The New Sudoku Players' Forum

[Laurence]

This XYZ solves the puzzle, but another XYZ does not ?.
E2,E5, F4 form XYZ on value 1 so eliminate from E4.
Then E4, D6 and D2 form an XY on value 2 so eliminate value 2 from E2, leaving D2 with value 2.
But this does not solve the puzzle.
Where is my logic, or lack of it wrong.

[JasonLion]

Your image shows D2, E2, and E5 eliminating 9 from E1. That doesn't correspond to either of the patterns you describe.

E2, E5, and F4 do form an XYZ-Wing eliminating 1 from E4, as you said.

E4, D6 and D2 do not form either an XY-Wing or an XYZ-Wing because they collectively contain four candidate digits (1239). An XY-Wing and an XYZ-Wing both require that the three cells share a total of 3 candidates. If E4 contained only 2 and 3 you would have an XY-Wing and the elimination you mentioned would result.

[Leren]

I'll show you how to solve the puzzle from the position shown. Our grid numbering sysem is a lot different from the one you use, so I won't confuse you by referring to to cells by grid references, except occasionally.

Firstly there is an XYZ wing in the cells marked * with elimination digit 9 as in the following diagram. That is the move that is shown on your diagram with the 3 blue cells and the red 9.

Code: Select all

*--------------------------------------------------------------*
| 2     8     1      | 5     37-9  4      | 6     379   379    |
| 5     7     3      |*29   *129   6      | 8     4     19     |
| 6     4     9      | 37    1378  18     | 5     2     137    |
|--------------------+--------------------+--------------------|
| 7     13    6      | 8     123   12     | 9     5     4      |
| 19    5     8      | 4    *19    7      | 3     6     2      |
| 4     39    2      | 39    6     5      | 7     1     8      |
|--------------------+--------------------+--------------------|
| 89    6     5      | 27    4     28     | 1     379   379    |
| 3     2     7      | 1     5     9      | 4     8     6      |
| 189   19    4      | 6     78    3      | 2     79    5      |
*--------------------------------------------------------------*

Removing the 9 from the cell shown leaves a pointing pair of 9's in the cells marked * removing the 9 as shown and solving the cell as 1 (I think you'd call that cell I2).

Code: Select all

*--------------------------------------------------------------*
| 2     8     1      | 5     37    4      | 6     379   379    |
| 5     7     3      |*29   *129   6      | 8     4     1-9    |
| 6     4     9      | 37    1378  18     | 5     2     137    |
|--------------------+--------------------+--------------------|
| 7     13    6      | 8     123   12     | 9     5     4      |
| 19    5     8      | 4     19    7      | 3     6     2      |
| 4     39    2      | 39    6     5      | 7     1     8      |
|--------------------+--------------------+--------------------|
| 89    6     5      | 27    4     28     | 1     379   379    |
| 3     2     7      | 1     5     9      | 4     8     6      |
| 189   19    4      | 6     78    3      | 2     79    5      |
*--------------------------------------------------------------*

This then results in a naked pair (37) in the cells marked * removing 37 from a cell as shown. I guess you'd call that cell E3.

Code: Select all

*--------------------------------------------------------------*
| 2     8     1      | 5     37    4      | 6     379   379    |
| 5     7     3      | 29    29    6      | 8     4     1      |
| 6     4     9      |*37    18-37 18     | 5     2    *37     |
|--------------------+--------------------+--------------------|
| 7     13    6      | 8     123   12     | 9     5     4      |
| 19    5     8      | 4     19    7      | 3     6     2      |
| 4     39    2      | 39    6     5      | 7     1     8      |
|--------------------+--------------------+--------------------|
| 89    6     5      | 27    4     28     | 1     379   379    |
| 3     2     7      | 1     5     9      | 4     8     6      |
| 189   19    4      | 6     78    3      | 2     79    5      |
*--------------------------------------------------------------*

There is then an XY wing in the 3 cells marked * with elimination digit 2 as shown.

Code: Select all

*--------------------------------------------------------------*
| 2     8     1      | 5     37    4      | 6     379   379    |
| 5     7     3      | 29   *29    6      | 8     4     1      |
| 6     4     9      | 37    18    18     | 5     2     37     |
|--------------------+--------------------+--------------------|
| 7     13    6      | 8     13-2 *12     | 9     5     4      |
| 19    5     8      | 4    *19    7      | 3     6     2      |
| 4     39    2      | 39    6     5      | 7     1     8      |
|--------------------+--------------------+--------------------|
| 89    6     5      | 27    4     28     | 1     379   379    |
| 3     2     7      | 1     5     9      | 4     8     6      |
| 189   19    4      | 6     78    3      | 2     79    5      |
*--------------------------------------------------------------*

You'll then find that the 2 in the top *'d cell is the only 2 in Column E as you'd call it, so it must be 2.

The puzzle is solved with singles from that point. Hope this helps.

Leren

[Laurence]

Hi, Leren,

I have already solved this puzzle exactly as you describe, but this was at my second attempt.
What is unclear to me is the other apparent XYZ from Row 2 Col5 [129] row5 col5[19], and row4 col6[12] on the value 1, thereby eliminating the 1 from row4 col5 [123]-1
This leaves an XY row 2 col4 [29],row6 col4 [39 and row4 col5 [23], which would eliminate the 2 from row2 col5, so that row2 col4 can only be a 2. The puzzle is invalid after this.

[JasonLion]

What is unclear to me is the other apparent XYZ from Row 2 Col5 [129] row5 col5[19], and row4 col6[12] on the value 1, thereby eliminating the 1 from row4 col5 [123]-1

That isn't an XYZ-Wing because one of the ends (pincers) has three candidate digits. An XYZ-Wing has three digits in the middle cell (the pivot), and two digits at either end (the pincers). Switching to a more standard notation for a moment, R2C5 and R4C6 both see (connect to) R5C5, so R5C5 is the pivot cell and R2C5 & R4C6 are the pincers.

The XY-Wing you describe after that (row 2 col4 [29],row6 col4 [39] and row4 col5 [23]) would be valid if it actually existed, but the previous step doesn't work, so row 4 col 5 never have the 1 removed, so the XY-Wing never comes into existence.

[Leren]

Code: Select all

*--------------------------------------------------------------*
| 2     8     1      | 5     37    4      | 6     379   379    |
| 5     7     3      | 29 XYZ129   6      | 8     4     1-9    |
| 6     4     9      | 37    1378  18     | 5     2     137    |
|--------------------+--------------------+--------------------|
| 7     13    6      | 8     123 XZ12     | 9     5     4      |
| 19    5     8      | 4   YZ19    7      | 3     6     2      |
| 4     39    2      | 39    6     5      | 7     1     8      |
|--------------------+--------------------+--------------------|
| 89    6     5      | 27    4     28     | 1     379   379    |
| 3     2     7      | 1     5     9      | 4     8     6      |
| 189   19    4      | 6     78    3      | 2     79    5      |
*--------------------------------------------------------------*

As I understand it you are proposing an XYZ Wing with Pivot cell r2c5 and Pincer cells r5c5 and r4c6, with X = 2, Y = 9 and Z = 1, which would remove the 1 from r4c5. The problem is that the Pivot cell doesn't see one of the Pincers, at r4c6.

What is supposed to happen with an XYZ wing is that if none of the 3 cells holds Z then the Pivot cell would be empty, so at least one of the three cells must hold Z. Any cell that can see all three cells can therefore have Z removed.

Since the Pivot cell doesn't see both Pincers you can't establish the required contradiction in the Pivot cell.

If none of your 3 cells held 1 then the Pivot cell r2c5 and one of the Pincers r4c6 would both have to be 2, but that's not a contradiction.

Now let's examine the XYZ Wing from my previous post to see why it works.

The Pivot cell is r2c5, the Pincers are r2c4 and r5c5, X = 1, Y = 2 and Z = 9. Crucially, the Pivot cell can see both Pincers. Now let's suppose that none of these three cells was Z = 9.

r2c4 would have to be 2, r5c5 would have to be 1 and the pivot cell r2c5 would have to be .... er ... nothing - aaaargh ! Can't have that can we ? So at least one of the three cells must be 9. Since r1c5 can see all three cells, it can't be 9.

You can see XYZ Wings described in graphical detail at good teaching sites here and here.

Leren

[Laurence]

Nicely explained.
Many thanks Leren and Jasonlion.
I had not understood the significance of the pivot cell having to have the XYZ till you pointed it out.
Cheers,
Laurence