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Minimizing the SOP

shamieh

Active member
Sep 13, 2013
539
I used a KMAP and got this as my expression

[tex]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}[/tex]


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: [tex] f = x_{1}x_{0} + !y_{1}!y_{0} + x_{1}!y_{0} + x_{0}!y_{1} + x_{1}!y_{1}[/tex]

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


Thanks in advance
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,493
I used a KMAP and got this as my expression

[tex]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}[/tex]

...

Some how the answer is this: [tex] f = x_{1}x_{0} + !y_{1}!y_{0} + x_{1}!y_{0} + x_{0}!y_{1} + x_{1}!y_{1}[/tex]
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
Sep 13, 2013
539
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
Jan 30, 2012
2,493
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
Sep 13, 2013
539
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.