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Math Help - Boole Algebra: DNF help

  1. #1
    Junior Member shinhidora's Avatar
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    Boole Algebra: DNF help

    Find the DNF for:

    f(x,y,z) = y\overline{z} + (y+z)(\overline{z+\overline{x}})

    so by using the axioms and stuff I found:

    f(x,y,z)= xy\overline{z}+ \overline{x}y\overline{z}

    but by putting everything into a table and filling in 1 for the 3 x,y,z,\overline{x},\overline{y},\overline{z} that give f = 1 I found:

    f(x,y,z) = \overline{x}y\overline{z} + x\overline{y}z + xy\overline{z} + xyz

    So what did I do wrong or did I overlook something?
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  2. #2
    MHF Contributor
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    Hello shinhidora
    Quote Originally Posted by shinhidora View Post
    Find the DNF for:

    f(x,y,z) = y\overline{z} + (y+z)(\overline{z+\overline{x}})

    so by using the axioms and stuff I found:

    f(x,y,z)= xy\overline{z}+ \overline{x}y\overline{z}

    but by putting everything into a table and filling in 1 for the 3 x,y,z,\overline{x},\overline{y},\overline{z} that give f = 1 I found:

    f(x,y,z) = \overline{x}y\overline{z} + x\overline{y}z + xy\overline{z} + xyz

    So what did I do wrong or did I overlook something?
    If the original expression was:
    f(x,y,z) = \bar{x}y\overline{z} + (y+z)(\overline{z+\overline{x}})?
    then I agree with your first answer:
    f(x,y,z)= xy\overline{z}+ \overline{x}y\overline{z}
    except that you can then simplify it further:
    f(x,y,z)= xy\overline{z}+ \overline{x}y\overline{z}
    = (x+\bar{x})y\bar{z}

    =y\bar{z}
    But if the original expression was correct, then I should write:
    f(x,y,z) = y\overline{z} + (y+z)(\overline{z+\overline{x}})
    = y\overline{z} + (y+z)(\overline{z}x)

    =y\bar{z}+xy\bar{z} +xz\bar{z}

    =y\bar{z}(1+x)

    =y\bar{z}
    which turns out to be the same anyway!

    Grandad
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  3. #3
    Junior Member shinhidora's Avatar
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    Quote Originally Posted by Grandad View Post
    But if the original expression was correct, then I should write:
    f(x,y,z) = y\overline{z} + (y+z)(\overline{z+\overline{x}})
    = y\overline{z} + (y+z)(\overline{z}x)

    =y\bar{z}+xy\bar{z} +xz\bar{z}

    =y\bar{z}(1+x)

    =y\bar{z}
    which turns out to be the same anyway!

    Grandad
    The original was correct...

    And I found the first answer in my Mathbook xy\overline{z}+\overline{x}y\overline{z} so I assume it's correct?

    While the second answer I found by putting everything into a table:

    x y z | \overline{x} \overline{y} \overline{z} | f = y\overline{z}+(y+z)(\overline{z+\overline{x}})
    000|111| 0
    001|110| 0
    010|101| 1
    100|011| 0
    011|100| 0
    101|010| 1
    110|001| 1
    111|000| 1

    so I get f = \overline{x}y\overline{z} + x\overline{y}z + xy\overline{z} + xyz
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  4. #4
    Junior Member shinhidora's Avatar
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    I know I'm not allowed to bump, but I have exam in 9 hours... :$

    and I'm really stuck :s
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  5. #5
    Moo
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    Quote Originally Posted by shinhidora View Post
    The original was correct...

    And I found the first answer in my Mathbook xy\overline{z}+\overline{x}y\overline{z} so I assume it's correct?

    While the second answer I found by putting everything into a table:

    x y z | \overline{x} \overline{y} \overline{z} | f = y\overline{z}+(y+z)(\overline{z+\overline{x}})
    000|111| 0
    001|110| 0
    010|101| 1
    100|011| 0
    011|100| 0
    101|010| 1
    110|001| 1
    111|000| 1

    so I get f = \overline{x}y\overline{z} + x\overline{y}z + xy\overline{z} + xyz
    If I'm not mistaking, the 2 red 1's are actually 0's. I didn't check the 0's.
    I think it's because you missed the line that overlaps z+\overline{x}

    Good luck for your exam
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  6. #6
    Junior Member shinhidora's Avatar
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    I fail at life... Thx Moo
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  7. #7
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    Hello shinhidora

    Sorry I wasn't able to get back to you last night. This may be a bit late for your immediate need - your exam - but let me see if I can clear this up.

    The answer f = xy\bar{z} + \bar{x}y\bar{z} is in full disjunctive normal form, which requires each variable to appear in every clause. However, this will clearly simplify to:
    f = xy\bar{z} + \bar{x}y\bar{z}
    =(x+\bar{x})y\bar{z}

    =1y\bar{z}

    =y\bar{z}
    which is also, of course, in DNF, but not full DNF because x has been eliminated.

    As far as your table is concerned, Moo is right: you have made an error in your working (although I'm not sure where, since you don't show it). Look at the attached diagram, where I have shown all the working. (Note: I have used x', etc to denote \bar{x} as it's easier to type.)

    You'll see that this does give the answer f = xy\bar{z} + \bar{x}y\bar{z}.

    Grandad
    Attached Thumbnails Attached Thumbnails Boole Algebra: DNF help-untitled.jpg  
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  8. #8
    Junior Member shinhidora's Avatar
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    Thx

    My Exam was ok The Boolean part was great (aiming for 19-20/20 )
    Though I'm afraid I messed up Matrices a bit... ah well
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