سوال ۷۸ کنکور ارشد مهندسی کامپیوتر ۱۴۰۵

Reading the specification

$f=\prod M(\dots)$ lists the maxterms (positions where $f=0$). With $d(0,1)$ marking don't-cares, and $a$ the MSB, $d$ the LSB:

• $f=0$ at $\{2,3,4,10,11,12,13,14\}$
• $d = X$ at $\{0,1\}$
• $f=1$ at the rest: $\{5,6,7,8,9,15\}$

Karnaugh map

Rows $= ab$, columns $= cd$ (Gray-code order):

	cd=00	01	11	10
ab=00	X	X	0	0
01	0	1	1	1
11	0	0	1	0
10	1	1	0	0

Minimum cover

Each 1-cell forces a product term:

• $m_{15}$ pairs only with $m_7 \Rightarrow bcd$
• $m_6$ pairs only with $m_7 \Rightarrow a'bc$
• $m_5$ pairs with $m_7 \Rightarrow a'bd$
• $m_8,m_9$ lie in $b=0,c=0$: the maximal group $b'c'$ also swallows the don't-cares $0,1$, whereas $ab'c'$ covers only $m_8,m_9$.

Either way the on-set is a four-term cover: $a'bd + a'bc + bcd + (b'c'\ \text{or}\ ab'c')$.

The hazard test — why $ab'c'$, not $b'c'$

A two-level SOP is static-1-hazard-free iff every pair of adjacent 1-cells shares a common product term.

With $b'c'$ (option 2) the don't-cares $m_0,m_1$ become 1. Now $m_1(0001)$ and $m_5(0101)$ are adjacent 1-cells, but $m_1$ is held only by $b'c'$ and $m_5$ only by $a'bd$ — no shared term, so the $b$-transition glitches: a static-1 hazard. Killing it needs the consensus term $a'c'd$, which is exactly the five-term option 4.

With $ab'c'$ (option 3) the don't-cares stay 0, so no new adjacency is introduced. The four on-set adjacencies are each already spanned by one term:

$5\!-\!7:\ a'bd,\qquad 6\!-\!7:\ a'bc,\qquad 7\!-\!15:\ bcd,\qquad 8\!-\!9:\ ab'c'$

Hence option 3 is hazard-free with only four terms — the simplest hazard-free SOP.

$f = a'bd + a'bc + bcd + ab'c'$

(Option 1 is invalid outright: $b'c$ covers $m_2,m_3,m_{10},m_{11}$, all of which must be 0.)

Exam trick: Enlarging a group by absorbing don't-cares can create a hazard. $b'c'$ is the cheaper term but turns $0,1$ into 1 and opens the $m_1$–$m_5$ transition; leaving those cells at 0 with $ab'c'$ keeps the cover hazard-free without paying for a consensus term.