Converting truth table to algebraic normal form

Question

Is there any efficient algorithm to convert a given truth table of a Boolean function to its equivalent algebraic normal form (ANF)?

I have seen that Sage has one implementation (official documentation):

sage: from sage.crypto.boolean_function import BooleanFunction
sage: B = BooleanFunction("12fe342a")
sage: B.algebraic_normal_form()
x0*x1*x2*x3*x4 + x0*x1*x2 + x0*x1*x3*x4 + x0*x1*x3 + x0*x1*x4 + x0*x2*x3*x4 + x0*x2*x4 + x0*x3*x4 + x0*x3 + x0 + x1*x2*x4 + x1*x3 + x1*x4 + x2*x3*x4 + x2*x3 + x2*x4

But, nowhere I could find the algorithm.

It would be helpful if someone kindly explains with a suitable example.

UPDATE The answer can be found here. It is copied below.

Answer 1

Source (Credit goes to user pico of crypto.stackexchange)

From TRUTH TABLE to ANF

First write [6, 4, 7, 8, 0, 5, 2, 10, 14, 3, 13, 1, 12, 15, 9, 11] in that way: the columns of matrix are those numbers in $\mathbb{F_2^4}$. $$ \begin{bmatrix} 0&0&1&0&0&1&0&0&0&1&1&1&0&1&1&1\\ 1&0&1&0&0&0&1&1&1&1&0&0&0&1&0&1\\ 1&1&1&0&0&1&0&0&1&0&1&0&1&1&0&0\\ 0&0&0&1&0&0&0&1&1&0&1&0&1&1&1&1 \end{bmatrix} $$ Then multiply it with Moebius transformation matrix :

$$ M_1 = \begin{bmatrix} 1 \end{bmatrix}, M_2 = \begin{bmatrix} 1&1\\ 0&1 \end{bmatrix}, \cdots, M_{2^k} = M_2 \otimes M_{2^{k-1}} = \begin{bmatrix} M_{2^{k-1}}&M_{2^{k-1}}\\ 0&M_{2^{k-1}} \end{bmatrix}. $$ So for $k=4$, the matrix is: $$ \begin{bmatrix} 1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 \\ 0 &1 &0 &1 &0 &1 &0 &1 &0 &1 &0 &1 &0 &1 &0 &1 \\ 0 &0 &1 &1 &0 &0 &1 &1 &0 &0 &1 &1 &0 &0 &1 &1\\ 0 &0 &0 &1 &0 &0 &0 &1 &0 &0 &0 &1 &0 &0 &0 &1\\ 0 &0 &0 &0 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1\\ 0 &0 &0 &0 &0 &1 &0 &1 &0 &1 &0 &1 &0 &1 &0 &1\\ 0 &0 &0 &0 &0 &0 &1 &1 &0 &0 &1 &1 &0 &0 &1 &1\\ 0 &0 &0 &0 &0 &0 &0 &1 &0 &0 &0 &1 &0 &0 &0 &1\\ 0 &0 &0 &0 &0 &0 &0 &0 &1 &1 &1 &1 &1 &1 &1 &1\\ 0 &0 &0 &0 &0 &0 &0 &0 &0 &1 &0 &1 &0 &1 &0 &1\\ 0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &1 &1 &0 &0 &1 &1\\ 0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &1 &0 &0 &0 &1\\ 0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &1 &1 &1 &1\\ 0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &1 &0 &1\\ 0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &1 &1\\ 0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 &1 \end{bmatrix} $$ Then you have this matrix: $$ \begin{bmatrix} 0&0&1&1&0&1&1&0&0&1&0&0&0&1&1&0\\ 1&1&0&0&1&1&1&0&0&1&1&0&0&0&0&0\\ 1&0&0&1&1&1&0&0&0&1&0&1&1&0&1&0\\ 0&0&0&1&0&0&0&0&1&1&0&1&0&1&0&0 \end{bmatrix} $$ Each row gives the coordinate function $S_1,S_2,S_3$and $S_4$ resp. The entries of each row are the coefficients of $1, x_0, x_1, x_0x_1, x_2, x_0x_2, x_1x_2, x_0x_1x_2, x_3, x_0x_3, x_1x_3, x_0x_1x_3, x_2x_3, x_0x_2x_3, x_1x_2x_3, x_0x_1x_2x_3$.

From ANF to TRUTH TABLE (TT)

Exactly the inverse of operations. Note that $M_{2^k}^{-1}=M_{2^k}$ for any $k$.

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i.e. [TT] * $[M]$ = [ANF] and [TT] = [ANF] * $[M]$.

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Note: The arithmetics are taken modulo 2.

Answer 2

The sage source code implies that algebraic normal form is the same as the Fourier expansion of the function (over the group $\mathbb{Z}_2^n$, where $n$ is the number of input bits). You can compute the Fourier transform using the well-known FFT algorithm(s).