[THOUGHT] Let me break down this problem step by step.

1. We're working over $\mathbb{F}_2^n$ where $n = 7$ (since the Hamming code $\mathcal{H}_7$ has length $2^7 - 1 = 127$).

Wait, let me reconsider. The Hamming code $\mathcal{H}_7$ is a $[7, 4, 3]$ code, meaning it has length 7, dimension 4, and minimum distance 3. So we're working in $\mathbb{F}_2^7$.

2. We need Boolean functions $f$ of degree 4 on $\mathbb{F}_2^7$.

3. A fast point of order 2 for $f$ is a nonzero $x \in \mathbb{F}_2^7$ such that $\deg(D_x f) \leq 4 - 1 - 2 = 1$.

4. The set of fast points of order 2 together with $\mathbf{0}$ should equal $\mathcal{C}$ (the Hamming code).

5. Two functions are equivalent with respect to 2 if their difference has degree at most 2.

Let me think about this more carefully. For a degree 4 Boolean function, the derivative $D_x f(u) = f(u+x) - f(u)$ has degree at most 3 in general. We want $\deg(D_x f) \leq 1$ for $x$ to be a fast point of order 2.

The Hamming code $\mathcal{H}_7$ has $2^4 = 16$ codewords (including $\mathbf{0}$), so we need exactly 15 nonzero fast points of order 2.

Let me write code to explore this problem. [/THOUGHT] [PYTHON] import numpy as np from itertools import combinations, product from functools import reduce

Work in F_2^7

n = 7

Generate all elements of F_2^7

elements = [tuple(int(b) for b in format(i, f'0{n}b')) for i in range(2**n)] elements_array = np.array(elements, dtype=np.uint8)

Generate the Hamming code H_7

H_7 is the [7,4,3] code. Its parity check matrix H has columns that are all nonzero binary vectors of length 3

Generator matrix G can be constructed from this

Parity check matrix for H_7 (columns are 1,2,3,4,5,6,7 in binary)

H = np.array([ [0, 0, 0, 1, 1, 1, 1], [0, 1, 1, 0, 0, 1, 1], [1, 0, 1, 0, 1, 0, 1] ], dtype=np.uint8)

Codewords are vectors c such that H @ c = 0 (mod 2)

hamming_code = [] for i in range(2**n): c = np.array([int(b) for b in format(i, f'0{n}b')], dtype=np.uint8) if np.all(H @ c % 2 == 0): hamming_code.append(tuple(c))

print(f"Hamming code has {len(hamming_code)} codewords") print("Codewords:", hamming_code[:5], "...") [/PYTHON]

[THOUGHT] Let me continue analyzing this problem more carefully. I need to run more code to actually solve the problem - the previous output just confirmed the Hamming code structure. Let me continue with the full computation. [/THOUGHT] [PYTHON] import numpy as np from itertools import combinations, product from functools import reduce

Work in F_2^7

n = 7

Generate all elements of F_2^7

elements = [tuple(int(b) for b in format(i, f'0{n}b')) for i in range(2**n)]

Parity check matrix for H_7

H = np.array([ [0, 0, 0, 1, 1, 1, 1], [0, 1, 1, 0, 0, 1, 1], [1, 0, 1, 0, 1, 0, 1] ], dtype=np.uint8)

Get Hamming code codewords

hamming_code = set() for i in range(2**n): c = np.array([int(b) for b in format(i, f'0{n}b')], dtype=np.uint8) if np.all(H @ c % 2 == 0): hamming_code.add(tuple(c))

For a Boolean function f of degree k, represented by its ANF (algebraic normal form)

The monomials of degree d in n variables correspond to subsets of {0,1,...,n-1} of size d

A degree 4 Boolean function has monomials of degree 0,1,2,3,4

We care about equivalence up to degree 2, so we only need to count degree 3 and 4 parts

For x to be a fast point of order 2, we need deg(D_x f) <= 1

D_x f(u) = f(u+x) + f(u) (in F_2)

For a monomial m = x_{i1} * x_{i2} * ... * x_{id},

D_a m(u) = m(u+a) + m(u)

The key insight: for a degree 4 function f, D_x f has degree <= 1 iff

the degree 4 and degree 3 parts of D_x f are zero.

Let's think about this algebraically. If f has ANF with degree 4 terms,

then D_x f involves derivatives of those terms.

For a monomial prod_{i in S} x_i where |S|=4,

D_a (prod_{i in S} x_i) = prod_{i in S}(x_i + a_i) + prod_{i in S} x_i

This has degree 3 if a has exactly one 1 in positions S, degree 2 if two 1s, etc.

Actually, for D_x f to have degree <= 1, we need:

- All degree 4 monomials: D_x of them should have degree <= 1

- All degree 3 monomials: D_x of them should have degree <= 1

For a degree 4 monomial m_S = prod_{i in S} x_i, D_x m_S has degree 4 - |S ∩ supp(x)|

where supp(x) is the support of x.

So D_x m_S has degree <= 1 iff |S ∩ supp(x)| >= 3

For a degree 3 monomial m_T, D_x m_T has degree <= 1 iff |T ∩ supp(x)| >= 2

So x is a fast point of order 2 iff:

- For every degree 4 monomial S in f: |S ∩ supp(x)| >= 3

- For every degree 3 monomial T in f: |T ∩ supp(x)| >= 2

We want the set of such x (plus 0) to equal the Hamming code.

Let's enumerate degree 4 monomials and degree 3 monomials

deg4_monomials = list(combinations(range(n), 4)) # 35 monomials deg3_monomials = list(combinations(range(n), 3)) # 35 monomials

print(f"Number of degree 4 monomials: {len(deg4_monomials)}") print(f"Number of degree 3 monomials: {len(deg3_monomials)}")

Convert Hamming code to supports (nonzero positions)

hamming_nonzero = [c for c in hamming_code if c != (0,)*n] print(f"Nonzero Hamming codewords: {len(hamming_nonzero)}") [/PYTHON]

[THOUGHT] I need to continue the computation to find the actual answer. Let me run more code to complete the analysis. [/THOUGHT] [PYTHON] import numpy as np from itertools import combinations, product

n = 7

Parity check matrix for H_7

H = np.array([ [0, 0, 0, 1, 1, 1, 1], [0, 1, 1, 0, 0, 1, 1], [1, 0, 1, 0, 1, 0, 1] ], dtype=np.uint8)

Get Hamming code codewords

hamming_code = set() for i in range(2**n): c = tuple(int(b) for b in format(i, f'0{n}b')) c_arr = np.array(c, dtype=np.uint8) if np.all(H @ c_arr % 2 == 0): hamming_code.add(c)

hamming_nonzero = [c for c in hamming_code if c != (0,)*n]

Get supports of nonzero Hamming codewords

def support(c): return frozenset(i for i in range(n) if c[i] == 1)

hamming_supports = [support(c) for c in hamming_nonzero] non_hamming_nonzero = [] for i in range(1, 2**n): c = tuple(int(b) for b in format(i, f'0{n}b')) if c not in hamming_code: non_hamming_nonzero.append(c)

non_hamming_supports = [support(c) for c in non_hamming_nonzero]

deg4_monomials = list(combinations(range(n), 4)) deg3_monomials = list(combinations(range(n), 3))

For x to be a fast point of order 2:

- For every deg 4 monomial S in f: |S ∩ supp(x)| >= 3

- For every deg 3 monomial T in f: |T ∩ supp(x)| >= 2

For x NOT to be a fast point:

- There exists deg 4 monomial S in f with |S ∩ supp(x)| <= 2, OR

- There exists deg 3 monomial T in f with |T ∩ supp(x)| <= 1

We need: fast points = hamming_nonzero (exactly)

For each x in hamming_nonzero: x IS a fast point

For each x not in hamming (and nonzero): x is NOT a fast point

Let's build constraint matrices

For deg4 monomial S and element x: constraint is |S ∩ supp(x)| >= 3 (for x to be fast)

This means: if |S ∩ supp(x)| <= 2, then S must NOT be in f (for x in Hamming)

And: if |S ∩ supp(x)| <= 2 for some S in f, then x is not fast (good for non-Hamming)

For Hamming codewords to be fast points:

For each x in Hamming nonzero, for each deg4 S: if |S ∩ supp(x)| <= 2, then S not in f

For each x in Hamming nonzero, for each deg3 T: if |T ∩ supp(x)| <= 1, then T not in f

Find forbidden deg4 monomials (those that would make some Hamming point not fast)

forbidden_deg4 = set() for x_supp in hamming_supports: for S in deg4_monomials: S_set = frozenset(S) if len(S_set & x_supp) <= 2: forbidden_deg4.add(S)

forbidden_deg3 = set() for x_supp in hamming_supports: for T in deg3_monomials: T_set = frozenset(T) if len(T_set & x_supp) <= 1: forbidden_deg3.add(T)

allowed_deg4 = [S for S in deg4_monomials if S not in forbidden_deg4] allowed_deg3 = [T for T in deg3_monomials if T not in forbidden_deg3]

print(f"Allowed deg4 monomials: {len(allowed_deg4)}") print(f"Allowed deg3 monomials: {len(allowed_deg3)}") print(f"Allowed deg4: {allowed_deg4}") print(f"Allowed deg3: {allowed_deg3}") [/PYTHON]

[THOUGHT] The computation shows that there are no allowed degree 4 or degree 3 monomials that satisfy the constraint that all Hamming codewords are fast points of order 2. This means there are no Boolean functions of degree 4 (up to equivalence with respect to 2) such that the set of fast points of order 2 together with 0 equals the Hamming code.

Wait, but the problem asks for degree 4 functions. If there are no allowed degree 4 monomials, then the only function would be one with no degree 4 or degree 3 terms, which would have degree at most 2, not degree 4.

This means there are no such Boolean functions of degree exactly 4 satisfying the given conditions. Therefore |F| = 0. [/THOUGHT]

\boxed{0}