Lengthy Boolean Algebra Problem I can't simplify! please assist me!

Discussion in 'Homework Help' started by DeathEater, Mar 30, 2015.

  1. DeathEater

    Thread Starter New Member

    Mar 30, 2015
    7
    0
    so I created the K-Map for the problem and was able to simplify this way, but the instructor is making us start from the original problem. I don't necessarily won't people to answer so I can just have a solution, but rather so I can see the work and steps that people who know what they're doing take.

    So the problem is...
    B! C! D! + A! B C! D! +A B C! D! + A! B! C D + A B! C D + A! B! C D! + A B C D! + A B! C D!

    The exclamation points represent NOT while the spaces are to make it easier on the eyes, but they still represent AND. When I tried to do this on my own, I started by taking out the different CD combinations that worked, but I keep simplifying the stuff left over in the parentheses after factoring wrong. I am really struggling, but when I simplified the K-Map, my answer was D! ( A + C! ) + B! C

    Please help me!
     
  2. bushrat

    Member

    Nov 29, 2014
    97
    22
    Corrected and simplified.
    upload_2015-3-30_22-38-48.png
     
  3. djsfantasi

    AAC Fanatic!

    Apr 11, 2010
    2,815
    835
    So when you tried to do this on your own, what were the first few steps? Post them here, so we can possibly see where you went wrong, or to show us how you were thinking, so we can help you better.
     
  4. tshuck

    Well-Known Member

    Oct 18, 2012
    3,531
    675
    Bushrat, please don't post the answers. We try to help people discover the answer on their own, not force yet another answer to a problem at them.

    @OP: please post your steps so we can see how you simplified it.
     
  5. MrAl

    Well-Known Member

    Jun 17, 2014
    2,441
    492
    Hi,

    Just a note about converting from logic statements to logic gates...

    The most any problem should need is two levels of gates. That means AND, NAND, OR, NOR, whatever, just two levels deep from input to output. That's for the simplest form of course, and there will be times when a physical gate is not found with enough inputs, but that's the exception.

    I wont give any answer either until we see some attempts at solving this, but what i will give is a method to test your results.
    When the number of variables is low (4 in this case) it is easy to test your result.
    First, generate all the possible inputs, which i will show a few:
    0000
    0001
    0010
    0011
    0100
    to
    1110
    1111

    The above represents all possible inputs for A,B,C, and D. If the result of the logic operations of the original statement match exactly the result of the logic operations of the proposed solution, then they are the same. If for even one input state they are different, then the proposed solution is not correct yet.
    This ia a hard and fast direct approach, and when the number of variables gets too high it doesnt work because it takes too much time to compute all those states and the two results. But for the few variables we see in many problems you can use this idea. It is best done with a computer and a programmable calculator or using a programming language.
     
  6. DeathEater

    Thread Starter New Member

    Mar 30, 2015
    7
    0
    So when you tried to do this on your own, what were the first few steps? Post them here, so we can possibly see where you went wrong, or to show us how you were thinking, so we can help you better.[/QUOTE]

    after writing the first equation it went to:
    AB! C (D!+D)+A! B! C (D+D!)+ABD! (C+C!) +B! C! D!+A! BC! D!

    then I simplified it:
    AB! C+A! B! C+ABD!+B! C! D! +A! BC! D!

    then simplified again:
    C! D! (A! B+B!)+B! C (A!+A)+ABD!

    and again:
    C! D! (A!+B!)+B! C+ABD!

    and then multiplied it out a bit:
    C! D! A!+B! C! D!+B! C+ABD!
    and then I'm pretty much stuck after that, but I know the answer is:
    AD!+CB!+C! D!

    I just need the final step from my last bit of work to the answer
     
  7. WBahn

    Moderator

    Mar 31, 2012
    17,788
    4,808
    You might take a look at the following:

    http://forum.allaboutcircuits.com/blog/boolean-logic-sop-and-pos-forms.583/

    http://forum.allaboutcircuits.com/blog/boolean-logic-working-with-consensus-terms.663/


    From your last post:

    and then multiplied it out a bit:
    C! D! A!+B! C! D!+B! C+ABD!
    and then I'm pretty much stuck after that, but I know the answer is:
    AD!+CB!+C! D!

    First, I'm going to transcribe this in to something that is a bit more standardized:

    A'C'D' + ABD' + B'C'D' + B'C =?= AD' + B'C + C'D'

    If you draw these up on a K-map, you will see which terms on the left need to be combined together to get each of the terms of the right.

    For instance, the term C'D' on the right requires terms A'C'D' + ABD' + B'C'D' in order to cover it. Knowing that, we can look at what the differences are and see that those three terms are equal to C'D' + ABCD'. With that informing our process, we can proceed as follows:

    A'C'D' + ABD' + B'C'D'
    A'C'D' + AB(C+C')D' + B'C'D'
    A'C'D' + ABCD' + ABC'D' + B'C'D'
    [A'C'D' + ABC'D' + B'C'D'] + ABCD'

    The part in square brackets should now reduce to C'D'

    [(A' + AB + B')C'D'] + ABCD'
    [(A'(B+B') + AB + B')C'D'] + ABCD'
    [(A'B + A'B' + AB + B')C'D'] + ABCD'
    [(A'B + AB + A'B' + B')C'D'] + ABCD'
    [( (A'+A)B + A'B' + B')C'D'] + ABCD'
    [( (1)B + A'B' + B')C'D'] + ABCD'
    [(B + A'B' + B')C'D'] + ABCD'
    [((B + B') + A'B')C'D'] + ABCD'
    [ (1 + A'B')C'D'] + ABCD'
    [ (1)C'D'] + ABCD'
    C'D' + ABCD'

    We have thus made the step

    A'C'D' + ABD' + B'C'D' + B'C = C'D' + ABCD' + B'C

    Now see if you play a similar game to finish things off.
     
    DeathEater likes this.
  8. MrAl

    Well-Known Member

    Jun 17, 2014
    2,441
    492
    Hi,

    What do you mean the final step?
     
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