Criptoanalisis

7/21/2019 Criptoanalisis

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= c1

= c2

= c3

= c4

= c5

= c2

= c3

= c4

n

n

n.

0010

00000010).

K.



M

n

K k

E (M, K ) : {0, 1}n × {0, 1}k → {0, 1}n

C

D(C, K ) = E −1(C, K ) : {0, 1}m × {0, 1}k → {0, 1}n

M

{0, 1}n {0, 1}n

K

{0, 1}n 2n!



n = 64

k = 64

48



n

M = m0m1 . . . mn−1

n

R0 = m0m1 . . . mn/2−1

L0 = mn/2mn/2+1 . . . mn−1

F

K i

K

K i

F

Li+1 = Ri

Ri+1 = L1 ⊗ F (K i, , Ri)

i = 0, ...r

r



r

C

n

C = c0c1 . . . cn−1 = Lr−1Rr−1

Lr−1 = c0c1 . . . cn/2−1 = Rr−2 = Lr−3 ⊗ F (K r−3, Rr−3) = . . .Rr−1 = cn/2 . . . cn−1 = Lr−2⊗F (K r−2, Rr−2) = Rr−3⊗F (K r−2, Lr−3⊗F (K r−3, Rr−3)) =

. . .

R0

L0 M

K 0, . . . , K r−2

K

r = 16

Ri Li 64/2 = 32

K i

48



K i

E.

M

E (M )

M = 10110110100001100110010101111010



K = 111010001110001101100011011011011011011000101101

K ⊗E (R) = 101100011110001101101100110110010111110010011000

S i

i

B

S i(B)

B = b0b1b2b3b4b5

b0b5 22 = 4

0 3

(24 = 16

0

15

4× 16

0

15

24 = 16



101100

2

10 = 2

0110 = 6)

S i(7) = 2

0010

P (L)

P (L) 16

7



F (R, K i) = P (S (E (R) ⊕K i)

F



k

j , j = 1 . . . 16

k

j = k16− j+1 ki



1/2



256

C 0

D0

C 0

D0



256

28



(R)

R

R

R = R ⊕ R

E

E (R) = E (R ⊕ R) = E (R) ⊕E (R) = A ⊕ A = A

(A ⊕K i) ⊕ (A ⊕K i) = A ⊕A ⊕ (K i ⊕ K i) = A ⊕A ⊕ 0 = A ⊕A = A



(B, B)

B = B⊕B

S (B) = S (B)⊕S (B)

i

j

1

0, 1, 2, 4, 8

10

14

64

∼ 22 %



(a, b)

(b, a)

52

8, 16, 6, 2, 12, 6, 8, 6

1, 2, 3, 4, 7, 8, 13, 15

E (R) = 1

E (R) = 53

K 1

B1 = E (R1) ⊕ K 1

B

1 = E (R

1) ⊕ K 1

B = B ⊗B

1 = E (R)1 ⊕ E (R)1 = E (R) = 52

B1 = E (R1)⊕K 1

K 1 = B1⊕E (R1)

K 1 = B

1⊕E (R

1)

B = 13

E (R1) E (R

1)

B1 B

1



1⊕ 6 = 7 53 ⊕ 6 = 51

1 ⊕ 16 = 17 53⊕ 16 = 37

1 ⊕ 22 = 23 53⊕ 22 = 35

1 ⊕ 28 = 29 53⊕ 28 = 41

1 ⊕ 34 = 35 53⊕ 34 = 23

1 ⊕ 36 = 37 53⊕ 36 = 17

1 ⊕ 40 = 41 53⊕ 40 = 29

1 ⊕ 50 = 51 53 ⊕ 50 = 7

{7, 17, 23, 29, 35, 37, 41, 51}

21 ⊕ 1 = 20 33 ⊕ 1 = 32

21 ⊕ 2 = 23 33 ⊕ 2 = 35

21 ⊕ 21 = 0 33 ⊕ 21 = 52

21 ⊕ 33 = 52 33 ⊕ 33 = 0

21 ⊕ 53 = 32 33 ⊕ 53 = 20

21 ⊕ 54 = 35 33 ⊕ 54 = 23

{20, 23, 0, 52, 32, 35}.

{7, 17, 23, 29, 35, 37, 41, 51} ∩ {20, 23, 0, 52, 32, 35} = {23, 35}



(R0)

R = 60 00 00 00

1

0

001100 = 12

12

14/64

P (L) = P (1400...) = 0080 82 00

L0)

L0⊕00808200 = R1 = 0

L0 = 008082 00



R0L0

L0 = 00 80 82 00R0 = 600000 00L1 = 60 00 00 00R1 = 000000 00

L2 = R1 = 00 00 0000R2 = L1 ⊕ 0 = 60 00 0000

M = M ⊕ M

= 00 80 82 00 60 00 00 00



(C, C )

0000000060000000

14

64

64/14

K 1



S 2 S 8



28 = 256

256



256

28

23 1

27 1

16

215 110486

226 2−24

234 2−32

235 2−32

242 2−40

243 2−40

250

2−48

251 2−48

257 2−56

257

247



243



http://sota.gen.nz/crypt_blues/biham91differentia.pdf

http://sota.gen.nz/crypt_blues/biham91differentia.pdf

http://cs.ucsb.edu/~koc/ac/notes/dc.pdf

http://www.cosic.esat.kuleuven.be/publications/article-2335.pdf

http://www.cosic.esat.kuleuven.be/publications/article-2335.pdf

http://en.wikipedia.org/wiki/Block_cipher_modes_of_operation

http://en.wikipedia.org/wiki/Block_cipher_modes_of_operation

http://ieeexplore.ieee.org/xpl/articleDetails.jsp?arnumber=5389567

http://ieeexplore.ieee.org/xpl/articleDetails.jsp?arnumber=5389567

http://epubs.siam.org/doi/abs/10.1137/0217022

http://en.wikipedia.org/wiki/Pseudorandom_function

http://en.wikipedia.org/wiki/Lucifer_(cipher)

http://en.wikipedia.org/wiki/National_Institute_of_Standards_and_Technology

http://en.wikipedia.org/wiki/National_Institute_of_Standards_and_Technology

http://en.wikipedia.org/wiki/Data_Encryption_Standard

http://en.wikipedia.org/wiki/Data_Encryption_Standard

http://netlab.cs.ucla.edu/wiki/files/shannon1949.pdf

http://netlab.cs.ucla.edu/wiki/files/shannon1949.pdf

http://tools.ietf.org/html/rfc2898

http://tools.ietf.org/html/rfc2898

http://en.wikipedia.org/wiki/Block_cipher

Criptoanalisis

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