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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 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 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 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 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   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   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   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   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