Question

What specific Boolean expression results from applying the Quine-McCluskey algorithm to a truth table where output is high only for minterms 0, 1, 4, 5, 8, and 9?

Accepted Answer

To apply the Quine-McCluskey algorithm for minterms 0, 1, 4, 5, 8, and 9, we represent these as four-variable binary numbers ABCD where A=8, B=4, C=2, and D=1. The minterms are 0000, 0001, 0100, 0101, 1000, and 1001. We group these by the number of ones they contain: Group 0 has 0000; Group 1 has 0001, 0100, and 1000; Group 2 has 0101 and 1001. We combine terms that differ by only one bit position. Comparing Group 0 and Group 1: 0000 and 0001 combine to 000-, 0000 and 0100 combine to 0-00, 0000 and 1000 combine to -000. Comparing Group 1 and Group 2: 0001 and 0101 combine to 0-01, 0001 and 1001 combine to -001, 0100 and 0101 combine to 010-, 1000 and 1001 combine to 100-. We then simplify these results. The prime implicants are 000-, -000, 0-01, -001, 010-, and 100-. Mapping these to the minterms, 000- covers 0 and 1, -000 covers 0 and 8, 0-01 covers 1 and 5, -001 covers 1 and 9, 010- covers 4 and 5, and 100- covers 8 and 9. Selecting a minimum set of prime implicants to cover all minterms results in the expression C&#x27;D&#x27; + A&#x27;C&#x27; + B&#x27;C&#x27;. Since C&#x27;D&#x27;, A&#x27;C&#x27;, and B&#x27;C&#x27; all cover the required minterms, the minimal Boolean expression is A&#x27;C&#x27; + B&#x27;C&#x27; + C&#x27;D&#x27;.

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What specific Boolean expression results from applying the Quine-McCluskey algorithm to a truth table where output is high only for minterms 0, 1, 4, 5, 8, and 9?