Boolean Algebra Simplification

[tommis15]

Hello all, I have a Boolean algebra question I need help with...

This is the truth table and the Karnaugh map associated. I got this from:
http://www.ee.calpoly.edu/media/uploads/resources/KarnaughExplorer_1.html



Now, for my assigment I need to simplify it, and this is what I've done.

F=A’B’C’D’ + A’B’CD’ + A’BC’D’ + A’BC’D + A’BCD’ + A’BCD + ABC’D’

F=A’B’D’ (C’+C) + A’BC’ (D’+D) + A’BC (D’+D) + ABC’D’

F=A’B’D’ (1) + A’BC’ (1) + A’BC (1) + ABC’D’

F=A’B’D’ + A’BC’ + A’BC + ABC’D’

F=A’B (C’+C) + A’B’D’ + ABC’D’

F=A’B (1) + A’B’D’ + ABC’D’

F=A’B + A’B’D’ + ABC’D’

Other useful info:
' = NOT, therefore
A' = NOT(A)

My answer is different than the one on the website... Is this normal? Am I making a mistake here or not using the appropriate method? Does the Karnaugh Map simplify the function more effectively than the written method?

Thanks in advance, all input appreciated!

tommis15

 

[Laplace]

All you need to do is draw three circles on the K-map and you get the answer.

 

[tommis15]

Yeah, I understand that. I wasn't clear... but... The assignment was to solve using boolean algebra formulas. I was using the Karnaugh map to double check my work.

 

[Laplace]

In that case, I would suggest simplifying the algebra in the same way that the K-map is simplified. In the initial sum of minterms, certain ones will be repeated in duplicate or in triplicate. Those that are within two circles in the map will appear twice; those within three circles appear thrice. Then group the minterms corresponding to the circles in the K-map. Simplify the Boolean groups from there.

 

[tommis15]

Thanks for your help... I handed the assignment in as it was already late a day and I couldn't hold off or else my teach would give me 0. All I could do is pray.