3.8 The 32-bit swap after the sixteenth iteration of the DES algorithm is needed to make the encryption process invertible by simply running the ciphertext back through the algorithm with the key order reversed. This was demonstrated in Problem 3.7. However, it still may not be entirely clear why the 32-bit swap is needed. To demonstrate why, solve the following exercises. First, some notation: A7B = the concatenation of the bit strings A and BT i(R7L) = the transformation defined by the ith iteration of the encryption algorithm for 1 <= I <= 16 TDi(R7L) = the transformation defined by the ith iteration of the encryption algorithm for 1 <=I<= 16 T17(R7L) = L7R, where this transformation occurs after the sixteenth iteration of the encryption algorithm a. Show that the composition TD1(IP(IP-1(T17(T16(L157R15))))) is equivalent to the transformation that interchanges the 32-bit halves, L15 and R15. That is, show that TD1(IP(IP-1(T17(T16(L157R15))))) = R157L15 b. Now suppose that we did away with the final 32-bit swap in the encryption algorithm. Then we would want the following equality to hold: TD1(IP(IP-1(T16(L157R15)))) = L157R15 Does it? Note: The following problems refer to details of DES that are described in Appendix S. | |
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