Coloring, Forcing Chains
This is a technique I only use on extremely difficult (Super-tough, Insane) puzzles once I’ve solved as much as I can with the prior techniques; having fully marked-up the remaining puzzle and reduced most of the cells to 2 or 3 choices.
The basic idea is to pick a cell that has just two possible numbers, and then use two different colored pencils (or highlighter-shades, in the case of an e-ink tablet), to work out the implications of both choices.
Circle the origin cell, so you can remember where you started. Then pick one of the two numbers and circle it. Then work out the implications of picking that number, following the chain of cancelations, and crossing out all the remaining choices in other cells that are ruled out. For example picking a 5 will cause all the 5s in the shared containers to be canceled out, which in turn forces other numbers (in all the cells with just two options), which leads to other cancelations, and so on.
Then use a different color or highlighter, and pick the second digit, and follow all the cancelations made by that second choice.
At this point, several things may happen, most of which are useful:
1. In one of the two chains, you may encounter a conflict or contradiction. For example, a cell may have all its possible values ruled out, or two cells in the same container may be forced to the same value. When this occurs, the particular number at the root of that chain is invalid and can be safely erased. Since you started with just two numbers, you’ve now solved your original square – it must be the other number!
2. In both chains, the same value may be forced for one or more cells. In this case you’ve solved those cells, and can probably make further progress using simpler techniques.
3. One of the chains may lead to solving the entire puzzle. This can’t happen with both, since the puzzle only has a single solution. The chain that solves the puzzle is obviously the correct choice.
4. It’s possible both chains won’t get very far and will both look reasonable. At that point, I would suggest erasing your marks and starting again with a different choice for root cell.
Here’s a puzzle where you can make some progress by following the chains made by the two choices in cell r2c1, which ultimate affects the cell in r1c2.
There are two candidates (1 and 2).
If r2c1=2, then r1c2=7.
If r2c1=1, then r5c1=2, and so r6c2=1, and so r6c8=3, and so r1c8=2, and so r1c2=7.
So whichever of the two values are placed into r2c1, we always get a 7 in r1c2.