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Extra info for The Complexity of Boolean Functions (Wiley Teubner on Applicable Theory in Computer Science)
Example text
We could cancel mi and would obtain a cheaper polynomial for f . ) . I(f) is the set of all implicants of f . We have already seen that minimal polynomials consist of implicants only. Obviously the sum of all implicants computes f . If m and m′ are implicants, but m is a proper submonom of m′ , m ∨ m′ = m by the law of simplification, and we may cancel m′ . 4 : An implicant m ∈ I(f) is called prime implicant if no proper submonom of m is an implicant of f . PI(f) is the set of all prime implicants of f .
35 Let t ∈ I(f) . 2) t contains either xn+1 or xn+2 or both of them. 1) t contains some xi if it contains xn+1 and xn+2 . If t contains xn+1 but not xn+2 , then t contains a full minterm ma . Otherwise t does not contain xi and xi for some i and we find vectors a and a′ differing only at position i such that t(a 0 1) = t(a′ 0 1) = 1 implying f(a 0 1) = f(a′ 0 1) = 1 . 4) |a| = |a′ | = 1 which is impossible if a and a′ differ only at one position. Altogether t contains xn+1 and ma for some a ∈ {0 1}n .
The following proof is long and technically involved, although the ideas are easy. 3 : Krapchenko’s adder S consists of five parts S1 S5 . In S1 uj = xj yj and vj = xj ⊕ yj are computed by n halfadders. The crucial part is the computation of the carry bits cj in S2 , S3 and S4 . Afterwards, it is easy to compute in S5 the outputs s0 = v0 , sj = vj ⊕ cj−1 for 1 ≤ j ≤ n − 1 and sn = cn−1 . 2) for j + 1 times we can compute cj from the u- and v-parameters. 8) ui vi+1 · · · vj This can be interpreted in the following way.