Question 62

We are given three constraints on the 4-variable function $f(w, x, y, z)$:

1. $f(w, 0, 0, z) = 1$:
   For $x=0, y=0$, the function is 1 for all $w, z$. This gives minterms:
   $m_0(0000) = 1, m_1(0001) = 1, m_8(1000) = 1, m_9(1001) = 1$.

2. $f(w, 1, y, z) = wz + y$:
   For $x=1$:
   • If $y=1$, $wz + 1 = 1$: minterms $m_6(0110), m_7(0111), m_{14}(1110), m_{15}(1111)$ are all 1.
   • If $y=0$, $wz + 0 = wz$: $m_{13}(1101) = 1$, and $m_4(0100) = 0, m_5(0101) = 0, m_{12}(1100) = 0$.

3. $f(1, x, 1, z) = x + z$:
   For $w=1, y=1$:
   • If $x=1$, $1 + z = 1$ (consistent with $m_{14}, m_{15} = 1$).
   • If $x=0$, $0 + z = z$: $m_{11}(1011) = 1$, and $m_{10}(1010) = 0$.

4. Minimizing on the K-map:
   • Group 1: $m_0, m_1, m_8, m_9$ forms a 4-cell group covering $x=0, y=0$, which gives $\overline{x}\,\overline{y}$ (2 literals).
   • Group 2: $m_6, m_7, m_{14}, m_{15}$ forms a 4-cell group covering $x=1, y=1$, which gives $xy$ (2 literals).
   • Group 3: The remaining 1-minterms $m_9, m_{11}, m_{13}, m_{15}$ form a 4-cell group covering $w=1, z=1$, which gives $wz$ (2 literals).
     The minimal SOP expression is:
   $$f(w, x, y, z) = \overline{x}\,\overline{y} + xy + wz$$

Counting literals: each of the three product terms contains 2 literals, giving a total of $2 + 2 + 2 = 6$ literals.