While doing a puzzle I had, twice, an X-wing come up where 3 of the boxes held the same 2 candidates and the 4th box contained the same 2 candidates plus one more. On a whim I tried making the box with the 3 candidates the number that was the odd one and it worked. Twice.
Example: r1c1 has candidates 1 and 5 ***** r1c3 has candidates 1 and 5 ***** r6c1 has candidates 1 and 5 ***** r6c3 has candidates 1, 5 and 9.
I made r6c3 a 9 and it worked (because the other 3 held identical candidates). It worked a 2nd time on another X-wing on the same puzzle.
Was this just a fluke or is this an actual solving technique?
Thank you.
Camchase
question on a solving technique
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question on a solving technique
by Camchase » Sat Feb 17, 2007 7:10 am
- Camchase
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by udosuk » Sat Feb 17, 2007 8:59 am
Congratulations, you're the 999876th individual who "discovered" this technique. It's academic name is "Unique Rectangle (Type 1)", and you can learn more in this thread.
Two particular cautions though.
1. This technique can only works on the premise that the puzzle has a unique solution (that's why it's named such). On puzzles with 2 or more solutions (and requiring you to find one of them), or on certain Sudoku variants which have extra constraints (such as the diagonals have no repeated numbers), applying this technique recklessly will run you into trouble (such as entering a state of "no solution").
2. This technique can only be applied when the 4 cells involved are spanning two rows, two columns and two 3x3 blocks. Your example works because (r1c1,r1c3) and (r6c1,r6c3) are spanning two different blocks. If, for example, they're (r1c1,r1c4) and (r6c1,r6c4), i.e. spanning four different blocks, you cannot apply this technique.
Two particular cautions though.
1. This technique can only works on the premise that the puzzle has a unique solution (that's why it's named such). On puzzles with 2 or more solutions (and requiring you to find one of them), or on certain Sudoku variants which have extra constraints (such as the diagonals have no repeated numbers), applying this technique recklessly will run you into trouble (such as entering a state of "no solution").
2. This technique can only be applied when the 4 cells involved are spanning two rows, two columns and two 3x3 blocks. Your example works because (r1c1,r1c3) and (r6c1,r6c3) are spanning two different blocks. If, for example, they're (r1c1,r1c4) and (r6c1,r6c4), i.e. spanning four different blocks, you cannot apply this technique.
- udosuk
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- Joined: 17 July 2005
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