# Boolean Practice Questions #2

Working on the second half of practice questions and can’t seem to understand this one.

1. I understand the “X” and being able to control that outcome, however, I don’t get why you wouldn’t group the “1” in the second row with the X that we turn into a “1” to its left. That way you have 3 groups instead of 2.

2. After you get the answer, I don’t get why on the gate model the “NOT” variables start with a 1 and not a 0.

Hi Natalie,

This is a great question, and the answer really comes down to the idea that there is flexibility in how you approach these problems, but some solutions are more efficient than others.

To answer question 1, there are usually multiple ways to make groupings that follow the rules, but it is best practice to choose the grouping strategy that results in the least possible number of groups. So that is why the solution focuses on 2 groups and not 3.

As for question #2, the values on the left are basically given to you for this problem; they could have been populated with any of the combinations of values from the k-map.

Take a look at this discussion to learn more: Logic Gates for Boolean Algebra

I hope this is helpful for you!
Jesse

Hi Jesse,

@jbyers For the first question, we seem to leave a “1” out (the second row) by only creating 2 groups. That’s why I was confused. Are you able to leave out a number If it doesn’t work? I thought you needed to account for every “1” in the table.

Thanks!

Hi Natalie,
You are exactly right that you need to account for all of the 1s.

In the solution, they are recommending that group 1 includes the 4 corners (red), and group 2 includes the block of 4 1s in top two rows, middle 2 columns (black). This would account for all of the 1s, and uses one of the Xs as a 1 in order to complete the block of 4 for group 2.

Let me know if you have any other questions!
Jesse

AHHH this makes so much more sense. Thank you!

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