Mathematical Foundation of Computer Science

DHARM

N-COM\APPENDIX.PM5

BOOLEAN ALGEBRA 371

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Fig. A.18
f(x, y, z) = (w + x + y′) (w + x′ + y)(x′ + y + z) (w′ + y′ z) (w + y + z′) (5 minterms)
Example A.8. Simplify the Boolean function f(x, y, z) = Σ (0, 2, 4, 6).
Solution: Here, summation of minterms are given by their equivalent decimal numbers. Since,
function f has three variables so K-map of three variables must be used to represent f. The
minterms of the function are marked by 1’s in the K-map shown in Fig. A.19.

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Fig. A.19
The adjacent cells marked by 1’s can be combined to form the term free from one variable
such as,
x′ y′ z′ + x y′ z′ = (x′ + x) y′ z′ = 1. y′ z′ = y′ z′ ;
and, x′ y z′ + x y z′ = (x′ + x) y z′ = 1. y z′ = y z′ ;
Further, these expression lies on the adjacent edges so it can be simplified as,
= y′ z′ + y z′ = (y′ + y) z′ = 1. z′ = z′ ....(1)
Cells holding 1’s can also be combined as, (see Fig. A.20)
x′ y′ z′ + x′ y z′ = x′ (y′ + y) z′ = x′. 1. z′ = x′ z′;
and, x y′ z′ + x y z′ = x (y′ + y) z′ = x. 1. z′ = x z′ ;

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Fig. A.20