Consider the following four variable Boolean function in sum-of-product form
\[ F(b_3, b_2, b_1, b_0) = \sum (0, 2, 4, 8, 10, 11, 12). \]
where the value of the function is computed by considering \(b_3 b_2 b_1 b_0\) as a 4-bit binary number, where \(b_3\) denotes the most significant bit and \(b_0\) denotes the least significant bit. Note that there are no don’t care terms. Which ONE of the following options is the CORRECT minimized Boolean expression for \(F\)?
Answer
(A)
Explanation
Write the minterms as \(b_3 b_2 b_1 b_0\): 0 = 0000, 2 = 0010, 4 = 0100, 8 = 1000, 10 = 1010, 11 = 1011, 12 = 1100. Group them on a K-map:
- \(\{0, 4, 8, 12\}\) all have \(b_1 = b_0 = 0\), giving \(\bar{b}_1 \bar{b}_0\).
- \(\{0, 2, 8, 10\}\) all have \(b_2 = b_0 = 0\), giving \(\bar{b}_2 \bar{b}_0\).
- 11 (1011) has only one neighbouring 1, namely 10 (1010), so it needs the pair \(\{10, 11\}\), giving \(b_3 \bar{b}_2 b_1\).
All three groups are essential and together cover exactly the seven minterms, so \(F = \bar{b}_1 \bar{b}_0 + \bar{b}_2 \bar{b}_0 + b_1 \bar{b}_2 b_3\). Option B misses minterm 11, C covers 14 and 15 (not in \(F\)), and D covers 1, 3, 5, 6 and 7.
Answer: A