Question 59

Given minterms $m(0, 2, 5, 7, 9, 11)$ and don't cares $d(3, 8, 10, 12, 14)$ on variables $P, Q, R, S$:

• Group 1 (corners and edges): $\{0, 2, 8, 10\}$ forms a quad covering minterms 0 and 2 using don't cares 8 and 10. The term is $\overline{Q}\,\overline{S}$. Minterm 0 cannot be combined into any other prime implicant, so $\overline{Q}\,\overline{S}$ is essential.
• Group 2 (row 01): $\{5, 7\}$ forms a pair covering minterms 5 and 7. Since neither 5 nor 7 has any other adjacent 1 or don't care, this pair is a prime implicant $\overline{P}QS$. Because minterm 5 is covered uniquely by it, $\overline{P}QS$ is essential.
• Group 3 (row 10): $\{8, 9, 10, 11\}$ forms a quad covering minterms 9 and 11 using don't cares 8 and 10. The term is $P\overline{Q}$. Minterm 9 cannot be grouped with any other 1 or don't care, so $P\overline{Q}$ is essential.

Minterm 2 is covered by $\overline{Q}\,\overline{S}$ (and $\{2, 3, 10, 11\}$), minterm 7 is covered by $\overline{P}QS$ (and $\{3, 7\}$), and minterm 11 is covered by $P\overline{Q}$ (and $\{2, 3, 10, 11\}$).

All 6 minterms are covered by these 3 essential prime implicants: $\overline{Q}\,\overline{S}$, $\overline{P}QS$, and $P\overline{Q}$. Hence, the number of essential prime implicants is 3.