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

the disjunctive normal form, and then replace partners by resolvents, until that is impossible.
The algorithm QMC1, for determining Prime Implicants
Definition 284 Let M be a set of monomials, then
R(M) := {m | (m x) ∈ M ∧ (m x) ∈ M} is called the set of resolvents of M
R(M) := {m ∈ M | m has a partner in M} (n xi and n xi are partners)
Definition 285 (Algorithm) Given f : B n → B
let M0 := DNF(f) and for all j > 0 compute (DNF as set of monomials)
Mj := R(Mj−1) (resolve to get sub-monomials)
Pj := Mj−1\ R(Mj−1) (get rid of redundant resolution partners)
terminate when Mj = ∅, return Pprime := n
j=1 Pj
c○: Michael Kohlhase 159
We will look at a simple example to fortify our intuition.
Example for QMC1
x1 x2 x3 f monomials
F F F T x1 0 x2 0 x3 0
F F T T x1 0 x2 0 x3 1
F T F F
F T T F
T F F T x1 1 x2 0 x3 0
T F T T x1 1 x2 0 x3 1
T T F F
T T T T x1 1 x2 1 x3 1
Pprime =
3
Pj = {x1 x3, x2}
j=1
M0 = {x1 x2 x3,
x1 x2 x3,
x1 x2 x3,
x1 x2 x3,
x1 x2 x3}
= : e 0 1
M 1 = { x1 x2
P 1 = ∅
R(e 0 1 ,e0 2 )
= : e 1 1
M 2 = { x2
P 2 = {x1 x3}
M 3 = ∅
P 3 = {x2}
= : e 0 2
, x2 x3
R(e 0 1 ,e0 3 )
= : e 1 2
, x2
R(e1 1 ,e1 4 ) R(e1 2 ,e1 3 )
= : e 0 3
}
, x2 x3
= : e 0 4
R(e 0 2 ,e0 4 )
= : e 1 3
= : e 0 5
, x1 x2
R(e 0 3 ,e0 4 )
= : e 1 4
, x1 x3
R(e 0 4 ,e0 5 )
But: even though the minimal polynomial only consists of prime implicants, it need not
contain all of them
c○: Michael Kohlhase 160
We now verify that the algorithm really computes what we want: all prime implicants of the
Boolean function we have given it. This involves a somewhat technical proof of the assertion
below. But we are mainly interested in the direct consequences here.
Properties of QMC1
Lemma 286 (proof by simple (mutual) induction)
1. all monomials in Mj have exactly n − j literals.
2. Mj contains the implicants of f with n − j literals.
3. Pj contains the prime implicants of f with n − j + 1 for j > 0 . literals
84
= : e 1 5
}