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

Karnaugh -Map , reduction of Boolean Expressions

  • 1.
    Minimization of SOPand POS Expressions using Karnaugh map Prepared by- Dr. Sutapa Mukherjee
  • 2.
    Different variable KMap 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 KMap 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 distributionsin 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) CD00 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 theexpression 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