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.

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Question

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(X₁)||C₁
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.

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2.) This question builds somewhat on top of the previous question. Instead of adding single bits at a time, let's scale this up to two bits. This is visually represented below: + Cout R1 RO For this problem, you are given a carry-in bit `Cin`, as well as the operands `A` and `B`, just as before. However, `A` and `B` are now two bits apiece, with A0 holding the bit at position 0 and `Al` holding the bit at position 1, and so on. This still produces `R` as a result, though the result is now two bits large, with `R0` holding the result value at position 0, and `R1` holding the result value at position 1. `Cout holds the carry-out bit, as before. As before, fill in the table below. Note the ordering of the inputs. The first few have already been completed for you. 2a) 0 2b) 0 2c) 0 2d) 0 Cin Al B1 A0 B0 | Cout R1 R0 2e) 0 2f) 0 2h) 0 21) 0 2k) 0 2g) 0 0 1 0 0 A1 A0 В1 ВО 2r) 1 2s) 1 2j) 0 1 0 2t) 1 2u) 1 2v) 2w) 1 2x) 1 2y) 1 2z) 1 21) 0 2m) 2n) 0 1 O O O O O O OOHHddd□□□ O O O O O O…
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