0

Here's the expression to simplify:

(A'BC') + (A'B'C) + (ABC)

Some answers I've seen have similar expressions but there's four sets of A's, B's, and C's so that changes the number of distributions and cancellations, like this one and this one.

Here's what I have:

(A'BC') + (A'B'C) + (ABC)

A'BC' + C(A'B' + AB)

I later used De Morgan's Law but I'm not sure if there's improvement:

B(A+C)' + C[(A+B)' + AB]

cocoakrispies98
  • 103
  • 2
  • Before I start my answer, are you familiar with the terms "Karnaugh map" or "Sum of Product Form"? – rhkoulen Nov 03 '21 at 16:37
  • Unless you can make use of something like an XOR, there is really no simplifying possible here. If you use a Karnaugh Map you will quickly see that. – Bram28 Nov 03 '21 at 16:46
  • I know sum of products, that I normally use with logic tables though @rhkoulen – cocoakrispies98 Nov 03 '21 at 16:59
  • If it's not simplifiable it may be a trick question, the question doesn't imply whether it is or isn't so if that's as far as I can go that's fine – cocoakrispies98 Nov 03 '21 at 17:00

1 Answers1

0

What you have been given to simplify is called a "Sum of Products Form" (named very creatively obviously).

These are quite often simplified using a Karnaugh Map, which allows you to represent all possibilities in a table and visually categorize outputs into blocks with common factors, allowing you to make a more slimmed down version of your expression. However, this method has it's limits, which you can see with this example.

Here is an empty Karnaugh Map in three variables: $$\begin{array} {|r|r|}\hline \downarrow A \hspace{0.5cm}BC\rightarrow& 00 & 01 & 11 & 10 \\ \hline 0 & 0 & 0 & 0 & 0 \\ \hline 1 & 0 & 0 & 0 & 0 \\ \hline \end{array}$$ The entries in the array indicate the result of $[col]$ AND $[row]$. So, for your example: $$\begin{array} {|r|r|}\hline \downarrow A \hspace{0.5cm}BC\rightarrow& 00 & 01 & 11 & 10 \\ \hline 0 & 0 & 0 & 0 & 1 \\ \hline 1 & 1 & 0 & 1 & 0 \\ \hline \end{array}$$ Consider the top-right example. When $A$ and $C$ are $0$ and $B$ is $1$, the result is $1$, which corresponds to the term $(A'BC')$ because when we plug in the values, this results in $(0' \cdot 1 \cdot 0') = (1 \cdot 1 \cdot 1) = 1$

Then, the goal of using this map is to draw rectangular regions (can wrap around edges) such that all internal entries are exclusively $1$. Now, it's clear that there's really no way to simplify this further with AND and OR, because there are no possible blocks we can make that only have $1$'s.

However, you can make use of the XOR gate.
From Wikipedia, $A \oplus B = (AB') + (A'B)$
This is relevant in our first two terms, where you can factor out $A'$. Observe: $$(A'BC') + (A'B'C) + (ABC)=$$ $$A'\cdot ((BC')+(B'C)) + (ABC)=$$ $$A'(B \oplus C) + ABC$$

And I think that's significant simplification for the problem in question.
If you are only allowed to use AND and OR, I'm fairly certain you cannot simplify further.

rhkoulen
  • 306