Original puzzle
- Code: Select all
*-----------*
|...|...|...|
|49.|.6.|.82|
|...|27.|..1|
|---+---+---|
|.56|..4|19.|
|..4|...|7..|
|.39|8..|24.|
|---+---+---|
|6..|.85|...|
|34.|.2.|.57|
|...|...|...|
*-----------*
Pencilmark grid showing a 10-cell XY-chain starting at r1c2 designated (A) and ending at r3c8 (K)
- Code: Select all
*--------------------------------------------------*
| 1 A26 23B | 45 45 8 | 369 7 69 |
| 4 9 7 | 1 6 3 | 5 8 2 |
| 5 68 38C | 2 7 9 | 4 36K 1 |
|----------------+----------------+----------------|
| 2 5 6 | 7 3 4 | 1 9 8 |
| 8 1 4 | 56H 9 2 | 7 36J 356 |
| 7 3 9 | 8 15G 16F | 2 4 56 |
|----------------+----------------+----------------|
| 6 7 12 | 349 8 5 | 39 12 349 |
| 3 4 18D | 69 2 16E | 689 5 7 |
| 9 28 5 | 346 14 7 | 368 12 346 |
*--------------------------------------------------*
To my knowledge, the notation for this XY-chain starting at A and ending at K with the common digit 6 is....
6-[r1c2]-2-[r1c3]-3-[r3c3]-8-[r8c3]-1-[r8c6]-6-[r6c6]-1-[r6c5]-5-[r5c4]-6-[r5c8]-3-[r3c8]-6
Because the two candidate 6s in Box3 (r1c79) and candidate 6 in Box1 (r3c2) "see" both the start (A) and end of the chain (K) then these three candidate 6s can be excluded leaving digit 6 to solve cell r1c2.
However, by setting 6 at the start of the chain (r1c2) the implications leave 3 at the end of the chain (r3c8) which then leaves row3 without a 6 because the row now has two 3s. Alternatively, setting 2 at r1c2 leaves 6 at the end of the chain (r3c8) which still leads to a conflict because there is no 3 in row3 (that's because there's two 8s).
I realize there is something wrong with my above analysis but I cannot see where I'm wrong. Thanks in advance of any help to explain my dilemma
Cec