General Computer Science 320201 GenCS I & II Lecture ... - Kwarc

The first result is a very naive counting argument based on the fact that we can always realize a
Boolean function via its DNF or CNF. The second result gives us a better complexity with a more
involved argument. Another difference between the proofs is that the first one is constructive,
i.e. we can read an algorithm that provides Boolean expressions of the complexity claimed by the
algorithm for a given Boolean function. The second proof gives us no such algorithm, since it is
non-constructive.
An Upper Bound for the Cost of BF with n variables
Idea: Every Boolean function has a DNF and CNF, so we compute its cost.
Example 267 Let us look at the size of the DNF or CNF for f ∈ (B 3 → B).
x1 x2 x3 f monomials clauses
0 0 0 1 x 0 1 x0 2 x0 3
0 0 1 1 x 0 1 x0 2 x1 3
0 1 0 0 x 1 1 + x0 2 + x1 3
0 1 1 0 x 1 1 + x0 2 + x0 3
1 0 0 1 x 1 1 x0 2 x0 3
1 0 1 1 x 1 1 x0 2 x1 3
1 1 0 0 x 0 1 + x0 2 + x1 3
1 1 1 1 x 1 1 x1 2 x1 3
Theorem 268 Any f : B n → B is realized by an e ∈ Ebool with C(e) ∈ O(n · 2 n ).
Proof: by counting (constructive proof (we exhibit a witness))
P.1 either en := CNF(f) has 2n
2
clauses or less or DNF(f) does monomials
take smaller one, multiply/sum the monomials/clauses at cost 2 n−1 − 1
there are n literals per clause/monomial ei, so C(ei) ≤ 2n − 1
so C(en) ≤ 2 n−1 − 1 + 2 n−1 · (2n − 1) and thus C(en) ∈ O(n · 2 n )
c○: Michael Kohlhase 148
For this proof we will introduce the concept of a “realization cost function” κ: N → N to save
space in the argumentation. The trick in this proof is to make the induction on the arity work
by splitting an n-ary Boolean function into two n − 1-ary functions and estimate their complexity
separately. This argument does not give a direct witness in the proof, since to do this we have to
decide which of these two split-parts we need to pursue at each level. This yields an algorithm for
determining a witness, but not a direct witness itself.
We can do better (if we accept complicated witness)
P.2 P.3 P.4 Theorem 269 Let κ(n) := max({C(f) | f : B n → B}), then κ ∈ O(2 n ).
Proof: we show that κ(n) ≤ 2 n + d by induction on n
P.1.1 base case:
We count the operators in all members: B → B = {f1, f0, fx1, fx1},
so κ(1) = 1 and thus κ(1) ≤ 2 1 + d for d = 0.
P.1.2 step case:
P.1.2.1 given f ∈ (B n → B), then f(a1, . . . , an) = 1, iff either
an = 0 and f(a1, . . . , an−1, 0) = 1 or
an = 1 and f(a1, . . . , an−1, 1) = 1
P.1.2.2 Let fi(a1, . . . , an−1) := f(a1, . . . , an−1, i) for i ∈ {0, 1},
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