Sum of products (SOP) form: Worked examples — Hardware deep dive

Before we start, one reminder so no symbol is unearned:

The scenario matrix

Every SOP question falls into one of these cells. The examples below are labelled with the cell(s) they cover.

#	Case class	What makes it tricky	Example
C1	Ordinary truth table, several 1-rows	routine but must not slip a bar	Ex 1
C2	Degenerate: all outputs 0	no minterms — what do you even write?	Ex 2
C3	Degenerate: all outputs 1	every row wins — does it collapse?	Ex 3
C4	Exactly one 1-row	a single minterm, no OR needed	Ex 4
C5	Expression in disguise (product-of-sums shape)	must distribute to reach SOP	Ex 5
C6	Non-canonical → canonical expansion	terms missing variables	Ex 6
C7	Real-world word problem	translate English → truth table → SOP	Ex 7
C8	Exam twist: given minterm list $\sum m (\cdot)$ , simplify	index notation + K-map thinking	Ex 8

We label the truth-table rows in binary counting order ( $000, 001, 010, \dots$ ). See the figure — the row number $i$ is just the binary value of $A B C$ .

Ex 1 — Ordinary truth table (cell C1)

Worked example Three winning rows

Build the canonical SOP for this function of $A, B, C$ .

$i$	A	B	C	F
0	0	0	0	0
1	0	0	1	1
2	0	1	0	0
3	0	1	1	1
4	1	0	0	0
5	1	0	1	0
6	1	1	0	1
7	1	1	1	0

Forecast: how many product terms will the answer have? (Count the 1-rows first.)

Step 1 — list the 1-rows. Rows $1, 3, 6$ . Why this step? SOP only lists the winners; ignore every 0-row.

Step 2 — turn each 1-row into a minterm. Rule: variable is 0 → write it barred; variable is 1 → write it plain (this makes the minterm equal 1 on exactly that row).

Row 1 ( $A = 0, B = 0, C = 1$ ) → $\overset{ˉ}{A} \overset{ˉ}{B} C$
Row 3 ( $A = 0, B = 1, C = 1$ ) → $\overset{ˉ}{A} B C$
Row 6 ( $A = 1, B = 1, C = 0$ ) → $A B \overset{ˉ}{C}$

Why this step? Each minterm is a "row-detector": it fires (=1) on its own row and nowhere else.

Step 3 — OR the minterms. $F = \overset{ˉ}{A} \overset{ˉ}{B} C + \overset{ˉ}{A} B C + A B \overset{ˉ}{C} = \sum m (1, 3, 6)$ Why this step? OR outputs 1 whenever any detector fires, exactly recreating the three 1-rows.

Verify: plug in row 3 ( $A = 0, B = 1, C = 1$ ): $\overset{ˉ}{A} \overset{ˉ}{B} C = 0$ , $\overset{ˉ}{A} B C = 1 \cdot 1 \cdot 1 = 1$ , $A B \overset{ˉ}{C} = 0$ . OR $= 1$ ✓. Plug in a 0-row, say row 7 ( $1, 1, 1$ ): $0 + 0 + 0 = 0$ ✓.

Ex 2 — All outputs 0 (cell C2, degenerate)

Ex 3 — All outputs 1 (cell C3, degenerate)

Ex 4 — Exactly one winning row (cell C4)

Ex 5 — Expression in disguise (cell C5)

Worked example Distribute to reach SOP

Put $F = (A + B) (\overset{ˉ}{A} + C)$ into SOP form.

Forecast: this is an AND-of-ORs (product-of-sums shape). How many product terms after we multiply out?

Step 1 — multiply out (distribute AND over OR), like FOIL but with OR/AND: $F = A \overset{ˉ}{A} + A C + B \overset{ˉ}{A} + B C$ Why this step? SOP is a sum of products; the outer operation must become OR. Distribution converts the outer AND into an OR of AND-terms — exactly the SOP shape. We use distribution (not, say, De Morgan) because we only need to change the grouping, not complement anything.

Step 2 — kill impossible terms with $A \overset{ˉ}{A} = 0$ . $F = 0 + A C + \overset{ˉ}{A} B + B C = A C + \overset{ˉ}{A} B + B C$ Why this step? $A \overset{ˉ}{A}$ can never be 1 (a thing and its opposite can't both be true), so it drops out — no row is lost.

Step 3 — remove the redundant term $B C$ (consensus theorem). With $A C$ and $\overset{ˉ}{A} B$ present, $B C$ is already covered: $F = A C + \overset{ˉ}{A} B$ Why this step? The consensus of $A C$ and $\overset{ˉ}{A} B$ (variable $A$ appearing true and complemented) is $B C$ , which is always redundant. Fewer gates, same output.

Verify: test $A = 0, B = 1, C = 0$ . Original: $(0 + 1) (1 + 0) = 1 \cdot 1 = 1$ . Our form: $A C + \overset{ˉ}{A} B = 0 + 1 \cdot 1 = 1$ ✓. Test $A = 1, B = 0, C = 0$ : original $(1) (0) = 0$ ; ours $0 + 0 = 0$ ✓.

Ex 6 — Non-canonical → canonical (cell C6)

Worked example Expand missing variables

$F (A, B, C) = A \overset{ˉ}{B} + C$ is valid SOP but not canonical (each term is missing a variable). Expand to minterms.

Forecast: term $A \overset{ˉ}{B}$ is missing $C$ ; term $C$ is missing $A$ and $B$ . How many minterms total?

Step 1 — expand $A \overset{ˉ}{B}$ by multiplying with $(C + \overset{ˉ}{C}) = 1$ . $A \overset{ˉ}{B} = A \overset{ˉ}{B} (C + \overset{ˉ}{C}) = A \overset{ˉ}{B} C + A \overset{ˉ}{B} \overset{ˉ}{C} = m (5) + m (4)$ Why this step? Multiplying by 1 changes no values, but $(C + \overset{ˉ}{C})$ splits the term across both values of the missing $C$ , giving full minterms.

Step 2 — expand $C$ by multiplying with $(A + \overset{ˉ}{A}) (B + \overset{ˉ}{B}) = 1$ . $C = (A + \overset{ˉ}{A}) (B + \overset{ˉ}{B}) C = A B C + A \overset{ˉ}{B} C + \overset{ˉ}{A} B C + \overset{ˉ}{A} \overset{ˉ}{B} C$ $= m (7) + m (5) + m (3) + m (1)$ Why this step? Two variables missing → multiply by two " $1$ "s to fan out over all four combinations.

Step 3 — OR everything and drop duplicates. Both terms produced $m (5)$ ; in Boolean $X + X = X$ , so keep one copy: $F = m (1) + m (3) + m (4) + m (5) + m (7) = \sum m (1, 3, 4, 5, 7)$ Why this step? Canonical SOP lists each winning row exactly once.

Verify: count — original $A \overset{ˉ}{B}$ covers rows 4,5; $C$ covers rows 1,3,5,7. Union $= {1, 3, 4, 5, 7}$ , five rows ✓. Spot-check row 4 ( $1, 0, 0$ ): $A \overset{ˉ}{B} = 1$ , $C = 0$ , so $F = 1$ ; and $m (4)$ is in our list ✓.

Ex 7 — Real-world word problem (cell C7)

Worked example The greenhouse fan

A fan should switch ON when it is Hot ( $H$ ) and the greenhouse is Occupied ( $O$ ), OR whenever the Manual override ( $M$ ) is pressed — regardless of anything else. Variables are 1 = true. Build the truth table and give the SOP.

Forecast: override alone forces ON — will many rows be 1?

Step 1 — translate English into a table over $H, O, M$ .

$i$	H	O	M	Fan
0	0	0	0	0
1	0	0	1	1
2	0	1	0	0
3	0	1	1	1
4	1	0	0	0
5	1	0	1	1
6	1	1	0	1
7	1	1	1	1

Why this step? "Hot AND Occupied" makes row 6 (and 7) fire; "Manual" alone makes every $M = 1$ row fire (rows 1,3,5,7).

Step 2 — canonical SOP from the 1-rows ${1, 3, 5, 6, 7}$ : $Fan = \overset{ˉ}{H} \overset{ˉ}{O} M + \overset{ˉ}{H} O M + H \overset{ˉ}{O} M + H O \overset{ˉ}{M} + H O M$

Step 3 — simplify back to the sentence. The four terms with $M = 1$ ( $m_{1}, m_{3}, m_{5}, m_{7}$ ) collapse: they cover all combinations of $H, O$ with $M = 1$ , so they sum to $M$ . The remaining $m_{6} = H O \overset{ˉ}{M}$ combines with $H O M$ (already inside $M$ ) to leave $H O$ : $Fan = M + H O$ Why this step? This is exactly the English: "ON if Manual, OR if Hot and Occupied." The algebra recovered the spec.

Verify: row 6 ( $H = 1, O = 1, M = 0$ ): $M + H O = 0 + 1 = 1$ ✓ (fan on, correct). Row 4 ( $1, 0, 0$ ): $0 + 0 = 0$ ✓ (hot but empty and no override → off). Row 1 ( $0, 0, 1$ ): $1 + 0 = 1$ ✓ (override wins).

Ex 8 — Exam twist: given minterm list, simplify (cell C8)

Worked example From

$\sum m$ to fewest gates A function is given directly as $F (A, B) = \sum m (0, 1, 2)$ . Write it out and simplify to minimum SOP.

Forecast: three of the four rows are 1 — the odd one out is row 3. Can you spot the shortcut?

Step 1 — decode the indices into minterms (2 variables, so $i$ = binary $A B$ ):

$m (0) = \overset{ˉ}{A} \overset{ˉ}{B}$ (row 00)
$m (1) = \overset{ˉ}{A} B$ (row 01)
$m (2) = A \overset{ˉ}{B}$ (row 10) $F = \overset{ˉ}{A} \overset{ˉ}{B} + \overset{ˉ}{A} B + A \overset{ˉ}{B}$ Why this step? The $\sum m$ notation is shorthand for "OR these minterms"; we make it explicit.

Step 2 — group cleverly. Pair for $\overset{ˉ}{A}$ , then pair for $\overset{ˉ}{B}$ , reusing $\overset{ˉ}{A} \overset{ˉ}{B}$ (allowed: $X + X = X$ lets you duplicate): $F = \overset{ˉ}{A} (\overset{ˉ}{B} + B) = \overset{ˉ}{A} (\overset{ˉ}{A} \overset{ˉ}{B} + \overset{ˉ}{A} B) + (\overset{ˉ}{A} + A) \overset{ˉ}{B} = \overset{ˉ}{B} (\overset{ˉ}{A} \overset{ˉ}{B} + A \overset{ˉ}{B})$ $F = \overset{ˉ}{A} + \overset{ˉ}{B}$ Why this step? Each parenthesis uses $X + \overset{ˉ}{X} = 1$ to erase one variable. Reusing $m (0)$ is legal because ORing the same term twice adds nothing.

Step 3 — sanity via the missing row. The only 0-row is row 3 ( $A = 1, B = 1$ ). And $\overset{ˉ}{A} + \overset{ˉ}{B} = 0$ exactly when both $A = 1$ and $B = 1$ — matches ✓. (This is De Morgan: $\overset{ˉ}{A} + \overset{ˉ}{B} = \overline{A B}$ , i.e. "not both".)

Verify: row 0 ( $0, 0$ ): $\overset{ˉ}{A} + \overset{ˉ}{B} = 1 + 1 = 1$ ✓. Row 3 ( $1, 1$ ): $0 + 0 = 0$ ✓. Two literals, one OR gate, two NOTs — far cheaper than three ANDs + OR.

Active Recall

Recall Cover every cell — can you name the SOP for each degenerate case?

All-0 function ::: $F = 0$ (empty OR, no minterms). All-1 function ::: canonical is all minterms; simplifies to $F = 1$ . Exactly one 1-row ::: a single full minterm, no OR, cannot be simplified. An AND-of-ORs expression ::: distribute (multiply out) to reach a sum of products.

Recall Why does multiplying a term by

$(X + \overset{ˉ}{X})$ expand it to canonical without changing it? Because $X + \overset{ˉ}{X} = 1$ , so we multiply by 1 (value unchanged) but the term splits into both values of the missing variable $X$ , producing full minterms.

Connections

Truth Tables — every example started here; the 1-rows are the SOP.
Minterms and Maxterms — the row-detectors we OR together.
Boolean Algebra Laws — distribution, $X + \overset{ˉ}{X} = 1$ , $X + X = X$ , consensus, absorption did all the simplifying.
Karnaugh Maps — the visual way to do Ex 8's grouping.
Product of Sums (POS) form — the dual: list the 0-rows instead.
Logic Gates (AND, OR, NOT) — each simplified form is a cheaper circuit.