Discussion
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
The minimized expression is \(\bar{b}_{1}\bar{b}_{0} + \bar{b}_{2}\bar{b}_{0} + b_{1}\bar{b}_{2}b_{3}\).
Group the minterms 0, 2, 4, 8, 10, 11, 12 on a Karnaugh map:
- \(\bar{b}_{1}\bar{b}_{0}\) covers 0, 4, 8, 12 (the whole column with \(b_{1}b_{0} = 00\)).
- \(\bar{b}_{2}\bar{b}_{0}\) covers 0, 2, 8, 10 (the four corners).
- \(b_{3}\bar{b}_{2}b_{1}\) covers 10 and 11, and is the only way to include 11.
All three are essential prime implicants, and together they cover every minterm exactly as required. Option B leaves out minterm 11. Option C's \(b_{1}b_{2}b_{3}\) covers 14 and 15, which are not in F, and misses 4, 11 and 12. Option D's \(\bar{b}_{3}\) would make F true for 1, 3, 5, 6 and 7, which are not minterms.
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