Question

The simplified form of the Boolean function $F(W,X,Y,Z)=\Sigma(4,5,10,11,12,13,14,15)$ with the minimum number of terms and literals is _____

• A. $WX+\overline{W}X\overline{Y}+W\overline{X}\overline{Y}$
• B. $WX+WY+X\overline{Y}$
• C. $X\overline{Y}+WY$
• D. $\overline{X}Y+\overline{W}\,\overline{Y}$

Hint:

In K-maps, look for groups spanning the $W$ dimension to drop $W$; the pair “$X\overline{Y}$ no matter $W,Z$” and the block “$WY$ no matter $X,Z$” jump out immediately.

Answer 1

The Correct Option is C

Step 1: List minterms (in $WXYZ$).
$4(0100),5(0101),10(1010),11(1011),12(1100),13(1101),14(1110),15(1111)$.
Step 2: Group logically.
- Minterms with $X=1,\,Y=0$ occur for both $W=0$ and $W=1$, independent of $Z$ $\Rightarrow X\overline{Y}$.
- For $W=1$ and $Y=1$ (regardless of $X,Z$) we have 10,11,14,15 $\Rightarrow WY$.
Step 3: Combine.
\[ F = X\overline{Y} + WY, \] which already covers all listed minterms with {two} terms and the fewest literals.