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 __________.

Let us analyze the given conditions to determine the minimal sum-of-products (SOP) expression for $f(w, x, y, z)$:

1. From $f(w, 0, 0, z) = 1$:When $x=0$ and $y=0$, the function is identically $1$, giving the product term $\overline{x}\overline{y}$.

2. From $f(w, 1, y, z) = wz + y$:When $x=1$, the function outputs $wz + y$. Multiplying by $x$ gives the terms $xy + wxz$.

3. From $f(1, x, 1, z) = x + z$:When $w=1$ and $y=1$, the function outputs $x + z$. This gives terms $wxy + wyz$.

Combining these conditions into a single Boolean expression:

Let us verify all three conditions with $f(w, x, y, z) = \overline{x}\overline{y} + xy + wz$:

• $f(w, 0, 0, z) = (1)(1) + (0)(0) + wz = 1$

• $f(1, x, 1, z) = \overline{x}(0) + x(1) + (1)z = x + z$

• $f(w, 1, y, z) = (0)\overline{y} + (1)y + wz = y + wz$

All conditions are satisfied.

In the minimal SOP expression:

• $\overline{x}\overline{y}$ contains 2 literals ($\overline{x}, \overline{y}$)

• $xy$ contains 2 literals ($x, y$)

• $wz$ contains 2 literals ($w, z$)