Boolean Algebra Simplification Help Needed

Discussion in 'Homework Help' started by wordy007, Dec 30, 2009.

  1. wordy007

    Thread Starter New Member

    Dec 30, 2009
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    0
    Hi im new here and have never used forums like this but so far it looks very good. Im having a spot of trouble simplifying a boolean algebra expression for my homework and tried everything i can think of but nothing seems to look right.

    The expression is: (p' = inverted) (++ = XOR)
    (p'qr's') + (pq'r's') + (pq'rs') + (pqr's)+(pqrs')

    Things i have tried so far:
    Distributive rule:
    (p'qr's') + pq's'(r'r) + pq(r'd+cd')
    Cancellation ([r]*r) = 1:
    (p'qr's') + pq's' + pq(r'd+cd')
    XOR - (r'd+cd'):
    (p'qr's') + pq's' + pq(c++d)

    Then that is as far as i can get it.

    I have also tried multiplying out the brackets at the beginning and using commutative law and cancellation to get to:
    q'+sr'+ps'+q'rp+s'q+p+r
    I would write the steps i took to get to this but its 3 pages long and more than likley i've done it all wrong.

    Would appreciate any help i can get as i just can't get it into it's simplest form!
     
    Last edited: Dec 30, 2009
  2. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    wordy007,

    No one like to read that clunky, clumsy notation you are using. Who taught you that, and where does it come from? I see it often on questions about Boolean algebra. Resubmit your expression using the following notation. It is so much easier to read.

    [p]*q*[r]*) ===> p'q'r's+pqrs+....

    Then I will help you.

    Ratch
     
  3. wordy007

    Thread Starter New Member

    Dec 30, 2009
    6
    0
    Sorry about that was just copying what i saw others were doing, i'll edit it now and again any help at all appreciated
     
  4. wordy007

    Thread Starter New Member

    Dec 30, 2009
    6
    0
    Again any help appreciated the more i look at the problem the harder it is to figure out.
     
  5. Fraser_Integration

    Member

    Nov 28, 2009
    142
    5
    I've just put it into a Karnaugh map (information in the online textbook if you need it), and it doesn't simplify much, I came out with:

    P'QRS + PQRS' + PQ'S' + PRS'
     
  6. JotDot

    New Member

    Jan 1, 2010
    3
    0
    I Haven't seen the notation specified by Ratch - I prefer what you submitted.

    Sorry Fraser_Integration but I can't seem to agree with your results - not saying I messed up :)

    Original Expression:
    (p'qr's') + (pq'r's') + (pq'rs') + (pqr's)+(pqrs')

    factor out the r's' and the rs' we get:
    (r's') (p'q + pq') + (rs')(pq' + pq) + pqr's

    Rewriting (pq' + pq) will give (p)(q'+q) or simply p
    Leaving us with (r's') (p'q + pq') + prs' + pqr's

    Note that p'q + pq' is (p xor q) but that does not give you much.

    Thus the final form would be a sum of four products:
    p'qr's' + pq'r's' + pqr's + pr's

    Making a Karnaugh Map for this shows there is not much possible.
     
    Last edited: Jan 1, 2010
  7. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    Fraser_Integration,

    Wrong answer if you are using JotDot's intrepretation of the original Boolean expression. See below.

    JotDot,

    I don't understand. First you say you say you have not seen the notation I recommended, and then you use it even though you prefer the OP's notation. Anyway, the answer you submitted is wrong. The correct answer is p'qr's'+pqr's+pq's'+prs' . How do I know that is the correct answer? I used a K-map and checked it with the program featured in the link below.

    http://forum.allaboutcircuits.com/showthread.php?t=22347&highlight=logic+lovers

    Ratch
     
  8. JotDot

    New Member

    Jan 1, 2010
    3
    0
    Sorry. I didn't realize the O/P edited the orignal post - I was expecting a second post so I wrongly assumed you didn't like how it was written. I should pay better attention.

    I was simply trying to work out the math for the O/P to show the simplifications - which is what I believe the O/P wanted - not just the final answer (which I disagreed with).
    I see I goofed writng down the last poduct; misplacing a symbol. It should be +prs' (not +pr's) Also I did miss one final reduction when I now look at the K-Map again.

    Your second posted answer I agree with: p'qr's'+pqr's+pq's'+prs'

    p.s.: That progam you linked is great for double checking results Would have been handy for me before I posted :)
     
  9. Fraser_Integration

    Member

    Nov 28, 2009
    142
    5
    I'm such a tool. I totally misread my own K-map. Shouldn't be doing boolean algebra after midnight!
     
  10. wordy007

    Thread Starter New Member

    Dec 30, 2009
    6
    0
    thanks very much guys for the help, it seems this boolean algebra doesn't simplify down much. Thank you all very much again all i have to do now is turn that into a logic circuit :)
     
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