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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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So after re-reading a few times it becomes clear. The key bits are static, so they won't change the probabilities. Now suppose that $\sum_K = 0$, then the equation in (5) will be true, with probability $\frac{15}{32}$ since it is the same equation. Suppose now that $\sum_K = 1$, the equation in (5) is now the inverse, and it will thus happen with ...

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If you include a key addition layer ($K_1$) at the output as well the key addition layer ($K_0$) at the input of the Sbox then you can perform linear cryptanalysis on this simple cipher. You shall have access to $P,C$ pairs but no keys, of course, and the $P/C$ relationship is $$C=K_1\oplus(S(P\oplus K_0))$$ and so your linear bias equations become something ...

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