The New Sudoku Players' Forum

[Yogi]

I’ll leave the programming and proving to others in a different thread. Let’s try to understand how a Suduku-solving PERSON might spot one.

57....9..........8.1.........168..42...1.28.9..2.9416.....26....6.9.82.4...41.6..

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+----------------------+----------------------+----------------------+
| 5      7      3468   | 238    346    13     | 9      123    136    |
| 23469  2349   3469   | 2357   34567  13579  | 3457   12357  8      |
| 234689 1      34689  | 23578  34567  3579   | 3457   2357   3567   |
+----------------------+----------------------+----------------------+
| 379    359    1      | 6      8      357    | 357    4      2      |
| 3467   345    34567  | 1      357    2      | 8      357    9      |
| 378    358    2      | 357    9      4      | 1      6      357    |
+----------------------+----------------------+----------------------+
| 134789 34589  345789 | 357    2      6      | 357    135789 1357   |
| 137    6      357    | 9      357    8      | 2      1357   4      |
| 23789  23589  35789  | 4      1      357    | 6      35789  357    |
+----------------------+----------------------+----------------------+


I had thought that tridagons were some kind of loop chain, but it seems that they are more like a pattern of cells with certain constraints. As I understand it now, it’s made up of 12 cells, each with the same 3 candidates, located within 4 boxes which are arranged in a rectangle, except that at least one of the cells has to have an extra candidate. In this way the situation is much like BUG+1, but it's a localized trivalue pattern rather than a bivalue pattern involving the whole puzzle. Again, without an extra candidate (or guardian) you would have a Deadly Pattern leading to multiple solutions or no solution. In this tridagon the boxes are 5689 and the candidates are 357. Here, r8c8 is the only cell in the pattern with the extra candidate, so it can be immediately solved for 1, simplifying the puzzle. Also, within the boxes each group of three cells must include all rows and columns, so they necessarily form a diagonal of some shape. I note that here 3 of the diagonals are ‘Falling,’ and one is ‘Rising.’ I wonder if this will always be the case. Is the odd one the one we call the antidiagonal?
If more than one of the cells has an extra candidate it is more complicated, just as BUG+2 requires more work to find a solution than BUG+1, but the tridagon still provides a path to advance the puzzle. I believe that it can also work if not all the 12 cells have all 3 of the candidates.
But we are just getting started here . . .

[Leren]

Well, I'm not a Trigadon expert, but here are pictures of the rising and falling box patterns, which I got off the Phil's Folly site.

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| .  .  * |    | *  .  . |    | .  *  . |            | *  .  . |    | .  .  * |    | .  *  . |
| .  *  . |    | .  .  * |    | *  .  . |            | .  *  . |    | *  .  . |    | .  .  * |
| *  .  . |    | .  *  . |    | .  .  * |            | .  .  * |    | .  *  . |    | *  .  . |
                 RISING                                               FALLING

A Trigadon is not a uniqueness pattern, if it is fully exposed, there will be not be a solution for the puzzle. The Trigadon must include 3 rising and one falling box patterns, or 3 falling and one rising box patterns.

AFAIK you can have missing digits in the 12 cells, but these degenerate Trigadons have not been studied much on this site, as there would be too many cases to consider.

Also a large number of similar impossible patterns have been discovered and catalogued. Well, that's it for me. If you want to learn more, go to the Trigadon thread and start reading.

Leren