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.
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.
14.
15.
16.
17.
18. • Use the rules of simplification and ‘ringing of adjacent cells’ in order
to make as many variables redundant.
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
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)
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
