by **Animator** » Fri Jun 03, 2005 2:50 pm

Actually Andrew, it was your reply that confused me...

First of all, there are two distinct groups: 4, 6, 7 and 2, 3, 8.

The group 4, 5, 6, 7 is incorrect.

Why? there are three cells that share the values 4, 6 and 7 (r1c3, r8c3, r9c3). This allows you to remove 4, 6 and 7 from r7c3 leaving only the number 5 as a candidate.

To Goldie, you should be looking for two things, just as you should for finding pairs...

First is a group of cells, which share the same candidates and where there are exactly the same number as cells as the number of candidates. In this case that would be the 4, 6, 7 group. In these cells you may not have any other candidates.

The other group of cells you should look for, is a group of cells which are the only possiblities for certain numbers. Again, the number of cells has to match the number of candidates. In this puzzle this would be the 2, 3, 8 group. These cells can have more candidates, but those are irrelevant, since these cells need to have the numbers 2, 3 8 (in some order).

Here is another example with only two numbers:

{1}, {2,3,4}, {3,4}, {3,4}, {2,3,4,5,6,7,8,9}, {2,3,4,5,6,7,8,9}, {2,3,4,5,6,7,8,9}, {2,3,4,5,6,7,8,9}, {2,3,4,5,6,7,8,9}

In this example there are cleary two cells that can only have the numbers 3 and 4. Therefor you eliminate them from the other cells, right? (or haven't you applied that technique yet?) (this would be the same as the first group I explained)

Another example:

{1}, {2,3,4,5,6,7,8,9}, {2,3,4,5,6,7,8,9}, {2,3,4,5,6,7}, {2,3,4,5,6,7}, {2,3,4,5,6,7}, {2,3,4,5,6,7}, {2,3,4,5,6,7}, {2,3,4,5,6,7}

In this second example there are only two cells that can have the numbers 8 and 9, but they have other candidates too... but because you know that these are the only two valid cells you know you can remove the other candidates from the cell. (this is the same as the second group)

Does this make sense to you?