Quine–McCluskey — worked example

Minimise F(A,B,C) = Σm(0, 1, 2, 5, 6, 7)

Six minterms over three variables. We’ll run all three QM stages by hand and arrive at the same answer the live reducer produces. The classroom-canonical result is $F = A'B' + AC + BC'$ — three two-cell implicants — which the chart will pare down to the minimum cover.

Stage 1 — list and group by ones-count

Minterm   Binary   Ones
m0        000      0
m1        001      1
m2        010      1
m5        101      2
m6        110      2
m7        111      3

Four groups by ones-count: {m0} (0 ones), {m1, m2} (1 one), {m5, m6} (2 ones), {m7} (3 ones). Combine pairs from adjacent groups that differ in exactly one bit.

Column 2 — first combinations

1. m0 (000) + m1 (001) differ in bit C → 00– (covers {0,1}).
2. m0 (000) + m2 (010) differ in bit B → 0–0 (covers {0,2}).
3. m1 (001) + m5 (101) differ in bit A → –01 (covers {1,5}).
4. m2 (010) + m6 (110) differ in bit A → –10 (covers {2,6}).
5. m5 (101) + m7 (111) differ in bit B → 1–1 (covers {5,7}).
6. m6 (110) + m7 (111) differ in bit C → 11– (covers {6,7}).

Every original minterm combined into something larger, so all of column 1 is ticked. Column 2 has six 2-cell implicants.

Column 3 — try to combine again

Two column-2 entries combine when they have the same dash position and differ in exactly one of the remaining bits.

00– (0,1)   vs  –01 (1,5)   different dash pos → no
00– (0,1)   vs  11– (6,7)   different bits, dash pos same but other bits differ in 2 → no
0–0 (0,2)   vs  –10 (2,6)   different dash pos → no
–01 (1,5)   vs  1–1 (5,7)   different dash pos → no
... (every other pair fails the same way)

Stage 2 — PI chart

PI         m0  m1  m2  m5  m6  m7
00– (A'B')  ×   ×
0–0 (A'C')  ×       ×
–01 (B'C)       ×       ×
–10 (BC')           ×       ×
1–1 (AC)                ×       ×
11– (AB)                    ×   ×

1. m0 is covered only by 00– and 0–0 — two PIs, so m0 alone doesn’t identify an essential.
2. m1 is covered only by 00– and –01 — same story.
3. No column has exactly one × in this chart. It’s cyclic — every minterm has two covers.
4. Petrick or by-eye covering: any cover of size 3 works. A clean choice is $00\text{-} + \text{-}10 + 1\text{-}1$ = $A'B' + BC' + AC$, three 2-literal terms covering all six minterms.

Combine

F = A'B' + BC' + AC     (3 terms, 6 literals)

From six 3-literal minterms (18 literals) down to a 6-literal SOP. Compare with the live answer on the Simulate page — the engine’s greedy cover may pick a different but equivalent 3-term combination.