Guide to Using Karnaugh Maps for Minimizing Boolean Expressions, Schemes and Mind Maps of Circuit Theory

This handout provides a step-by-step guide on how to use karnaugh maps to minimize boolean expressions with 2, 3, or 4 variables. The concept of k-maps, their forms (sum of product and product of sum), and the process of solving expressions using the k-map. It also covers the k-map fill order, grouping rules, and the concept of minterms and maxterms.

Typology: Schemes and Mind Maps

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STEM Success Center
CS 231
Handout
Guide To The K-Map (Karnaugh Map)
In many digital circuits and practical problems we need to find expression with minimum
variables. We can minimize Boolean expressions of 2, 3, or 4 variables very easily using the K-
map without using any Boolean algebra theorems. The K-map can take two forms Sum of
Product (SOP) and Product of Sum (POS) according to the needs of the problem. The K-map is
table-like representation but it gives more information than TRUTH TABLE. We fill the grid of K-
map with 0โ€™s and 1โ€™s then solve it by making groups.
Steps to solve expression using the K-map
1. Select K-map according to the number of variables.
2. Identify minterms or maxterms as given in the problem.
3. For SOP put 1โ€™s in blocks of K-map respective to the minterms (0โ€™s elsewhere).
4. For POS put 0โ€™s in blocks of K-map respective to the maxterms(1โ€™s elsewhere).
5. Make rectangular groups containing total terms in power of two like 2,4,8 ..(except 1)
and try to cover as many elements as you can in one group.
6. From the groups made in step 5 find the product terms and sum them up for SOP form.
The K-map Fill Order
2 - Variable Map
3- Variable Map
4 - Variable Map
A\B
0
1
0
0
1
1
2
3
A\BC
00
01
11
10
0
0
1
3
2
1
4
5
7
6
AB\CD
00
01
11
10
Grouping Rules
The Karnaugh map uses the following rules for the simplification of expressions by grouping
together adjacent cells containing ones
1. No zeros allowed.
2. No diagonals.
3. Only power of 2 number
of cells in each group.
4. Groups should be as
large as possible.
5. Everyone must be in at
least one group.
6. Overlapping allowed.
7. Wrap around is allowed.
8. Get the fewest number
of groups possible.
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STEM Success Center

CS 231

Handout

Guide To The K-Map (Karnaugh Map)

In many digital circuits and practical problems we need to find expression with minimum

variables. We can minimize Boolean expressions of 2, 3, or 4 variables very easily using the K-

map without using any Boolean algebra theorems. The K-map can take two forms Sum of

Product (SOP) and Product of Sum (POS) according to the needs of the problem. The K-map is

table-like representation but it gives more information than TRUTH TABLE. We fill the grid of K-

map with 0โ€™s and 1โ€™s then solve it by making groups.

Steps to solve expression using the K-map

1. Select K-map according to the number of variables.

2. Identify minterms or maxterms as given in the problem.

3. For SOP put 1โ€™s in blocks of K-map respective to the minterms (0โ€™s elsewhere).

4. For POS put 0โ€™s in blocks of K-map respective to the maxterms(1โ€™s elsewhere).

5. Make rectangular groups containing total terms in power of two like 2,4,8 ..(except 1)

and try to cover as many elements as you can in one group.

6. From the groups made in step 5 find the product terms and sum them up for SOP form.

The K-map Fill Order

2 - Variable Map 3 - Variable Map 4 - Variable Map

A\B 0 1

A\BC 00 01 11 10

AB\CD 00 01 11 10

Grouping Rules

The Karnaugh map uses the following rules for the simplification of expressions by grouping

together adjacent cells containing ones

1. No zeros allowed.

2. No diagonals.

3. Only power of 2 number

of cells in each group.

4. Groups should be as

large as possible.

5. Everyone must be in at

least one group.

6. Overlapping allowed.

7. Wrap around is allowed.

8. Get the fewest number

of groups possible.

STEM Success Center

CS 231

Handout

We perform the Sum of minterm also known as Sum of products (SOP).

โ— The minterm for each combination of the variables that produce a 1 in the function and

then taking the OR of all those terms.

We perform the Product of Maxterm also known as Product of sum (POS).

โ— The maxterm for each combination of the variables that produce a 0 in the function and

then taking the AND of all those terms.

Truth table representing minterm and maxterm

Minterms Maxterms X Y Z Product Terms Sum Terms 0 0 0 m 0 = ๐‘‹ โ‹… ๐‘Œ โ‹… ๐‘ = min( ๐‘‹, ๐‘Œ, ๐‘ ) M 0 = ๐‘‹ + ๐‘Œ + ๐‘Œ = max( ๐‘‹, ๐‘Œ, ๐‘ ) 0 0 1 m 1 = ๐‘‹ โ‹… ๐‘Œ โ‹… ๐‘ = min( ๐‘‹, ๐‘Œ, ๐‘ ) M 1 = ๐‘‹ + ๐‘Œ + ๐‘ = max( ๐‘‹, ๐‘Œ, ๐‘ ) 0 1 0 m 2 = ๐‘‹ โ‹… ๐‘Œ โ‹… ๐‘ = min( ๐‘‹, ๐‘Œ, ๐‘ ) M 2 = ๐‘‹ + ๐‘Œ + ๐‘ = max( ๐‘‹, ๐‘Œ, ๐‘ ) 0 1 1 m 3 = ๐‘‹ โ‹… ๐‘Œ โ‹… ๐‘ = min( ๐‘‹, ๐‘Œ, ๐‘ ) M 3 = ๐‘‹ + ๐‘Œ + ๐‘ = max( ๐‘‹, ๐‘Œ, ๐‘ ) 1 0 0 m 4 = ๐‘‹ โ‹… ๐‘Œ โ‹… ๐‘ = min( ๐‘‹, ๐‘Œ, ๐‘ ) M 4 = ๐‘‹ + ๐‘Œ + ๐‘ = max( ๐‘‹, ๐‘Œ, ๐‘ ) 1 0 1 m 5 = ๐‘‹ โ‹… ๐‘Œ โ‹… ๐‘ = min( ๐‘‹, ๐‘Œ, ๐‘ ) M 5 = ๐‘‹ + ๐‘Œ + ๐‘ = max( ๐‘‹, ๐‘Œ, ๐‘ ) 1 1 0 m 6 = ๐‘‹ โ‹… ๐‘Œ โ‹… ๐‘ = min( ๐‘‹, ๐‘Œ, ๐‘ ) M 6 = ๐‘‹ + ๐‘Œ + ๐‘ = max( ๐‘‹, ๐‘Œ, ๐‘ ) 1 1 1 m 7 = ๐‘‹ โ‹… ๐‘Œ โ‹… ๐‘ = min( ๐‘‹, ๐‘Œ, ๐‘ ) M 7 = ๐‘‹ + ๐‘Œ + ๐‘ = max( ๐‘‹, ๐‘Œ, ๐‘ ) From the table above, it is clear that minterm is expressed in product format and maxterm is expressed in sum format.