Karnaugh problems

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

  1. lusito92

    Thread Starter New Member

    Jan 29, 2013
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    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



    Answer : wx + vw'

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

    Well-Known Member

    Oct 18, 2012
    3,531
    675
    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
     
  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



    Answer : wx + vw'


    what is your question?
     
  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
    So, what is your question:confused:

    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...

    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...
     
  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

    =ad'
    were it applicable
     
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