Simplify the following Boolean expression to a minimum number of literals: (a + b + c') (a' b' + c)?

Discussion in 'Homework Help' started by vlin2, Feb 28, 2015.

  1. vlin2

    Thread Starter New Member

    Feb 28, 2015
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    (a+b+c')(a'b'+c)

    =aa' ab' ac + ba bb' bc + c'a' c'b' c'c

    =1 ab' ac + ba 1 bc + c'a' c'b' 1

    =1 + 1 + a' + b' + c' + ac + bc + 1

    =a'b'c' + ac + bc

    =a'b'c'+c(a+b)

    Did I do this problem correctly? I'm having trouble reducing the literals any further.
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    Your final result is correct (in that it is equal to the starting expression), but I don't follow your work. For instance:

    =1 + 1 + a' + b' + c' + ac + bc + 1

    Is simply 1 since 1 + anything is 1. So it would appear that you are using a very unusual notation.

    Also, aa' is 0, not 1. That is assuming that you are using the normal convention of 0 for False, 1 for True, + for OR and * for AND.
     
  3. vlin2

    Thread Starter New Member

    Feb 28, 2015
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    redid the problem

    aa'b' + ac + ba'b' + bc + c'a'b' + c'c

    = 0b' + ac + a'b + bc + a'b'c' + 0

    = 0 + 0 + a' + b' + c' +ac + bc + 0 <- for this part why did the 0b' and the a'b turn to 0?

    = a'b'c' + ac +bc + 0

    = a'b'c' + c(a+b)
     
  4. WBahn

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    Mar 31, 2012
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    Who's work is this? It would appear that it is not yours.

    In both of your "solutions", one line doesn't lead to another: Your first line is correct this time, but your second line has an error in it, your third line has numerous inconsistencies and does not follow in any way from the second. Your fourth line doesn't follow from your third, and yet is correct as is the final line. If I were a grader, I would conclude that you had spiked the answer from some source and were just trying to get the first and last line of work to appear consistent with the solution hoping the grader wouldn't actually look at your work.
     
  5. vlin2

    Thread Starter New Member

    Feb 28, 2015
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    Got the answer online, trying to figure out how to get to it now.
     
    Last edited: Mar 1, 2015
  6. vlin2

    Thread Starter New Member

    Feb 28, 2015
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    aa'b' + ac + ba'b' + bc + c'a'b' + c'c

    aa' goes to 0
    bb' goes to 0
    c'c goes to 0

    (0)b' +ac + (0)a' + bc + a'b'c' + 0

    a'b'c' +c(a+b) after the zeros drop
     
  7. WBahn

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    Mar 31, 2012
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    Much better.

    Now, as to whether it is the minimum number of literals, that depends on the rules. Notice that the original expression contains six literals. So does your solution. So by this metric neither is better than the other. If you are expected to use SOP or POS form, then neither is acceptable regardless of how many literals they have. If you are allowed to use the exclusive-OR function, you can cut that number in half.
     
  8. vlin2

    Thread Starter New Member

    Feb 28, 2015
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    Yeah I learned about XOR but the professor said to only use AND, and that XOR wouldn't be on the first exam.
     
  9. WBahn

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    Mar 31, 2012
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    Okay, so XOR is off the table. But if you are only allowed to use AND, then how are you going to do it since all of those expressions also use OR? They also use NOT, but that can be considered buried in the literals since a literal can either be the straight or the complemented version of a signal.
     
  10. vlin2

    Thread Starter New Member

    Feb 28, 2015
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    whoops I should have said all the basic ones like AND OR NOT
     
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