I am working on multi-output boolean function i.e. a function that takes an $n$-bit input and gives an $m$-bit output for all possible inputs (i.e. $2^n$ inputs).

To be more exact: I am trying to do a cryptanalysis of the AES 8x8 s-box, with the goal to learn the whole procedure of finding/calculating non-linearity.

Doing some research, I learned that I need to find the minimum hamming distance from all $n$ variable affine functions. Sadly enough, I don't really grasp all of the math which would be required to find the non-linearity.

In the end, I’ld like to be able to calculate the Linear Approximation Table of the AES s-box. Can someone please explain how to build a LAT of the AES Sbox. How can I practically calculate the non-linearity of a multi-output boolean function like the AES s-box?


1 Answer 1


Before we start with vectorial Boolean functions, let's recall the definition of the nonlinearity of a Boolean function:

$$\mathcal{NL}(f) = \min_{a \in \mathbb{F}_2^n} d_H(f, \ell_a \oplus b),$$

where $\ell_a \oplus b$ represents the affine Boolean function defined by the bitvector $a$: $\ell_a(x) = a \cdot x$ ($\cdot$ is the dot product). The above equation pretty much defines the nonlinearity of a Boolean function as the minimum Hamming distance $d_H$ to some affine function.

This distance $d_H$ can be expressed using the Walsh-Spectrum of $f$. That is, $d_H(f, \ell_a) = 2^{n - 1} - \frac{1}{2}\mathcal{W}_f(a)$. To see why this is true, the definition of the Walsh transform should help: $$\mathcal{W}_f(a) = \sum_{x \in \mathbb{F}_2^n} (-1)^{f(x) \oplus \ell_a(x)}.$$ (and note that $(-1)^{g(x)} = 1 - 2g(x)$, $\sum_x g(x) = w_H(x))$)

Minimizing the distance hence corresponds to maximizing the Walsh-Spectrum: $$\mathcal{NL}(f) = 2^{n - 1} - \frac{1}{2} \max_{a \in \mathbb{F}_2^n} \left|\mathcal{W_f(a)}\right|.$$

Why the absolute value? Remember that we're measuring the distance to affine functions and note that $d_H(f, \ell_a \oplus 1) = 2^{n - 1} + \frac{1}{2}\mathcal{W}_f(a)$.

Why is this useful? Well, it turns out there is a rather efficient algorithm to compute the Walsh-Spectrum of some Boolean function. See for example here.

Now that we can compute the nonlinearity of a Boolean function, it's easy to define that of a vectorial Boolean function $F$:

$$\mathcal{NL}(F) = \min_{a \in \mathbb{F}_2^n} \mathcal{NL}(F \cdot a).$$

In other words, the nonlinearity is the minimum of the nonlinearities of the Boolean functions which are linear combinations of the coordinates (outputs) of $F$.

Given what we know about the nonlinearity of Boolean functions, we can compute this as

$$\mathcal{NL}(F) = 2^{n - 1} - \frac{1}{2} \max_{a \in \mathbb{F}_2^n, b \in \mathbb{F}_2^m} \left|\mathcal{W}_{F\cdot b}(a)\right|.$$

Sometimes the expression in the absolute value is called the Walsh transform of $F$: $$\mathcal{W}_F(a, b) = \mathcal{W}_{F\cdot b}(a)$$

To compute this, you can use the fast Walsh-Hadamard transform (FWHT) algorithm again. Note also this corresponds directly to the LAT (depending on conventions, the term $2^{n - 1}$ is added or not).

  • 1
    $\begingroup$ I see your questions has been marked as a duplicate now. At first I interpreted the question as asking for an explanation of the more theoretical aspects of nonlinearity. Maybe this isn't the right type of answer after all, but I'll leave it here just in case it's useful to someone. $\endgroup$
    – Aleph
    Jul 19, 2016 at 20:49
  • 1
    $\begingroup$ @Aleph To rescue the question (and your answer) I edited the Q and re-opened it. Btw: nice answer. $^{+1}$ $\endgroup$
    – e-sushi
    Jul 19, 2016 at 21:42
  • $\begingroup$ I see nothing in this answer about masks. $\endgroup$
    – Melab
    Dec 22, 2017 at 20:32

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