The minimum number of gates needed to implement the Boolean function f(x, y, z)=z(x+y)+\overline{(\ov{z}+x+y)(\overline{x}+\overline{y})} is

NIMCET 2008 Question Paper with solutions

Question : 104 of 120

Marks: +1, -0

The minimum number of gates needed to implement the Boolean function

f(x, y, z)=z(x+y)+\ov{(\ov{z}+x+y)(\ov{x}+\ov{y})}

Solution:

f(x, y, z)=z(x+y)+(\ov{\ov{z}+x+y)({x}↖{→}+\ov{y})}

(given)

=z(x+y)+(z \ov{x} \ov{y})+(x y)

(by De Morgan's law)

=x y+y z+z x+z {x}↖{→} ˜{y}

(by Distributive law)

=y z+z x+(x y+z \ov{x} \ov{y})

=y z+2 x+(x y+z)(x y+\ov{x} {y}↖{→}) \;\;

(by Distributive law)

=y z+z x+(x y+z) ⋅ 1

(by Complement law)

=2 x+x y+z(y+1)

=2 x+x y+z ⋅ 1

(by Absorption law)

=x y+z(x+1)=x y+z ⋅ 1=x y+z

⇒ f(x, y, z)=x y+z

It requires two gates.

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