Karnaugh map exercises with answers: 3 variables

Ten functions over A, B and C as minterm lists, to be grouped on a two-row, four-column Karnaugh map. Draw the map in Gray-code order, circle the groups, write the sum of products and compare with the map and the derived expression under each problem.

The engine's layout puts A on the rows and B, C on the columns in the order 00, 01, 11, 10, so the top row reads m0, m1, m3, m2 and the bottom row m4, m5, m7, m6. Horizontal neighbours differ in one variable, and so do the two end columns, which wrap around: m0 and m2 are adjacent, as are m4 and m6. A group of two drops one variable, a group of four drops two, and the whole map drops all three.

The set includes a wraparound pair, a four-cell block across the two end columns, a single cell, a checkerboard with no pair at all (the parity function, where every cell stays a full minterm), and a function with six prime implicants of which three suffice. For that last one the map shows one of two equally short covers; both are correct, and the exact check accepts either.

1. Group the Karnaugh map of Σm(A, B, C; 0, 2, 4, 6) and write the result. Σm(A, B, C; 0, 2, 4, 6)

#	A	B	C	f
0	FALSE	FALSE	FALSE	TRUE
1	FALSE	FALSE	TRUE	FALSE
2	FALSE	TRUE	FALSE	TRUE
3	FALSE	TRUE	TRUE	FALSE
4	TRUE	FALSE	FALSE	TRUE
5	TRUE	FALSE	TRUE	FALSE
6	TRUE	TRUE	FALSE	TRUE
7	TRUE	TRUE	TRUE	FALSE

The engine found no law that shortens this expression; it is already as short as its rewriting laws can make it.

Σm(A, B, C; 0, 2, 4, 6)

Classification: contingent

A \ BC

1m0

0m1

0m3

1m2

1m4

0m5

0m7

1m6

cells m0, m2, m4, m6NOT C

NOT C · Derived NOT C. Exactly equivalent on every assignment.

The four cells sit in the two end columns, BC = 00 and BC = 10, which are adjacent by wraparound: one group of four with A and B changing, so both drop and the result is NOT C.

2. Group the Karnaugh map of Σm(A, B, C; 3, 7) and write the result. Σm(A, B, C; 3, 7)

#	A	B	C	f
0	FALSE	FALSE	FALSE	FALSE
1	FALSE	FALSE	TRUE	FALSE
2	FALSE	TRUE	FALSE	FALSE
3	FALSE	TRUE	TRUE	TRUE
4	TRUE	FALSE	FALSE	FALSE
5	TRUE	FALSE	TRUE	FALSE
6	TRUE	TRUE	FALSE	FALSE
7	TRUE	TRUE	TRUE	TRUE

The engine found no law that shortens this expression; it is already as short as its rewriting laws can make it.

Σm(A, B, C; 3, 7)

Classification: contingent

A \ BC

0m0

0m1

1m3

0m2

0m4

0m5

1m7

0m6

cells m3, m7B AND C

B AND C · Derived B AND C. Exactly equivalent on every assignment.

m3 and m7 are the BC = 11 column: A changes and drops, result B AND C.

3. Group the Karnaugh map of Σm(A, B, C; 0, 1, 4, 5) and write the result. Σm(A, B, C; 0, 1, 4, 5)

#	A	B	C	f
0	FALSE	FALSE	FALSE	TRUE
1	FALSE	FALSE	TRUE	TRUE
2	FALSE	TRUE	FALSE	FALSE
3	FALSE	TRUE	TRUE	FALSE
4	TRUE	FALSE	FALSE	TRUE
5	TRUE	FALSE	TRUE	TRUE
6	TRUE	TRUE	FALSE	FALSE
7	TRUE	TRUE	TRUE	FALSE

The engine found no law that shortens this expression; it is already as short as its rewriting laws can make it.

Σm(A, B, C; 0, 1, 4, 5)

Classification: contingent

A \ BC

1m0

1m1

0m3

0m2

1m4

1m5

0m7

0m6

cells m0, m1, m4, m5NOT B

NOT B · Derived NOT B. Exactly equivalent on every assignment.

The left half of the map, columns 00 and 01, is a block of four with A and C changing: NOT B.

4. Group the Karnaugh map of Σm(A, B, C; 0, 2) and write the result. Σm(A, B, C; 0, 2)

#	A	B	C	f
0	FALSE	FALSE	FALSE	TRUE
1	FALSE	FALSE	TRUE	FALSE
2	FALSE	TRUE	FALSE	TRUE
3	FALSE	TRUE	TRUE	FALSE
4	TRUE	FALSE	FALSE	FALSE
5	TRUE	FALSE	TRUE	FALSE
6	TRUE	TRUE	FALSE	FALSE
7	TRUE	TRUE	TRUE	FALSE

The engine found no law that shortens this expression; it is already as short as its rewriting laws can make it.

Σm(A, B, C; 0, 2)

Classification: contingent

A \ BC

1m0

0m1

0m3

1m2

0m4

0m5

0m7

0m6

cells m0, m2NOT A AND NOT C

NOT A AND NOT C · Derived NOT A AND NOT C. Exactly equivalent on every assignment.

m0 and m2 are at the two ends of the top row; they wrap around and form a pair in which only B changes: NOT A AND NOT C.

5. Group the Karnaugh map of Σm(A, B, C; 0, 2, 4, 5, 6) and write the result. Σm(A, B, C; 0, 2, 4, 5, 6)

#	A	B	C	f
0	FALSE	FALSE	FALSE	TRUE
1	FALSE	FALSE	TRUE	FALSE
2	FALSE	TRUE	FALSE	TRUE
3	FALSE	TRUE	TRUE	FALSE
4	TRUE	FALSE	FALSE	TRUE
5	TRUE	FALSE	TRUE	TRUE
6	TRUE	TRUE	FALSE	TRUE
7	TRUE	TRUE	TRUE	FALSE

The engine found no law that shortens this expression; it is already as short as its rewriting laws can make it.

Σm(A, B, C; 0, 2, 4, 5, 6)

Classification: contingent

A \ BC

1m0

0m1

0m3

1m2

1m4

1m5

0m7

1m6

cells m4, m5A AND NOT B
cells m0, m2, m4, m6NOT C

(A AND NOT B) OR NOT C · Derived (A AND NOT B) OR NOT C. Exactly equivalent on every assignment.

Two groups: the wraparound block of four m0, m2, m4, m6 gives NOT C, and the pair m4, m5 (A = TRUE, B = FALSE) gives A AND NOT B; m4 is used twice. Result NOT C OR (A AND NOT B).

6. Group the Karnaugh map of Σm(A, B, C; 2, 3, 6, 7) and write the result. Σm(A, B, C; 2, 3, 6, 7)

#	A	B	C	f
0	FALSE	FALSE	FALSE	FALSE
1	FALSE	FALSE	TRUE	FALSE
2	FALSE	TRUE	FALSE	TRUE
3	FALSE	TRUE	TRUE	TRUE
4	TRUE	FALSE	FALSE	FALSE
5	TRUE	FALSE	TRUE	FALSE
6	TRUE	TRUE	FALSE	TRUE
7	TRUE	TRUE	TRUE	TRUE

The engine found no law that shortens this expression; it is already as short as its rewriting laws can make it.

Σm(A, B, C; 2, 3, 6, 7)

Classification: contingent

A \ BC

0m0

0m1

1m3

1m2

0m4

0m5

1m7

1m6

cells m2, m3, m6, m7B

B · Derived B. Exactly equivalent on every assignment.

The right half, columns 11 and 10, is a block of four with A and C changing: B.

7. Group the Karnaugh map of Σm(A, B, C; 5) and write the result. Σm(A, B, C; 5)

#	A	B	C	f
0	FALSE	FALSE	FALSE	FALSE
1	FALSE	FALSE	TRUE	FALSE
2	FALSE	TRUE	FALSE	FALSE
3	FALSE	TRUE	TRUE	FALSE
4	TRUE	FALSE	FALSE	FALSE
5	TRUE	FALSE	TRUE	TRUE
6	TRUE	TRUE	FALSE	FALSE
7	TRUE	TRUE	TRUE	FALSE

The engine found no law that shortens this expression; it is already as short as its rewriting laws can make it.

Σm(A, B, C; 5)

Classification: contingent

A \ BC

0m0

0m1

0m3

0m2

0m4

1m5

0m7

0m6

cells m5A AND NOT B AND C

A AND NOT B AND C · Derived A AND NOT B AND C. Exactly equivalent on every assignment.

A single cell with no 1 beside it keeps all three variables: A AND NOT B AND C.

8. Group the Karnaugh map of Σm(A, B, C; 1, 2, 4, 7) and write the result. Σm(A, B, C; 1, 2, 4, 7)

#	A	B	C	f
0	FALSE	FALSE	FALSE	FALSE
1	FALSE	FALSE	TRUE	TRUE
2	FALSE	TRUE	FALSE	TRUE
3	FALSE	TRUE	TRUE	FALSE
4	TRUE	FALSE	FALSE	TRUE
5	TRUE	FALSE	TRUE	FALSE
6	TRUE	TRUE	FALSE	FALSE
7	TRUE	TRUE	TRUE	TRUE

The engine found no law that shortens this expression; it is already as short as its rewriting laws can make it.

Σm(A, B, C; 1, 2, 4, 7)

Classification: contingent

A \ BC

0m0

1m1

0m3

1m2

1m4

0m5

1m7

0m6

cells m1NOT A AND NOT B AND C
cells m2NOT A AND B AND NOT C
cells m4A AND NOT B AND NOT C
cells m7A AND B AND C

(NOT A AND NOT B AND C) OR (NOT A AND B AND NOT C) OR (A AND NOT B AND NOT C) OR (A AND B AND C) · Derived (NOT A AND NOT B AND C) OR (NOT A AND B AND NOT C) OR (A AND NOT B AND NOT C) OR (A AND B AND C). Exactly equivalent on every assignment.

A checkerboard: no two 1-cells are adjacent, so there is no group larger than one and the sum of four full minterms is the result. The function is parity, A XOR B XOR C, which the exact check accepts as the same function.

9. Group the Karnaugh map of Σm(A, B, C; 0, 1, 2, 5, 6, 7) and write the result; how many prime implicants are there? Σm(A, B, C; 0, 1, 2, 5, 6, 7)

#	A	B	C	f
0	FALSE	FALSE	FALSE	TRUE
1	FALSE	FALSE	TRUE	TRUE
2	FALSE	TRUE	FALSE	TRUE
3	FALSE	TRUE	TRUE	FALSE
4	TRUE	FALSE	FALSE	FALSE
5	TRUE	FALSE	TRUE	TRUE
6	TRUE	TRUE	FALSE	TRUE
7	TRUE	TRUE	TRUE	TRUE

The engine found no law that shortens this expression; it is already as short as its rewriting laws can make it.

Σm(A, B, C; 0, 1, 2, 5, 6, 7)

Classification: contingent

A \ BC

1m0

1m1

0m3

1m2

0m4

1m5

1m7

1m6

cells m0, m1NOT A AND NOT B
cells m5, m7A AND C
cells m2, m6B AND NOT C

(NOT A AND NOT B) OR (A AND C) OR (B AND NOT C) · Derived (NOT A AND NOT B) OR (A AND C) OR (B AND NOT C). Exactly equivalent on every assignment.

Six pairs are prime implicants, in a ring; three of them cover everything: m0, m1 gives NOT A AND NOT B, m5, m7 gives A AND C, m2, m6 gives B AND NOT C. The other three pairs (m1, m5; m6, m7; m0, m2) form the second cover, (NOT B AND C) OR (A AND B) OR (NOT A AND NOT C), equally correct.

10. Group the Karnaugh map of Σm(A, B, C; 1, 3, 5, 6, 7) and write the result. Σm(A, B, C; 1, 3, 5, 6, 7)

#	A	B	C	f
0	FALSE	FALSE	FALSE	FALSE
1	FALSE	FALSE	TRUE	TRUE
2	FALSE	TRUE	FALSE	FALSE
3	FALSE	TRUE	TRUE	TRUE
4	TRUE	FALSE	FALSE	FALSE
5	TRUE	FALSE	TRUE	TRUE
6	TRUE	TRUE	FALSE	TRUE
7	TRUE	TRUE	TRUE	TRUE

The engine found no law that shortens this expression; it is already as short as its rewriting laws can make it.

Σm(A, B, C; 1, 3, 5, 6, 7)

Classification: contingent

A \ BC

0m0

1m1

1m3

0m2

0m4

1m5

1m7

1m6

cells m6, m7A AND B
cells m1, m3, m5, m7C

(A AND B) OR C · Derived (A AND B) OR C. Exactly equivalent on every assignment.

The middle columns 01 and 11 form a block of four, m1, m3, m5, m7, with A and B changing: C. The pair m6, m7 adds A AND B. Result C OR (A AND B).

Questions

Why does the engine report more prime implicants than groups it used?

A prime implicant is any group that cannot be enlarged; a cover picks the ones needed to reach every 1-cell. Σm(A, B, C; 0, 1, 2, 5, 6, 7) has six prime implicants arranged in a ring and needs three of them, so the map lists six and uses three. The derived expression is irreducible: no term of it can be removed.

Worked examples

Four-variable K-map: Σm(0, 2, 5, 7, 8, 10, 13, 15) simplified

Glossary

Tools

Karnaugh map solver: 2 to 4 variables, groups, SOP and POS, don’t-cares from assumptions

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Karnaugh map vs Quine–McCluskey

Practice

Generated from the Tangibool engine, semantics tangibool-semantics/1, engine 1.0.0, build 2026-11-07. Every table, map, simplification and verdict on these pages was computed by the Tangibool engine when the site was built, and every claim was re-checked by the same exact checker the app uses. Limits are stated where they apply. © 2026 Foreranger LTD.

Karnaugh map exercises with answers: 3 variables

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