Question 3 of 6: Variable-Entered Maps

Question 3: Variable-Entered Maps (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 $F(A..H)$ where $B,C,E,H$ appear as a literal in EVERY product term; (b) a 4-variable truth table for $F_1(A,B,C,D)$ with three don't-care rows.

Find. (a) $F$ entered into a variable-entered map; (b) $F_1$ entered into a variable-entered map, and its simplified read-out.

Approach. A variable-entered map (VEM) plots only a subset of the variables on its axes and "enters" the remaining literals directly into each cell as an algebraic expression (0, 1, a literal, or a product of the leftover variables). Pick axis variables that appear (as a literal) in every product term, so every term maps to exactly one cell; when two terms land on the same cell, the cell's entry is the OR of what each contributes — and a plain (unnegated, no-leftover-variable) term landing on a cell that also receives that same variable's complement immediately collapses the cell to the constant 1.

1. Part (a) — choose $B,C,E,H$ as the map's 4 axis variables (each appears, true or complemented, in all six terms), leaving $A,D,F,G$ as the entered (leftover) variables. Reading each term's axis literals fixes its cell; its entered literals (if any) are what gets written inside that cell:

   Term-by-term breakdown

   Term	Axis cell $(B,C,E,H)$	Entered literal(s)
   $\overline{BC}EFH$	$(0,0,1,1)$	$F$
   $BC\overline{D}EH$	$(1,1,1,1)$	$\overline D$
   $\overline{BC}EH\overline{F}$	$(0,0,1,1)$	$\overline F$
   $ABCEH\overline{G}$	$(1,1,1,1)$	$A\overline G$
   $BCEH$	$(1,1,1,1)$	(none — constant 1)
   $A\overline{BC}\ \overline{EH}$	$(0,0,0,0)$	$A$

   Two pairs of terms collide on the same axis cell.
2. Merge the colliding cells. Cell $(0,0,1,1)$ receives $F$ (term 1) OR $\overline F$ (term 3): $F+\overline F=\boxed{1}$, a constant. Cell $(1,1,1,1)$ receives $\overline D$ (term 2) OR $A\overline G$ (term 4) OR the constant $1$ (term 5, "$BCEH$" with no leftover variable) — and OR-ing anything with a bare $1$ collapses the whole cell to $\boxed{1}$ regardless of $D,A,G$. Cell $(0,0,0,0)$ has only term 6, so it is entered as $\boxed{A}$ directly. Every other one of the 16 axis cells receives no term at all and is entered $0$.
3. Read the map back out (bonus check). Only three cells are non-zero, so the SOP read directly off the completed map is $$F=\boxed{\overline{BC}EH+BCEH+A\overline{BC}\ \overline{EH}}$$ which was verified to reproduce the original 6-term expression across all $2^8=256$ input combinations — the map-entered-variable technique has folded six terms over eight variables into three prime implicants without ever building a 256-row truth table by hand.
4. Part (b) — axis $A,B$ (top) and $C$ (side), enter $D$. For each $(A,B,C)$ combination the table gives a $(D{=}0,D{=}1)$ pair; the cell entry follows the map-entered-variable rule (both cells 0 $\Rightarrow$ 0; both 1 $\Rightarrow$ 1; one is a don't-care $\Rightarrow$ take the other value, since $\Phi$ imposes no constraint; genuine $0/1$ conflict $\Rightarrow$ enter $D$ or $\overline D$):

   F1 entered into the VEM (rows $C=0,1$; columns $AB=00,01,11,10$)

   C∖AB	00	01	11	10
   0	Φ→0	0	$\overline D$→$D$	0
   1	1	0	0	1

   (cell $AB{=}11,C{=}0$: rows $(A,B,C,D){=}(1,1,0,0){\to}0$ and $(1,1,0,1){\to}1$ genuinely conflict, so the entry is $D$.)
5. Read out $F_1$. The two $1$-cells at $C{=}1,AB{=}00$ and $C{=}1,AB{=}10$ both have $B{=}0$ and differ only in $A$ — they combine into the 2-literal term $\overline BC$. The lone $D$-cell at $A{=}1,B{=}1,C{=}0$ contributes the 4-literal term $ABC\overline{\,}\,D$ (i.e. $AB\overline CD$). The single don't-care cell ($A{=}B{=}C{=}D{=}0$ family) is set to $0$ since it does not extend either group: $$F_1=\boxed{\overline BC+ABC'D}$$ verified against every specified (non-don't-care) row of the given truth table.