# Thread: tricky karnaugh method question

1. ## 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

2. Originally Posted by khonics89
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?

3. see the karnaugh table.. when you draw it up- you go by those and then insert 1's and 0's ??

4. anyone!!!!!!!!!!!

5. im so confused, it doesn't seem to work. have you gotten further

6. ## Here it is

(w ∨ x ∨ y ∨ z)(w' ∨ x' ∨ y' ∨ z')=

W'X
W'Y
W'Z
WX'
WY'
WZ'

7. 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?