# On Converting CNF to DNF

• Peter Bro Miltersen
• Ingo Wegener
Conference paper
Part of the Lecture Notes in Computer Science book series (LNCS, volume 2747)

## 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 $${\mathsf{dnfsize}} (n^c,n) = 2^{n -\theta(c^{-1}\frac{n}{\log n})}$$.

## Keywords

Boolean Function Vertex Cover Disjunctive Normal Form Binary Decision Diagram Small Clause
These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.

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## Authors and Affiliations

• Peter Bro Miltersen
• 1