# Finding minimisation as the sum of 3 terms from K-Map

I was given the following Boolean function:

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

And from it, I was able to produce the following truth table and K-Map (which I assume both are correct):

However, now I am asked to find a minimisation as the sum of the three terms in the function F(x, y, z) given above. I'm really stuck here and unfortunately the lectures were of no help.

The problem you are having is that your assumption that the Karnough map you created is correct was false. You have placed a $$0$$ in the $$x'y'z'$$ box where there should in fact be a $$1$$.

Once you have corrected this, follow the procedure bellow to minimize the function.

The way to minimize the function, is to circle neighboring terms in the Karnaugh map which are true. By circling two terms which are next to another, in a function which is of three variables, you will produce a two variable term.

As an example, the leftmost blue circle I made contains the term $$y'z'$$, as it is true regardless of the truth of $$x$$.

Once you have circled every box that is true, simply OR the terms together to produce the final expression. The expression the question requires should only contain three terms, so try to create three circles. I had success with $$x'y+xz+y'z'$$, but other combinations exist.

Could you explain the logic behind F(x,y,z) = y'z' + xz + x'y a little more? How do they tie to the circling you just did? How did you decide that they were xz and x'y?

EDIT: Onur Ozbek asked a few question in the comments addressing how the circle placement is decided. I hope the following image clears things up.

If $$F(x,y,z)$$ is true outside the $$z$$ column, then it is true for $$z'$$. The same goes for the variables $$x$$ and $$y$$.

• Could you explain the logic behind F(x,y,z) = y'z' + xz + x'y a little more? How do they tie to the circling you just did? Nov 20, 2019 at 6:14
• Notice that F(x,y,z) is true whenever y'z' is true, which is the first circle, its also true whenever the second circle is true, and for the third. Also note that F(x,y,z) is not true for any other term in the Karnough map, other than the ones circled. This implies that F(x,y,z) is true for the first circle x'y, OR the second one, xz, OR the third y'z'. So the final expression will be F(x,y,z) = y'z' + xz + x'y. Nov 20, 2019 at 6:18
• Yea but your circles are between y'z, yz and yz'. How did you decide that they were xz and x'y? Nov 20, 2019 at 6:21
• I knew that the second circle was xz because it was in the x row, and z column. Maybe an image will help this explanation more. I'll add it to the answer shortly. Nov 20, 2019 at 6:28

Alternative solution:

$$\begin{array}{ccccc}&x'y'&x'y&xy&xy'\\z&0&\color{red}{1}&\color{red}{1}&\color{blue}{1}\\z'&\color{orange}{1}&\color{orange}{1}&0&\color{blue}{1}\end{array}$$

$$\color{orange}{x'z'}+\color{red}{yz}+\color{blue}{xy'}$$

• I like your formatting Nov 20, 2019 at 6:34