On Converting CNF to DNF

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