Question 3 of 6: Variable-Entered Map and Karnaugh Map Minimization

Question 3: Variable-Entered Map and Karnaugh Map Minimization (20 marks)

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) an 8-variable function $F(A,B,C,D,E,F,G,H)$ — note the exam reuses letter "F" for both the function name and one of its input literals, so this solution writes the input literal as $F_v$ to avoid confusion. (b) the 16-row truth table above for a 4-variable function $F1(A,B,C,D)$ with five required 1s and five don't-cares.

Find. (a) A compact Variable-Entered Map (VEM) representation of $F$; (b) the minimal sum-of-products for $F1$, read off a pair of stacked 3-variable K-maps.

Approach. (a) Scan every product term for a literal common to all six — $H$ appears positively in every term — factor it out, then pick two of the remaining variables ($B,C$) as the map's row/column axes and enter the leftover variables' sub-expression into each of the 4 cells (the essence of a VEM: more variables than a plain K-map can show are pushed into the cell contents instead of the axes). (b) Split the 4-variable truth table into two 3-variable K-maps by the extra variable $D$, plot $C$ (rows) vs. $AB$ (Gray-code columns) in each, then group 1s and don't-cares (taking only the don't-cares that help) into the largest possible power-of-2 rectangles.

1. Part (a) — factor the common literal $H$. Every one of the six terms contains $H$ uncomplemented, so: $$F=H\cdot\big[B\bar CE F_v+BC\bar DE+\bar BCEF_v+ABCE\bar G+B\bar CE+A\bar B\bar C\bar E\big]$$
2. Group the bracket by $(B,C)$ and simplify each cell. Sorting the six bracketed terms by their $B,C$ literals: $\bar B\bar C$ gets $A\bar E$ (from the last term); $\bar BC$ gets $EF_v$; $B\bar C$ gets $EF_v+E=E$ (the two-literal $E$ term absorbs the three-literal $EF_v$ term, $X+X\bar Y=X$ with $X=E$); $BC$ gets $\bar DE+AE\bar G=E(\bar D+A\bar G)$: $$\text{cell}(\bar B\bar C)=A\bar E,\ \ \text{cell}(\bar BC)=EF_v,\ \ \text{cell}(B\bar C)=\boxed{E},\ \ \text{cell}(BC)=E(\bar D+A\bar G)$$
3. Assemble the VEM. Using $B$ as the row variable and $C$ as the column variable, with every cell entry additionally ANDed by the common factor $H$:

4. Part (b) — fill the two stacked K-maps from the truth table. Splitting by $D$: the $D{=}0$ map holds minterms 0,2,4,6,8,10,12,14 and the $D{=}1$ map holds 1,3,5,7,9,11,13,15, each plotted at row $C$, column $AB$ (Gray order 00,01,11,10).
5. Group 1s and don't-cares into the largest legal rectangles. Column $AB{=}00$ and $AB{=}10$ are adjacent by the map's left–right wraparound (Gray-code ends), so the four cells with $B{=}0,C{=}1$ across BOTH $D$-maps (minterms 2, 10 real 1s; 3, 11 real 1s) form one 4-cell group independent of $A$ and $D$: $$\boxed{B'C}\ \text{(4 cells: }m_2,m_3,m_{10},m_{11}\text{)}$$ Separately, the $D{=}1$ map's $AB{=}11$ column (both $C$ rows) covers the real 1 at $m_{13}$ and the don't-care at $m_{15}$, giving a second 2-cell group independent of $C$: $$\boxed{ABD}\ \text{(2 cells: }m_{13},m_{15}\text{)}$$ No don't-care outside these two groups is needed, and every specified 0 (rows 4,5,7,9,12,14) lies outside both groups, so together they already cover all five required 1s with no extra terms.
6. Read out the minimal SOP. Summing the two groups: $$F1=\boxed{\bar BC+ABD}$$ A single 5-literal, 2-term expression — the smallest cover consistent with the table (confirmed: no single product term reproduces all five required 1s without also covering a specified 0).