LogFAQs > #958071638

Sahuagin posted...

I just don't know how they determine that those two 4s both belong in the same group despite colliding. and why wouldn't the colliding 4 just make a new third group by itself?

This solution only works for two groups at a time. And the process to identify groups for this solution must see only two groups. The point is that selecting one number means the other number is excluded. It's binary XOR logic.

As I stated in previous post, find a number that appears exactly twice in the same region (row, column, or box). That means they are in two different groups. Follow one of these numbers but across a different region. Here you see that 4 in B1 has collisions in all three regions: same row B7; same column G1; same box C3. That means B1 is in one group, and its collisions will all be in the second group.

Now follow one of the second group. Like G1. It has another collision at G5. Thus G5 is opposite group to G1, means same group as B1. This is where you're confused? Do a simple check. If G1 is chosen, then B1 and G5 must be eliminated. The reverse is also true. If G1 is eliminated, B1 and G5 must be chosen. Thus you should come to understand that B1 and G5 are the same group, and G1 is the opposite group.

Now keep following G5. It has second collision at C5; hence C5 belongs to second group (along with B7, G1, and C3). As I stated in previous post, each sequential cell linked this way will alternate into yellow or green groups. Keep going until you find all valid members of both groups using the criterion that exactly two of that number in the same region.

Even though by now we have already identified the problem with the yellow group at row C. Two yellows at C3 and C5 means we can eliminate the yellow group. Actually another similar problem with yellow can be seen in the top-right box.
---

Ergonomic keyboard layouts for Android https://goo.gl/KR1vK6
Shena'Fu's Online Card Creator https://bit.ly/cardcre