ECE 3110: Introduction to Digital
Systems
Simplifying Sum of Products
using Karnaugh Maps
Previous…
Simplifying SOP:
Draw K-map
Find prime implicants
Find distinguished 1-cell
Determine essential prime implicants if available
Select all essential prime implicants and the minimal set
of the remaining prime implicants that cover the
remaining 1’s.
2
Example 2
 - Combining (0,2)
Product term : X’Z’
- Combining (2,3)
Product term : X’Y
- Combining (3,7)
Product term :YZ
XY
X
00
Z
0
1
0
1
1
0
01
2
3
1
1
11
6
7
0
1
10
4
5
0
Z
0
 X’Z’, X’Y, and YZ are prime implicants
Y
 X’Z’, YZ are essential prime implicants
 X’Y is non-essential prime implicant (redundant) because all its
minterms are covered in the other essential prime implicants
 F= X’Z’+X’Y+YZ (complete sum)
 OR:
F = X’Z’+YZ ( the minimal sum of F )
3
Example 3
The prime implicants :
- Cells(0,1,2,3) : W’X’
- Cells(0,1,4,5) : W’Y’
- Cells(1,3,5,7) : W’Z
- Cells(7,15)
: XYZ
- Cells(14,15) : WXY
W
WX

00
YZ
00
01
0
1
3
1
1
01
4
5
7
11
1
11
12
0
13
0
15
 The essential prime implicants:
1
1
1
Y
- W’X’
2 1 6
14
10
1
- W’Y’
0
1
- WXY
 Cell 7 is not covered by any of the
X
essential prime implicants. Its covered by
two non-essential prime implicant. We choose
the one with the less number of variables which is W’Z
 F= W’X’+W’Y’+WXY+W’Z
11
10
8
9
11
0
0
Z
0
10
0
4
eclipse
Given two prime implicants P and Q in a
reduced map, P is said to eclipse Q if P
covers at least all the 1-cells covered by
Q. (P…Q).
Removing Q because P is at least as good
as Q.
5
Exercise
 Row
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
W X Y Z
0 0 0 0
0 0 0 1
0 0 1 0
0 0 1 1
0 1 0 0
0 1 0 1
0 1 1 0
0 1 1 1
1 0 0 0
1 0 0 1
1 0 1 0
1 0 1 1
1 1 0 0
1 1 0 1
1 1 1 0
1 1 1 1
F
1
1
1
0
1
1
1
0
1
1
0
0
1
1
1
0
W
WX
00
YZ
00
01
01
11
10
0
4
12
8
1
5
13
9
Z
11
Y
10
3
7
15
11
2
6
14
10
X
6
Exercise Solution:
 Essential prime implicants:
- cells (0,1,4,5,8,9,12,13)
The product term : Y’
- cells (2,6,0,4)
The product term : W’Z’
- Cells (4,6,12,14)
The product term : XZ’
 F=Y’+W’Z’+XZ’
W
WX
00
YZ
00
01
11
Y
10
0
1
3
2
1
1
0
1
01
4
5
7
6
1
1
0
1
11
12
1
13
1
15
0
14
1
10
8
9
11
1
1
Z
0
10
0
X
7
Another Example
F=  (0,2,3,4,5,7)
x, y , z
 The prime implicants:
1- (0,2) X’Z’
2- (0,4) Y’Z’
3- (2,3) X’Y
4- (3,7) YZ
5- (4,5) XY’
6 -(5,7) XZ
 No essential prime implicant!
XY
X
00
Z
0
1
0
1
1
0
01
2
3
11
6
1
7
1
0
1
10
4
5
1
Z
1
Y
 Two possible minimal sums :
1- Using the prime implicants 1,4,and 5 , F= X’Z’+YZ+XY’
2- Using the prime implicants 2,3,and 6 , F= Y’Z’+ X’Y+XZ
8
Yet another Example
WX
W
 - Cells (5,13,7,15) can be combined
00
01
11
YZ
to form an essential prime implicant.
0
4
12
W & Y change
00
1
0
0
X & Z remain constant, X=1,Z=1
1
5
13
- The product term : XZ
01
0
1
1
 - Cells (0,8,2,10) can be combined
3
7
15
to form an essential prime implicant.
11
0
1
1
W & Y change
Y
Z & X remain constant, X=0, Z=0
2
6
14
10
1
0
0
- The product term : X’Z’
 F= XZ + X’Z’
 Note that the corner cells (0,2),(0,8),(8,10),(2,10)
X
can be combined to form the implicants :
W’X’Z’ , X’Y’Z’, WX’Z’, X’YZ’ but, they are not prime implicants.
10
8
9
11
1
0
Z
0
10
1
9
Next…
Simplifying PoS
Read Chapter 4.4---4.7
10