[/spoiler]
but I did reach a point where there seemed to be a cyclic dependency, and to make any further progress I had to guess, expecting to backtrack (it was a 50% guess and I got it right). That’s not normal in Sudoku so I don’t know if I’d overlooked something.

Are you familiar with the “BUG removal” technique? From the point you reached, you should be able to eliminate all but two possibilities from all but two of the cells; those two cells have three possibilities. If there were just one, you’d be able to apply BUG removal. Can you see how to apply it anyway?[spoiler]In the bottom center-block, the possibilities for the four corner squares are

[149] [12]
[14] [128]

If the 1 is in the upper-right or lower-left corner, then we can eliminate it from upper-left and lower-right. But this leaves us a case in which all unfilled squares have exactly two possibilities, which is forbidden by BUG. So the 1 must be in upper-left or lower-right, and hence can be eliminated from upper-right and lower-left.[/spoiler]That’s one way to proceed, anyway.

I’ve never liked techniques that rest on the assumption that there is exactly one valid solution. It seems to me that a good technique should be able to confirm the existence of a unique solution. I’ll use such methods to find wrong guesses, to use in a guess-and-check method, but not for the actual solution (that is to say, “It looks like this one is probably wrong, so I’ll see where that leads, in the hope of proving it wrong”).

I don’t like them either, and clearly they aren’t strictly necessary in this case (it in fact has only one solution); but I hadn’t found any other common patterns that apply. Now I’ve found a nicer solution, though.From your partial solution, R7C3 must be [19], R7C4 must be [149], and R9C4 must be [14]. Since all three of these must contain some values in [149], R7C6 cannot contain any of [149], so it must be 2.

For the OP’s partial solution, it’s easy to see that R7C6 must be [12]. But my point is thatat least one of R7C3=[19], R7C4=[149], and R9C4=[14] must be a 1, and if any one of these is 1 it rules out R7C6=1. So R7C6=2. (BrainBashers calls this an “XYZ wing.”)

Wow. I think I understand this. An interesting extension of the exclusion rules I am used to. Whether I can apply it in the future is another question entirely, because I am at a loss as to how one would spot it in the first place.

Oh, and you are all right that R2C9 = 8. I just knew I couldn’t do that without at least one screw up. :smack:

A properly constructed Sudoku has one, and only one correct solution. Suppose you have a situation like the following, four cells and you’ve narrowed each one down to two possibilities:

[1,2] … [1,2]

[1,2] … [1,2]

Those four cells can be anywhere in the puzzle as long as they form a square, and not so far apart that they are in four separate blocks. (They can’t all be in the same block, either.)

This puzzle must have two solutions. Either

1 … 2

2 … 1

or

2 … 1

1 … 2

Which means it’s not a properly formed Sudoku. That’s a BUG pattern and it should never happen.

Some Sudokus are about eliminating wrong possibilites as much as filling in right ones. If there’s a 3 in the first column, I cross out the 3’s in all the other first-column cells because two 3’s in a column Can’t Happen. This is the same reasoning. If you see a number that would lead to a BUG pattern, you can cross it out because a BUG pattern Can’t Happen.

Of course, I’d probably try every other method first. Even if my description is correct, I’m not sure I could pick that out in a real puzzle.