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Assignment 4 Solutions

# Assignment 4 Solutions - 1 a We can choose which of the 56...

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1) a) We can choose which of the 56 bits to be ones The answer is C(56, 26) = 6.65x1015 = 252.4. b) The original key space was 256. That’s a decrease of 256-252.54= 23.6, which means it will take less than 8.24% of the time. 2) a)

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b) Lets send C to the encryption oracle. The double encryption of C results the plaintext. Let M = L0 || R0. Now we show that the encryption of C gives M. For the first round: L4 = R3 R4 = L3 XOR F(R3, K1) L1’ = R0’ = L4 = R3 R1’ = L0’ XOR F(R1’, K1) = R4 XOR F(R3, K1) = (L3 XOR F(R3, K1)) XOR F(R3, K1) = L3 First round of re-encryption exactly undoes one round of the first encryption. This step repeats for all the rounds. At the end, we get L4’ = R0 and R4’ = L0. We do the final swap and get M’ = R4’ || L4’ = L0 || R0 = M. 3) a) Since bit can be 0 or 1, there are 2 n different plaintext blocks. Since we are doing one-to-one mapping, total possible mappings are (2 n )! Since the key is n bits long we have 2 n b) It can be equated as Pr[K != K’] = 1 – Pr[K = K′], given that the mappings are the same .
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