Between the cells weak links are sufficient, linked cells have at least 1 candidate in common.

If that candidate k is ON in one endpoint and OFF in the other endpoint than all cells that 'see' the 2 endpoint cells can eliminate candidate k.

I want to check the puzzle below using the XYchain method

- Code: Select all

+-----------+------------+-----------+

| 1 8 4 | 5 79 3 | 69 67 2 |

| 79 3 2 | 6 4 8 | 19 5 17 |

| 579 57 6 | 2 79 1 | 4 3 8 |

+-----------+------------+-----------+

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

| 2 67 5 | 9 8 4 | 36 1 37 |

| 67 4 3 | 1 5 2 | 8 67 9 |

+-----------+------------+-----------+

| 56 56 8 | 3 1 9 | 7 2 4 |

| 4 2 7 | 8 6 5 | 13 9 13 |

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

+-----------+------------+-----------+

**Hidden Text:**Show

Lots of bi-value cells and r3c1 has 3 candidates. The solution of the puzzle is known and r3c1 must be a 7.

But I can make a XYchain (57)r3c2-(79)r3c5-(97)r1c5-(69)r1c7-(76)r1c8-(67)r6c8-(76)r6c1.

r3c1 sees both end-cells of the chain and cannot be a 7 and so ruins the puzzle.

What is wrong here?