سوال ۸۲ کنکور ارشد مهندسی کامپیوتر ۱۴۰۰

Build the 4-variable K-map (rows $ab$, columns $cd$). Ones are $\{0,1,4,5,10,15\}$; don't-cares ($X$) are $\{7,14\}$.

A prime implicant (PI) is a maximal group of $1$/$X$ cells; it is essential (EPI) iff it is the only PI that covers some required minterm (a $1$, never a don't-care).

Minterm 0: its only $1$-neighbours are $m_1$ and $m_4$, both inside the quad $\{0,1,4,5\}$. That quad is the PI $a'c'$ (all cells have $a=0,\ c=0$) and it cannot grow further. Since $m_0$ sits in no other PI, $a'c'$ is essential.

Minterm 10: its only $1$/$X$ neighbour is the don't-care $m_{14}$, giving the pair $\{10,14\}=acd'$ (here $a=1,\ c=1,\ d=0$). No larger group is possible ($m_{11}=0$), so $m_{10}$ forces $acd'$ — also essential.

Minterm 15: it can be covered by either $abc=\{14,15\}$ or $bcd=\{7,15\}$. Two competing PIs cover it, so neither is essential; we may pick either to finish the cover.

The remaining required minterms $1,4,5$ are already inside the essential $a'c'$.

Thus the essential prime implicants are exactly $a'c' \quad\text{and}\quad acd',$ so the count is $2$.

Exam trick: A cell with a single $1$/$X$ neighbour forces a unique PI ⇒ that PI is essential. Don't-cares may help form PIs but never force essentiality, since only required $1$s decide EPIs.