K2 Choosing Prime Implicants

# 1 1 bc 00 1 00 1 1 1 01 0 1 1 01 1 1 01 1 1 01 1 1 01

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1 1 BC 00 0 0 1 0 00 0 1 1 1 01 0 0 1 1 01 0 0 1 1 01 0 0 1 1 01 0 0 1 1 01 0 0 1 1 01 0 0 1 1 11 1 0 0 1 11 1 1 1 1 11 1 0 0 1 11 1 1 1 1 11 1 0 0 1 11 1 1 1 1 10 0 1 1 0 10 1 1 0 1 10 0 1 1 0 10 1 1 0 1 10 0 1 1 0 10 1 1 0 1 A = 0 A = 1 A = 0 A = 1 A = 0 A = 1 AC′D′E (17,25) A′C′DE (3,11) BC′D′E (9,25) Remember, our minimal cover will only consist of prime implicants. Specifically, it will consist of all of the essential prime implicants, and a minimal number of prime implicants that cover the minterms not covered by essential prime implicants. To best achieve the task of covering minterms, we will focus on covering the minterms not already covered after each choice that we make. Start by noting that prime implicant BCE′ is essential, as it is the only prime implicant that covers m 12 . DE 00 01 11 10 00 01 11 10 BC 00 1 00 1 1 1 01 1 1 01 1 1 11 1 1 11 1 1 1 1 10 1 1 10 1 1 1 A = 0 A = 1 Having covered those minterms, we need not cover them again. It will be okay if we choose other prime implicants that happen to overlap with the prime implicant that is already in the cover, but given a choice

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between covering those minterms and covering other uncovered minterms, we should work on adding uncovered minterms to the cover. With that in mind, we can remove those minterms from the map. DE 00 01 11 10 00 01 11 10 BC 00 1 00 1 1 1 01 1 1 01 1 1 11 11 1 1 10 1 1 10 1 1 1 A = 0 A = 1 Now, compare the remains of two prime implicants: CDE′ and B′CD: DE 00 01 11 10 00 01 11 10 BC 00 1 00 1 1 1 01 1 1 01 1 1 11 11 1 1 10 1 1 10 1 1 1 A = 0 A = 1 Because we had to choose BCE′ to cover m 12 , we also covered m 14 , m 28 , and m 30 . Because of that, CDE′ only covers two minterms that aren’t already covered: m 6 and m 22 . B′CD covers those two minterms and two others: m 7 and m 23 . Given a choice between choosing a prime implicant that covers m 6 and m 22 , and one that covers those two minterms and two others, we should choose the prime implicant that covers more uncovered minterms. Here, we say that B′CD dominates CDE′. Prime implicant A dominates prime implicant B when A covers all of B’s minterms, and other minterms. The only reason we have to choose a dominated prime implicant is when it costs less than the prime implicant that dominates it. If it doesn’t, then we have no reason ever to choose it.
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