The New Sudoku Players' Forum

[Yogi]

In 'Stuck Again,' Leren recently wrote:
*--------------------------------------------------------------------------------*
| 289 3 28 | 5 1 4 | 279 27 6 |
| 569 7 56 | 69 3 2 | 14 14 8 |
| 269 4 1 | 69 7 8 | 259 3 259 |
|-----------------------------+--------------------------+---------------------------|
| 12578 1258 2578 | 4 6 19 | 12579 1257 3 |
| 4 *15 9 | 3 2 7 | 8 6 *15 |
| 3 6 27 | 8 5 19 | 12479 1247 1249 |
|--------------------------+--------------------------+--------------------------|
| 2568 9 3 | 1 48 56 | 2456 245 7 |
|f1256 *125 4 | 7 9 56 | 3 8 *125 |
| 15678 58-1 5678 | 2 48 3 | 1456 9 145 |
*--------------------------------------------------------------------------------*

"There is a finned XWing in 1's in r58/c29 with a fin at r8c1, which removes the 1 from r9c2."

For me, the elimination logic works simply enough, although mine seems to remove 1 as a candidate from both r8c1 and r9c2 in all cases:

1) 1r5c9 => r5c2<>1 => r8c2 OR r9c2 =1 => r8c1<>1
2) OR 1r8c9 => r8c1<>1 Delete candidate 1 from r8c1
3) Now, 1r5c9 => r8c9<>1 => 1r8c2 => r9c2<>1
4) OR 1r5c2 => r89c1<>1 Delete candidate 1 from r9c2

Does that mean both r8c1 AND r9c2 should be regarded as fins?

Whatever, this is all academic if you don't know how to spot such an animal. In this example, I would see Row5 as having a 15 Locked Pair, but I would not have noted that Row8 was so closely related to it. Row8 has matching candidates in the same columns, but there are other candidates as well, and other possible sites for the 1 and 5 in that row, and there are other possible sites for the 1 and 5 in Columns 2 and 9.
Is the start-point simply the fact that there are four cells in a rectangle r58c29 which all have the candidates 1 and 5, which COULD become an XWing if certain conditions were met? What other conditions would be required? And if it works for 1's, why does it not work for 5's, which are also candidates in these four corners?

Sorry, I worked real hard to get the diagram all pretty in my draft, but the program seems to have over-ruled my tidy-ups.

[Leren]

Yogi wrote : 1) 1r5c9 => r5c2<>1 => r8c2 OR r9c2 =1 => r8c1<>1

You've made a mistake here because there is a 1 in r4c2, which invalidates the red part of your chain.

Here is the relevant PM displayed correctly and I've marked the 1 in r4c2 with an X:

Code: Select all

*--------------------------------------------------------------------------------*
| 289     3       28       | 5       1       4        | 279     27      6        |
| 569     7       56       | 69      3       2        | 14      14      8        |
| 269     4       1        | 69      7       8        | 259     3       259      |
|--------------------------+--------------------------+--------------------------|
| 12578  X1258    2578     | 4       6       19       | 12579   1257    3        |
| 4      *15      9        | 3       2       7        | 8       6      *15       |
| 3       6       27       | 8       5       19       | 12479   1247    1249     |
|--------------------------+--------------------------+--------------------------|
| 2568    9       3        | 1       48      56       | 2456    245     7        |
|f1256   *125     4        | 7       9       56       | 3       8      *125      |
| 15678   58-1    5678     | 2       48      3        | 1456    9       145      |
*--------------------------------------------------------------------------------*

To make the diagram display correctly, select it and click on the Code button at the top of the Reply box.

Leren

[Yogi]

Code: Select all

*--------------------------------------------------------------------------------*
| 289     3       28       | 5       1       4        | 279     27      6        |
| 569     7       56       | 69      3       2        | 14      14      8        |
| 269     4       1        | 69      7       8        | 259     3       259      |
|--------------------------+--------------------------+--------------------------|
| 12578  X1258    2578     | 4       6       19       | 12579   1257    3        |
| 4      *15      9        | 3       2       7        | 8       6      *15       |
| 3       6       27       | 8       5       19       | 12479   1247    1249     |
|--------------------------+--------------------------+--------------------------|
| 2568    9       3        | 1       48      56       | 2456    245     7        |
|f1256   *125     4        | 7       9       56       | 3       8      *125      |
| 15678   58-1    5678     | 2       48      3        | 1456    9       145      |
*--------------------------------------------------------------------------------*

OK That Code trick worked.
Anyway, I see that either option of r5c9 eliminates 1 from r9c2, but initially I did not see how it related to the fin cell r8c1.
1r5c9 => 1r9c7 => r9c2<>1 OR r5c9<>1 => 1r5c2 => r9c2<>1 so candidate 1 is eliminated from r9c2 anyway.
So what IS the importance of what's going on in r8c1 in this case? Well it does seem to fit with what Andrew Stuart said about it:
"If you can form an X-Wing by ignoring the fin cells, then you can keep your elimination of any cell that shares the same Unit as all the cells in the fin." That does it for r9c2 anyway.
I take it that the so-called fin cell(s) must be in line with the potential second column or row of the XWing, and the first row or column must have only two possible candidate cells.

[Leren]

There are a few different theoretical explanations as to how Finned XWings (or any other moves) work, but this is possibly the simplest explanation.

Suppose the 1 in r8c1 was False. Then there would be an X Wing in the four * cells, and you could eliminate the 1's in r4c2, r9c2, r6c9 and r9c9. You can call these Potential Eliminations, or PE's for short.

Now suppose the 1 in r8c1 was True. Then the 1 in r9c2 could be eliminated. Since r8c1 can't see any other of the X Wing PE's then you can't deduce anything about them.

So, since the 1 in r8c1 can only be True or False, and either way the 1 in r9c2 is False, then it is False and can be eliminated.

So in general, look for the X Wing PE's and any that can see the fin can be eliminated.

Compare this explanation with Andrew Stuart's here or Hodoku's here to see how I did.

Leren

[Yogi]

Thanx. Your explanation is quite simple to follow and Andrew's outline is in the same vein. I find Hodoku a bit off-planet.
I think that's because what works and when it works (the right scenario) is much more important to me than why it works.

As the cop with the notebook said, "Jes' gimme the fax, ma'm."