Sudoku Players' Forum (clone)

Can the two ends of an XY chain inhabit the same box and validly eliminate one or more candidates in that box?

I can't see why not though the absence of a reply to this stage doesn't make me feel too confident with my response .

In the following random example showing say Box1, candidates [26] [68] and [28] comprise an XY-Wing with candidates [26] being the "Pivot" cell (r1c2). The two "Stem" cells are r1c3 and r3c2 and include two candidate 8's. As another candidate 8 in r3c3 "sees" both ends of the two "stem" cells (ie. it "sees" the two candidate 8's in r1c3 and r3c2) then that candidate 8 can be removed from r3c3.

Code: Select all

 *--------*
 | 1|26|68|
 |--|--|--|
 | 4| 5| 9|
 |--|--|--|
 | 7|28|38| 
 *--------*


The XY-Wing chain would be...

8-[r3c2]-2-[r1c2]-6-[r1c3]-8

Cec

sure they can, but then they become a naked triple (as in Cec's example) or a naked quad!

On second thoughts these aren't the only possibilities. The chain can start and end in the same box but go outside of the box. I'm sure I've come across examples of this. I'll post them if I can find them!

here you go:

Code: Select all

 . . . | 2 . 4 | . . .
 4 . . | . . . | . . 1
 . . 1 | . . . | 3 . .
 ------+-------+------
 . . 3 | . . . | 9 . .
 . 9 . | 6 5 2 | . 8 .
 6 . 8 | . . . | 4 . 2
 ------+-------+------
 . . 5 | . . . | 8 . .
 . . . | 3 6 5 | . . .
 . 6 . | 1 . 9 | . 3 .



After singles and locked candidates you reach here:

Code: Select all

 . 3 7 | 2 1 4 | 6 . 8   
 4 . 6 | . . . | 2 7 1   
 . . 1 | . 7 6 | 3 4 .   
 ------+-------+------
 . . 3 | . 4 . | 9 . .   
 1 9 4 | 6 5 2 | 7 8 3   
 6 . 8 | . . . | 4 . 2   
 ------+-------+------
 3 1 5 | 4 2 7 | 8 . .   
 8 4 9 | 3 6 5 | 1 2 7   
 7 6 2 | 1 8 9 | 5 3 4   


Now there's an xy chain starting at r3c4 finishing at r2c6 which eliminates 8 from r2c4.

Thanks ab for confirming my thoughts. I did realize the [268] candidates formed a naked triple but wanted to "keep it simple" to answer
caraemily's query.

Cec