Simplifying a boolean expression to a minimum number of literals

Discussion in 'Homework Help' started by mcc123pa, Sep 12, 2010.

  1. mcc123pa

    Thread Starter Member

    Sep 12, 2010
    40
    0
    Hi everyone:

    I was assigned the following problem for homeowork:

    A'B'D+A'C'D+BD (a ' mark after a letter means a bar)

    The directions read simplify the expression to an expression containing a minimum number of literals.

    Here is my attempt, could someone please tell me if it is right? If I'm wrong, could you please post the correct way to do this?

    A'B'D+BD+A'C'D (original function re-arranged)
    (B+B')(A'D+D)+A'C'D
    (1)(A'D+D)+A'C'D
    A'D+D+A'C'D
    A'D+A'C'D+D
    A'D(1+C')+D
    A'D(1)+D
    D(A'+1)
    D(1)
    D

    Thanks!
     
  2. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
    Unfortunately, it's not quite correct. The error is in the second line where you rush to make the factorization. Take your time and make one factorization at a time. Try it once more. If you find it too hard, feel free to ask again.
     
  3. mcc123pa

    Thread Starter Member

    Sep 12, 2010
    40
    0
    Hi, thanks for the response!

    Here is my latest attempt:

    A'B'D+A'C'D+BD (original function)

    D(A'B'+A'C'+B) (factored out a D)

    D((B+B')(A'+1)+A'C') (factored out B+B')

    D(A'+1+A'C') (after converting B+B' to 1)

    Is this as far as I can go with this? Is D(A'+1+A'C') my final answer?
     
  4. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
    You can't just factor out B+B' out of two terms and hope to get it right. Factor out one letter at a time to ensure you don't make logic mistakes.
    Here's how it goes:

    A'B'D+BD+A'C'D =\\<br />
D(A'B'+B) +A'C'D=\ \ \ \text{% Factoring out D}\\<br />
D(A'+B)(B'+B)+A'C'D=\ \ \ \text{% Using the identity X+YZ=(X+Y)(X+Z)}\\<br />
D(A'+B)+A'C'D=\ \ \ \text{% B+B'=1 wich is a neutral term}\\<br />
D(A'+B+A'C')=\ \ \ \text{% Factoring out D}\\<br />
D(A'(C'+1)+B)=\ \ \ \text{% Factoring out A'}\\<br />
D(A'+B)=\ \ \ \text{% C'+1=1}\\<br />
A'D+BD

    This is the final result.
     
  5. mcc123pa

    Thread Starter Member

    Sep 12, 2010
    40
    0
    Thanks so much for your help!!
     
  6. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
    The "Thanks" button is down on the right!
    (Yeah, we all have our weaknesses... :p)
     
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