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

Proof: by contradiction: let p /∈ pmin
P.1 We know that f = fpmin and pmin = n j=1 pj for some n ∈ N and prime implicants pj.
P.2 so for all c ∈ Bn with f(c) = T there is a j ≤ n with fpj (c) = T.
P.3 so p cannot be essential
c○: Michael Kohlhase 164
Let us now apply the optimized algorithm to a slightly bigger example.
A complex Example for QMC (Function and DNF)
x1 x2 x3 x4 f monomials
F F F F T x1 0 x2 0 x3 0 x4 0
F F F T T x1 0 x2 0 x3 0 x4 1
F F T F T x1 0 x2 0 x3 1 x4 0
F F T T F
F T F F F
F T F T T x1 0 x2 1 x3 0 x4 1
F T T F F
F T T T F
T F F F F
T F F T F
T F T F T x1 1 x2 0 x3 1 x4 0
T F T T T x1 1 x2 0 x3 1 x4 1
T T F F F
T T F T F
T T T F T x1 1 x2 1 x3 1 x4 0
T T T T T x1 1 x2 1 x3 1 x4 1
c○: Michael Kohlhase 165
A complex Example for QMC (QMC1)
M0 = {x1 0 x2 0 x3 0 x4 0 , x1 0 x2 0 x3 0 x4 1 , x1 0 x2 0 x3 1 x4 0 ,
x1 0 x2 1 x3 0 x4 1 , x1 1 x2 0 x3 1 x4 0 , x1 1 x2 0 x3 1 x4 1 ,
x1 1 x2 1 x3 1 x4 0 , x1 1 x2 1 x3 1 x4 1 }
M1 = {x1 0 x2 0 x3 0 , x1 0 x2 0 x4 0 , x1 0 x3 0 x4 1 , x1 1 x2 0 x3 1 ,
x1 1 x2 1 x3 1 , x1 1 x3 1 x4 1 , x2 0 x3 1 x4 0 , x1 1 x3 1 x4 0 }
P1 = ∅
M2 = {x1 1 x3 1 }
P2 = {x1 0 x2 0 x3 0 , x1 0 x2 0 x4 0 , x1 0 x3 0 x4 1 , x2 0 x3 1 x4 0 }
M3 = ∅
P3 = {x1 1 x3 1 }
Pprime = {x1 x2 x3, x1 x2 x4, x1 x3 x4, x2 x3 x4, x1 x3}
c○: Michael Kohlhase 166
A better Mouse-trap for QMC1: optimizing the data structure
Idea: Do the calculations directly on the DNF table
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