Code: Select all

```
1 2 567 | 49 3 49 | 56 67 8
8 C69 356 | 2 1 7 | 45 D36 469
C379 C79 4 | 6 5 8 | 1 237 A29
--------------+---------------+--------------
5 14 9 | 3 7 26 | 8 246 A126
26 3 68 | 5 48 1 | 9 246 7
267 14 678 | 48 9 26 | 3 5 A126
--------------+---------------+--------------
B69 5 1 | 7 2 49 | B46 8 3
367 678 367 | 1 48 5 | 2 9 A46
4 89 2 | 89 6 3 | 7 1 5
```

1. Unkx80 pointed out the four cells (marked C & D above in boxes 1 & 3) that form an ALS-XZ or the equivalent WXYZ-Wing (C+D) that eliminates the 3 in r2c3.

Now, it also seems that one can apply Subset Counting to these same four cells in which candidates 6,7,9 each have a multiplicity of one, while candidate 3 has a multiplicity of two, giving a total subset multiplicity of 5. So, a 3 placed in r2c3 would eliminate both of the 3-candidates in the subset and thereby reduce the total multiplicity to 3, which is one below the cell count. Hence, r2c3 <> 3, once again. These results clearly show that at least three different named solution techniques will eliminate the 3 in r2c3 using the same four cells. This is also in keeping with Ruud’s 4-technique example grid used in the SudoCue Solving Guide’s sections on XYZ-Wing, APE, ALS-XZ, and Subset Counting.

2. Ruud pointed out a different elimination using another ALS-XZ rule applied to the sets marked A and B above in c9 and r7, respectively. With X=4 as the restricted common, and Z=9, then r3c1 <> 9.

One would now hope that Subset Counting could also be applied to cells A+B, although it is not immediately obvious how to do this, since both the 9’s and 6’s have a multiplicity of two, and there are also six cells with five candidates.

So, the key question is...should Subset Counting and the ALS-XZ rule both provide the same elimination when applied to the same subset of cells? And, if so, how would it work with the subset A+B above?