# a question on Karno table transformation...

Discussion in 'Homework Help' started by transgalactic, May 2, 2008.

1. ### transgalactic Thread Starter Active Member

Apr 29, 2008
74
0
the question asks me to implement a certain function in a minimal way
using a MUX
i have constructed a Karno table
and found the minimal function which consists only from the variables a,c,d

now in the solution they constructed a table only from a,c,d varibles

http://s290.photobucket.com/albums/ll279/transgalactic/?action=view&current=IMG_8818.jpg

i am having trouble to understand it because for every value of a,c,d
we have two cells in the original karno table
so how they deside which value belongs to each cell in the a,c,d table
???????

2. ### Caveman Active Member

Apr 15, 2008
471
0
The picture is not really clear enough. Just show the original Karnaugh Map

Apr 29, 2008
74
0
4. ### Caveman Active Member

Apr 15, 2008
471
0
That would help, but I'll take a whack at it anyway.

I'm guessing the phi looking symbols are don't cares as opposed to zeros.
Look at the left two columns of the first table. These columns are only different by the b input. Notice that in every row, at most one column has a value, and the other is a don't care.
Now look at the right two columns. The same thing applies. This shows that the value of b doesn't matter, so just remove it.

5. ### transgalactic Thread Starter Active Member

Apr 29, 2008
74
0
but how they desided what value to put in each cell of the small table???

i cant see the guide line for which they carved out this small table ou of the original Ktable
because for every cell in the small table we have a representation of three variables
we in the original Ktable takes 2 cells(two values)

?????????

6. ### Caveman Active Member

Apr 15, 2008
471
0
First push, the two left columns together, then push the two right columns together. If they both have don't cares the result is don't care. If one has a value, but the other is don't care, then the value is the result.
Do you have that?

It will look like this.
___ 0 1
00| 0 1
01| 1 d
11| d 1
10| 0 0

So you have a 2 column, 4 row table. The columns are for the two possible values of a. The rows are the same as before.

Now you just rewrite the table using the 8 possible values.

7. ### transgalactic Thread Starter Active Member

Apr 29, 2008
74
0
i have understood your folding method
you have folded the columns of ab in order get rid of the "b" variable.

what hapeens when i have a '0' in one cell and '1' in the other

but your resolt table doesnt look like the minimal resolt table i was given.

i dont understand "rewrite the table using 8 possible values"

how to rewrite them?????
we already reached to size of our resolt table
but the values are not the same
the number of 1"s in your table are two in my table its 4
there is no dontcares in my table

what do i need to do??
????????

8. ### Caveman Active Member

Apr 15, 2008
471
0
Then you can't fold it down.

If you followed the step I said you should get the table I showed. If you don't understand that step, then stop. You must understand that first!

Now to get to the output table you were given, look at the resulting table, you will see there are 8 values
acd out
000 0
001 1
010 0
011 d
100 1
101 d
110 0
111 1

Now write a new Karnaugh map off of this, except put ac in the columns and d in the rows:
ac$\;\;$00 01 11 10
d$\,$0| 0$\;$ 0 $\;$ 0 $\;$ 1
$\;$ 1| 1$\,$ d $\,$ 1 $\,$ d

By replacing the don't cares in this with 1s, the result is even more simplified to the final result given.

9. ### transgalactic Thread Starter Active Member

Apr 29, 2008
74
0
ok.. 2 question on this:
1.you are correct that if we substitute all the don't cares with 1 we get my minimal table
but in the test i wont have the answer in front of me
and i will need to decide which value to put in each don't care
what is the reason behind putting '1' instead of don't cares
2.what should i do in case the folding fails because of 0 and 1 encounter