# K-map grouping methods?

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?

### Screenshots

Hi Jason,
You nailed it with the question quoted above. In solving these problems there are generally multiple grouping strategies that can work. Ideally, we try to come up with a strategy that includes the smallest total number of larger groupings that include all of the 1s. By focusing on a smaller number of groupings, it really just makes the next step of simplification a little easier and less prone to bugs.

So, while it is possible to achieve a solution with the “extra” groupings, it does make the simplification more complex and tends to lead to errors. I would recommend trying out the simplification with the recommended 4 groups and see what you come up with using that strategy.

Jesse

Thank you that makes sense now, the smallest number of the largest groups needed to select every 1. Thank you!

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