Why is the nonlinearity of this Boolean function evaluating to $\frac12$?

Question

I am using the method presented in this paper to find the nonlinearity of the function

$$ f: \mathbb{F}^1_2 \to \mathbb{F}^1_2 \\ f(x) = x$$

The truth table is $f = [0 \space \space 1]$. Now, I read from the paper by Terry Ritter that

Nonlinearity is the number of bits which must change in the truth table of a Boolean function to reach the closest affine function.

This means the nonlinearity value should be a whole number.

The algorithm to calculate nonlinearity is to first use the Fast Walsh Transform to find the Walsh spectrum, then use the formula

$$Nl(f_k) = 2^{k-1} - \dfrac12 \cdot\max_{a\in\mathbb{F_2^{2^k}}} |W_f(a)| $$

where the Walsh spectrum is calculated by multiplying the truth table of the function by the corresponding Hadamard matrix.

So, since $k = 1$, we use the Hadamard matrix of size $2^1$ giving the following Walsh spectrum:

$$ \begin{bmatrix}0 & 1\end{bmatrix} \begin{bmatrix}1 & 1\\1 & -1\end{bmatrix} = \begin{bmatrix}1 & -1\end{bmatrix} \implies \max_{a\in\mathbb{F_2^{2^k}}} |W_f(a)| = |-1| = 1 $$

Therefore

$$ Nl(f_{k=1}) = 2^{0} - \dfrac12 \cdot 1 = \dfrac12$$

What am I missing?

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In case the links are dead, the linked papers are:

1. Calculating Nonlinearity of Boolean Functions with Walsh-Hadamard Transform by Pedro Miguel Sosa
2. Measuring Boolean Function Nonlinearity by Walsh Transform by Terry Ritter

Answer 1

In this formulation you need to convert your function's output range to $\{-1,+1\}$ via $$f`(x)=(-1)^{f(x)}$$ and apply the Walsh Hadamard to the new function $f`(x)$. Using the zero one formulation means you are off by a constant depending on the number of variables since

$$ (-1)^u=1-2u $$ for $u\in \{0,1\}.$

See my answer below on Boolean functions and crypto, it may be useful given your recent questions.

How are boolean functions used in cryptography?

Answer 2

In addition to the answer by kodlu, after carefully re-reading the papers, I was able to figure it out. Key things to note:

1. If we use the Fast Walsh Transform on Boolean functions consisting of $\{0,1\}$ then the formula for nonlinearity is

... half the number of bits in the function, less the absolute value of the unexpected distance.

That is $$ Nl(f) = \dfrac12 \cdot 2^k - \max_{a\in\mathbb{F}_2^{2^k}} |W_f(a)|\\ = 2^{k-1} - \max_{a\in\mathbb{F}_2^{2^k}} |W_f(a)| $$

Therefore, for the question in the original post we have

$$Nl(f) = 2^{0} - |1| = 0$$

Alternatively, page 20 here (alt link) suggests to proceed as follows: After finding the Fast Walsh transform,

1. Add $2^{k-1}$ to each entry in the row except the first entry. This gives us a new row, call it $FHT'$

2. If an entry in less than $2^{k-1}$ it remains unchanged. Otherwise, if an entry of $FHT'$ is greater than $2^{k-1}$ then subtract it from $2^k$.

3. Finally, the nonlinearity is the smallest of these adjusted elements.

2. If we use the Fast Walsh Transform on Boolean functions consisting of $\{1,-1\}$ then the formula for nonlinearity is

$$ Nl(f) = 2^{k-1} - \dfrac12 \cdot\max_{a\in\mathbb{F}_2^{2^k}} |W_f(a)| $$

Because

Using real values $\{1,-1\}$ doubles the magnitude and changes the sign of the FWT results

Source