Hi,

I have an expression of nine literals as follows:

BC+AC'+AB+BCD.

I have to reduce it to four literals.

This is the way I proceeded.

BC+AC'+AB+BCD = BC(1+D)+AC'+AB
= BC+AC'+AB

I have reduced it to six literals. But I am not able to proceed further.

Please Help!!!


Thanks and Regards,
Ramla

 

It cannot be simplified more than that, I think. But wait for a confirmation from an expert

 

I think using K-map we will get a more simplified answer.

 

ramla,

BC+AC'+AB+BCD = BC(1+D)+AC'+AB
= BC+AC'+AB

I have reduced it to six literals. But I am not able to proceed further.

Please Help!!!

chadchoud,

It cannot be simplified more than that, I think. But wait for a confirmation from an expert

vvkannan,

I think using K-map we will get a more simplified answer.

Correct, a K-map or one of the tabulation methods is the way to go. Doing it with Boolean algebra is the pits. Once you have done it with the K-map or any other closed method, you will easily be able see what terms make up the final solution, and then back compute your Boolean algebraic method to get the minimal solution.

Behold:

BC+AC'+AB
BC+AC'+AB(C+C')
BC+AC'+ABC+ABC'
BC(1+A)+AC'(1+B)
BC+AC'

Ratch

 

Thanks a lot ratch..