Karnaugh MAP VS Boolean Algebra as Boolean Simplification Methods. Same Answer after simplification??
I've been having problems simplifying Boolean expressions with Boolean algebra and Karnaugh maps, I tried to simplify the same Boolean Expression using both ways but the two answers I got were different. Can that happen? After any Boolean expression simplification whether we use Karnaugh maps or Boolean algebra, the answers should be same, right? Or can the answers be different?
I tried simplifying the expression $A'B'C'+A'B'C+A'BC'+ABC'+ABC$. With boolean algebra I got $A'B'C+AB$, and with Karnaugh maps I got $A' + B$.
Both Karnaugh Map and Boolean Algebra Simplification need not to give same answer. The answer may differ.
The Boolean Algebra Simplification is sometimes tricky because we need smart use of Properties (Absorption and Distributive) and Theorems (Redundancy Theorem)
Even K-Map Solutions are not unique. The answer may differ based on your choice of pairing. The same set of minterms can be simplified in two ways as shown in this attached image.
Now, considering your problem, all the three expressions are not equivalent.
Let three expressions be
F: A′B′C′+A′B′C+A′BC′+ABC′+ABC
K: A′B′C+AB
S: A′+B
| A | B | C | F | K | S |
|---|---|---|---|---|---|
| 0 | 0 | 0 | 1 | 0 | 1 |
| 0 | 0 | 1 | 1 | 1 | 1 |
| 0 | 1 | 0 | 1 | 0 | 1 |
| 0 | 1 | 1 | 0 | 0 | 1 |
| 1 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 | 0 | 0 |
| 1 | 1 | 0 | 1 | 1 | 1 |
| 1 | 1 | 1 | 1 | 1 | 1 |
Clearly, F, K and S aren't equivalent.
Thus, the claimed simplified expression (using K-Map and Boolean) are actually incorrect.
The actual simplified expressions are :
AB+A′B′+A′C′
AB+A′B′+BC′