Boolean Algebra:Sum of Products

Discussion in 'Math' started by philrox, Mar 29, 2011.

  1. philrox

    Thread Starter New Member

    Mar 29, 2011
    Hello everyone. I have been working on this problem for a lifetime.. and I can't seem to figure it out. There are several questions to this equation.. but essentially it's boggling my mind.

    We were given the following information below.

    X Y Z q4 q5 q6 q7
    0 0 0 0 1 1 1
    0 0 1 1 0 1 0
    0 1 0 1 1 0 0
    0 1 1 0 0 0 0
    1 0 0 1 0 0 0
    1 0 1 0 0 1 0
    1 1 0 0 0 1 0
    1 1 1 1 1 0 1

    One of the questions was:
    Select all of the terms below that would be used to form a "sum of products" representation for q5.
    X' * Y' * Z'
    X' * Y' * Z
    X' * Y * Z'
    X' * Y * Z
    X * Y' * Z'
    X * Y' * Z
    X * Y * Z'
    X * Y * Z
    These were the options given to us to check off^^^ (its an online quiz, we have as long until next class to complete this.)
    Now, I successfully completed this for q4, but he would also like us to figure out q6, q7, and obviously q5. I'm not sure about what I am supposed to do exactly. Any help would be really appreciated.

    And in case anybody is confused by the apostrophes, that is to symbolize the compliment of X, Y, or Z. The professor didn't even explain half of this in class, and it's difficult for everybody. I'm not horrible at math, in fact I took AP Calculus in high school. However, this problem is just killing me right now.
  2. Georacer


    Nov 25, 2009
    What did you do to simplify q4? How is q4 different that the rest of the outputs?

    Please post any work you have done so far or your thoughts for the solution, in order to help the community find the error in your line of thoughts.
  3. philrox

    Thread Starter New Member

    Mar 29, 2011
    The answer I received for q4 is respectively:

    q4 = X’ * Y’ * Z + X’ * Y * Z’ + X *Y’ *Z’ + X * Y * Z

    That is the most I have completed so far. I'm not sure how to arrive at q5 through q7 though.

    I simply did math in my head to arrive at the q4 answer.

    I figured the if there were an odd number of one's, it was a 1. If there were an even number of one's, it was a 0.

    It worked for q4, but not the rest.
  4. Georacer


    Nov 25, 2009
    Oh, now I see why you couldn't answer the question. Here's how it is done:

    For each output (qi) you see that for each ace (1) in the corresponding column, there is a 3-bit combination of XYZ in front of it. That means that if this combination appears, you must get a 1 from your output.

    For example, in q4, the first 1 appears for XYZ=000. The minterm that will give a 1 for XYZ=000 is X'Y'Z'.
    The second 1 of q4 appears for XYZ=001. The corresponding minterm is X'Y'Z.

    Try it once more and correct q4 too, which is wrong.
  5. philrox

    Thread Starter New Member

    Mar 29, 2011
    It took me a second to figure out what you were saying, but it really helped once I understood it! Thanks a lot for the helping. It saved my friend and I from failing a quiz, and a test for later this week.

    I appreciate it a lot!