Problem 13

One can also construct a product-of-sums expression from a Karnaugh map. We begin by finding rectangles of the types that form implicants, but with all points 0, instead of all points 1. Call such a rectangle an anti-implicant. We can construct for each anti-implicant a sum of literals that is 1 on all points but those of the anti-implicant. For each variable x, this sum has literal x if the anti-implicant includes only points for which x = 0, and it has literal x if the anti-implicant has only points for which x = 1. Otherwise, the sum does not have a literal involving x. Find all the prime anti-implicants for your Karnaugh maps of Problem 10.

Answer: output(p,q,r,s) = p's + q'r + qs' + pr'

output(p, q, r, s) = p'q' + p'r' + p's' + q'r' + q's' + r's'

output(p, q, r, s) = p'q'r's + p'q'rs' + p'qr's' + qrs + pq'r's' + prs + pqs + pqr

output (p,q,r,s) = p' + q' + r' + s

e. answer: output(p, q, r, s) = p' + q'r'

Karnaugh Map output rs 00 01 11 10 2,900 111 01 11 11 111 101 10 1 1 1 1 Karnaugh Map output rs 00 01 11 10 p,9 00 1 1 1 1 _011101 11|1|ooo 10 1 1 0 1 Karnaugh Map output ris 00 01_11 10 'p,q 00 0101 01 10 10 11 0 1 1 1 10 1 0 1 0 Karnaugh Map output ris 00 01 11 10 p,q 00| 1|1|1|1 o1|1|1|1|1 111110 101111 Karnaugh Map output ris 00 01 11 10 p,q 00 1 1 1 1 011 111 11 0 0 0 0 10 1 1 0 0