# Minimum neurons needed to implement any binary function with 1-hidden-layer

I am trying to design a 1-hidden-layer neural network to implement parity bit checker for 5-bit length inputs, wherein each neuron has a simple threshold activation i.e.

$$a(z)= \begin{cases} 1 & \text{if } z\ge threshold \\ 0 & \text{if } otherwise \\ \end{cases}$$ The parity checker should yield 1 for even number of ON bits, and zero otherwise.

The answer, I referred to, has $$2^{(5-1)}=16$$ neurons for the hidden layer. Whereas, I could design a network (see below) using only 5 neurons for the same layer. Wondering, what does this $$2^{(n-1)}$$ signify? Is it the max bound of hidden neurons for a 1-hidden-layer neural network? What is the minimum bound expression, if any?

Note: I know the concept behind the $$2^{(n-1)}$$, as in, how any binary function can be expressed as DNF (disjunctive normal form) of $$2^n$$ terms, and how they can be simplified to $$\frac{2^n}{2}$$ terms where each term can be implemented with a single neuron with threshold activation.

Note: I also found that my design with $$5 (=\log_2{2^{n=5}} \div \log_2{2^{m=1}})$$ hidden neurons matches with this paper. Is the lowest bound expression from this paper accepted widely!