Jump to content
  • Advertisement
Sign in to follow this  
Alatar

More Boolean Algebra

This topic is 2476 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

If you intended to correct an error in the post then please contact us.

Recommended Posts

I'm still trying to understand how to simplify Boolean algebra expressions.

Can someone help me understand how to simplify these pelase?

1. A.B + B.!C + !A.!C

2. D + !B.!C + C.!D + !A.!B.C

Thanks. I would appreciate any help.

Share this post


Link to post
Share on other sites
Advertisement
Those took me a while.


A.B + B.!C + !A.!C

A.B + !A.!C + B.!C // reordered -> the last one is some sort of linear combination of the other two
A.B + !A.!C + (B.!C).(A+!A) // augmented with something true (A+!A)
A.B + !A.!C + B.!C.A + B.!C.!A // expanded
A.B + B.!C.A + !A.!C + B.!C.!A // reordered
(A.B).(1+!C) + (!A.!C).(1+B) // factored out our "basis" (we knew that the third term in the beginning was a combination of those two)
(A.B) + (!A.!B) // done.


D + !B.!C + C.!D + !A.!B.C

(D + C.!D) + (!B.!C + !A.!B.C) // reordered. I placed brackets around the two groups we simplify next.
(D+C).(D+!D) + (!B.(!C + (!A.C))) // factor out
(D+C) + (!B.(!C+!A).(!C+C)) // left: identity, right: expanded
(D+C) + (!B.(!C+!A)) // right: identity removed,
D+C + !B.!C + !B.!A // expanded
D+(C+ !B.!C) + !B.!A // again I placed brackets to show what will be simplified next
D+(C+!B).(C+!C)) +!B.!A // once more we found an identity
D+C+ (!B+ !B.A) // and finally the last simplification brings us to:
D+C+!B // done.

Cheers!

PS: You can verify the steps with wolframalpha.com!

Share this post


Link to post
Share on other sites
The way I approach these problems, I try to visualize the truth table of the expression and then identify large chunks that are 1. For instance,

A.B + B.!C + !A.!C


C=0:
B=0 B=1
A=0 1 1
A=1 0 1

C=1:
B=0 B=1
A=0 1 0
A=1 0 0


Try to visualize one table in front of the other, forming a cube. You can now group the ones in two groups: one is A.B and the other is B.!C. So the expression is simply A.B + B.!C

Share this post


Link to post
Share on other sites
Sign in to follow this  

  • Advertisement
×

Important Information

By using GameDev.net, you agree to our community Guidelines, Terms of Use, and Privacy Policy.

GameDev.net is your game development community. Create an account for your GameDev Portfolio and participate in the largest developer community in the games industry.

Sign me up!