Part 4: Minterms and Maxterms

Source: Part-4+Minterms+and+Maxterms.pdf

1. Minterms (Standard Products)
💡 Simple Explanation: A minterm is one specific row in a truth table where the output is 1. For 2 variables A and B, the minterm for A=1,B=0 is $A\overline{B}$ — it's only true for that exact combination. We label minterms as $m_0, m_1, m_2, ...$ based on the binary value of the inputs.

Formal Definition: A minterm is a product (AND) term containing all $n$ variables of the function, where each variable appears exactly once — either in normal or complemented form. For $n$ variables, there are $2^n$ minterms.

Each minterm corresponds to exactly one row of the truth table — the row where that minterm equals 1.

For $n$ variables, there are $2^n$ possible minterms.

Naming convention: $m_i$ where $i$ is the decimal equivalent of the binary input combination. If the variable is 1, it appears normal. If it's 0, it appears complemented.

Example for 3 variables (A, B, C): $m_5 = 5_{10} = 101_2$ → A=1, B=0, C=1 → $A\overline{B}C$

Key Points

  • Minterm = product (AND) of ALL variables — each appears once
  • Variable is normal (uncomplemented) when the corresponding bit is 1
  • Variable is complemented when the corresponding bit is 0
  • $m_i$: $i$ = decimal value of the input combination
  • For $n$ variables: $2^n$ minterms total
  • Each minterm is 1 for exactly ONE input combination and 0 for all others
  • The canonical SOP = sum of minterms where F=1

Formulas

$$F = \sum m(i, j, k, ...)$$
Canonical SOP notation: F equals the sum (OR) of minterms at positions i, j, k, ... (where F=1)

Worked Examples

Example 1

Problem: For 3 variables (A, B, C), what is minterm $m_3$?

  1. $3_{10} = 011_2$
  2. So A=0, B=1, C=1
  3. A=0 → complemented: $\overline{A}$
  4. B=1 → normal: $B$
  5. C=1 → normal: $C$
  6. Minterm $m_3 = \overline{A}BC$

Answer: $m_3 = \overline{A}BC$

Example 2

Problem: Express $F(A,B,C) = \sum m(1, 4, 5, 7)$ as a Boolean expression.

  1. $m_1 = 001 → \overline{A}\overline{B}C$
  2. $m_4 = 100 → A\overline{B}\overline{C}$
  3. $m_5 = 101 → A\overline{B}C$
  4. $m_7 = 111 → ABC$
  5. $F = \overline{A}\overline{B}C + A\overline{B}\overline{C} + A\overline{B}C + ABC$

Answer: $F = \overline{A}\overline{B}C + A\overline{B}\overline{C} + A\overline{B}C + ABC$

⚠️ Common Traps

Minterm $m_0$ is NOT undefined — it's $\overline{A}\overline{B}\overline{C}$ (all variables complemented).
The number $i$ in $m_i$ corresponds to the binary input where each variable is 1 (normal) or 0 (complemented) — don't reverse the convention!
Make sure to include ALL variables in each minterm. If you have 3 variables but wrote $AB$ (missing C), that's NOT a minterm.

🧪 Quick Self-Test

For variables A, B, C: what Boolean expression is minterm $m_6$?
$6 = 110_2$ → A=1, B=1, C=0 → $AB\overline{C}$
How many minterms exist for a 4-variable Boolean function?
For $n$ variables: $2^n$ minterms. $2^4 = 16$ minterms.
A minterm can be true (equal to 1) for more than one input combination.
Each minterm equals 1 for exactly ONE input combination and 0 for all others. That's the defining property of a minterm.
2. Maxterms (Standard Sums)
💡 Simple Explanation: A maxterm is the opposite of a minterm. It's a sum (OR) term that's 0 for exactly one row of the truth table. For A=1,B=0 the maxterm is $\overline{A}+B$ — notice the convention is REVERSED compared to minterms!

Formal Definition: A maxterm is a sum (OR) term containing all $n$ variables, where each variable appears exactly once. A maxterm is 0 for exactly one input combination and 1 for all others.

Maxterms are the dual of minterms:

Where minterms use AND and are 1 for one combination, maxterms use OR and are 0 for one combination.

CRITICAL: The complement convention is REVERSED!
In a maxterm $M_i$: if the bit is 1, the variable is complemented. If the bit is 0, the variable is normal.

This makes sense because: an OR term is 0 only when ALL its literals are 0. So if A=1, we need $\overline{A}$ (which is 0 when A=1).

Key Points

  • Maxterm = sum (OR) of ALL variables — each appears once
  • ⚠️ OPPOSITE convention to minterms: variable is COMPLEMENTED when bit is 1, NORMAL when bit is 0
  • $M_i$: $i$ = decimal value of the input combination where it equals 0
  • Each maxterm is 0 for exactly ONE input combination
  • Maxterm $M_i$ is the complement of minterm $m_i$: $M_i = \overline{m_i}$
  • For $n$ variables: $2^n$ maxterms total
  • The canonical POS = product of maxterms where F=0

Formulas

$$F = \prod M(i, j, k, ...)$$
Canonical POS notation: F equals the product (AND) of maxterms at positions i, j, k, ... (where F=0)
$$M_i = \overline{m_i}$$
A maxterm is the complement of the corresponding minterm

Worked Examples

Example 1

Problem: For 3 variables (A, B, C), what is maxterm $M_3$?

  1. $3_{10} = 011_2$
  2. A=0 → normal (uncomplemented): $A$
  3. B=1 → complemented: $\overline{B}$
  4. C=1 → complemented: $\overline{C}$
  5. Maxterm $M_3 = A + \overline{B} + \overline{C}$

Answer: $M_3 = A + \overline{B} + \overline{C}$

Example 2

Problem: Express $F(A,B,C) = \prod M(0, 2, 3, 6)$ as a Boolean expression.

  1. $M_0 = 000 → A+B+C$
  2. $M_2 = 010 → A+\overline{B}+C$
  3. $M_3 = 011 → A+\overline{B}+\overline{C}$
  4. $M_6 = 110 → \overline{A}+\overline{B}+C$
  5. $F = (A+B+C)(A+\overline{B}+C)(A+\overline{B}+\overline{C})(\overline{A}+\overline{B}+C)$

Answer: $F = (A+B+C)(A+\overline{B}+C)(A+\overline{B}+\overline{C})(\overline{A}+\overline{B}+C)$

⚠️ Common Traps

⚠️ #1 EXAM TRAP: Maxterm complement convention is OPPOSITE to minterm! In $m_i$: 1→normal, 0→complemented. In $M_i$: 1→COMPLEMENTED, 0→NORMAL.
If $F = \sum m(1,4,5,7)$ then $F = \prod M(0,2,3,6)$ — the maxterm list is ALL indices NOT in the minterm list.
Don't confuse $M$ (capital - maxterm) with $m$ (lowercase - minterm).

🧪 Quick Self-Test

For variables A, B, C: what is maxterm $M_5$?
$5 = 101_2$ → A=1→$\overline{A}$, B=0→$B$, C=1→$\overline{C}$ → $M_5 = \overline{A}+B+\overline{C}$
In maxterms, a variable is complemented when its corresponding bit is 0.
OPPOSITE! In maxterms: bit=1 → complemented, bit=0 → normal. This is reversed from minterms.
3. Converting Between SOP (Σm) and POS (ΠM) Forms
💡 Simple Explanation: SOP and POS are two sides of the same coin. If you know which rows give F=1 (minterms), the remaining rows give F=0 (maxterms). For 3 variables with 8 total rows: if F=1 for {1,4,5,7}, then F=0 for {0,2,3,6}.
The key insight: minterms where F=1 and maxterms where F=0 are complementary sets.

For $n$ variables, the total set of indices is $\{0, 1, 2, ..., 2^n - 1\}$.

If $F = \sum m(\text{set A})$, then $F = \prod M(\text{set B})$ where B = complement of A (all indices NOT in A).

Similarly, to find the complement of F ($\overline{F}$): $\overline{F} = \sum m(\text{set B})$ where B contains the indices NOT in the original minterm list.

Key Points

  • SOP to POS: take the complement of the index set. $\sum m(1,4,5,7) = \prod M(0,2,3,6)$
  • The union of minterm indices and maxterm indices = all possible indices
  • For $n$ variables, total indices = $\{0, 1, ..., 2^n-1\}$
  • To find $\overline{F}$: use the indices NOT in F's minterm list
  • $\overline{F}$ in SOP uses the indices from F's maxterm list, and vice versa

Formulas

$$F = \sum m(i,j,...) \iff F = \prod M(\text{all indices not in } \{i,j,...\})$$
Converting canonical SOP ↔ POS: complement the index set
$$\overline{F} = \sum m(\text{maxterm indices of } F) = \prod M(\text{minterm indices of } F)$$
The complement of F uses the opposite index sets

Worked Examples

Example 1

Problem: Given $F(A,B,C) = \sum m(0, 2, 5, 7)$, express F in POS form.

  1. 3 variables → 8 indices total: {0,1,2,3,4,5,6,7}
  2. Minterm indices: {0, 2, 5, 7}
  3. Maxterm indices = complement: {1, 3, 4, 6}
  4. $F = \prod M(1, 3, 4, 6)$

Answer: $F(A,B,C) = \prod M(1, 3, 4, 6)$

Example 2

Problem: Given $F(A,B) = \sum m(1,2)$, find $\overline{F}$ in SOP form.

  1. 2 variables → 4 indices total: {0,1,2,3}
  2. F's minterms: {1, 2}
  3. $\overline{F}$'s minterms = remaining: {0, 3}
  4. $\overline{F} = \sum m(0, 3) = \overline{A}\overline{B} + AB$

Answer: $\overline{F} = \sum m(0, 3) = \overline{A}\overline{B} + AB$

⚠️ Common Traps

When converting SOP ↔ POS, you swap the INDEX SET, not the form of each term! Don't try to complement individual minterms.
Make sure you account for ALL indices. For 3 variables = 8 indices (0-7), for 4 variables = 16 indices (0-15).
The complement of F is NOT just changing Σ to Π with the same indices! You need the COMPLEMENTARY index set.

🧪 Quick Self-Test

If $F(A,B,C) = \sum m(0,1,3,7)$, what are the maxterm indices for the POS form?
All 3-variable indices: {0,1,2,3,4,5,6,7}. Remove {0,1,3,7} → remaining = {2,4,5,6}.
For a 3-variable function, how many total minterms or maxterms exist?
$2^n = 2^3 = 8$ total minterms (or maxterms).
4. Don't Care Conditions
💡 Simple Explanation: Sometimes certain input combinations can never happen (like entering month 13 on a calendar). Since those cases will never occur, we 'don't care' whether the output is 0 or 1 — and we can choose whichever helps us simplify the circuit more.
In real circuits, some input combinations may be impossible or irrelevant. These are called don't care conditions, denoted by 'd' or 'X'.

When simplifying, you can treat don't cares as either 0 or 1 — whichever leads to a simpler expression. This is especially useful in Karnaugh map simplification.

Notation: $F(A,B,C) = \sum m(1,5,7) + d(2,6)$ — means F=1 at {1,5,7}, F=X (don't care) at {2,6}, F=0 at all other indices.

Key Points

  • Don't care = input combination that can never happen OR we don't care about its output
  • Denoted by 'd', 'X', or 'φ' in truth tables and expressions
  • Can be treated as 0 or 1 during simplification — choose what gives simpler circuit
  • Common example: BCD input — combinations 1010-1111 never occur
  • Don't cares are listed separately: $F = \sum m(...) + d(...)$
  • In K-maps, don't cares can be included in groups to make larger groupings

Worked Examples

Example 1

Problem: A BCD-to-7-segment decoder receives 4-bit BCD input. What are the don't care conditions?

  1. BCD uses only 0-9, which is $0000$ to $1001$ in 4-bit binary
  2. 4-bit binary can represent 0-15 ($0000$ to $1111$)
  3. Combinations 10-15 ($1010$ to $1111$) can never occur in valid BCD
  4. These 6 combinations are don't cares: $d(10, 11, 12, 13, 14, 15)$

Answer: Don't care conditions: $d(10, 11, 12, 13, 14, 15)$ — these BCD inputs are invalid.

⚠️ Common Traps

Don't cares are NOT always 0 and NOT always 1 — you CHOOSE the best value during simplification.
In the final implementation, don't care outputs might actually produce 0 or 1, but since those inputs can't occur, it doesn't matter.
When listing minterms, don't mix don't cares into the minterm list — keep them separate!

🧪 Quick Self-Test

A don't care condition means:
Don't care means the output doesn't matter — it can be 0 or 1, whichever helps simplify the expression.
In a BCD system, the input combination 1100 (decimal 12) is a don't care condition.
BCD only uses 0-9. Input 1100 = 12 can never occur in valid BCD, so it's a don't care.