k-maps

1. K-maps
BY UNSA SHAKIR

2. Example
• Minimize the following Boolean function using sum of
products (SOP):
• f(a,b,c,d) = m(3,7,11,12,13,14,15)
abcd
3 0011 a`b`cd
7 0111 a`bcd
11 1011 ab`cd
12 1100 abc`d`
13 1101 abc`d
14 1110 abcd`
15 1111 abcd

3. Example
f(a,b,c,d) = m(3,7,11,12,13,14,15)
=a`b`cd + a`bcd + ab`cd + abc`d`+ abc`d + abcd` + abcd
=cd(a`b` + a`b + ab`) + ab(c`d` + c`d + cd` + cd )
=cd(a`[b` + b] + ab`) + ab(c`[d` + d] + c[d` + d])
=cd(a`[1] + ab`) + ab(c`[1] + c[1])
=ab+ab`cd + a`cd
=ab+cd(ab` + a`)
=ab+ cd(a + a`)(a`+b`)
= ab + a`cd + b`cd
= ab +cd(a` + b`)

4. Example
Maximize the following Boolean
function using product of sum
(POS):
f(a,b,c,d)=M(0,1,2,4,5,6,8,9,10)
=m(3,7,11,12,13,14,15)
=[(a+b+c+d)(a+b+c+d`)(a+b`+c`+d`)
(a`+b+c`+d`)(a`+b`+c+d)(a`+b`+c+d`)
(a`+b`+c`+d)(a`+b`+c`+d`)]

5. Karnaugh Maps(K-maps)
• Karnaugh maps -- A tool for representing Boolean
functions of up to six variables.
• K-maps are tables of rows and columns with entries
represent 1`s or 0`s of SOP and POS representations
respectively.

6. Karnaugh Maps(K-maps)
• An n-variable K-map has 2n cells with each cell
corresponding to an n-variable truth table
value.
• K-map cells are labeled with the corresponding
truth-table row.
• K-map cells are arranged such that adjacent
cells correspond to truth rows that differ in
only one bit position (logical adjacency).

7. Karnaugh Maps(K-maps)
• If mi is a minterm of f, then place a 1 in cell i of the
K-map.
• If Mi is a maxterm of f, then place a 0 in cell i.
• If di is a don’t care of f, then place a d or x in cell i.

8. Examples
• Two variable K-map f(A,B)=m(0,1,3)=A`B`+A`B+AB
1 0
1 1
A 0 1
B01

9. Three variablemap
• f(A,B,C) = m(0,3,5)= A`B`C`+A`BC+AB`C
A`B`
0 0
A`B
0 1
A B
1 1
A B`
1 0
1
A`B`C’ ` `
1
A`BC
1
AB`CC 1
C’ 0

10. Maxtermexample
0 0 0
0 0
A`B` A`B AB AB`
C`
C
(A+B) (A+B`) (A`+B`) (A`+B)
C
C`
f(A,B,C) = M(1,2,4,6,7)
=(A+B+C`)(A+B`+C)(A`+B+C) )(A`+B`+C) (A`+B`+C`)

11. Fourvariableexample
(a) Minterm form.(b) Maxtermform.
f(a,b,Q,G) = m(0,3,5,7,10,11,12,13,14,15) = M(1,2,4,6,8,9)

13. SimplificationofBooleanFunctions UsingK-maps
• K-map cells that are physically adjacent are also logically
adjacent. Also, cells on an edge of a K-map are logically
adjacent to cells on the opposite edge of the map.
• If two logically adjacent cells both contain logical 1s, the two
cells can be combined to eliminate the variable that has value
1 in one cell’s label and value 0 in the other.

18. • Use the rules of simplification and ‘ringing of adjacent cells’ in order
to make as many variables redundant.

20. Karnaugh Mapping - Another Worked
Z f A,B,C A.B B.C B.C A.B.C

21. Example
Simplifyf=A`BC`+ABC`+ABCusing;
(a) Sum of minterms. (b)Maxterms.
• Each cell of an n-variable K-map has n logically adjacent
cells.
C
AB
10
0 2 6 4
1 3 7 5
1
B
0
0C
C
AB
00 11 10 00 01 11
0 2 6 4
1 3 7 5
0
1C
A A
B
AB
BC
01
1 1
1 0
0 0
0
a-
b-
f(A,B,C) = AB + BC
f(A,B,C) = B(A + C)

22. ExampleSimplify
CD
AB
00 01 11
00
01
11
10
D
C
AB
10 CD 00 01 11 10
00
01
11
10
D
A A
C
B
(a)
B
(b)
CD
AB
00 01 11 10
00
01
11
10
D
A
C
CD
AB
00 01 11 10
0 4 12 8
1 5 13 9
3 7 15 11
2 6 14 10
00
01
11
10
D
A
C
B
(c)
B
(d)
1 1
1 1 1
1
1 1
1
0 4
1
12 8
1
1 5
1
13
1
9
3
1
7
1
15
1
11
2
1
6 14 10
1
0 4
1
12 8
1
1 5
1
13
1
9
3
1
7
1
15
1
11
2
1
6 14 10
1
0 4
1
12 8
1
1 5
1
13
1
9
3
1
7
1
15
1
11
2
1
6 14 10
1
f(A,B,C,D) = m(2,3,4,5,7,8,10,13,15)

23. ExampleRedundantselections
f(A,B,C,D) =m(0,5,7,8,10,12,14,15)
1 1
1
CD
AB
00 01 11 10
00
01
11
10
D
A
C
CD
AB
00 01 11 10
0 4 12 8
1 5 13 9
3 7 15 11
2 6 14 10
01
11
10
D
A
C
B
(a)
B
(b)
CD
AB
00 01 11 10
0 4 12 8
1 5 13 9
3 7 15 11
2 6 14 10
00
01
11
10
D
A
C
CD
AB
00 01 11 10
0 4 12 8
1 5 13 9
3 7 15 11
2 6 14 10
00
01
11
10
D
A
C
B
(c)
B
(d)
1
1 1
1
1
0
1
4 12
1
8
1
1 5
1
13 9
3 7
1
15
1
11
2 6 14
1
10
1
1
1 1
1
00 1 11
1
1
1 1
1
1 11
1

24. Example

25. Example

26. Example

27. Example

28. Condition A
Condition B
Condition C

29. The output of the circuit can be expressed as
f = AB + AC + BC

30. The output of the circuit can be expressed as
f = AB + AC + BC

31. The output of the circuit can be expressed as
f = AB + AC + BC

32. Finally, we get

33. •Minterms that may produce either 0 or 1 for the function.
• They are marked with an ´ in the K-map.
•This happens, for example, when we don’t input certain
minterms to the Boolean function.
• These don’t-care conditions can be used to provide further
simplification of the algebraic expression.
(Example) F = A`B`C`+A`BC` +ABC`
d=A`B`C +A`BC +AB`C
F = A` + BC`
Don’t-care condition

34. Example
Design the minimum-cost SOP and POS expression for the
function
f(x1, x2, x3, x4) = Σ m(4, 6, 8, 10, 11, 12, 15) + D(3, 5, 7, 9)

35. Let’s Use a K-Map
f(x1, x2, x3, x4) = Σ m(4, 6, 8, 10, 11, 12, 15) + D(3, 5, 7, 9)
x1
x2
x3
x4 00 01 11 10
00
01
11
10
x2
x4
x1
x3
m0
m1 m5
m4 m12
m13
m8
m9
m3
m2 m6
m7 m15
m14
m11
m10

36. f(x1, x2, x3, x4) = Σ m(4, 6, 8, 10, 11, 12, 15) + D(3, 5, 7, 9)
x1
x2
x3
x4 00 01 11 10
00
01
11
10
x2
x4
x1
x3
m0
m1 m5
m4 m12
m13
m8
m9
m3
m2 m6
m7 m15
m14
m11
m10
0
0
1
d
d
0
d
1
1
0
1
d
1
0
1
1

37. The SOP Expression

38. What about the POS Expression?
f(x1, x2, x3, x4) = Σ m(4, 6, 8, 10, 11, 12, 15) + D(3, 5, 7, 9)
x1
x2
x3
x4 00 01 11 10
00
01
11
10
x2
x4
x1
x3
m0
m1 m5
m4 m12
m13
m8
m9
m3
m2 m6
m7 m15
m14
m11
m10
0
0
1
d
d
0
d
1
1
0
1
d
1
0
1
1

39. The POS Expression

40. Example
Use K-maps to find the minimum-cost SOP and POS expression
for the function

41. Let’s map the expression to the K-Map
x1
x2
x3
x4 00 01 11 10
00
01
11
10
x2
x4
x1
x3
m0
m1 m5
m4 m12
m13
m8
m9
m3
m2 m6
m7 m15
m14
m11
m10

42. x1
x2
x3
x4 00 01 11 10
00
01
11
10
x2
x4
x1
x3
m0
m1 m5
m4 m12
m13
m8
m9
m3
m2 m6
m7 m15
m14
m11
m10
d
d
d

43. x1
x2
x3
x4 00 01 11 10
00
01
11
10
x2
x4
x1
x3
m0
m1 m5
m4 m12
m13
m8
m9
m3
m2 m6
m7 m15
m14
m11
m10
d
d
d

44. The SOP Expression

45. What about the POS Expression?
x1
x2
x3
x4 00 01 11 10
00
01
11
10
x2
x4
x1
x3
m0
m1 m5
m4 m12
m13
m8
m9
m3
m2 m6
m7 m15
m14
m11
m10
1
1
1
0
1
0
1
0
d
1
0
d
1
d
1
0

46. The POS Expression

47. Example
Derive the minimum-cost SOP expression for

48. First, expand the expression

49. Construct the K-Map for this expression
x1
x2
x3 00 01 11 10
0
1
(b) Karnaughmap
x2
x3
0 0
0 1
1 0
1 1
m0
m1
m3
m2
0
0
0
0
0 0
0 1
1 0
1 1
1
1
1
1
m4
m5
m7
m6
x1
(a) Truthtable
m0
m1 m3
m2 m6
m7
m4
m5
s1 s2 s3
s1 s2
s3

51. More Examples
f1(x, y, z) = ∑ m(2,3,5,7)
f2(x, y, z) = ∑ m (0,1,2,3,6)
 f1(x, y, z) = x’y + xz
f2(x, y, z) = x’+yz’
yz
X 00 01 11 10
0 1 1
1 1 1
1 1 1 1
1
x
yz

52. Example
Simplify the following Boolean function (A,B,C,D) =
∑m(0,1,2,4,5,7,8,9,10,12,13).
cd
ab
111
11
111
111
g(A,B,C,D) = c’+b’d’+a’bd
111
11
111
111

53. Example
Simplify the function f(a,b,c,d)
whose K-map is shown at the right.
f = a’c’d+ab’+cd’+a’bc’
or
f = a’c’d+ab’+cd’+a’bd’
xx11
xx00
1011
1010
xx11
xx00
1011
1010
0 1 0 1
1 1 0 1
0 0 x x
1 1 x x
ab
cd
00
01
11
10
00 01 11 10

54. Another Example
Simplify the function g(a,b,c,d) whose
K-map is shown at right.
g = a’c’+ ab
or
g = a’c’+b’d
x 1 0 0
1 x 0 x
1 x x 1
0 x x 0
x 1 0 0
1 x 0 x
1 x x 1
0 x x 0
x 1 0 0
1 x 0 x
1 x x 1
0 x x 0
ab
cd