TON RI RC Lecturer: PhD. Ngo Huu Phuc Tel: 0438 326 077 Mob: 098 5696 580 Email: ngohuuphuc76@gmail.com CHNG 1: KHI NIM C BNL thuyt s v h m 1 @Copyrights by Dr. Ngo Huu Phuc, Le Quy Don Technical University NI DUNG @Copyrights by Dr. Ngo Huu Phuc, Le Quy Don Technical University 2 1. Cc php ton trn s nguyn. 2. Biu din cc s nguyn. 3. nh l v s d Trung Quc v ng dng. 4. Cc h m. 1. Cc php ton trn s nguyn (1/5) @Copyrights by Dr. Ngo Huu Phuc, Le Quy Don Technical University 3 1.1. Php chia nguyn. Cho hai s nguyn n v m ta ni n chia ht cho m nu tn ti s nguyn k sao cho n = k.m v k hiu l m|n. nh l 1. Cho n, m v k l cc s nguyn. Khi a- Nu k|n v k|m th k|(n + m). b- Nu k|n th k|n m vi mi s nguyn m . c- Nu k|n v n|m th k|m. 1. Cc php ton trn s nguyn (2/5) @Copyrights by Dr. Ngo Huu Phuc, Le Quy Don Technical University 4 1.1. Php chia nguyn (tip) nh l 2. Mi s nguyn dng u c th c vit duy nht di dng tch ca cc s nguyn t. nhl3.Choalmtsnguynvdlsnguyn dng. Khi tn ti cc s q v r duy nht, vi 0 s r < d, sao choa = dq + r. Haisnguynnvmgilnguyntcngnhaunu USCLN(n,m) = 1. Cc s nguyn a1, a2, . . . , an c gi l i mt nguyn t cng nhau nu USCLN(ai, aj) =1 vi mi 1 s i, j sn.1. Cc php ton trn s nguyn (3/5) @Copyrights by Dr. Ngo Huu Phuc, Le Quy Don Technical University 5 1.1. Php chia nguyn (tip) nh l 4. Cho n, m l hai s nguyn dng. Khi ab = USCLN(n,m) BSCNN(n,m) Hai s nguyn n v m gi l ng d theo modulo k nu n mod k = m mod k, ta k hiu n m (mod k). nh l 5. Nu n m (mod k) v p q (mod k). Khi : a)n+p m + q (mod k) b) np m q (mod k) Phn t b c gi l phn t nghch o ca a theo modulo m nu ab 1 (mod m) v k hiu l a -1 , khi aa -1 1 (mod m). 1. Cc php ton trn s nguyn (4/5) @Copyrights by Dr. Ngo Huu Phuc, Le Quy Don Technical University 6 1.2. Thut ton Euclid. B : Cho a = b q + r trong a, b, q, r l cc s nguyn dng. Khi USCLN(a,b) = USCLN(b,r) Chng minh. Vi mi c s chung d ca a v b khi a - bXq = r, suy ra d cng l c s ca r, tc l dcng l c s chung ca b v r vy USCLN(a,b) = USCLN(b,r). Thut ton Euclid. Input. a, b (a >b) t r0 = a v r1 = b. Bc 1. r0 = r1 q1 + r20 s r2 < r1 Bc 2. Nu r2 =0 th r0 = r1v r1 = r2quay li bc 1 ngc li sang bc 3. Output.r1. 1. Cc php ton trn s nguyn (5/5) @Copyrights by Dr. Ngo Huu Phuc, Le Quy Don Technical University 7 1.2. Thut ton Euclid (tip) Thut ton Euclid c dng tm c s chung ln nht ca hai s nguyn. V d tm USCLN(91,287). Trc ht ly s ln hn 287 chia cho s nh 91 ta c 287 =91X 3 + 14 bt k c s chung no ca 287 v 91 cng l c s ca 287 -91X 3 = 14. V cng nh vy, bt k c s chung no ca 91 v 14 cng l c s ca 287 =91X 3 + 14 . Do USCLN ca 91 v 14 cng l USCLN ca 287 v 91. T c USCLN(91,287) = USCLN(91,14) Tng t nh vy v 91 = 14X 6 + 7 ta c USCLN(91,14) = USCLN(14,7) = 7 2. Biu din cc s nguyn (1/2) @Copyrights by Dr. Ngo Huu Phuc, Le Quy Don Technical University 8 nh l 6. Cho b l mt s nguyn dng ln hn 1. Khi nu n l mt s nguyn dng th n c th c biu din mt cch duy nht di dng: n = akbk + ak-1bk-1 + . . . .+ a1b1 + a0 Trong kl s nguyn khng m, a0, a1, a2,. . . ak l cc s nguyn khng m nh hn b v ak =0. Biu din n trong nh l trn c gi l trin khai c s b ca n. 2. Biu din cc s nguyn (2/2) @Copyrights by Dr. Ngo Huu Phuc, Le Quy Don Technical University 9 V d: V d: Cho n = 165, b = 8 ta c 165 = 2X 82 + 4X 81 + 5 Trong v d ny ta c th biu din nh sau (245)8 gi l cch biu din theo h bt phn. V d: Cho n = 351, b = 2 ta c 351 = 1X 28 + 0X 27 + 1X 26 + 0X 25 + 1X 24 + 1X 23 +1X 22 +1X 21 + 0X 20 ta nhn c dy {ak} sau (101011111)2gi l biu din nh phn ca s 351. 3. nh l v s d Trung Quc v ng dng (1/13) @Copyrights by Dr. Ngo Huu Phuc, Le Quy Don Technical University 10 S d Trung Quc: nh l v s d Trung Quc. Gi s m1, m2,. . ., mn l cc s nguyn dng, nguyn t cng nhau tng i mt v a1, a2,. . ., an l cc s nguyn. Khi h n phng trnh ng d x ai (mod mi) vi1s isn s c mt nghim duy nht theo modulo M = m1 m2 . . . mn c cho theo cng thc sau: Trong Mi = M/mi v yi = Mi-1 mod mi vi 1 si sn. M mod an1 ii ==i iy M X3. nh l v s d Trung Quc v ng dng (2/13) @Copyrights by Dr. Ngo Huu Phuc, Le Quy Don Technical University 11 ng dng Gi s m1, m2,. . . , mn l cc s nguyn t cng nhau tng i mt, tc l USCLN(mi,mj)=1 vi mi i= j . Gi s rng a1, a2,. . . , an l cc s nguyn, xt h cc phng trnh ng d sau: x a1 (mod m1) x a2 (mod m2)(1) . . .x an (mod mn) Khi nh l v s d Trung Quc khng nh rng h ny c nghim duy nht theo Modulo M = m1 X m2 X. . . X mn . 3. nh l v s d Trung Quc v ng dng (3/13) @Copyrights by Dr. Ngo Huu Phuc, Le Quy Don Technical University 12 ng dng (tip) K hiu nh x: t : ZM Zm1 X Zm2 X . . . Zmn nh x ny c nh ngha nh sau: t(x) = (x mod m1, x mod m2,. . . ,x mod mn) V d: Cho n = 2, m1= 5, m2= 3 t M = 15.Khi t(x) nh x c cc gi tr nh sau: t(0) = (0,0) t(1) = (1,1)t(2) = (2,2) t(3) = (3,0) t(4) = (4,1)t(5) = (0,2) t(6) = (1,0) t(7) = (2,1)t(8) = (3,2) t(9) = (4,0) t(10) = (0,1) t(11) = (1,2) t(12) = (2,0) t(13) = (3,1) t(14) = (4,2) 3. nh l v s d Trung Quc v ng dng (4/13) @Copyrights by Dr. Ngo Huu Phuc, Le Quy Don Technical University 13 ng dng (tip) chng minh nh l v s d Trung Quc, cn chng minh t l mt song nh. iu ny c th thy d dng qua v d trn. Ni cch khc, cn ch ra cng thc ca nh x ngc t -1: Vi 1 si sn, nh ngha: Khi d dng thy rngUSCLN(Mi,mi) = 1 , vi 1 si sn Ta nh ngha yi = Mi-1 mod mi phn t nghch o ny tn ti do USCLN(Mi,mi) = 1 v c th tm c bng thut ton Euclid m rng. iimMM =3. nh l v s d Trung Quc v ng dng (5/13) @Copyrights by Dr. Ngo Huu Phuc, Le Quy Don Technical University 14 ng dng (tip) Theo nh ngha ta c Miyi 1 (mod mi) , vi 1 si sn. nh ngha: : Zm1 Zm2 . . . Zmn ZM

Ta s chng t rng = t -1 , tc l n s cho ta mt cng thc tng minh gii h ng d ban u. M mod a ) a ,..., a , (an1 ii n 2 1 ==i iy M 3. nh l v s d Trung Quc v ng dng (6/13) @Copyrights by Dr. Ngo Huu Phuc, Le Quy Don Technical University 15 ng dng (tip) K hiu X = (aj,. . ., an) v cho 1 sj sn. Xt s hng ai Miyi trong tng trn khi rt gn theo modulo mj .Nu i = jth ai Miyi ai (mod mi) v Miyi 1 (mod mi) Nu i =jthai Miyi 0 (mod mi) do mi|M trong trng hp ny. T ta c: Do iu ny ng i vi mi i, 1 si sn nn Xl nghim ca h phng trnh ng d. ) (mod a) M (mod a Xin1 iiii imy M=3. nh l v s d Trung Quc v ng dng (7/13) @Copyrights by Dr. Ngo Huu Phuc, Le Quy Don Technical University 16 ng dng (tip) CnphichngminhnghimXlduynhtcah phng trnh ng d. V: t l nh x t tpZMc lc lng l M sang tp Zm1 Zm2 . . . Zmn cng c lc lng M, v t l ton nh t suy ra t l n nh (xc nh php tng ng 1-1), iu ny ko theo t l mt song nh v t -1 = . Ch l t -1 l mt hm tuyn tnh ca cc bin (aj,. . ., an). 3. nh l v s d Trung Quc v ng dng (8/13) @Copyrights by Dr. Ngo Huu Phuc, Le Quy Don Technical University 17 Thut ton Euclid m rng: Gii thut sau ch thc hin vi cc s nguyn m>a>0, biu din bng gi m: Procedure Euclid_Extended (a,m)int y0=0, y1:=1;While a>0 do{ r:= m mod aif r=0 then Breakq:= m div a y:= y0-y1*qm:=aa:=ry0:=y1

y1:=y }If a>1 Then Return "A khng kh nghch theo moun m"else Return " Nghch o modulo m ca a l y" 3. nh l v s d Trung Quc v ng dng (9/13) @Copyrights by Dr. Ngo Huu Phuc, Le Quy Don Technical University 18 V d v tm nghch o theo Modulo: Cho a=143, m=7, tm nghch o ca a. Gii: V 143 mod 7 = 3, nn cn tm nghch o ca 3 modulo 7. Bcmarqy0 y1 y 0731201-2 1310........ Kt qu tnh ton trong bng cho ta 2. Ly s i ca 2 theo modulo 7 c 5. Vy: 3-1 mod 7 = 5 3. nh l v s d Trung Quc v ng dng (10/13) @Copyrights by Dr. Ngo Huu Phuc, Le Quy Don Technical University 19 V d v tm nghch o theo Modulo: Cho a=30, m=101, tm nghch o ca a. Gii: Bcmarqy0 y1 y 01013011301-3 13011821-37 211831-37-10 383227-1027 43211-1027-37 5210........ Kt qu tnh ton trong bng cho ta 37. Ly s i ca 37 theo modulo 101 c 64. Vy: 30-1 mod 101 = 64 3. nh l v s d Trung Quc v ng dng (11/13) @Copyrights by Dr. Ngo Huu Phuc, Le Quy Don Technical University 20 V d v h phng trnh ng d: Cho h phng trnh ng d: x 5 (mod 7) x 3 (mod 11) x 10 (mod 13) 3. nh l v s d Trung Quc v ng dng (12/13) @Copyrights by Dr. Ngo Huu Phuc, Le Quy Don Technical University 21 V d v h phng trnh ng d (tip): Tnh: M = 7 11 13 = 1001, M1 = 11 13 = 143, M2 = 7 13 = 91, M3 = 7 11 = 77, y1 = 143 -1 mod 7= 5 theo Euclid m rng y2 = 91-1 mod 11= 4 theo Euclid m rng v y3 = 77 -1 mod 13 = 12 theo Euclid m rng 3. nh l v s d Trung Quc v ng dng (13/13) @Copyrights by Dr. Ngo Huu Phuc, Le Quy Don Technical University 22 V d v h phng trnh ng d (tip): Khi =t-1: Z7 Z11 Z13 ZM c dng: t-1 (a1, a2, a3) = (5 143 a1 + 4 91 a2 + 12 77 a3) mod 1001 Khi vi a1 = 5 , a2 = 3 v a3 = 10 nghim ca h phng trnh l: X = (5 143 5 + 3 91 4 + 10 77 12) mod 1001= (3 575 + 1 092 + 9 240) mod 1001 = 13 907 mod 1001 = 894 mod 1001 = 894 4. Cc h m (1/5) @Copyrights by Dr. Ngo Huu Phuc, Le Quy Don Technical University 23 Xem xt mt s h m: 1. H m thp phn. 2. H m nh phn. 3. H m bt phn (Octal). 4. h m thp lc phn (Hexa). 4. Cc h m (2/5) @Copyrights by Dr. Ngo Huu Phuc, Le Quy Don Technical University 24 1. H m thp phn. Biu din s n bt k trong h thp phn theo cng thc: n = ak10k + ak-110k-1 + . . . .+ a1101 + a0100

trong 0 s ai s9,i = 1, 2, 3, . . . k 4. Cc h m (3/5) @Copyrights by Dr. Ngo Huu Phuc, Le Quy Don Technical University 25 2. H m nh phn. Biu din s n bt k trong h nh phn theo cng thc: n = ak2k + ak-12k-1 + . . . .+ a121 + a020

trong 0 s ai s1, i = 1, 2, 3, . . . k 4. Cc h m (4/5) @Copyrights by Dr. Ngo Huu Phuc, Le Quy Don Technical University 26 3. H m bt phn (Octal). S n bt k c biu din trong h bt phn theo cng thc: n = ak8k + ak-18k-1 + . . . .+ a181 + a080

trong 0 s ai s7, i = 1, 2, 3, . . . k4. Cc h m (5/5) @Copyrights by Dr. Ngo Huu Phuc, Le Quy Don Technical University 27 4. H m thp lc phn (Octal). S n bt k c biu din trong thp lc phn theo cng thc: n = ak16k + ak-116k-1 + . . . .+ a1161 + a0160

trong 0 s ai s15, i = 1, 2, 3, . . . ktc l ai e {0, 1, 2, . . . , A, B, . . .,F}