0 · Recall Warm-up
Answer from memory first. Answers are in the Answer Vault at the bottom.
- What is an essential prime implicant? from Lesson 4.1.2
- Can a group in a K-map consist only of don't-cares? from Lesson 4.1.3
- Which rule lets two terms differing in one variable merge? from Lesson 3.1.2
1 · The Hook
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K-maps rely on your eyes, and eyes fail beyond 5 or 6 variables. In the 1950s Quine and McCluskey turned the same idea, "merge terms that differ in one bit", into a table a person (or program) can grind through. GATE names it in the syllabus as the tabular method and asks questions about its steps and its results.
2 · Predict First
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Can the minterms 0101 (5) and 0111 (7) be merged into one term? What about 0101 (5) and 0110 (6)?
Make a guess in your head. The answer opens Section 3.
3 · Core Concept
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Predict First — the answer
5 and 7 differ only in the third bit → merge to 01-1 (A'BD). 5 and 6 differ in two bits → cannot merge.
🗣️
In plain words
Write every minterm in binary. Repeatedly merge pairs that differ in exactly one bit, replacing that bit with a dash (-). Terms that can no longer be merged are the prime implicants. Then use a chart to choose the fewest prime implicants that cover all the minterms.
1. Step 1: group by number of 1s
Only terms in neighbouring groups (k ones and k + 1 ones) can differ in exactly one bit, so group them first.
2. Step 2: merge, column after column
- Compare each term with each term of the next group. If they differ in exactly one bit, write the merged term with a dash there, and tick both.
- In later columns, two terms may merge only if their dashes are in the same positions and they differ in one other bit.
- Repeat until nothing merges. Unticked terms are the prime implicants.
- Don't-cares are included in this step like minterms.
3. Step 3: prime implicant chart
Rows: the PIs. Columns: the minterms (not the don't-cares). Put an X where a PI covers a minterm.
- A column with a single X marks an essential PI.
- Remove the columns covered by EPIs; choose the fewest remaining PIs to cover the rest (Petrick's method when there is a real choice).
4. Worked example
F(A, B, C, D) = Σm(0, 1, 2, 5, 6, 7, 8, 9, 10, 14).
| Column 1 | Column 2 | Column 3 |
|---|---|---|
| 0 0000 ✓ | 0,1 000- ✓ · 0,2 00-0 ✓ · 0,8 -000 ✓ | 0,1,8,9 -00- |
| 1 0001 ✓ · 2 0010 ✓ · 8 1000 ✓ | 1,5 0-01 · 1,9 -001 ✓ · 2,6 0-10 ✓ · 2,10 -010 ✓ · 8,9 100- ✓ · 8,10 10-0 ✓ | 0,2,8,10 -0-0 |
| 5 0101 ✓ · 6 0110 ✓ · 9 1001 ✓ · 10 1010 ✓ | 5,7 01-1 · 6,7 011- · 6,14 -110 ✓ · 10,14 1-10 ✓ | 2,6,10,14 --10 |
| 7 0111 ✓ · 14 1110 ✓ |
Prime implicants (unticked): -00- = B'C', -0-0 = B'D', --10 = CD', 0-01 = A'C'D, 01-1 = A'BD, 011- = A'BC. Chart: m₉ is covered only by B'C', and m₁₄ only by CD', so both are essential. They cover 0, 1, 2, 6, 8, 9, 10, 14. Left: 5 and 7, both covered by A'BD. F = B'C' + CD' + A'BD (3 terms, 7 literals).
4 · Visual
Merging rules
0101 (5) and 0111 (7) -> differ in 1 bit -> 01-1 OK
0101 (5) and 0110 (6) -> differ in 2 bits -> no
0-01 and 0-11 -> dashes aligned, 1 other bit differs -> 0--1 OK
0-01 and 00-1 -> dashes in different places -> no
5 · Worked Examples
Example 1 Easy
Can 1-10 and 1-11 be merged? Into what?
Solution
Dashes match, last bit differs → 1-1- (AC).
Example 2 Medium
After the merging steps, a term is left unticked. What is it?
Solution
A prime implicant (it cannot be enlarged).
Example 3 GATE
In the worked example, how many prime implicants are found, and how many are essential?
Solution
6 prime implicants, 2 essential (B'C' and CD').
6 · GATE Traps
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- Merge only terms differing in exactly one bit, with dashes in the same places.
- Don't-cares join the merging but are not columns of the chart.
- Every unticked term (in any column) is a prime implicant, not just those in the last column.
- The same merged term can arise from two different pairs; list it once.
7 · Memory Hook
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"Group by count of 1s, merge one-bit differences, dashes must line up; the unticked are prime; the chart picks the cover."
8 · Connect the Dots
Phase 4 is complete: we can find the cheapest two-level form of any function. Phase 5 builds real combinational circuits: adders, comparators, multiplexers and decoders.
9 · Self-Test
GATE format. Try every question before opening the vault.
Q1MCQ1 markMerging
Which pair can be merged in the Quine–McCluskey method?
- (A) 0-01 and 0-11
- (B) 1-0- and 1--0
- (C) 0-01 and 00-1
- (D) 0101 and 0110
Q2NAT1 markMerged term
The minterms 6 (0110) and 14 (1110) are merged. How many literals does the merged term have?
Numerical answer: ________
Q3NAT2 marksPrime implicants
F(A, B, C, D) = Σm(0, 1, 2, 5, 6, 7, 8, 9, 10, 14). How many prime implicants does the tabular method find?
Numerical answer: ________
Q4MCQ2 marksResult
The minimal SOP of F(A, B, C, D) = Σm(0, 1, 2, 5, 6, 7, 8, 9, 10, 14) is:
- (A) B'D' + A'C'D + A'BC
- (B) B'C' + B'D' + CD' + A'BD
- (C) B'C' + CD' + A'C'D + A'BC
- (D) B'C' + CD' + A'BD
Q5MCQ1 markChart
In the prime implicant chart, an essential prime implicant is found by looking for:
- (A) the longest term
- (B) a column with exactly one X
- (C) a row with no Xs
- (D) a row with the most Xs
10 · Rule Card
- Group minterms (and don't-cares) by number of 1s.
- Merge pairs differing in exactly one bit (dashes aligned); tick merged terms; repeat.
- Unticked terms = prime implicants.
- Chart (minterms only): single-X columns → EPIs; then cover the rest with the fewest PIs (Petrick if needed).
11 · Concept Ledger
Unlocked in this lesson. From here on, these terms may be used.
Quine–McCluskey methodgrouping by number of onesmerging terms with dashesticking merged termsprime implicants from the tableprime implicant chartessential PIs from single-X columnsPetrick's methoddon't-cares in the tabular method
12 · Answer Vault
Answer Vault
Try the questions first. Open when you are ready to check.
Recall Warm-up
- A PI that is the only cover of some 1.
- No.
- XY + XY' = X.
Self-Test
Q1 Answer: (A)1 mark
Dashes aligned, one other bit differs.
- (B) is wrong: Dashes in different positions.
- (C) is wrong: Dashes in different positions.
- (D) is wrong: Two bits differ.
Re-read: Lesson 4.2.1 — The Quine–McCluskey tabular method
Q2 Answer: 31 mark
-110 = BCD'.
Re-read: Lesson 4.2.1 — The Quine–McCluskey tabular method
Q3 Answer: 62 marks
B'C', B'D', CD', A'C'D, A'BD, A'BC.
Re-read: Lesson 4.2.1 — The Quine–McCluskey tabular method
Q4 Answer: (D)2 marks
EPIs B'C' and CD', then A'BD covers 5 and 7.
- (A) is wrong: Misses 8, 9 and 14.
- (B) is wrong: B'D' is redundant.
- (C) is wrong: Four terms where three suffice.
Re-read: Lesson 4.2.1 — The Quine–McCluskey tabular method
Q5 Answer: (B)1 mark
That minterm has only one possible cover.
- (A) is wrong: No.
- (C) is wrong: Such a row wouldn't be a PI of the function.
- (D) is wrong: Size doesn't decide essentiality.
Re-read: Lesson 4.2.1 — The Quine–McCluskey tabular method