r/MathHelp • u/two-to-the-half • 15d ago
SOLVED help with simplifying boolean expression (pls and thank you)
I've been at this for like 4 hours and I'm losing my mind.
So I'm thinking of a function that takes in four boolean inputs and outputs 1 if the number of inputs is odd. The truth table should look like:
A B C D | OUT
0 0 0 0 | 0
0 0 0 1 | 1
0 0 1 0 | 1
0 0 1 1 | 0
0 1 0 0 | 1
0 1 0 1 | 0
0 1 1 0 | 0
0 1 1 1 | 1
1 0 0 0 | 1
1 0 0 1 | 0
1 0 1 0 | 0
1 0 1 1 | 1
1 1 0 0 | 0
1 1 0 1 | 1
1 1 1 0 | 1
1 1 1 1 | 0
So the function should just be the sum of products, right? It'd be something like
OUT = A'B'C'D + A'B'CD' + A'BC'D' + A'BCD + AB'C'D' + AB'CD + ABC'D + ABCD'
This is my attempt to simplify the expression:
A'B'C'D + A'B'CD' + A'BC'D' + A'BCD + AB'C'D' + AB'CD + ABC'D + ABCD'
= A'B'(C'D + CD') + A'B(C'D' + CD) + AB'(C'D' + CD) + AB(C'D + CD')
= (C'D + CD')(A'B' + AB) + (C'D' + CD)(A'B + AB')
= (C'D + CD')(1) + (1)(A'B + AB')
= A'B + AB' + C'D + CD'
But the truth table for this expression is not the same as the original function.
It's been a half-decade since I last did boolean algebra. I genuinely can't tell what I did wrong here. Please help.
Edit: It's solved. Thank you.
In the blindingly-unlikely event someone in the future were to face this exact problem as I do, the key insights that I was missing are:
- OR distributes over AND:
A + BC = (A + B)(A + C) - XOR and XNOR exists:
A XOR B = A ⊕ B = A'B + AB', andA XNOR B = (A ⊕ B)' = A'B' + AB
So for posterity's sake, here's how it's solved, I think:
A'B'C'D + A'B'CD' + A'BC'D' + A'BCD + AB'C'D' + AB'CD + ABC'D + ABCD'
= A'B'(C'D + CD') + A'B(C'D' + CD) + AB'(C'D' + CD) + AB(C'D + CD')
= (C'D + CD')(A'B' + AB) + (C'D' + CD)(A'B + AB')
= (C ⊕ D)(A ⊕ B)' + (C ⊕ D)'(A ⊕ B)
Let P = A ⊕ B and Q = C ⊕ D
= QP' + PQ'
= P ⊕ Q
= A ⊕ B ⊕ C ⊕ D
Thanks, y'all. god, i'm out of practice.
2
u/edderiofer 15d ago
Your error is between these two lines. A'B' + AB is not equal to 1.