Let and be two Boolean functions defined on for some . We say and are equivalent with respect to if .
For a Boolean function of degree , a fast point of order for is defined as a nonzero element such that where is the map on defined as .
Let be the binary Hamming code of length . We define be the set of the Boolean functions of degree up to equivalence with respect to , such that the set of fast points of order with is equal to . Find .
[THOUGHT] Let be a Boolean function of degree 4. A fast point of order for is a nonzero element such that , where . We are given that , and is the binary Hamming code of length . We are looking for the number of Boolean functions of degree 4 such that the set of fast points of order 2 with is equal to . Since has degree 4, we have . A fast point of order 2 is a nonzero element such that . We are given that the set of fast points of order 2 is . This means for any , we have . Let , where and is a degree 4 polynomial. Then, Since , we have . Since is a degree 4 polynomial, . Thus, . Since , we have . Let , where . Since , we have for some . We are given that for all . Let be the Hamming code. Then, for any , we have . We are looking for the number of Boolean functions such that for all . The number of elements in is . Since the degree of is at most 1 for all , we have that the derivative of must be zero at 127 points. Since , the derivative of has degree 3. Thus, we have at most 8 points where the derivative is zero. However, we are given that for all . Consider the case where for all . This means that is constant for all . Let . Then, is constant for all . The number of functions such that for all is equal to the number of functions such that . Since we are given that the degree of is exactly 4, we have . The number of functions such that for all is equal to the number of functions such that .
Final Answer: The final answer is