On Converting CNF to DNF

Abstract

We study how big the blow-up in size can be when one switches between the CNF and DNF representations of boolean functions. For a function f:{0,1}ⁿ →{0,1}, ${\mathsf{cnfsize}}\left(f\right)$ denotes the minimum number of clauses in a CNF for f; similarly, ${\mathsf{dnfsize}}\left(f\right)$ denotes the minimum number of terms in a DNF for f. For 0≤ m ≤ 2^n − 1, let ${\mathsf{dnfsize}}\left(m,n\right)$ be the maximum ${\mathsf{dnfsize}}\left(f\right)$ for a function f:{0,1}ⁿ →{0,1} with ${\mathsf{cnfsize}}\left(f\right) \leq m$. We show that there are constants c ₁,c ₂ ≥ 1 and ε > 0, such that for all large n and all $m \in [ \frac{1}{\epsilon}n,2^{\epsilon{n}}]$, we have

$$ 2^{n -- c_1\frac{n}{\log(m/n)}}~ \leq~ {\mathsf{dnfsize}}(m,n) ~\leq~ 2^{n-c_2 \frac{n}{\log(m/n)}}.$$

In particular, when m is the polynomial n ^c, we get ${\mathsf{dnfsize}} (n^c,n) = 2^{n -\theta(c^{-1}\frac{n}{\log n})}$.