# karnaugh maps by lovishbajaj

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```									Presentation on K-maps

Submitted by :Lovish bajaj

Karnaugh maps
 Graphical method for simplification of boolean expression.  It is a graphical chart which contain boxes.  The information contained in truth table or available in SOP or POS form is represented in k-map.  K-maps can be written for 2,3,4…. Upto 6 variables . Beyond that the technique becomes very cumbersome.

K-map structure
2- variable k-map :
B
• A two variable k-map consist of 4 boxes. • A and B are two input variables.
0

A
0 1

0

2

1 1 3

• 0 and 1 are the value of A or B.

3- variable k-map :
A B

The sequence follows gray code and not the binary.

C

00 0 0 1

01

11

1 0

• The three variable k-map consist

of 8 boxes. • A , B and C are the three variables .

2

6

4

1

3

7

5

4 - variable k-map :
A B C D
00 0 • Four variable k- map 4 5 7 6 12 13 15 14 8 9 11 10

0 0

0 1

1 1

1 0

consist of sixteen boxes. • It has four variables A,B,C&D.

01

1 3 2

11

10

PLOTTING A K-MAP
• Whenever a k-map is to be practically used for simplification, the entries inside the boxes are to be done from given truth table. • For example :A B 0 1

A
0 0 1 1

B
0 1 0 1

Y
0 0 0 1 1 0

0
0

0
2

0
1

1
3

A & B are the input variables

Output value of Y is entered into the boxes

Representation of standard SOP form on the k-maps:• The logical expression in standard SOP form can be represented on k-map by simply entering 1’s in the boxes of the k-map corresponding to each minterm present in the equation. • The remaining boxes are to be filled with zero. • For example :A B

Q.

Draw k- map for the following expression :

C

00

01

11

1 0

1

0
0 2

1
6

1
4

Y=  m 0,1,4,6,7
1

0
1 3

1
7

0
5

Representation of standard POS form on k-map:• Logical expressions in the standard POS form can be represented on k-map by entering 0’s in the boxes of k-map corresponding to each maxterm present in the given equation • The remaining boxes are filled with one. • For example :-

Q.

Draw k-map for following expression:

00

01

11

1 0

M 0,2,6

0
0

0
2

0
6

1
4

1
1

1
3

1
7

1
5

How does simplification takes place ?
• Once we plot the logic function or truth table on a k-map , we have to use the grouping technique for simplifying the logic function. • Grouping means combining the terms in the adjacent boxes. • Grouping of adjacent 1’s is done for simplification of SOP form. • Grouping of adjacent 0’s is done for simplification of POS form.

Way of grouping
• The grouping follows the binary rule i.e. we can group 1,2,,4,8,16,32… number of 1’s or 0’s. We cannot group 3,5,7,9…. Number of 1’s or 0’s. • Pairs : A group of adjacent 1’s or 0’s is called a pair. • Quad : A group of four adjacent 1’s or 0’s is called a quad. • Octet : A group of eight adjacent 1’s or 0’s is called a octet.

Grouping of two adjacent I’s (pair):
A B

ABC’
A’B’ A’B 0 0 AB 1 0

AB’C’
AB’ 1 0

C
C’ C

0 0

SIMPLIFICATION :Y = ABC’ + AB’C’ = AC’(B + B’) = AC’ …. Since (B+B’) = 1
Conclusion :- By pairing two adjacent 1’s we can eliminate one variable.

A’B A B C Simplification:Y = A’BC’ + A’BC = A’B(C’ + C) = A’B …. SINCE C’+C = 1 C’ C A’B’ 0 A’B 1 AB 0 AB’ 0

0

1

0

0

A B

C
Simplification:Y = A’B’C + AB’C = B’C(A’ + A) = B’C …. SINCE A’+A = 1 C’ C

A’B’ 0 1

A’B 0 0

AB 0 0

AB’ 0 1

Conclusion :- By pairing two adjacent 1’s we can eliminate one variable.

B’C

• If we group four 1’s from the adjacent cells of k-map then the group is called a quad.
• After forming a quad , the simplification takes place in such a way that the two variables which are not same will be eliminated . • Thus quad will eliminate two variables.

For eg:• The two variables which are same in all minterms are C and D’ . • The variables which are not same in all variables are A and B .

A B C D A’B’ C’D’ C’D CD 0 0 0 1 A’B 0 0 0 1 AB 0 0 0 1 AB’ 0 0 0 1

• A and B will be eliminated and the output will be : Y = C D’ Simplification :-

CD’

Y = CD’

Y = A’B’CD’ + A’BCD’ + ABCD’ + AB’CD’ = CD’ ( A’B’ + A’B + AB + AB’ ) = CD’ *A’(B’+B) + A(B+B’)+ = CD’*A+A’+ = CD’ ……Proved. Conclusion:Two variables are eliminated

Top and bottom 1’s forming a quad

Four adjacent 1’s forming a square. A B

C D C’D’ C’D CD

A B A’B’ 0 0 0 0 A’B 1 0 0 1 AB 1 0 0 1 AB’ 0 0 0 0

C D C’D’ C’D CD

A’B’ 0 1 1 0

A’B 0 1 1 0

AB 0 0 0 0

AB’ 0 0 0 0

CD’

CD’

Y = BD’

Y = A’D

A and C are changing so they are eliminated

B and C are changing so they are eliminated

1’s corresponding to the corners A B C D A’B’ 1 A’B 0 0 0 0 AB 0 0 0 0 AB’ 1 0 0 1

C’D’
C’D CD 0 0 1

CD’

Y = B’D’ ( A and C are eliminated )

Grouping of eight adjacent 1’s(octet):• A group of eight adjacent 1’s is called octet.

• When an octet is formed three variables will change and only one variable will remain same in all minterms .

• The three variables which will change will be eliminated and which does not change wil appear as output.

A B C D A’B’ Simplification:C’D’ C’D Y = A’B’C’D’ + A’B’C’D + A’B’CD + A’B’CD’ + A’BC’D’ + A’BC’D + A’BCD + A’BCD’ CD 1 1 1 1 A’B 1 1 1 1 AB 0 0 0 0 AB’ 0 0 0 0

= A’B’C’ ( D+D’ ) + A’B’C ( D + D’ ) + A’BC’ ( D’+D ) + A’BC ( D + D’ )
= A’B’ ( C+C’ ) + A’B ( C’ + C ) = A’ ( B + B’ ) = A’

CD’

Y = A’

THREE VARIABLES ARE ELIMINATED

Leftmost and the rightmost forming a octet

A B C D A’B’ 1 C’D’ C’D 1 0 0 1 A’B 0 AB 0 AB’ 1

CD
CD’

1
1

0
0

0
0

1
1

Y = B’

Minimization of SOP expressions :Minimization procedure:Step 1 :- Prepare the k-map and place 1’s according to the given truth table or logical expressions. Fill the remaining cells with 0’s.

Step 2:- Locate the isolated I’s i.e. the 1’s which are not combined with other .

Step 3:- Identify the 1’s which can be combined to form a pair in only one way and then encircle them.

Step 4:- Identify the 1’s which can be combined to form a quad in only one way and encircle them.

Step 5:- Identify the 1’s which can be combined to form an octet in only one way and encircle them.

Step 6:- after identifying the pairs, quads and octets ,check whether if any one is yet their to be encircled. If yes then encircle them with each other o rwith already encircled 1’s (by means of overlapping).
 The number of groups should be minimum.

For eg:Solve the expression Y =  m(0,1,2,5,13,15). • Prepare the k-map and place 1’s and 0’s at their positions.
A B C D Pair 2:A’BC’ A’B’ 1 0 0 0 CD’ Pair 3: BCD A’B 1 1 1 0 AB 0 0 1 0 Pair 1: B’C’D’ AB’ 1 0 0 0

As there are no isolated 1’s , quads and octets so the minimized expression is :

C’D’

C’D CD

Y = B’C’D’ + A’BC’ + BCD

Don’t care conditions:-

(



)

• For SOP form we enter 1’s corresponding to the combinations of the input variables which produce the high input. And we enter 0’s for the remaining cells of k-map.

• For POS form we enter 0’s corresponding to the combinations of the input variables which produce the high input. And we enter 1’s for the remaining cells of k-map.
• But it is not always true that the cells not containing 1’s (in SOP) will contain 0’s some combination of input variables do not occur .

• Also for some functions the outputs corresponding to certain combinations of input variables do not matter

• In such conditions we have freedom to assume a 0 or 1 as output for each of these combination
• These conditions are known as don’t care conditions in k-map and in k- map is represented with a cross mark in the corresponding cell.



The don’t care condition () may be assumed to be 0 or 1 as per the need of simplification .

e.g. :Simplify the k-map :Y =  m (1,3,7,11,15) + d(0,2,5)
C D A B A’B’ C’D’ 0 0 0 CD’ Y = AB A’B 1 AB 1 1 1 1 AB’

Y = C’D’

Don’t care conditions are treated as 1’s .




0 0 0

C’D CD


1 0

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