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Math Help - tricky karnaugh method question

  1. #1
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    tricky karnaugh method question

    Apply the Karnaugh map method to find all simple Boolean expressions for the Boolean function (w ∨ x ∨ y ∨ z)(w' ∨ x' ∨ y' ∨ z').

    I started off like this... with the truth table

    w x y z and everything and then.. when I reach to the table where it goes something like this.. wx,wx',w'x'and w'x and yz,yz',y'z' y'z

    It sort of stuffs up from there... :\

    can some one shed some light on this thanks
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  2. #2
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    Quote Originally Posted by khonics89 View Post
    apply the karnaugh map method to find all simple boolean expressions for the boolean function (w ∨ x ∨ y ∨ z)(w' ∨ x' ∨ y' ∨ z').

    I started off like this... With the truth table

    w x y z and everything and then.. When i reach to the table where it goes something like this.. Wx,wx',w'x'and w'x and yz,yz',y'z' y'z

    it sort of stuffs up from there... :\

    can some one shed some light on this thanks

    wait what are you saying, is wx, wx' going down?

    and yz, yz'.... up top for the map?
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  3. #3
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    see the karnaugh table.. when you draw it up- you go by those and then insert 1's and 0's ??
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  4. #4
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    anyone!!!!!!!!!!!
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  5. #5
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    im so confused, it doesn't seem to work. have you gotten further
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  6. #6
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    Here it is

    (w ∨ x ∨ y ∨ z)(w' ∨ x' ∨ y' ∨ z')=

    W'X
    W'Y
    W'Z
    WX'
    WY'
    WZ'
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  7. #7
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    I would do the following in your situation: you've got an expression in conjunctive normal form (CNF). Karnaugh maps are much better suited to working in disjunctive normal form (DNF). So, the first step I would take is to convert your expression into DNF. The link I just gave you is the reverse of the process you need to do, but the reverse (converting into DNF from CNF) is entirely analogous.

    Then, once you have your DNF, I would write out your Karnaugh map, and find the simplest covering expression. Make sense?
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