simplify boolean expression using k' map, f(x,y,z) = xyz + xyz’ + xy’z + x’yz + x’yz’ + xy’z’ + x’y’z’

Question

f(x,y,z) = xyz + xyz’ + xy’z + x’yz + x’yz’ + xy’z’ + x’y’z’

here's my work. is it correct?

cuz I don't think my answer is correct:/

Answer 1

If you look at your K-map, your eventual answer $z'$ cannot be correct, because that would correspond to having all $1$'s at the bottom (corresponding to $z'$), but not having $1$'s in the $z$ row.

Your expression $x + y'z'+x'y$ is correct, but from that you should proceed:

$x + y'z'+x'y = x +y'z'+y=x+z'+y$

Note that by DeMorgan this is the same as $(x'zy')'$ ... which makes sense, since the function is only $0$ in the case of $x'zy'$. Indeed, when you have far more $1$'s than $0$'s in your K-map, it can be easier to focus on the $0$'s and generate an expression for those ... you just need to negate that term at the end.

Answer 2

All of the rectangles of this particular kmap can be expanded to contain 4 cells. An optimal kmap is made when all the zones chosen are as large as possible.

The bottom zone can be expanded to include all the bottom 4 cells.

The rightmost zone can be expanded to contain the left cells.

So it is like

$$\begin{array} {|c|c|c|c|} \hline \color{red}{\blacksquare} \color{green}{\blacksquare} & \color{green}{\blacksquare}& & \color{red}{\blacksquare} \\ \hline \color{red}{\blacksquare} \color{green}{\blacksquare} \color{purple}{\blacksquare} & \color{green}{\blacksquare} \color{purple}{\blacksquare} & \color{purple}{\blacksquare} & \color{purple}{\blacksquare} \color{red}{\blacksquare} \\ \hline \end{array}$$

Where you have the greenzone, redzone, and purple zone.

So the simplified expression is $\color{green}{x} + \color{red}{\text{something}} + \color{purple}{\text{something}}$.

The second line of your solution is correct but not optimal. The 3rd and 4th lines are wrong. If you get an optimal solution you should not be able to simplify it into a shorter sum-of-products.

Alternatively, you could see that the missing square makes the equation $(x'y'z)'$ and use Demorgan's to simplify that.