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

Answer : wx + vw'

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

 

do it like this:
[ table]header 1 | header two | header three
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

 

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

Answer : wx + vw'


what is your question?

 

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

 

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

So, what is your question

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

 

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'

 

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'

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...

Your logic seems sound and your process correct.

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...

 

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...

Your logic seems sound and your process correct.

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...

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.

 

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.

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

=ad'
were it applicable