Question 1 of 6: K-map minimisation and static-hazard analysis of a 4-variable function

Question 1: K-map minimisation and static-hazard analysis of a 4-variable function [3+3+3+3 = 12]

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $F=0$ at the maxterms $M(0,2,6,7,8,10,14,15)$, so $F=1$ at the complementary minterm set $\Sigma(1,3,4,5,9,11,12,13)$, over variables $A,B,C,D$ (A most significant).

Find. The minimal PoS and minimal SoP expressions for $F$, and for each, whether it is free of static hazards under a single-input-change transition; if not, the smallest hazard-free version.

Approach. Plot $F$ on one 4-variable K-map, group the 1-cells for the minimal SoP and the 0-cells for the minimal PoS, then check every pair of physically-adjacent same-value cells (Hamming distance 1) for a shared group — an adjacent pair with no common group is a static hazard, fixed by adding the consensus term that covers exactly that pair.

1. (a) Minimal PoS — group the 0-cells. The eight 0s form two quads (each wraps around the K-map's own edges, which is valid because Gray-code ends $00$ and $10$ are themselves adjacent): rows $AB{=}00,10$ (both have $B{=}0$) with columns $CD{=}00,10$ (both have $D{=}0$) give the quad $B{=}0,D{=}0$, i.e. the sum term $(B+D)$ (zero exactly when $B{=}0$ and $D{=}0$); rows $AB{=}01,11$ (both $B{=}1$) with columns $CD{=}11,10$ (both $C{=}1$) give the quad $B{=}1,C{=}1$, i.e. the sum term $(\bar B+\bar C)$. $$\boxed{F=(B+D)(\bar B+\bar C)}.$$
2. (b) Hazard check on the minimal PoS. Adjacent 0-cell pairs (Hamming distance 1) are $(0,2),(0,8),(2,6),(2,10),(6,7),(6,14),(7,15),(8,10),(10,14),(14,15)$. Term $(B+D)$ is 0 exactly on $\{0,2,8,10\}$ and term $(\bar B+\bar C)$ is 0 exactly on $\{6,7,14,15\}$; checking each pair against these two zero-groups, the pairs $(2,6)$ and $(10,14)$ share NO common term (2 and 10 lie only in the first zero-group, 6 and 14 only in the second) — not hazard-free. Both hazards sit on the transition $C{:}0\to1$ with $B{=}0\to1$ simultaneously possible; the consensus of $(B+D)$ and $(\bar B+\bar C)$ on variable $B$ is $(D+\bar C)$, whose zero-group is exactly $\{2,6,10,14\}$ — it covers both gaps. $$\boxed{F_{\text{hazard-free PoS}}=(B+D)(\bar B+\bar C)(\bar C+D)}.$$
3. (c) Minimal SoP — group the 1-cells. The eight 1s form two quads: rows $AB{=}01,11$ ($B{=}1$) with columns $CD{=}00,01$ ($C{=}0$) give the product term $B\bar C$; rows $AB{=}00,10$ ($B{=}0$) with columns $CD{=}01,11$ ($D{=}1$) give the product term $\bar BD$. $$\boxed{F=B\bar C+\bar BD}.$$
4. (d) Hazard check on the minimal SoP. Adjacent 1-cell pairs are $(1,3),(1,5),(1,9),(3,11),(4,5),(4,12),(5,13),(9,11),(9,13),(12,13)$. Term $B\bar C$ covers $\{4,5,12,13\}$ and term $\bar BD$ covers $\{1,3,9,11\}$; the pairs $(1,5)$ and $(9,13)$ have no common covering term (1 and 9 only in the second group, 5 and 13 only in the first) — not hazard-free. The consensus of $B\bar C$ and $\bar BD$ on $B$ is $\bar CD$, covering $\{1,5,9,13\}$ exactly. $$\boxed{F_{\text{hazard-free SoP}}=B\bar C+\bar BD+\bar CD}.$$