by **Lummox JR** » Sun Oct 02, 2005 6:37 am

To get a full understanding of coloring, benjam, it's important to know that there are several ways it can eliminate candidates. Also, it's not so much that you can't color a cell if there are more than 2 candidates in the box; the rule is only that it must have a conjugate. In this case you were able to follow a chain of conjugates right along, but coloring won't always work that way.

The method you saw here showed that you can start off with one square and follow conjugates until one color eliminates itself. In this case you ended up with two blue cells in the same box.

You can also find cells where two conjugate colors intersect. If that cell "touches" both blue and green, it must be false.

However far more commonly you'll find multiple colors, which can still be useful. To represent colors I usually go with letters, such as aA for the first pair of conjugates, bB for the second pair, etc. With multiple colors you may be able to connect them, if for example you find that a and b are in the same house, and A and B are in the same house, in which case a=B and b=A. You may also find that a particular color is impossible because it connects to two conjugates, such as if a and b connect, and A and b connect; there b must be false.

When working with multiple colors I now know the best method is not to try to prove several colors are equivalent, but rather to prove that one excludes another. Nick70 did some good work in this regard, proving you can say that if a excludes b, and B (b's conjugate) excludes c, a excludes c. If you know that two colors exclude each other, at least one if not both of their conjugates must be true. Therefore you can eliminate any cells that intersect those colors. (E.g., if a excludes b, then any intersection of A and B is false.) I don't think most people take coloring that far, but it's very worth it because it will show eliminations that otherwise will not appear.