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 ≤ 2n − 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 \({\mathsf{dnfsize}} (n^c,n) = 2^{n -\theta(c^{-1}\frac{n}{\log n})}\).
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Miltersen, P.B., Radhakrishnan, J., Wegener, I. (2003). On Converting CNF to DNF. In: Rovan, B., Vojtáš, P. (eds) Mathematical Foundations of Computer Science 2003. MFCS 2003. Lecture Notes in Computer Science, vol 2747. Springer, Berlin, Heidelberg. https://doi.org/10.1007/978-3-540-45138-9_55
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DOI: https://doi.org/10.1007/978-3-540-45138-9_55
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