ISI2011-PCB-CS-6b

Question

For the function given by the Karnaugh map shown below, you can change at most one $1$ or one $0$ entry to a DON'T CARE. Determine what single change of this kind produces the simplest two-level AND-OR realization. Assume both uncomplemented and complemented inputs are available.

Comments

Since we need to find simplest AND-OR realization, we need to find the minimal SOP possible.

1. Making the bottom-left corner 0 as don't care, results in 2 sub-cubes of size 4, 1 sub-cube of size 2 and 1 sub-cube size of size 1

2. Making the 1st row third column 0 as don't care OR making the 2nd row fourth column 0 as don't care results in 1 sub-cube of size 4 and 3 sub-cubes of size 2

Considering sub-cubes size (max as easier to implement), Option 1 is answer.

---

Make 2 row 3 rd column as don't care

---

That will result in POS. Simplifying the function in terms of SOP will result in more min-terms than the no. of min-terms in Option 1

Assume wx as row & yz as column of K map, then f = (w+x) (y'+z) (w'+z) (w'+x'+z) and upon turning it into SOP, I got 12 min-terms.