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

Hint:

When multiple functional constraints are given, use Karnaugh maps to systematically minimize the expression.

Answer 1

Correct Answer: 6

To find the minimal sum-of-products expression for the Boolean function \(f(w,x,y,z)\), we analyze the given conditions:

• \(f(w,0,0,z) = 1\)
• \(f(1,x,1,z) = x + z\)
• \(f(w,1,y,z) = wz + y\)

Let's consider each condition:

1. Condition 1: When \(x=y=0\), regardless of \(w\) and \(z\), \(f(w,0,0,z)\) evaluates to 1. Thus, the corresponding minterms cover all combinations of \(w\) and \(z\) with \(x=0\) and \(y=0\): \(w\bar{x}\bar{y}\bar{z}\), \(w\bar{x}\bar{y}z\), \(\bar{w}\bar{x}\bar{y}\bar{z}\), and \(\bar{w}\bar{x}\bar{y}z\).
2. Condition 2: For \(w=1\) and \(y=1\), function simplifies to \(x+z\). This case contributes the terms \(x\) and \(z\) when \(w=1\) and \(y=1\).
3. Condition 3: For \(x=1\), function simplifies to \(wz+y\). This implies \(wyz\) and \(wz\) terms are relevant when \(w=1\).

The potential sum-of-products expression, analyzing these conditions, is:

\(f(w,x,y,z) = w\bar{x}\bar{y}\bar{z} + w\bar{x}\bar{y}z + \bar{w}\bar{x}\bar{y}\bar{z} + \bar{w}\bar{x}\bar{y}z + x + z + wyz + wz\)

To minimize, note:

• \(\bar{x}\bar{y}\) condition can be simplified further to account for shared terms with \(x + z\) and \(wz\).

Minimize using Boolean algebra:

1. Combine terms: \(x + z\) covers various conditions.
2. Minimized expression: \(z + x + wy\).

Thus, the minimal sum-of-products expression has 6 literals: \[z + x + wy\]. Calculating, we see it uses direct literals from each relevant term.

Finally, the number of literals in the minimal sum-of-products expression is \(6\), which fits the specified range of (6,6).