denis_berthier wrote:mith wrote:

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`1.3....89.57.8.....68..2....8.....4.......3...34....123....14.....92...3....43..1;15082;482867;3`

The first has a relatively tame TOFC from the 6 guardian candidate choice which reduces it to a couple skyscrapers, but the second looks completely disgusting. Curious what your solver will make of this one, Denis.

SER = 11.7

Harder, but solved in W8+OR6W9

..............

hidden-pairs-in-a-column: c5{n1 n3}{r3 r4} ==> r4c5≠9, r4c5≠7, r4c5≠6, r4c5≠5

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

| 1 2-4 3 | 4567 567 567 | 257 8 9 |

| 24 5 7 | 134 8 9 | 126 236 46 |

| 9 6 8 | 13457 13 2 | 157 357 457 |

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

| 2567 8 12569 | 123 13 567 | 5679 4 567 |

| 567 129 12569 | 12 #5679 4 | 3 #5679 8 |

| 567 3 4 | 567 5679 8 |#5679 1 2 |

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

| 3 279 2569 | 8 567 1 | 4 25679 567 |

| 4568 147 156 | 9 2 567 | 5678 567 3 |

| 2568 279 2569 | 567 4 3 |#256789 25679 1 |

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

I see that, from here can elimination 4r1c2 as below: Tridagon (567)B5689 => (9)r5c5=(9)r5c8/r69c7=(2)r9c7=(8)r9c7

Present by diagram: => r1c2<>4

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`(8)r9c7-r9c1=(8-4)r8c1=(4)r2c1*`

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(9)r5c5-r6c5=r6c7--(9=567)r4c679-(567=2)r4c1--(2=4)r2c1*

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(9)r5c8/r69c7----- |

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(2)r9c7-r12c7=r2c8---------------------------

Thanks for the puzzle!

totuan