Can the following K-map be grouped with 1s in the corners of a subsection of a 16-cell K-map?
Electronic – Is this K-map grouping legal
karnaugh map
Related Solutions
If we look at the top left corner, where a=0 and c=0, we see that s follows b. This gives us a partial result of s=b. Moving down to the bottom left corner, we see that s is inverted when a=1. This is the XOR operation, and gives us s=a^b. Moving over to the right side, we see that the output is inverted when c=1. This gives us our final result of s=a^b^c.
To construct the truth table, you need to manually assess each combination. A table works well, hence the name "truth table"! I assume you understand logical ANDs and ORs, to make sense of this answer.
First, you want to solve each ANDed group separately. Boolean algebra has the same order of precedence as standard algebra, with AND treated like multiplication, and OR treated like addition. Put these answers in a table. Don't worry, I'll attach a picture to demonstrate. Once you have all of these statements figured out, then you can OR them together. Follow the red lines on the following table:
Now that the table is completed, you can build a map. One of the standard configurations is shown below. You have two bits defining the columns, and the other two bits defining the rows. Find the square that intersects the binary inputs (A, B, C, and D), and fill in the answer from your truth table. I've done two of them, in Purple and Orange:
I'll leave the rest for you! You didn't ask how to solve the K-Map. I assume you know how?
Take care!
(P.S. I've included a typo in the truth table. Can you find it?)
Best Answer
Yes, K-maps wrap around like this. Moving from the top row to the bottom row is still only changing one bit.
The ones in the corners of your left rectangle are:
$$\overline{WXY}Z + \overline{WX}YZ + W\overline{XY}Z + W\overline{X}YZ$$
You can factor out the \$\overline X\$ and the \$Z\$:
$$\overline{X}Z(\overline{WY} + \overline{W}Y + W\overline{Y} + WY)$$
As you can see, the values in the parentheses cover all possible combinations of \$W\$ and \$Y\$, so they reduce to 1:
$$\overline{X}Z(1) = \overline{X}Z$$