Bi tp ton ri rc BI TP CHNG IBi 1:S m vng cn thit nh nht l bao nhiu m bo 25 triu my in thoi khc nhau. Mi in thoi c 9 ch s c dng 0XX-8XXXXX vi X nhn gi tr t 0 n 9.GiiV s m vng c dng: 0XX-8XXXXX, vi X nhn cc gi tr t 0 n 9 (10s), c07ktXdovysc107trnghp. Do, theonguynl Dirichlet vi 10 triu my in thoi th s m vng cn thit l: ] [ 3 5 , 2000 . 000 . 10000 . 000 . 25

]]]. Vy s m vng cn thit tha yu cu bi ton l 3.Bi 2:Bin s xe gm 8 k t, dng NN-NNNN-XN, v d 75_1576_F1. Hai s u l m tnh, X l ch ci (26 ch ci). N gm cc s 0, 1, , 9. Hi mt tnh no cn ng k cho 10 triu xe th cn bao nhiu serial (X).GiiBi ton ny c 02 cch hiu: serial y c th l 02 k t NN u tin hoc l 02 k t XN cui cng.Cch hiu 1: (serial l 02 k t XN cui cng).Hai s NN u l m tnh, do nh nc quy nh nn khng nh hng n kt qu bi ton.Suktcnli c5ktlN, nhvyc510 trnghp. Theo nguynlDirichlet, sserial Xti thiuphi thamn:100000 . 100000 . 000 . 10

]]]. iu ny khng hp l v s k t ch ci ch l 26. Do vy, nu bi ton sa li l 1 triu bng s xe th kt qu hp l hn, khi s serial l: 10000 . 100000 . 000 . 1

]]].Cch hiu 2: (serial l 02 k t NN u tin)Bn k t NNNN s c 104 trng hp, 02 k t XN s c 26*10 = 260 trng hp. Theo quy tc nhn, tng s trng hp s l: 104*260 = 2.600.000. Do , theo nguyn l Dirichlet, s serial ti thiu phi l: ] [ 4 84 , 3000 . 600 . 2000 . 000 . 10

]]].Vy cn 04 s serial ng k cho 10 triu xe.Trng CSP Qung Tr 1Bi tp ton ri rc Bi 3:C bao nhiu xu nh phn c di 10:a. Bt u bng 00 hoc kt thc bng 11.b. Bt u bng 00 v kt thc bng 11.Giia. Bt u bng 00 hoc kt thc bng 11.Xu nh phn bt u bng 00 c dng: 00.xxxx.xxxx. K t x c th l 0 hoc 1, c 8 k t x do vy c 82xu.Xu nh phn kt thc bng 11 c dng: xx.xxxx.xx11. Tng t ta cng tnh c c 82xu.Xu nh phn bt u bng 00 v kt thc bng 11 c dng 00.xxxx.xx11. Tng t nh trn, ta cng tnh c c 62xu.Vy s xu nh phn bt u bng 00 hay kt thc bng 11 l:448 64 512 2 2 * 26 8 nxu.b. Bt u bng 00 v kt thc bng 11.Xu nh phn tha mn bi phi c dng: 00.xxxx.xx11. Hai k t u v 02 k t cui l khng i, do vy ch cn 06 k t gia. Do s xu nh phn tha mn bi l: 26 xu.Bi 4:Kha 29 CNTT c 150 SV hc NNLT Java, 160 SV hoc Delphi, 40 SV hc c hai mn trn.a. Tm tt c SV ca kha 29 bit rng SV no cng phi hc t nht 01 mn.b. BittngsSVl285, hi cbaonhiuSVkhnghcJavahoc Delphi.GiiGi J:SV hc JavaD: SV hc Delphia. S SV ca kha 29 l: 270 40 160 1501 + + D J D J D J n SVb. Cu b c 02 cch hiu: Cch 01: khng hc t nht 01 mn.S SV khng hc Java hoc Delphi l (p dng nguyn l b tr) ta tnh c: 245 40 2852 D J n n SVCch 02: khng hc Java cng chng hc Delphi:Trng CSP Qung Tr 2Bi tp ton ri rc Theo cch hiu ny, p dng nguyn l b tr ta tnh c s SV nh sau: 15 40 160 150 285'2 + + D J D J n D J n SVBi 5:Mi ngi s dng my tnh dng password c 6 -> 8 k t. Cc k t c th l ch s hoc ch ci, mi password phi c t nht 01 ch s. Tm tng s password c th c.GiiBi ton ny cng c th c hiu theo 02 cch.Cch 01: phn bit ch thng vi ch hoa. Ch ci thng: 26Ch ci hoa: 26Ch s: 10Do , tng cng c 26 + 26 + 10 = 62 k t khc nhau.Nu password c n k t.Tng s trng hp:n62S password khng c ch s: n52Suy ra s password c t nht 01 ch s: n nnn 52 62 p dng cho cc trng hp n =6, 7, 8. Tng s password tha yu cu bi l:040 . 583 . 949 . 410 . 167 52 62 52 62 52 628 8 7 7 6 68 7 6 + + + + n n n nCch 02: khng phn bit ch thng vi ch hoa:Cch lm hon ton tng t, nhng thay v s dng cc s 62 v 52 th y s dng 02 s: 36 v 26. Kt qu s l:063.360 2.684.483. 26 36 26 36 26 368 8 7 7 6 68 7 6 + + + + n n n nBi 6:C n l th b vo n b th. Hi xc sut xy ra trng hp khng c l th no b ng c b th ca n.GiiV c n phong b v n b th nn c tt c N = n! cch b th khc nhau. m s cch b th sao cho khng l th no ng a ch, ta p dng nguyn l b tr:N = n! N1 + N2 ... + (1)nNn,Trng CSP Qung Tr 3Bi tp ton ri rc trong Nm (1 m n) l s cch b th sao cho c t nht m l th ng a ch, Nm l s cch ly m l th t n l, vi mi cch ly m l th, c (n-m)! cch b m l th ny ng a ch, nh vy:Nm = mnC(n - m)! = !!kndo vy N = n!(1 ! 11 + ! 21 ... + (1)n !1n),Do xc sut tha bi ton: k1 1 1 11 + - +...+(-1)! 1! 2! 3! k!N NpN n Bi 7:Ch ra rng nu chn 5 s t tp 8 s {1, 2, , 7, 8} th bao gi cng c t nht 01 cp s c tng l 9.GiiT 8 s trn, ta chia thnh 04 cp: {1, 8}, {2, 7}, {3, 6}, {4, 5} v tng ca mi cp u bng 9. Nh vy, bi s tr thnh chn 5 s t 4 cp s trn. Theo nguyn l Dirichlet, phi c t nht 01 cp s c chn ht. Vy bi ton c chng minh.Bi 8:Chng minh rng trong bt k mt nhm 27 t ting Anh no cng c t nht 2 t bt u t cng 01 ch ci.GiiBng ch ci ca ting anh gm 26 k t: a, b, c, , x, y, z. V c 27 t ting Anh v mi t bt u bng 01 ch ci nn theo nguyn l Dirichlet phi c t nht 02 t bt u bng cng 01 ch ci.Bi 9:Cn phi c bao nhiu SV ghi tn vo lp TRR chc chn c t nht 65 SV t cng im thi, gi s thang im thi gm 10 bc.GiiGi n l s sinh vin ti thiu tha mn bi, theo nguyn l Dirichlet th ] [ 6510n. Do vy 641 1 64 * 10 + n SV.Bi 10:Trng CSP Qung Tr 4Bi tp ton ri rc Tm h thc truy hi v cho iu kin u tnh s cc xu nh phn c di n v khng c 2 s 0 lin tip.C bao nhiu xu nh phn nh th c di bng 5.GiiVi xu nh phn c di n, ta chia thnh 02 trng hp:Nu k t cui cng l 1 th k t trc (k t th n 1) c th l 1 hay l 0 u c.Nu k t cui cng l 0 th k t trc (k t th n 1) ch c th l 1 (v nu l 0 th vi phm yu cu bi ton) nhng k t trc na (th n 2) c th l 0 hay 1 u c.T 02 trng hp trn ta suy ra c:2 1 + n n nf f fCc iu kin u: 21 f, 32 fC 13 xu nh phn c di 5 v khng c 2 s 0 lin tip.Bi 11:Dy cc s Fibonacci tha2 1 + n n nf f f, cho iu kin u: '1010ff. Hy tm h thc truy hi ca Fibonacci.GiiPhng trnh c trng: 21 0 x x c cc nghim l: r1 = 25 1+ v r2 = 25 1.Do cc s Fibonacci tng qut s c dng:1 21 5 1 52 2n nnf | ` | `+ + . , . ,vi cc iu kin ban u : 1 2101 2 1210051 5 1 51 1 12 25ff + | ` | ` + ' ' '+ . , . , Do cc s Fibonacci c cho bi cng thc nh sau:1 1 5 1 1 52 2 5 5n nnf| ` | `+ . , . ,Trng CSP Qung Tr 5Bi tp ton ri rc Bi 12:Tm nghim ca h thc truy hi sau:3 2 16 5 2 + n n n na a a a trong cc iu kin u l: 70 a, 41 a, 82 a.GiiPhng trnh c trng 0 ) 6 )( 1 ( 0 6 5 22 2 3 + x x x x x xCc nghim ca phng trnh c trng: ' 321210xxxDo , h thc truy hi s c dng: n n nna 3 ) 2 ( 13 2 1 + + Vi cc iu kin u c cho: 70 a, 41 a, 82 a. Ta c h phng trnh nh sau:' '+ + + + + 1359 4 83 2 473213 2 13 2 13 2 1 Vy nghim ca h thc truy hi l: n nna 3 ) 2 ( 3 5 + B i 13: Tm h thc truy hi vnr. Vinr l s min ca mt phng b phn chia bi n ng thng. Bit rng khng c 2 ng thng no song song v cng khng c 03 ng thng no i qua cng 1 im.GiiVi n ng thng, theo bi th ng thng th n s ct n 1 ng thng cn li ti n 1 im, tc l s ct n 1 + 1 = n phn mt phng. Do , s phn mt phng tng ln l n. T , ta c c h thc truy hi: n r rn n+ 1 .Cc iu kin u l:Trng CSP Qung Tr 6Bi tp ton ri rc n = 0: r0 = 1.n = 1: r1 = 2.BI TP CHNG IIBi 14Chngminhrngtrongmtnthlunctnht02nhc cng bc.GiiTrong th n, s bc ti a cungTH1: Gi s th khng c nh treo, do s bc ti thiu ca cc nh l 1, s bc ti a ca cc nh l n-1 (v l n th). C n nh, s bc ca cc nh i t 1 n n-1 (n-1) gi tr. Do theo nguyn l Dirichlet phi c t nht 02 nh c cng bc.TH2: Gi s th c t nht 01 nh treo, khi s bc ti thiu ca cc nh l 0, v s bc ti a ch l n-2 (v l n th, ng thi c nh treo). C n nh, s bc ca cc nh ch c th i t 0 n n-2 (n-1) gi tr. Do theo nguyn l Dirichlet phi c t nht 02 nh c cng bc.Bi 15:Tnh tng s bc canK (n th ).GiiVi th th mi nh u ni vi cc nh cn li. Do vy, khi c n nh th mi nh u ni vi n -1 nh cn li, tc l bc ca mi nh u bng n 1.Vy, tng s bc ca c th l: n*(n 1) bc.II. Cc bi tp trong giy kim tra ln 1.Bi 16: (ging bi 12 phn trc).Tm nghim ca h thc truy hi sau:3 2 16 5 2 + n n n na a a atrong cc iu kin u l: 70 a, 41 a, 82 a.GiiPhng trnh c trng 0 ) 6 )( 1 ( 0 6 5 22 2 3 + x x x x x xTrng CSP Qung Tr 7Bi tp ton ri rc Cc nghim ca phng trnh c trng: ' 321210xxxDo , h thc truy hi s c dng: n n nna 3 ) 2 ( 13 2 1 + + Vi cc iu kin u c cho: 70 a, 41 a, 82 a. Ta c h phng trnh nh sau:' '+ + + + + 1359 4 83 2 473213 2 13 2 13 2 1 Vy h thc truy hi l: n nna 3 ) 2 ( 3 5 + Bi 17:Trong tng s 2504 sinh vin ca mt khoa cng ngh thng tin, c 1876 theo hc mn NNLT Pascal, 999 hc mn ngn ng Fortran v 345 hc mn ngn ng C. Ngoi ra cn bit 876 sinh vin hc c Pascal v Fortran, 232 hc c Fortran v C, 290 hc c Pascal v C. Nu 189 sinh vin hc c 03 mn Psacal, Fortran v C th trong trng hp c bao nhiu sinh vin khng hc mn no trong c 03 mn ni trn.GiiGiP: l tp gm cc SV hc PascalF: l tp gm cc SV hc FortranC: l tp gm cc SV hc CN: l tng s SV (2504 SV)Gi K l s SV hc t nht 01 mnTheo nguyn l b tr, ta c: K P F C P F C P F F C C P P F C + + + U U I I I I I493 2011 2504 2011 189 290 232 876 345 999 1876 + + + K N K KSVVy c 493 SV khng hc mn no trong 03 mn: Pascal, Fortran v C.Trng CSP Qung Tr 8Bi tp ton ri rc Bi 18 : Hy tm s nh, s cnh, s bc ca mi nh v xc nh cc nh c lp, nh treo, ma trn lin k, ma trn lin thuc trong mi th v hng sau:GiiCu 18.1.S nh: 8S cnh: 11nh c lp: Dnh treo: khng cTn nh a b C d e g h iBc ca nh 3 2 4 0 5 3 2 3Ma trn lin k:

,`

.|0 1 0 0 0 1 0 11 0 0 0 0 0 1 00 0 0 2 0 1 0 00 0 2 0 0 1 1 10 0 0 0 0 0 0 01 0 1 1 0 0 0 00 1 0 1 0 0 0 01 0 0 1 0 1 0 0, th t nh: a, b, c, d, e, g, h, iMa trn lin thuc:

,`

.|1 0 0 1 0 0 0 0 1 0 01 0 0 0 0 0 1 0 0 0 00 1 1 0 1 0 0 0 0 0 00 1 1 0 0 1 0 1 0 1 00 0 0 0 0 0 0 0 0 0 00 0 0 1 1 1 0 0 0 0 10 0 0 0 0 0 1 1 0 0 00 0 0 0 0 0 0 0 1 1 111 10 9 8 7 6 5 4 3 2 1IHGEDCBAe e e e e e e e e e eTrng CSP Qung Tr 9Bi tp ton ri rc trong : ') , () , () , () , (4321e b ei a ee a ec a e') , () , () , () , (8765i c eg c ee c eh b e') , () , () , (1 11 09i h eg e eg e eCu 18.2.

S nh: 5S cnh: 12nh c lp: khng cnh treo: khng cTn nh a b c d eBc ca nh 6 5 5 5 3Ma trn lin k:

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.|0 1 0 1 11 0 3 1 00 3 1 0 01 1 0 0 31 0 0 3 1, th t nh: a, b, c, d, Ma trn lin thuc:

,`

.|1 0 0 0 0 0 1 1 0 0 0 01 1 1 1 0 1 0 0 0 0 0 00 1 1 1 1 0 0 0 0 0 0 00 0 0 0 0 1 1 0 1 1 1 00 0 0 0 0 0 0 1 1 1 1 112 11 10 9 8 7 6 5 4 3 2 1edcbae e e e e e e e e e e eTrng CSP Qung Tr 10abcdeBi tp ton ri rc trong :') , () , () , () , (4321b a eb a eb a ea a e') , () , () , () , (8765c c ed b ee b ee a e') , () , () , () , (1 21 11 09e d ed c ed c ed c eBi 19 : Hai n th vi ma trn lin k sau y c l ng cu khng?

,`

.|0 1 1 11 0 0 01 0 0 11 0 1 0,

,`

.|0 1 1 11 0 0 11 0 0 11 1 1 0.GiiDa vo ma trn lin k ca hai n th ta c th v li cc th bng hnh v:Theo hnh v ca hai n th ta thy chng khng c cng s cnh, mt bn c 4 cnh v mt bn c 5 cnh. Vy hai th c ma trn lin k cho trn khng ng cu.Bi ton ny c th khng cn v hnh li cng c, t ma trn k ta cng c th d dng xc nh c s cnh ca mi th ln lt l 4 v 5. Do vy chng khng th ng cu.Bi 20 : Xt xem cc th cho sau y c ng cu vi nhau khng?GiiTrng CSP Qung TrU1U2U3U4 11V1V2V3V4Bi tp ton ri rc a. Hnh 01. Hai th cho trn c: s nh, s cnh, tng s bc v s bc ca mi nh bng nhau. c bit, cc nh ca th th nht v th hai khi sp theo th t sau y th chng hon ton tng ng v mi mt: th th nht u1U2u3u4u5u6 th th hai v5V6v3v2v1v4S bc ca mi nh 3 4 4 3 5 5Chnh v vy, hai th trn l ng cu.b. Hnh 02.Hai th c hng cho trn khi sp theo th t sau y v cc nh th chng tng ng v tt c cc mt: t s nh, tng s bc, bc vo, bc ra ca mi nh, tng s cnh, th tv chiu ca cc cnh u tng ng:Trng CSP Qung Tr 12u11u22u33u44u55u66v11v22v44v33v55v66u11u22u33u44u55u66v11v22v66v33v55v44Bi tp ton ri rc th th nht u1u2u3u4u5u6 th th hai v3v5v1v2v4v6Bc vo: deg-(X) 1 2 1 2 2 1Bc ra: deg+(X) 2 1 2 1 1 2V vy, hai th c hng trn l ng cu vi nhau.Bi 21: (3.1)Cho G l th c v nh v e cnh, cn m v M tng ng l bc nh nht v ln nht cc nh ca G. Chng t rng:2em Mv GiiV m v M tng ng l bc nh nht v ln nht cc nh ca G, do ta d dng c c:11deg( ) .deg( ) , 1,deg( ) .viiiviiv vmm v M i vv vM '2 .2. 2 .2 .e v mev m e v M m Me v M v '(pcm)Bi22:(3.2)Chng minh rng nu G l n th phn i c v nh v e cnh, khi chng minh bt ng thc sau y:2(1)4ve GiiGi n1, n2 ln lt l s nh ca mi phn (n1 + n2 = v). V l n th phn i nn s cnh nhiu nht khi n l n th phn i , tc l: 1 2, n nK.Khi , s cnh nhiu nht s l: 1 2 1 2(2) n n n e n n .Ta d dng c c:2 2 2 2 21 2 1 1 2 2 1 1 2 2 1 2( ) 0 2 0 2 4 n n n n n n n n n n n n + + + Trng CSP Qung Tr 13Bi tp ton ri rc 2 2( 2 )1 21 2( )4 4n n vn n e e+ (pcm).Bi23:(3.4)Hy v cc th v hng biu din bi cc ma trn sau:a.1 2 32 0 43 4 0| ` . ,b.1 2 0 12 0 3 00 3 1 11 0 1 0| ` . ,c.0 1 3 0 41 2 1 3 03 1 1 0 10 3 0 0 24 0 1 2 3| ` . ,GiiBi24:(3.6)Tm ma trn lin k cho cc th sau:Trng CSP Qung Tr 14ABCh.aA BCDh.bABCDEh.cBi tp ton ri rc a.Knb.Cnc.Wnd.Km,ne.QnGii0 1 1 1 ... 1 11 0 1 1 ... 1 11 1 0 1 ... 1 1. : 1 1 1 0 ... 1 1... ... ... ... ... ... ...1 1 1 1 ... 0 11 1 1 1 ... 1 0naK| ` . ,

0 1 0 0 ... 0 11 0 1 0 ... 0 00 1 0 1 ... 0 0. : 0 0 1 0 ... 0 0... ... ... ... ... ... ...0 0 0 0 ... 0 11 0 0 0 ... 1 0nbC| ` . ,0 1 0 0 ... 1 11 0 1 0 ... 0 10 1 0 1 ... 0 1. : 0 0 1 0 ... 0 1... ... ... ... ... ... ...1 0 0 0 ... 0 11 1 1 1 ... 1 0ncW| ` . , ,0 ... 0 1 ... 1... ... ...... ... ...0 ... 0 1 ... 1. :1 ... 1 0 ... 0... ... ...... ... ...1 ... 1 0 ... 0mnm nmnd K| ` ' ' . ,647 48647 48647 48647 48Bi 25: (3.8)Hai th vi ma trn lin k sau y c ng cu vi nhau khng?0 1 0 11 0 0 1( .1)0 0 0 11 1 1 0h| ` . ,0 1 1 11 0 0 1( .2)1 0 0 11 1 1 0h| ` . ,GiiHaithvimatrnlinktrnkhngthngcuvinhauv: chng c s cnh khc nhau: th th nht c 4 cnh, th th hai c 5 cnh.Trng CSP Qung Tr 15Bi tp ton ri rc Bi 26: (3.9)Hai th vi ma trn lin k sau y c ng cu vi nhau khng?1 1 0 0 01 0 1 0 1( .1)0 0 0 1 10 1 1 1 0h| ` . ,0 1 0 0 10 1 1 1 0( .2)1 0 0 1 01 0 1 0 1h| ` . ,GiiTha thy, theo em ngh th y l hai ma trn lin thuc ch khng phi l hai ma trn lin k. V nu l hai ma trn lin thuc th chng ng cu vi nhau v:1 2 3 4 51234e e e e ev 1 1 0 0 0( .1') v 1 0 1 0 1v 0 0 0 1 1v 0 1 1 1 0h| ` . ,' ' ' ' '1 2 3 4 5'1'2'3'40 1 0 0 1( .2') 0 1 1 1 01 0 0 1 01 0 1 0 1e e e e evh vvv| ` . ,Xt nh x f t V1 ln V2sao cho:' '1 1 2 2' '3 3 4 4( ) _&_ ( ) ;( ) _&_ ( ) ;f v v f v vf v v f v v ' , ngthi biu din li th ca ma trn lin thuc hnh (h.2). Trong , cc cnh c sp theo th t:' ' ' ' '2 5 3 1 4e e e e e . Lc ny, hai ma trn lin thuc hon ton ging nhau. V vy chng ng cu vi nhau. ' ' ' ' ' ' ' ' ' '1 2 3 4 5 2 5 3 1 4' '1 1' '2 2' '3 3' '4 40 1 0 0 1 1 1 0 0 0( .2') ( .2'') 0 1 1 1 0 1 0 1 0 11 0 0 1 0 0 0 0 1 11 0 1 0 1 0 1 1 1 0e e e e e e e e e ev vh h v vv vv v| ` | ` . , . ,Bi 27: (3.10)Trng CSP Qung Tr 16Bi tp ton ri rc Cc cp th sau c ng cu vi nhau khng?GiiBi ny hon ton ging bi s 20 gii trn.Bi 28: (3.11)Cho V = {2, 3, 4, 5, 6, 7, 8} v E l tp hp cc cp phn t (u, v) ca V sao cho u < v v u vi v l cc s nguyn t cng nhau. Hy v th c hng ( ) , G VE .Tm s ng i phn bit di 3 t nh 2 ti nh 8.GiiBi 29: (3.12)Hy tm s ng i di n gia hai nh lin k (t. khng lin k) ty trong K3,3 vi mi gi tr ca n sau:a. n = 2, b. n = 3, c. n = 4, d. n = 5.GiiTrng CSP Qung Tr 177243568123564(II) (I)Bi tp ton ri rc Cch 1:Hai nh lin k phi 2 phn khc nhau. Mt cnh ch c th ni t 1 nh phn (I) n 1 nh phn (II) v ngc li. Gi m l s ng i gia 2 nh bt k trong K3,3 c di n. TH1: n chn.Nu n chn th nh u v nh cui ca ng i phi cng 1 phn, do vy chng khng th lin k. TH2: n l.Nu n l th nh u v nh cui ca ng i phi trn 2 phn khc nhau, do vy chng phi lin k (v y l K3,3).Mc khc mi mt nh phn ny lun c 3 phng n i qua 1 nh phn kia. Do vy ta c c cc kt lun sau y:o Hai nh lin k, n chn: m = 0,o Hai nh lin k, n l: m = 3n-1,o Hai nh khng lin k, n chn: m = 3n-1,o Hai nh khng lin k, n l: m = 0.p dng cho cc trng hp: di ng i n = 2 n = 3 n = 4 n = 5S ng i (gia 2 nh lin k) 0 9 0 81S ng i (2 gia nh khng lin k)3 0 27 0Cch 2:K3,3 c ma trn lin k:0 0 0 1 1 10 0 0 1 1 10 0 0 1 1 11 1 1 0 0 01 1 1 0 0 01 1 1 0 0 0| ` . ,Ta d dng chng minh c (bng quy np):Trng CSP Qung Tr 18Bi tp ton ri rc 1 1 11 1 11 1 1n1 1 11 1 11 1 13 3 3 0 0 03 3 3 0 0 03 3 3 0 0 0A , 20 0 0 3 3 30 0 0 3 3 30 0 0 3 3 3n n nn n nn n nn n nn n nn n nn k | ` . ,1 1 11 1 11 1 1n1 1 11 1 11 1 10 0 0 3 3 30 0 0 3 3 30 0 0 3 3 3A , 2 13 3 3 0 0 03 3 3 0 0 03 3 3 0 0 0n n nn n nn n nn n nn n nn n nn k | ` + . ,Theo nh l 2 [543], s ng i c di n t nh i n nh j l gi tr ca phn t ai,j ca ma trn An. Do vy ta c c kt qu sau: di ng i n = 2 n = 3 n = 4 n = 5S ng i (gia 2 nh lin k) 0 9 0 81S ng i (gia 2 nh khng lin k)3 0 27 0BI TP CHNG IIIBi 30: (4.1).Vi gi tr no ca n th cc th sau y l th Euler?a. Knb. Cnc. Wnd. QnGiiiu kin cn v mt th l th Euler khi v ch khi tt c cc nh ca n u c bc chn. a. Kn:V mi nh ca Knc s bc bng nhau v bng n 1. Knl th Euler th n 1 =chn. Do vy n phi l (n >= 3).Trng CSP Qung Tr 19Bi tp ton ri rc b. Cn:V mi nh ca Cn u c bc l 2 nn Cn lun l th Euler. c. Wn:Tr mt nh c bc l n 1, cn li cc nh khc u c bc l 3, do vy y khng th l th Euler.d. Qn:V mi nh u c bc l n, do vy Qn l th Euler th n chn.Bi 31: (4.2)Vi cc gi tr no ca m v n th th phn i y Km,n c:a. Chu trnh Euler.b. ng i Euler.Giia. V cc nh ca th phn i Km,n c bc l m hoc n. Do vy, n l th Euler th m v n u phi chn.b. mt th c ng i Euler th phi c ng 2 nh bc l, cc nh cn li c bc chn. Do vy mt trong 2 gi tr m, n phi l 2, gi tr cn li phi l s l.Bi 3 2:(4.3)Vi gi tr no ca m v n th th phn i y Km,n c chu trnh Hamilton.GiiCch 1:Theo nh l irc, nu G l n th c n nh v mi nh ca G u c bc khng nh hn 2n th G l mt th Hamilton. Vi Km,n, cc nh c bc m hoc n, y l th Hamilton th: ( ) / 2( ) / 2n n mm n m n mm n m + ' +Cch 2:Theo nh l Ore nu G l mt n th c n nh v bt k hai nh no khng k nhau cng c tng s bc khng nh hn n th G l th Hamilton. Trng CSP Qung Tr 20Bi tp ton ri rc Vi Km,n 2 nh khng k nhau s c bc cng l n hoc cng l m. Do , l th Hamilton th: ( ) ( )( ) ( )n n n mm n m n mm m n m+ + '+ +Cch 3:Theo nh l 4.2.6 nu G l th phn i vi hai tp nh l V1, V2 c s nh cng bng n (n 2) v bc ca mi nh ln hn 2nth G l mt th Hamilton. Do vy, nu m = n th hin nhin Km,n l th Hamilton.Bi 33: (4.4)Chng minh rng th lp phng Qnl mt th Hamilton. V cy lit k tt c cc chu trnh Hamilton ca th lp phng Q3.Giia. Chng minh Qn l th Hamilton:Dng m Gray gii bi ton ny, ta s chng minh bng quy np ton hc nh sau:Vi n = 1: hin nhin ng vi 2 m: 0, 1.Gi s bi ton ng vi n = k, tc l ta c m:1 2 3 1a ... ( 2 )kq qa a a a q Ta phi chng minh bi ton ng vi n = k + 1. iu ny hin nhin ng khi ta gn thm 2 k t 0 v 1 vo cc m ny theo cch sau:

1 2 3 1 q 1 3 2 10a 0 0 ... 0 0 1 1 ... 1 1 1 ( 2 )kq q qa a a a a a a a a q Vy bi ton ng vi mi n, hay l bi ton c chng minh.b. V cy lit k tt c cc chu trnh Hamilton ca Q3:Trng CSP Qung Tr 21010000001010100011101011110110101111 100 111 001 111 100 111 001 111 010 111110 101 110 011 101 110 101 011 011 110 011 101111 100 111 010 111 100 111 001 111 010 111 001101 110 011 110 110 101 011 101 110 011 101 011100 010 010 100 100 001 001 100 010 001 001 010000 000 000 000 000 000 000 000 000 000 000 0002 134 5 678 91011 12 13 141516 17 1819 20 21Bi tp ton ri rc Bi 34: (4.5)Trong mt cuc hp c 15 ngi mi ngy ngi vi nhau quanh mt bn trn mt ln. Hi c bao nhiu cch sp xp sao cho mi ln ngi hp, mi ngi c hai ngi bn cnh l bn mi, v sp xp nh th no.GiiBi 35 :(4.6)Hiu trng mi 2n (n 2) sinh vin gii n d tic. Mi sinh vin gii quen t nht n sinh vin gii khc n d tic. Chng minh rng lun lun c th xp tt c cc sinh vin gii ngi xung quanh mt bn trn, mi ngi ngi gia hai ngi m sinh vin quen.GiiXt th G = (V, E), trong : V l tp cc sinh vin n d tic; E = (u,v) vi u, v thuc V v u, v c quen bit nhau.Theo cch xc lp th trn th y l n th c 2n nh, mi nh c bc ti thiu l n (v mi sinh vin quen vi t nht n sinh vin khc). Do , theo nh l irac (4.2.3) th G l th Hamilton. Mc khc, y l th v hng nn bi ton c chng minh.Bi 36 :(4.7)Mt ng vua xy dng mt lu i ct bu vt. Ngi ta tm thy s ca lu i (hnh sau) vi li dn: mun tm bu vt, ch cn t mt trong cc phng bn ngoi cng (s 1, 2, 6, 10, ...), i qua tt c cc ca phng, mi ca ch mt ln; bu vt c giu sau ca cui cng. Hy tm ni giu vt.Trng CSP Qung Tr 22Bi tp ton ri rc GiiBu vt s c giu sau ca t phng 19 qua phng 18.Sau y l ng i tm bu vt: 6-2-1-4-3-7-11-12-8-13-12-17-16-20-21-17-18-13-14-9-5-4-2-5-6-10-15-14-19-18. Bi 3 7:(4.8) th cho trong hnh sau gi l th Peterson P.a. Tm mt ng i Hamilton trong P.b. Chng minh rng P \ {v}, vi v l mt nh bt k ca P, l mt th Hamilton.Giia.Tm mt ng i Hamilton trong G:a b c d e f h k g i b.Ta chia G thnh 2 nhm: a, b, c, d, e v f, g, h, i, k. Cc phn t trong 2 nhm ny c vai tr ging nhau, do vy bi ton c 2 trng hp tng qut.TH1: b im a:i f e d c b h k g i TH2: b im f:Trng CSP Qung Tr 23aek ib gf hd cBi tp ton ri rc h k d e a g i c b h Bi 38 :(4.9)GiibitonngiuathTrungHoavithcchotrong hnh sau y:GiiBi 39 :(4.10)Chng minh rng th G cho trong hnh sau c ng i Hamilton (t s n r) nhng khng c chu trnh Hamilton.Giing i Hamilton t r n s:s a b c e d g f h r Gi s c chu trnh Hamilton th i qua nh s phi i qua 02 nh a v c, ng thi i qua nh b cng phi i qua 02 nh a v c. Nh vy 02 nh a v c u phi i qua 02 ln. iu ny khng ng nu y l th Hamilton v ngoi tr nh u v nh cui th khng c nh no chu trnh i qua 02 ln (bi ton c chng minh).Trng CSP Qung Tr 24acbsrfedghBi tp ton ri rc Bi 40 :(4.11) Cho th d v:1. th c mt chu trnh va l chu trnh Euler va l chu trnh Hamilton.2. th c mt chu trnh Euler v mt chu trnh Hamilton, nhng hai chu trnh khng trng nhau.3. th c 6 nh, l th Hamilton, nhng khng phi l th Euler.4. th c 6 nh, l th Euler, nhng khng phi l th Hamilton.Gii1. Mt chu trnh va l chu trnh Euler, va l chu trnh Hamilton.2. Mt chu trnh Euler, mt chu trnh Hamilton nhng khng trng nhau.Chu trnh Euler:a b c e a c d e f a Chu trinh Hamilton:a b c d e f a 3. th Hamilton 6 nh nhng khng phi l th Euler (h.1).4. th Euler 6 nh nhng khng phi l th Hamilton (h.2).Bi 41 :(4.12)Trng CSP Qung Tr 25acdebfacdebfh.1h.2Bi tp ton ri rc Chng minh rng con m khng th i qua tt c cc ca mt bn c c 4 x 4 hoc 5 x 5 vung, mi ch mt ln, ri tr v ch c.GiiTa c th xem y l mt th vi mi c l mt nh, mi (u,v) l mt cnh nu con m c th i t u n v. Bi ton tr thnh xc nh xem cc th G1 v G2 tng ng vi bn c 4x4 v 5x5 c phi l th Hamilton khng?Ta s chng minh bi ton bng phng php phn chng (tng t nh bi 39 - 4.10), tc l gi s cc th G1 v G2 l cc th Hamilton.T G1 ta nhn thy, chu trnh Hamilton mun i qua nh 4 phi i qua 02 nh 6 v 11, ng thi mun i qua nh 13 cng phi i qua 02 nh 6 v 11. Nh vy,nh 6v11 phic lpli tnht 02ln, iunytrivi gi thuyt G1 l th Hamilton (v ch c nh u v nh cui mi c lp li). Tc l G1 khng phi l th Hamilton (pcm).Tng t i vi G2, chu trnh Hamilton (nu c): i qua nh 1 th phi i qua nh 8 v 12, mun i qua nh 5 th phi i qua nh 8 v 14, mun i qua nh 25 phi i qua nh 14 v 18, mun i qua nh 12 phi i qua nh 18 v 12. Nh vy cc nh 8, 14, 18, 12 u phi i qua hai ln, iu ny khng ng vi th Hamilton. Vy G2 khng phi l th Hamilton (pcm).Trng CSP Qung Tr 261 2 3 487 6 59 10 11 1216 15 14 131 2 3 4 510 8 9 7 6111621 221712 13 141520 19 1823 2425