How to bound the size of Boolean circuits?

Question

Every function $f\colon \{ 0, 1 \}^n \to \{ 0, 1 \}$ can be computed by a circuit over the standard unbounded fan-in basis $\mathcal{B}_1 = \{ \neg, (\vee^n)_{n \in \mathbb{N}}, (\wedge^n)_{n \in \mathbb{N}} \}$ of depth $3$ and size $\mathcal{O}(2^n)$ via the disjunctive or conjunctive normal form. Now, we can transform each circuit over $\mathcal{B}_1$ into a circuit over the standard bounded fan-in basis $\mathcal{B}_0 = \{ \neg, \vee, \wedge \}$ by substituting a gate of fan-in $k$ by a circuit of depth $\mathcal{O}(\log k)$ and size $\mathcal{O}(k)$ and, since each gate of the original circuit has a fan-in smaller than $\mathcal{O}(2^n)$, we get a circuit of depth $\mathcal{O}(n)$ and size $\mathcal{O}(2^{2n})$. How does http://www.cs.umd.edu/~jkatz/complexity/f05/lecture4.pdf obtain a size bounded by $\mathcal{O}(n \cdot 2^n)$?

Answer 1

Every function can be computed by a CNF (or a DNF). A CNF is the conjunction of up to $2^n$ clauses. Each clause involves up to $n$ variables. You can express each clause using a circuit of size $O(n)$, and the entire CNF using a circuit of size $O(n 2^n)$. In fact, this constructs a formula of size $O(n2^n)$.

The conversion from $\mathcal{B}_1$ to $\mathcal{B}_0$ doesn't increase the number of leaves in the circuit, and only increases the number of wires by a constant factor. It could increase the number of gates by an unbounded amount: for example, a fan-in $k$ AND gates translates to $k-1$ fan-in $2$ AND gates.

Lupanov improved the above bound to $O(2^n/n)$, which is tight due to a counting argument; see for example lecture notes of Jayalal Sarma. In the case of formulas, the tight bound is $O(2^n/\log n)$.