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5

It is of course possible to write DES or any block cipher as a system of non-linear equations involving the plaintext bits, the ciphertext bits, and the key bits, which hold with probability 1. In principle, cracking the cipher would then merely involve collecting enough linearly independent equations (e.g. from a couple different known plaintexts) and then ...


4

The S-Boxes are lossy. They map 6-bit inputs to 4-bit outputs, so for a given 4-bit output there are several possible inputs. Considering that there are 8 S-boxes, that's 16 bits of information lost per round, or 256 bits for all 16 rounds. It's much easier to exhaustively search the 56-bit keyspace than try to work backwards against that kind of information ...


0

The idea in [1] (disregarding reception errors) is to first remove the impact of the message. This is done by computing the syndrome $\qquad\qquad\mathbf{H} \mathbf{v} = \mathbf{H} (\mathbf{m} \oplus \mathbf{k}) = \mathbf{H}\mathbf{k}$. In the case of a ${1}/{5}$ repetition code, this could correspond to adding even number of (bold) codeword symbols ...



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