ronk wrote:So one needs to consider ...

- only those candidates outside the uniquenss pattern which destroy the deadly pattern, and
- of those, only the candidates which have complements, and
- of those, only one of each complementary pair (pair?) of candidates.

No. If you choose to consider candidates outside the uniqueness pattern which destroy the deadly pattern it is enough to consider only the candidates in either all rows, columns or boxes involved. In this MUG case the column alternative is more complicated (as there is more than 2 candidate values for each cell) so let's forget about that for now.

- Code: Select all
`*-----------------------------------------------------------*`

| 356 4 #1356 |*689 *159 *1689 |#578 2 567 |

| 9 578 567 | 24 245 +68 | 1 4568 3 |

| 56 18 2 | 3 +14 7 | 9 48 56 |

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

| 2567 9 3567 | 1 2368 4 | 23578 3568 2567 |

| 8 12357 13567 | 269 236 39 | 2357 1356 4 |

| 2346 123 346 | 7 2368 5 | 238 9 126 |

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

| 47 37 8 | 5 +3479 2 | 6 13 19 |

| 1 2357 9 | 46 467 36 | 2345 35 8 |

| 2345 6 345 |*89 *1349 *18 | 234 7 #259 |

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

Looking at the extra values in the two rows we have the possibilities: r1c3=1, r1c7=8 and r9c9=9. Now, if r1c3<>1 and r1c7<>8 then r1c456 is a hidden triplet 189 and if r9c9<>9 then r9c456 is a hidden triplet 189 => Nothing you can do could possibly avoid the deadly pattern anymore! Therefore one of the three mentioned cellvalues must be true.

Looking at the boxes, we also have three candidates outside the pattern: r2c6=8, r3c5=1 and r7c5=9. If r2c6<>8 and r3c5<>1 then r1c456 is a hidden triplet 189 and if r7c5<>9 then r9c456 is a hidden triplet 189 => at least one of the three cellvalues must be true. This would btw let us make the elimination

if r1c6=1 (=>r3c5<>1) => r9c6=8 (=>r2c6<>8) => r9c4=9 (=>r7c5<>9)

=> r1c6<>1

RW