GATE 2021 CS (CS2) – Question 62
Boolean f(w,x,y,z) satisfies f(w,0,0,z)=1, f(1,x,1,z)=x+z, f(w,1,y,z)=wz+y. Find the number of literals in a minimal sum-of-products expression.
Practise this question in The GATE Grind →
Show answer and explanation
Correct answer: 6 to 6
Explanation
For y=0,x=0 the function is one. For y=1,x=1 it is one. For x=1,y=0 it is wz, and for w=1,y=1 it is x+z. Combining the cases gives f=x′y′+xy+wz, with unspecified input combinations treated as don’t-cares. This uses three two-literal terms, six literals total. A Karnaugh map requires distinct groups covering the x=y=0 region, x=y=1 region, and remaining w=z=1 region; none expands to a one-literal implicant without covering a specified zero. Thus six is minimal.