Simplifying a Boolean function from a Kernaugh Map

Question

Given the three variable Karnaugh Map:

x\yz  00  01   11   10
  \___________________
  0 | 0   1    1    0
  1 | 1   0    0    1

I am supposed to write a simplified expression for the Boolean function. Based on this KMap, I figured there should be 2 terms after simplified. I can't figure out how to simplify them though. Here is what I'm stuck at:

F(xyz) = x(yz)' + x'(yz) + (xy)'z + (xy)z'

Hint: $y$ serves no purpose. Carefully consider the relationship between $x$, $z$, and the output. — Ken, Jun 03 '15 at 02:16
F(xyz) = xy'z' + x'yz + x'y'z + xyz' is correct. You cannot put brackets around lows. (y'z') ≠ (yz)' Nothing to do with kmap answer. — StainlessSteelRat, Jun 04 '15 at 22:56

score 1 · Accepted Answer · answered Jun 03 '15 at 02:20

The products that are adjacent on the Karnaugh map are $(100,110)$ and $(001,011)$. In the first pair, $x$ and $z$ are set, but $y$ can take either value. So, the pair of products $xy'z' + xyz'$ can be "factored" into $xz'$.

Similarly, the other pair can be simplified into $x'z$.

So, we find $$ F(x,y,z) = x'z + xz' $$ (which is to say, $x$ XOR $z$).

Simplifying a Boolean function from a Kernaugh Map

1 Answers1