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Simplifying and K-Map

  1. Sep 10, 2009 #1
    1. The problem statement, all variables and given/known data

    Okay, I have question I was pondering about for a while...

    If I had the equation below and wanted to transform it into a K-Map, is the only way I can do it to expand it completely? Are there any shortcuts?

    2. Relevant equations

    This is just a random equation out of my mind :

    (A+B+C)(A+B'+C)(A+B+C')(A'+B+C)(A'+B')

    So I been taught when handling K-Map if I get a SOP, I would make all of portions similar to include all 3 variables eg.(A+B+C) then convert them to binary, invert them and take their sum to map it into the K-Map as a short cut.

    Example for (A + B + C) I would take 111, invert it 000, the decimal will be 0, and then I map this as a 0(Maxterm or was it minterm @_@) into the 0 portion of the K-Map.

    But I was thinking is there anyway to convert the portions of my above equation to a (A+B+C) form easily or do I have to expand them to make it a POS then map it into a K-Map.
     
  2. jcsd
  3. Sep 10, 2009 #2

    LCKurtz

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    When you only have three variables A, B, and C, there are only the 8 states:

    ABC
    000
    001
    011
    010
    110
    111
    101
    100

    If any of your factors is 0 in the product of sums, the result is 0. And in any individual factor in any term is 1 the sum is 1. It only takes a minute or so to just plug them in to fill out the K-table.
     
  4. Sep 12, 2009 #3
    You are telling me to pluck in the individual values 1 by 1 in the equation than sort it out in a K-Map which would cost a lot of time in an exam...

    I am looking for shortcuts to save time like the one I posted above...

    I am asking if besides sorting it out manually and expanding the above equation to a POS than map it. Is there anyway to map the above into K-map quick or change the portions to include all 3 A, B, C using any techniques?

    Yeah, but i c ur point, it seems quickest in this case.
     
  5. Sep 12, 2009 #4

    LCKurtz

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    Well, if time was of the essence, I might do it this way for a product of sums. Here's your example:

    (A+B+C)(A+B'+C)(A+B+C')(A'+B+C)(A'+B')

    This will be 0 if and only if one of the factors is zero, and each factor will be zero only if each term is zero. So for each factor, note what makes it zero:

    (A+B+C) 0,0,0
    (A+B'+C) 0,1,0
    (A+B+C') 0,0,1
    (A'+B+C) 1.0.0
    (A'+B') 1,1,X (here it doesn't matter what X is)

    The above can be read off immediately from your formula. Put in 0's accordingly in your K map and 1's everywhere else. Should be fast enough for an exam.
     
  6. Sep 12, 2009 #5
    I see what you mean, but I can't map the above... see the way I been taugh a K-map can be divided into portions of numbers.

    0 1 3 2
    4 5 7 6
    12 13 15 16
    8 9 11 10

    So by adding up the binary numbers of the portions I can immediately know where to map it.. example (A'+B+C) 1.0.0 gives me 4... so it goes in 4.

    Yep, so I suppose from the equation I gave, doing it manually maybe the fastest way.
     
  7. Sep 12, 2009 #6

    LCKurtz

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    But you don't have to do the binary conversion to fill in your k-table. Don't you label the top and left side with the reflected Gray code, like this?:

    --- 00 01 11 10
    ---------------------
    00|
    01| 0
    11|
    10|

    Your example only has three bits, so you would only need two rows. But the point is that you can read the binary right off the table and put the 0's in. For example, I have put the 0 in the 100 spot; I don't have to think about the fact that is 4.
     
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