Proof: Let N =
Lemma 4 plays a central role in our analysis. It gives a bound on the number of terms which are almost satissed by an assignment a, in terms of k and j. It is natural to ask how good is this bound. The following example, due to Galia Givaty (private communication), shows that it is essentially optimal (up to constants). More precisely, for any values of k and j we give a function F k;j with k j variables x 0 ; x 2 ; : : :; x kj?1 , and an assignment a such that jY (a)j = k j. The function F k;j has kj terms: t i = x i x i+1 : : :x i+k?1modkj (0 i kj ? 1) It can be easily veriied that the function is Read-k (the variable x i appears only in the terms t i?k+1modkj ; : : :; t i). It is also Satisfy-j as if an assignment satisses a term t i it cannot satisfy any of the k ? 1 terms t i?k+1modkj ; : : :; t i?1modkj (as they all contain the variable x i while t i contains x i). Now, let a be the all \1" assignment. Each term t i is almost satissed by a with respect to index i. Hence, jY (a)j = kj. 23 Bsh95] N. H. Bshouty. Exact learning boolean functions via the monotone theory. Jac94] J. Jackson. An eecient membership-query algorithm for learning DNF with respect to the uniform distribution. 22 Ang87] D. Angluin. Learning k-term DNF formulas using queries and counterexamples. learning DNF and characterizing statistical query learning using fourier analysis. 21 representation, so there is still a term t C that can be satissed. Therefore, we can deene a to be the assignment in which all the literals in C are set to zero, all the literals in t C are set to one, and all other literals are set arbitrarily. We thus have that C(a) = 0 but t C (a) = 1 (and hence F n (a) = 1), a contradiction. 7 Bibliographic Remarks The work presented here appeared in preliminary forms as AP92] and BKK + 94]. The algorithm for constant k and j is from AP92] and the extension to kj = O(log n= log log n) from BKK + 94]. The family of functions given in Section 6 was deened …