eleven wrote:Again very easy to solve. After clearing the diagonal there is a UR 1.1 (12 or 16), then only 2 pairs to the end.Mauricio wrote:One more automorphic puzzle

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`+-------+-------+-------+`

| . . . | . . 1 | . . 2 |

| . 1 . | . 3 . | . 4 . |

| . . 5 | 6 . . | . . . |

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

| . . 2 | 7 . . | 8 . . |

| . 8 . | . . 4 | . 3 . |

| 1 . . | . 9 . | . . . |

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

| . . . | 3 . . | 5 . . |

| . 9 . | . 8 . | . 1 . |

| 6 . . | . . . | . . 7 |

+-------+-------+-------+ ER=11.3/11.3/10.5

There are 2 paths to solve this puzzle.

Path 1 - using a "pseudo UR" (sorry, forgot the official name of this technique), i.e. the path adopted by eleven:

After filling the leading diagonal with the 3 self-mapping digits {1,5,7}, and applying the ensuring singles:

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`+----------------------+----------------------+----------------------+`

| 7 346 34689 | 459 45 1 | 369 5689 2 |

| 289 1 689 | 59 3 2789 | 679 4 5689 |

| 23489 234 5 | 6 247 2789 | 1 789 389 |

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

| 459 45 2 | 7 6 3 | 8 59 1 |

| 59 8 679 |*2 *1 4 | 679 3 569 |

| 1 3467 3467 | 8 9 5 | 2467 267 46 |

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

| 248 247 1 | 3 247 2679 | 5 2689 4689 |

| 2345 9 347 | 45 8 267 | 2346 1 346 |

| 6 2345 348 |*1 -245 29 | 2349 289 7 |

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

Consider r59c45: these 4 cells are not part of the initial givens, so can't form a deadly pattern (otherwise the original puzzle must have multiple solutions). Therefore r9c5 can't be 2.

Thus r19c5={45} (naked pair @ c5).

4 @ r3,b1 locked @ r3c12, can't be @ r1c23.

4 @ r4,b4 locked @ r4c12, can't be @ r6c23.

4 @ c3,b7 locked @ r89c3, can't be @ r789c12.

Hidden single @ r7: r7c9=4.

Symmetry: r9c7=9.

All naked singles from here.

(eleven seems to suggest that this puzzle can be solved by 2 pairs and singles only after the pseudo UR. I don't see how to do this. I have to use 1 naked pair and 3 intersections (aka locked candidates) plus singles to finish it. )

Path 2 - without using UR-related techniques:

From the above pencilmark grid:

r1c45+r2c4={459} (naked triple @ b2).

r4c12+r5c1={459} (naked triple @ b4).

4 @ r3,b1 locked @ r3c12, can't be @ r1c23.

9 @ c3,b1 locked @ r12c3, can't be @ r23c1.

9 @ r3,b3 locked @ r3c89, can't be @ r12c789.

r1c27={36} (naked pair @ r1).

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`+-------------------+-------------------+-------------------+`

| 7 36 89 | 459 45 1 | 36 *58 2 |

| 28 1 689 | 59 3 278 | 67 4 568 |

| 2348 234 5 | 6 27 278 | 1 789 389 |

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

| 459 45 2 | 7 6 3 | 8 *59 1 |

| 59 8 67 | 2 1 4 | 679 3 569 |

| 1 367 367 | 8 9 5 | 2467 267 46 |

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

| 248 247 1 | 3 247 2679 | 5 *2689 4689 |

| 2345 9 347 | 45 8 267 | 2346 1 346 |

| 6 2345 348 | 1 245 29 | 2349 *289 7 |

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

Now consider r79c8.

By symmetry these 2 cells can't be {26}.

So r79c8 must have at least one of {89}.

Thus r1479c8 form a killer naked triple {589} @ c8.

Therefore r36 can't have {589}.

Hence, r3c8=7.

Symmetry: r8c3=7.

All naked singles from here.