5 variable karnaugh-map, witch variable do you take "out"?

Discussion in 'Homework Help' started by JacobK, Oct 9, 2016.

  1. JacobK

    Thread Starter New Member

    Oct 9, 2016
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    0
    Hi,
    So I have a truth table with a function x and all combinations not in the truth table is don't care and i should minimize the function x.
    abefg x
    11110 0
    01000 0
    11101 0
    11001 0
    01011 1
    10011 1
    10111 1
    11000 1
    11111 0
    11011 0

    (If my description is unclear please look at the pdf where i scanned the task)
    I do a 5 variable karnuagh map that have this variables as columns and rows then i fill in the karnaugh map, you can se my rows and columns below:
    g = 0 g = 1
    ab\ef ab\ef
    After i fill in the map and group them i get x = b' + a'g + ae'g'

    In the solution they do the 5 variable karnuagh map like this:
    a = 0 a = 1
    be\fg be\fg

    They get x = b' + af'g' + a'f

    Even if i have different columns and rows should I not get the same answer?

    My question is is there a specific rule for what the columns and the rows should be? Why do they take out a?
    In the 4 variable k-maps its often a natural rows and columns like ab cd.

    Thanks for all help!
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    There are often multiple solutions to a minimization problem, particularly if you omit consensus terms.
     
  3. MrChips

    Moderator

    Oct 2, 2009
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    To create a 5-function Karnaugh map, I would draw two 4-function Karnaugh maps, one map for /E and a second for E.
     
  4. WBahn

    Moderator

    Mar 31, 2012
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    That's what he's doing.

    He chose to do it with 'g' as the variable that distinguished the two 4-variable maps while the solution chose to use 'a' (for whatever reason, you've chosen 'e'. He didn't get the same expression and he's asking if that is because of which variable was chosen and, if so, how to decide which variable to use. The answer is that it shouldn't matter and, if he did it correctly, the two expressions should be equivalent even though that might not be obvious and might take some manipulating to prove that they are.
     
  5. MrChips

    Moderator

    Oct 2, 2009
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    As you said, it doesn't matter which variable you single out, the results will be the same.
    The reason for my choosing variable E is because it is the MSB. Map for /E will be for binary values 0-15 and the map for E would be 16-31.

    If I had singled out variable A (LSB), then the map for /A would be all even numbers and map for A would be all odd numbers.

    Six of one and half a dozen of the other, same difference.
     
  6. WBahn

    Moderator

    Mar 31, 2012
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    How did you determine that, of variables {a, b, e, f, g} that E is the MSB? (or that A is the LSB, for that matter)?

    Leaving that aside, I completely agree that the (correct) results will be equivalent (as far as the patterns that we care about), but they may not look the same. I think that is what is tripping up the TS.

    The two results offered:

    x = b' + a'g + ae'g' (The TS's solution)
    x = b' + af'g' + a'f (The official solution)

    appear quite different -- and, indeed, they are very different. But the real question is whether they are equivalent for the ten input combinations that matter. With so many "don't care" rows in the truth table, there is a LOT of flexibility.

    The easiest way to check if they are the same (as far as the problem is concerned) is to see if each term turns on any patterns that it shouldn't and to verify that, between them, they turn on all the patterns that they should.

    For instance, (a'g) will turn on any pattern of the form 0xxx1. This turns on the 5th pattern and nothing else (that we care about).

    Using this approach, the TS should be able to establish whether the two solutions are equivalent (for the patterns of concern) within just a couple of minutes.
     
  7. MrChips

    Moderator

    Oct 2, 2009
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    Sorry, I did not look at the specific problem of the TS. I am using the common notation of binary variables E-D-C-B-A in order from MSB to LSB.
     
  8. JacobK

    Thread Starter New Member

    Oct 9, 2016
    2
    0
    Thanks for all the help!
    Now i get that the variables that distinguish the k maps should not matter.
    I did as you said WBahn and checked if the two results are equivalent for the ten input combinations that matter and they where equivalent so i guess my solution is also correct.

    Thanks again.
     
  9. WBahn

    Moderator

    Mar 31, 2012
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    4,805
    Yep, that's the result I got, too.
     
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