The dual of a Boolean function $F(x_1, x_2, \dots, x_n, +, \cdot, ')$, written as $F^D$, is the same expression as that of $F$ with $+$ and $\cdot$ swapped. $F$ is said to be self-dual if $F = F^D$. The number of self-dual functions with $n$ Boolean variables is

A Boolean function $F$ of $n$ variables has $2^n$ possible input combinations in its truth table.

For a function to be self-dual, the condition $F(x_1, x_2, \dots, x_n) = \overline{F(\overline{x_1}, \overline{x_2}, \dots, \overline{x_n})}$ must hold for all input combinations.

This condition groups the $2^n$ minterms into $2^{n-1}$ complementary pairs $(m_i, m_{2^n - 1 - i})$. For each pair:

• Choosing the function value for $m_i$ uniquely fixes the value for the complementary minterm $m_{2^n - 1 - i}$ (if $F(m_i) = 0$, then $F(m_{2^n - 1 - i}) = 1$, and vice versa).

Since there are $2^{n-1}$ independent pairs and 2 independent choices ($0$ or $1$) for each pair, the total number of self-dual functions is:

Hence, option (D) is correct.