Colouring: Minimum number of candidate cells

Ruud · Site Admin

I'm trying to figure out what the minimum number of candidate cells for a digit would be, to avoid useless colouring.

Here's what I've figured out so far:

When a digit is bound to 9 cells in the grid, that digit is completely solved.

There is no situation with 8 bound digits, because the last one would be a single, and therefore bound.

With 7 bound digits, the 4 candidates will form an X-wing shape, because there are only 2 rows and 2 columns that allow the remaining digits. Colouring would be useless.

With 6 bound digits, there can be between 6 and 9 candidates for the remaining 3 digits. With less than 6 candidates, at least one of the digits will be a hidden single. However, I can see no scenario where colouring would be able to eliminate candidates.

So this is my provisionary minimum:
1. At least 4 unbound digits
2. At least 9 candidate cells (4 pairs + 1 extra)

I would appreciate an expert opinion.

Lummox JR · Posted: Mon Oct 17, 2005 12:03 am Post subject:

I believe you may be correct. Coloring with 3 unbound digits is certainly impossible, but with 4 you can find a workable pattern. It did also require 9 cells for me to build a pattern.