Karnaugh map exercises with answers: 4 variables

Ten functions over A, B, C and D as minterm lists, to be grouped on the four-by-four Karnaugh map. Draw the map with AB on the rows and CD on the columns in Gray-code order, circle the groups, and compare with the engine's map and derived expression under each problem.

In the engine's layout the rows are AB = 00, 01, 11, 10 and the columns CD = 00, 01, 11, 10, so the first row is m0, m1, m3, m2 and the last row is m8, m9, m11, m10. Both edges wrap: the top and bottom rows are neighbours, and so are the left and right columns, which makes the four corners m0, m2, m8, m10 one group. Groups of two, four, eight and sixteen cells drop one, two, three and four variables.

The set includes the four corners, a full row, a full column, a block of eight, a wraparound pair of rows, the classic two-group function from the worked examples, a single cell, and a ten-cell function that needs a group of eight and a group of four sharing cells. Each solution names the cells of each group, says which variables change inside it, and gives the sum of products.

1. Group the Karnaugh map of Σm(A, B, C, D; 0, 2, 8, 10) and write the result. Σm(A, B, C, D; 0, 2, 8, 10)

#	A	B	C	D	f
0	FALSE	FALSE	FALSE	FALSE	TRUE
1	FALSE	FALSE	FALSE	TRUE	FALSE
2	FALSE	FALSE	TRUE	FALSE	TRUE
3	FALSE	FALSE	TRUE	TRUE	FALSE
4	FALSE	TRUE	FALSE	FALSE	FALSE
5	FALSE	TRUE	FALSE	TRUE	FALSE
6	FALSE	TRUE	TRUE	FALSE	FALSE
7	FALSE	TRUE	TRUE	TRUE	FALSE
8	TRUE	FALSE	FALSE	FALSE	TRUE
9	TRUE	FALSE	FALSE	TRUE	FALSE
10	TRUE	FALSE	TRUE	FALSE	TRUE
11	TRUE	FALSE	TRUE	TRUE	FALSE
12	TRUE	TRUE	FALSE	FALSE	FALSE
13	TRUE	TRUE	FALSE	TRUE	FALSE
14	TRUE	TRUE	TRUE	FALSE	FALSE
15	TRUE	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, D; 0, 2, 8, 10)

Classification: contingent

AB \ CD

1m0

0m1

0m3

1m2

0m4

0m5

0m7

0m6

0m12

0m13

0m15

0m14

1m8

0m9

0m11

1m10

cells m0, m2, m8, m10NOT B AND NOT D

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

m0, m2, m8, m10 are the four corners, adjacent by wraparound on both edges: A and C change, B and D are FALSE throughout. Result NOT B AND NOT D.

2. Group the Karnaugh map of Σm(A, B, C, D; 0, 1, 2, 3) and write the result. Σm(A, B, C, D; 0, 1, 2, 3)

#	A	B	C	D	f
0	FALSE	FALSE	FALSE	FALSE	TRUE
1	FALSE	FALSE	FALSE	TRUE	TRUE
2	FALSE	FALSE	TRUE	FALSE	TRUE
3	FALSE	FALSE	TRUE	TRUE	TRUE
4	FALSE	TRUE	FALSE	FALSE	FALSE
5	FALSE	TRUE	FALSE	TRUE	FALSE
6	FALSE	TRUE	TRUE	FALSE	FALSE
7	FALSE	TRUE	TRUE	TRUE	FALSE
8	TRUE	FALSE	FALSE	FALSE	FALSE
9	TRUE	FALSE	FALSE	TRUE	FALSE
10	TRUE	FALSE	TRUE	FALSE	FALSE
11	TRUE	FALSE	TRUE	TRUE	FALSE
12	TRUE	TRUE	FALSE	FALSE	FALSE
13	TRUE	TRUE	FALSE	TRUE	FALSE
14	TRUE	TRUE	TRUE	FALSE	FALSE
15	TRUE	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, D; 0, 1, 2, 3)

Classification: contingent

AB \ CD

1m0

1m1

1m3

1m2

0m4

0m5

0m7

0m6

0m12

0m13

0m15

0m14

0m8

0m9

0m11

0m10

cells m0, m1, m2, m3NOT A AND NOT B

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

The whole first row, AB = 00, with C and D taking all four combinations: NOT A AND NOT B.

3. Group the Karnaugh map of Σm(A, B, C, D; 5, 7, 13, 15) and write the result. Σm(A, B, C, D; 5, 7, 13, 15)

#	A	B	C	D	f
0	FALSE	FALSE	FALSE	FALSE	FALSE
1	FALSE	FALSE	FALSE	TRUE	FALSE
2	FALSE	FALSE	TRUE	FALSE	FALSE
3	FALSE	FALSE	TRUE	TRUE	FALSE
4	FALSE	TRUE	FALSE	FALSE	FALSE
5	FALSE	TRUE	FALSE	TRUE	TRUE
6	FALSE	TRUE	TRUE	FALSE	FALSE
7	FALSE	TRUE	TRUE	TRUE	TRUE
8	TRUE	FALSE	FALSE	FALSE	FALSE
9	TRUE	FALSE	FALSE	TRUE	FALSE
10	TRUE	FALSE	TRUE	FALSE	FALSE
11	TRUE	FALSE	TRUE	TRUE	FALSE
12	TRUE	TRUE	FALSE	FALSE	FALSE
13	TRUE	TRUE	FALSE	TRUE	TRUE
14	TRUE	TRUE	TRUE	FALSE	FALSE
15	TRUE	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, D; 5, 7, 13, 15)

Classification: contingent

AB \ CD

0m0

0m1

0m3

0m2

0m4

1m5

1m7

0m6

0m12

1m13

1m15

0m14

0m8

0m9

0m11

0m10

cells m5, m7, m13, m15B AND D

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

The central block of four, rows AB = 01 and 11 by columns CD = 01 and 11: A and C change, B = TRUE, D = TRUE throughout. Result B AND D.

4. Group the Karnaugh map of Σm(A, B, C, D; 0, 4, 8, 12) and write the result. Σm(A, B, C, D; 0, 4, 8, 12)

#	A	B	C	D	f
0	FALSE	FALSE	FALSE	FALSE	TRUE
1	FALSE	FALSE	FALSE	TRUE	FALSE
2	FALSE	FALSE	TRUE	FALSE	FALSE
3	FALSE	FALSE	TRUE	TRUE	FALSE
4	FALSE	TRUE	FALSE	FALSE	TRUE
5	FALSE	TRUE	FALSE	TRUE	FALSE
6	FALSE	TRUE	TRUE	FALSE	FALSE
7	FALSE	TRUE	TRUE	TRUE	FALSE
8	TRUE	FALSE	FALSE	FALSE	TRUE
9	TRUE	FALSE	FALSE	TRUE	FALSE
10	TRUE	FALSE	TRUE	FALSE	FALSE
11	TRUE	FALSE	TRUE	TRUE	FALSE
12	TRUE	TRUE	FALSE	FALSE	TRUE
13	TRUE	TRUE	FALSE	TRUE	FALSE
14	TRUE	TRUE	TRUE	FALSE	FALSE
15	TRUE	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, D; 0, 4, 8, 12)

Classification: contingent

AB \ CD

1m0

0m1

0m3

0m2

1m4

0m5

0m7

0m6

1m12

0m13

0m15

0m14

1m8

0m9

0m11

0m10

cells m0, m4, m8, m12NOT C AND NOT D

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

The whole first column, CD = 00: A and B change, result NOT C AND NOT D.

5. Group the Karnaugh map of Σm(A, B, C, D; 0, 1, 2, 4, 5, 6, 8, 9, 12, 13) and write the result. Σm(A, B, C, D; 0, 1, 2, 4, 5, 6, 8, 9, 12, 13)

#	A	B	C	D	f
0	FALSE	FALSE	FALSE	FALSE	TRUE
1	FALSE	FALSE	FALSE	TRUE	TRUE
2	FALSE	FALSE	TRUE	FALSE	TRUE
3	FALSE	FALSE	TRUE	TRUE	FALSE
4	FALSE	TRUE	FALSE	FALSE	TRUE
5	FALSE	TRUE	FALSE	TRUE	TRUE
6	FALSE	TRUE	TRUE	FALSE	TRUE
7	FALSE	TRUE	TRUE	TRUE	FALSE
8	TRUE	FALSE	FALSE	FALSE	TRUE
9	TRUE	FALSE	FALSE	TRUE	TRUE
10	TRUE	FALSE	TRUE	FALSE	FALSE
11	TRUE	FALSE	TRUE	TRUE	FALSE
12	TRUE	TRUE	FALSE	FALSE	TRUE
13	TRUE	TRUE	FALSE	TRUE	TRUE
14	TRUE	TRUE	TRUE	FALSE	FALSE
15	TRUE	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, D; 0, 1, 2, 4, 5, 6, 8, 9, 12, 13)

Classification: contingent

AB \ CD

1m0

1m1

0m3

1m2

1m4

1m5

0m7

1m6

1m12

1m13

0m15

0m14

1m8

1m9

0m11

0m10

cells m0, m2, m4, m6NOT A AND NOT D
cells m0, m1, m4, m5, m8, m9, m12, m13NOT C

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

Two groups: the block of eight in columns CD = 00 and 01 gives NOT C, and the four cells m0, m2, m4, m6 (rows AB = 00 and 01, columns CD = 00 and 10, wrapping) give NOT A AND NOT D. They share m0 and m4. Result NOT C OR (NOT A AND NOT D).

6. Group the Karnaugh map of Σm(A, B, C, D; 0, 1, 4, 5, 8, 9, 12, 13) and write the result. Σm(A, B, C, D; 0, 1, 4, 5, 8, 9, 12, 13)

#	A	B	C	D	f
0	FALSE	FALSE	FALSE	FALSE	TRUE
1	FALSE	FALSE	FALSE	TRUE	TRUE
2	FALSE	FALSE	TRUE	FALSE	FALSE
3	FALSE	FALSE	TRUE	TRUE	FALSE
4	FALSE	TRUE	FALSE	FALSE	TRUE
5	FALSE	TRUE	FALSE	TRUE	TRUE
6	FALSE	TRUE	TRUE	FALSE	FALSE
7	FALSE	TRUE	TRUE	TRUE	FALSE
8	TRUE	FALSE	FALSE	FALSE	TRUE
9	TRUE	FALSE	FALSE	TRUE	TRUE
10	TRUE	FALSE	TRUE	FALSE	FALSE
11	TRUE	FALSE	TRUE	TRUE	FALSE
12	TRUE	TRUE	FALSE	FALSE	TRUE
13	TRUE	TRUE	FALSE	TRUE	TRUE
14	TRUE	TRUE	TRUE	FALSE	FALSE
15	TRUE	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, D; 0, 1, 4, 5, 8, 9, 12, 13)

Classification: contingent

AB \ CD

1m0

1m1

0m3

0m2

1m4

1m5

0m7

0m6

1m12

1m13

0m15

0m14

1m8

1m9

0m11

0m10

cells m0, m1, m4, m5, m8, m9, m12, m13NOT C

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

The left half of the map, columns CD = 00 and 01, is a block of eight: A, B and D change, only C is fixed at FALSE. Result NOT C.

7. Group the Karnaugh map of Σm(A, B, C, D; 0, 2, 5, 7, 8, 10, 13, 15) and write the result. Σm(A, B, C, D; 0, 2, 5, 7, 8, 10, 13, 15)

#	A	B	C	D	f
0	FALSE	FALSE	FALSE	FALSE	TRUE
1	FALSE	FALSE	FALSE	TRUE	FALSE
2	FALSE	FALSE	TRUE	FALSE	TRUE
3	FALSE	FALSE	TRUE	TRUE	FALSE
4	FALSE	TRUE	FALSE	FALSE	FALSE
5	FALSE	TRUE	FALSE	TRUE	TRUE
6	FALSE	TRUE	TRUE	FALSE	FALSE
7	FALSE	TRUE	TRUE	TRUE	TRUE
8	TRUE	FALSE	FALSE	FALSE	TRUE
9	TRUE	FALSE	FALSE	TRUE	FALSE
10	TRUE	FALSE	TRUE	FALSE	TRUE
11	TRUE	FALSE	TRUE	TRUE	FALSE
12	TRUE	TRUE	FALSE	FALSE	FALSE
13	TRUE	TRUE	FALSE	TRUE	TRUE
14	TRUE	TRUE	TRUE	FALSE	FALSE
15	TRUE	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, D; 0, 2, 5, 7, 8, 10, 13, 15)

Classification: contingent

AB \ CD

1m0

0m1

0m3

1m2

0m4

1m5

1m7

0m6

0m12

1m13

1m15

0m14

1m8

0m9

0m11

1m10

cells m0, m2, m8, m10NOT B AND NOT D
cells m5, m7, m13, m15B AND D

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

Two groups of four: the corners m0, m2, m8, m10 give NOT B AND NOT D, and the centre m5, m7, m13, m15 gives B AND D. Result (NOT B AND NOT D) OR (B AND D), which is B <-> D; A and C are irrelevant.

8. Group the Karnaugh map of Σm(A, B, C, D; 1, 3, 9, 11) and write the result. Σm(A, B, C, D; 1, 3, 9, 11)

#	A	B	C	D	f
0	FALSE	FALSE	FALSE	FALSE	FALSE
1	FALSE	FALSE	FALSE	TRUE	TRUE
2	FALSE	FALSE	TRUE	FALSE	FALSE
3	FALSE	FALSE	TRUE	TRUE	TRUE
4	FALSE	TRUE	FALSE	FALSE	FALSE
5	FALSE	TRUE	FALSE	TRUE	FALSE
6	FALSE	TRUE	TRUE	FALSE	FALSE
7	FALSE	TRUE	TRUE	TRUE	FALSE
8	TRUE	FALSE	FALSE	FALSE	FALSE
9	TRUE	FALSE	FALSE	TRUE	TRUE
10	TRUE	FALSE	TRUE	FALSE	FALSE
11	TRUE	FALSE	TRUE	TRUE	TRUE
12	TRUE	TRUE	FALSE	FALSE	FALSE
13	TRUE	TRUE	FALSE	TRUE	FALSE
14	TRUE	TRUE	TRUE	FALSE	FALSE
15	TRUE	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, D; 1, 3, 9, 11)

Classification: contingent

AB \ CD

0m0

1m1

1m3

0m2

0m4

0m5

0m7

0m6

0m12

0m13

0m15

0m14

0m8

1m9

1m11

0m10

cells m1, m3, m9, m11NOT B AND D

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

Rows AB = 00 and 10 are neighbours by wraparound, and columns CD = 01 and 11 are adjacent: a block of four with A and C changing, B = FALSE, D = TRUE throughout. Result NOT B AND D.

9. Group the Karnaugh map of Σm(A, B, C, D; 6) and write the result. Σm(A, B, C, D; 6)

#	A	B	C	D	f
0	FALSE	FALSE	FALSE	FALSE	FALSE
1	FALSE	FALSE	FALSE	TRUE	FALSE
2	FALSE	FALSE	TRUE	FALSE	FALSE
3	FALSE	FALSE	TRUE	TRUE	FALSE
4	FALSE	TRUE	FALSE	FALSE	FALSE
5	FALSE	TRUE	FALSE	TRUE	FALSE
6	FALSE	TRUE	TRUE	FALSE	TRUE
7	FALSE	TRUE	TRUE	TRUE	FALSE
8	TRUE	FALSE	FALSE	FALSE	FALSE
9	TRUE	FALSE	FALSE	TRUE	FALSE
10	TRUE	FALSE	TRUE	FALSE	FALSE
11	TRUE	FALSE	TRUE	TRUE	FALSE
12	TRUE	TRUE	FALSE	FALSE	FALSE
13	TRUE	TRUE	FALSE	TRUE	FALSE
14	TRUE	TRUE	TRUE	FALSE	FALSE
15	TRUE	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, D; 6)

Classification: contingent

AB \ CD

0m0

0m1

0m3

0m2

0m4

0m5

0m7

1m6

0m12

0m13

0m15

0m14

0m8

0m9

0m11

0m10

cells m6NOT A AND B AND C AND NOT D

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

A single cell, m6, with no 1 beside it: nothing drops and the result is the full minterm NOT A AND B AND C AND NOT D.

10. Group the Karnaugh map of Σm(A, B, C, D; 0, 1, 2, 3, 8, 9, 10, 11) and write the result. Σm(A, B, C, D; 0, 1, 2, 3, 8, 9, 10, 11)

#	A	B	C	D	f
0	FALSE	FALSE	FALSE	FALSE	TRUE
1	FALSE	FALSE	FALSE	TRUE	TRUE
2	FALSE	FALSE	TRUE	FALSE	TRUE
3	FALSE	FALSE	TRUE	TRUE	TRUE
4	FALSE	TRUE	FALSE	FALSE	FALSE
5	FALSE	TRUE	FALSE	TRUE	FALSE
6	FALSE	TRUE	TRUE	FALSE	FALSE
7	FALSE	TRUE	TRUE	TRUE	FALSE
8	TRUE	FALSE	FALSE	FALSE	TRUE
9	TRUE	FALSE	FALSE	TRUE	TRUE
10	TRUE	FALSE	TRUE	FALSE	TRUE
11	TRUE	FALSE	TRUE	TRUE	TRUE
12	TRUE	TRUE	FALSE	FALSE	FALSE
13	TRUE	TRUE	FALSE	TRUE	FALSE
14	TRUE	TRUE	TRUE	FALSE	FALSE
15	TRUE	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, D; 0, 1, 2, 3, 8, 9, 10, 11)

Classification: contingent

AB \ CD

1m0

1m1

1m3

1m2

0m4

0m5

0m7

0m6

0m12

0m13

0m15

0m14

1m8

1m9

1m11

1m10

cells m0, m1, m2, m3, m8, m9, m10, m11NOT B

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

The first and last rows, AB = 00 and AB = 10, are neighbours by wraparound and together form a block of eight: A, C and D change and only B is fixed. Result NOT B.

Questions

How do I spot wraparound groups quickly?

Look at the values rather than the drawing: a set of 2, 4 or 8 cells is a group exactly when some variables are fixed across all of them and the others take every combination. The four corners all have B = FALSE, D = FALSE while A and C take all four combinations, which is why they group; the first and last rows share B = FALSE, which is why together they form a block of eight.

Laws

Karnaugh map grouping rule: adjacent cells drop the variable that changes

Glossary

Practice

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Karnaugh map exercises with answers: 4 variables

Questions

Related

Laws

Glossary

See also

Practice