# Boolean function complexity by Jukna S. By Jukna S.

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This requires additional 2n/2 · 2n/2 = 2n gates, and the entire circuit has size at most 2n + n2n/2 To include the case when n is odd, we just multiply the last term by 2. ⊔ ⊓ We now turn to the actual construction of an eﬃcient circuit for a given boolean function f (x) of n variables. Let 1 ≤ k, m ≤ n be integer parameters (to be speciﬁed latter). 5), we can write f (x) as a disjunction 2n /m f (x) = a Ka (x1 , . . , xk ) ∧ fa,i (xk+1 , . . , xn ) , i=1 where a ranges over {0, 1}k , and each fa,i belongs to Hn−k,m (i).

The operator L(x) is just a set of m ≤ n parity functions, and hence, can be computed by a trivial circuit of size O(n2 ), which is o(N/n) because log N = Ω(n), by our assumption. The function h can be computed by a small circuit just because it accepts at most N/n3 vectors x ∈ D. Indeed, h(x) = 0 for all x ∈ D0 because then L(x) ∈ L(D0 ). Hence, h can accept a vector x ∈ D only if x ∈ D1 and g(L(x)) = 0, that is, if x ∈ D1 and L(x) = L(y) for some y ∈ D0 . Since the operator L is almost injective, and since 2m ≥ N n3 , there are at most 2−m N2 ≤ N/n3 pairs (y, x) ∈ D0 × D1 such that L(x) = L(y).

0) = 1 if and only if {u, v} ∈ E . If the graph is bipartite then we only require that this must hold for vertices u and v from diﬀerent color classes. Note that in both cases (bipartite or not), on input vectors with fewer than two 1s as well as on vectors with more than two 1s the function g can take arbitrary values! Another way to treat this concept is to view edges as 2-element sets of vertices, and boolean functions (or circuits) as accepting/rejecting subsets 44 1 Our Adversary: The Circuit B B B B B A A A A A A B Fig.