TITLE: What is the number of self dual boolean functions?
QUESTION [2 upvotes]: The dual of a Boolean function $F(x_1,x_2 \dots x_n,+,\bullet)$, written as $F^D$, is the same expression as that of $F$ with $+$ and $\bullet$ swapped. $F$ is said to be self-dual if $F=F^D$. What is the number of self-dual functions with $n$ Boolean variables?
I have no clue where to begin with. Any subtle hint would be great.
Thanks !

REPLY [1 votes]: For any self-dual function, the number of it's minterms must be equal to it's number of maxterms. Also, no two complementary minterms must be present in the function. By complementary minterms, I mean those whose sum is equal to $n$, the number of variables of the function.
Hence there will be $2^{n-1}$ such complementary minterm pairs, one can select atmost one minterm from each of the pairs and should select one from all pairs to get a self dual function. That can be done in $2^{2^{n-1}}$ ways, which gives the total number of self dual functions.
{"set_name": "stack_exchange", "score": 2, "question_id": 781885}