# karnaugh map problem

Discussion in 'Homework Help' started by poindexter, Aug 26, 2006.

1. ### poindexter Thread Starter Member

Aug 26, 2006
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Hi folks,I'm stuck on one particular problem and have wasted nearly a whole day on it.I'm sure with a little help I can get a breakthrough.Her it is.Simplify this expression using a karnaugh map:AC[B+A(B+C')].Hope someone can help me Thanx.Poindexter

2. ### poindexter Thread Starter Member

Aug 26, 2006
10
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the problem is I cant get it to a sum of products

3. ### Papabravo Expert

Feb 24, 2006
10,340
1,850
You just work from the inside out using:
AA = A and
CC' = 0
Code ( (Unknown Language)):
1.
2. AC[B + A(B + C')]    =
3. AC[B + AB + AC']     =
4. ACB + ACAB + ACAC'   =
5. ACB + ACB  + 0       =
6. ACB
7.
Not much of a Karnaugh map problem after all -- is it?

4. ### poindexter Thread Starter Member

Aug 26, 2006
10
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I'm supposed to used a k map to simplify it so Im not sure if we've done it right.But i did learn something from you .thanx for the time to help

5. ### Papabravo Expert

Feb 24, 2006
10,340
1,850
As you correctly pointed out, the beginning of a Karnaugh map problem is a sum of products expression. If getting to that point leaves you with only one term then the map doesn't do much to help. It is also not the only technique that can be used. It works best with three or four variables. For a large number of variables other techniques need to be used because the maps loose their intuitive value.

6. ### Papabravo Expert

Feb 24, 2006
10,340
1,850
Yo! Poindexter

Did you solve the other problem or did you wimp out?

7. ### poindexter Thread Starter Member

Aug 26, 2006
10
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I'm supposed to use the karnaugh map to simplify.But it seems that getting to the sum of products already simplifies it and eliminates the need to use the karnaugh map.I dunno what the solution is so will just submit it and let the teacher decide what it is they want.What other problem are you talking about?

8. ### Papabravo Expert

Feb 24, 2006
10,340
1,850
I could have sworn there was another problem involving x's and y's in your original post, but I don't see it now.

Mar 15, 2007
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Mar 18, 2007
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11. ### terrytapp Member

Mar 30, 2006
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My question after reading this post is this.
I can make a Karnaugh map from a truth table and reduce everything.
(easy enough) BUT how do you do it backwards.
Say you have the answer of the sum of your product A C [ (B+A) (B+C) ]
AS Poindexter originally posted, how do you build up from that? I am not clear on this.

_______A ___B __C

A

B
_
A B
___
A B
_
A B

I mean how do you graph it out?

12. ### jpates New Member

May 15, 2007
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I know it is a bit late...and I am really new to this material, but maybe I can help. The other post about solving this one algebraically is definitely a whole lot easier, but since you had to solve using a K-Map, here ya go.

To solve this using a Karnaugh Map, step 1 would be to create the Truth Table.
(using a bit of logic, you can see if A or C is 0, then the whole answer is 0, so you only need to solve for 101 and 111)

*Attaching a Word document since it is a pain to format on this forum.

From there, you can see that the answer is ABC from the map...and the previously posted algebraic solution.

File size:
27 KB
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37
13. ### recca02 Senior Member

Apr 2, 2007
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or may be we can do it with product of sum (not sure though how it wud turn out)

14. ### Papabravo Expert

Feb 24, 2006
10,340
1,850
No to both. A Karnaugh map with a single term will have a single 1. It is both trivial and useless. The whole purpose of a Karnaugh map is to spot overlapping terms which result in simplified solution. No matter how you try to dress this pig there is no point in using a Karnaugh map for a single term boolean equation.