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}, cnfsize(f) denotes the minimum number of clauses in a CNF for f; similarly, dnfsize(f) denotes the minimum number of terms in a DNF for f. For 0 ≤ m ≤ 2n-1, let dnfsize(m, n) be the maximum dnfsize(f) for a function f : {0, 1}n → {0, 1} with cnfsize(f) ≤ m. We show that there are constants c1, c2 ≥ 1 and ∈ > 0, such that for all large n and all m ∈ [1/∈n, 2∈n], we have 2n-c1 n/log(m/n) ≤ dnfsize(m, n) ≤ 2n-c2 n/log(m/n) In particular, when m is the polynomial nc, we get dnfsize(nc, n) = 2n-θ(c-1 n/log n). © Springer-Verlag Berlin Heidelberg 2003.