I'd like to simplify this boolean function: f(a,b,c,d)=∑(1,3,5,8,9,11,15) to its minimal SOP and POS forms. My solution is: SOP: A'·B'·C'·D + A'·B'·C·D + A'·B·C'·D + A·B'·C'·D' + A·B'·C'·D' + A·B'·C·D + A·B·C·D POS: (A+B+C+D)·(A+B+C'+D')·(A+B'+C+D')·(A'+B+C+D)·(A'+B+C+D')·(A'+B+C'+D')·(A'+B+C'+D') Is it right? Is there more to do?
834 2 2 gold badges 14 14 silver badges 36 36 bronze badges asked Nov 23, 2016 at 6:13 Nani Bhavani Nani Bhavani 15 1 1 silver badge 4 4 bronze badgesThe two expressions do not correspond to each other. For instance, the first term in (1) is A'B'C'D and in (2) A+B+C+D . It should be A'B'C'D' or A+B+C+D' otherwise.
Commented Nov 27, 2016 at 16:19The minimal SOP (sum of products) and the minimal POS (product of sums) of the given boolean function are depicted in these two Karnaugh maps.
Each of the necessary terms corresponds by color with the graphic representation.
Using the Karnaugh map for only four variables is really quick, but I could also use the Quine–McCluskey algorithm or rules of Boolean algebra for the minterms and maxterms in the truth table:
index A B C D output minterms maxterms -------+---------+--------+-------------+-------------------- 0 0 0 0 0 0 ~ M0: (a+b+c+d) 1 0 0 0 1 1 ~ m1: a'·b'·c'·d 2 0 0 1 0 0 ~ M2: (a+b+c'+d) 3 0 0 1 1 1 ~ m3: a'·b'·c·d 4 0 1 0 0 0 ~ M4: (a+b'+c+d) 5 0 1 0 1 1 ~ m5: a'·b·c'·d 6 0 1 1 0 0 ~ M6: (a+b'+c'+d) 7 0 1 1 1 0 ~ M7: (a+b'+c'+d') 8 1 0 0 0 1 ~ m8: a·b'·c'·d' 9 1 0 0 1 1 ~ m9: a·b'·c'·d 10 1 0 1 0 0 ~ M10: (a'+b+c'+d) 11 1 0 1 1 1 ~ m11: a·b'·c·d 12 1 1 0 0 0 ~ M12: (a'+b'+c+d) 13 1 1 0 1 0 ~ M13: (a'+b'+c+d') 14 1 1 1 0 0 ~ M14: (a'+b'+c'+d) 15 1 1 1 1 1 ~ m15: a·b·c·d
You can also check the correctness of your future solutions by using wolfram alpha.
