The New Sudoku Players' Forum

[eleven]

Thanks, champagne.

The pairs exclusions are very similar to, how i verified the eliminations.
I also remembered now, that i have seen the abi loop, but i forgot about it, because i have never used it myself.

[Cenoman]

My solver found this DBJE

I was surprised at first with the qualification "Junior" for these exocets. I assume that the rationales used by David P. Bird in his post above are referred to.

I can not tell where are the corresponding rules in the JExocet Compendium

I have processed this puzzle manually, with help of David' compendium, to be sure I catched correctly the concerns in the posts above.

Code: Select all

 +--------------------------+----------------------------+-------------------------+
 |  9      8      B123x     |  7        1345     12345   |  6      1235   1245     |
 |  7      12346X B1236x    |  34569    134569   12345   |  2458   1235   12458    |
 | t1346y  12346X  5        |  346      8        1234    |  247    9      1247     |
 +--------------------------+----------------------------+-------------------------+
 |  5     T12369x t123689z  |  38       7        138     |  289    4      12689    |
 |  168    7       1268     |  458      145      9       |  258    1256   3        |
 |  138    139     4        |  2        135      6       |  5789   157    15789    |
 +--------------------------+----------------------------+-------------------------+
 |  2     b3469yz  36789Z   |  345689   34569    34578   |  1      567    45679    |
 | T1346   5       13679Z   |  3469     2        347     |  479    8      4679     |
 |  468Y  b469yz   6789Z    |  1        4569     4578    |  3      2567   245679   |
 +--------------------------+----------------------------+-------------------------+

To me, manual JE solver, I just would say that, for both these two potential exocets, the fundamental property is proved by colouring digits failing the two cover houses rule:
For exocet (1236)r12c3, r4c2, r8c1, the 2s in the base cells have the same parity as 2r4c2 target cell (tags x, X).
Likewise, for exocet (3469)r89c2, r3c1, r4c3, the 4s and the 6s in the base cells have the same parity as 4r3c1 and 9r4c3 target cells (tags y, Y and z, Z) resp.
The exocet pattern being proved for both exocets, the elimination criteria listed in David's JE compendium can be applied:
- non-base digits in target cells: -9r4c2, -4r8c1, -1r3c1, -128r4c3
- base digits in target cells that can't be true in mirror nodes: -3r3c1, -36r4c3, -13r4c2, -6r8c1
- non-base locked digit in one mirror node cell, other cell restricted to the base digits in opposite target cell: -9r7c2

These are the eliminations found by yzfwsf solver.

The PM is now:

Code: Select all

 +-----------------------+----------------------------+-------------------------+
 |  9     8       123    |  7        1345     12345   |  6      1235   1245     |
 |  7     12346   1236   |  34569    134569   12345   |  2458   1235   12458    |
 |  46    12346   5      |  346      8        1234    |  247    9      1247     |
 +-----------------------+----------------------------+-------------------------+
 |  5     26      9      |  38       7        138     |  28     4      1268     |
 |  168   7       1268   |  458      145      9       |  258    1256   3        |
 |  138   13      4      |  2        135      6       |  5789   157    15789    |
 +-----------------------+----------------------------+-------------------------+
 |  2     469     3678   |  345689   34569    34578   |  1      567    45679    |
 |  13    5       1367   |  3469     2        347     |  479    8      4679     |
 |  468   469     678    |  1        4569     4578    |  3      2567   245679   |
 +-----------------------+----------------------------+-------------------------+

where it is obvious that:
- 9 is a true base digit for for exocet (3469)r89c2, r3c1, r4c3, and 3 is a false base digit (not true in any target, therefore withdrawn from the base cells)
- for exocet (1236)r12c3, r4c2, r8c1, (1,3) and (2,6) are incompatible base pairs, due to target cell configuration.
No use of a third exocet, no use of abi loops.

From there, the puzzle is solved with AIC's only (maybe a long sequence, but no need of any complex step).
Hope this helps.

[icecream17]

I stumbled upon this because of https://taupierbw.be/SudokuCoach/SC_Exocet.shtml and I think Elimination 8 and 9 (hopefully the site numbering is correct) are partially invalid.

It is not the case that one mirror node is a base candidate while the other is not a base candidate. Instead, it is possible for the other to be a false base candidate.

(This would completely invalidate Elimination 8, but it can be resurrected in an even more rare scenario. Remember that a false base candidate must appear in the escape squares once. So, if all escape squares that don't see the mirror node don't have any base candidates (kinda like companion cells), then Elimination 8 does work.)

Elimination 9 works when a non-base-candidate is locked. It's not guaranteed when a base-candidate is locked.

[rybinpv]

Hi all!
Maybe I'm wrong, but I think the pictures can better reflect the situation.
For example, this example is from the document "13 JE1 (Single Target) Example" in the group.
If anyone has a desire, I will give other examples from the documentation as I study.