First of all, I'm new at this crazy game. How'd I get roped into this, anyway??? Oh, yeah. I did it to myself, go figger.
What I know of this technique comes from Scanraid,
http://www.scanraid.com/AdvanStrategies.htm#XYZ
The Scanraid web page shows only the "hinge" square as containing all three of the possible candidates, or XYZ. The other squares contain only XZ and YZ.
Today I found a puzzle in which the "hinge" candidates were duplicated in one of the arms, all three. My question is, with this technique can you have more than two candidates in the "outer cells?" In this case, it allowed me to eliminate 2 candidates instead of just one.
Here's the germaine portion of the puzzle:
* | * | * || * | * | * ||12379| 5 | 139 ||
* | * | * || * | * | * || 46 | 27 | 46 ||
127 | 4 | 39 || 27 | 56 | 56 ||12379| 8 | 139 ||
The "pivot" square is R3C9 (139). The "outer" cells are R3C3 (39) and R1C9 (139 again). It seems to me that there exists no scenario in which R3C7 can contain 3 or 9, since all three cells can "see" the contents of R3C7.
In most XYZ wings the outer cells have only 2 candidates. In other words,
R1C9 would have only the candidates 13 or 19, not 139. I have never seen an example like I've shown here on any web page.
Am I missing something?
Luke in Ca