Why is the highlighted area where D is constant is not considered in this Karnaugh map?

Question

Here is a link to the K map: http://tma.main.jp/logic/logic.php?lang=en&type=eq&eq=%28%7EA%7EBC%7ED%29+%2B+%28%7EA%7EBCD%29+%2B+%28%7EAB%7ECD%29+%2B+%28%7EABCD%29+%2B+%28A%7EB%7EC%7ED%29+%2B+%28A%7EB%7ECD%29+%2B+%28A%7EBC%7ED%29+%2B+%28A%7EBCD%29+%2B+%28AB%7EC%7ED%29+%2B+%28AB%7ECD%29+%2B+%28ABC%7ED%29+%2B+%28ABCD%29

I've also attached the screenshot below.

My questions is, if the groups should be large enough, why the highlighted area is not considered but only a subset is considered to get BD instead of D?

Thank you in advance.

Answer 1

The groups should be "large enough" so they cancel a variable X and NOT X out. This only works when a complete X or NOT X is hit by your selection. When you select your six selected cells you will hit the AD area completely, but not the A'D area. This means you cannot cancel them out and get just D as you want to. When you hit the whole AD and A'D area this has the meaning as: "Well, the condition depends on D and is totally unrelated to the value of A."

When written as boolean algebra you get:

AD OR A'D
(A OR A') AND D
( true  ) AND D
D

That would be the goal. But since you don't cover the whole A'D area you can't make this simplification here.

To check if you have selected the correct numbers of cell in a selected area, the numbers of cells must be a value of power of 2. So it must be one of 1, 2, 4, 18, 16, ...

In this case you have the regions:

• A (2*4=8)
• BD (2*2=4)
• B'C (2*2=4, notice how the region "wraps" around the edge of the K-map)

Keep in mind that the result for A'B'C'D is false, not true. But when you have just D in your boolean algebra like in A+D+B'C, the result would be true.