Consider a Boolean function f(w,x,y,z) such that
f(w,0,0,z) = 1 f(1,x,1,z) = x+z f(w,1,y,z) = wz+y
The number of literals in the minimal sum-of-products expression of f is _________ .
(A) 6
(B) 3
(C) 8
(D) 1
Answer: (A)
Explanation: f(w,0,0,z) = 1
When x = 0, and y = 0, the min-term will be 1 irrespective of w and z.
f(1,x,1,z) = x+z
When w = 1 and y = 1, the min-term will be x or’d with z.
Similarly for the last one.
f(w, x, y, z) = sigma(0,1,6,7,8,9,11,13,14,15) + don’t(2,3)
f(w, x, y, z) = x’y’ + wz + xy
Thus, the number of literals will be 6.
Quiz of this Question
Article Tags :
Recommended Articles