11. CU HI N TP CHNG 2 1.Pht biu cc nguyn l c bn ca php m: Nguyn l cng; Nguyn l nhn; Cho cc th d minh ho. 2.Nu s khc nhau gia t hp v chnh hp chp k t n phn t. Mi quan h gia t hp v chnh hp? Mi quan h gia chnh hp v hon v?3.Hy gii thch cc tm cng thc tnh s cch chn c lp k phn t t n phn t cho? Gii thch ti sao c th dng t lp tnh s nghim nguyn khng m ca phng trnh x1 + x2 + + xn = k, trong k l s nguyn dng. 4.Hy lp cng thc tnh s hon v ca cc tp hp c cc phn t ging nhau v s cch phn chia mt tp hp c n phn t thnh k tp con. Cho th d minh ho. 5.Phtbiunguynlbtr.ngdngnguynlbtrtmsnghimnguynkhng m ca phng trnh x1 + x2 + x3 + x4 = 20 tho mn x1 < 7, x2 < 5 v x3 < 4. 7.Trnh by cc phng php gii cc h thc truy hi. Cho th d minh ho 8.Pht biu nguyn l Di-ric-l. ng dng nguyn l Di-ric-l chng minh rng trong bt k 91 s nguyn no cng c t nht 10 s c cng ch s cui cng. 9.Trnh by thut ton quay lui gii bi ton lit k. Cho th d minh ho. 10. Cho mt vi th d gii bi ton tn ti bng phn chng. Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...69 CHNG 3. CC KHI NIM C BN V TH 1. Cc nh ngha v th v biu din hnh hc ca th 1.1.Cc nh ngha v cc loi th 1.2.Bc ca nh ca th 1.3.Mt s dng th c bit 1.4.Mt vi ng dng ca cc th c bit 2. Biu din th bng i s 2.1.Danh sch k2.2.Danh sch cnh 2.3.Ma trn k 2.4.Ma trn lin thuc 3. S ng cu ca cc th 4. Tnh lin thng trong th 4.1.ng i, chu trnh 4.2. th lin thng 5.S n nh trong, s n nh ngoi v nhn ca th. 5.1.S n nh trong5.2.S n nh ngoi 5.3.Nhn ca th6.Sc s ca th 6.1.nh ngha 6.2.Mt s nh l v sc s 6.3.Vi th d ng dng sc s Lthuytth raitthk18binh tonhcLeonhardEuler(17071783) gc Thy s nhng ch yu sng v lm vic Nga v c. N c ng dng gii cc bi ton trong nhiu lnh vc khc nhau, chng hn l thuyt mng in, mng giao thng, cu trc ho hc, , v c bit trong l thuyt thng tin v iu khin hc. 1.Cc nh ngha v th v biu din hnh hc ca th th l mt cu trc ri rc gm cc nh v cc cnh ni cc nh. Ngi ta phn loi th theo c tnh v s cc cnh ni gia cc nh. Trng i hc Nng nghip H Ni Gio trnh Gio trnh Ton Ri rc...70 1.1.Cc nh ngha v cc loi th a. nh ngha th nh ngha:Cho X l mt tp hp khng rng cc i tng no v U X2 . B G = (X, U) c gi l mt th. Nu X l tp hu hn th gi l th hu hn. Nu X l tp v hn th gi l th v hn. Gio trnh ny ch xt th hu hn v gi tt l th. Mi phn t xX c gi l mt nh ca th.Tp X gi l tp cc nh ca th. Nu mi phn t u = (x, y)U l khng phn bit th t th th c gi l th v hng. Mi phn t uU gi l mt cnh ca th. Tp U gi l tp cc cnh ca th. Numiphntu=(x,y)Ucathcsptht(xtrc,ysau)th