thám mã des

N BO MT THNG TIN H M DES

NG TH TUYT H T012825

ti:

Tm hiu h m chun ci t des v

thm m 3 vng



MC LC

I .1 Gii thiu........................................................................................................ 4

I.2 Cc H M Thng Dng: ................................................................................... 5

e. Phng php Affine ............................................................................................ 7

f. Phng php Vigenere ........................................................................................ 8

I.2 LP M DES ................................................................................................. 22

I. 3 THM M DES ............................................................................................ 27

I.3.1. Thm m h DES - 3 vng ................................................................... 32

II.3.2. Thm m h DES 6-vng........................................................................ 38

II.3. 3 Cc thm m vi sai khc ........................................................................ 44

III. CI T THM M DES 3 VNG ....................................................... 45

III.1 Giao Din . .................................................................................................... 45

III.2 X L .............................................................................................................



LI NI U

Hin nay, nc ta ang trong giai on tin hnh cng nghip ha, hin i ha t nc.

Tin hc c xem l mt trong nhng ngnh mi nhn. Tin hc v ang ng gp rt nhiu

cho x hi trong mi kha cnh ca cuc sng.

M ha thng tin l mt ngnh quan trng v c nhiu ng dng trong i sng x hi.

Ngy nay, cc ng dng m ha v bo mt thng tin ang c s dng ngy cng ph bin

hn trong cc lnh vc khc nhau trn Th gii, t cc lnh vc an ninh, qun s, quc

phng, cho n cc lnh vc dn s nh thng mi in t, ngn hng

ng dng m ha v bo mt thng tin trong cc h thng thng mi in t, giao dch

chng khn, tr nn ph bin trn th gii v s ngy cng tr nn quen thuc vi ngi

dn Vit Nam. Thng 7/2000, th trng chng khn ln u tin c hnh thnh ti Vit

Nam; cc th tn dng bt u c s dng, cc ng dng h thng thng mi in t ang

bc u c quan tm v xy dng. Do , nhu cu v cc ng dng m ha v bo mt

thng tin tr nn rt cn thit.



I. MT S PHNG PHP M HA

I .1 Gii thiu

nh ngha 1.1: Mt h m mt (cryptosystem) l mt b-nm (P, C, K, E, D) tha mn

cc iu kin sau:

1. P l khng gian bn r. tp hp hu hn tt c cc mu tin ngun cn m ha c th c

2. C l khng gian bn m. tp hp hu hn tt c cc mu tin c th c sau khi m ha

3. K l khng gian kh. tp hp hu hn cc kha c th c s dng

4. Vi mi kha kK, tn ti lut m ha ekE v lut gii m dkD tng ng. Lut m

ha ek: P C v lut gii m ek: C P l hai nh x tha mn ,k kd e x x x P

Tnh cht 4. l tnh cht chnh v quan trng ca mt h thng m ha. Tnh cht ny bo

m vic m ha mt mu tin xP bng lut m ha ekE c th c gii m chnh xc

bng lut dkD.



nh ngha 1.2: Zm c nh ngha l tp hp {0, 1, ..., m-1}, c trang b php cng (k

hiu +) v php nhn (k hiu l ). Php cng v php nhn trong Zm c thc hin tng

t nh trong Z, ngoi tr kt qu tnh theo modulo m

V d: Gi s ta cn tnh gi tr 11 13 trong Z16. Trong Z, ta c kt qu ca php nhn

1113=143. Do 14315 (mod 16) nn 1113=15 trong Z16.

Mt s tnh cht ca Zm

1. Php cng ng trong Zm, i.e., a, b Zm, a+b Zm

2. Tnh giao hn ca php cng trong Zm, i.e., a, b Zm, a+b =b+a

3. Tnh kt hp ca php cng trong Zm, i.e., a, b, c Zm, (a+b)+c =a+(b+c)

4. Zm c phn t trung ha l 0, i.e., a Zm, a+0=0+a=a

5. Mi phn t a trong Zm u c phn t i l m a

6. Php nhn ng trong Zm, i.e., a, b Zm, ab Zm

7. Tnh giao hn ca php cng trong Zm, i.e., a, b Zm, ab=ba

8. Tnh kt hp ca php cng trong Zm, i.e., a, b, c Zm, (ab)c =a(bc)

9. Zm c phn t n v l 1, i.e., a Zm, a1=1a=a

10. Tnh phn phi ca php nhn i vi php cng, i.e., a, b, c Zm, (a+b)c

=(ac)+(bc)

11. Zm c cc tnh cht 1, 3 5 nn to thnh 1 nhm. Do Zm c tnh cht 2 nn to thnh

nhm Abel. Zm c cc tnh cht (1) (10) nn to thnh 1 vnh

I.2 Cc H M Thng Dng:

a. H M y (Shift Cipher )

Shift Cipher l mt trong nhng phng php lu i nht c s dng m

ha. Thng ip c m ha bng cch dch chuyn (xoay vng) tng k t i k v tr trong

bng ch ci.

Phng php Shift Cipher

Cho P = C = K = Z26. Vi 0 K 25, ta nh ngha



eK = x + K mod 26

v

dK = y - K mod 26

(x,y Z26)

trong 26 l s k t trong bng ch ci La tinh, mt cch tng t cng c th

nh ngha cho mt bng ch ci bt k. ng thi ta d dng thy rng m y l mt h mt

m v dK(eK(x)) = x vi mi xZ26.

b. H KEYWORD-CEASAR

Trong h m ny kha l mt t no c chn trc, v d PLAIN. T ny xc

nh dy s nguyn trong Z26 (15,11,0,8,13) tng ng vi v tr cc ch ci ca cc ch

c chn trong bng ch ci. By gi bn r s c m ha bng cch dng cc hm lp

m theo th t:

e15, e11, e0, e8, e13, e15, e11, e0, e8, e,...

vi eK l hm lp m trong h m chuyn.

c. H M Vung (SQUARE)

Trong h ny cc t kha c dng theo mt cch khc hn. Ta dng bng ch ci

ting Anh (c th b i ch Q, nu mun tng s cc ch s l mt s chnh phng) v i

hi mi ch trong t kha phi khc nhau. By gi mi ch ca bng ch ci c vit

di dng mt hnh vung, bt u bng t kha v tip theo l nhng ch ci cn li theo

th t ca bng ch.

d. M th v

Mt h m khc kh ni ting . H m ny c s dng hng trm nm nay.

Phng php :

Cho P = C = Z26. K gm tt c cc hn v c th c ca 26 k hiu 0,...,25.

Vi mi hn v K, ta nh ngha:



e(x) = (x)

v nh ngha d(y) = -1(y)

vi -1 l hn v ngc ca hn v .

Trong m th v ta c th ly P v C l cc bng ch ci La tinh. Ta s dng Z26 trong

m y v lp m v gii m u l cc php tn i s.

e. Phng php Affine

Cho P = C = Z26 v cho

K = {(a,b) Z26 Z26 : gcd(a,26) = 1}

Vi K = (a,b) K, ta xc nh

eK(x) = ax+b mod 26

v

dK = a-1(y-b) mod 26

(x,y Z26)

Phng php Affine li l mt trng hp c bit khc ca Substitution Cipher.

c th gii m chnh xc thng tin c m ha bng hm ek E th ek phi l mt

song nh. Nh vy, vi mi gi tr yZ26, phng trnh ax+by (mod 26) phi c nghim

duy nht xZ26.

Phng trnh ax+by (mod 26) tng ng vi ax(yb ) (mod 26). Vy, ta ch cn kho

st phng trnh ax(yb ) (mod 26)

nh l1.1: Phng trnh ax+by (mod 26) c nghim duy nht xZ26 vi mi gi tr bZ26 khi

v ch khi a v 26 nguyn t cng nhau.

Vy, iu kin a v 26 nguyn t cng nhau bo m thng tin c m ha bng hm ek c

th c gii m v gii m mt cch chnh xc.

Gi (26) l s lng phn t thuc Z26 v nguyn t cng nhau vi 26.



nh l 1.2: Nu

m

i

ei

ipn1

vi pi l cc s nguyn t khc nhau v ei Z+, 1 i m th

m

i

ei

ei

ii ppn1

1

Trong phng php m ha Affine , ta c 26 kh nng chn gi tr b, (26) kh nng chn gi

tr a. Vy, khng gian kha K c tt c n(26) phn t.

Vn t ra cho phng php m ha Affine Cipher l c th gii m c thng tin

c m ha cn phi tnh gi tr phn t nghch o a1 Z26.

f. Phng php Vigenere

phng php m ha Vigenere s dng mt t kha (keyword) c di m. C th xem

nh phng php m ha Vigenere Cipher bao gm m php m ha Shift Cipher c p dng

lun phin nhau theo chu k.

Khng gian kha K ca phng php Vigenere c s phn t l 26, ln hn hn phng

php s lng phn t ca khng gian kha K trong phng php Shift Cipher. Do , vic tm

ra m kha k gii m thng ip c m ha s kh khn hn i vi phng php Shift

Cipher.

Phng php m ha Vigenere Cipher

Chn s nguyn dng m. nh ngha P = C = K = (Z26)m

K = { (k0, k1, ..., kr-1) (Z26)r}

Vi mi kha k = (k0, k1, ..., kr-1) K, nh ngha:

ek(x1, x2, ..., xm) = ((x1+k1) mod 26, (x2+k2) mod n, ..., (xm+km) mod 26)

dk(y1, y2, ..., ym) = ((y1k1) mod n, (y2k2) mod n, ..., (ymkm) mod 26)

vi x, y (Z26)m



g. H m Hill

Phng php Hill Cipher c Lester S. Hill cng b nm 1929: Cho s nguyn dng m,

nh ngha P = C = (Z26)m. Mi phn t xP l mt b m thnh phn, mi thnh phn thuc Z26.

tng chnh ca phng php ny l s dng m t hp tuyn tnh ca m thnh phn trong

mi phn t xP pht sinh ra m thnh phn to thnh phn t yC.

Phng php m ha Hill Cipher

Chn s nguyn dng m. nh ngha:

P = C = (Z26)m v K l tp hp cc ma trn mm kh nghch

Vi mi kha K

kkk

kkkkk

k

mmmm

m

m

,2,1,

,21,2

,12,11,1

, nh ngha:

mmmm

m

m

mk

kkk

kkkkk

xxxxkxe

,2,1,

,21,2

,12,11,1

21 ,...,,

vi x=(x1, x2, ..., xm) P

v dk(y) = yk1 vi y C

Mi php tn s hc u c thc hin trn Zn

h. M hn v

Nhng phng php m ha nu trn u da trn tng chung: thay th mi k t trong

thng ip ngun bng mt k t khc to thnh thng ip c m ha. tng chnh

ca phng php m hn v l vn gi nguyn cc k t trong thng ip ngun m ch thay

i v tr cc k t; ni cch khc thng ip ngun c m ha bng cch sp xp li cc k

t trong .



Phng php m ha m hn v

Chn s nguyn dng m. nh ngha:

P = C = (Z26)m v K l tp hp cc hn v ca m phn t {1, 2, ..., m}

Vi mi kha K, nh ngha:

mm xxxxxxe ,...,,...,, 2121 v

mm yyyyyyd 111 ,...,,...,, 2121

vi 1 hn v ngc ca

Phng php m hn v chnh l mt trng hp c bit ca phng php Hill. Vi mi

hn v ca tp hp {1, 2, ..., m} , ta xc nh ma trn k = (ki, j ) theo cng thc sau:

lai ngc hptrng trong

neu,0,1

,ji

k ji

Ma trn k l ma trn m mi dng v mi ct c ng mt phn t mang gi tr 1, cc phn

t cn li trong ma trn u bng 0. Ma trn ny c th thu c bng cch hn v cc hng hay

cc ct ca ma trn n v Im nn k l ma trn kh nghch. R rng, m ha bng phng php

Hill vi ma trn k hn tn tng ng vi m ha bng phng php m hn v vi hn v .

d. M vng

Trong cc h trc u cng mt cch thc l cc phn t k tip nhau ca bn r u

c m ha vi cng mt kha K. Nh vy xu m y s c dng sau:

y = y1y2... = eK(x1) eK(x2)...

Cc h m loi ny thng c gi l m khi (block cipher).

Cn i vi cc h m dng. tng y l sinh ra mt chui kha z = z1z2..., v s

dng n m ha xu bn r x = x1x2...theo qui tc sau:



)...()(... 2121 21 xexeyyy zz

I.3 Quy trnh thm m:

C mi phng php m h ta li c mt phng php thm m tng ng nhng

nguyn tc chung vic thm m c thnh cng th yu cu ngi thm m phi

bit h m no c dng h. Ngi ra ta cn phi bit c bn m v bn r ng.

nhn chung cc h m i xng l d ci t vi tc thc thi nhanh.

Tnh an tn ca n ph thuc vo cc yu t :

Khng gian kh phi ln

vi cc php trn thch hp cc h m i xng c th to ra c mt h m

mi c tnh an tn cao.

bo mt cho vic truyn kha cng cn c x l mt cch nghim tc.

V mt h m h d liu ra i (DES). DES c xem nh l chun m ha d liu

cho cc ng dng t ngy 15 thng 1 nm 1977 do y ban Quc gia v Tiu chun ca M

xc nhn v c 5 nm mt ln li c chnh sa, b sung.

DES l mt h m c trn bi cc php th v hn v. vi php trn thch hp th

vic gii m n li l mt bi tn kh kh. ng thi vic ci t h m ny cho nhng ng

dng thc t li kh thun li. Chnh nhng l do n c ng dng rng ri ca DES

trong sut hn 20 nm qua, khng nhng ti M m cn l hu nh trn khp th gii. Mc

d theo cng b mi nht (nm 1998) th mi h DES, vi nhng kh nng ca my tnh hin

nay, u c th b kha trong hn 2 gi. Tuy nhin DES cho n nay vn l mt m hnh

chun cho nhng ng dng bo mt trong thc t.

II. H M CHUN DES (Data Encryption Standard)

II.1 c t DES

Phng php DES m ha t x c 64 bit vi kha k c 56 bit thnh mt t c y 64 bit.

Thut tn m ha bao gm 3 giai on:

1. Vi t cn m ha x c di 64 bit, to ra t x0 (cng c di 64 bit) bng cch hn

v cc bit trong t x theo mt hn v cho trc IP (Initial Permutation). Biu din x0 = IP(x)

= L0R0, L0 gm 32 bit bn tri ca x0, R0 gm 32 bit bn phi ca x0



L0 R0

x0

Hnh.1 Biu din dy 64 bit x thnh 2 thnh phn L v R

2. Xc nh cc cp t 32 bit Li, Ri vi 1 i 16theo quy tc sau:

Li = Ri-1

Ri = Li-1 f (Ri-1, Ki)

vi biu din php tn XOR trn hai dy bit, K1, K2, ..., K16 l cc dy 48 bit pht sinh

t kha K cho trc (Trn thc t, mi kha Ki c pht sinh bng cch hn v cc bit

trong kha K cho trc).

L i-1 Ri-1

f Ki

L i Ri

Hnh.2 Quy trnh pht sinh dy 64 bit LiRi t dy 64 bit Li-1Ri-1v kha Ki

3. p dng hn v ngc IP-1 i vi dy bit R16L16, thu c t y gm 64 bit. Nh vy, y

= IP-1 (R16L16)



Hm f c s dng bc 2 l

A

B1 B2 B3 B4 B5 B6 B7 B8

J

E(A)

E

+



Hm f c gm 2 tham s: Tham s th nht A l mt dy 32 bit, tham s th hai J l mt

dy 48 bit. Kt qu ca hm f l mt dy 32 bit. Cc bc x l ca hm f(A, J)nh sau:

Tham s th nht A (32 bit) c m rng thnh dy 48 bit bng hm m rng E. Kt qu

ca hm E(A) l mt dy 48 bit c pht sinh t A bng cch hn v theo mt th t

nht nh 32 bit ca A, trong c 16 bit ca A c lp li 2 ln trong E(A).



Thc hin php tn XOR cho 2 dy 48 bit E(A) v J, ta thu c mt dy 48 bit B. Biu

din B thnh tng nhm 6 bit nh sau:B = B1B2B3B4B5B6B7B8

S dng 8 ma trn S1, S2,..., S8, mi ma trn Si c kch thc 416 v mi dng ca ma trn

nhn 16 gi tr t 0 n 15. Xt dy gm 6 bit Bj = b1b2b3b4b5b6, Sj(Bj)

c xc nh bng gi tr ca phn t ti dng r ct c ca Sj, trong , ch s dng r c

biu din nh phn l b1b6, ch s ct c c biu din nh phn l b2b3b4b5. Bng cch

ny, ta xc nh c cc dy 4 bit Cj = Sj(Bj), 1 j 8.

Tp hp cc dy 4 bit Cj li. ta c c dy 32 bit C = C1C2C3C4C5C6C7C8. Dy 32 bit thu

c bng cch hn v C theo mt quy lut P nht nh chnh l kt qu ca hm F(A,

J)

cc hm c s dng trong DES.

Hn v khi to IP s nh 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



iu ny c ngha l bit th 58 ca x l bit u tin ca IP(x); bit th 50 ca x l bit

th hai ca IP(x) v.v.

Hn v ngc IP-1 s l:

IP-1

40

39

38

37

36

35

34

33

8

7

6

5

4

3

2

1

48

47

46

45

44

43

42

41

16

15

14

13

12

11

10

9

56

55

54

53

52

51

50

49

24

23

22

21

20

19

18

17

64

63

62

61

60

59

58

57

32

31

30

29

28

27

26

25

Hm m rng E c c t theo bng sau:

E bng chn bit

32

4

8

12

1

5

9

13

2

6

10

14

3

7

11

15

4

8

12

16

5

9

13

17



16

20

24

28

17

21

25

29

18

22

26

30

19

23

27

31

20

24

28

32

21

25

29

1

Tm S-hp v hn v P s c biu din nh sau:

S1

14

0

4

15

4

15

1

12

13

7

14

8

1

4

8

2

2

14

13

4

15

2

6

9

11

13

2

1

8

1

11

7

3

10

15

5

10

6

12

11

6

12

9

3

12

11

7

14

5

9

3

10

9

5

10

0

0

3

5

6

7

8

0

13

S2

15

3

0

13

1

13

14

8

8

4

7

10

14

7

11

1

6

15

10

3

11

2

4

15

3

8

13

4

4

14

1

2

9

12

5

11

7

0

8

6

2

1

12

7

13

10

6

12

12

6

9

0

0

9

3

5

5

11

2

14

10

5

15

9

S3

10

13

0

7

9

0

14

9

6

3

3

4

15

6

5

10

1

2

13

8

12

5

7

14

11

12

4

11

2

15

8

1



13

1

6

10

4

13

9

0

8

6

15

9

3

8

0

7

11

4

1

15

2

14

12

3

5

11

10

5

14

2

7

12

S4

7

13

10

3

13

8

6

15

14

11

9

0

3

5

0

6

0

6

12

10

6

15

11

1

9

0

7

13

10

3

13

8

1

4

15

9

2

7

1

4

8

2

3

5

5

12

14

11

11

1

5

12

12

10

2

7

4

14

8

2

15

9

4

14

S5

2

14

4

11

12

11

2

8

4

2

1

12

1

12

11

7

7

4

10

0

10

7

13

14

11

13

7

2

6

1

8

13

8

5

15

6

5

0

9

15

3

15

12

0

15

10

5

9

13

3

6

10

0

9

3

4

14

8

0

5

9

6

14

3

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

0

9

3

4

8

1

7

10

13

10

14

7

3

14

10

9

12

3

15

5

9

5

6

0

7

12

8

15

5

2

0

14

10

15

5

2

6

8

9

3

1

6

2

12

S8

13

1

7

2

2

15

11

1

8

13

4

14

4

8

1

7

6

10

9

4

15

3

12

10

11

7

14

8

1

4

2

13

10

12

0

15

9

5

6

12

3

6

10

9

14

11

13

0

5

0

15

3

0

14

3

5

12

9

5

6

7

2

8

11

P

16

29

1

5

2

32

19

22

7

12

15

18

8

27

13

11

20

28

23

31

24

3

30

4

21

17

26

10

14

9

6

25



K l xu c di 64 bit, trong c 56 bit dng lm kha v 8 bit dng kim tra

s bng nhau ( pht hin li). Cc bit cc v tr 8, 16, ..., 64 c xc nh, sao cho mi

byte cha s l cc s 1. V vy, tng li c th c pht hin trong mi 8 bit. Cc bit kim

tra s bng nhau l c b qua khi tnh lch kha.

1. Cho kha 64 bit K, loi b cc bit kim tra v hn v cc bit cn li ca K tng

ng vi hn v (c nh) PC-1. Ta vit PC-1(K) = C0D0, vi C0 bao gm 28 bit u tin ca

PC-1(K) v D0 l 28 bit cn li.

2. Vi i nm trong khong t 1 n 16, ta tnh

Ci = LSi(Ci-1)

Di = LSi(Di-1)

v Ki = PC-2(CiDi), LSi biu din php chuyn chu trnh (cyclic shift) sang tri hoc ca mt

hoc ca hai v tr ty thuc vo tr ca i; y mt v tr nu i = 1, 2, 9 hoc 16 v y 2 v tr

trong nhng trng hp cn li. PC-2 l mt hn v c nh khc.

Vic tnh lch kha c minh ha nh hnh v sau:

K

PC-1

C0 D0

C1 D1 PC-2 K1

LS1LS1

LS2 LS2

...

LS16 LS16

C16 D16 PC-2 K16



Cc hn v PC-1 v PC-2 c s dng trong vic tnh lch kha l nh sau:

PC-1

57

1

10

19

63

7

14

21

49

58

2

11

55

62

6

13

41

50

59

34

7

54

61

5

33

42

51

60

39

46

53

28

25

34

43

52

31

38

45

20

17

26

35

44

23

30

37

12

9

18

27

36

15

22

29

4

PC-2

14

3

23

17

28

19

11

15

12

24

6

4

1

21

26

5

10

8



16

41

30

44

46

7

50

40

49

42

27

31

51

39

50

20

37

45

56

36

13

47

33

34

29

2

55

48

53

32

By gi ta s hin th kt qu vic tnh lch kha. Nh nhn xt trn, mi vng s

dng kha 48 bit tng ng vi 48 bit trong K. Cc thnh phn trong cc bng sau s ch ra

cc bit trong K c s dng trong cc vng khc nhau.

I.2 LP M DES

y l v d v vic lp m s dng DES. Gi s ta m ha bn r sau trong dng thp

lc phn (Hexadecimal)

0123456789ABCDEF

s dng kha thp lc phn

133457799BBCDFF1

Kha trong dng nh phn khng c cc bit kim tra s l:

00010010011010010101101111001001101101111011011111111000.

Ap dng IP, ta nhn c L0 v R0 (trong dng nh phn) :

L0

L1 = R0

=

=

11001100000000001100110011111111

11110000101010101111000010101010

16 vng lp m c th hin nh sau:

E(R0)

K1

=

=

011110100001010101010101011110100001010101010101

000110110000001011101111111111000111000001110010



E(R0) K1

Output S-hp

f(R0,K1)

L2 = R1

=

=

=

=

011000010001011110111010100001100110010100100111

01011100100000101011010110010111

00100011010010101010100110111011

11101111010010100110010101000100

E(R1)

K2

E(R1) K2

Output S-hp

f(R1, K2)

L3 = R2

=

=

=

=

=

=

011101011110101001010100001100001010101000001001

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)

=

=

=

=

=

010100000100001011111000000001010111111110101001

011100101010110111010110110110110011010100011101

001000101110111100101110110111100100101010110100

00100001111011011001111100111010

10111011001000110111011101001100



L5 = R4 = 011101110

E(R4)

K5

E(R4) K5

Xut S-hp

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)

K8

E(R7) K8

S-box output

f(R7, K8)

L9 = R8

=

=

=

=

=

=

000000001100001001010101010111110100000010100000

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

=

=

=

=

010110101111111010101011111010101111110110100101

001000010101111111010011110111101101001110000110

011110111010000101111000001101000010111000100011

01110011000001011101000100000001



f(R10, K11)

L12 = R11

=

=

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

S-box output

f(R13, K14)

L15 = R14

=

=

=

=

=

=

000011110001011000000110100010101010101011110100

010111110100001110110111111100101110011100111010

010100000101010110110001011110000100110111001110

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

Cui cng, p dng IP-1 cho R16L16 ta nhn c bn m trong dng thp lc phn

nh sau:

85E813540F0AB405

I. 3 THM M DES

Mt phng php rt ni ting trong thm m DES l thm m vi sai (differential

cryptanalysic) do Biham v Shamir xut. l phng php thm vi bn r c chn.

N khng c s dng trong thc t thm m DES 16 vng, m ch c s dng

thm cc h DES c t vng hn.

By gi ta s m t nhng tng c bn ca k thut ny. t mc ch thm

m, ta c th b qua hn v khi to IP v hn v o ca n (bi v iu khng cn thit

cho vic thm m). Nh nhn xt trn, ta xt cc h DES n vng, vi n 16. Trong ci

t ta c th coi L0R0 l bn r v LnRn nh l bn m.



Thm m vi sai i hi phi so snh x-or (exclusive-or) ca hai bn r vi x-or ca

hai bn m tng ng. Ni chung, ta s quan st hai bn r L0R0 v L0*R0* vi tr x-or c

c t L0R0 = L0R0 L0*R0*. Trong nhng tho lun sau ta s s dng k hiu () ch x-

or ca hai xu bit.

nh ngha 3.1: Cho Sj l mt S-hp (1 j 8). Xt mt cp xu 6-bit l (Bj,Bj* ). Ta

ni rng, xu nhp x-or (ca Sj) l Bj Bj* v xu xut x-or (ca Sj) l Sj(Bj) Sj(Bj*).

Ch l xu nhp x-or l xu bit c di 6, cn xu xut x-or c di 4.

nh ngha 3.2: Vi bt k Bj (Z2) 6, ta nh ngha tp (Bj) gm cc cp (Bj,Bj*)

c x-or nhp l Bj.

D dng thy rng, bt k tp (Bj) no cng c 26 = 64 cp, v do

(Bj) = {(Bj, Bj Bj) : Bj (Z2) 6 }

Vi mi cp trong (Bj), ta c th tnh xu x-or xut ca Sj v lp c phn b kt

qu. C 64 xu xut x-or, c phn b trong 24 = 16 gi tr c th c. Tnh khng ng u

ca cc phn b l c s m thm.

V d 3.1: Gi s ta xt S1 l S-hp u tin v xu nhp x-or l 110100. Khi

(110100) = {(000000, 110100), (000001, 110101), ..., (111111, 001011)}

Vi mi cp trong tp (110100), ta tnh xu xut x-or ca S1. Chng hn,

S1(000000) = E16 = 1110, S1(110100) = 1001,

nh vy xu xut x-or cho cp (000000,110100) l 0111.

Nu thc hin iu cho 64 cp trong (110100) th ta nhn c phn b ca cc

xu x-or xut 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 v d 3.1, ch c 8 trong s 16 xu x-or xut c th c xut hin tht s. V d

c th ny ch ra s phn b rt khng u ca cc xu x-or xut. Ni chung, nu ta c

nh S-hp Sj v xu nhp x-or Bj, th trung bnh c khong 75 - 80% cc xu x-or xut c

th c xut hin thc s.

m t cc phn b ta a ra nh ngha sau.

nh ngha 3.3: Vi 1 j 8 v vi cc xu bit Bj di 6 v Cj di 4, ta nh

ngha:

INj(Bj,Cj) = {Bj (Z2)6 : Sj(Bj) Sj(Bj Bj) = Cj}

v

Nj(Bj, Cj) = INj(Bj, Cj)

Bng sau s cho cc xu nhp c th c vi xu x-or nhp 110100

Xu xut x-or Cc xu nhp c th c

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

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) tnh s cc cp vi xu nhp x-or bng Bj c xu xut x-or bng Cj vi

S-hp Sj. Cc cp c cc xu nhp x-or c c t v a ra cch tnh cc xu xut x-or

c th nhn c t tp INj(Bj, Cj). rng, tp ny c th phn thnh Nj(Bj, Cj) /2 cp,

mi cp c xu x-or nhp bng Bj.

Phn b trong v d 3.1 cha cc tr N1(110100, C1), C1 (Z2)4. Trong bng trn

cha cc tp IN(110100, C1).

Vi mi tm S-hp, c 64 xu nhp x-or c th c. Nh vy, c 512 phn b c th

tnh c. Nhc li l, xu nhp cho S-hp vng th i l B= E J, vi E = E(Ri-1) l m



rng ca Ri-1 v J = Ki gm cc bit kha ca vng i. By gi xu nhp x-or (cho tt c tm S-

hp) c th tnh c nh sau:

B B* = (E J) (E* J) = E E*

iu ny rt quan trng thy rng, xu nhp x-or khng ph thuc vo cc bit kha

J. (Do , xu xut x-or cng khng ph thuc vo cc bit kha.)

Ta s vit mi B, E v J nh l ni ca tm xu 6-bit:

B = B1B2B3B4B5B6B7B8

E = E1E2E3E4E5E6E7E8

J = J1J2J3J4J5J6J7J8

v ta cng s vit B* v E* nh vy. By gi gi s l ta bit cc tr Ej v Ej* vi mt j no

, 1 j 8, v tr ca xu xut x-or cho Sj, Cj = Sj(Bj) Sj(Bj* ). Khi s l:

Ej Jj INj(Ej, Cj),

vi Ej = Ej Ej*.

nh ngha 3.4: Gi s Ej v Ej* l cc xu bit di 6, v Cj l xu bit di 4. Ta nh

ngha:

testj(Ej, Ej*, Cj) = { Bj Ej : Bj INj(Ej, Cj) },

vi Ej = Ej Ej*.

nh l 3.1:

Gi s Ej v Ej* l hai xu nhp cho S-hp Sj, v xu xut x-or cho Sj l Cj. K hiu

Ej = Ej Ej* . Khi cc bit kha Jj c trong tp testj(Ej, Ej*, Cj).

, chnh l cc xu bit Nj(Ej, Cj) di 6 trong tp testj(Ej, Ej*, Cj); gi tr

chnh xc ca Jj phi l mt trong s .

V d 3.2:

Gi s E1 = 000001, E1*= 110101 v C1= 1101. Do N1(110101,1101) = 8, ng bng 8

xu bit trong tp test1(000001, 110101, 1101). T bng trn ta thy rng



IN1(110100, 1101) = {000110, 010000, 010110, 011100, 100010, 100100, 101000, 110010}

Cho nn

test1(000001, 110101,1101) = {000111, 010001, 010111, 011101, 100011, 100101, 101001,

110011}

Nu ta c mt b ba th hai nh th E1, E1*, C1, khi ta s nhn c tp th hai

test1 ca cc tr cho cc bit kha trong J1. Tr ng ca J1 cn phi nm trong giao ca cc S-

hp. Nu ta c mt vi b ba nh vy, khi ta c th mau chng tm c cc bit kha

trong J1. Mt cch r rng hn thc hin iu l lp mt bng ca 64 b m biu din

cho 64 kh nng ca ca 6 kha bit trong J1. B m s tng mi ln, tng ng vi s xut

hin ca cc bit kha trong tp test1 cho mt b ba c th. Cho t b ba, ta hy vng tm c

duy nht mt b m c tr t; tr s tng ng vi tr ng ca cc bit kha trong J1.

I.3.1. Thm m h DES - 3 vng

By gi ta s xt tng va trnh by cho vic thm m h DES - ba vng. Ta s bt

u vi cp bn r v cc bn m tng ng: L0R0, L0*R0*, L3R3 v L3*R3*. Ta c th biu

din R3 nh sau:

R3 = L2 f(R2, K3)

= R1 f(R2, K3)

= L0 f(R0, K1) f(R2, K3)

R3* c th biu din mt cch tng t , do vy:

R3 = L0 f(R0, K1) f(R0*, K1) f(R2, K3) f(R2*, K3)

By gi, gi s ta chn c cc bn r sao cho R0 = R0*, chng hn:

R0 = 00...0

Khi f(R0, K1) = f(R0*, K1), v do :

R3 = L0 f(R2, K3) f(R2*, K3)

im ny R3 l c bit khi n c th tnh c t hai bn m, v L0 l bit c

khi n c th tnh c t hai bn r. Ngha l ta c th tnh c f(R2,K3)f(R2*,K3) t

phng trnh:



f(R2, K3) f(R2*, K3) = R3 L0

By gi f(R2, K3) = P(C) v f(R2*, K3) = P(C*), vi C v C* tng ng l k hiu

ca hai xu xut ca tm S-hp (nhc li, P l c nh, l hn v c bit cng khai). Nn:

P(C) P(C*) = R3 L0

v kt qu l:

C = C C* = P-1(R3 L0) (1)

l xu xut x-or cho tm S-hp trong vng ba.

By gi, R2 = L3 v R2* = L3* l bit (chng l mt phn ca cc bn m). T

y ta c th tnh:

E = E(L3) (2)

v

E* = E(L3*) (3)

s dng hm m rng E c bit cng khai. Chng l nhng xu nhp cho cc S-hp cho

vng ba. Nh vy gi ta bit E, E*, v C cho vng ba v ta c th tip tc xy dng cc

tp test1, ..., test8 ca cc tr c th c cho cc bit kha trong J1, ..., J8.

Gii thut va xt c th biu din bi cc m sau:

Input: L0R0, L0*R0*, L3R3 v L3*R3*, vi R0 = R0*

1. Tnh C = P-1(R3 L0)

2. Tnh E = E(L3) v E* = E(L*)

3. for j = 1 to 8 do

compute testj(Ej, Ej*, Cj)

Vic m thm s s dng mt s b ba E, E*, C nh vy. Ta s lp tm bng cc b

m v do xc nh c 48 bit trong K3, l kha cho vng ba. 56 bit trong kha khi c

th tm c hn tn t 28 = 256 kh nng cho 8 bit kha.

By gi ta s minh ha iu qua v d sau.



V d 3.3

Gi s ta c ba cp bn r v bn m, vi cc bn m cng c cc xu x-or c m

ha bi cng mt kha. ngn gn ta s dng h thp lc phn:

Bn r Bn m

748502CD38451097

3874756438451097

03C70306D8A09F10

78560A0960E6D4CB

486911026ACDFF31

375BD31F6ACDFF31

45FA285BE5ADC730

134F7915AC253457

357418DA013FEC86

12549847013FEC86

D8A31B2F28BBC5CF

0F317AC2B23CB944

T cp u tin ta tnh cc xu nhp ca S-hp (cho vng 3) t cc phng trnh (2)

v (3). Chng l:

E = 000000000111111000001110100000000110100000001100

E* = 101111110000001010101100000001010100000001010010

Xu xut x-or ca S-hp c tnh bng phng trnh (1) s l:

C = 10010110010111010101101101100111

T cp th hai, ta tnh c cc xu nhp cho S-hp l:

E = 101000001011111111110100000101010000001011110110

E* = 100010100110101001011110101111110010100010101001

v xu xut x-or ca S-hp l:

C = 10011100100111000001111101010110

T cp th ba, cc xu nhp cho S-hp s l:

E = 111011110001010100000110100011110110100101011111



E* = 000001011110100110100010101111110101011000000100

v xu xut x-or ca S-hp l:

C = 11010101011101011101101100101011

Tip theo, ta lp bng cc tr trong tm mng b m cho mi cp. Ta s minh ha th

tc vi cc mng m cho J1 t cp u tin. Trong cp ny, ta c E1= 101111 v C1 =

1001. Tp:

IN1(101111, 1001) = {000000, 000111, 101000, 101111}

Do E1 = 000000 ta c:

J1 test1(000000, 101111, 1001) = {000000, 000111, 101000, 101111}

Do ta tng cc tr 0, 7, 40 v 47 trong cc mng m cho J1.

Cui cng ta s trnh by cc bng. Nu ta xem cc xu bit di 6 nh l biu din

ca cc s nguyn trong khong 0-63, th 64 tr s tng ng vi 0, 1, ..., 63. Cc mng m

s l nh 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

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 mi tm mng m, c duy nht mt b m c tr l 3. V tr ca cc b m

xc nh cc bit kha trong J1, ..., J8. Cc v tr l: 47, 5, 19, 0, 24, 7, 7, 49. Chuyn cc

s nguyn sang dng nh phn, ta nhn c J1, ..., J8:

J1 = 101111

J2 = 000101

J3 = 010011

J4 = 000000



J5 = 011000

J6 = 000111

J7 = 000111

J8 = 110001

By gi ta c th to ra 48 bit kha, bng cch quan st lch kha cho vng ba. Suy ra

l K c dng:

0001101 0110001 01?01?0 1?00100

0101001 0000??0 111?11? ?100011

vi cc bit kim tra c loi b v ? k hiu bit kha cha bit. Kha y (trong

dng thp lc phn, gm c bit kim tra) s l:

1A624C89520DEC46

II.3.2. Thm m h DES 6-vng

By gi ta s m t vic m rng tng trn cho vic thm m trn h DES 6-vng.

tng y l la chn mt cch cn thn cp bn r vi xu x-or c th v sau xc nh

cc xc sut ca cc dy c th ca cc xu x-or qua cc vng lp m. By gi ta cn nh

ngha mt khi nim quan trng sau.

nh ngha 3.5: Cho n 1 l s nguyn. c trng ca vng th n l mt danh sch cc dng

L0, R0, L1, R1, p1, ..., Ln, Rn, pn

tha mn cc iu kin sau:

1. Li = Ri-1 vi 1 i n

2. Cho 1 i n v Li-1, Ri-1 v L*i-1, R*i-1 l c chn sao cho Li-1 L*i-1 = Li-1 v Ri-1

R*i-1 = Ri-1. Gi s Li, Ri v Li* , Ri* l tnh c nh vic p dng mt vng lp m

DES. Khi xc sut Li L*i = Li v Ri R*i = Ri chnh xc bng pi. (Ch l, xc

sut ny c tnh trn tt c cc b c th c ca J = J1...J8) .

Xc sut c trng c nh ngha bng tch p = p1 ... pn.

Nhn xt: Gi s ta chn L0, R0 v L0*, R0* sao cho L0 L0* = L0 v R0 R0*= R0 v ta

p dng n vng lp m ca DES, nhn c L1. ..., Ln v R1, ..., Rn. Khi ta khng th i



hi xc sut Li Li* = Li v Ri Ri* = Ri cho tt c i ( 1 i n) l p1 ... pn. Bi v

cc b -48 trong lch kha K1, ..., Kn khng phi l c lp ln nhau. (Nu n b-48 ny uc

chn c lp mt cch ngu nhin, th iu xc nhn l ng). Nhng ta s coi rng p1 ...

pn chnh xc l xc xut .

Ta cn cn xc nhn l, cc xc sut pi trong c trng l cc cp bn r c xc nh ty

(nhng c nh) c c t bng xu x-or, vi 48 bit kha cho mt vng lp m DES l c

248 kh nng. Do vic thm m s nhm vo vic xc nh kha c nh (nhng cha bit).

Do cn c chn cc bn m ngu nhin (nhng chng c cc xu x-or c c t), hy

vng rng cc xc sut cc xu x-or trong n vng lp m trng hp vi cc xu x-or, c

c t trong c trng, tng i mt p1, ..., pn tng ng.

Trong v d sau y, ta s trnh by c trng vng 1 lm c s cho vic thm m DES ba

vng trong hnh sau (nh trn, ta s s dng cch biu din theo h thp lc phn).

L0 = bt k R0 = 0000000016

L1 = 0000000016 R1 = L0 p = 1

Ta cng s m t mt c trng vng 1 khc nh sau

L0 = 0000000016 R0 = 6000000016

L1 = 6000000016 R1 = 0080820016 p = 14/64

Ta hy xt c trng sau mt cch chi tit hn. Khi f(R0, K1) v f(R0*, K1) c tnh, bc

u tin l m rng R0 v R0*. Xu x-or kt qu ca hai m rng l:

001100...0



Tc l xu x-or nhp cho S1 l 001100 v cc xu x-or cho by S-hp khc u l 000000.

Cc xu xut x-or cho S2 n S8 u l 0000. Xu xut x-or cho S1 l 1110 vi xc sut 14/64

(do N1(001100, 1110) = 14). Nn ta nhn c:

C = 11100000000000000000000000000000

vi xc sut 14/64. Ap dng P, ta nhn c:

P(C) P(C*) = 00000000100000001000001000000000

trong dng thp lc phn s l 0080820016. Khi xu ny cng x-or vi L0, ta nhn c R1

vi xc sut 14/64. Do L1 = R0.

Vic thm m DES su vng da trn c trng ba vng c cho trong hnh sau.

Trong thm m 6-vng, ta bt u vi L0R0. L0*R0*, L6R6 v L6*R6*, m ta phi chn bn r

sao cho L0= 4008000016 v R.0= 0400000016, ta c th biu din R0 nh 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* cng c th biu din tng t, ta c

R0 = L3 f(R3, K4) f(R3*, K4) f(R5, K6) f(R5*, K6) (4)

( l tng t nh thm m 3-vng)

R6 l c bit. T c trng ta tnh L3 = 0400000016 v R3 = 4008000016 vi xc sut

1/16. Nu nh vy, th xu nhp x-or cho S-hp trong vng 4 c th tnh c nh hm m

rng phi l:



001000000000000001010000...0

Cc xu x-or cho S2, S5, S6, S7 v S8 tt c u bng 000000, v v th xu xut x-or l 0000

cho tt c nm S-hp trong vng 4. iu ny c ngha l, ta c th tnh c cc xu xut

x-or cho nm S-hp trong vng 6 nh phng trnh (4). Do gi s ta tnh:

C1C2C3C4C5C6C7C8 = P-1(R6 04000000)

mi Ci l xu bit c di 4. Khi vi xc sut 1/16, th s dn n l C2, C5, C6, C7 v

C8 tng ng l cc xu x-or xut ca S2, S5, S6, S7 v S8 trong vng 6. Cc xu nhp cho

cc S-hp trong vng 6 c th tnh c l E2, E5, E6, E7 v E8; v E2*, E5*, E6*, E7* v E8*,

vi

E1E2E3E4E5E6E7E8 = E(R5) = E(L6)

v

E1*E2*E3*E4*E5*E6*E7*E8* = E(R5*) = E(L6*)

c th tnh c t cc bn r nh sau:

Input: L0R0, L0*R0*, L6R6 v L6*R6*; vi L0 = 4008000016

v R0 = 0400000016.

1. Tnh C = P-1(R6 0400000016)

2. Tnh E = E(L6) v E* = E(L6*)

3. for j {2,5,6,7,8} do

tnh testj( Ej, Ej*, Cj)

Ta cng s xc nh 30 bit kha trong J2, J5, J6, J7 v J8 nh trong thm m 3-vng.

Bi tn, xu xut x-or gi nh cho vng 6 l chnh xc ch vi xc sut 1/16. Cn 15/16

phn cn li ta s thng nhn c nhng xu v dng ngu nhin hn l cc bit kha.



nh ngha 3.6: Gi s L0 L0* = L0 v R0 R0*= R0. Ta ni rng, cp bn r L0R0 v L0*

R0* l ng (right) ng vi c trng nu Li Li* = Li v Ri Ri*= Ri cho mi i, 1 i n.

Cp tri vi cp c nh ngha gi l cp sai (wrong).

Ta mong rng, khong 1/16 s cp ca ta l ng, cn cc cp cn li l cp sai ng vi c

trng vng ba ca ta.

Chin lc ca ta l tnh Ej. Ej* v Cjnh m t trn v sau xc nh testj(Ej, Ej*, Cj)

vi j = 2,5,6,7,8. Nu ta bt u vi mt cp ng, th th cc bit kha chnh xc cho mi Jj s

nm trong tp testj. Nu cp l sai, th tr Cj s khng ng, v l nguyn do gi nh

rng, mi tp testj thc cht l ngu nhin.

Ta c th nhn ra cp ng theo phng php sau: Nu testj= 0, vi bt k j {2,5,6,7,8},

khi ta tt yu c c cp ng. By gi cho mt cp sai, ta c th hy vng rng, xc sut

testj= 0 cho mt j c th l xp x 1/5. l l do gi nh l, Nj(Ej, Cj) = testj v

nh nhn xt t trc, xc sut Nj(Ej, Cj) = 0 l xp x 1/5. Xc sut c nm testj

u dng l vo khong 0.85 0.33, qu vy xc sut t nht mt testj bng 0 l vo

khong 0.67. Nn ta c khong 2/3 s cp l sai, nh vo mt nhn xt n gin, c gi l

php lc (filtering operation). T s ca cc cp ng trn cc cp cn li sau php lc l vo

khong:

61311615161

161

V d 3.4: Gi s ta c cp bn r - bn m sau:

Bn r Bn m

86FA1C2B1F51D3BE

C6F21C2B1B51D3BE

1E23ED7F2F553971

296DE2B687AC6340



Ch l, L0 = 4008000016 v R0 = 0400000016. Xu nhp v xu xut ca S-hp cho vng 6

c tnh nh sau:

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 cc tp testj s l nh 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 thy rng, hai tp test5 v test7 l rng , nn cp ny l cp sai v n b loi b bng php

lc.

By gi gi s ta c cp sao cho testj> 0 vi j = 2,5,6,7,8 l nhng tp cn li sau php

lc.(Bi v ta khng bit c l cp no ng, cp no sai.) Ta ni rng, xu bit J2J5J6J7J8

di 30 l c xut bi cp nu Jj testj vi j = 2,5,6,7,8. S cc cp c xut l:



8,7,6,5,2j

jtest

l bnh thng vi s xu bit c xut l kh ln. (Chng hn. ln hn 80000)

Gi s, ta lp bng cho tt c cc xu c xut nhn c t N cp, m khng b loi bi

php lc. Vi mi cp ng, th xu bit ng J2J5J6J7J8 s l xu c xut. Xu bit ng

s c tnh khong 3N/16 ln. Xu bit sai thng xut hin t hn, bi v chng xut hin

ngu nhin v c khong 230 kh nng. (L mt s rt ln.)

Ta nhn c mt bng cc ln tt c cc xu c xut, nn ta s dng mt thut

tn ch i hi mt khng gian v thi gian t nht. Ta c th m ha bt k mt tp testj no

thnh mt vc t Tj c di 64, vi ta th i ca Tj c t bng 1 (0 i63), nu xu

bit di 6 l biu din ca i trong tp testj; v ta th i c t bng 0 trong trng

hp ngc li ( iu ny ging nh mng cc b m m ta s dng trong thm m DES

ba vng).

Vi mi cp cn li, ta xy dng cc vc t nh trn v gi chng l Tji, j=2,5,6,7,8; 1

i N. Vi I {1, ..., N} ta ni rng I l chp nhn c (allowable) nu vi mi j

{2,5,6,7,8} c t nht mt ta bng I trong vc t

Ii

ijT

Nu cp th i l cp ng cho mi iI, th tp I l chp nhn c. Do ta cho rng

tp chp nhn c c kch thc (xp x) 3N/16, l tp xut v ta hy vng l ch gm cc

bit kha ng ch khng c cc xu khc. iu ny lm n gin ha cho vic xy dng tt

c cc tp chp nhn c I bng mt thut tn qui.

II.3. 3 Cc thm m vi sai khc

Phng php thm m vi sai cn c th p dng thm cc h DES nhiu vng hn.

Vi h DES 8-vng i hi 214 bn r chn v cc h 10-, 12-, 14- v 16-vng i hi c

tng ng 224, 231, 239 v 247 bn m chn. Nn ni chung l kh phc tp.

Cc k thut thm m vi sai c Biham v Shamir pht trin. Cc phng php thm m

DES khc c Matsui s dng nh l thm m tuyn tnh.



III. H M DES 3 VNG

Chng trnh gm hai phn:

Phn Giao Din (cha trong th mc GiaoDien): C chc nng x l

giao din.

Phn X L (cha trong th mc XuLy): c chc nng h tr cc hm

x l.

III.1 Giao Din ( Package GiaoDien).

a. Mn hnh chnh (Mainform.vb)

Form lp m v gii m DES(Des.vb)



Source code mt s hm chnh trong form giai m 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

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

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

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


For i = 0 To 15



Next

For i = 0 To 3




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



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

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

Ham khoi tao s2



Sub khoitao_s2()

Dim i As Integer


For i = 0 To 15



Next

For i = 0 To 3


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

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

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

Hm khi to s3

Sub khoitao_s3()

Dim i As Integer


For i = 0 To 15



Next

For i = 0 To 3


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

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

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

s(3).Rows(3).Item(13) = 7

s(3).Rows(3).Item(14) = 2

s(3).Rows(3).Item(15) = 14


End Sub

Hm khi to s4

Sub khoitao_s4()

Dim i As Integer


For i = 0 To 15





Next

For i = 0 To 3


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

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

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


End Sub

Hm khi to S5

Sub khoitao_s5()

Dim i As Integer


For i = 0 To 15



Next

For i = 0 To 3


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

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

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

Hm khi to S6

Sub khoitao_s6()

Dim i As Integer


For i = 0 To 15



Next

For i = 0 To 3


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

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

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

Hm khi to S7

Sub khoitao_s7()



Dim i As Integer


For i = 0 To 15



Next

For i = 0 To 3


s(7).Rows.Add(row)

Next

s(7).Rows(0).Item(0) = 13

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

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

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

Khi to gi tr bin

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

Ct bit cui

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)

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

H v pc2

Function hvpc2(ByVal str As String) As String

Dim tam(47) As Char

Dim i As Integer

For i = 0 To 47

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

Hm to dy kh

Sub taodaykhoa()

Dim khoa as String =

"00010011001101000101011101111001100110111011110011011111111

10001"

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)

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

Mt s hm x l chui nhp

Nhp nh phn

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 += " "

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

Mt s hm m h

Hm hn v ip

Function hvip(ByVal x As String) As String

Dim tam(63) As Char



Dim i As Integer

For i = 0 To 63

tam(i) = x.Substring(ip(i) - 1, 1)

Next

Return tam

End Function

Hm hn v 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



Hm hn v ip tr

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)

Next

Return tam

End Function

M h

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, d

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