bctntlvn (42).pdf
Transcript of bctntlvn (42).pdf
![Page 1: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/1.jpg)
ĐỒ ÁN TỐT NGHIỆP - BẢO MẬT THÔNG TIN
![Page 2: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/2.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
MUÏC LUÏC
I .1 Giôùi thieäu....................................................................................................... 3 I.2 Caùc Heä Maõ Thoâng Duïng:................................................................................. 3 e. Phöông phaùp Affine........................................................................................... 4 f. Phöông phaùp Vigenere....................................................................................... 5 I.2 LAÄP MAÕ DES................................................................................................. 14 I. 3 THAÙM MAÕ DES............................................................................................ 17
I.3.1. Thaùm maõ heä DES - 3 voøng................................................................. 20 II.3.2. Thaùm maõ heä DES 6-voøng ..................................................................... 24 II.3. 3 Caùc thaùm maõ vi sai khaùc ...................................................................... 28
III. CAØI ÑAËT THAÙM MAÕ DES 3 VOØNG...................................................... 28 III.1 Giao Dieän .................................................................................................... 28 III.2 XÖÛ LYÙ ............................................................................................................
![Page 3: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/3.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
LÔØI NOÙI ÑAÀU Hieän nay, nöôùc ta ñang trong giai ñoaïn tieán haønh coâng nghieäp hoùa, hieän ñaïi hoùa ñaát
nöôùc. Tin hoïc ñöôïc xem laø moät trong nhöõng ngaønh muõi nhoïn. Tin hoïc ñaõ vaø ñang ñoùng goùp raát nhieàu cho xaõ hoäi trong moïi khía caïnh cuûa cuoäc soáng.
Maõ hoùa thoâng tin laø moät ngaønh quan troïng vaø coù nhieàu öùng duïng trong ñôøi soáng xaõ hoäi.
Ngaøy nay, caùc öùng duïng maõ hoùa vaø baûo maät thoâng tin ñang ñöôïc söû duïng ngaøy caøng phoå bieán hôn trong caùc lónh vöïc khaùc nhau treân Theá giôùi, töø caùc lónh vöïc an ninh, quaân söï, quoác phoøng…, cho ñeán caùc lónh vöïc daân söï nhö thöông maïi ñieän töû, ngaân haøng…
ÖÙng duïng maõ hoùa vaø baûo maät thoâng tin trong caùc heä thoáng thöông maïi ñieän töû, giao dòch
chöùng khoaùn,… ñaõ trôû neân phoå bieán treân theá giôùi vaø seõ ngaøy caøng trôû neân quen thuoäc vôùi ngöôøi daân Vieät Nam. Thaùng 7/2000, thò tröôøng chöùng khoaùn laàn ñaàu tieân ñöôïc hình thaønh taïi Vieät Nam; caùc theû tín duïng baét ñaàu ñöôïc söû duïng, caùc öùng duïng heä thoáng thöông maïi ñieän töû ñang ôû böôùc ñaàu ñöôïc quan taâm vaø xaây döïng. Do ñoù, nhu caàu veà caùc öùng duïng maõ hoùa vaø baûo maät thoâng tin trôû neân raát caàn thieát.
![Page 4: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/4.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
I. MOÄT SOÁ PHÖÔNG PHAÙP MAÕ HOÙA
I .1 Giôùi thieäu
Ñònh nghóa 1.1: Moät heä maõ maät (cryptosystem) laø moät boä-naêm (P, C, K, E, D) thoûa maõn caùc ñieàu kieän sau: 1. P laø khoâng gian baûn roõ. taäp hôïp höõu haïn taát caû caùc maåu tin nguoàn caàn maõ hoùa coù theå coù 2. C laø khoâng gian baûn maõ. taäp hôïp höõu haïn taát caû caùc maåu tin coù theå coù sau khi maõ hoùa 3. K laø khoâng gian khoaù. taäp hôïp höõu haïn caùc khoùa coù theå ñöôïc söû duïng 4. Vôùi moãi khoùa k∈K, toàn taïi luaät maõ hoùa ek∈E vaø luaät giaûi maõ dk∈D töông öùng. Luaät maõ hoùa ek: P → C vaø luaät giaûi maõ ek: C → P laø hai aùnh xaï thoûa maõn
( )( ) ,k kd e x x x P= ∀ ∈
Tính chaát 4. laø tính chaát chính vaø quan troïng cuûa moät heä thoáng maõ hoùa. Tính chaát naøy baûo ñaûm vieäc maõ hoùa moät maåu tin x∈P baèng luaät maõ hoùa ek∈E coù theå ñöôïc giaûi maõ chính xaùc baèng luaät dk∈D. Ñònh nghóa 1.2: Zm ñöôïc ñònh nghóa laø taäp hôïp {0, 1, ..., m-1}, ñöôïc trang bò pheùp coäng (kyù hieäu +) vaø pheùp nhaân (kyù hieäu laø ×). Pheùp coäng vaø pheùp nhaân trong Zm ñöôïc thöïc hieän töông töï nhö trong Z, ngoaïi tröø keát quaû tính theo modulo m Ví duï: Giaû söû ta caàn tính giaù trò 11 × 13 trong Z16. Trong Z, ta coù keát quaû cuûa pheùp nhaân 11×13=143. Do 143≡15 (mod 16) neân 11×13=15 trong Z16. Moät soá tính chaát cuûa Zm 1. Pheùp coäng ñoùng trong Zm, i.e., ∀ a, b ∈ Zm, a+b ∈ Zm 2. Tính giao hoaùn cuûa pheùp coäng trong Zm, i.e., ∀ a, b ∈ Zm, a+b =b+a 3. Tính keát hôïp cuûa pheùp coäng trong Zm, i.e., ∀ a, b, c ∈ Zm, (a+b)+c =a+(b+c) 4. Zm coù phaàn töû trung hoøa laø 0, i.e., ∀ a ∈ Zm, a+0=0+a=a 5. Moïi phaàn töû a trong Zm ñeàu coù phaàn töû ñoái laø m – a 6. Pheùp nhaân ñoùng trong Zm, i.e., ∀ a, b ∈ Zm, a×b∈ Zm 7. Tính giao hoaùn cuûa pheùp coäng trong Zm, i.e., ∀ a, b ∈ Zm, a×b=b×a 8. Tính keát hôïp cuûa pheùp coäng trong Zm, i.e., ∀ a, b, c ∈ Zm, (a×b)×c =a×(b×c) 9. Zm coù phaàn töû ñôn vò laø 1, i.e., ∀ a ∈ Zm, a×1=1×a=a 10. Tính phaân phoái cuûa pheùp nhaân ñoái vôùi pheùp coäng, i.e., ∀ a, b, c ∈ Zm, (a+b)×c
=(a×c)+(b×c) 11. Zm coù caùc tính chaát 1, 3 – 5 neân taïo thaønh 1 nhoùm. Do Zm coù tính chaát 2 neân taïo thaønh
nhoùm Abel. Zm coù caùc tính chaát (1) – (10) neân taïo thaønh 1 vaønh
I.2 Caùc Heä Maõ Thoâng Duïng:
a. Heä Maõ Ñaày (Shift Cipher )
![Page 5: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/5.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
Shift Cipher laø moät trong nhöõng phöông phaùp laâu ñôøi nhaát ñöôïc söû duïng ñeå maõ hoùa. Thoâng ñieäp ñöôïc maõ hoùa baèng caùch dòch chuyeån (xoay voøng) töøng kyù töï ñi k vò trí trong baûng chöõ caùi.
Phöông phaùp Shift Cipher
Cho P = C = K = Z26. Vôùi 0 ≤ K ≤ 25, ta ñònh nghóa
eK = x + K mod 26 vaø dK = y - K mod 26 (x,y ∈ Z26)
trong ñoù 26 laø soá kyù töï trong baûng chöõ caùi La tinh, moät caùch töông töï cuõng coù theå
ñònh nghóa cho moät baûng chöõ caùi baát kyø. Ñoàng thôøi ta deã daøng thaáy raèng maõ ñaåy laø moät heä maät maõ vì dK(eK(x)) = x vôùi moïi x∈Z26.
b. Heä KEYWORD-CEASAR Trong heä maõ naøy khoùa laø moät töø naøo ñoù ñöôïc choïn tröôùc, ví duï PLAIN. Töø naøy
xaùc ñònh daõy soá nguyeân trong Z26 (15,11,0,8,13) töông öùng vôùi vò trí caùc chöõ caùi cuûa caùc chöõ ñöôïc choïn trong baûng chöõ caùi. Baây giôø baûn roõ seõ ñöôïc maõ hoùa baèng caùch duøng caùc haøm laäp maõ theo thöù töï:
e15, e11, e0, e8, e13, e15, e11, e0, e8, e,... vôùi eK laø haøm laäp maõ trong heä maõ chuyeån.
c. Heä Maõ Vuoâng (SQUARE) Trong heä naøy caùc töø khoùa ñöôïc duøng theo moät caùch khaùc haún. Ta duøng baûng chöõ
caùi tieáng Anh (coù theå boû ñi chöõ Q, neáu muoán toång soá caùc chöõ soá laø moät soá chính phöông) vaø ñoøi hoûi moïi chöõ trong töø khoùa phaûi khaùc nhau. Baây giôø moïi chöõ cuûa baûng chöõ caùi ñöôïc vieát döôùi daïng moät hình vuoâng, baét ñaàu baèng töø khoùa vaø tieáp theo laø nhöõng chöõ caùi coøn laïi theo thöù töï cuûa baûng chöõ.
d. Maõ theá vò Moät heä maõ khaùc khaù noåi tieáng . Heä maõ naøy ñaõ ñöôïc söû duïng haøng traêm naêm nay.
Phöông phaùp :
Cho P = C = Z26. K goàm taát caû caùc hoaùn vò coù theå coù cuûa 26 kyù hieäu 0,...,25. Vôùi moãi hoaùn vò π∈K, ta ñònh nghóa:
eπ(x) = π(x) vaø ñònh nghóa dπ(y) = π-1(y) vôùi π -1 laø hoaùn vò ngöôïc cuûa hoaùn vò π.
Trong maõ theá vò ta coù theå laáy P vaø C laø caùc baûng chöõ caùi La tinh. Ta söû duïng Z26 trong maõ ñaåy vì laäp maõ vaø giaûi maõ ñeàu laø caùc pheùp toaùn ñaïi soá.
e. Phöông phaùp Affine
![Page 6: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/6.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
Cho P = C = Z26 vaø cho
K = {(a,b) ∈ Z26 × Z26 : gcd(a,26) = 1}
Vôùi K = (a,b) ∈ K, ta xaùc ñònh eK(x) = ax+b mod 26
vaø dK = a-1(y-b) mod 26 (x,y ∈ Z26)
Phöông phaùp Affine laïi laø moät tröôøng hôïp ñaëc bieät khaùc cuûa Substitution Cipher. Ñeå coù theå giaûi maõ chính xaùc thoâng tin ñaõ ñöôïc maõ hoùa baèng haøm ek∈ E thì ek phaûi laø moät song aùnh. Nhö vaäy, vôùi moãi giaù trò y∈Z26, phöông trình ax+b≡y (mod 26) phaûi coù nghieäm duy nhaát x∈Z26. Phöông trình ax+b≡y (mod 26) töông ñöông vôùi ax≡(y–b ) (mod 26). Vaäy, ta chæ caàn khaûo saùt phöông trình ax≡(y–b ) (mod 26)
Ñònh lyù1.1: Phöông trình ax+b≡y (mod 26) coù nghieäm duy nhaát x∈Z26 vôùi moãi giaù trò b∈Z26
khi vaø chæ khi a vaø 26 nguyeân toá cuøng nhau. Vaäy, ñieàu kieän a vaø 26 nguyeân toá cuøng nhau baûo ñaûm thoâng tin ñöôïc maõ hoùa baèng haøm ek coù theå ñöôïc giaûi maõ vaø giaûi maõ moät caùch chính xaùc. Goïi φ(26) laø soá löôïng phaàn töû thuoäc Z26 vaø nguyeân toá cuøng nhau vôùi 26.
Ñònh lyù 1.2: Neáu ∏=
=m
i
ei
ipn1
vôùi pi laø caùc soá nguyeân toá khaùc nhau vaø ei ∈ Z+, 1 ≤ i ≤ m thì
( ) ( )∏=
−−=m
i
ei
ei
ii ppn1
1φ
Trong phöông phaùp maõ hoùa Affine , ta coù 26 khaû naêng choïn giaù trò b, φ(26) khaû naêng choïn giaù trò a. Vaäy, khoâng gian khoùa K coù taát caû nφ(26) phaàn töû. Vaán ñeà ñaët ra cho phöông phaùp maõ hoùa Affine Cipher laø ñeå coù theå giaûi maõ ñöôïc thoâng tin ñaõ ñöôïc maõ hoùa caàn phaûi tính giaù trò phaàn töû nghòch ñaûo a–1 ∈ Z26.
f. Phöông phaùp Vigenere phöông phaùp maõ hoùa Vigenere söû duïng moät töø khoùa (keyword) coù ñoä daøi m. Coù theå xem
nhö phöông phaùp maõ hoùa Vigenere Cipher bao goàm m pheùp maõ hoùa Shift Cipher ñöôïc aùp duïng luaân phieân nhau theo chu kyø.
Khoâng gian khoùa K cuûa phöông phaùp Vigenere coù soá phaàn töû laø 26, lôùn hôn haún phöông phaùp soá löôïng phaàn töû cuûa khoâng gian khoùa K trong phöông phaùp Shift Cipher. Do ñoù, vieäc tìm ra maõ khoùa k ñeå giaûi maõ thoâng ñieäp ñaõ ñöôïc maõ hoùa seõ khoù khaên hôn ñoái vôùi phöông phaùp Shift Cipher.
Phöông phaùp maõ hoùa Vigenere Cipher Choïn soá nguyeân döông m. Ñònh nghóa P = C = K = (Z26)m
![Page 7: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/7.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
K = { (k0, k1, ..., kr-1) ∈ (Z26)r} Vôùi moãi khoùa k = (k0, k1, ..., kr-1) ∈ K, ñònh nghóa: ek(x1, x2, ..., xm) = ((x1+k1) mod 26, (x2+k2) mod n, ..., (xm+km) mod 26) dk(y1, y2, ..., ym) = ((y1–k1) mod n, (y2–k2) mod n, ..., (ym–km) mod 26) vôùi x, y ∈ (Z26)m g. Heä maõ Hill Phöông phaùp Hill Cipher ñöôïc Lester S. Hill coâng boá naêm 1929: Cho soá nguyeân döông
m, ñònh nghóa P = C = (Z26)m. Moãi phaàn töû x∈P laø moät boä m thaønh phaàn, moãi thaønh phaàn thuoäc Z26. YÙ töôûng chính cuûa phöông phaùp naøy laø söû duïng m toå hôïp tuyeán tính cuûa m thaønh phaàn trong moãi phaàn töû x∈P ñeå phaùt sinh ra m thaønh phaàn taïo thaønh phaàn töû y∈C.
Phöông phaùp maõ hoùa Hill Cipher Choïn soá nguyeân döông m. Ñònh nghóa: P = C = (Z26)m vaø K laø taäp hôïp caùc ma traän m×m khaû nghòch
Vôùi moãi khoùa K
kkk
kkkkk
k
mmmm
m
m
∈
⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜
⎝
⎛
=
,2,1,
,21,2
,12,11,1
L
MMM
LL
L
, ñònh nghóa:
( ) ( )⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜
⎝
⎛
==
mmmm
m
m
mk
kkk
kkkkk
xxxxkxe
,2,1,
,21,2
,12,11,1
21 ,...,,
L
MMM
LL
L
vôùi x=(x1, x2, ..., xm) ∈ P
vaø dk(y) = yk–1 vôùi y∈ C Moïi pheùp toaùn soá hoïc ñeàu ñöôïc thöïc hieän treân Zn h. Maõ hoaùn vò Nhöõng phöông phaùp maõ hoùa neâu treân ñeàu döïa treân yù töôûng chung: thay theá moãi kyù töï
trong thoâng ñieäp nguoàn baèng moät kyù töï khaùc ñeå taïo thaønh thoâng ñieäp ñaõ ñöôïc maõ hoùa. YÙ töôûng chính cuûa phöông phaùp maõ hoaùn vò laø vaãn giöõ nguyeân caùc kyù töï trong thoâng ñieäp nguoàn maø chæ thay ñoåi vò trí caùc kyù töï; noùi caùch khaùc thoâng ñieäp nguoàn ñöôïc maõ hoùa baèng caùch saép xeáp laïi caùc kyù töï trong ñoù.
Phöông phaùp maõ hoùa maõ hoaùn vò Choïn soá nguyeân döông m. Ñònh nghóa: P = C = (Z26)m vaø K laø taäp hôïp caùc hoaùn vò cuûa m phaàn töû {1, 2, ..., m} Vôùi moãi khoùa π ∈ K, ñònh nghóa:
![Page 8: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/8.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
( ) ( ) ( ) ( )( )mm xxxxxxe ππππ ,...,,...,, 2121 = vaø
( ) ( ) ( ) ( )( )mm yyyyyyd 111 ,...,,...,,
2121 −−−=ππππ
vôùi π–1 hoaùn vò ngöôïc cuûa π
Phöông phaùp maõ hoaùn vò chính laø moät tröôøng hôïp ñaëc bieät cuûa phöông phaùp Hill. Vôùi moãi hoaùn vò π cuûa taäp hôïp {1, 2, ..., m} , ta xaùc ñònh ma traän kπ = (ki, j ) theo coâng thöùc sau:
( )
⎩⎨⎧ =
=laïi ngöôïc hôïptröôøng trong
neáu,0,1
,ji
k jiπ
Ma traän kπ laø ma traän maø moãi doøng vaø moãi coät coù ñuùng moät phaàn töû mang giaù trò 1, caùc phaàn töû coøn laïi trong ma traän ñeàu baèng 0. Ma traän naøy coù theå thu ñöôïc baèng caùch hoaùn vò caùc haøng hay caùc coät cuûa ma traän ñôn vò Im neân kπ laø ma traän khaû nghòch. Roõ raøng, maõ hoùa baèng phöông phaùp Hill vôùi ma traän kπ hoaøn toaøn töông ñöông vôùi maõ hoùa baèng phöông phaùp maõ hoaùn vò vôùi hoaùn vò π.
d. Maõ voøng
Trong caùc heä tröôùc ñeàu cuøng moät caùch thöùc laø caùc phaàn töû keá tieáp nhau cuûa baûn roõ ñeàu ñöôïc maõ hoùa vôùi cuøng moät khoùa K. Nhö vaäy xaâu maõ y seõ coù daïng sau:
y = y1y2... = eK(x1) eK(x2)... Caùc heä maõ loaïi naøy thöôøng ñöôïc goïi laø maõ khoái (block cipher).
Coøn ñoái vôùi caùc heä maõ doøng. YÙ töôûng ôû ñaây laø sinh ra moät chuoãi khoùa z = z1z2..., vaø söû duïng noù ñeå maõ hoùa xaâu baûn roõ x = x1x2...theo qui taéc sau:
)...()(... 2121 21xexeyyy zz==
I.3 Quy trình thaùm maõ: Cöù moãi phöông phaùp maõ hoaù ta laïi coù moät phöông phaùp thaùm maõ töông öùng nhöng nguyeân taéc chung ñeå vieäc thaùm maõ ñöôïc thaønh coâng thì yeâu caàu ngöôøi thaùm maõ phaûi bieát heä maõ naøo ñöôïc duøng hoaù. Ngoaøi ra ta coøn phaûi bieát ñöôïc baûn maõ vaø baûn roõ öùng.
nhìn chung caùc heä maõ ñoái xöùng laø deã caøi ñaët vôùi toác ñoä thöïc thi nhanh. Tính an toaøn cuûa noù phuï thuoäc vaøo caùc yeáu toá :
• Khoâng gian khoaù phaûi ñuû lôùn • vôùi caùc pheùp troän thích hôïp caùc heä maõ ñoái xöùng coù theå taïo ra ñöôïc moät heä
maõ môùi coù tính an toaøn cao. • baûo maät cho vieäc truyeàn khoùa cuõng caàn ñöôïc xöû lyù moät caùch nghieâm tuùc.
Vaø moät heä maõ hoaù döõ lieäu ra ñôøi (DES). DES ñöôïc xem nhö laø chuaån maõ hoùa döõ lieäu cho caùc öùng duïng töø ngaøy 15 thaùng 1 naêm 1977 do UÛy ban Quoác gia veà Tieâu chuaån cuûa Myõ xaùc nhaän vaø cöù 5 naêm moät laàn laïi coù chænh söûa, boå sung.
DES laø moät heä maõ ñöôïc troän bôûi caùc pheùp theá vaø hoaùn vò. vôùi pheùp troän thích hôïp thì vieäc giaûi maõ noù laïi laø moät baøi toaùn khaù khoù. Ñoàng thôøi vieäc caøi ñaët heä maõ naøy cho nhöõng öùng duïng thöïc teá laïi khaù thuaän lôïi. Chính nhöõng lyù do ñoù noù ñöôïc öùng duïng roäng raõi cuûa DES trong suoát hôn 20 naêm qua, khoâng nhöõng taïi Myõ maø coøn laø haàu nhö treân khaép theá giôùi. Maëc duø theo coâng boá môùi nhaát (naêm 1998) thì moïi heä DES, vôùi nhöõng khaû naêng
![Page 9: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/9.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
cuûa maùy tính hieän nay, ñeàu coù theå beû khoùa trong hôn 2 giôø. Tuy nhieân DES cho ñeán nay vaãn laø moät moâ hình chuaån cho nhöõng öùng duïng baûo maät trong thöïc teá.
II. HEÄ MAÕ CHUAÅN DES (Data Encryption Standard)
II.1 Ñaëc taû DES Phöông phaùp DES maõ hoùa töø x coù 64 bit vôùi khoùa k coù 56 bit thaønh moät töø coù y 64 bit. Thuaät toaùn maõ hoùa bao goàm 3 giai ñoaïn: 1. Vôùi töø caàn maõ hoùa x coù ñoä daøi 64 bit, taïo ra töø x0 (cuõng coù ñoä daøi 64 bit) baèng caùch hoaùn vò caùc bit trong töø x theo moät hoaùn vò cho tröôùc IP (Initial Permutation). Bieåu dieãn x0 = IP(x) = L0R0, L0 goàm 32 bit beân traùi cuûa x0, R0 goàm 32 bit beân phaûi cuûa x0
L0 R0
x0 Hình.1 Bieåu dieãn daõy 64 bit x thaønh 2 thaønh phaàn L vaø R
2. Xaùc ñònh caùc caëp töø 32 bit Li, Ri vôùi 1≤ i ≤ 16theo quy taéc sau:
Li = Ri-1
Ri = Li-1⊕ f (Ri-1, Ki) vôùi ⊕ bieåu dieãn pheùp toaùn XOR treân hai daõy bit, K1, K2, ..., K16 laø caùc daõy 48 bit phaùt sinh töø khoùa K cho tröôùc (Treân thöïc teá, moãi khoùa Ki ñöôïc phaùt sinh baèng caùch hoaùn vò caùc bit trong khoùa K cho tröôùc).
Li-1 Ri-1
f Ki
⊕
Li Ri Hình.2 Quy trình phaùt sinh daõy 64 bit LiRi töø daõy 64 bit Li-1Ri-1vaø khoùa Ki
3. AÙp duïng hoaùn vò ngöôïc IP-1 ñoái vôùi daõy bit R16L16, thu ñöôïc töø y goàm 64 bit. Nhö vaäy, y = IP-1 (R16L16)
![Page 10: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/10.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
Haøm f ñöôïc söû duïng ôû böôùc 2 laø
A
B1 B2 B3 B4 B5 B6 B7 B8
S1
J
E(A)
S2 S3 S4 S5 S6 S7 S8
C1 C2 C3 C4 C5 C6 C7 C8
f(A,J)
E
+
P
![Page 11: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/11.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
Haøm f coù goàm 2 tham soá: Tham soá thöù nhaát A laø moät daõy 32 bit, tham soá thöù hai J laø
moät daõy 48 bit. Keát quaû cuûa haøm f laø moät daõy 32 bit. Caùc böôùc xöû lyù cuûa haøm f(A, J)nhö sau:
• Tham soá thöù nhaát A (32 bit) ñöôïc môû roäng thaønh daõy 48 bit baèng haøm môû roäng E. Keát quaû cuûa haøm E(A) laø moät daõy 48 bit ñöôïc phaùt sinh töø A baèng caùch hoaùn vò theo moät thöù töï nhaát ñònh 32 bit cuûa A, trong ñoù coù 16 bit cuûa A ñöôïc laäp laïi 2 laàn trong E(A).
• Thöïc hieän pheùp toaùn XOR cho 2 daõy 48 bit E(A) vaø J, ta thu ñöôïc moät daõy 48 bit B. Bieåu dieãn B thaønh töøng nhoùm 6 bit nhö sau:B = B1B2B3B4B5B6B7B8
• Söû duïng 8 ma traän S1, S2,..., S8, moãi ma traän Si coù kích thöôùc 4×16 vaø moãi doøng cuûa ma traän nhaän ñuû 16 giaù trò töø 0 ñeán 15. Xeùt daõy goàm 6 bit Bj = b1b2b3b4b5b6, Sj(Bj) ñöôïc xaùc ñònh baèng giaù trò cuûa phaàn töû taïi doøng r coät c cuûa Sj, trong ñoù, chæ soá doøng r coù bieåu dieãn nhò phaân laø b1b6, chæ soá coät c coù bieåu dieãn nhò phaân laø b2b3b4b5. Baèng caùch naøy, ta xaùc ñònh ñöôïc caùc daõy 4 bit Cj = Sj(Bj), 1 ≤ j ≤ 8.
• Taäp hôïp caùc daõy 4 bit Cj laïi. ta coù ñöôïc daõy 32 bit C = C1C2C3C4C5C6C7C8. Daõy 32 bit thu ñöôïc baèng caùch hoaùn vò C theo moät quy luaät P nhaát ñònh chính laø keát quaû cuûa haøm F(A, J) caùc haøm ñöôïc söû duïng trong DES. Hoaùn vò khôûi taïo IP seõ nhö sau:
IP 58 50 42 34 26 18 10 2 60 52 44 36 28 20 12 4 62 54 46 38 30 22 14 6 64 56 48 40 32 24 16 8 57 49 41 33 25 17 9 1 59 51 43 35 27 19 11 3 61 53 45 37 29 21 13 5 63 55 47 39 31 23 15 7
Ñieàu naøy coù nghóa laø bit thöù 58 cuûa x laø bit ñaàu tieân cuûa IP(x); bit thöù 50 cuûa x laø bit thöù hai cuûa IP(x) v.v. Hoaùn vò ngöôïc IP-1 seõ laø:
IP-1 40 39 38 37
8 7 6 5
48 47 46 45
16 15 14 13
56 55 54 53
24 23 22 21
64 63 62 61
32 31 30 29
![Page 12: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/12.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
36 35 34 33
4 3 2 1
44 43 42 41
12 11 10 9
52 51 50 49
20 19 18 17
60 59 58 57
28 27 26 25
Haøm môû roäng E ñöôïc ñaëc taû theo baûng sau:
E – baûng choïn bit 32 4 8
12 16 20 24 28
1 5 9
13 17 21 25 29
2 6 10 14 18 22 26 30
3 7 11 15 19 23 27 31
4 8 12 16 20 24 28 32
5 9 13 17 21 25 29 1
Taùm S-hoäp vaø hoaùn vò P seõ ñöôïc bieåu dieãn nhö sau:
S1 14
0 4
15
4 15
1 12
13 7
14 8
1 4 8 2
2 14 13
4
15269
1113
21
81
117
31015
5
106
1211
612
93
1211
714
593
10
9 5
10 0
0 3 5 6
780
13
S2 15
3 0
13
1 13 14
8
8 4 7
10
14 7
11 1
6 15 10
3
1124
15
38
134
414
12
912
511
7086
21
127
1310
612
12690
0 9 3 5
5 11
2 14
105
159
S3
10 13 13
1
0 7 6
10
9 0 4
13
14 9 9 0
6 3 8 6
34
159
15638
510
07
12
114
1381
15
1252
14
71412
3
1112
511
4 11 10
5
2 15 14
2
817
12
S4
![Page 13: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/13.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
7 13 10 3
13 8 6
15
14 11
9 0
3 5 0 6
0 6
12 10
61511
1
907
13
103
138
14
159
2714
8235
5121411
1115
12
12 10
2 7
4 14
8 2
1594
14
S5 2
14 4
11
12 11
2 8
4 2 1
12
1 12 11
7
7 4
10 0
107
1314
1113
72
618
13
85
156
509
15
31512
0
1510
59
1336
10
0 9 3 4
14 8 0 5
96
143
S6
12 10 9 4
1 15 14 3
10 4 15 2
15 2 5 12
9 7 2 9
2 12 8 5
6 9 12 15
8 5 3 10
0 6 7 11
13 1 0 14
3 13 4 1
4 14 10 7
14 0 1 6
7 11 13 0
5 3 11 8
11 8 6 13
S7
4 13
1 6
11 0 4
11
2 11 11 13
14 7
13 8
15 4
12 1
0934
817
10
131014
7
31410
9
123
155
9560
712
815
520
14
10 15
5 2
6 8 9 3
162
12
S8 13
1 7 2
2 15 11
1
8 13
4 14
4 8 1 7
6 10
9 4
153
1210
117
148
142
13
1012
015
956
12
36
109
141113
0
50
153
0 14
3 5
12 9 5 6
728
11
P 1629
152
321922
7121518
8271311
2028233124
330
4
2117261014
96
25
![Page 14: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/14.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
K laø xaâu coù ñoä daøi 64 bit, trong ñoù coù 56 bit duøng laøm khoùa vaø 8 bit duøng ñeå kieåm tra söï baèng nhau (ñeå phaùt hieän loãi). Caùc bit ôû caùc vò trí 8, 16, ..., 64 ñöôïc xaùc ñònh, sao cho moãi byte chöùa soá leû caùc soá 1. Vì vaäy, töøng loãi coù theå ñöôïc phaùt hieän trong moãi 8 bit. Caùc bit kieåm tra söï baèng nhau laø ñöôïc boû qua khi tính lòch khoùa.
1. Cho khoùa 64 bit K, loaïi boû caùc bit kieåm tra vaø hoaùn vò caùc bit coøn laïi cuûa K töông öùng vôùi hoaùn vò (coá ñònh) PC-1. Ta vieát PC-1(K) = C0D0, vôùi C0 bao goàm 28 bit ñaàu tieân cuûa PC-1(K) vaø D0 laø 28 bit coøn laïi.
2. Vôùi i naèm trong khoaûng töø 1 ñeán 16, ta tính Ci = LSi(Ci-1) Di = LSi(Di-1)
vaø Ki = PC-2(CiDi), LSi bieåu dieãn pheùp chuyeån chu trình (cyclic shift) sang traùi hoaëc cuûa moät hoaëc cuûa hai vò trí tuøy thuoäc vaøo trò cuûa i; ñaåy moät vò trí neáu i = 1, 2, 9 hoaëc 16 vaø ñaåy 2 vò trí trong nhöõng tröôøng hôïp coøn laïi. PC-2 laø moät hoaùn vò coá ñònh khaùc.
Vieäc tính lòch khoùa ñöôïc minh hoïa nhö hình veõ sau:
Caùc hoaùn vò PC-1 vaø PC-2 ñöôïc söû duïng trong vieäc tính lòch khoùa laø nhö sau:
PC-1
57 49 41 33 25 17 9
K
PC-1
C0 D0
C1 D1 PC-2 K1
LS1LS1
LS2 LS2
...
LS16 LS16
C16 D16 PC-2 K16
![Page 15: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/15.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
1 10 19 63
7 14 21
58 2
11 55 62
6 13
505934
75461
5
42516039465328
34435231384520
26354423303712
182736152229
4
PC-2
14 3
23 16 41 30 44 46
172819
750404942
1115122731513950
2464
2037455636
121261347333429
510
82
55485332
Baây giôø ta seõ hieån thò keát quaû vieäc tính lòch khoùa. Nhö ñaõ nhaän xeùt ôû treân, moãi
voøng söû duïng khoùa 48 bit töông öùng vôùi 48 bit trong K. Caùc thaønh phaàn trong caùc baûng sau seõ chæ ra caùc bit trong K ñöôïc söû duïng trong caùc voøng khaùc nhau.
I.2 LAÄP MAÕ DES
Ñaây laø ví duï veà vieäc laäp maõ söû duïng DES. Giaû söû ta maõ hoùa baûn roõ sau trong daïng thaäp luïc phaân (Hexadecimal)
0123456789ABCDEF söû duïng khoùa thaäp luïc phaân
133457799BBCDFF1 Khoùa trong daïng nhò phaân khoâng coù caùc bit kieåm tra seõ laø: 00010010011010010101101111001001101101111011011111111000.
Aùp duïng IP, ta nhaän ñöôïc L0 vaø R0 (trong daïng nhò phaân) :
L0
L1 = R0 = =
11001100000000001100110011111111 11110000101010101111000010101010
16 voøng laäp maõ ñöôïc theå hieän nhö sau:
E(R0) K1
E(R0) ⊕ K1 Output S-hoäp
f(R0,K1) L2 = R1
= = = = = =
011110100001010101010101011110100001010101010101 000110110000001011101111111111000111000001110010 011000010001011110111010100001100110010100100111 01011100100000101011010110010111 00100011010010101010100110111011 11101111010010100110010101000100
E(R1) = 011101011110101001010100001100001010101000001001
![Page 16: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/16.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
K2 E(R1) ⊕ K2
Output S-hoäp f(R1, K2)
L3 = R2
= = = = =
011110011010111011011001110110111100100111100101 000011000100010010001101111010110110001111101100 11111000110100000011101010101110 00111100101010111000011110100011 11001100000000010111011100001001
E(R2)
K3 E(R2) ⊕ K3
S-box output f(R2, K3)
L4 = R3
= = = = = =
111001011000000000000010101110101110100001010011 010101011111110010001010010000101100111110011001 101100000111110010001000111110000010011111001010 00100111000100001110000101101111 01001101000101100110111010110000 10100010010111000000101111110100
E(R3)
K4 E(R3) ⊕ K4
S-box output f(R3, K4)
L5 = R4
= = = = = =
010100000100001011111000000001010111111110101001 011100101010110111010110110110110011010100011101 001000101110111100101110110111100100101010110100 00100001111011011001111100111010 10111011001000110111011101001100 011101110
E(R4)
K5 E(R4) ⊕ K5 Xuaát S-hoäp
f(R4, K5) L6 = R5
= = = = = =
101110101110100100000100000000000000001000001010 011111001110110000000111111010110101001110101000 110001100000010100000011111010110101000110100010 01010000110010000011000111101011 00101000000100111010110111000011 10001010010011111010011000110111
E(R5)
K6 E(R5) ⊕ K6
S-box output f(R5, K6)
L7 = R6
= = = = = =
110001010100001001011111110100001100000110101111 011000111010010100111110010100000111101100101111 101001101110011101100001100000001011101010000000 01000001111100110100110000111101 10011110010001011100110100101100 11101001011001111100110101101001
E(R6)
K7 E(R6) ⊕ K7
S-box output f(R6, K7)
L8 = R7
= = = = = =
111101010010101100001111111001011010101101010011 111011001000010010110111111101100001100010111100 000110011010111110111000000100111011001111101111 00010000011101010100000010101101 10001100000001010001110000100111 00000110010010101011101000010000
E(R7) = 000000001100001001010101010111110100000010100000
![Page 17: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/17.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
K8 E(R7) ⊕ K8
S-box output f(R7, K8)
L9 = R8
= = = = =
111101111000101000111010110000010011101111111011 111101110100100001101111100111100111101101011011 01101100000110000111110010101110 00111100000011101000011011111001 11010101011010010100101110010000
E(R8) K9 E(R8) ⊕ K9 S-box output f(R8, K9) L10 = R9
= = = = = =
011010101010101101010010101001010111110010100001 111000001101101111101011111011011110011110000001 100010100111000010111001010010001001101100100000 00010001000011000101011101110111 00100010001101100111110001101010 00100100011111001100011001111010
E(R9)
K10 E(R9) ⊕ K10
S-box output f(R9, K10) L11 = R10
= = = = = =
000100001000001111111001011000001100001111110100 101100011111001101000111101110100100011001001111 101000010111000010111110110110101000010110111011 11011010000001000101001001110101 01100010101111001001110000100010 10110111110101011101011110110010
E(R10) K11
E(R10) ⊕ K11 S-box output
f(R10, K11) L12 = R11
= = = = = =
010110101111111010101011111010101111110110100101 001000010101111111010011110111101101001110000110 011110111010000101111000001101000010111000100011 01110011000001011101000100000001 11100001000001001111101000000010 11000101011110000011110001111000
E(R11)
K12 E(R11) ⊕ K12 S-box output
f(R11, K12) L13 = R12
011000001010101111110000000111111000001111110001 011101010111000111110101100101000110011111101001 000101011101101000000101100010111110010000011000 01111011100010110010011000110101 11000010011010001100111111101010 01110101101111010001100001011000
E(R12)
K13 E(R12)⊕ K13
S-box output f(R12, K13)
L14 = R13
= = = = = =
001110101011110111111010100011110000001011110000 100101111100010111010001111110101011101001000001 101011010111100000101011011101011011100010110001 10011010110100011000101101001111 11011101101110110010100100100010 00011000110000110001010101011010
E(R13)
K14 E(R13)⊕ K14
= = =
000011110001011000000110100010101010101011110100 010111110100001110110111111100101110011100111010 010100000101010110110001011110000100110111001110
![Page 18: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/18.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
S-box output f(R13, K14)
L15 = R14
= = =
01100100011110011001101011110001 10110111001100011000111001010101 11000010100011001001011000001101
E(R14)
K15 E(R14)⊕ K15
S-box output f(R14, K15)
L16 = R15
= = = = = =
111000000101010001011001010010101100000001011011 101111111001000110001101001111010011111100001010 010111111100010111010100011101111111111101010001 10110010111010001000110100111100 01011011100000010010011101101110 01000011010000100011001000110100
E(R15)
K16 E(R15)⊕ K16
S-box output f(R15, K16)
R16
= = = = = =
001000000110101000000100000110100100000110101000 110010110011110110001011000011100001011111110101 111010110101011110001111000101000101011001011101 10100111100000110010010000101001 11001000110000000100111110011000 00001010010011001101100110010101
Cuoái cuøng, aùp duïng IP-1 cho R16L16 ta nhaän ñöôïc baûn maõ trong daïng thaäp luïc phaân nhö sau:
85E813540F0AB405
I. 3 THAÙM MAÕ DES
Moät phöông phaùp raát noåi tieáng trong thaùm maõ DES laø “thaùm maõ vi sai“ (differential cryptanalysic) do Biham vaø Shamir ñeà xuaát. Ñoù laø phöông phaùp thaùm vôùi baûn roõ ñöôïc choïn. Noù khoâng ñöôïc söû duïng trong thöïc teá ñeå thaùm maõ DES 16 voøng, maø chæ ñöôïc söû duïng ñeå thaùm caùc heä DES coù ít voøng hôn.
Baây giôø ta seõ moâ taû nhöõng yù töôûng cô baûn cuûa kyõ thuaät naøy. Ñeå ñaït muïc ñích thaùm maõ, ta coù theå boû qua hoaùn vò khôûi taïo IP vaø hoaùn vò ñaûo cuûa noù (bôûi vì ñieàu ñoù khoâng caàn thieát cho vieäc thaùm maõ). Nhö ñaõ nhaän xeùt ôû treân, ta xeùt caùc heä DES n voøng, vôùi n ≤ 16. Trong caøi ñaët ta coù theå coi L0R0 laø baûn roõ vaø LnRn nhö laø baûn maõ.
Thaùm maõ vi sai ñoøi hoûi phaûi so saùnh x-or (exclusive-or) cuûa hai baûn roõ vôùi x-or cuûa hai baûn maõ töông öùng. Noùi chung, ta seõ quan saùt hai baûn roõ L0R0 vaø L0
*R0* vôùi trò x-or
ñöôïc ñaëc taû L0’R0’ = L0R0 ⊕ L0*R0
*. Trong nhöõng thaûo luaän sau ta seõ söû duïng kyù hieäu (‘) ñeå chæ x-or cuûa hai xaâu bit.
Ñònh nghóa 3.1: Cho Sj laø moät S-hoäp (1 ≤ j ≤ 8). Xeùt moät caëp xaâu 6-bit laø (Bj,Bj*
). Ta noùi raèng, xaâu nhaäp x-or (cuûa Sj) laø Bj ⊕ Bj
* vaø xaâu xuaát x-or (cuûa Sj) laø Sj(Bj) ⊕ Sj(Bj*).
Chuù yù laø xaâu nhaäp x-or laø xaâu bit coù ñoä daøi 6, coøn xaâu xuaát x-or coù ñoä daøi 4.
Ñònh nghóa 3.2: Vôùi baát kyø Bj ’ ∈ (Z2) 6, ta ñònh nghóa taäp Δ(Bj’) goàm caùc caëp (Bj,Bj
*) coù x-or nhaäp laø Bj’. Deã daøng thaáy raèng, baát kyø taäp Δ(Bj’) naøo cuõng coù 26 = 64 caëp, vaø do ñoù
![Page 19: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/19.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
Δ(Bj’) = {(Bj, Bj ⊕ Bj’) : Bj ∈ (Z2) 6 }
Vôùi moãi caëp trong Δ(Bj’), ta coù theå tính xaâu x-or xuaát cuûa Sj vaø laäp ñöôïc phaân boá
keát quaû. Coù 64 xaâu xuaát x-or, ñöôïc phaân boá trong 24 = 16 giaù trò coù theå coù. Tính khoâng ñoàng ñeàu cuûa caùc phaân boá ñoù laø cô sôû ñeå maõ thaùm.
Ví duï 3.1: Giaû söû ta xeùt S1 laø S-hoäp ñaàu tieân vaø xaâu nhaäp x-or laø 110100. Khi ñoù Δ(110100) = {(000000, 110100), (000001, 110101), ..., (111111, 001011)} Vôùi moãi caëp trong taäp Δ(110100), ta tính xaâu xuaát x-or cuûa S1. Chaúng haïn,
S1(000000) = E16 = 1110, S1(110100) = 1001, nhö vaäy xaâu xuaát x-or cho caëp (000000,110100) laø 0111.
Neáu thöïc hieän ñieàu ñoù cho 64 caëp trong Δ(110100) thì ta nhaän ñöôïc phaân boá cuûa caùc xaâu x-or xuaát sau:
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
0 8 16 6 2 0 0 12 6 0 0 0 0 8 0 6
Trong ví duï 3.1, chæ coù 8 trong soá 16 xaâu x-or xuaát coù theå coù xuaát hieän thaät söï. Ví duï cuï theå naøy ñaõ chæ ra söï phaân boá raát khoâng ñeàu cuûa caùc xaâu x-or xuaát. Noùi chung, neáu ta coá ñònh S-hoäp Sj vaø xaâu nhaäp x-or Bj’, thì trung bình coù khoaûng 75 - 80% caùc xaâu x-or xuaát coù theå coù xuaát hieän thöïc söï.
Ñeå moâ taû caùc phaân boâ ñoù ta ñöa ra ñònh nghóa sau.
Ñònh nghóa 3.3: Vôùi 1 ≤ j ≤ 8 vaø vôùi caùc xaâu bit Bj’ ñoä daøi 6 vaø Cj’ ñoä daøi 4, ta ñònh nghóa:
INj(Bj’,Cj’) = {Bj ∈ (Z2)6 : Sj(Bj) ⊕ Sj(Bj ⊕ Bj’) = Cj’} vaø
Nj(Bj’, Cj’) = ⎮INj(Bj’, Cj’)⎮
Baûng sau seõ cho caùc xaâu nhaäp coù theå coù vôùi xaâu x-or nhaäp 110100
Xaâu xuaát x-or Caùc xaâu nhaäp coù theå coù 0000 0001 000011, 001111, 011110, 011111
101010, 101011, 110111, 111011
0010 000100, 000101, 001110, 010001 010010, 010100, 011010, 011011 100000, 100101, 010110, 101110 101111, 110000, 110001, 111010
0011 000001, 000010, 010101, 100001 110101, 110110
0100 010011, 100111
![Page 20: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/20.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
0101 0110
0111
000000, 001000, 001101, 010111 011000, 011101, 100011, 101001 101100, 110100, 111001, 111100
1000 001001, 001100, 011001, 101101 111000, 111101
1001 1010 1011 1100 1101 000110, 010000, 010110, 011100
110010, 100100, 101000, 110010 1110 1111 000111, 001010, 001011, 110011
111110, 111111 Nj(Bj’, Cj’) tính soá caùc caëp vôùi xaâu nhaäp x-or baèng Bj’ coù xaâu xuaát x-or baèng Cj’
vôùi S-hoäp Sj. Caùc caëp ñoù coù caùc xaâu nhaäp x-or ñöôïc ñaëc taû vaø ñöa ra caùch tính caùc xaâu xuaát x-or coù theå nhaän ñöôïc töø taäp INj(Bj’, Cj’). Ñeå yù raèng, taäp naøy coù theå phaân thaønh Nj(Bj’, Cj’) /2 caëp, moãi caëp coù xaâu x-or nhaäp baèng Bj’.
Phaân boá trong ví duï 3.1 chöùa caùc trò N1(110100, C1’), C1’∈ (Z2)4. Trong baûng treân chöùa caùc taäp IN(110100, C1’).
Vôùi moãi taùm S-hoäp, coù 64 xaâu nhaäp x-or coù theå coù. Nhö vaäy, coù 512 phaân boá coù theå tính ñöôïc. Nhaéc laïi laø, xaâu nhaäp cho S-hoäp ôû voøng thöù i laø B= E⊕ J, vôùi E = E(Ri-1) laø môû roäng cuûa Ri-1 vaø J = Ki goàm caùc bit khoùa cuûa voøng i. Baây giôø xaâu nhaäp x-or (cho taát caû taùm S-hoäp) coù theå tính ñöôïc nhö sau:
B ⊕ B* = (E ⊕ J) ⊕ (E* ⊕ J) = E ⊕ E* Ñieàu naøy raát quan troïng ñeå thaáy raèng, xaâu nhaäp x-or khoâng phuï thuoäc vaøo caùc bit
khoùa J. (Do ñoù, xaâu xuaát x-or cuõng khoâng phuï thuoäc vaøo caùc bit khoùa.)
Ta seõ vieát moãi B, E vaø J nhö laø noái cuûa taùm xaâu 6-bit:
B = B1B2B3B4B5B6B7B8 E = E1E2E3E4E5E6E7E8
J = J1J2J3J4J5J6J7J8 vaø ta cuõng seõ vieát B* vaø E* nhö vaäy. Baây giôø giaû söû laø ta ñaõ bieát caùc trò Ej vaø Ej
* vôùi moät j naøo ñoù, 1 ≤ j ≤ 8, vaø trò cuûa xaâu xuaát x-or cho Sj, Cj’ = Sj(Bj) ⊕ Sj(Bj
* ). Khi ñoù seõ laø: Ej ⊕ Jj ∈ INj(Ej’, Cj’),
vôùi Ej’ = Ej ⊕ Ej*.
Ñònh nghóa 3.4: Giaû söû Ej vaø Ej* laø caùc xaâu bit ñoä daøi 6, vaø Cj’ laø xaâu bit ñoä daøi 4. Ta ñònh
nghóa:
testj(Ej, Ej*, Cj’) = { Bj ⊕ Ej : Bj ∈ INj(Ej’, Cj’) },
vôùi Ej’ = Ej ⊕ Ej*.
![Page 21: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/21.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
Ñònh lyù 3.1: Giaû söû Ej vaø Ej
* laø hai xaâu nhaäp cho S-hoäp Sj, vaø xaâu xuaát x-or cho Sj laø Cj’. Kyù hieäu Ej’ = Ej ⊕ Ej
* . Khi ñoù caùc bit khoùa Jj coù trong taäp testj(Ej, Ej*, Cj’). Ñeå yù, ñoù chính laø caùc xaâu bit Nj(Ej’, Cj’) ñoä daøi 6 trong taäp testj(Ej, Ej
*, Cj’); giaù trò chính xaùc cuûa Jj phaûi laø moät trong soá ñoù. Ví duï 3.2: Giaû söû E1 = 000001, E1
*= 110101 vaø C1’= 1101. Do ñoù N1(110101,1101) = 8, ñuùng baèng 8 xaâu bit trong taäp test1(000001, 110101, 1101). Töø baûng treân ta thaáy raèng IN1(110100, 1101) = {000110, 010000, 010110, 011100, 100010, 100100, 101000, 110010} Cho neân test1(000001, 110101,1101) = {000111, 010001, 010111, 011101, 100011, 100101, 101001, 110011}
Neáu ta coù moät boä ba thöù hai nhö theá E1, E1*, C1’, khi ñoù ta seõ nhaän ñöôïc taäp thöù
hai test1 cuûa caùc trò cho caùc bit khoùa trong J1. Trò ñuùng cuûa J1 caàn phaûi naèm trong giao cuûa caùc S-hoäp. Neáu ta coù moät vaøi boä ba nhö vaäy, khi ñoù ta coù theå mau choùng tìm ñöôïc caùc bit khoùa trong J1. Moät caùch roõ raøng hôn ñeå thöïc hieän ñieàu ñoù laø laäp moät baûng cuûa 64 boä ñeám bieåu dieãn cho 64 khaû naêng cuûa cuûa 6 khoùa bit trong J1. Boä ñeám seõ taêng moãi laàn, töông öùng vôùi söï xuaát hieän cuûa caùc bit khoùa trong taäp test1 cho moät boä ba cuï theå. Cho t boä ba, ta hy voïng tìm ñöôïc duy nhaát moät boä ñeám coù trò t; trò ñoù seõ töông öùng vôùi trò ñuùng cuûa caùc bit khoùa trong J1.
I.3.1. Thaùm maõ heä DES - 3 voøng
Baây giôø ta seõ xeùt yù töôûng vöøa trình baøy cho vieäc thaùm maõ heä DES - ba voøng. Ta seõ baét ñaàu vôùi caëp baûn roõ vaø caùc baûn maõ töông öùng: L0R0, L0
*R0*, L3R3 vaø L3
*R3*. Ta coù theå
bieåu dieãn R3 nhö sau: R3 = L2 ⊕ f(R2, K3) = R1 ⊕ f(R2, K3) = L0 ⊕ f(R0, K1) ⊕ f(R2, K3)
R3* coù theå bieåu dieãn moät caùch töông töï , do vaäy: R3’ = L0’ ⊕ f(R0, K1) ⊕ f(R0
*, K1) ⊕ f(R2, K3) ⊕ f(R2*, K3)
Baây giôø, giaû söû ta ñaõ choïn ñöôïc caùc baûn roõ sao cho R0 = R0*, chaúng haïn:
R0’ = 00...0 Khi ñoù f(R0, K1) = f(R0
*, K1), vaø do ñoù: R3’ = L0’⊕ f(R2, K3) ⊕ f(R2
*, K3) ÔÛ ñieåm naøy R3’ laø ñöôïc bieát khi noù coù theå tính ñöôïc töø hai baûn maõ, vaø L0’ laø bieát
ñöôïc khi noù coù theå tính ñöôïc töø hai baûn roõ. Nghóa laø ta coù theå tính ñöôïc f(R2,K3)⊕f(R2*,K3)
töø phöông trình: f(R2, K3) ⊕ f(R2
*, K3) = R3’ ⊕ L0’ Baây giôø f(R2, K3) = P(C) vaø f(R2
*, K3) = P(C*), vôùi C vaø C* töông öùng laø kyù hieäu cuûa hai xaâu xuaát cuûa taùm S-hoäp (nhaéc laïi, P laø coá ñònh, laø hoaùn vò ñöôïc bieát coâng khai). Neân:
![Page 22: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/22.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
P(C) ⊕ P(C*) = R3’ ⊕ L0’ vaø keát quaû laø:
C’ = C ⊕ C* = P-1(R3’ ⊕ L0’) (1)
Ñoù laø xaâu xuaát x-or cho taùm S-hoäp trong voøng ba.
Baây giôø, R2 = L3 vaø R2* = L3* laø ñaõ bieát (chuùng laø moät phaàn cuûa caùc baûn maõ). Töø ñaây ta coù theå tính:
E = E(L3) (2) vaø E* = E(L3
*) (3) söû duïng haøm môû roäng E ñöôïc bieát coâng khai. Chuùng laø nhöõng xaâu nhaäp cho caùc S-hoäp cho voøng ba. Nhö vaäy giôø ta ñaõ bieát E, E*, vaø C’ cho voøng ba vaø ta coù theå tieáp tuïc xaây döïng caùc taäp test1, ..., test8 cuûa caùc trò coù theå coù cho caùc bit khoùa trong J1, ..., J8.
Giaûi thuaät vöøa xeùt coù theå bieåu dieãn bôûi caùc maõ sau:
Input: L0R0, L0*R0
*, L3R3 vaø L3*R3
*, vôùi R0 = R0*
1. Tính C’ = P-1(R3’ ⊕ L0’) 2. Tính E = E(L3) vaø E* = E(L*) 3. for j = 1 to 8 do
compute testj(Ej, Ej*, Cj’)
Vieäc maõ thaùm seõ söû duïng moät soá boä ba E, E*, C’ nhö vaäy. Ta seõ laäp taùm baûng caùc
boä ñeám vaø do ñoù xaùc ñònh ñöôïc 48 bit trong K3, laø khoùa cho voøng ba. 56 bit trong khoùa khi ñoù coù theå tìm ñöôïc hoaøn toaøn töø 28 = 256 khaû naêng cho 8 bit khoùa.
Baây giôø ta seõ minh hoïa ñieàu ñoù qua ví duï sau. Ví duï 3.3
Giaû söû ta coù ba caëp baûn roõ vaø baûn maõ, vôùi caùc baûn maõ cuøng coù caùc xaâu x-or ñöôïc maõ hoùa bôûi cuøng moät khoùa. Ñeå ngaén goïn ta söû duïng heä thaäp luïc phaân:
Baûn roõ Baûn maõ
748502CD38451097 3874756438451097
03C70306D8A09F10 78560A0960E6D4CB
486911026ACDFF31 375BD31F6ACDFF31
45FA285BE5ADC730 134F7915AC253457
357418DA013FEC86 12549847013FEC86
D8A31B2F28BBC5CF 0F317AC2B23CB944
Töø caëp ñaàu tieân ta tính caùc xaâu nhaäp cuûa S-hoäp (cho voøng 3) töø caùc phöông trình
(2) vaø (3). Chuùng laø: E = 000000000111111000001110100000000110100000001100 E* = 101111110000001010101100000001010100000001010010 Xaâu xuaát x-or cuûa S-hoäp ñöôïc tính baèng phöông trình (1) seõ laø:
C’ = 10010110010111010101101101100111
![Page 23: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/23.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
Töø caëp thöù hai, ta tính ñöôïc caùc xaâu nhaäp cho S-hoäp laø: E = 101000001011111111110100000101010000001011110110 E* = 100010100110101001011110101111110010100010101001
vaø xaâu xuaát x-or cuûa S-hoäp laø: C’ = 10011100100111000001111101010110
Töø caëp thöù ba, caùc xaâu nhaäp cho S-hoäp seõ laø:
E = 111011110001010100000110100011110110100101011111 E* = 000001011110100110100010101111110101011000000100
vaø xaâu xuaát x-or cuûa S-hoäp laø: C’ = 11010101011101011101101100101011
Tieáp theo, ta laäp baûng caùc trò trong taùm maûng boä ñeám cho moãi caëp. Ta seõ minh hoïa thuû tuïc vôùi caùc maûng ñeám cho J1 töø caëp ñaàu tieân. Trong caëp naøy, ta coù E1’= 101111 vaø C1’ = 1001. Taäp:
IN1(101111, 1001) = {000000, 000111, 101000, 101111} Do E1 = 000000 ta coù:
J1 ∈ test1(000000, 101111, 1001) = {000000, 000111, 101000, 101111} Do ñoù ta taêng caùc trò 0, 7, 40 vaø 47 trong caùc maûng ñeám cho J1.
Cuoái cuøng ta seõ trình baøy caùc baûng. Neáu ta xem caùc xaâu bit ñoä daøi 6 nhö laø bieåu dieãn cuûa caùc soá nguyeân trong khoaûng 0-63, thì 64 trò seõ töông öùng vôùi 0, 1, ..., 63. Caùc maûng ñeám seõ laø nhö sau:
J1
1 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 1 1 0 0 0 0 1 0 0 0 1 0 0 1 0 0 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
J2
0 0 0 1 0 3 0 0 1 0 0 1 0 0 0 0 0 1 0 0 0 2 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 1 0 1 0 0 0 1 0 0 0 1 1 0 0 0 0 1 0 1 0 2 0 0 0
J3
0 0 0 0 1 1 0 0 0 0 0 0 0 0 1 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0 1 1 0 2 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0
J4
3 1 0 0 0 0 0 0 0 0 2 2 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 0 1 1
![Page 24: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/24.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
1 1 1 0 1 0 0 0 0 1 1 1 0 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 2 1
J5
0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 2 0 0 0 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 1 0 0 0 0 2 0
J6
1 0 0 1 1 0 0 3 0 0 0 0 1 0 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 1 0 0 0 0 0 0 0 0 1 0 0 1 1 0 1 1 0 0 0 0 0 0 0 0
J7 0 0 2 1 0 1 0 3 0 0 0 1 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 2 0 0 0 2 0 0 0 0 1 2 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1
J8
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 1 0 1 0 3 0 0 0 0 1 0 0 0 0 0 0 0 0 0
Trong moãi taùm maûng ñeám, coù duy nhaát moät boä ñeám coù trò laø 3. Vò trí cuûa caùc boä
ñeám ñoù xaùc ñònh caùc bit khoùa trong J1, ..., J8. Caùc vò trí ñoù laø: 47, 5, 19, 0, 24, 7, 7, 49. Chuyeån caùc soá nguyeân ñoù sang daïng nhò phaân, ta nhaän ñöôïc J1, ..., J8:
J1 = 101111 J2 = 000101 J3 = 010011 J4 = 000000 J5 = 011000 J6 = 000111 J7 = 000111 J8 = 110001
Baây giôø ta coù theå taïo ra 48 bit khoùa, baèng caùch quan saùt lòch khoùa cho voøng ba. Suy ra laø K coù daïng:
0001101 0110001 01?01?0 1?00100 0101001 0000??0 111?11? ?100011
vôùi caùc bit kieåm tra ñaõ ñöôïc loaïi boû vaø “?” kyù hieäu bit khoùa chöa bieát. Khoùa ñaày ñuû (trong daïng thaäp luïc phaân, goàm caû bit kieåm tra) seõ laø:
![Page 25: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/25.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
1A624C89520DEC46 II.3.2. Thaùm maõ heä DES 6-voøng
Baây giôø ta seõ moâ taû vieäc môû roäng yù töôûng treân cho vieäc thaùm maõ treân heä DES 6-voøng. YÙ töoûng ôû ñaây laø löïa choïn moät caùch caån thaän caëp baûn roõ vôùi xaâu x-or ñaëc thuø vaø sau ñoù xaùc ñònh caùc xaùc suaát cuûa caùc daõy ñaëc thuø cuûa caùc xaâu x-or qua caùc voøng laäp maõ. Baây giôø ta caàn ñònh nghóa moät khaùi nieäm quan troïng sau.
Ñònh nghóa 3.5: Cho n ≥ 1 laø soá nguyeân. Ñaëc tröng cuûa voøng thöù n laø moät danh saùch caùc daïng
L0’, R0’, L1’, R1’, p1, ..., Ln’, Rn’, pn thoûa maõn caùc ñieàu kieän sau: 1. Li’ = Ri-1’ vôùi 1 ≤ i ≤ n 2. Cho 1 ≤ i ≤ n vaø Li-1, Ri-1 vaø L*
i-1, R*i-1 laø ñaõ ñöôïc choïn sao cho Li-1 ⊕ L*
i-1 = L’i-1 vaø Ri-1 ⊕ R*
i-1 = R’i-1. Giaû söû Li, Ri vaø Li* , Ri
* laø tính ñöôïc nhôø vieäc aùp duïng moät voøng laäp maõ DES. Khi ñoù xaùc suaát ñeå Li ⊕ L*
i = Li’ vaø Ri ⊕ R*i = Ri’ chính xaùc baèng pi.
(Chuù yù laø, xaùc suaát naøy ñöôïc tính treân taát caû caùc boä coù theå coù cuûa J = J1...J8) . Xaùc suaát ñaëc tröng ñöôïc ñònh nghóa baèng tích p = p1 × ...× pn.
Nhaän xeùt: Giaû söû ta choïn L0, R0 vaø L0
*, R0* sao cho L0 ⊕ L0
* = L0’ vaø R0 ⊕ R0*= R0’ vaø
ta aùp duïng n voøng laäp maõ cuûa DES, nhaän ñöôïc L1. ..., Ln vaø R1, ..., Rn. Khi ñoù ta khoâng theå ñoøi hoûi xaùc suaát ñeå Li ⊕ Li
* = Li’ vaø Ri ⊕ Ri* = Ri’ cho taát caû i ( 1 ≤ i ≤ n) laø p1 × ...× pn.
Bôûi vì caùc boä -48 trong lòch khoùa K1, ..., Kn khoâng phaûi laø ñoäc laäp laãn nhau. (Neáu n boä-48 naøy ñuôïc choïn ñoäc laäp moät caùch ngaãu nhieân, thì ñieàu xaùc nhaän laø ñuùng). Nhöng ta seõ coi raèng p1 × ... × pn chính xaùc laø xaùc xuaát ñoù.
Ta coøn caàn xaùc nhaän laø, caùc xaùc suaát pi trong ñaëc tröng laø caùc caëp baûn roõ ñöôïc xaùc ñònh tuøy yù (nhöng coá ñònh) ñöôïc ñaëc taû baèng xaâu x-or, vôùi 48 bit khoùa cho moät voøng laäp maõ DES laø coù 248 khaû naêng. Do ñoù vieäc thaùm maõ seõ nhaèm vaøo vieäc xaùc ñònh khoùa coá ñònh (nhöng chöa bieát). Do ñoù caàn coá choïn caùc baûn maõ ngaãu nhieân (nhöng chuùng coù caùc xaâu x-or ñöôïc ñaëc taû), hy voïng raèng caùc xaùc suaát ñeå caùc xaâu x-or trong n voøng laäp maõ truøng hôïp vôùi caùc xaâu x-or, ñöôïc ñaëc taû trong ñaëc tröng, töøng ñoâi moät p1, ..., pn töông öùng.
Trong ví duï sau ñaây, ta seõ trình baøy ñaëc tröng voøng 1 ñeå laøm cô sôû cho vieäc thaùm maõ DES ba voøng trong hình sau (nhö ôû treân, ta seõ söû duïng caùch bieåu dieãn theo heä thaäp luïc phaân).
L’0 = baát kyø R’0 = 0000000016 L’1 = 0000000016 R’1 = L’0 p = 1
Ta cuõng seõ moâ taû moät ñaëc tröng voøng 1 khaùc nhö sau
L’0 = 0000000016 R’0 = 6000000016 L’1 = 6000000016 R’1 = 0080820016 p = 14/64
![Page 26: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/26.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
Ta haõy xeùt ñaëc tröng sau moät caùch chi tieát hôn. Khi f(R0, K1) vaø f(R0*, K1) ñöôïc tính, böôùc
ñaàu tieân laø môû roäng R0 vaø R0*. Xaâu x-or keát quaû cuûa hai môû roäng laø:
001100...0 Töùc laø xaâu x-or nhaäp cho S1 laø 001100 vaø caùc xaâu x-or cho baûy S-hoäp khaùc ñeàu laø 000000. Caùc xaâu xuaát x-or cho S2 ñeán S8 ñeàu laø 0000. Xaâu xuaát x-or cho S1 laø 1110 vôùi xaùc suaát 14/64 (do N1(001100, 1110) = 14). Neân ta nhaän ñöôïc:
C’ = 11100000000000000000000000000000 vôùi xaùc suaát 14/64. Aùp duïng P, ta nhaän ñöôïc:
P(C) ⊕ P(C*) = 00000000100000001000001000000000 trong daïng thaäp luïc phaân seõ laø 0080820016. Khi xaâu naøy coäng x-or vôùi L0’, ta nhaän ñöôïc R1’ vôùi xaùc suaát 14/64. Do ñoù L1’ = R0’.
Vieäc thaùm maõ DES saùu voøng döïa treân ñaëc tröng ba voøng ñöôïc cho trong hình sau. Trong thaùm maõ 6-voøng, ta baét ñaàu vôùi L0R0. L0
*R0*, L6R6 vaø L6
*R6*, maø ta phaûi choïn baûn
roõ sao cho L0’= 4008000016 vaø R.0’= 0400000016, ta coù theå bieåu dieãn R0 nhö sau:
L0’ L1’ L2’ L3’
= = = =
4008000016
0400000016 0000000016 0400000016
R0’R1’R2’R3’
= = = =
0400000016 0000000016 0400000016 4008000016
p = 1/4 p = 1 p = 1/4
R6 = L5 ⊕ f(R5, K6) = R4 ⊕ f(R5, K6) = L3 ⊕ f(R3, K4) ⊕ f(R5, K6)
R6* cuõng coù theå bieåu dieãn töông töï, ta coù
R0’ = L3’ ⊕ f(R3, K4) ⊕ f(R3*, K4) ⊕ f(R5, K6) ⊕ f(R5
*, K6) (4) (Ñeå yù laø töông töï nhö thaùm maõ 3-voøng) R6’ laø ñöôïc bieát. Töø ñaëc tröng ta tính L3’ = 0400000016 vaø R3’ = 4008000016 vôùi xaùc suaát 1/16. Neáu nhö vaäy, thì xaâu nhaäp x-or cho S-hoäp trong voøng 4 coù theå tính ñöôïc nhôø haøm môû roäng phaûi laø:
001000000000000001010000...0 Caùc xaâu x-or cho S2, S5, S6, S7 vaø S8 taát caû ñeàu baèng 000000, vaø vì theá xaâu xuaát x-or laø 0000 cho taát caû naêm S-hoäp ñoù trong voøng 4. Ñieàu naøy coù nghóa laø, ta coù theå tính ñöôïc caùc xaâu xuaát x-or cho naêm S-hoäp ñoù trong voøng 6 nhôø phöông trình (4). Do ñoù giaû söû ta tính:
C1’C2’C3’C4’C5’C6’C7’C8’ = P-1(R6’ ⊕ 04000000) moãi Ci laø xaâu bit coù ñoä daøi 4. Khi ñoù vôùi xaùc suaát 1/16, thì seõ daãn ñeán laø C2’, C5’, C6’, C7’ vaø C8’ töông öùng laø caùc xaâu x-or xuaát cuûa S2, S5, S6, S7 vaø S8 trong voøng 6. Caùc xaâu nhaäp cho caùc S-hoäp ñoù trong voøng 6 coù theå tính ñöôïc laø E2, E5, E6, E7 vaø E8; vaø E2
*, E5*, E6
*, E7*
vaø E8*, vôùi
E1E2E3E4E5E6E7E8 = E(R5) = E(L6) vaø E1
*E2*E3
*E4*E5
*E6*E7
*E8* = E(R5
*) = E(L6*)
coù theå tính ñöôïc töø caùc baûn roõ nhö sau:
![Page 27: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/27.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
Input: L0R0, L0*R0
*, L6R6 vaø L6*R6
*; vôùi L0’ = 4008000016 vaø R0’ = 0400000016. 1. Tính C’ = P-1(R6’ ⊕ 0400000016) 2. Tính E = E(L6) vaø E* = E(L6
*) 3. for j ∈ {2,5,6,7,8} do tính testj( Ej, Ej*, Cj’)
Ta cuõng seõ xaùc ñònh 30 bit khoùa trong J2, J5, J6, J7 vaø J8 nhö trong thaùm maõ 3-voøng.
Baøi toaùn, ñeå xaâu xuaát x-or giaû ñònh cho voøng 6 laø chính xaùc chæ vôùi xaùc suaát 1/16. Coøn 15/16 phaàn coøn laïi ta seõ thöôøng nhaän ñöôïc nhöõng xaâu voâ duïng ngaãu nhieân hôn laø caùc bit khoùa.
Ñònh nghóa 3.6: Giaû söû L0 ⊕ L0
* = L0’ vaø R0 ⊕ R0*= R0’. Ta noùi raèng, caëp baûn roõ L0R0 vaø
L0* R0
* laø ñuùng (right) öùng vôùi ñaëc tröng neáu Li ⊕ Li* = Li’ vaø Ri ⊕ Ri
*= Ri’ cho moïi i, 1 ≤ i ≤ n. Caëp traùi vôùi caëp ñöôïc ñònh nghóa goïi laø caëp sai (wrong).
Ta mong raèng, khoaûng 1/16 soá caëp cuûa ta laø ñuùng, coøn caùc caëp coøn laïi laø caëp sai öùng vôùi ñaëc tröng voøng ba cuûa ta.
Chieán löôïc cuûa ta laø tính Ej. Ej* vaø Cj’nhö ñaõ moâ taû ôû treân vaø sau ñoù xaùc ñònh testj(Ej, Ej
*, Cj’) vôùi j = 2,5,6,7,8. Neáu ta baét ñaàu vôùi moät caëp ñuùng, thì thì caùc bit khoùa chính xaùc cho moãi Jj seõ naèm trong taäp testj. Neáu caëp laø sai, thì trò Cj’ seõ khoâng ñuùng, vaø ñoù laø nguyeân do ñeå giaû ñònh raèng, moãi taäp testj thöïc chaát laø ngaãu nhieân.
Ta coù theå nhaän ra caëp ñuùng theo phöông phaùp sau: Neáu ⎮testj⎮= 0, vôùi baát kyø j∈ {2,5,6,7,8}, khi ñoù ta taát yeáu coù ñöôïc caëp ñuùng. Baây giôø cho moät caëp sai, ta coù theå hy voïng raèng, xaùc suaát ñeå ⎪testj⎪= 0 cho moät j cuï theå laø xaáp xæ 1/5. Ñoù laø lyù do ñeå giaû ñònh laø, Nj(Ej’, Cj’) = ⎪testj⎪ vaø nhö ñaõ nhaän xeùt töø tröôùc, xaùc suaát ñeå Nj(Ej’, Cj’) = 0 laø xaáp xæ 1/5. Xaùc suaát ñeå caû naêm testj ñeàu döông laø vaøo khoaûng 0.85 ≈ 0.33, quaû vaäy xaùc suaát ñeå ít nhaát moät testj baèng 0 laø vaøo khoaûng 0.67. Neân ta coù khoaûng 2/3 soá caëp laø sai, nhôø vaøo moät nhaän xeùt ñôn giaûn, ñöôïc goïi laø pheùp loïc (filtering operation). Tyû soá cuûa caùc caëp ñuùng treân caùc caëp coøn laïi sau pheùp loïc laø vaøo khoaûng:
61311615161
161=
×+
Ví duï 3.4: Giaû söû ta coù caëp baûn roõ - baûn maõ sau:
Baûn roõ Baûn maõ 86FA1C2B1F51D3BE C6F21C2B1B51D3BE
1E23ED7F2F553971 296DE2B687AC6340
Chuù yù laø, L0’ = 4008000016 vaø R0’ = 0400000016. Xaâu nhaäp vaø xaâu xuaát cuûa S-hoäp cho voøng 6 ñöôïc tính nhö sau:
![Page 28: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/28.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
j Ej Ej
* Cj’ 2 5 6 7 8
111100 111101 011010 101111 111110
010010 111100 000101 010110 101100
1101 0001 0010 1100 1101
Khi ñoù caùc taäp testj seõ laø nhö sau:
j testj 2 14, 15,26, 30, 32, 33, 48, 52 5 6 7, 24, 36, 41, 54, 59 7 8 34, 35, 48, 49
Ta thaáy raèng, hai taäp test5 vaø test7 laø roãng , neân caëp naøy laø caëp sai vaø noù bò loaïi boû baèng pheùp loïc. Baây giôø giaû söû ta coù caëp sao cho ⎪testj⎪> 0 vôùi j = 2,5,6,7,8 laø nhöõng taäp coøn laïi sau pheùp loïc.(Bôûi vì ta khoâng bieát ñöôïc laø caëp naøo ñuùng, caëp naøo sai.) Ta noùi raèng, xaâu bit J2J5J6J7J8 ñoä daøi 30 laø ñöôïc ñeà xuaát bôûi caëp neáu Jj ∈ testj vôùi j = 2,5,6,7,8. Soá caùc caëp ñöôïc ñeà xuaát laø:
∏∈ 8,7,6,5,2j
jtest
Ñoù laø bình thöôøng vôùi soá xaâu bit ñöôïc ñeà xuaát laø khaù lôùn. (Chaúng haïn. lôùn hôn
80000) Giaû söû, ta laäp baûng cho taát caû caùc xaâu ñöôïc ñeà xuaát nhaän ñöôïc töø N caëp, maø khoâng bò loaïi bôûi pheùp loïc. Vôùi moãi caëp ñuùng, thì xaâu bit ñuùng J2J5J6J7J8 seõ laø xaâu ñöôïc ñeà xuaát. Xaâu bit ñuùng seõ ñöôïc tính khoaûng 3N/16 laàn. Xaâu bit sai thöôøng xuaát hieän ít hôn, bôûi vì chuùng xuaát hieän ngaãu nhieân vaø coù khoaûng 230 khaû naêng. (Laø moät soá raát lôùn.)
Ta nhaän ñöôïc moät baûng cöïc lôùn taát caû caùc xaâu ñöôïc ñeà xuaát, neân ta söû duïng moät thuaät toaùn chæ ñoøi hoûi moät khoâng gian vaø thôøi gian ít nhaát. Ta coù theå maõ hoùa baát kyø moät taäp testj naøo thaønh moät veùc tô Tj coù ñoä daøi 64, vôùi toïa ñoä thöù i cuûa Tj ñöôïc ñaët baèng 1 (0≤ i≤63), neáu xaâu bit ñoä daøi 6 laø bieåu dieãn cuûa i ôû trong taäp testj; vaø toïa ñoä thöù i ñöôïc ñaët baèng 0 trong tröôøng hôïp ngöôïc laïi ( ñieàu naøy gioáng nhö maûng caùc boä ñeám maø ta ñaõ söû duïng trong thaùm maõ DES ba voøng).
Vôùi moãi caëp coøn laïi, ta xaây döïng caùc veùc tô nhö treân vaø goïi chuùng laø Tji,
j=2,5,6,7,8; 1 ≤ i≤ N. Vôùi I ⊆ {1, ..., N} ta noùi raèng I laø chaáp nhaän ñöôïc (allowable) neáu vôùi moãi j ∈ {2,5,6,7,8} coù ít nhaát moät toïa ñoä baèng ⎪I⎪ trong veùc tô
![Page 29: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/29.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
∑∈ Ii
ijT
Neáu caëp thöù i laø caëp ñuùng cho moãi i∈I, thì taäp I laø chaáp nhaän ñöôïc. Do ñoù ta cho raèng taäp chaáp nhaän ñöôïc coù kích thöôùc (xaáp xæ) 3N/16, laø taäp ñeà xuaát vaø ta hy voïng laø chæ goàm caùc bit khoùa ñuùng chöù khoâng coù caùc xaâu khaùc. Ñieàu naøy laøm ñôn giaûn hoùa cho vieäc xaây döïng taát caû caùc taäp chaáp nhaän ñöôïc I baèng moät thuaät toaùn ñeä qui. II.3. 3 Caùc thaùm maõ vi sai khaùc
Phöông phaùp thaùm maõ vi sai coøn coù theå aùp duïng ñeå thaùm caùc heä DES nhieàu voøng hôn. Vôùi heä DES 8-voøng ñoøi hoûi 214 baûn roõ choïn vaø caùc heä 10-, 12-, 14- vaø 16-voøng ñoøi hoûi coù töông öùng 224, 231, 239 vaø 247 baûn maõ choïn. Neân noùi chung laø khaù phöùc taïp.
Caùc kyõ thuaät thaùm maõ vi sai ñöôïc Biham vaø Shamir phaùt trieån. Caùc phöông phaùp thaùm maõ DES khaùc ñaõ ñöôïc Matsui söû duïng nhö laø thaùm maõ tuyeán tính.
III. HEÄ MAÕ DES 3 VOØNG
Chöông trình goàm hai phaàn:
• Phaàn Giao Dieân (chöùa trong thö muïc GiaoDien): Coù chöùc naêng xöû lyù giao dieän.
• Phaàn Xöû Lyù (chöùa trong thö muïc XuLy): coù chöùc naêng hoä trôï caùc haøm xöû lyù.
III.1 Giao Dieän ( Package GiaoDien).
a. Maøn hình chính (Mainform.vb)
Form laäp maõ vaø giaûi maõ DES(Des.vb)
![Page 30: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/30.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
Source code moät soá haøm chính trong form giai maõ Des
Imports System.IO
Public Class des Inherits System.Windows.Forms.Form
khai bao bien
Dim str As String
Dim s(7) As DataTable
Dim ip() As String
'Dim iptru() As String
Dim e() As String
Dim p() As String
Dim pc1() As String
Dim pc2() As String
Dim daykhoa(15) As String
Dim x As String
Dim daynhap(29) As String
![Page 31: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/31.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
Dim daybanma(29) As String
khoi tao
Sub khoitao_s0()
Dim i As Integer
s(0) = New DataTable
For i = 0 To 15
Dim col As DataColumn = New DataColumn
s(0).Columns.Add(col)
Next
For i = 0 To 3
Dim row As DataRow = s(0).NewRow
s(0).Rows.Add(row)
Next
s(0).Rows(0).Item(0) = 14
s(0).Rows(0).Item(1) = 4
s(0).Rows(0).Item(2) = 13
s(0).Rows(0).Item(3) = 1
s(0).Rows(0).Item(4) = 2
s(0).Rows(0).Item(5) = 15
s(0).Rows(0).Item(6) = 11
s(0).Rows(0).Item(7) = 8
s(0).Rows(0).Item(8) = 3
s(0).Rows(0).Item(9) = 10
s(0).Rows(0).Item(10) = 6
s(0).Rows(0).Item(11) = 12
s(0).Rows(0).Item(12) = 5
s(0).Rows(0).Item(13) = 9
s(0).Rows(0).Item(14) = 0
![Page 32: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/32.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
s(0).Rows(0).Item(15) = 7
s(0).Rows(1).Item(0) = 0
s(0).Rows(1).Item(1) = 15
s(0).Rows(1).Item(2) = 7
s(0).Rows(1).Item(3) = 4
s(0).Rows(1).Item(4) = 14
s(0).Rows(1).Item(5) = 2
s(0).Rows(1).Item(6) = 13
s(0).Rows(1).Item(7) = 1
s(0).Rows(1).Item(8) = 10
s(0).Rows(1).Item(9) = 6
s(0).Rows(1).Item(10) = 12
s(0).Rows(1).Item(11) = 11
s(0).Rows(1).Item(12) = 9
s(0).Rows(1).Item(13) = 5
s(0).Rows(1).Item(14) = 3
s(0).Rows(1).Item(15) = 8
s(0).Rows(2).Item(0) = 4
s(0).Rows(2).Item(1) = 1
s(0).Rows(2).Item(2) = 14
s(0).Rows(2).Item(3) = 8
s(0).Rows(2).Item(4) = 13
s(0).Rows(2).Item(5) = 6
s(0).Rows(2).Item(6) = 2
s(0).Rows(2).Item(7) = 11
s(0).Rows(2).Item(8) = 15
s(0).Rows(2).Item(9) = 12
s(0).Rows(2).Item(10) = 9
![Page 33: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/33.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
s(0).Rows(2).Item(11) = 7
s(0).Rows(2).Item(12) = 3
s(0).Rows(2).Item(13) = 10
s(0).Rows(2).Item(14) = 5
s(0).Rows(2).Item(15) = 0
s(0).Rows(3).Item(0) = 15
s(0).Rows(3).Item(1) = 12
s(0).Rows(3).Item(2) = 8
s(0).Rows(3).Item(3) = 2
s(0).Rows(3).Item(4) = 4
s(0).Rows(3).Item(5) = 9
s(0).Rows(3).Item(6) = 1
s(0).Rows(3).Item(7) = 7
s(0).Rows(3).Item(8) = 5
s(0).Rows(3).Item(9) = 11
s(0).Rows(3).Item(10) = 3
s(0).Rows(3).Item(11) = 14
s(0).Rows(3).Item(12) = 10
s(0).Rows(3).Item(13) = 0
s(0).Rows(3).Item(14) = 6
s(0).Rows(3).Item(15) = 13
dgs0.DataSource = s(0)
End Sub
Ham khoi tao s1
Sub khoitao_s1()
Dim i As Integer
s(1) = New DataTable
For i = 0 To 15
![Page 34: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/34.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
Dim col As DataColumn = New DataColumn
s(1).Columns.Add(col)
Next
For i = 0 To 3
Dim row As DataRow = s(1).NewRow
s(1).Rows.Add(row)
Next
s(1).Rows(0).Item(0) = 15
s(1).Rows(0).Item(1) = 1
s(1).Rows(0).Item(2) = 8
s(1).Rows(0).Item(3) = 14
s(1).Rows(0).Item(4) = 6
s(1).Rows(0).Item(5) = 11
s(1).Rows(0).Item(6) = 3
s(1).Rows(0).Item(7) = 4
s(1).Rows(0).Item(8) = 9
s(1).Rows(0).Item(9) = 7
s(1).Rows(0).Item(10) = 2
s(1).Rows(0).Item(11) = 13
s(1).Rows(0).Item(12) = 12
s(1).Rows(0).Item(13) = 0
s(1).Rows(0).Item(14) = 5
s(1).Rows(0).Item(15) = 10
s(1).Rows(1).Item(0) = 3
s(1).Rows(1).Item(1) = 13
s(1).Rows(1).Item(2) = 4
s(1).Rows(1).Item(3) = 7
s(1).Rows(1).Item(4) = 15
![Page 35: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/35.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
s(1).Rows(1).Item(5) = 2
s(1).Rows(1).Item(6) = 8
s(1).Rows(1).Item(7) = 14
s(1).Rows(1).Item(8) = 12
s(1).Rows(1).Item(9) = 0
s(1).Rows(1).Item(10) = 1
s(1).Rows(1).Item(11) = 10
s(1).Rows(1).Item(12) = 6
s(1).Rows(1).Item(13) = 9
s(1).Rows(1).Item(14) = 11
s(1).Rows(1).Item(15) = 5
s(1).Rows(2).Item(0) = 0
s(1).Rows(2).Item(1) = 14
s(1).Rows(2).Item(2) = 7
s(1).Rows(2).Item(3) = 11
s(1).Rows(2).Item(4) = 10
s(1).Rows(2).Item(5) = 4
s(1).Rows(2).Item(6) = 13
s(1).Rows(2).Item(7) = 1
s(1).Rows(2).Item(8) = 5
s(1).Rows(2).Item(9) = 8
s(1).Rows(2).Item(10) = 12
s(1).Rows(2).Item(11) = 6
s(1).Rows(2).Item(12) = 9
s(1).Rows(2).Item(13) = 3
s(1).Rows(2).Item(14) = 2
s(1).Rows(2).Item(15) = 15
s(1).Rows(3).Item(0) = 13
![Page 36: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/36.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
s(1).Rows(3).Item(1) = 8
s(1).Rows(3).Item(2) = 10
s(1).Rows(3).Item(3) = 1
s(1).Rows(3).Item(4) = 3
s(1).Rows(3).Item(5) = 15
s(1).Rows(3).Item(6) = 4
s(1).Rows(3).Item(7) = 2
s(1).Rows(3).Item(8) = 11
s(1).Rows(3).Item(9) = 6
s(1).Rows(3).Item(10) = 7
s(1).Rows(3).Item(11) = 12
s(1).Rows(3).Item(12) = 0
s(1).Rows(3).Item(13) = 5
s(1).Rows(3).Item(14) = 14
s(1).Rows(3).Item(15) = 9
dgs1.DataSource = s(1)
End Sub
Ham khoi tao s2
Sub khoitao_s2()
Dim i As Integer
s(2) = New DataTable
For i = 0 To 15
Dim col As DataColumn = New DataColumn
s(2).Columns.Add(col)
Next
For i = 0 To 3
Dim row As DataRow = s(2).NewRow
s(2).Rows.Add(row)
![Page 37: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/37.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
Next
s(2).Rows(0).Item(0) = 10
s(2).Rows(0).Item(1) = 0
s(2).Rows(0).Item(2) = 9
s(2).Rows(0).Item(3) = 14
s(2).Rows(0).Item(4) = 6
s(2).Rows(0).Item(5) = 3
s(2).Rows(0).Item(6) = 15
s(2).Rows(0).Item(7) = 5
s(2).Rows(0).Item(8) = 1
s(2).Rows(0).Item(9) = 13
s(2).Rows(0).Item(10) = 12
s(2).Rows(0).Item(11) = 7
s(2).Rows(0).Item(12) = 11
s(2).Rows(0).Item(13) = 4
s(2).Rows(0).Item(14) = 2
s(2).Rows(0).Item(15) = 8
s(2).Rows(1).Item(0) = 13
s(2).Rows(1).Item(1) = 7
s(2).Rows(1).Item(2) = 0
s(2).Rows(1).Item(3) = 9
s(2).Rows(1).Item(4) = 3
s(2).Rows(1).Item(5) = 4
s(2).Rows(1).Item(6) = 6
s(2).Rows(1).Item(7) = 10
s(2).Rows(1).Item(8) = 2
s(2).Rows(1).Item(9) = 8
s(2).Rows(1).Item(10) = 5
![Page 38: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/38.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
s(2).Rows(1).Item(11) = 14
s(2).Rows(1).Item(12) = 12
s(2).Rows(1).Item(13) = 11
s(2).Rows(1).Item(14) = 15
s(2).Rows(1).Item(15) = 1
s(2).Rows(2).Item(0) = 13
s(2).Rows(2).Item(1) = 6
s(2).Rows(2).Item(2) = 4
s(2).Rows(2).Item(3) = 9
s(2).Rows(2).Item(4) = 8
s(2).Rows(2).Item(5) = 15
s(2).Rows(2).Item(6) = 3
s(2).Rows(2).Item(7) = 0
s(2).Rows(2).Item(8) = 11
s(2).Rows(2).Item(9) = 1
s(2).Rows(2).Item(10) = 2
s(2).Rows(2).Item(11) = 12
s(2).Rows(2).Item(12) = 5
s(2).Rows(2).Item(13) = 10
s(2).Rows(2).Item(14) = 14
s(2).Rows(2).Item(15) = 7
s(2).Rows(3).Item(0) = 1
s(2).Rows(3).Item(1) = 10
s(2).Rows(3).Item(2) = 13
s(2).Rows(3).Item(3) = 0
s(2).Rows(3).Item(4) = 6
s(2).Rows(3).Item(5) = 9
s(2).Rows(3).Item(6) = 8
![Page 39: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/39.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
s(2).Rows(3).Item(7) = 7
s(2).Rows(3).Item(8) = 4
s(2).Rows(3).Item(9) = 15
s(2).Rows(3).Item(10) = 14
s(2).Rows(3).Item(11) = 3
s(2).Rows(3).Item(12) = 11
s(2).Rows(3).Item(13) = 5
s(2).Rows(3).Item(14) = 3
s(2).Rows(3).Item(15) = 12
dgs2.DataSource = s(2)
End Sub
Haøm khôûi taïo s3
Sub khoitao_s3()
Dim i As Integer
s(3) = New DataTable
For i = 0 To 15
Dim col As DataColumn = New DataColumn
s(3).Columns.Add(col)
Next
For i = 0 To 3
Dim row As DataRow = s(3).NewRow
s(3).Rows.Add(row)
Next
s(3).Rows(0).Item(0) = 7
s(3).Rows(0).Item(1) = 13
s(3).Rows(0).Item(2) = 14
s(3).Rows(0).Item(3) = 3
s(3).Rows(0).Item(4) = 0
![Page 40: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/40.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
s(3).Rows(0).Item(5) = 6
s(3).Rows(0).Item(6) = 9
s(3).Rows(0).Item(7) = 10
s(3).Rows(0).Item(8) = 1
s(3).Rows(0).Item(9) = 2
s(3).Rows(0).Item(10) = 8
s(3).Rows(0).Item(11) = 5
s(3).Rows(0).Item(12) = 11
s(3).Rows(0).Item(13) = 12
s(3).Rows(0).Item(14) = 4
s(3).Rows(0).Item(15) = 15
s(3).Rows(1).Item(0) = 13
s(3).Rows(1).Item(1) = 8
s(3).Rows(1).Item(2) = 11
s(3).Rows(1).Item(3) = 5
s(3).Rows(1).Item(4) = 6
s(3).Rows(1).Item(5) = 15
s(3).Rows(1).Item(6) = 0
s(3).Rows(1).Item(7) = 3
s(3).Rows(1).Item(8) = 4
s(3).Rows(1).Item(9) = 7
s(3).Rows(1).Item(10) = 2
s(3).Rows(1).Item(11) = 12
s(3).Rows(1).Item(12) = 1
s(3).Rows(1).Item(13) = 10
s(3).Rows(1).Item(14) = 14
s(3).Rows(1).Item(15) = 9
s(3).Rows(2).Item(0) = 10
![Page 41: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/41.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
s(3).Rows(2).Item(1) = 6
s(3).Rows(2).Item(2) = 9
s(3).Rows(2).Item(3) = 0
s(3).Rows(2).Item(4) = 12
s(3).Rows(2).Item(5) = 11
s(3).Rows(2).Item(6) = 7
s(3).Rows(2).Item(7) = 13
s(3).Rows(2).Item(8) = 15
s(3).Rows(2).Item(9) = 1
s(3).Rows(2).Item(10) = 3
s(3).Rows(2).Item(11) = 14
s(3).Rows(2).Item(12) = 5
s(3).Rows(2).Item(13) = 2
s(3).Rows(2).Item(14) = 8
s(3).Rows(2).Item(15) = 4
s(3).Rows(3).Item(0) = 3
s(3).Rows(3).Item(1) = 15
s(3).Rows(3).Item(2) = 0
s(3).Rows(3).Item(3) = 6
s(3).Rows(3).Item(4) = 10
s(3).Rows(3).Item(5) = 1
s(3).Rows(3).Item(6) = 13
s(3).Rows(3).Item(7) = 8
s(3).Rows(3).Item(8) = 9
s(3).Rows(3).Item(9) = 4
s(3).Rows(3).Item(10) = 5
s(3).Rows(3).Item(11) = 11
s(3).Rows(3).Item(12) = 12
![Page 42: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/42.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
s(3).Rows(3).Item(13) = 7
s(3).Rows(3).Item(14) = 2
s(3).Rows(3).Item(15) = 14
dgs3.DataSource = s(3)
End Sub
Haøm khôûi taïo s4
Sub khoitao_s4()
Dim i As Integer
s(4) = New DataTable
For i = 0 To 15
Dim col As DataColumn = New DataColumn
s(4).Columns.Add(col)
Next
For i = 0 To 3
Dim row As DataRow = s(4).NewRow
s(4).Rows.Add(row)
Next
s(4).Rows(0).Item(0) = 2
s(4).Rows(0).Item(1) = 12
s(4).Rows(0).Item(2) = 4
s(4).Rows(0).Item(3) = 1
s(4).Rows(0).Item(4) = 7
s(4).Rows(0).Item(5) = 10
s(4).Rows(0).Item(6) = 11
s(4).Rows(0).Item(7) = 6
s(4).Rows(0).Item(8) = 8
s(4).Rows(0).Item(9) = 5
s(4).Rows(0).Item(10) = 3
![Page 43: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/43.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
s(4).Rows(0).Item(11) = 15
s(4).Rows(0).Item(12) = 13
s(4).Rows(0).Item(13) = 0
s(4).Rows(0).Item(14) = 14
s(4).Rows(0).Item(15) = 9
s(4).Rows(1).Item(0) = 14
s(4).Rows(1).Item(1) = 11
s(4).Rows(1).Item(2) = 2
s(4).Rows(1).Item(3) = 12
s(4).Rows(1).Item(4) = 4
s(4).Rows(1).Item(5) = 7
s(4).Rows(1).Item(6) = 13
s(4).Rows(1).Item(7) = 1
s(4).Rows(1).Item(8) = 5
s(4).Rows(1).Item(9) = 0
s(4).Rows(1).Item(10) = 15
s(4).Rows(1).Item(11) = 10
s(4).Rows(1).Item(12) = 3
s(4).Rows(1).Item(13) = 9
s(4).Rows(1).Item(14) = 8
s(4).Rows(1).Item(15) = 6
s(4).Rows(2).Item(0) = 4
s(4).Rows(2).Item(1) = 2
s(4).Rows(2).Item(2) = 1
s(4).Rows(2).Item(3) = 11
s(4).Rows(2).Item(4) = 10
s(4).Rows(2).Item(5) = 13
s(4).Rows(2).Item(6) = 7
![Page 44: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/44.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
s(4).Rows(2).Item(7) = 8
s(4).Rows(2).Item(8) = 15
s(4).Rows(2).Item(9) = 9
s(4).Rows(2).Item(10) = 12
s(4).Rows(2).Item(11) = 5
s(4).Rows(2).Item(12) = 6
s(4).Rows(2).Item(13) = 3
s(4).Rows(2).Item(14) = 0
s(4).Rows(2).Item(15) = 14
s(4).Rows(3).Item(0) = 11
s(4).Rows(3).Item(1) = 8
s(4).Rows(3).Item(2) = 12
s(4).Rows(3).Item(3) = 7
s(4).Rows(3).Item(4) = 0
s(4).Rows(3).Item(5) = 14
s(4).Rows(3).Item(6) = 2
s(4).Rows(3).Item(7) = 13
s(4).Rows(3).Item(8) = 6
s(4).Rows(3).Item(9) = 15
s(4).Rows(3).Item(10) = 0
s(4).Rows(3).Item(11) = 9
s(4).Rows(3).Item(12) = 10
s(4).Rows(3).Item(13) = 4
s(4).Rows(3).Item(14) = 5
s(4).Rows(3).Item(15) = 3
dgs4.DataSource = s(4)
End Sub
Haøm khôûi taïo S5
![Page 45: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/45.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
Sub khoitao_s5()
Dim i As Integer
s(5) = New DataTable
For i = 0 To 15
Dim col As DataColumn = New DataColumn
s(5).Columns.Add(col)
Next
For i = 0 To 3
Dim row As DataRow = s(5).NewRow
s(5).Rows.Add(row)
Next
s(5).Rows(0).Item(0) = 12
s(5).Rows(0).Item(1) = 1
s(5).Rows(0).Item(2) = 10
s(5).Rows(0).Item(3) = 15
s(5).Rows(0).Item(4) = 9
s(5).Rows(0).Item(5) = 2
s(5).Rows(0).Item(6) = 6
s(5).Rows(0).Item(7) = 8
s(5).Rows(0).Item(8) = 0
s(5).Rows(0).Item(9) = 13
s(5).Rows(0).Item(10) = 3
s(5).Rows(0).Item(11) = 4
s(5).Rows(0).Item(12) = 14
s(5).Rows(0).Item(13) = 7
s(5).Rows(0).Item(14) = 5
s(5).Rows(0).Item(15) = 11
s(5).Rows(1).Item(0) = 10
![Page 46: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/46.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
s(5).Rows(1).Item(1) = 15
s(5).Rows(1).Item(2) = 4
s(5).Rows(1).Item(3) = 2
s(5).Rows(1).Item(4) = 7
s(5).Rows(1).Item(5) = 12
s(5).Rows(1).Item(6) = 9
s(5).Rows(1).Item(7) = 5
s(5).Rows(1).Item(8) = 6
s(5).Rows(1).Item(9) = 1
s(5).Rows(1).Item(10) = 13
s(5).Rows(1).Item(11) = 14
s(5).Rows(1).Item(12) = 0
s(5).Rows(1).Item(13) = 11
s(5).Rows(1).Item(14) = 3
s(5).Rows(1).Item(15) = 8
s(5).Rows(2).Item(0) = 9
s(5).Rows(2).Item(1) = 14
s(5).Rows(2).Item(2) = 15
s(5).Rows(2).Item(3) = 5
s(5).Rows(2).Item(4) = 2
s(5).Rows(2).Item(5) = 8
s(5).Rows(2).Item(6) = 12
s(5).Rows(2).Item(7) = 3
s(5).Rows(2).Item(8) = 7
s(5).Rows(2).Item(9) = 0
s(5).Rows(2).Item(10) = 4
s(5).Rows(2).Item(11) = 10
s(5).Rows(2).Item(12) = 1
![Page 47: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/47.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
s(5).Rows(2).Item(13) = 13
s(5).Rows(2).Item(14) = 11
s(5).Rows(2).Item(15) = 6
s(5).Rows(3).Item(0) = 4
s(5).Rows(3).Item(1) = 3
s(5).Rows(3).Item(2) = 2
s(5).Rows(3).Item(3) = 12
s(5).Rows(3).Item(4) = 9
s(5).Rows(3).Item(5) = 5
s(5).Rows(3).Item(6) = 15
s(5).Rows(3).Item(7) = 10
s(5).Rows(3).Item(8) = 11
s(5).Rows(3).Item(9) = 14
s(5).Rows(3).Item(10) = 1
s(5).Rows(3).Item(11) = 7
s(5).Rows(3).Item(12) = 6
s(5).Rows(3).Item(13) = 0
s(5).Rows(3).Item(14) = 8
s(5).Rows(3).Item(15) = 13
dgs5.DataSource = s(5)
End Sub
Haøm khôûi taïo S6
Sub khoitao_s6()
Dim i As Integer
s(6) = New DataTable
For i = 0 To 15
Dim col As DataColumn = New DataColumn
s(6).Columns.Add(col)
![Page 48: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/48.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
Next
For i = 0 To 3
Dim row As DataRow = s(6).NewRow
s(6).Rows.Add(row)
Next
s(6).Rows(0).Item(0) = 4
s(6).Rows(0).Item(1) = 11
s(6).Rows(0).Item(2) = 2
s(6).Rows(0).Item(3) = 14
s(6).Rows(0).Item(4) = 15
s(6).Rows(0).Item(5) = 0
s(6).Rows(0).Item(6) = 8
s(6).Rows(0).Item(7) = 13
s(6).Rows(0).Item(8) = 3
s(6).Rows(0).Item(9) = 12
s(6).Rows(0).Item(10) = 9
s(6).Rows(0).Item(11) = 7
s(6).Rows(0).Item(12) = 5
s(6).Rows(0).Item(13) = 10
s(6).Rows(0).Item(14) = 6
s(6).Rows(0).Item(15) = 1
s(6).Rows(1).Item(0) = 13
s(6).Rows(1).Item(1) = 0
s(6).Rows(1).Item(2) = 11
s(6).Rows(1).Item(3) = 7
s(6).Rows(1).Item(4) = 4
s(6).Rows(1).Item(5) = 9
s(6).Rows(1).Item(6) = 1
![Page 49: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/49.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
s(6).Rows(1).Item(7) = 10
s(6).Rows(1).Item(8) = 14
s(6).Rows(1).Item(9) = 3
s(6).Rows(1).Item(10) = 5
s(6).Rows(1).Item(11) = 12
s(6).Rows(1).Item(12) = 2
s(6).Rows(1).Item(13) = 15
s(6).Rows(1).Item(14) = 8
s(6).Rows(1).Item(15) = 6
s(6).Rows(2).Item(0) = 1
s(6).Rows(2).Item(1) = 4
s(6).Rows(2).Item(2) = 11
s(6).Rows(2).Item(3) = 13
s(6).Rows(2).Item(4) = 12
s(6).Rows(2).Item(5) = 3
s(6).Rows(2).Item(6) = 7
s(6).Rows(2).Item(7) = 14
s(6).Rows(2).Item(8) = 10
s(6).Rows(2).Item(9) = 15
s(6).Rows(2).Item(10) = 6
s(6).Rows(2).Item(11) = 8
s(6).Rows(2).Item(12) = 0
s(6).Rows(2).Item(13) = 5
s(6).Rows(2).Item(14) = 9
s(6).Rows(2).Item(15) = 2
s(6).Rows(3).Item(0) = 6
s(6).Rows(3).Item(1) = 11
s(6).Rows(3).Item(2) = 13
![Page 50: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/50.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
s(6).Rows(3).Item(3) = 8
s(6).Rows(3).Item(4) = 1
s(6).Rows(3).Item(5) = 4
s(6).Rows(3).Item(6) = 10
s(6).Rows(3).Item(7) = 7
s(6).Rows(3).Item(8) = 9
s(6).Rows(3).Item(9) = 5
s(6).Rows(3).Item(10) = 0
s(6).Rows(3).Item(11) = 15
s(6).Rows(3).Item(12) = 14
s(6).Rows(3).Item(13) = 2
s(6).Rows(3).Item(14) = 3
s(6).Rows(3).Item(15) = 12
dgs6.DataSource = s(6)
End Sub
Haøm khôûi taïo S7
Sub khoitao_s7()
Dim i As Integer
s(7) = New DataTable
For i = 0 To 15
Dim col As DataColumn = New DataColumn
s(7).Columns.Add(col)
Next
For i = 0 To 3
Dim row As DataRow = s(7).NewRow
s(7).Rows.Add(row)
Next
s(7).Rows(0).Item(0) = 13
![Page 51: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/51.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
s(7).Rows(0).Item(1) = 2
s(7).Rows(0).Item(2) = 8
s(7).Rows(0).Item(3) = 4
s(7).Rows(0).Item(4) = 6
s(7).Rows(0).Item(5) = 15
s(7).Rows(0).Item(6) = 11
s(7).Rows(0).Item(7) = 1
s(7).Rows(0).Item(8) = 10
s(7).Rows(0).Item(9) = 9
s(7).Rows(0).Item(10) = 3
s(7).Rows(0).Item(11) = 14
s(7).Rows(0).Item(12) = 5
s(7).Rows(0).Item(13) = 0
s(7).Rows(0).Item(14) = 12
s(7).Rows(0).Item(15) = 7
s(7).Rows(1).Item(0) = 1
s(7).Rows(1).Item(1) = 15
s(7).Rows(1).Item(2) = 13
s(7).Rows(1).Item(3) = 8
s(7).Rows(1).Item(4) = 10
s(7).Rows(1).Item(5) = 3
s(7).Rows(1).Item(6) = 7
s(7).Rows(1).Item(7) = 4
s(7).Rows(1).Item(8) = 12
s(7).Rows(1).Item(9) = 5
s(7).Rows(1).Item(10) = 6
s(7).Rows(1).Item(11) = 11
s(7).Rows(1).Item(12) = 0
![Page 52: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/52.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
s(7).Rows(1).Item(13) = 14
s(7).Rows(1).Item(14) = 9
s(7).Rows(1).Item(15) = 2
s(7).Rows(2).Item(0) = 7
s(7).Rows(2).Item(1) = 11
s(7).Rows(2).Item(2) = 4
s(7).Rows(2).Item(3) = 1
s(7).Rows(2).Item(4) = 9
s(7).Rows(2).Item(5) = 12
s(7).Rows(2).Item(6) = 14
s(7).Rows(2).Item(7) = 2
s(7).Rows(2).Item(8) = 0
s(7).Rows(2).Item(9) = 6
s(7).Rows(2).Item(10) = 10
s(7).Rows(2).Item(11) = 13
s(7).Rows(2).Item(12) = 15
s(7).Rows(2).Item(13) = 3
s(7).Rows(2).Item(14) = 5
s(7).Rows(2).Item(15) = 8
s(7).Rows(3).Item(0) = 2
s(7).Rows(3).Item(1) = 1
s(7).Rows(3).Item(2) = 14
s(7).Rows(3).Item(3) = 7
s(7).Rows(3).Item(4) = 4
s(7).Rows(3).Item(5) = 10
s(7).Rows(3).Item(6) = 8
s(7).Rows(3).Item(7) = 13
s(7).Rows(3).Item(8) = 15
![Page 53: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/53.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
s(7).Rows(3).Item(9) = 12
s(7).Rows(3).Item(10) = 9
s(7).Rows(3).Item(11) = 0
s(7).Rows(3).Item(12) = 3
s(7).Rows(3).Item(13) = 5
s(7).Rows(3).Item(14) = 6
s(7).Rows(3).Item(15) = 11
dgs7.DataSource = s(7)
End Sub
Khôûi taïo giaù trò bieán
Sub khoitao()
ip = txtip.Text.Split(" ", ";", ":", ".")
'iptru = txtiptru.Text.Split(" ", " ", ";", ":", ".")
e = txte.Text.Split(" ", ";", ":", ".")
p = txtp.Text.Split(" ", ";", ":", ".")
pc1 = txtpc1.Text.Split(" ", ";", ":", ".")
pc2 = txtpc2.Text.Split(" ", ";", ":", ".")
khoitao_s0()
khoitao_s1()
khoitao_s2()
khoitao_s3()
khoitao_s4()
khoitao_s5()
khoitao_s6()
khoitao_s7()
End Sub
Caét bit cuoái
Function catbitcuoi(ByVal k As String) As String 'dua vao 64 bit tra ra 56 bit
![Page 54: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/54.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
Dim i As Integer = 0
Dim j As Integer
Dim tam As String
While i < 63
For j = i To i + 6
tam += k.Substring(j, 1)
Next
i = i + 8
End While
Return tam
End Function
Function hvpc1(ByVal k As String) As String
Dim tam(63) As Char
Dim i As Integer
For i = 0 To 63
tam(i) = k.Substring(i, 1)
Next
tam = catbitcuoi(tam)
For i = 0 To 55
tam(i) = k.Substring(Integer.Parse(pc1(i) - 1), 1)
Next
Return tam
End Function
Hoaù vò pc2
Function hvpc2(ByVal str As String) As String
Dim tam(47) As Char
Dim i As Integer
For i = 0 To 47
![Page 55: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/55.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
tam(i) = str.Substring(Integer.Parse(pc2(i) - 1), 1)
Next
Return tam
End Function
Function ls(ByVal s As String, ByVal n As Integer) As String
Return s.Substring(n, s.Length - n) + s.Substring(0, n)
End Function
Haøm taïo daõy khoaù
Sub taodaykhoa()
Dim khoa as String =
"0001001100110100010101110111100110011011101111001101111111110001"
Dim khoa As String = txtkhoak.Text
Dim j As Integer
If khoa.Length > 8 Then
khoa = txtkhoak.Text.Remove(8, khoa.Length - 8)
txtkhoak.Text = khoa
End If
Dim tam As String
For j = 0 To khoa.Length - 1
tam += bi_acsii(Asc(khoa.Substring(j, 1)))
Next
khoa = tam
Dim khoa1 As String = hvpc1(khoa)
Dim d(16) As String
Dim c(16) As String
c(0) = khoa1.Substring(0, 28)
c(0) = ls(c(0), 1)
d(0) = khoa1.Substring(28, 28)
![Page 56: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/56.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
d(0) = ls(d(0), 1)
daykhoa(0) = hvpc2(c(0) + d(0))
txtdaykhoa.Text += daykhoa(0) + Chr(9)
Dim i As Integer
For i = 1 To 15
If i = 2 - 1 Or i = 9 - 1 Or i = 16 - 1 Then
c(i) = ls(c(i - 1), 1)
d(i) = ls(d(i - 1), 1)
Else
c(i) = ls(c(i - 1), 2)
d(i) = ls(d(i - 1), 2)
End If
daykhoa(i) = hvpc2(c(i) + d(i))
txtdaykhoa.Text += daykhoa(i) + Chr(9)
Next i
End Sub
Moät soá haøm xöû lyù chuoãi nhaäp
Nhaäp nhò phaân
Sub binarynhap()
x = txtchuoinhap.Text
Dim y As String
Dim i As Integer
Dim j As Integer
Dim sokitudu As Integer = x.Length Mod 8
If sokitudu > 0 Then
Dim sokituthem As Integer = 8 - sokitudu
For i = 1 To sokituthem
x += " "
![Page 57: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/57.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
Next
End If
Dim sodaynhap As Integer = x.Length \ 8
ReDim daynhap(sodaynhap - 1)
For i = 0 To sodaynhap - 1
daynhap(i) = x.Substring(i * 8, 8)
y = ""
For j = 0 To daynhap(i).Length - 1
y += bi_acsii(Asc(daynhap(i).Substring(j, 1)))
Next
daynhap(i) = y
Next
End Sub
Function bi_acsii(ByVal int As Integer) As String
Dim tam(7) As Char
Dim i As Integer
For i = 0 To 7
tam(i) = (int Mod 2).ToString
int \= 2
Next
Array.Reverse(tam)
Return tam
End Function
Moät soá haøm maõ hoaù
Haøm hoaùn vò ip
Function hvip(ByVal x As String) As String
Dim tam(63) As Char
Dim i As Integer
![Page 58: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/58.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
For i = 0 To 63
tam(i) = x.Substring(ip(i) - 1, 1)
Next
Return tam
End Function
Haøm hoaùn vò e
Function hve(ByVal r As String) As String
Dim tam(47) As Char
Dim i As Integer
For i = 0 To 47
tam(i) = r.Substring(e(i) - 1, 1)
Next
Return tam
End Function
Function hvp(ByVal c As String) As String
Dim tam(31) As Char
Dim i As Integer
For i = 0 To 31
tam(i) = c.Substring(p(i) - 1, 1)
Next
Return tam
End Function
Haøm hoaùn vò ip tröø
Function hviptru(ByVal c As String) As String
Dim tam(63) As Char
Dim i As Integer
For i = 0 To 63
tam(ip(i) - 1) = c.Substring(i, 1)
![Page 59: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/59.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
Next
Return tam
End Function
Maõ hoaù
Function mahoa() As String
binarynhap()
Dim k As Integer
Dim y As String
For k = 0 To daynhap.Length - 1
'x += "0000000100100011010001010110011110001001101010101100110111101111"
x = daynhap(k)
Dim x0 As String = hvip(x)
Dim l(15) As String
Dim r(15) As String
Dim i, j As Integer
Dim l0 As String = x0.Substring(0, 32)
Dim r0 As String = x0.Substring(32, 32)
l(0) = r0
For i = 0 To 31
r(0) += (l0.Substring(i, 1) Xor f(r0, daykhoa(0)).Substring(i, 1)).ToString
Next
For i = 1 To 15
l(i) = r(i - 1)
Dim a As String = f(r(i - 1), daykhoa(i))
For j = 0 To 31
r(i) += (l(i - 1).Substring(j, 1) Xor a.Substring(j, 1)).ToString
Next j
![Page 60: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/60.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
Next i
Dim v As String = hviptru(r(15) + l(15))
y += v
Next k
Return y
End Function
Haøm f
Function f(ByVal r As String, ByVal k As String) As String
Dim i As Integer
Dim e As String
Dim hv As String = hve(r)
For i = 0 To 47
e += (hv.Substring(i, 1) Xor k.Substring(i, 1)).ToString
Next
Dim b(7) As String
Dim c As String
For i = 0 To 7
b(i) = e.Substring(i * 6, 6)
Dim haibitcuoi As String = b(i).Substring(0, 1) + b(i).Substring(5, 1)
Dim bonbitgiua As String = b(i).Substring(1, 4)
Dim srow = thapphan(haibitcuoi)
Dim scol = thapphan(bonbitgiua)
Dim sij As Integer = s(i).Rows(srow).Item(scol)
c += binary(sij)
Next
c = hvp(c)
Return c
End Function
![Page 61: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/61.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
Haøm thaäp phaân
Function thapphan(ByVal b As String) As Integer
Dim i As Integer
Dim tam As Integer = 0
For i = 0 To b.Length - 1
If b.Substring(i, 1) = 1 Then
tam += 2 ^ (b.Length - 1 - i)
End If
Next
Return tam
End Function
Haøm nhò phaân
Function binary(ByVal a As Integer) As String
Dim i As Integer
Dim tam(3) As Char
For i = 0 To 3
tam(i) = (a Mod 2).ToString
a = a \ 2
Next
Array.Reverse(tam)
Return tam
End Function
Haøm ñoåi ra chöõ
Function doirachu(ByVal y As String) As String
Dim tam As String = y
Dim tam1 As String = ""
Dim so As Integer
Dim i As Integer
![Page 62: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/62.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
Dim j As Integer
While i < tam.Length - 1
so = 0
For j = i To i + 7
If tam.Substring(j, 1) = 1 Then
so += 2 ^ (7 - (j - i))
End If
Next
tam1 += Chr(so)
i = i + 8
End While
Return tam1
End Function
#End Region
moät soá haøm duøng ñeå giaûi maõ
taïo baûn maõ nhò phaân
Sub binaybanma()
Dim sodaybanma = txtbanma.Text.Length \ 8
Dim i, j As Integer
Dim tam As String
ReDim daybanma(sodaybanma - 1)
For i = 0 To sodaybanma - 1
daybanma(i) = txtbanma.Text.Substring(i * 8, 8)
tam = ""
For j = 0 To daybanma(i).Length - 1
tam += bi_acsii(Asc(daybanma(i).Substring(j, 1)))
Next
daybanma(i) = tam
![Page 63: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/63.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
Next i
End Sub
Haøm giaûi maõ
Function giaima() As String
binaybanma()
Dim k As Integer
x = ""
For k = 0 To daybanma.Length - 1
Dim y0 As String = hvip(daybanma(k))
Dim rr As String = y0.Substring(0, 32)
Dim ll As String = y0.Substring(32, 32)
Dim i, j As Integer
Dim l(15) As String
Dim r(15) As String
r(15) = ll
For i = 0 To 31
l(15) += (rr.Substring(i, 1) Xor f(ll, daykhoa(15)).Substring(i, 1)).ToString
Next
For i = 14 To 0 Step -1
r(i) = l(i + 1)
Dim a As String = f(l(i + 1), daykhoa(i))
For j = 0 To 31
l(i) += (r(i + 1).Substring(j, 1) Xor a.Substring(j, 1)).ToString
Next j
Next i
x += hviptru(l(0) + r(0))
Next k
![Page 64: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/64.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
Return x
End Function
Moät soá söï kieän giuùp cho vieäc thöïc thi
Private Sub des_Load(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles MyBase.Load
khoitao()
End Sub
Private Sub btlapma_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles btlapma.Click
khoitao()
If txtchuoinhap.Text = "" Then
MessageBox.Show("B�n hãy nh�p vào chu�i c�n mã hoá")
txtchuoinhap.Focus()
Return
End If
If txtkhoak.Text = "" Or txtkhoak.Text.Length < 8 Then
MessageBox.Show("B�n hãy nh�p vào khóa :8 kí t�")
txtkhoak.Focus()
Return
End If
txtdaykhoa.Text = ""
taodaykhoa()
txtbanmabit.Text = mahoa()
txtbanma.Text = doirachu(mahoa())
End Sub
Private Sub btthoat_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles btthoat.Click
Me.Close()
End Sub
![Page 65: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/65.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
Private Sub btgiaima_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles btgiaima.Click
If txtkhoak.Text = "" Or txtkhoak.Text.Length < 8 Then
MessageBox.Show("B�n hãy nh�p khóa : 8 kí t�")
txtkhoak.Focus()
Return
End If
If txtbanma.Text = "" Then
MessageBox.Show("B�n hãy nh�p vào chu�i c�n gi�i mã")
txtbanma.Focus()
Return
End If
txtdaykhoa.Text = ""
taodaykhoa()
txtbanrobit.Text = giaima()
txtbanro.Text = doirachu(giaima())
End Sub
End Class
Form thaùm maõ Des (thamma.vb)
![Page 66: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/66.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
Imports System.Windows.Forms
Public Class thammades Inherits System.Windows.Forms.Form
Khai baùo moät soá bieán vaø haøm
Dim banro() As String
Dim banma() As String
Dim hve(47) As Integer
Dim hvp(31) As Integer
Dim hvpc2(47) As Integer
Dim hvpc1(55) As Integer
Dim hvip(63) As Integer
Dim e() As String
Dim esao() As String
Dim ephay() As String
Dim cphay() As String
![Page 67: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/67.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
Dim jhop(7, 63) As Integer
Dim s(7) As DataTable
Dim chuoikhoa(255) As String
Dim daykhoa(2) As String
Dim n As Integer '6 ban ro va 6 ban ma
Const m = 16 '8 ki tu
Const hebit = 4 'he 256
Dim flag As Boolean
Khôûi taïo
Sub khoitao_s0()
Dim i As Integer
s(0) = New DataTable
For i = 0 To 15
Dim col As DataColumn = New DataColumn
s(0).Columns.Add(col)
Next
For i = 0 To 3
Dim row As DataRow = s(0).NewRow
s(0).Rows.Add(row)
Next
s(0).Rows(0).Item(0) = 14
s(0).Rows(0).Item(1) = 4
s(0).Rows(0).Item(2) = 13
s(0).Rows(0).Item(3) = 1
s(0).Rows(0).Item(4) = 2
s(0).Rows(0).Item(5) = 15
s(0).Rows(0).Item(6) = 11
s(0).Rows(0).Item(7) = 8
![Page 68: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/68.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
s(0).Rows(0).Item(8) = 3
s(0).Rows(0).Item(9) = 10
s(0).Rows(0).Item(10) = 6
s(0).Rows(0).Item(11) = 12
s(0).Rows(0).Item(12) = 5
s(0).Rows(0).Item(13) = 9
s(0).Rows(0).Item(14) = 0
s(0).Rows(0).Item(15) = 7
s(0).Rows(1).Item(0) = 0
s(0).Rows(1).Item(1) = 15
s(0).Rows(1).Item(2) = 7
s(0).Rows(1).Item(3) = 4
s(0).Rows(1).Item(4) = 14
s(0).Rows(1).Item(5) = 2
s(0).Rows(1).Item(6) = 13
s(0).Rows(1).Item(7) = 1
s(0).Rows(1).Item(8) = 10
s(0).Rows(1).Item(9) = 6
s(0).Rows(1).Item(10) = 12
s(0).Rows(1).Item(11) = 11
s(0).Rows(1).Item(12) = 9
s(0).Rows(1).Item(13) = 5
s(0).Rows(1).Item(14) = 3
s(0).Rows(1).Item(15) = 8
s(0).Rows(2).Item(0) = 4
s(0).Rows(2).Item(1) = 1
s(0).Rows(2).Item(2) = 14
s(0).Rows(2).Item(3) = 8
![Page 69: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/69.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
s(0).Rows(2).Item(4) = 13
s(0).Rows(2).Item(5) = 6
s(0).Rows(2).Item(6) = 2
s(0).Rows(2).Item(7) = 11
s(0).Rows(2).Item(8) = 15
s(0).Rows(2).Item(9) = 12
s(0).Rows(2).Item(10) = 9
s(0).Rows(2).Item(11) = 7
s(0).Rows(2).Item(12) = 3
s(0).Rows(2).Item(13) = 10
s(0).Rows(2).Item(14) = 5
s(0).Rows(2).Item(15) = 0
s(0).Rows(3).Item(0) = 15
s(0).Rows(3).Item(1) = 12
s(0).Rows(3).Item(2) = 8
s(0).Rows(3).Item(3) = 2
s(0).Rows(3).Item(4) = 4
s(0).Rows(3).Item(5) = 9
s(0).Rows(3).Item(6) = 1
s(0).Rows(3).Item(7) = 7
s(0).Rows(3).Item(8) = 5
s(0).Rows(3).Item(9) = 11
s(0).Rows(3).Item(10) = 3
s(0).Rows(3).Item(11) = 14
s(0).Rows(3).Item(12) = 10
s(0).Rows(3).Item(13) = 0
s(0).Rows(3).Item(14) = 6
s(0).Rows(3).Item(15) = 13
![Page 70: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/70.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
End Sub
Khôûi taïo haøm s1
Sub khoitao_s1()
Dim i As Integer
s(1) = New DataTable
For i = 0 To 15
Dim col As DataColumn = New DataColumn
s(1).Columns.Add(col)
Next
For i = 0 To 3
Dim row As DataRow = s(1).NewRow
s(1).Rows.Add(row)
Next
s(1).Rows(0).Item(0) = 15
s(1).Rows(0).Item(1) = 1
s(1).Rows(0).Item(2) = 8
s(1).Rows(0).Item(3) = 14
s(1).Rows(0).Item(4) = 6
s(1).Rows(0).Item(5) = 11
s(1).Rows(0).Item(6) = 3
s(1).Rows(0).Item(7) = 4
s(1).Rows(0).Item(8) = 9
s(1).Rows(0).Item(9) = 7
s(1).Rows(0).Item(10) = 2
s(1).Rows(0).Item(11) = 13
s(1).Rows(0).Item(12) = 12
s(1).Rows(0).Item(13) = 0
s(1).Rows(0).Item(14) = 5
![Page 71: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/71.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
s(1).Rows(0).Item(15) = 10
s(1).Rows(1).Item(0) = 3
s(1).Rows(1).Item(1) = 13
s(1).Rows(1).Item(2) = 4
s(1).Rows(1).Item(3) = 7
s(1).Rows(1).Item(4) = 15
s(1).Rows(1).Item(5) = 2
s(1).Rows(1).Item(6) = 8
s(1).Rows(1).Item(7) = 14
s(1).Rows(1).Item(8) = 12
s(1).Rows(1).Item(9) = 0
s(1).Rows(1).Item(10) = 1
s(1).Rows(1).Item(11) = 10
s(1).Rows(1).Item(12) = 6
s(1).Rows(1).Item(13) = 9
s(1).Rows(1).Item(14) = 11
s(1).Rows(1).Item(15) = 5
s(1).Rows(2).Item(0) = 0
s(1).Rows(2).Item(1) = 14
s(1).Rows(2).Item(2) = 7
s(1).Rows(2).Item(3) = 11
s(1).Rows(2).Item(4) = 10
s(1).Rows(2).Item(5) = 4
s(1).Rows(2).Item(6) = 13
s(1).Rows(2).Item(7) = 1
s(1).Rows(2).Item(8) = 5
s(1).Rows(2).Item(9) = 8
s(1).Rows(2).Item(10) = 12
![Page 72: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/72.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
s(1).Rows(2).Item(11) = 6
s(1).Rows(2).Item(12) = 9
s(1).Rows(2).Item(13) = 3
s(1).Rows(2).Item(14) = 2
s(1).Rows(2).Item(15) = 15
s(1).Rows(3).Item(0) = 13
s(1).Rows(3).Item(1) = 8
s(1).Rows(3).Item(2) = 10
s(1).Rows(3).Item(3) = 1
s(1).Rows(3).Item(4) = 3
s(1).Rows(3).Item(5) = 15
s(1).Rows(3).Item(6) = 4
s(1).Rows(3).Item(7) = 2
s(1).Rows(3).Item(8) = 11
s(1).Rows(3).Item(9) = 6
s(1).Rows(3).Item(10) = 7
s(1).Rows(3).Item(11) = 12
s(1).Rows(3).Item(12) = 0
s(1).Rows(3).Item(13) = 5
s(1).Rows(3).Item(14) = 14
s(1).Rows(3).Item(15) = 9
End Sub
Khôûi taïo haøm s2
Sub khoitao_s2()
Dim i As Integer
s(2) = New DataTable
For i = 0 To 15
Dim col As DataColumn = New DataColumn
![Page 73: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/73.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
s(2).Columns.Add(col)
Next
For i = 0 To 3
Dim row As DataRow = s(2).NewRow
s(2).Rows.Add(row)
Next
s(2).Rows(0).Item(0) = 10
s(2).Rows(0).Item(1) = 0
s(2).Rows(0).Item(2) = 9
s(2).Rows(0).Item(3) = 14
s(2).Rows(0).Item(4) = 6
s(2).Rows(0).Item(5) = 3
s(2).Rows(0).Item(6) = 15
s(2).Rows(0).Item(7) = 5
s(2).Rows(0).Item(8) = 1
s(2).Rows(0).Item(9) = 13
s(2).Rows(0).Item(10) = 12
s(2).Rows(0).Item(11) = 7
s(2).Rows(0).Item(12) = 11
s(2).Rows(0).Item(13) = 4
s(2).Rows(0).Item(14) = 2
s(2).Rows(0).Item(15) = 8
s(2).Rows(1).Item(0) = 13
s(2).Rows(1).Item(1) = 7
s(2).Rows(1).Item(2) = 0
s(2).Rows(1).Item(3) = 9
s(2).Rows(1).Item(4) = 3
s(2).Rows(1).Item(5) = 4
![Page 74: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/74.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
s(2).Rows(1).Item(6) = 6
s(2).Rows(1).Item(7) = 10
s(2).Rows(1).Item(8) = 2
s(2).Rows(1).Item(9) = 8
s(2).Rows(1).Item(10) = 5
s(2).Rows(1).Item(11) = 14
s(2).Rows(1).Item(12) = 12
s(2).Rows(1).Item(13) = 11
s(2).Rows(1).Item(14) = 15
s(2).Rows(1).Item(15) = 1
s(2).Rows(2).Item(0) = 13
s(2).Rows(2).Item(1) = 6
s(2).Rows(2).Item(2) = 4
s(2).Rows(2).Item(3) = 9
s(2).Rows(2).Item(4) = 8
s(2).Rows(2).Item(5) = 15
s(2).Rows(2).Item(6) = 3
s(2).Rows(2).Item(7) = 0
s(2).Rows(2).Item(8) = 11
s(2).Rows(2).Item(9) = 1
s(2).Rows(2).Item(10) = 2
s(2).Rows(2).Item(11) = 12
s(2).Rows(2).Item(12) = 5
s(2).Rows(2).Item(13) = 10
s(2).Rows(2).Item(14) = 14
s(2).Rows(2).Item(15) = 7
s(2).Rows(3).Item(0) = 1
s(2).Rows(3).Item(1) = 10
![Page 75: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/75.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
s(2).Rows(3).Item(2) = 13
s(2).Rows(3).Item(3) = 0
s(2).Rows(3).Item(4) = 6
s(2).Rows(3).Item(5) = 9
s(2).Rows(3).Item(6) = 8
s(2).Rows(3).Item(7) = 7
s(2).Rows(3).Item(8) = 4
s(2).Rows(3).Item(9) = 15
s(2).Rows(3).Item(10) = 14
s(2).Rows(3).Item(11) = 3
s(2).Rows(3).Item(12) = 11
s(2).Rows(3).Item(13) = 5
s(2).Rows(3).Item(14) = 3
s(2).Rows(3).Item(15) = 12
End Sub
Khôûi taïo haøm s3
Sub khoitao_s3()
Dim i As Integer
s(3) = New DataTable
For i = 0 To 15
Dim col As DataColumn = New DataColumn
s(3).Columns.Add(col)
Next
For i = 0 To 3
Dim row As DataRow = s(3).NewRow
s(3).Rows.Add(row)
Next
s(3).Rows(0).Item(0) = 7
![Page 76: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/76.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
s(3).Rows(0).Item(1) = 13
s(3).Rows(0).Item(2) = 14
s(3).Rows(0).Item(3) = 3
s(3).Rows(0).Item(4) = 0
s(3).Rows(0).Item(5) = 6
s(3).Rows(0).Item(6) = 9
s(3).Rows(0).Item(7) = 10
s(3).Rows(0).Item(8) = 1
s(3).Rows(0).Item(9) = 2
s(3).Rows(0).Item(10) = 8
s(3).Rows(0).Item(11) = 5
s(3).Rows(0).Item(12) = 11
s(3).Rows(0).Item(13) = 12
s(3).Rows(0).Item(14) = 4
s(3).Rows(0).Item(15) = 15
s(3).Rows(1).Item(0) = 13
s(3).Rows(1).Item(1) = 8
s(3).Rows(1).Item(2) = 11
s(3).Rows(1).Item(3) = 5
s(3).Rows(1).Item(4) = 6
s(3).Rows(1).Item(5) = 15
s(3).Rows(1).Item(6) = 0
s(3).Rows(1).Item(7) = 3
s(3).Rows(1).Item(8) = 4
s(3).Rows(1).Item(9) = 7
s(3).Rows(1).Item(10) = 2
s(3).Rows(1).Item(11) = 12
s(3).Rows(1).Item(12) = 1
![Page 77: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/77.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
s(3).Rows(1).Item(13) = 10
s(3).Rows(1).Item(14) = 14
s(3).Rows(1).Item(15) = 9
s(3).Rows(2).Item(0) = 10
s(3).Rows(2).Item(1) = 6
s(3).Rows(2).Item(2) = 9
s(3).Rows(2).Item(3) = 0
s(3).Rows(2).Item(4) = 12
s(3).Rows(2).Item(5) = 11
s(3).Rows(2).Item(6) = 7
s(3).Rows(2).Item(7) = 13
s(3).Rows(2).Item(8) = 15
s(3).Rows(2).Item(9) = 1
s(3).Rows(2).Item(10) = 3
s(3).Rows(2).Item(11) = 14
s(3).Rows(2).Item(12) = 5
s(3).Rows(2).Item(13) = 2
s(3).Rows(2).Item(14) = 8
s(3).Rows(2).Item(15) = 4
s(3).Rows(3).Item(0) = 3
s(3).Rows(3).Item(1) = 15
s(3).Rows(3).Item(2) = 0
s(3).Rows(3).Item(3) = 6
s(3).Rows(3).Item(4) = 10
s(3).Rows(3).Item(5) = 1
s(3).Rows(3).Item(6) = 13
s(3).Rows(3).Item(7) = 8
s(3).Rows(3).Item(8) = 9
![Page 78: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/78.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
s(3).Rows(3).Item(9) = 4
s(3).Rows(3).Item(10) = 5
s(3).Rows(3).Item(11) = 11
s(3).Rows(3).Item(12) = 12
s(3).Rows(3).Item(13) = 7
s(3).Rows(3).Item(14) = 2
s(3).Rows(3).Item(15) = 14
End Sub
Khôûi taïo haøm s4
Sub khoitao_s4()
Dim i As Integer
s(4) = New DataTable
For i = 0 To 15
Dim col As DataColumn = New DataColumn
s(4).Columns.Add(col)
Next
For i = 0 To 3
Dim row As DataRow = s(4).NewRow
s(4).Rows.Add(row)
Next
s(4).Rows(0).Item(0) = 2
s(4).Rows(0).Item(1) = 12
s(4).Rows(0).Item(2) = 4
s(4).Rows(0).Item(3) = 1
s(4).Rows(0).Item(4) = 7
s(4).Rows(0).Item(5) = 10
s(4).Rows(0).Item(6) = 11
s(4).Rows(0).Item(7) = 6
![Page 79: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/79.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
s(4).Rows(0).Item(8) = 8
s(4).Rows(0).Item(9) = 5
s(4).Rows(0).Item(10) = 3
s(4).Rows(0).Item(11) = 15
s(4).Rows(0).Item(12) = 13
s(4).Rows(0).Item(13) = 0
s(4).Rows(0).Item(14) = 14
s(4).Rows(0).Item(15) = 9
s(4).Rows(1).Item(0) = 14
s(4).Rows(1).Item(1) = 11
s(4).Rows(1).Item(2) = 2
s(4).Rows(1).Item(3) = 12
s(4).Rows(1).Item(4) = 4
s(4).Rows(1).Item(5) = 7
s(4).Rows(1).Item(6) = 13
s(4).Rows(1).Item(7) = 1
s(4).Rows(1).Item(8) = 5
s(4).Rows(1).Item(9) = 0
s(4).Rows(1).Item(10) = 15
s(4).Rows(1).Item(11) = 10
s(4).Rows(1).Item(12) = 3
s(4).Rows(1).Item(13) = 9
s(4).Rows(1).Item(14) = 8
s(4).Rows(1).Item(15) = 6
s(4).Rows(2).Item(0) = 4
s(4).Rows(2).Item(1) = 2
s(4).Rows(2).Item(2) = 1
s(4).Rows(2).Item(3) = 11
![Page 80: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/80.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
s(4).Rows(2).Item(4) = 10
s(4).Rows(2).Item(5) = 13
s(4).Rows(2).Item(6) = 7
s(4).Rows(2).Item(7) = 8
s(4).Rows(2).Item(8) = 15
s(4).Rows(2).Item(9) = 9
s(4).Rows(2).Item(10) = 12
s(4).Rows(2).Item(11) = 5
s(4).Rows(2).Item(12) = 6
s(4).Rows(2).Item(13) = 3
s(4).Rows(2).Item(14) = 0
s(4).Rows(2).Item(15) = 14
s(4).Rows(3).Item(0) = 11
s(4).Rows(3).Item(1) = 8
s(4).Rows(3).Item(2) = 12
s(4).Rows(3).Item(3) = 7
s(4).Rows(3).Item(4) = 0
s(4).Rows(3).Item(5) = 14
s(4).Rows(3).Item(6) = 2
s(4).Rows(3).Item(7) = 13
s(4).Rows(3).Item(8) = 6
s(4).Rows(3).Item(9) = 15
s(4).Rows(3).Item(10) = 0
s(4).Rows(3).Item(11) = 9
s(4).Rows(3).Item(12) = 10
s(4).Rows(3).Item(13) = 4
s(4).Rows(3).Item(14) = 5
s(4).Rows(3).Item(15) = 3
![Page 81: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/81.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
End Sub
Khôûi taïo haøm s5
Sub khoitao_s5()
Dim i As Integer
s(5) = New DataTable
For i = 0 To 15
Dim col As DataColumn = New DataColumn
s(5).Columns.Add(col)
Next
For i = 0 To 3
Dim row As DataRow = s(5).NewRow
s(5).Rows.Add(row)
Next
s(5).Rows(0).Item(0) = 12
s(5).Rows(0).Item(1) = 1
s(5).Rows(0).Item(2) = 10
s(5).Rows(0).Item(3) = 15
s(5).Rows(0).Item(4) = 9
s(5).Rows(0).Item(5) = 2
s(5).Rows(0).Item(6) = 6
s(5).Rows(0).Item(7) = 8
s(5).Rows(0).Item(8) = 0
s(5).Rows(0).Item(9) = 13
s(5).Rows(0).Item(10) = 3
s(5).Rows(0).Item(11) = 4
s(5).Rows(0).Item(12) = 14
s(5).Rows(0).Item(13) = 7
s(5).Rows(0).Item(14) = 5
![Page 82: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/82.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
s(5).Rows(0).Item(15) = 11
s(5).Rows(1).Item(0) = 10
s(5).Rows(1).Item(1) = 15
s(5).Rows(1).Item(2) = 4
s(5).Rows(1).Item(3) = 2
s(5).Rows(1).Item(4) = 7
s(5).Rows(1).Item(5) = 12
s(5).Rows(1).Item(6) = 9
s(5).Rows(1).Item(7) = 5
s(5).Rows(1).Item(8) = 6
s(5).Rows(1).Item(9) = 1
s(5).Rows(1).Item(10) = 13
s(5).Rows(1).Item(11) = 14
s(5).Rows(1).Item(12) = 0
s(5).Rows(1).Item(13) = 11
s(5).Rows(1).Item(14) = 3
s(5).Rows(1).Item(15) = 8
s(5).Rows(2).Item(0) = 9
s(5).Rows(2).Item(1) = 14
s(5).Rows(2).Item(2) = 15
s(5).Rows(2).Item(3) = 5
s(5).Rows(2).Item(4) = 2
s(5).Rows(2).Item(5) = 8
s(5).Rows(2).Item(6) = 12
s(5).Rows(2).Item(7) = 3
s(5).Rows(2).Item(8) = 7
s(5).Rows(2).Item(9) = 0
s(5).Rows(2).Item(10) = 4
![Page 83: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/83.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
s(5).Rows(2).Item(11) = 10
s(5).Rows(2).Item(12) = 1
s(5).Rows(2).Item(13) = 13
s(5).Rows(2).Item(14) = 11
s(5).Rows(2).Item(15) = 6
s(5).Rows(3).Item(0) = 4
s(5).Rows(3).Item(1) = 3
s(5).Rows(3).Item(2) = 2
s(5).Rows(3).Item(3) = 12
s(5).Rows(3).Item(4) = 9
s(5).Rows(3).Item(5) = 5
s(5).Rows(3).Item(6) = 15
s(5).Rows(3).Item(7) = 10
s(5).Rows(3).Item(8) = 11
s(5).Rows(3).Item(9) = 14
s(5).Rows(3).Item(10) = 1
s(5).Rows(3).Item(11) = 7
s(5).Rows(3).Item(12) = 6
s(5).Rows(3).Item(13) = 0
s(5).Rows(3).Item(14) = 8
s(5).Rows(3).Item(15) = 13
End Sub
Khôûi taïo haøm s6
Sub khoitao_s6()
Dim i As Integer
s(6) = New DataTable
For i = 0 To 15
Dim col As DataColumn = New DataColumn
![Page 84: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/84.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
s(6).Columns.Add(col)
Next
For i = 0 To 3
Dim row As DataRow = s(6).NewRow
s(6).Rows.Add(row)
Next
s(6).Rows(0).Item(0) = 4
s(6).Rows(0).Item(1) = 11
s(6).Rows(0).Item(2) = 2
s(6).Rows(0).Item(3) = 14
s(6).Rows(0).Item(4) = 15
s(6).Rows(0).Item(5) = 0
s(6).Rows(0).Item(6) = 8
s(6).Rows(0).Item(7) = 13
s(6).Rows(0).Item(8) = 3
s(6).Rows(0).Item(9) = 12
s(6).Rows(0).Item(10) = 9
s(6).Rows(0).Item(11) = 7
s(6).Rows(0).Item(12) = 5
s(6).Rows(0).Item(13) = 10
s(6).Rows(0).Item(14) = 6
s(6).Rows(0).Item(15) = 1
s(6).Rows(1).Item(0) = 13
s(6).Rows(1).Item(1) = 0
s(6).Rows(1).Item(2) = 11
s(6).Rows(1).Item(3) = 7
s(6).Rows(1).Item(4) = 4
s(6).Rows(1).Item(5) = 9
![Page 85: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/85.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
s(6).Rows(1).Item(6) = 1
s(6).Rows(1).Item(7) = 10
s(6).Rows(1).Item(8) = 14
s(6).Rows(1).Item(9) = 3
s(6).Rows(1).Item(10) = 5
s(6).Rows(1).Item(11) = 12
s(6).Rows(1).Item(12) = 2
s(6).Rows(1).Item(13) = 15
s(6).Rows(1).Item(14) = 8
s(6).Rows(1).Item(15) = 6
s(6).Rows(2).Item(0) = 1
s(6).Rows(2).Item(1) = 4
s(6).Rows(2).Item(2) = 11
s(6).Rows(2).Item(3) = 13
s(6).Rows(2).Item(4) = 12
s(6).Rows(2).Item(5) = 3
s(6).Rows(2).Item(6) = 7
s(6).Rows(2).Item(7) = 14
s(6).Rows(2).Item(8) = 10
s(6).Rows(2).Item(9) = 15
s(6).Rows(2).Item(10) = 6
s(6).Rows(2).Item(11) = 8
s(6).Rows(2).Item(12) = 0
s(6).Rows(2).Item(13) = 5
s(6).Rows(2).Item(14) = 9
s(6).Rows(2).Item(15) = 2
s(6).Rows(3).Item(0) = 6
s(6).Rows(3).Item(1) = 11
![Page 86: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/86.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
s(6).Rows(3).Item(2) = 13
s(6).Rows(3).Item(3) = 8
s(6).Rows(3).Item(4) = 1
s(6).Rows(3).Item(5) = 4
s(6).Rows(3).Item(6) = 10
s(6).Rows(3).Item(7) = 7
s(6).Rows(3).Item(8) = 9
s(6).Rows(3).Item(9) = 5
s(6).Rows(3).Item(10) = 0
s(6).Rows(3).Item(11) = 15
s(6).Rows(3).Item(12) = 14
s(6).Rows(3).Item(13) = 2
s(6).Rows(3).Item(14) = 3
s(6).Rows(3).Item(15) = 12
End Sub
Khôûi taïo haøm s7
Sub khoitao_s7()
Dim i As Integer
s(7) = New DataTable
For i = 0 To 15
Dim col As DataColumn = New DataColumn
s(7).Columns.Add(col)
Next
For i = 0 To 3
Dim row As DataRow = s(7).NewRow
s(7).Rows.Add(row)
Next
s(7).Rows(0).Item(0) = 13
![Page 87: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/87.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
s(7).Rows(0).Item(1) = 2
s(7).Rows(0).Item(2) = 8
s(7).Rows(0).Item(3) = 4
s(7).Rows(0).Item(4) = 6
s(7).Rows(0).Item(5) = 15
s(7).Rows(0).Item(6) = 11
s(7).Rows(0).Item(7) = 1
s(7).Rows(0).Item(8) = 10
s(7).Rows(0).Item(9) = 9
s(7).Rows(0).Item(10) = 3
s(7).Rows(0).Item(11) = 14
s(7).Rows(0).Item(12) = 5
s(7).Rows(0).Item(13) = 0
s(7).Rows(0).Item(14) = 12
s(7).Rows(0).Item(15) = 7
s(7).Rows(1).Item(0) = 1
s(7).Rows(1).Item(1) = 15
s(7).Rows(1).Item(2) = 13
s(7).Rows(1).Item(3) = 8
s(7).Rows(1).Item(4) = 10
s(7).Rows(1).Item(5) = 3
s(7).Rows(1).Item(6) = 7
s(7).Rows(1).Item(7) = 4
s(7).Rows(1).Item(8) = 12
s(7).Rows(1).Item(9) = 5
s(7).Rows(1).Item(10) = 6
s(7).Rows(1).Item(11) = 11
s(7).Rows(1).Item(12) = 0
![Page 88: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/88.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
s(7).Rows(1).Item(13) = 14
s(7).Rows(1).Item(14) = 9
s(7).Rows(1).Item(15) = 2
s(7).Rows(2).Item(0) = 7
s(7).Rows(2).Item(1) = 11
s(7).Rows(2).Item(2) = 4
s(7).Rows(2).Item(3) = 1
s(7).Rows(2).Item(4) = 9
s(7).Rows(2).Item(5) = 12
s(7).Rows(2).Item(6) = 14
s(7).Rows(2).Item(7) = 2
s(7).Rows(2).Item(8) = 0
s(7).Rows(2).Item(9) = 6
s(7).Rows(2).Item(10) = 10
s(7).Rows(2).Item(11) = 13
s(7).Rows(2).Item(12) = 15
s(7).Rows(2).Item(13) = 3
s(7).Rows(2).Item(14) = 5
s(7).Rows(2).Item(15) = 8
s(7).Rows(3).Item(0) = 2
s(7).Rows(3).Item(1) = 1
s(7).Rows(3).Item(2) = 14
s(7).Rows(3).Item(3) = 7
s(7).Rows(3).Item(4) = 4
s(7).Rows(3).Item(5) = 10
s(7).Rows(3).Item(6) = 8
s(7).Rows(3).Item(7) = 13
s(7).Rows(3).Item(8) = 15
![Page 89: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/89.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
s(7).Rows(3).Item(9) = 12
s(7).Rows(3).Item(10) = 9
s(7).Rows(3).Item(11) = 0
s(7).Rows(3).Item(12) = 3
s(7).Rows(3).Item(13) = 5
s(7).Rows(3).Item(14) = 6
s(7).Rows(3).Item(15) = 11
End Sub
Khôûi taïo hoaùn vò e
Sub khoitao_hve()
hve(0) = 32
hve(1) = 1
hve(2) = 2
hve(3) = 3
hve(4) = 4
hve(5) = 5
hve(6) = 4
hve(7) = 5
hve(8) = 6
hve(9) = 7
hve(10) = 8
hve(11) = 9
hve(12) = 8
hve(13) = 9
hve(14) = 10
hve(15) = 11
hve(16) = 12
hve(17) = 13
![Page 90: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/90.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
hve(18) = 12
hve(19) = 13
hve(20) = 14
hve(21) = 15
hve(22) = 16
hve(23) = 17
hve(24) = 16
hve(25) = 17
hve(26) = 18
hve(27) = 19
hve(28) = 20
hve(29) = 21
hve(30) = 20
hve(31) = 21
hve(32) = 22
hve(33) = 23
hve(34) = 24
hve(35) = 25
hve(36) = 24
hve(37) = 25
hve(38) = 26
hve(39) = 27
hve(40) = 28
hve(41) = 29
hve(42) = 28
hve(43) = 29
hve(44) = 30
hve(45) = 31
![Page 91: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/91.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
hve(46) = 32
hve(47) = 1
End Sub
Khôûi taïo hoaùn vò p
Sub khoitao_hvp()
hvp(0) = 16
hvp(1) = 7
hvp(2) = 20
hvp(3) = 21
hvp(4) = 29
hvp(5) = 12
hvp(6) = 28
hvp(7) = 17
hvp(8) = 1
hvp(9) = 15
hvp(10) = 23
hvp(11) = 26
hvp(12) = 5
hvp(13) = 18
hvp(14) = 31
hvp(15) = 10
hvp(16) = 2
hvp(17) = 8
hvp(18) = 24
hvp(19) = 14
hvp(20) = 32
hvp(21) = 27
hvp(22) = 3
![Page 92: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/92.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
hvp(23) = 9
hvp(24) = 19
hvp(25) = 13
hvp(26) = 30
hvp(27) = 6
hvp(28) = 22
hvp(29) = 11
hvp(30) = 4
hvp(31) = 25
End Sub
Khôûi taïo hoaùn vò pc2
Sub khoitao_hvpc2()
hvpc2(0) = 14
hvpc2(1) = 17
hvpc2(2) = 11
hvpc2(3) = 24
hvpc2(4) = 1
hvpc2(5) = 5
hvpc2(6) = 3
hvpc2(7) = 28
hvpc2(8) = 15
hvpc2(9) = 6
hvpc2(10) = 21
hvpc2(11) = 10
hvpc2(12) = 23
hvpc2(13) = 19
hvpc2(14) = 12
hvpc2(15) = 4
![Page 93: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/93.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
hvpc2(16) = 26
hvpc2(17) = 8
hvpc2(18) = 16
hvpc2(19) = 7
hvpc2(20) = 27
hvpc2(21) = 20
hvpc2(22) = 13
hvpc2(23) = 2
hvpc2(24) = 41
hvpc2(25) = 52
hvpc2(26) = 31
hvpc2(27) = 37
hvpc2(28) = 47
hvpc2(29) = 55
hvpc2(30) = 30
hvpc2(31) = 40
hvpc2(32) = 51
hvpc2(33) = 45
hvpc2(34) = 33
hvpc2(35) = 48
hvpc2(36) = 44
hvpc2(37) = 49
hvpc2(38) = 39
hvpc2(39) = 56
hvpc2(40) = 34
hvpc2(41) = 53
hvpc2(42) = 46
hvpc2(43) = 42
![Page 94: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/94.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
hvpc2(44) = 50
hvpc2(45) = 36
hvpc2(46) = 29
hvpc2(47) = 32
End Sub
Khôûi taïo hoaù vò pc1
Sub khoitao_hvpc1()
hvpc1(0) = 57
hvpc1(1) = 49
hvpc1(2) = 41
hvpc1(3) = 33
hvpc1(4) = 25
hvpc1(5) = 17
hvpc1(6) = 9
hvpc1(7) = 1
hvpc1(8) = 58
hvpc1(9) = 50
hvpc1(10) = 42
hvpc1(11) = 34
hvpc1(12) = 26
hvpc1(13) = 18
hvpc1(14) = 10
hvpc1(15) = 2
hvpc1(16) = 59
hvpc1(17) = 51
hvpc1(18) = 43
hvpc1(19) = 35
hvpc1(20) = 27
![Page 95: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/95.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
hvpc1(21) = 19
hvpc1(22) = 11
hvpc1(23) = 3
hvpc1(24) = 60
hvpc1(25) = 52
hvpc1(26) = 44
hvpc1(27) = 36
hvpc1(28) = 63
hvpc1(29) = 55
hvpc1(30) = 47
hvpc1(31) = 39
hvpc1(32) = 31
hvpc1(33) = 23
hvpc1(34) = 15
hvpc1(35) = 7
hvpc1(36) = 62
hvpc1(37) = 54
hvpc1(38) = 46
hvpc1(39) = 38
hvpc1(40) = 30
hvpc1(41) = 22
hvpc1(42) = 14
hvpc1(43) = 6
hvpc1(44) = 61
hvpc1(45) = 53
hvpc1(46) = 45
hvpc1(47) = 37
hvpc1(48) = 29
![Page 96: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/96.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
hvpc1(49) = 21
hvpc1(50) = 13
hvpc1(51) = 5
hvpc1(52) = 28
hvpc1(53) = 20
hvpc1(54) = 12
hvpc1(55) = 4
End Sub
Khôûi taïo hoaù vò ip
Sub khoitao_hvip()
hvip(0) = 58
hvip(1) = 50
hvip(2) = 42
hvip(3) = 34
hvip(4) = 26
hvip(5) = 18
hvip(6) = 10
hvip(7) = 2
hvip(8) = 60
hvip(9) = 52
hvip(10) = 44
hvip(11) = 36
hvip(12) = 28
hvip(13) = 20
hvip(14) = 12
hvip(15) = 4
hvip(16) = 62
hvip(17) = 54
![Page 97: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/97.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
hvip(18) = 46
hvip(19) = 38
hvip(20) = 30
hvip(21) = 22
hvip(22) = 14
hvip(23) = 6
hvip(24) = 64
hvip(25) = 56
hvip(26) = 48
hvip(27) = 40
hvip(28) = 32
hvip(29) = 24
hvip(30) = 16
hvip(31) = 8
hvip(32) = 57
hvip(33) = 49
hvip(34) = 41
hvip(35) = 33
hvip(36) = 25
hvip(37) = 17
hvip(38) = 9
hvip(39) = 1
hvip(40) = 59
hvip(41) = 51
hvip(42) = 43
hvip(43) = 35
hvip(44) = 27
hvip(45) = 19
![Page 98: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/98.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
hvip(46) = 11
hvip(47) = 3
hvip(48) = 61
hvip(49) = 53
hvip(50) = 45
hvip(51) = 37
hvip(52) = 29
hvip(53) = 21
hvip(54) = 13
hvip(55) = 5
hvip(56) = 63
hvip(57) = 55
hvip(58) = 47
hvip(59) = 39
hvip(60) = 31
hvip(61) = 23
hvip(62) = 15
hvip(63) = 7
End Sub
Khôûi taïo caùc haøm
Sub khoitao()
khoitao_s0()
khoitao_s1()
khoitao_s2()
khoitao_s3()
khoitao_s4()
khoitao_s5()
khoitao_s6()
![Page 99: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/99.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
khoitao_s7()
khoitao_hve()
khoitao_hvp()
khoitao_hvpc2()
khoitao_hvpc1()
khoitao_hvip()
End Sub
caùc haøm hoaùn vò
Haøm hoaùn vò ip
Function hoanvi_ip(ByVal x As String) As String
Dim tam(63) As Char
Dim i As Integer
For i = 0 To 63
tam(i) = x.Substring(hvip(i) - 1, 1)
Next
Return tam
End Function
Haøm hoaùn vò iptru
Function hoanvi_iptru(ByVal c As String) As String
Dim tam(63) As Char
Dim i As Integer
For i = 0 To 63
tam(hvip(i) - 1) = c.Substring(i, 1)
Next
Return tam
End Function
Haøm hoaùn vò e
Function hoanvi_e(ByVal r As String) As String
![Page 100: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/100.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
Dim tam(47) As Char
Dim i As Integer
For i = 0 To 47
tam(i) = r.Substring(hve(i) - 1, 1)
Next
Return tam
End Function
Haøm hoaùn vò p
Function hoanvi_p(ByVal c As String) As String
Dim tam(31) As Char
Dim i As Integer
For i = 0 To 31
tam(i) = c.Substring(hvp(i) - 1, 1)
Next
Return tam
End Function
Haøm hoaùn vò ptru
Function hoanvi_ptru(ByVal c As String) As String
Dim tam(31) As Char
Dim i As Integer
For i = 0 To 31
tam(hvp(i) - 1) = c.Substring(i, 1)
Next
Return tam
End Function
Haøm hoaùn vò pc1
Function hoanvi_pc1(ByVal k As String) As String
Dim tam(63) As Char
![Page 101: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/101.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
Dim i As Integer
For i = 0 To 63
tam(i) = k.Substring(i, 1)
Next
tam = catbitcuoi(tam)
For i = 0 To 55
tam(i) = k.Substring(Integer.Parse(hvpc1(i) - 1), 1)
Next
Return tam
End Function
Haøm hoaùn vò pc1tru
Function hoanvi_pc1tru(ByVal c As String) As String
Dim tam(63) As Char
Dim i As Integer
For i = 0 To 63
tam(i) = "#"
Next
For i = 0 To 55
tam(hvpc1(i) - 1) = c.Substring(i, 1)
Next
Dim tam1 As String
Return tam
End Function
Haøm hoaùn vò pc2
Function hoanvi_pc2(ByVal str As String) As String
Dim tam(47) As Char
Dim i As Integer
![Page 102: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/102.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
For i = 0 To 47
tam(i) = str.Substring(Integer.Parse(hvpc2(i) - 1), 1)
Next
Return tam
End Function
Haøm hoaùn vò pc2tru
Function hoanvi_pc2tru(ByVal c As String) As String
Dim tam(55) As Char
Dim i As Integer
For i = 0 To 55
tam(i) = "?"
Next
For i = 0 To 47
tam(hvpc2(i) - 1) = c.Substring(i, 1)
Next
Return tam
End Function
gan va kiem tra du lieu nhap
haøm gaùn döõ lieäu
Function gandulieu() As Boolean
Try
Dim i As Integer = 0
Dim j As Integer
Dim strbanro As String = txtbanro.Text.Replace(Chr(10), "")
Dim strbanma As String = txtbanma.Text.Replace(Chr(10), "")
banro = strbanro.Split(Chr(13))
banma = strbanma.Split(Chr(13))
n = banro.Length - 1
![Page 103: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/103.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
ReDim e((n - 1) / 2)
ReDim esao((n - 1) / 2)
ReDim ephay((n - 1) / 2)
ReDim cphay((n - 1) / 2)
If (banro.Length - 1 < n Or banma.Length - 1 < n) Then
MessageBox.Show("thi�u b�n rõ hay b�n mã")
Return False
Else
Return True
End If
Catch ex As Exception
MessageBox.Show("D� li�u nh�p không h�p l�")
Return False
End Try
End Function
Haøm kieåm tra
Function kiemtra() As Boolean
Dim i As Integer
For i = 0 To n
If banro(i) = "" Then
MessageBox.Show("baïn nhaäp chöa ñuû " & (i + 1))
Return False
End If
If banro(i).Length <> m Then
MessageBox.Show( (i + 1) & m )
Return False
End If
Next
![Page 104: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/104.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
For i = 0 To n
If banma(i) = "" Then
MessageBox.Show("B�n hãy nh�p vào b�n mã th� " & (i + 1))
Return False
End If
If banma(i).Length <> m Then
MessageBox.Show("B�n mã " & (i + 1) & " ph�i có " & m & " kí t�")
Return False
End If
Next
i = 0
While i <= n
If banro(i).Substring(m / 2, m / 2) <> banro(i + 1).Substring(m / 2, m / 2) Then
MessageBox.Show("D� li�u b�n rõ " & (i + 1) & " , " & (i + 2) & " nh�p không h�p l�")
Return False
End If
i = i + 2
End While
Return True
End Function
xu li e_esao_ephay_cphay
xöû lyù e, e sao, e phaåy
Sub xuli_e_esao_ephay()
Dim i As Integer
'e(0)=
![Page 105: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/105.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
"000000000111111000001110100000000110100000001100"
'e(1)= "101000001011111111110100000101010000001011110110"
'e(2)= "111011110001010100000110100011110110100101011111"
'esao(0)="101111110000001010101100000001010100000001010010"
'esao(1)="100010100110101001011110101111110010100010101010"
'esao(2) = "000001011110100110100010101111110101011000000100"
For i = 0 To (n - 1) / 2
Dim l3 = banma(i * 2).Substring(0, 32)
Dim l3sao = banma(i * 2 + 1).Substring(0, 32)
e(i) = hoanvi_e(l3)
esao(i) = hoanvi_e(l3sao)
ephay(i) = phay(e(i), esao(i))
Next
End Sub
Function phay(ByVal a As String, ByVal b As String) As String
Dim i As Integer
Dim c As String
For i = 0 To a.Length - 1
c += (a.Substring(i, 1) Xor b.Substring(i, 1)).ToString
Next
Return c
End Function
Xöû lyù c phaåy
Sub xuli_cphay()
Dim i, j As Integer
For i = 0 To (n - 1) / 2
Dim r3 As String = banma(i * 2).Substring(32, 32)
![Page 106: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/106.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
Dim r3sao As String = banma(i * 2 + 1).Substring(32, 32)
Dim l0 As String = banro(i * 2).Substring(0, 32)
Dim l0sao As String = banro(i * 2 + 1).Substring(0, 32)
Dim r3phay As String = ""
Dim l0phay As String = ""
For j = 0 To 31
r3phay += (r3.Substring(j, 1) Xor r3sao.Substring(j, 1)).ToString
l0phay += (l0.Substring(j, 1) Xor l0sao.Substring(j, 1)).ToString
cphay(i) += (r3phay.Substring(j, 1) Xor l0phay.Substring(j, 1)).ToString
Next
cphay(i) = hoanvi_ptru(cphay(i))
Next
'cphay(0) = "10010110010111010101101101100111"
'cphay(1) = "10011100100111000001111101010110"
'cphay(2) = "11010101011101011101101100101011"
End Sub
cac ham chuyen doi
Function bi_str(ByVal a As String) As String
Dim i As Integer
Dim b As String
For i = 0 To a.Length - 1
b += binary(asc(a.Substring(i, 1)), hebit)
Next
Return b
End Function
Sub bi_banro_banma()
Dim i As Integer
![Page 107: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/107.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
For i = 0 To n
banro(i) = bi_str(banro(i))
banma(i) = bi_str(banma(i))
Next
End Sub
Function bi_so(ByVal a As Integer) As String
Dim i As Integer
Dim tam(5) As Char
For i = 0 To 5
tam(i) = (a Mod 2).ToString
a \= 2
Next
Array.Reverse(tam)
Return tam
End Function
Function thapphan(ByVal b As String) As Integer
Dim i As Integer
Dim tam As Integer = 0
For i = 0 To b.Length - 1
If b.Substring(i, 1) = 1 Then
tam += 2 ^ (b.Length - 1 - i)
End If
Next
Return tam
End Function
Function binary(ByVal a As Integer, ByVal n As Integer) As String
Dim i As Integer
Dim tam(n - 1) As Char
![Page 108: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/108.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
For i = 0 To n - 1
tam(i) = (a Mod 2).ToString
a = a \ 2
Next
Array.Reverse(tam)
Return tam
End Function
Haøm taïo taäp test
Function tap_test() As String()
Dim i, j, k As Integer
Dim ee(7) As String
Dim eephay(7) As String
Dim ccphay(7) As String
Dim test(63) As String
For i = 0 To (n - 1) / 2
For j = 0 To 7
ee(j) = e(i).Substring(j * 6, 6)
eephay(j) = ephay(i).Substring(j * 6, 6)
ccphay(j) = cphay(i).Substring(j * 4, 4)
test = tap_in(ee(j), eephay(j), ccphay(j), s(j))
For k = 0 To 63
If test(k) = 1 Then
jhop(j, k) += 1
End If
Next
Next
Next
End Function
![Page 109: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/109.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
Haøm taïo khoaù voøng 3
Function khoavong3() As String
Dim khoav3 As String
Dim i, j As Integer
Dim count As Integer = 0
Dim vitrimax As Integer
For i = 0 To 7
count = 0
Dim max As Integer = (n + 1) / 2
For j = 0 To 63
If jhop(i, j) = max Then
count += 1
vitrimax = j
End If
Next
If count > 1 Then
lbthongbao.Text = "Nh�p thêm b�n mã và b�n rõ, vì ch�a xác đ�nh đ��c ph�n t� max trong jh�p"
btthamma.Enabled = False
txtbanro.Focus()
n += 2
ReDim banro(n)
ReDim banma(n)
ReDim e((n - 2) / 2)
ReDim esao((n - 2) / 2)
ReDim ephay((n - 2) / 2)
ReDim cphay((n - 2) / 2)
flag = False
![Page 110: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/110.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
Exit Function
End If
khoav3 += binary(vitrimax, 6)
Next
txtkhoav3.Text = khoav3
Return khoav3
End Function
Haøm Xöû lyù khoaù
Function xulikhoa() As String
Dim khoa As String = khoavong3()
If flag = False Then
Exit Function
End If
khoa = hoanvi_pc2tru(khoa)
Dim haitambitdau As String = khoa.Substring(0, 28)
Dim haitambitcuoi As String = khoa.Substring(28, 28)
haitambitdau = dichphai(haitambitdau, 4)
haitambitcuoi = dichphai(haitambitcuoi, 4)
khoa = haitambitdau + haitambitcuoi
Dim i As Integer
khoa = hoanvi_pc1tru(khoa)
txtkhoa.Text = khoa
Return khoa 'khoa 56 bit
End Function
Haøm taïo chuoãi khoaù
Sub taochuoikhoa(ByVal c As String)
If flag = False Then
Exit Sub
![Page 111: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/111.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
End If
Dim chuoibinary(255) As String
Dim i As Integer
For i = 0 To 255
chuoibinary(i) = binary(i, 8)
chuoikhoa(i) = taotungkhoa(c, chuoibinary(i))
Next
End Sub
Haøm taïo töøng khoaù
Function taotungkhoa(ByVal a As String, ByVal b As String) As String
Dim i As Integer
Dim j As Integer = 0
Dim tam(63) As Char
For i = 0 To 63
tam(i) = a.Substring(i, 1)
If tam(i) = "?" Then
tam(i) = b.Substring(j, 1)
j += 1
End If
Next
Return tam
End Function
Haøm dòch phaûi
Function dichphai(ByVal s As String, ByVal n As Integer) As String
Return s.Substring(s.Length - n, n) + s.Substring(0, s.Length - n)
End Function
Function tap_in(ByVal e As String, ByVal ephay As String, ByVal cphay As String, ByVal s As DataTable) As String()
![Page 112: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/112.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
Dim tam(63), b, bsao, c, csao, cphaytam As String
Dim i As Integer
For i = 0 To 63
b = bi_so(i)
Dim haibitcuoi As String = b.Substring(0, 1) + b.Substring(5, 1)
Dim bonbitgiua As String = b.Substring(1, 4)
Dim srow = thapphan(haibitcuoi)
Dim scol = thapphan(bonbitgiua)
Dim sij As Integer = s.Rows(srow).Item(scol)
c = binary(sij, 4)
bsao = phay(ephay, b)
haibitcuoi = bsao.Substring(0, 1) + bsao.Substring(5, 1)
bonbitgiua = bsao.Substring(1, 4)
srow = thapphan(haibitcuoi)
scol = thapphan(bonbitgiua)
sij = s.Rows(srow).Item(scol)
csao = binary(sij, 4)
cphaytam = phay(c, csao)
If cphaytam = cphay Then
tam(thapphan(phay(e, b))) = 1
End If
Next
Return tam
End Function
Moät soá Haøm cho xöû lyù söï kieän
Private Sub btthamma_Click(ByVal sender As System.Object, ByVal ee As System.EventArgs) Handles btthamma.Click
ReDim banro(n)
![Page 113: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/113.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
ReDim banma(n)
ReDim e((n - 1) / 2)
ReDim esao((n - 1) / 2)
ReDim ephay((n - 1) / 2)
ReDim cphay((n - 1) / 2)
flag = True
If gandulieu() = False Then
Exit Sub
End If
If kiemtra() = False Then
Exit Sub
End If
bi_banro_banma()
xuli_e_esao_ephay()
xuli_cphay()
tap_test()
taochuoikhoa(xulikhoa())
If flag = False Then
Exit Sub
End If
chonkhoadung()
End Sub
Private Sub btthoat_Click(ByVal sender As System.Object, ByVal ee As System.EventArgs) Handles btthoat.Click
Me.Close()
End Sub
Private Sub thammades_Load(ByVal sender As System.Object, ByVal ee As System.EventArgs) Handles MyBase.Load
![Page 114: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/114.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
Dim strbanro As String
strbanro += "748502CD38451097" + Chr(13) + Chr(10)
strbanro += "3874756438451097" + Chr(13) + Chr(10)
strbanro += "486911026ACDFF31" + Chr(13) + Chr(10)
strbanro += "375BD31F6ACDFF31" + Chr(13) + Chr(10)
strbanro += "357418DA013FEC86" + Chr(13) + Chr(10)
strbanro += "12549847013FEC86"
txtbanro.Text = strbanro
Dim strbanma As String
strbanma += "03C70306D8A09F10" + Chr(13) + Chr(10)
strbanma += "78560A0960E6D4CB" + Chr(13) + Chr(10)
strbanma += "45FA285BE5ADC730" + Chr(13) + Chr(10)
strbanma += "134F7915AC253457" + Chr(13) + Chr(10)
strbanma += "D8A31B2F28BBC5CF" + Chr(13) + Chr(10)
strbanma += "0F317AC2B23CB944"
txtbanma.Text = strbanma
khoitao()
End Sub
'des
Haøm caét bit cuoái
Function catbitcuoi(ByVal k As String) As String 'dua vao 64 bit tra ra 56 bit
Dim i As Integer = 0
Dim j As Integer
Dim tam As String
While i < 63
For j = i To i + 6
tam += k.Substring(j, 1)
![Page 115: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/115.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
Next
i = i + 8
End While
Return tam
End Function
Function ls(ByVal s As String, ByVal n As Integer) As String
Return s.Substring(n, s.Length - n) + s.Substring(0, n)
End Function
Haøm taïo daõy khoaù
Sub taodaykhoa(ByVal khoa As String)
khoa = hoanvi_pc1(khoa)
Dim d(2) As String
Dim c(2) As String
c(0) = khoa.Substring(0, 28)
c(0) = ls(c(0), 1)
d(0) = khoa.Substring(28, 28)
d(0) = ls(d(0), 1)
daykhoa(0) = hoanvi_pc2(c(0) + d(0))
Dim i As Integer
For i = 1 To 2
If i = 1 Then
c(i) = ls(c(i - 1), 1)
d(i) = ls(d(i - 1), 1)
Else
c(i) = ls(c(i - 1), 2)
d(i) = ls(d(i - 1), 2)
End If
daykhoa(i) = hoanvi_pc2(c(i) + d(i))
![Page 116: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/116.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
Next i
End Sub
Haøm xöû lyù chuoãi nhaäp
Function bi_acsii(ByVal int As Integer) As String
Dim tam(7) As Char
Dim i As Integer
For i = 0 To 7
tam(i) = (int Mod 2).ToString
int \= 2
Next
Array.Reverse(tam)
Return tam
End Function
Haøm maõ hoaù
Function mahoa(ByVal x As String) As String
Dim i, j As Integer
Dim l(2) As String
Dim r(2) As String
'x = hoanvi_ip(x)
Dim l0 As String = x.Substring(0, 32)
Dim r0 As String = x.Substring(32, 32)
l(0) = r0
For i = 0 To 31
r(0) += (l0.Substring(i, 1) Xor f(r0, daykhoa(0)).Substring(i, 1)).ToString
Next
For i = 1 To 2
l(i) = r(i - 1)
![Page 117: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/117.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
Dim a As String = f(r(i - 1), daykhoa(i))
For j = 0 To 31
r(i) += (l(i - 1).Substring(j, 1) Xor a.Substring(j, 1)).ToString
Next j
Next i
Dim t As String = l(2) + r(2)
Return t
'Return hoanvi_iptru(r(2) + l(2))
End Function
Haøm taïo haøm f
Function f(ByVal r As String, ByVal daykhoa_k As String) As String
Dim i As Integer
Dim e As String
Dim hv As String = hoanvi_e(r)
For i = 0 To 47
e += (hv.Substring(i, 1) Xor daykhoa_k.Substring(i, 1)).ToString
Next
Dim b(7) As String
Dim c As String
For i = 0 To 7
b(i) = e.Substring(i * 6, 6)
Dim haibitdaucuoi As String = b(i).Substring(0, 1) + b(i).Substring(5, 1)
Dim bonbitgiua As String = b(i).Substring(1, 4)
Dim srow = thapphan(haibitdaucuoi)
Dim scol = thapphan(bonbitgiua)
Dim sij As Integer = s(i).Rows(srow).Item(scol)
c += binary(sij, 4)
![Page 118: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/118.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
Next
Return hoanvi_p(c)
End Function
Haøm ñoåi ra chöõ
'Function doirachu(ByVal y As String) As String
'Dim tam As String = y
'Dim tam1 As String = ""
'Dim so As Integer
'Dim i As Integer
'Dim j As Integer
'While i < tam.Length - 1
'so = 0
'For j = i To i + 7
'If tam.Substring(j, 1) = 1 Then
'so += 2 ^ (7 - (j - i))
'End If
'Next
'tam1 += Chr(so)
'i = i + 8
'End While
'Return tam1
'End Function
Function doirachu(ByVal y As String) As String
Dim tam As String = y
Dim tam1 As String = ""
Dim so As Integer
Dim i, j As Integer
While i < tam.Length - 1
![Page 119: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/119.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
so = 0
For j = i To i + 3
If tam.Substring(j, 1) = 1 Then
so += 2 ^ (3 - (j - i))
End If
Next
tam1 += chucai(so)
i = i + 4
End While
Return tam1
End Function
Chöõ caùi
Function chucai(ByVal so As Integer) As Char
Select Case so
Case 0
Return "0"
Case 1
Return "1"
Case 2
Return "2"
Case 3
Return "3"
Case 4
Return "4"
Case 5
Return "5"
Case 6
Return "6"
![Page 120: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/120.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
Case 7
Return "7"
Case 8
Return "8"
Case 9
Return "9"
Case 10
Return "A"
Case 11
Return "B"
Case 12
Return "C"
Case 13
Return "D"
Case 14
Return "E"
Case 15
Return "F"
End Select
End Function
Haøm chuyeån maõ asc
Function asc(ByVal a As Char)
Select Case a
Case "0"
asc = 0
Case "1"
asc = 1
Case "2"
![Page 121: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/121.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
asc = 2
Case "3"
asc = 3
Case "4"
asc = 4
Case "5"
asc = 5
Case "6"
asc = 6
Case "7"
asc = 7
Case "8"
asc = 8
Case "9"
asc = 9
Case "A"
asc = 10
Case "B"
asc = 11
Case "C"
asc = 12
Case "D"
asc = 13
Case "E"
asc = 14
Case "F"
asc = 15
End Select
![Page 122: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/122.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
End Function
Haøm choïn khoaù
Sub chonkhoadung()
Dim i As Integer
Dim x As String = banro(0)
'For i = 0 To banro(0).Length - 1
'x += binary(asc(banro(0).Substring(i, 1)), hebit)
'Next
For i = 0 To 255
taodaykhoa(chuoikhoa(i))
Dim banma1 As String = doirachu(mahoa(x))
If banma1 = doirachu(banma(0)) Then
chuoikhoa(i) = chuoikhoa(i).Replace("#", "0")
txtkhoa.Text = doirachu(chuoikhoa(i))
Exit Sub
End If
Next
MessageBox.Show("khong tim duoc khoa")
End Sub
Private Sub txtbanro_TextChanged(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles txtbanro.TextChanged
btthamma.Enabled = True
lbthongbao.Text = ""
End Sub
Private Sub txtbanma_TextChanged(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles txtbanma.TextChanged
![Page 123: bctntlvn (42).pdf](https://reader034.fdocuments.net/reader034/viewer/2022042507/55824eabd8b42a213a8b534f/html5/thumbnails/123.jpg)
ÑOÀ AÙN BAÛO MAÄT THOÂNG TIN HEÄ MAÕ DES
NGOÂ THÒ TUYEÁT HAØ – T012825
btthamma.Enabled = True
lbthongbao.Text = ""
End Sub
End Class