Simpler answer from truth table?

Discussion in 'Feedback and Suggestions' started by calamari, Jan 28, 2004.

  1. calamari

    Thread Starter New Member

    Jan 28, 2004
    1
    0
    Hi,

    First of all, let me thank you guys for an excellent site! Here's a minor improvement I've come up with for the second "Converting truth tables into Boolean expressions" example.

    In that second example the truth table is converted into the Product-Of-Sums expression:

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

    This should be able to be simplified to

    (a(+)b)+(a(+)c)+(b(+)c)

    where (+) represents exclusive-or.

    Here is my proof:
    (a+b+c)(a'b'c') Given
    =(a+b+c)a'+(a+b+c)b'+(a+b+c)c' Distributive
    =aa'+ba'+ca'+ab'+bb'+cb'+ac'+bc'+cc' Distributive
    =0+ba'+ca'+ab'+0+cb'+ac'+bc'+0 Identity or Complement
    =ba'+ca'+ab'+cb'+ac'+bc' Identity
    =ab'+ba'+ac'+ca'+bc'+cb' Commutative
    =ab'+a'b+ac'+a'c+bc'+b'c Commutative
    =(ab'+a'b)+(ac'+a'c)+(bc'+b'c) Associative
    =(a(+)b)+(a(+)c)+(b(+)c) Exclusive-or

    Drawing this out saves a couple circuit components (4 vs 6).
    I also compared the truth tables (to double check) and interestingly enough, in the process I found that only two of the three exclusive-or gates are necessary to generate the correct bit pattern (i.e.):

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

    This brings the number of circuit components down to 3 vs 6. I don't have a proof of that except the truth table. It also seems that a'bc+ab'c'=a(+)b+a(+)c. Some proofs would be great.

    calamari
     
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