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I read a paper of linear cryptanalysis that is J.Daemen et al's Probability distribution of correlation and differentials in block cipher.

In this paper's lemma 8's proof, I can't understand this formula (page.13).

The number of balanced Boolean functions $g(a)$ for a given value of $x$ is :

$$ {{2^{n-1}} \choose {2^{n-2}+x}} {{2^{n-1}} \choose {2^{n-2}-x}} = {{2^{n-1}} \choose {2^{n-2}+x}}^2$$ and

If we divide this by the total number of balanced Boolean functions,..."

Please, tell me how this formula is derived.

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So $x$ measures the departure from balance in the left half of the truth table, say it is the excess number of 1's in the left half of length $2^{n-1}$. By the definition of binomial coefficients there are $$\binom{2^{n-1}}{2^{n-2}+x}$$ such vectors. This is the first factor. The second factor is the count of vectors missing $x$ 1's, since the overall truth table is balanced.

Using $$\binom{a}{b}=\binom{a}{a-b}$$ completes the argument.

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  • $\begingroup$ Why the probability is computed by only g(a)? u, v, alpha don't affect probability? $\endgroup$
    – jyj
    Dec 7 '18 at 2:44
  • $\begingroup$ Yes it is. Read the discussion preceding equation (20) starting in the previous page. $\endgroup$
    – kodlu
    Dec 7 '18 at 3:45
  • $\begingroup$ Glad it was helpful. Perhaps you can upvote the answer, if you liked it :-) $\endgroup$
    – kodlu
    Dec 8 '18 at 9:01

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