Karnaugh problems

Discussion in 'Homework Help' started by lusito92, Feb 12, 2013.

1. lusito92 Thread Starter New Member

Jan 29, 2013
25
0
 possible wx wx' w'x' w'x uv 1 0 1 1 uv' 1 0 0 0 u'v' 1 0 0 0 u'v 1 0 1 1

sorry I'm not sure how to do tables :/

2. tshuck Well-Known Member

Oct 18, 2012
3,531
675
do it like this:
row1 entry 1 | row1 entry 2 | row1 entry 3
row2 entry 1 | row2 entry 2 | row2 entry 3
[ /table]

...turns into:
 header 1 header two header three row1 entry 1 row1 entry 2 row1 entry 3 row2 entry 1 row2 entry 2 row2 entry 3

3. tshuck Well-Known Member

Oct 18, 2012
3,531
675
So, like this, right?
 wx wx' w'x' w'x uv 1 0 1 1 uv' 1 0 0 0 u'v' 1 0 0 0 u'v 1 0 1 1

4. lusito92 Thread Starter New Member

Jan 29, 2013
25
0
I need to simplify this : UVWX + UV'WX + U'V'WX + U'VWX + UVW'X' + UVW'X + U'VW'X' + U'VW'X by using karnaugh maps

5. tshuck Well-Known Member

Oct 18, 2012
3,531
675

You have what looks to be a karnaugh map above, no?....

6. lusito92 Thread Starter New Member

Jan 29, 2013
25
0
yeah but I dont know if is right? can you please tell me if I did right, and I got the right simplification
 wx wx' w'x' w'x uv 1 0 1 1 uv' 1 0 0 0 u'v' 1 0 0 0 u'v 1 0 1 1

Simplification : wx + vw'

7. tshuck Well-Known Member

Oct 18, 2012
3,531
675
They are typically written as:
 uv\wx 00 01 11 10 00 0 0 1 0 01 1 1 1 0 11 1 1 1 0 10 0 0 1 0

Simplification : wx + vw'
but as long as each header does not differ from a neighboring header by more than 1 bit, it will work...

As you can see, there is another way to group 4 1's, but then you have another, superfluous term. This would appear to be the reduced version...

You should also note that grouping the max-terms will be the same number of gates...

8. lusito92 Thread Starter New Member

Jan 29, 2013
25
0
So is correct? because I dont know if I can group 4 1's even if they are parallel and not square shape, like the 1's on red.

9. tshuck Well-Known Member

Oct 18, 2012
3,531
675
Yes, you can group them provided they are next to each other: line, square, across boundaries, and even at the corners...

you could even do it like this:
 ab\cd 00 01 11 10 00 1 0 0 1 01 0 0 0 0 11 0 0 0 0 10 1 0 0 1

= b'd'
or
 ab\cd 00 01 11 10 00 0 0 0 0 01 0 0 0 0 11 1 0 0 1 10 1 0 0 1