In many digital circuits and practical problems we need to find expression with minimum variables. We can minimize Boolean expressions of 3, 4 variables very easily using K-map without using any Boolean algebra theorems. K-map can take two forms Sum of Product (SOP) and Product of Sum (POS) according to the need of problem. K-map is table like representation but it gives more information than TRUTH TABLE. We fill grid of K-map with 0’s and 1’s then solve it by making groups.

__Steps to solve expression using K-map-__

- Select K-map according to the number of variables.
- Identify minterms or maxterms as given in problem.
- For SOP put 1’s in blocks of K-map respective to the minterms (0’s elsewhere).
- For POS put 0’s in blocks of K-map respective to the maxterms(1’s elsewhere).
- 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.
- From the groups made in step 5 find the product terms and sum them up for SOP form.

__SOP FORM__

__SOP FORM__

__K-map of 3 variables-__

Z= ∑A,B,C(1,3,6,7)

From **red** group we get product term—

A’C

From** green** group we get product term—

AB

Summing these product terms we get- **Final expression (A’C+AB)**

__K-map for 4 variables__

F(P,Q,R,S)=∑(0,2,5,7,8,10,13,15)

From **red** group we get product term—

QS

From **green** group we get product term—

Q’S’

Summing these product terms we get- **Final expression (QS+Q’S’)**

__POS FORM__

__POS FORM__

__K-map of 3 variables-__

F(A,B,C)=π(0,3,6,7)

From **red **group we find terms

A B C’

Taking complement of these two

A’ B’ C

Now **sum** up them

(A’ + B’ + C)

From** green** group we find terms

B C

Taking complement of these two terms

B’ C’

Now sum up them

(B’+C’)

From **brown **group we find terms

A’ B’ C’

Taking complement of these two

A B C

Now **sum** up them

(A + B + C)

We will take product of these three terms :**Final expression (A’ + B’ + C) (B’ + C’) (A + B + C)**

__2. K-map of 4 variables-__

F(A,B,C,D)=π(3,5,7,8,10,11,12,13)

From **green** group we find terms

C’ D B

Taking their complement and summing them

(C+D’+B’)

From **red** group we find terms

C D A’

Taking their complement and summing them

(C’+D’+A)

From **blue** group we find terms

A C’ D’

Taking their complement and summing them

(A’+C+D)

From **brown ** group we find terms

A B’ C

Taking their complement and summing them

(A’+B+C’)

Finally we express these as product –**(C+D’+B’).(C’+D’+A).(A’+C+D).(A’+B+C’)**

** PITFALL–** *Always remember

**POS ≠ (SOP)’***The correct form is (**POS of F)=(SOP of F’)’**

This article is contributed by Anuj Bhatam. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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