My sudoku brain broke

Stuck on LA Times Sudoku page (no registration required), a 5-star puzzle from Friday December 26th. Link. Given numbers are in green; my inserts are in black.

I always seem to get to this point and get stuck. There’s obviously a technique for getting past it, but the techniques (X-wings and swordfish) are hard to grasp. What am I looking for here?


6 2 7 | _ _ _ | 9 _ 5
9 5 _ | _ 2 _ | _ 7 _
8 _ 3 | 9 5 7 | 2 _ _
- - - - - - - - - - -
4 _ _ | 5 _ _ | 7 _ 9
5 _ 9 | 7 1 4 | _ 2 _
7 _ _ | _ 3 9 | _ 5 4
- - - - - - - - - - -
3 _ 5 | _ _ _ | 6 _ 7
2 9 _ | _ 7 5 | _ 8 3
1 7 _ | _ _ _ | 5 _ 2

Here’s a few clues:

R2C9 cannot be 6, because the only places that can be 6 in the second 3x3 square are in row2. That means that R2C9 can only be 1 or 8.

It is then possible to show by a chain of either/ors that if R6C7 is not 1, then R2C3 has to be 1, and vice-versa. This means that the two squares at R2C7 and R6C3 cannot be 1.

This opens up a chain of either/ors for the 1s that eliminate further candidates from being 1s. Some of the eliminated squares than have only two numbers left and can be scouted for either/or chains for other numbers, etc.

P.S., I couldn’t do this without a computer aid that shows me remaining candidates for each square.

In square 3 on the top section, 8 can only go in R2, C7 or 9. That eliminates 8 from R2 in square 2 ; 8 can only go in R1.

Actually, the clues I gave above aren’t that helpful, since they dead-end. Here’s a much better one:

R5C2 cannot be 8, since either R7C2 is 8, or if it’s not, then by a chain of either/ors R5C9 has to be 8.

That opens up the rest of the puzzle.

Since row 5 contains a 1, the only possible 1’s in column 9 are in row 2 or 3, so other 1’s can be excluded in square 3.

Lumpy - could you explain the either/or thing to me? Also, what calculator are you using? Simple Sudoku got stuck at the same place.

Ditto square 2 - the one’s in row 2 can be excluded and the 6’s in row 2 of square 3.

I was using Simple Sudoku to display the candidates but it doesn’t solve it automatically. When I plug in Fish’s numbers I get the following results:

R7C2 is either 4 or 8. R3C2 is 1 or 4. R3C9 is 1 or 6. R5C9 is 6 or 8.

R5C2 can’t be 8 if R7C2 is 8. If it’s 4, then R3C2 has to be 1; R3C9 has to be 6; and R5C9 has to be 8. In which case R5C2 can’t be 8 anyway. I believe most Sudoku references call this type of proof “chaining”.

Ah, thank you Lumpy. I see where you’re going with that.

If only there were a simple way to see such chains without all the “if this, then that” throughout the puzzle.

Errrm, I’m still not getting it.

The five squares in Lumpy’s example make a shape like a P, like this:


1 or 4 .... 1 or 6
R5C2   .... 6 or 8
4 or 8

We can test the possibilities 4 and 8 in that one box. If we choose 8, then R5C2 cannot contain 8.

If we choose 4, then you follow the possibilities around the chain and end up with 8 in the “6 or 8” box.

In both cases you have ruled out 8 in R5C2.