# Minimizing the SOP

#### shamieh

##### Active member
I used a KMAP and got this as my expression

$$f = !y_{1}!y_{0} + !x_{1}x_{0}!y_{1} + x_{1}!y_{1} + x_{1}!y_{0} + x_{1}x_{0}y_{1}$$

Is there any way I can minimize this? Maybe I'm just not seeing it. I thought the whole point of a KMAP was to minimize the expression?

Some how the answer is this: $$f = x_{1}x_{0} + !y_{1}!y_{0} + x_{1}!y_{0} + x_{0}!y_{1} + x_{1}!y_{1}$$

If anyone is interested I had to design a circuit with output f with 4 inputs. I'm supposed to be showing the simplest sum of product expression for f. My f row for my truth table was this:
1
0
0
0

1
1
0
0

1
1
1
0

1
1
1
1

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
I used a KMAP and got this as my expression

$$f = !y_{1}!y_{0} + !x_{1}x_{0}!y_{1} + x_{1}!y_{1} + x_{1}!y_{0} + x_{1}x_{0}y_{1}$$

...

Some how the answer is this: $$f = x_{1}x_{0} + !y_{1}!y_{0} + x_{1}!y_{0} + x_{0}!y_{1} + x_{1}!y_{1}$$
You can turn the three-variable minterms into two-variable ones, namely, remove $!x_1$ from $x_0!x_1!y_1$ and $y_1$ from $x_0x_1y_1$.

#### shamieh

##### Active member
You can turn the three-variable minterms into two-variable ones, namely, remove $!x_1$ from $x_0!x_1!y_1$ and $y_1$ from $x_0x_1y_1$.

Well I can't combine them because they don't differ by two variables correct? So what minimization "tool" should I use? Should I factor something? I mean how do you just "get rid of them". See what I'm saying? What method I should I be using? Sorry if I sound ignorant.

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
When you remove a variable from a three-variable minterm, its representation in the Karnaugh map grows from 2 to 4 cells. However, if the added two cells are already covered by other minterms, then the Boolean function does not change.

In this case, the minterm $x_0y_1y_1$ represents two cells in the middle of column 3 ($x_0=x_1=1$). When you remove $y_1$, the result is the complete column 3. But the top cell of column 3 is already covered by $!y_0!y_1$, and the bottom cell of column 3 is covered by $x_1!y_1$. So removing $y_1$ from $x_0y_1y_1$ does not change the function. A similar thing happens with turning $x_0!x_1!y_1$ into $x_0!y_1$.

When reading off a minimal formula from a Karnaugh map, the temptation is always to break the cells corresponding to 1 into disjoint regions. But this results in smaller regions and therefore larger minterms. Instead, one must make regions as large as possible by using the fact that overlap is allowed.

#### shamieh

##### Active member
Awesome explanation! Thanks so much. So it looks like I probably missed a overlapping grouping I could of put together then - thus getting not exactly the complete minimization.