Activity 2.2.1 Circuit Simpification:
Karnaugh Mapping
Write the simplified sum-of-products (SOP) logic expression for the K-Maps shown below
1. F1 =
2. F2 =
B
B
A
1
1
A
0
0
A
3. F3=
C
C
CD
CD
CD
CD
AB
0
1
AB
0
1
0
1
AB
0
0
AB
0
1
0
1
AB
1
1
AB
1
1
0
0
AB
1
0
AB
1
1
0
0
AB+ABC+ABC
CD+ACD+ACD
After transferring the truth table data into the K-Maps, write the simplified sum-of-products
(SOP) logic expression for the K-Maps shown below.
4. F4 =
Q
R
F4
0
0
0
0
1
0
1
0
1
1
1
1
R
R
Q
0
0
Q
1
1
Q
© 2014 Project Lead The Way, Inc.
Digital Electronics Activity 2.2.1 Circuit Simplification: Karnaugh Mapping – Page 1
5. F5 =
Q
R
S
F5
0
0
0
0
0
0
1
1
0
1
0
1
0
1
1
0
1
0
0
0
1
0
1
1
1
1
0
1
1
1
1
0
S
S
QR
0
1
QR
1
0
QR
1
0
QR
0
1
ST
ST
ST
ST
QR
0
0
0
0
QR
0
1
1
1
QR
0
1
1
0
QR
1
1
0
1
6. F6 =
Q
R
S
T
F6
0
0
0
0
0
0
0
0
1
0
0
0
1
0
0
0
0
1
1
0
0
1
0
0
0
0
1
0
1
1
0
1
1
0
1
0
1
1
1
1
1
0
0
0
1
1
0
0
1
1
1
0
1
0
1
1
0
1
1
0
1
1
0
0
0
1
1
0
1
1
1
1
1
0
0
1
1
1
1
1
QRST+1+QRST+QRST
© 2014 Project Lead The Way, Inc.
Digital Electronics Activity 2.2.1 Circuit Simplification: Karnaugh Mapping – Page 2
After labeling the K-Map and transferring the truth table data into it, write the simplified
sum-of-products (SOP) logic expression for the K-Maps shown below.
7. F7 =
W
X
F7
0
0
0
0
1
1
1
0
1
1
1
1
W
W
X
1
1
X
1
0
Y
Y
WX
1
1
WX
1
0
WX
1
1
WX
1
0
8. F8 =
W
X
Y
F8
0
0
0
1
0
0
1
1
0
1
0
1
0
1
1
0
1
0
0
1
1
0
1
0
1
1
0
1
1
1
1
1
© 2014 Project Lead The Way, Inc.
Digital Electronics Activity 2.2.1 Circuit Simplification: Karnaugh Mapping – Page 3
9. F9 =
W
X
Y
Z
F9
0
0
0
0
1
0
0
0
1
1
0
0
1
0
0
0
0
1
1
0
0
1
0
0
0
0
1
0
1
1
0
1
1
0
1
0
1
1
1
0
1
0
0
0
0
1
0
0
1
1
1
0
1
0
1
1
0
1
1
0
1
1
0
0
0
1
1
0
1
1
1
1
1
0
0
1
1
1
1
0
YZ
YZ
YZ
YZ
WX
1
1
0
0
WX
0
1
0
1
WX
0
1
0
0
WX
0
1
0
1
YZ+WXYZ+WXYZ+WXYZ
© 2014 Project Lead The Way, Inc.
Digital Electronics Activity 2.2.1 Circuit Simplification: Karnaugh Mapping – Page 4
Write the simplified sum-of-products (SOP) logic expression for the K-Maps shown below.
Be sure to take advantage of any don’t care conditions.
10. F10 =
L
L
K
X
1
K
1
0
KL+L
11. F11 =
M
M
KL
0
0
KL
1
X
KL
1
1
KL
X
0
KLM
12. F12 =
MN
MN
MN
MN
KL
0
1
1
0
KL
1
1
0
X
KL
0
X
0
X
KL
0
1
0
1
MN+KLMN+KLMN+KLMN
© 2014 Project Lead The Way, Inc.
Digital Electronics Activity 2.2.1 Circuit Simplification: Karnaugh Mapping – Page 5
Conclusion
1. Give three advantages of using K-mapping to simplify logic expressions over Boolean
algebra.
It is quicker, simpler, and graphic.
2. The three variable K-maps shown below can be completed with three groups of two.
The two groups shown (cells #1 & #3; cells #4 & #6) are required. The third group,
needed to cover the one in cell #2, could be cells #2 & #3 or cells #2 & #6.
Write the two possible logic expressions for the function F1.
3. These logic expressions are considered equal and equivalent but they do not look the
same. Explain why these two expressions can be considered equal and equivalent
even though they are not identical.
© 2014 Project Lead The Way, Inc.
Digital Electronics Activity 2.2.1 Circuit Simplification: Karnaugh Mapping – Page 6
Going Further – Optional
1. The following four variable K-Maps can be solved using the traditional method of
grouping the ones (Identify the 3 groups of 8).
F1  ?
CD
CD
CD
CD
AB
1
1
1
1
AB
1
0
0
1
AB
1
1
1
1
AB
1
1
1
1
2. Rather than taking this approach, let’s get creative and take advantage of the fact that
the K-Map contains only two zeros. Group these zeros and write the logic expression.
Since you grouped the zeroes, this is the logic expression for F1 .
3. Now apply DeMorgan’s Theorem to get the logic expression for F1 .
4. What is the advantage of taking this approach over the traditional approach of circling
the ones?
5. Are there any disadvantages?
© 2014 Project Lead The Way, Inc.
Digital Electronics Activity 2.2.1 Circuit Simplification: Karnaugh Mapping – Page 7