Solutions_HW4_547 (1)

510 a each round of des is now given by the following

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Unformatted text preview: 5. Now store the pairs (k1 , k2 ) in S , if (z1 , k1 ) and (z2 , k2 ) are such that z1 = z2 . At the end of this algorithm S will contain all the possible pairs that satisfy ￿ y = DESk (x). Given another two plaintext/ciphertext pairs and trying all the elements of S with respect to these pairs will enable the adversary to identify the correct (k1 , k2 ) with high probability. 5.10 (a) Each round of DES is now given by the following rule: Li Ri 1 Mote = = that z = (L8 , R8 ) R i− 1 L i− 1 ⊕ f k i ( R i− 1 ) = L i − 1 ⊕ ( R i − 1 ⊕ k i ) Fall 2010 Comp 547: Cryptography and Data Security 4 Thus, for i = 1 we have: L1 = R0 R1 = L 0 ⊕ R0 ⊕ k 1 ⇒ L 1 ⊕ R1 = L 0 ⊕ k 1 For i = 2, we have: L2 = R2 = = R 1 = L 0 ⊕ R0 ⊕ k 1 L 1 ⊕ R1 ⊕ k 2 L0 ⊕ k1 ⊕ k2 ⇒ L 2 ⊕ R 2 = R0 ⊕ k 2 For i = 3, we have: L3 = R3 = = R2 = L 0 ⊕ k 1 ⊕ k 2 L 2 ⊕ R2 ⊕ k 3 R0 ⊕ k 2 ⊕ k 3 ⇒ L 3 ⊕ R3 = L 0 ⊕ R0 ⊕ k 1 ⊕ k 3 and so on. At the 16th round we get: L16 = R0 ⊕ k15 ⊕ k14 ⊕ k12 ⊕ ... ⊕ k5 ⊕ k3 ⊕ k2 R16 = L0 ⊕ R0 ⊕ k16 ⊕ k15 ⊕ k13 ⊕ ... ⊕ k4 ⊕ k3 ⊕ k1 that we write more concisely as L16 = R0 ⊕ K1 and R16 = L0 ⊕ R0...
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This document was uploaded on 02/24/2014.

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