Edhesive intro to cs term 2 8.9 practice
2 answers:
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Answer:
(1)Minterms complement = XYZ (2) Compliment of Minterms = Σm(0,1, 2 , 4 , 6) (3) (X+Y+Z) (4) Minimized SOP = Z + XY
Manterms = πM
Explanation:
Solution
Recall that:
Given the function F (X, Y , Z)=Σm(0,1, 2 , 4 , 6)
(1) Canonical Disjunctive Normal Form: In boolean algebra, the boolean function can be expressed as Canonical Disjunctive form known as minterms
In Minterm we assign 'I' to each uncomplimented variable and '0' to each complemented/complementary variable
For the given question stated we ave the following:
Minterms = XYZ, XYZ, XYZ, XYZ, XYZ.
(2) Canonical Conjunctive Normal Form: In boolean algebra, the boolean function can be expressed as Canonical Disjunctive form known as maxterms.
In Maxterms we assign '0' to each uncomplimented variable and '1' to each complemented/complementary variable
Compliment of Minterms = Σm(0,1, 2 , 4 , 6)
Maxterms = πM
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