there is a naked pair r9c2 and r9c8, 2 cells containing the candidates {2,9}, that can eliminate the 2's and 9's elsewhere on row 9, specifically r9c3, r9c6 and r9c9
There is a row claim on the box for all the candidates {3}: all the 3's in row 8 are also in box 7, so the positions in box 7 not on row 8 can be elimintated. If it were not so, row 8 wouldn't have any candidate 3. This eliminates the 3 as a candidate in r9c3
There is a rc x-wing on the 7's row 5 and row 8 column 1 and column 4 that permits a removal of the candidate 7's in r8c3 and r8c6
Code:
Hugo's Puzzle
`8 ``3 ```4 | ``6 7 ``1 | `2 `5 ``9
`1 ``6 ```2 | ``4 9 ``5 | `7 `3 ``8
`9 ``7 ```5 | ``2 8 ``3 | `1 `6 ``4
------------+-----------+-----------
`2 ``5 `139 | `89 4 689 | 39 `7 136
37 `19 ```6 | `79 5 ``2 | `4 `8 `13
`4 ``8 ``79 | ``1 3 679 | `5 29 `26
------------+-----------+-----------
`5 129 1789 | ``3 6 789 | 89 `4 `27
37 ``4 `389 | 789 2 `89 | `6 `1 ``5
`6 `29 ``78 | ``5 1 ``4 | 38 29 `37
There is an ALS-XZ move that proves r9c2 cannot be {9}
one of the sets defining the move lies r5.29 or r5c2 and r5c9. This set holds three candidates {139} in two positions.
the second set defining the move is b9.389 or the third, eighth and ninth positions of box number 9. or r7c9, r9c8 and r9c9. Here there are four candidates {279} in three positions.
There is a restrictive common candidate {3} between the two sets. That is to say the candidate {3} is either in the first set, or the second set, or none of the sets.
The {9} in r9c2 contacts all the candidate {9}'s in both sets. If r9c2 was a {9}, both sets would revert to naked subsets, violating that {3} must be in one or the other set. I abbreviate this move
r5.29 -3- b9.389 {9} r9c2
the -3- is a "weak link" notation for the connection betwen the two sets.
Code:
Sudoku Grid
`8 `3 ``4 | `6 7 `1 | 2 5 `9
`1 `6 ``2 | `4 9 `5 | 7 3 `8
`9 `7 ``5 | `2 8 `3 | 1 6 `4
----------+---------+--------
`2 `5 `13 | `8 4 `6 | 9 7 13
37 19 ``6 | 79 5 `2 | 4 8 13
`4 `8 `79 | `1 3 79 | 5 2 `6
----------+---------+--------
`5 19 179 | `3 6 79 | 8 4 `2
37 `4 `39 | 79 2 `8 | 6 1 `5
`6 `2 ``8 | `5 1 `4 | 3 9 `7
next, r5c4 cannot be a {9} because r6 and r8 would only have {9} then in column 3. I abbreviate this
ff {9} r5c4 % r68 > c3
thenseforth hidden singles are enough to complete the solution.
