Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I apologize if this is a dumb question, but I'm having some trouble seeing how we can go from the Boolean equation

P = (P1 + P2)(P3 + P4)(P1 + P3)(P5 + P6)(P2 + P5)(P4 + P6)

to

P = P1*P4*P5 + P1*P3*P5*P6 + P2*P3*P4*P5 + P2*P3*P5*P6 + 
      P1*P2*P4*P6 + P1*P2*P3*P6 + P2*P3*P4*P6 + P2*P3*P6

I keep getting P = P1 P4 P5 + P2 P3 P6 + P2 P3 P4 P5 + P1 P2 P4 P6 + P1 P3 P5 P6 as my minimization.

share|improve this question
    
What exactly are you asking about? How to get symbol expression or how to find boolean value once you have it ? –  vittore Jan 26 '13 at 3:28
6  
Remember that in a Boolean algebra, P1*P1=P1, P2*P2=P2, etc. I haven't checked out every term but I think this is just a case of distributing and then simplifying. –  Brett Frankel Jan 26 '13 at 3:29
    
Why don't you post this as answer, @BrettFrankel –  Rustyn Jan 26 '13 at 4:02
    
I keep getting P = P1 P4 P5 + P2 P3 P6 + P2 P3 P4 P5 + P1 P2 P4 P6 + P1 P3 P5 P6 as my minimization of this. –  John Roberts Jan 26 '13 at 4:16
    
If there is no other conditions but the fact that Pi is bool variable, both your expressions are wrong –  vittore Jan 26 '13 at 4:27
show 1 more comment

2 Answers

up vote 3 down vote accepted

The second answer is just a simplification of the first answer using $Q+1=1$

$P =P_1P_4P_5 + P_1P_3P_5P_6 + P_2P_3P_4P_5 + P_2P_3P_5P_6 + P_1P_2P_4P_6 + P_1P_2P_3P_6 + P_2P_3P_4P_6 + P_2P_3P_6$

$P =P_1P_4P_5 + P_1P_3P_5P_6 + P_2P_3P_4P_5 + P_1P_2P_4P_6 + (P_2P_3P_6)(P_5+P_1+P_4+1)$

$P =P_1P_4P_5 + P_1P_3P_5P_6 + P_2P_3P_4P_5 + P_1P_2P_4P_6 + P_2P_3P_6$

share|improve this answer
    
So both answers are right. Nice observation. +1 –  coffeemath Jan 26 '13 at 9:28
add comment

Sorry for the bad format.

P = (P1+P2)(P3+P4)(P1+P3)(P5+P6)(P2+P5)(P4+P6)
P = (P1+P2)(P1+P3)(P4+P3)(P4+P6)(P5+P6)(P5+P2)         (rearranging terms)
P = (P1+P2*P3)(P4+P3*P6)(P5+P6*P2)                     (X+A)(X+B)=X+AB
P = (P1+P2*P3)(P4+P3*P6)P5+(P1+P2*P3)(P4+P3*P6)P6*P2   (distributive)
P = (P1+P2*P3)P4*P5+(P1+P2*P3)P3*P6*P5+(P1+P2*P3)P4*P6*P2+(P1+P2*P3)P3*P6*P2  (X*X=X)
P =  P1*P4*P5 + P2*P3*P4*P5 + P1*P3*P6*P5 + P2*P3*P6*P5 + P1*P4*P6*P2 + P2*P3*P4*P6 + P1*P3*P6*P2 + P2*P3*P6
share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.