wapati wrote:This is a puzzle I like, not too hard yet hard enough.

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

. . . | 6 . . | 2 9 .

5 9 . | . . 4 | . . 3

---------------------

8 . . | . 4 . | . . .

. . 3 | . . 9 | . 1 2

6 . . | . 3 . | . . .

---------------------

9 1 . | . . 8 | . . 7

. . . | 9 . . | 4 8 .

. . 5 | . . 7 | . . 9

Ooh, that one is tricky with the x-wing coming so early. It seems one can solve it with that x-wing, an xy-wing, a skyscraper and a UR. But here is a different approach using BUG-lites. First, grant me the early x-wing so that I can get to here:

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` *-----------------------------------------------------------*`

| 12 6 8 | 12(5) 9 3 | 157* 57* 4 |

| 13 37 4 | 6 1578 15 | 2 9 158 |

| 5 9 27 | 12(8) 178 4 | 18 6 3 |

|-------------------+-------------------+-------------------|

| 8 25 19& | 157& 4 12(5)6| 379*& 37* 56 |

| 7 4 3 | 58% (58)6 9 | 568 1 2 |

| 6 25 19& | 1578& 3 12(5) | 79& 4 58 |

|-------------------+-------------------+-------------------|

| 9 1 6 | 4 2 8 | 35* 35* 7 |

| 23 37 27 | 9 15 156 | 4 8 16 |

| 4 8 5 | 3 16 7 | 16 2 9 |

*-----------------------------------------------------------*

To avoid a deadly pattern in r147c78, we must have r1c7=1 or r4c7=9. On the other hand, to avoid a deadly pattern in r46c347, we must have r4c4=5, r4c7=3, r6c4=5 or r6c4=8. The key cell is now r5c4=[58]. Take any cell that sees r456c4. If it contains a 5 or 8, then r5c4 will contain the other and hence we must have r4c7=3. But an easy argument (using the strong link on 7 in row 1) then shows that r1c7 = 7, a contradiction.

We conclude that r1c4, r5c5, r46c6<>5 and r3c4, r5c5<>8, solving the puzzle. I'm not sure how this should be written up as a chain.