Boolean Simplification Methods. : Karnaugh Maps vs Boolean Algebra

Question

I am practicing problems on simplifying Boolean Expressions with Boolean Algebra and Karnaugh Maps.

• I tried to simplify the same Boolean Expression using both ways but I got two different answers. Can that happen?
• After any Boolean Expression simplification, whatever may be the method, the answers should be same, right? Or can the answers be different?

I tried simplifying the expression $$\mathcal{A'B'C'+A'B'C+A'BC'+ABC'+ABC}$$ With boolean algebra I got $\mathcal{A'B'C+AB}$, and with Karnaugh Maps I got $\mathcal{A'+B}$

Answer 1

Boolean algebra and Karnaugh map simplification should give the same or very similar answers, though without some creativity the boolean algebra one may end up not getting quite as simple an answer because things that can be used to simplify some parts may have been absorbed by other parts.

Some expressions have multiple simplified forms that are equivalent! In the specific case of $A'B'C' + A'B'C + A'BC'+ABC' + ABC$, there are two equivalent fully simplified forms:

• $AB + A'B' + A'C'$
• $AB + A'B' + BC'$

and a small variety of slightly less simplified forms (note the additional symbol on the last term):

• $AB + A'B' + A'BC$
• $AB + A'C' + A'B'C$
• $A'B' + BC' + ABC$

Unfortunately, neither of your two proposed solutions shown in the comments are correct, as can be seen by simply counting 'on' bits: $A'+B$ has $6$ on bits, and $A'B'C+AB$ has $3$, where the original has $5$. I'm not sure how you got either of those, unfortunately: I can't find a place in my own workings where an error would create these.

Answer 2

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
$\mathcal{F=A′B′C′+A′B′C+A′BC′+ABC′+ABC}$
$\mathcal{K=A′B′C+AB}$
$\mathcal{S=A′+B}$

$\mathcal{A}$	$\mathcal{B}$	$\mathcal{C}$	$\mathcal{F}$	$\mathcal{K}$	$\mathcal{S}$
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, $\mathcal{F}$, $\mathcal{K}$ and $\mathcal{S}$ aren't equivalent.

Thus, the claimed simplified expression (using K-Map and Boolean) are actually incorrect.

The actual simplified expressions are :

$\mathcal{AB+A′B′+A′C′}$

$\mathcal{AB+A′B′+BC′}$