Results 1 to 10 of 10

Math Help - Simplify Boolean Function

  1. #1
    Junior Member
    Joined
    Mar 2010
    From
    Malaysia
    Posts
    26

    Simplify Boolean Function

    1. p= x'y' + (xy')' (z(x + z'))'

    my problem was from here im stuck and does not know to using Boolean algebra or karnaugh maps for getting simplify this question. Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor undefined's Avatar
    Joined
    Mar 2010
    From
    Chicago
    Posts
    2,340
    Awards
    1
    Quote Originally Posted by wkn0524 View Post
    1. p= x'y' + (xy')' (z(x + z'))'
    I'm pretty unfamiliar with the subject, so my reply may not be very helpful, but I'm wondering: is addition the same as XOR and multiplication the same as AND? This would lead to

    0 + 0 = 0
    0 + 1 = 1
    1 + 0 = 1
    1 + 1 = 0

    0 * 0 = 0
    0 * 1 = 0
    1 * 0 = 0
    1 * 1 = 1

    If so, then a (probably slow) way of solving is to construct an 8-row truth table and then find a simpler expression from that. I'm curious to see what people more experienced with these problems post.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Just to use slightly more standard syntax: you want to simplify

    p=\overline{x}\,\overline{y}+\overline{x\overline{  y}}\:\overline{z(x+\overline{z})}, correct?

    Are you wanting to put the result in disjunctive or conjunctive normal form, or are you just trying to minimize the number of logic gates required?

    Note to undefined: in digital logic, at least, the plus symbol is usually the inclusive or, and you use something like \otimes for exclusive or.

    I would probably use Karnaugh maps or DeMorgan's rules to simplify.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Incidentally, this question really ought to be in the Discrete Mathematics, Set Theory and Logic forum, not in Calculus.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Mar 2010
    From
    Malaysia
    Posts
    26
    Ya, that the method of using Boolean algebra, but i got compare to the Boolean algebra with Karnaugh maps, the result are not the same.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Show us what you did. I'm a fan of disjunctive normal form myself. What did you get for that?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Mar 2010
    From
    Malaysia
    Posts
    26
    Quote Originally Posted by Ackbeet View Post
    Just to use slightly more standard syntax: you want to simplify

    p=\overline{x}\,\overline{y}+\overline{x\overline{  y}}\:\overline{z(x+\overline{z})}, correct?

    Are you wanting to put the result in disjunctive or conjunctive normal form, or are you just trying to minimize the number of logic gates required?

    Note to undefined: in digital logic, at least, the plus symbol is usually the inclusive or, and you use something like \otimes for exclusive or.

    I would probably use Karnaugh maps or DeMorgan's rules to simplify.
    Ya, at first De morgan's rules is needed.
    There are two steps to simplify are using the rules of Boolean Algebra and karnaugh map.
    its mean that i only need to simplify the question.

    Here my step:
    i got change the question.

    <br />
p=\overline{(\overline{x} + y)(x\overline{y}+z(x + \overline{z})}<br />
    After DeMorgan rules:

    <br />
p= x\overline{y} + (\overline{x} + y)(\overline{z} + \overline{x}z)<br />

    <br />
p= x\overline{y} + (\overline{x}\,\overline{z} +\overline{x}z + y\overline{z} + \overline{x}yz)<br />

    Maybe im stuck at here.
    Last edited by wkn0524; June 9th 2010 at 10:17 AM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor undefined's Avatar
    Joined
    Mar 2010
    From
    Chicago
    Posts
    2,340
    Awards
    1
    Quote Originally Posted by Ackbeet View Post
    Just to use slightly more standard syntax: you want to simplify

    p=\overline{x}\,\overline{y}+\overline{x\overline{  y}}\:\overline{z(x+\overline{z})}, correct?

    Are you wanting to put the result in disjunctive or conjunctive normal form, or are you just trying to minimize the number of logic gates required?

    Note to undefined: in digital logic, at least, the plus symbol is usually the inclusive or, and you use something like \otimes for exclusive or.

    I would probably use Karnaugh maps or DeMorgan's rules to simplify.
    Thanks for the info. I guessed XOR because then we would have normal addition and multiplication in \mathbb{Z}/2\mathbb{Z}. Hopefully I'll get to studying the field soon.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    For disjunctive normal form, I got

    <br />
p=<br />
\overline{x}\,\overline{y}<br />
+\overline{x}\,\overline{z}<br />
+\overline{x}\,z<br />
+y\,\overline{z}<br />
+\overline{x}\,y\,z<br />
.

    Is that what you got? If so, what did you do then?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Nov 2009
    Posts
    130
    <br />
p= x'y' + (x'+y) (z'+(x + z')') =x'y'+(x'+y)(z'+x'z)=<br />
    =x'y'+(x'z'+x'x'z+yz'+yx'z)=x'y'+x'z'+x'z+yz'+x'yz

    Now here is the thing you need to do so you can create the karnogh map:

    x'y'(z+z')+x'z'(y+y')+x'z(y+y')+yz'(x+x')+x'yz=
    x'y'z'+x'y'z+x'y'z+x'yz'+x'y'z+x'yz+xyz'+x'yz'+x'y  z=
    Remove the duplicates:
    =x'y'z'+x'y'z+x'yz'+x'yz+xyz'

    Karnough map:
    1 1 1 1
    0 0 0 1

    Take the 4 neighbour '1' (1st row of karnough map) and 2 neighbour '1' in the right angle.

    The minimization is: yz'+x'
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. boolean function
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: March 7th 2011, 01:09 PM
  2. Simplify boolean expression.
    Posted in the Discrete Math Forum
    Replies: 8
    Last Post: March 1st 2011, 05:57 PM
  3. Simplify the expression using Boolean algebra?
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: August 31st 2010, 09:27 PM
  4. Simplify boolean algebra
    Posted in the Discrete Math Forum
    Replies: 8
    Last Post: February 27th 2010, 07:46 PM
  5. I need to simplify these boolean expressions.
    Posted in the Algebra Forum
    Replies: 0
    Last Post: April 30th 2009, 07:21 AM

Search Tags


/mathhelpforum @mathhelpforum