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I have a problem about the formula in Matsui's paper about linear cryptanalysis.

Let $N$ be the number of given random plaintexts, p be the probability that equation $P[i_1,i_2,\cdots,i_a]\oplus C[j_1,j_2,\cdots,j_b]\oplus F_n(C_L,K_n)[l_1,l_2,\cdots,l_d]=K[k_1,k_2,\cdots,k_c]$ holds, and assume $|p-1/2|$ is sufficiently small. Let $q^{(i)}$ be the probability that the following equation holds for a subkey candidate $K_n^{(i)}$ and a random variable $X$: $F_n(X,K_n)[l_1,l_2,\cdots,l_d]=F_n(X,K_n^{(i)})[l_1,l_2,\cdots,l_d]$. Then if $q^{(i)}$'s are independent, the success rate of Algorithm 2 is $\int_{x=-2\sqrt{N}|p-1/2|}^{\infty}(\prod\limits_{K_n^{(i)}\ne K_n}\int_{-x-4\sqrt{N}(p-1/2)q^{(i)}}^{x+4\sqrt{N}(p-1/2)(1-q^{(i)})}\frac{1}{\sqrt{2\pi}}e^{-y^2/2}dy)\frac{1}{\sqrt{2\pi}}e^{-x^2/2}dx$.

I don't know why $y$ ranges from $-x-4\sqrt{N}(p-1/2)q^{(i)}$ to $x+4\sqrt{N}(p-1/2)(1-q^{(i)})$. I am trying to understand whether $4\sqrt{N}(p-1/2)q^{(i)}$ is more likely a retrodictive guess based on experimental results or based on a specific mathematical model.

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2 Answers 2

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Ali Aydın Selçuk wrote a comprehensive article in the Journal of Cryptology on success probability in differential and linear cryptanalysis.

The limits have to do with using the Gaussian approximation, then replacing it with a Folded normal distribution (since we tack absolute deviations not signed deviations), and then asymptotic analysis backed by experimental verification.

See the arguments culminating in Theorem 2. I believe the limits are the same, up to notational changes. If not, this is a rigorous analysis while Matsui's was somewhat handwavy, though a great advance in cryptanalysis.

The paper concludes

In this paper, we gave a formal probabilistic model of success in linear and differential cryptanalysis. We also provided efficient formulations that can be used to estimate the success probability of a given attack or to find its plaintext requirement to achieve a certain success level. Experimental results show that the formulas developed for LC are quite precise, especially when a success probability of 90% or higher is of interest

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  • $\begingroup$ Thank you for your suggestion. I am curious about why Matsui uses $-x-4\sqrt{N}(p-1/2)q^{(i)}$ to $x+4\sqrt{N}(p-1/2)(1-q^{(i)})$ as the range. It seems to me that the expression is not very intuitive mathematically. I am trying to understand whether $4\sqrt{N}(p-1/2)q^{(i)}$ is more likely a rough assumption or based on a specific mathematical model. $\endgroup$ Commented Jun 5 at 0:41
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I want to present my proof, hoping that it is correct.

There are two assumptions which are heuristic in nature:

  • [Assumption 1] $|\frac{T_i}{N}-\frac{1}{2}|$ of a linear expression evaluated with wrong subkey candidates has a distribution independent of the key value.
  • [Assumption 2] $|\frac{T_0}{N}-\frac{1}{2}|$​ of a linear expression evaluated with the right subkey candidates has a distribution independent of the distribution defined in Assumption 1 and independent of the key value.

Based on the law of large numbers, we could get that the distributions of $\frac{T_0}{N}$ and $\frac{T_i}{N}$ are well approximated by a normal law. Specifically, $X_0=\frac{T_0}{N}-\frac{1}{2}\sim\mathcal{N}(p-\frac{1}{2},\frac{1}{4N})$ for right key candidates. Because $q^{(i)}$ is the probability that $F_n(X,K_n)[l_1,l_2,\cdots,l_d]=F_n(X,K_n^{(i)})[l_1,l_2,\cdots,l_d]$ holds, the probability that $P[i_1,i_2,\cdots,i_a]\oplus C[j_1,j_2,\cdots,j_b]\oplus F_n(C_L,K_n^{(i)})[l_1,l_2,\cdots,l_d]=K[k_1,k_2,\cdots,k_c]$ holds is $pq^{(i)}+(1-p)(1-q^{(i)})$. Then $X_i=\frac{T_i}{N}-\frac{1}{2}\sim\mathcal{N}(pq^{(i)}+(1-p)(1-q^{(i)})-\frac{1}{2},\frac{1}{4N})$ for wrong key candidates.

We assume for simplicity that $p>\frac{1}{2}$ (when $p<\frac{1}{2}$, it could be obtained by substituting $-X_0$ for $X_0$).

$P_S=\int_0^{\infty}\left(\prod\limits_{K_n^{(i)}\ne K_n}\int_{-x}^xf_{(i)}(y)dy\right)f_0(x)dx\\=\int_{-2\sqrt{N}|p-1/2|}^{\infty}\left(\prod\limits_{K_n^{(i)}\ne K_n}\int_{-\frac{x}{2\sqrt{N}}-|p-\frac{1}{2}|}^{\frac{x}{2\sqrt{N}}+|p-\frac{1}{2}|}f_{(i)}(y)dy\right)\phi(x)dx\\=\int_{x=-2\sqrt{N}|p-1/2|}^{\infty}(\prod\limits_{K_n^{(i)}\ne K_n}\int_{-x-4\sqrt{N}(p-1/2)q^{(i)}}^{x+4\sqrt{N}(p-1/2)(1-q^{(i)})}\frac{1}{\sqrt{2\pi}}e^{-y^2/2}dy)\frac{1}{\sqrt{2\pi}}e^{-x^2/2}dx$

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