6. Consider the following simplified version of the CFB mode. The plaintext is broken into 32-bit pieces: P = [P₁, P2, ,], where each P; has 32 bits, rather than the 8 bits used in CFB. Encryption proceeds as follows. An initial 64-bit X₁ is chosen. Then for j = 1,2,..., the following is performed: CjPjL32 (Ek (Xj)) Xj+1 = R32(Xj)||C; where L32(X) denotes the 32 leftmost bits of X, R32 (X) denotes the rightmost 32 bits of X, and XY denotes the string obtained by writing X followed by Y. Find the decryption algorithm.
6. Consider the following simplified version of the CFB mode. The plaintext is broken into 32-bit pieces: P = [P₁, P2, ,], where each P; has 32 bits, rather than the 8 bits used in CFB. Encryption proceeds as follows. An initial 64-bit X₁ is chosen. Then for j = 1,2,..., the following is performed: CjPjL32 (Ek (Xj)) Xj+1 = R32(Xj)||C; where L32(X) denotes the 32 leftmost bits of X, R32 (X) denotes the rightmost 32 bits of X, and XY denotes the string obtained by writing X followed by Y. Find the decryption algorithm.
Chapter8: Data And Network Communication Technology
Section: Chapter Questions
Problem 41VE
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