Question 1 of 6: PoS to truth table, canonical SoP, minimal SoP and hazard removal

Question 1 (12 marks) — PoS to truth table, canonical SoP, minimal SoP and hazard removal

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. A four-variable function as a product of three sum terms (maxterm factors). Find. its truth table, the canonical minterm list, the minimal SoP cover, and a static-hazard-free cover.

Approach. Each sum factor is zero for exactly one setting of the variables it names (a maxterm); collect those zero-rows, the remaining rows are the minterms, then group on a K-map and test adjacent 1-cells for static-1 hazards.

1. Locate the zero-rows of each factor. A sum term is 0 only when every literal is 0. $B+\overline{C}=0$ needs $B=0,\,C=1$ → minterms $\{2,3,10,11\}$. $\overline{A}+\overline{B}+C=0$ needs $A=1,\,B=1,\,C=0$ → $\{12,13\}$. $B+C+D=0$ needs $B=C=D=0$ → $\{0,8\}$. So $f=0$ on $\{0,2,3,8,10,11,12,13\}$.
2. Canonical SoP (part b). The complementary set is the minterm list: $$\boxed{f=\sum m(1,4,5,6,7,9,14,15)}$$
3. Truth table (part a). Writing $f=1$ on exactly the minterms above gives the table below.

Minimal SoP (part c). Grouping the 1-cells on the K-map: the whole $A=0,B=1$ row is a quad ($\overline{A}B$), the $B=1,C=1$ column is a quad ($BC$), and $m_1,m_9$ pair as $\overline{B}\,\overline{C}D$. Hence $$\boxed{f=\overline{A}B+BC+\overline{B}\,\overline{C}D}$$ (3 terms, 7 literals).

Hazard analysis (part d). A static-1 hazard occurs when two adjacent 1-cells are covered by different product terms with no common term. Cells $m_1=0001$ (in $\overline{B}\,\overline{C}D$) and $m_5=0101$ (in $\overline{A}B$) are adjacent — they differ only in $B$ — and lie in different loops, so the cover is not hazard-free. Adding the consensus term $\overline{A}\,\overline{C}D$, which spans both cells, bridges the transition:

$$\boxed{f_{\text{hazard-free}}=\overline{A}B+BC+\overline{B}\,\overline{C}D+\overline{A}\,\overline{C}D}$$

All other adjacent 1-cell pairs already share a loop ($m_6,m_7$ share $\overline{A}B$ and $BC$; $m_7,m_{15}$ and $m_6,m_{14}$ share $BC$; $m_1,m_9$ share $\overline{B}\,\overline{C}D$), so this single added term makes the network glitch-free for all single-input changes.