The document describes the Karnaugh map method for minimizing Boolean functions with various numbers of variables. It includes: - Introduction to K-maps and their use for simplifying logic functions - Construction of 2, 3, and 4 variable K-maps and the relationship between variable assignments and map cells - Examples of simplifying 2-variable and 3-variable logic functions using K-maps - Rules for grouping cells in K-maps to minimize logic functions

1. G H PATEL COLLEGE OF ENGINEERING AND TECHNOLOGY
DEPARTMENT OF INFORMATION TECHNOLOGY
Subject : 2131004 (Digital Electronics)
K-map Method
Preparad By:
Harekrushna Patel (130110116035)

2. Contents
• Introduction
• Two variable maps
• Three variable maps
• Four variable maps
• Five variable maps
• Six variable maps

3. Introduction
• The map method provides a simple straight
forward procedure for minimizing Boolean
functions.
• This method may be regarded either as a
pictorial form of a truth table or as an
extension of the Venn diagram.
• The map method, first proposed by Veitch (1)
and slightly modify by Karnaugh (2), is also
known as the ‘Veitch diagram’ or the
‘Karnaugh map’.

4. Cont.
Minterm
• Standard Product Term
• For n – variable function → 2n minterm
• Sum of all minterms = 1 i.e. Σmi = 1

5. Cont.
Maxterm
• Standard Sum Term
• For n – variable function → 2n maxterm
• Product of all maxterms = 1 i.e. ΠMj = 1

6. Cont.
• Forms of Boolean function:
– Sum of Product(SOP) form
– Product of Sum(POS) form

7. Cont.
• SOP Form:
– AND - OR Logic or NAND - NAND Logic

8. Cont.
• POS Form:
– OR - AND Logic or NOR - NOR Logic

9. Rules
• No zeros allowed.
• No diagonals.
• Only power of 2 number of cells in each
group.
• Groups should be as large as possible.
• Every 1 must be in at least one group.
• Overlapping allowed.
• Wrap around allowed.
• Fewest number of groups possible.

10. Two variable K-map
• There are four minterms for two variables;
hence the map consists of four squares, one
for each minterm.
• The 0’s and 1’s marked for each row and each
column designate the values of variables x and
y, respectively.
mo m1
m2 m3

11. Cont.
mo m1
m2 m3

12. Cont.
mo m1
m2 m3
x
y
• Take two variables x and y

13. Cont.
y’ y
0 1
mo m1
m2 m3
x
y
0
1
X’
X
• Relation between squares & two variables

14. Cont.
y’ y
0 1
x’y’
x
y
0
1
X’
X
• Relation between squares & two variables

15. Cont.
y’ y
0 1
x’y’ x’y
x
y
0
1
X’
X
• Relation between squares & two variables

16. Cont.
y’ y
0 1
x’y’ x’y
xy’
x
y
0
1
X’
X
• Relation between squares & two variables

17. Cont.
y’ y
0 1
x’y’ x’y
xy’ xy
x
y
0
1
X’
X
• Relation between squares & two variables

18. Example
• Simplify following two Boolean functions:
– F1 = xy
– F2 = x+y

19. Cont.
mo m1
m2 m3
x
y
0
1
• F1 = xy……????
0 1
X’
X
y’ y

20. Cont.
0 0
0 1
x
• F1 = xy
y
0
1
0 1
X’
X
y’ y

21. Cont.
y’ y
0 1
mo m1
m2 m3
x
y
0
1
X’
X
• F2 = x + y……????

22. Cont.
y’ y
0 1
0 1
1 1
x
y
0
1
X’
X
• F2 = x + y = x’y + xy’ + xy = m1 + m2 + m3

23. Three variable K-map
• There eight minterms for three binary
variables. Therefore, a map consists of eight
squares.
m0 m1 m3 m2
m4 m5 m7 m6

24. Cont.
m0 m1 m3 m2
m4 m5 m7 m6

25. Cont.
m0 m1 m3 m2
m4 m5 m7 m6
x
yz
• Take three variables x, y and z

26. Cont.
x
yz
0
1
y’z’ y’z y z y z’
00 01 11 10
x’
x
m0 m1 m3 m2
m4 m5 m7 m6
• Relation between squares & three variables

27. Cont.
x
yz
y’z’ y’z y z y z’
x’ 0
x’y’z’
1
00 01 11 10
x
• Relation between squares & three variables

28. Cont.
x
yz
y’z’ y’z y z y z’
x’ 0
x’y’z’ x’y’z
1
00 01 11 10
x
• Relation between squares & three variables

29. Cont.
x
yz
y’z’ y’z y z y z’
x’ 0
x’y’z’ x’y’z x’yz
1
00 01 11 10
x
• Relation between squares & three variables

30. Cont.
x
yz
y’z’ y’z y z y z’
x’ 0
x’y’z’ x’y’z x’yz x’yz’
1
00 01 11 10
x
• Relation between squares & three variables

31. Cont.
x
yz
y’z’ y’z y z y z’
x’ 0
x’y’z’ x’y’z x’yz x’yz’
1
00 01 11 10
xy’z’
x
• Relation between squares & three variables

32. Cont.
x
yz
y’z’ y’z y z y z’
x’ 0
x’y’z’ x’y’z x’yz x’yz’
1
00 01 11 10
xy’z’ xy’z
x
• Relation between squares & three variables

33. Cont.
x
yz
y’z’ y’z y z y z’
x’ 0
x’y’z’ x’y’z x’yz x’yz’
1
00 01 11 10
xy’z’ xy’z xyz
x
• Relation between squares & three variables

34. Cont.
x
yz
0
x 1
xy’z’ xy’z xyz xyz’
y’z’ y’z y z y z’
00 01 11 10
x’
x’y’z’ x’y’z x’yz x’yz’
• Relation between squares & three variables

35. Example
• Simplify the Boolean function:
– F = x’yz + xy’z’ + xyz + xyz’
• Ans.:
– x’yz = m3
– xy’z’ = m4
– xyz = m7
– xyz’ = m6

36. Cont.
x
yz
0
1
y’z’ y’z y z y z’
00 01 11 10
x’
x
m0 m1 m3 m2
m4 m5 m7 m6
• F = x’yz + x’yz’ + xy’z’ + xy’z

37. Cont.
x
yz
0
1
y’z’ y’z y z y z’
00 01 11 10
x’
x
0 0 1 0
1 0 1 1
• F = x’yz + x’yz’ + xy’z’ + xy’z

38. Cont.
x
yz
0
1
y’z’ y’z y z y z’
00 01 11 10
x’
x
0 0 1 0
1 0 1 1
• Final Ans.
F = yz + xz’

39. Four Variable K-map
• There sixteen minterms for four binary
variables. Therefore, a map consists of sixteen
squares.
m0 m1 m3 m2
m4 m5 m7 m6
m12 m13 m15 m14
m8 m9 m11 m10

40. Cont.
C’D’ C’D C D C D’
00 01 11 10
m0 m1 m3 m2
m4 m5 m7 m6
m12 m13 m15 m14
m8 m9 m11 m10
00
01
11
10
AB
A’B’
A’B
A B
A B’
CD
• Take four variables A,B,C and D

41. Cont.
C’D’ C’D C D C D’
00 01 11 10
A’B’C’D’
00
01
11
10
AB
A’B’
A’B
A B
A B’
CD
• Relation between squares & four variables

42. Cont.
C’D’ C’D C D C D’
00 01 11 10
A’B’C’D’ A’B’C’D
00
01
11
10
AB
A’B’
A’B
A B
A B’
CD
• Relation between squares & four variables

43. Cont.
C’D’ C’D C D C D’
00 01 11 10
A’B’C’D’ A’B’C’D A’B’CD
00
01
11
10
AB
A’B’
A’B
A B
A B’
CD
• Relation between squares & four variables

44. Cont.
C’D’ C’D C D C D’
00 01 11 10
A’B’C’D’ A’B’C’D A’B’CD A’B’CD’
00
01
11
10
AB
A’B’
A’B
A B
A B’
CD
• Relation between squares & four variables

45. Cont.
C’D’ C’D C D C D’
00 01 11 10
A’B’C’D’ A’B’C’D A’B’CD A’B’CD’
A’BC’D’
00
01
11
10
AB
A’B’
A’B
A B
A B’
CD
• Relation between squares & four variables

46. Cont.
C’D’ C’D C D C D’
00 01 11 10
A’B’C’D’ A’B’C’D A’B’CD A’B’CD’
A’BC’D’ A’BC’D
00
01
11
10
AB
A’B’
A’B
A B
A B’
CD
• Relation between squares & four variables

47. Cont.
C’D’ C’D C D C D’
00 01 11 10
A’B’C’D’ A’B’C’D A’B’CD A’B’CD’
A’BC’D’ A’BC’D A’BCD
00
01
11
10
AB
A’B’
A’B
A B
A B’
CD
• Relation between squares & four variables

48. Cont.
C’D’ C’D C D C D’
00 01 11 10
A’B’C’D’ A’B’C’D A’B’CD A’B’CD’
A’BC’D’ A’BC’D A’BCD A’BCD’
00
01
11
10
AB
A’B’
A’B
A B
A B’
CD
• Relation between squares & four variables

49. Cont.
C’D’ C’D C D C D’
00 01 11 10
A’B’C’D’ A’B’C’D A’B’CD A’B’CD’
A’BC’D’ A’BC’D A’BCD A’BCD’
ABC’D’
00
01
11
10
AB
A’B’
A’B
A B
A B’
CD
• Relation between squares & four variables

50. Cont.
C’D’ C’D C D C D’
00 01 11 10
A’B’C’D’ A’B’C’D A’B’CD A’B’CD’
A’BC’D’ A’BC’D A’BCD A’BCD’
ABC’D’ ABC’D
00
01
11
10
AB
A’B’
A’B
A B
A B’
CD
• Relation between squares & four variables

51. Cont.
C’D’ C’D C D C D’
00 01 11 10
A’B’C’D’ A’B’C’D A’B’CD A’B’CD’
A’BC’D’ A’BC’D A’BCD A’BCD’
ABC’D’ ABC’D ABCD
00
01
11
10
AB
A’B’
A’B
A B
A B’
CD
• Relation between squares & four variables

52. Cont.
C’D’ C’D C D C D’
00 01 11 10
A’B’C’D’ A’B’C’D A’B’CD A’B’CD’
A’BC’D’ A’BC’D A’BCD A’BCD’
ABC’D’ ABC’D ABCD ABCD’
00
01
11
10
AB
A’B’
A’B
A B
A B’
CD
• Relation between squares & four variables

53. Cont.
C’D’ C’D C D C D’
00 01 11 10
A’B’C’D’ A’B’C’D A’B’CD A’B’CD’
A’BC’D’ A’BC’D A’BCD A’BCD’
ABC’D’ ABC’D ABCD ABCD’
AB’C’D’
00
01
11
10
AB
A’B’
A’B
A B
A B’
CD
• Relation between squares & four variables

54. Cont.
C’D’ C’D C D C D’
00 01 11 10
A’B’C’D’ A’B’C’D A’B’CD A’B’CD’
A’BC’D’ A’BC’D A’BCD A’BCD’
ABC’D’ ABC’D ABCD ABCD’
AB’C’D’ AB’C’D
00
01
11
10
AB
A’B’
A’B
A B
A B’
CD
• Relation between squares & four variables

55. Cont.
C’D’ C’D C D C D’
00 01 11 10
A’B’C’D’ A’B’C’D A’B’CD A’B’CD’
A’BC’D’ A’BC’D A’BCD A’BCD’
ABC’D’ ABC’D ABCD ABCD’
AB’C’D’ AB’C’D AB’CD
00
01
11
10
AB
A’B’
A’B
A B
A B’
CD
• Relation between squares & four variables

56. Cont.
C’D’ C’D C D C D’
00 01 11 10
A’B’C’D’ A’B’C’D A’B’CD A’B’CD’
A’BC’D’ A’BC’D A’BCD A’BCD’
ABC’D’ ABC’D ABCD ABCD’
AB’C’D’ AB’C’D AB’CD AB’CD’
00
01
11
10
AB
A’B’
A’B
A B
A B’
CD
• Relation between squares & four variables

57. Example
• Simplify the Boolean function:
– F(w, x, y, z) = Σ(1,5,12,13)

58. Cont.
m0 m1 m3 m2
m4 m5 m7 m6
m12 m13 m15 m14
m8 m9 m11 m10
W
Z
X
Y
00 01 11 10
00
01
11
10
F(w, x, y, z) = Σ(1,5,12,13)

59. Cont.
0 1 0 0
0 1 0 0
1 1 0 0
0 0 0 0
W
Z
X
Y
00 01 11 10
00
01
11
10
F(w, x, y, z) = Σ(1,5,12,13)
Put 1 in place of
m1, m5, m12, m13

60. Cont.
0 1 0 0
0 1 0 0
1 1 0 0
0 0 0 0
W
Z
X
Y
00 01 11 10
00
01
11
10
F(w, x, y, z) = Σ(1,5,12,13)
Put 1 in place of
m1, m5, m12, m13
Making pairs

61. Cont.
0 1 0 0
0 1 0 0
1 1 0 0
0 0 0 0
W
Z
X
Y
00 01 11 10
00
01
11
10
F(w, x, y, z) = Σ(1,5,12,13)
Put 1 in place of
m1, m5, m12, m13
Making pairs
Hence the simplified
Expression is
F = WY’Z + W’Y’Z

62. Five variable K-map
• There thirty two minterms for five binary
variables. Therefore, a map consists of thirty
two squares.
m16 m17 m19 m18
m20 m21 m23 m22
M28 m29 M31 m30
m24 m25 m27 m26
m0 m1 m3 m2
m4 m5 m7 m6
m12 m13 m15 m14
m8 m9 m11 m10

63. Cont.
000 001 011 010 110 111 101 100
m16 m17 m19 m18
m20 m21 m23 m22
m28 m29 m31 m30
m24 m25 m27 m26
E
m0 m1 m3 m2
m4 m5 m7 m6
m12 m13 m15 m14
m8 m9 m11 m10
CD
AB
00
01
11
10
• Relation between squares & five variables

64. Cont.
• Example:
– Design a circuit of 5 input variables that generates
output 1 if and only if the number of 1’s in the
input is prime (i.e., 2, 3 or 5).
• Ans.:
– The minterms can easily be found from Karnaugh
Map where addresses of 2,3 or 5 numbers of 1.

65. Cont.

66. Cont.

67. Cont.
• Hence the simplified expression becomes
BC’D’E + A’BC’D + AC’DE’ + AB’C’D + A’B’CE + A’CDE’
+ A’BCD + AB’CD’ + ABD’E’ + AB’DE’ + A’B’DE +
ABCDE

68. 6 variable K-map
• A 6-variable K-Map will have 26 = 64 cells. A
function F which has maximum decimal value
of 63, can be defined and simplified by a 6-
variable Karnaugh Map.

69. Cont.

70. Cont.
• Boolean table for 6 variables is quite big, so
we have shown only values, where there is a
noticeable change in values which will help us
to draw the K-Map.
• A = 0 for decimal values 0 to 31 and A = 1 for
31 to 63.
• B = 0 for decimal values 0 to 15 and 32 to 47.
B = 1 for decimal values 16 to 31 and 48 to 63.

71. Cont.
No. A B C D E F Minterm
m0 0 0 0 0 0 0 A’B’C’D’E’F’
m15 0 0 1 1 1 1 A’B’CDEF
m16 0 1 0 0 0 0 A’BC’D’E’F’
m31 0 1 1 1 1 1 A’BCDEF
m32 1 0 0 0 0 0 AB’C’D’E’F’
m47 1 0 1 1 1 1 AB’CDEF
m48 1 1 0 0 0 0 ABC’D’E’F’
m63 1 1 1 1 1 1 ABCDEF

72. Cont.
• Example:
– F = Σ (0, 2, 4, 8, 10, 13, 15, 16, 18, 20, 23, 24, 26,
32, 34, 40, 41, 42, 45, 47, 48, 50, 56, 57, 58, 60,
61)
• Ans.:
– Since, the biggest number is 61, we need to have
6 variables to define this function.

73. F = Σ (0, 2, 4, 8, 10, 13, 15, 16, 18, 20, 23, 24, 26, 32, 34, 40, 41, 42, 45, 47,
48, 50, 56, 57, 58, 60, 61)

74. Cont.
• Hence the simplified expression becomes
F = D’F’ + ACE’F + B’CDF + A’C'E’F’ + ABCE’ +
A’BC’DEF

75. Cont.
• Example:
– F = Σ (0, 1, 2, 3, 4, 5, 8, 9, 12, 13, 16, 17, 18, 19,
24, 25, 36, 37, 38, 39, 52, 53, 60, 61)
• Ans.:
– Since, the biggest number is 61, we need to have
6 variables to define this function.

76. Cont.

77. Cont.
• Hence the simplified expression becomes
F = A’B'E’ + A’C'D’ + A’D'E’ + AB’C'D + ABCE’

78. THANK YOU...
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