Problem Analysis #6

[THOUGHT] Let $f$ be a Boolean function of degree 4. A fast point of order $\ell$ for $f$ is a nonzero element $x \in \mathbb{F}_2^n$ such that $\deg(D_xf) \le k-1-\ell$, where $D_xf(u) = f(u+x)-f(u)$. We are given that $n=7$, and $\mathcal{C}$ is the binary Hamming code $\mathcal{H}_7$ of length $2^7-1=127$. We are looking for the number of Boolean functions $f$ of degree 4 such that the set of fast points of order 2 with $\mathbf{0}$ is equal to $\mathcal{C}$. Since $f$ has degree 4, we have $k=4$. A fast point of order 2 is a nonzero element $x \in \mathbb{F}_2^7$ such that $\deg(D_xf) \le 4-1-2 = 1$. We are given that the set of fast points of order 2 is $\mathcal{C}$. This means for any $x \in \mathcal{C}$, we have $\deg(D_xf) \le 1$. Let $f(u) = g(u)+h(u)$, where $\deg(g)\le 3$ and $h(u)$ is a degree 4 polynomial. Then,

Since $\deg(g)\le 3$, we have $\deg(g(u+x)-g(u))\le 3$. Since $h$ is a degree 4 polynomial, $\deg(h(u+x)-h(u))\le 3$. Thus, $\deg(D_xf)\le 3$. Since $x\in \mathcal{C}$, we have $\deg(D_xf)\le 1$. Let $f(u) = \sum_{S\subseteq\{1,\dots,7\}} a_Su_S$, where $a_S \in \{0,1\}$. Since $\deg(f)=4$, we have $|S|=4$ for some $a_S=1$. We are given that $\deg(D_xf)\le 1$ for all $x\in \mathcal{C}$. Let $\mathcal{C}$ be the Hamming code. Then, for any $x\in \mathcal{C}$, we have $\deg(D_xf)\le 1$. We are looking for the number of Boolean functions $f$ such that $\deg(D_xf)\le 1$ for all $x\in \mathcal{C}$. The number of elements in $\mathcal{C}$ is $2^7-1=127$. Since the degree of $D_xf$ is at most 1 for all $x\in \mathcal{C}$, we have that the derivative of $f$ must be zero at 127 points. Since $\deg(f)=4$, the derivative of $f$ has degree 3. Thus, we have at most 8 points where the derivative is zero. However, we are given that $\deg(D_xf)\le 1$ for all $x\in \mathcal{C}$. Consider the case where $\deg(D_xf)=0$ for all $x\in \mathcal{C}$. This means that $f(u+x)-f(u)$ is constant for all $x\in \mathcal{C}$. Let $f(0)=0$. Then, $f(u+x)-f(u)$ is constant for all $x\in \mathcal{C}$. The number of functions $f$ such that $\deg(D_xf)\le 1$ for all $x\in \mathcal{C}$ is equal to the number of functions such that $\deg(f)\le 4$. Since we are given that the degree of $f$ is exactly 4, we have $\deg(D_xf)\le 3$. The number of functions $f$ such that $\deg(D_xf)\le 1$ for all $x\in \mathcal{C}$ is equal to the number of functions such that $\deg(f)\le 4$.

Final Answer: The final answer is $\boxed{0}$