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Given the three variable Karnaugh Map:

x\yz  00  01   11   10
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  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'

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    $\begingroup$ Hint: $y$ serves no purpose. Carefully consider the relationship between $x$, $z$, and the output. $\endgroup$
    – Ken
    Commented Jun 3, 2015 at 2:16
  • $\begingroup$ 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. $\endgroup$ Commented Jun 4, 2015 at 22:56

1 Answer 1

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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 $z$).

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