tdyj fyuk fy

1. Simplification of Boolean
functions
Tabular method

2. Tabular method
Step 1 − group the minterms of the function according to the ones
count in their binary representation in an ascending order.
Gi contains the minterms with i ones in their binary presentation

3. Minterm A B C D
0 0 0 0 0
1 0 0 0 1
3 0 0 1 1
7 0 1 1 1
8 1 0 0 0
9 1 0 0 1
11 1 0 1 1
15 1 1 1 1
Example1:

Group Minterms A B C D
G0 0 0 0 0 0
G1 1 0 0 0 1
8 1 0 0 0
G2 3 0 0 1 1
9 1 0 0 1
G3 7 0 1 1 1
11 1 0 1 1
G4 15 1 1 1 1

4. Step2:
Check the minterms in a group with only the minterms in the next
group for one bit difference
• G0 is checked with G1, G1 with G2, G2 with G3, and so on.
• The Different digit is represented with a -
• repeat

5. Group Minterms A B C D
G0 0 0 0 0 0
G1 1 0 0 0 1
8 1 0 0 0
G2 3 0 0 1 1
9 1 0 0 1
G3 7 0 1 1 1
11 1 0 1 1
G4 15 1 1 1 1

Group Minterms A B C D
G0,G1 0, 1 0 0 0 -
0, 8 - 0 0 0
G1,G2 1, 3 0 0 - 1
1, 9 - 0 0 1
8, 9 1 0 0 -
G2, G3 3, 7 0 - 1 1
3,11 - 0 1 1
9,11 1 0 - 1
G3,G4 7, 15 - 1 1 1
11, 15 1 - 1 1

6. Group Minterms A B C D
G0,G1 0, 1 0 0 0 -
0, 8 - 0 0 0
G1,G2 1, 3 0 0 - 1
1, 9 - 0 0 1
8, 9 1 0 0 -
G2, G3 3, 7 0 - 1 1
3,11 - 0 1 1
9,11 1 0 - 1
G3,G4 7, 15 - 1 1 1
11, 15 1 - 1 1
Group Minterms A B C D
G0,G1
G1, G2
0, 1, 8, 9 - 0 0 -
0, 8, 1, 9 - 0 0 -
G1,G2
G2, G3
1, 9, 3, 11 - 0 - 1
G2, G3
G3,G4
3,7, 11,15 - - 1 1
3,11,7,15 - - 1 1


7. Group Minterms A B C D
G0,G1
G1, G2
0, 1, 8, 9 - 0 0 -
G1,G2
G2, G3
1, 9, 3, 11 - 0 - 1
G2, G3
G3,G4
3,7, 11,15 - - 1 1
0 1 3 7 8 9 11 15
x x x x
x x x x
x x x x
Step3: identify essential prime implicants
And minimal covers
Minterms of function
𝑓 (𝐴,𝐵,𝐶,𝐷)=∑(0,1,3,7,8,9,11,15)
B’D
B’C’
CD
B’C’ is EPI for m0 , m8
CD is EPI for m7 , m15
F=B’C’+CD

8. Minterm A B C D
2 0 0 1 0
5 0 1 0 1
6 0 1 1 0
7 0 1 1 1
10 1 0 1 0
13 1 1 0 1
15 1 1 1 1
Example2:

Group Minterms A B C D
G1 2 0 0 1 0
G2 5 0 1 0 1
6 0 1 1 0
10 1 0 1 0
G3 7 0 1 1 1
13 1 1 0 1
G4 15 1 1 1 1

9. 
Group Minterms A B C D
G1,G2 2, 6 0 - 1 0
2, 10 - 0 1 0
G2,G3 5, 7 0 1 - 1
5, 13 - 1 0 1
6, 7 0 1 1 -
G3, G4 7, 15 - 1 1 1
13,15 1 1 - 1
Group Minterms A B C D
G1 2 0 0 1 0
G2 5 0 1 0 1
6 0 1 1 0
10 1 0 1 0
G3 7 0 1 1 1
13 1 1 0 1
G4 15 1 1 1 1

10. Group Minterms A B C D
G2,G3
G3, G4
5,7,13, 15 - 1 - 1

Group Minterms A B C D
G1,G2 2, 6 0 - 1 0
2, 10 - 0 1 0
G2,G3 5, 7 0 1 - 1
5, 13 - 1 0 1
6, 7 0 1 1 -
G3, G4 7, 15 - 1 1 1
13,15 1 1 - 1

11. 2 5 6 7 10 13 15
x x x x
Minterms of function
BD
BD is EPI for m5 , m7, m13, m15
𝑓 (𝐴,𝐵,𝐶,𝐷)=∑(2,5,6,7,10,13,15)
Group Minterms A B C D
G2,G3
G3, G4
5,7,13, 15 - 1 - 1

12. 2 5 6 7 10 13 15
x x
x x
x x
x x
x x
x x
x x
Step3: identify essential prime implicants
And all minimal covers
Minterms of function
A’BD
A’CD’
ABD
B’CD’ is EPI for m10
𝑓 (𝐴,𝐵,𝐶,𝐷)=∑(2,5,6,7,10,13,15)
Group Minterms A B C D
G1,G2 2, 6 0 - 1 0
2, 10 - 0 1 0
G2,G3 5, 7 0 1 - 1
5, 13 - 1 0 1
6, 7 0 1 1 -
G3, G4 7, 15 - 1 1 1
13,15 1 1 - 1
B’CD’
A’BC
BCD
BC’D

13. 2 5 6 7 10 13 15
x x
x x
x x
x x
x x
x x
x x
Step3: identify essential prime implicants
And all minimal covers
Minterms of function
A’BD
A’CD’
ABD
A’CD’ is PI for m6
A’BC is PI for m6
𝑓 (𝐴,𝐵,𝐶,𝐷)=∑(2,5,6,7,10,13,15)
Group Minterms A B C D
G1,G2 2, 6 0 - 1 0
2, 10 - 0 1 0
G2,G3 5, 7 0 1 - 1
5, 13 - 1 0 1
6, 7 0 1 1 -
G3, G4 7, 15 - 1 1 1
13,15 1 1 - 1
B’CD’
A’BC
BCD
BC’D

14. Find all the minimal covers
BD is EPI for m5 , m7, m13, m15
B’CD’ is EPI for m10
A’CD’ is PI for m6
A’BC is PI for m6
F(A,B,C,D)=BD+B’CD’+A’CD’
OR
F(A,B,C,D)=BD+B’CD’+A’BC