Boolean Algebra and Simplification Techniques
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1
1
1
A B
A B
A B
A B
C
C
1
1
1
1
A B
A B
A B
A B
C
C
1
1
1
A B
A B
A B
A B
C
C
1
Y 2
Y 2-1
Y 1
Figure 6.19 Example 6.12.
The minimized expressions for Y 1 and Y 2 are as follows:
Y 1 = BBC + AAC + AABBC
(6.64)
Y 2 = AAB + AABBC + BBC
(6.65)
Example 6.13
Write the simplified Boolean expression given by the Karnaugh map shown in Fig. 6.20.
Solution
• The Karnaugh map is shown in Fig. 6.21.
• Consider the group of four 1s at the top left of the map. It yields a term AAC.
• Consider the group of four 1s, two on the extreme left and two on the extreme right. This group
yields a term AAD.
• The third group of two 1s is in the third row of the map. The third row corresponds to the intersection
of A and B, as is clear from the map. Therefore, this group yields a term ABC.
• The simplified Boolean expression is given by AAC + AAD + AABBC.
Figure 6.20 Example 6.13.