Did it at last with coloring, no guess. Coloring is about like doing two chains simultaneously, one red chain, one green chain. Eventually -- but you have to be very careful -- one of the colors has a conflict and you can enter all of the non-conflicting colors at once.
EZPZ, never a rest from clicking.
Difficulty score 18. No green.
Since the subject of coloring came up, here is my idea of how to do it. It only take 1 digit, and either a cell can or cannot be that digit. You construct a chain starting with the assumption that the first cell cannot be the digit. That lead to a logical deduction that another cell must be the digit in question. From that you find a way to infer that a third cell must be the selected digit, and so on. You end the chain with a cell the must be the digit. If there are any cells that can "see" both the starting cell of the chain and the ending cell, then the digit can be eliminated as a possibility in those cells.
Oops, the third cell in the chain cannot be the digit. The must be and cannot be condition alternates down the chain starting with cannot be and ending with must be.
Since either the first cell in the chain is the digit then none of the cells that can see the first cell can be the digit, or the last cell in the chain must be the digit eliminating all cells that can see the last cell in the chain. So cells that can see both the first and last cells in the chain cannot be the digit in either case.
I can do it so much faster without the pressure of having to GO!!!
Only 475 more puzzles until I am fully caught up! At the rate I'm going, I'll be Elite again in 2017.