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 $\underline{\hspace{2cm}}$.

Question

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 $\underline{\hspace{2cm}}$.

cdquestions Admin · Accepted Answer

The given conditions define the function values for overlapping regions of the input space. By constructing the Karnaugh map using the given constraints and simplifying the resulting Boolean expression, the minimal sum-of-products form is obtained. Counting the literals (variables or their complements) appearing in all product terms of the minimal SOP expression gives: $$ 6 $$ Final Answer: $$ \boxed{6} $$

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 \(\underline{\hspace{2cm}}\).

Show Hint

Correct Answer: 6

Solution and Explanation

Questions Asked in GATE CS exam

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 \(\underline{\hspace{2cm}}\).

Show Hint

Correct Answer: 6

Solution and Explanation

Questions Asked in GATE CS exam

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 \(\underline{\hspace{2cm}}\).