ppt on K-Map Karnaugh Map is a method of simplifying Boolean expressions. The Karnaugh map reduces the need for extensive calculations by taking advantage of humans' pattern-recognition capability. The number of cells in the K-map is determined by the number of input variables and is mathematically expressed as two raised to the power of the number of input variables, i.e., 2 to the power n, where the number of input variables is n. Thus, to simplify a logical expression with two inputs, we require a K-map with 4 (= 22) cells. So, for 2 variable K-Map we need 4 cells, for 3 variable K-Map we need 8 cells and for 4 variable K-Map we need 16 cells . In K-Map representation each cell follows Gray code. There is one bit difference in two adjacent cells. In my PPT I have clearly shown an example to reduce a Boolean expression using K-Map. I have shown 2,3 and 4 variables K-Map . after putting the position of variables , how to pair up the cells is clearly drawn in PPT . So it is very easy to get the reduced term of a Boolean algebraic expression using K-Map.

1. Minimization of SOP and POS
Expressions using Karnaugh
map
Prepared by- Dr. Sutapa Mukherjee

2. Different variable K Map
0 1
0 A B A B
1 A B A B
A
B 00 01 11 10
0 A B C A B C A B C A B C
1 A B C A B C A B C A B C
A
BC
Two Variable Map Three Variable Map

3. Four variable K Map
00 01 11 10
00 A B C D A B C D A B C D A B C D
01 A B C D A B C D A B C D A B C D
11 A B C D A B C D A B C D A B C D
10 A B C D A B C D A B C D A B C D
AB
CD 00 01 11 10
00 A+B+C+D A+B+C+D A+B+C+D A+B+C+D
01 A+B+C+D A+B+C+D A+B+C+D A+B+C+D
11 A+B+C+D A+B+C+D A+B+C+D A+B+C+D
10 A+B+C+D A+B+C+D A+B+C+D A+B+C+D
CD
AB
POS Form
SOP Form

4. Min term distributions in K- map
CD 00 01 11 10
00
m0 m1 m3 m2
01
m4 m5 m7 m6
11
m12 m13 m15 m14
10
m8 m9 m11 m10
AB

5. Reduce the Expression
f=∑m(1,5,6,12,13,14)+d(2,4)
CD 00 01 11 10
00
m0
1
m1 m3
X
m2
01 X
m4
1
m5 m7
1
m6
11 1
m12
1
m13 m15
1
m14
10
m8 m9 m11 m10
AB
2
3
1
fmin=BC+A CD+BD

6. Problem
• Reduce the expression f=∑m(1,5,6,12,13,14)+d(2,4)
• The given expression in the POS form is f=∏M(0,3,7,8,9,10,11,15)∏d(2,4)
00 01 11 10
00 0
M0 M1
0
M3
X
M2
01 X
M4 M5
0
M7 M6
11
M12 M13
0
M15 M14
10 0
M8
0
M9
0
M11
0
M10
AB CD
F=(A+B)(C+D)(B+D)
2
3 1