### Learning Challenge

Boolean Algebra

### Question

This is regarding the 4th K-map on the first set of K-map problems (first Boolean Algebra Project section). I’m confused as to how K-map groupings should be done. In the solution sheet, it lists the groupings as follows:

Group 1 (top row): w’x’

Group 2 (four corners): w’x’y’z’ + w’x’yz’ + wx’y’z’ + wx’yz’ = x’z’

Group 3 (upper right): w’x’yz + w’x’yz’ + w’xyz + w’xyz’ = w’y

Group 4 (third row): wx

F = w’x’ + x’z’ + w’y + wx

### What I’ve tried

The way I grouped it when I worked it was into 6 groups:

Group 1 (top row): w’x’

Group 2 (four corners): w’x’y’z’ + w’x’yz’ + wx’y’z’ + wx’yz’ = x’z’

Group 3 (upper right): w’x’yz + w’x’yz’ + w’xyz + w’xyz’ = w’y

Group 4 (third row): wx

Group 5 (middle right): w’xyz + w’xyz’ + wxyz + wxyz’ = xy

Group 6 (right column): yz’

F = w’x’ + x’z’ + w’y + wx + xy + yz’

Checking the two groups I included that weren’t in the solution–the middle-right square grouping (Group 5), and the right column (Group 6)–both appear to yield 1 when I plug them into the K map, so they seem to be true (whenever x and y are both true, there is a 1, and whenever y is true and z is false, there is a 1). Would including those two groups cause an incorrect solution? I don’t think I can simplify it anymore since they are in the sum-of-products format. Are groups 5 and 6 unnecessary since the other 4 groups include every square being compared in groups 5 and 6? Wouldn’t it be useful to know every possibility that leads to a true value?