## Almost XY Wing

Advanced methods and approaches for solving Sudoku puzzles

### Almost XY Wing

Code: Select all
` *-----------*Start of the puzzle: |...|...|...| |2..|...|..9| |...|754|...| |---+---+---| |5.4|...|6.1| |.2.|...|.5.| |.31|...|49.| |---+---+---| |..9|8.2|5..| |.5.|4.6|.1.| |.6.|975|.2.| *-----------*Current Position: *-----------* |.8.|29.|..5| |2.5|68.|..9| |91.|754|2..| |---+---+---| |594|32.|6.1| |62.|149|.5.| |.31|56.|492| |---+---+---| |..9|812|5..| |.52|436|91.| |16.|975|.24| *-----------* Current Position with PM: *--------------------------------------------------* | 34   8    367  | 2    9    13   | 17   346  5    | | 2    47   5    | 6    8    13   | 17   34   9    | | 9    1    36   | 7    5    4    | 2    368  368  | |----------------+----------------+----------------| | 5    9    4    | 3    2    78   | 6    78   1    | | 6    2    78   | 1    4    9    | 38   5    378  | | 78   3    1    | 5    6    78   | 4    9    2    | |----------------+----------------+----------------| | 34   47   9    | 8    1    2    | 5    367  367  | | 78   5    2    | 4    3    6    | 9    1    78   | | 1    6    38   | 9    7    5    | 38   2    4    | *--------------------------------------------------*`

Almost XY-Wing

I have sometimes been using a this step in the process of sudoku solving. I have a feeling that this may be a very frequently used technique.

Have a look at r4c8 and then at r5c7 and r7c8. If r7c8 did not have the candidate 6, we could eliminate 3 from r9c7 by XY-wing.

As it stands, we cannot do so. But we can see that, if we did eliminate the 3, on solving the other squares of block nine, r7c8 would become 6. So we can safely eliminate 7 from r7c8.

Now this solves the puzzle as it leaves us with only one 7 in column 8.

I did not think I had to be this clear. Sorry for that. Let me now go on and make my point.

If r7c8<>6, then by XY wing, r9c7<>3,=8 and then r8c9=7, r7c8<>7.
If r7c8=6, then r7c8<>7.
Therfore, r7c8<>7
Last edited by Finlip on Sun Oct 15, 2006 8:29 am, edited 1 time in total.
Finlip

Posts: 49
Joined: 15 July 2005

### Re: Almost XY Wing

Finlip wrote:
Code: Select all
` *--------------------------------------------------* | 34   8    367  | 2    9    13   | 17   346  5    | | 2    47   5    | 6    8    13   | 17   34   9    | | 9    1    36   | 7    5    4    | 2    368  368  | |----------------+----------------+----------------| | 5    9    4    | 3    2    78   | 6    78   1    | | 6    2    78   | 1    4    9    | 38   5    378  | | 78   3    1    | 5    6    78   | 4    9    2    | |----------------+----------------+----------------| | 34   47   9    | 8    1    2    | 5    367  367  | | 78   5    2    | 4    3    6    | 9    1    78   | | 1    6    38   | 9    7    5    | 38   2    4    | *--------------------------------------------------*`

As it stands, we cannot do so. But we can see that, if we did eliminate the 3, on solving the other squares of block nine, r7c8 would become 6. So we can safely eliminate 7 from r7c8.

Okay, I see it now... But it needs some clarification

If r9c7<>3, r9c7=8 => r8c9=7 => r7c8<>7

Since "r7c8<>6 => r9c7<>3" we could deduce "r9c7=3 => r7c8=6".
(Alternatively, r9c7=3 => r5c7=8 => r4c8=7 => r7c8<>7)

So no matter what r7c8<>7.

This feels like one of Carcul's mysterious move...
Last edited by udosuk on Fri Oct 13, 2006 2:54 pm, edited 2 times in total.
udosuk

Posts: 2698
Joined: 17 July 2005

[Edited:] Finlip, Nice catch on the UR Type 2. See later message for best analysis from me.
Last edited by daj95376 on Fri Oct 13, 2006 10:10 pm, edited 1 time in total.
daj95376
2014 Supporter

Posts: 2624
Joined: 15 May 2006

The almost xy-wing in question seems like a roundabout way of coming up with the bivalue chain

(7=8)r4c8 - (8=3)r5c7 - (3=8)r9c7 - (8=7)r8c9

which eliminates every seven seeing both sevens in r4c8 and r8c9
Myth Jellies

Posts: 593
Joined: 19 September 2005

[Edited:] There's a 6-node Remote Pairs on <78> that leads to [r5c9]=3 and a cascade of Singles to solve the puzzle? This eliminates the need for an X-Wing in <8> and also explains [r7c8]<>7.

Code: Select all
` *--------------------------------------------------* | 34   8    367  | 2    9    13   | 137  346  5    | | 2    47   5    | 6    8    13   | 137  34   9    | | 9    1    36   | 7    5    4    | 2    368  368  | |----------------+----------------+----------------| | 5    9    4    | 3    2   *78   | 6   *78   1    | | 6    2    78   | 1    4    9    | 38   5   -378  | |*78   3    1    | 5    6   *78   | 4    9    2    | |----------------+----------------+----------------| | 34   47   9    | 8    1    2    | 5   -367  367  | |*78   5    2    | 4    3    6    | 9    1   *78   | | 1    6    38   | 9    7    5    | 38   2    4    | *--------------------------------------------------*`
daj95376
2014 Supporter

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Finlip wrote:As it stands, we cannot do so. But we can see that, if we did eliminate the 3, on solving the other squares of block nine, r7c8 would become 6. So we can safely eliminate 7 from r7c8.

I cannot see how you make that conclusion. Much better than that:

if r7c8 is not "6" then {XY-Wing: [r9c7]-3-[r7c8]-7-[r4c8]-8-[r5c7]-3-
-[r9c7]} => r9c7=8 (r5c7=3) => r8c9=7 => r5c9=3.

So the conclusion is: r7c8 must be "6".
However, colors also solve the puzzle at this stage.

Carcul
Carcul

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Joined: 04 November 2005

Carcul wrote:if r7c8 is not "6" then {XY-Wing: [r9c7]-3-[r7c8]-7-[r4c8]-8-[r5c7]-3-[r9c7]} => r9c7=8 (r5c7=3) => r8c9=7 => r5c9=3.

Doesn't look right to me!

After r8c9=7 why must r5c9=3?

Like Daj said the UR r59c37 (r5c3=7) solves it easily. It's just a moral issue whether we use it or not...
udosuk

Posts: 2698
Joined: 17 July 2005

udosuk wrote:After r8c9=7 why must r5c9=3?
E.g. r8c9=7 => r8c1=8 => r6c1=7 => r5c3=8 => r5c9=3
ravel

Posts: 998
Joined: 21 February 2006

Udosuk wrote:Like Daj said the UR r59c37 (r5c3=7) solves it easily.

There is no UR in those cells. In a UR, four cells must share 2 columns, 2 rows, and 2 boxes. In your case, we have 2 rows, 2 columns, and 4 boxes. Wrong logic, but a coincidently correct deduction.

Udosuk wrote:After r8c9=7 why must r5c9=3?

Because due to an X-Wing, there is no "8" in r5c9.

Daj95376 wrote:There's a 6-node Remote Pairs on <78> that leads to [r5c9]=3 and a cascade of Singles to solve the puzzle?

That move only solves the puzzle after the UR in cells r12c67.

Carcul
Carcul

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Joined: 04 November 2005

### A Simpler Solution

Code: Select all
` *----------------------------------------------------* | 347   8     367 | 2     9     13 | 137   3467  5   | | 2     47    5   | 6     8     13 | 137   347   9   | | 9     1     36  | 7     5     4  | 2     368   368 | |-----------------+----------------+-----------------| | 5     9     4   | 3     2     78 | 6     78    1   | | 6     2     78  | 1     4     9  | 378   5     378 | | 78    3     1   | 5     6     78 | 4     9     2   | |-----------------+----------------+-----------------| | 347   47    9   | 8     1     2  | 5     367   367 | | 78    5     2   | 4     3     6  | 9     1     78  | | 1     6     38  | 9     7     5  | 38    2     4   | *----------------------------------------------------*`

[r9c3]-3-[r7c1]=3=[r1c1]=4=[r1c8]=6=[r3c89]-6-[r3c3]-3-[r9c3],

and so r9c3<>3 solving the puzzle.

Carcul
Carcul

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Joined: 04 November 2005

Carcul wrote:
Udosuk wrote:Like Daj said the UR r59c37 (r5c3=7) solves it easily.
There is no UR in those cells. In a UR, four cells must share 2 columns, 2 rows, and 2 boxes. In your case, we have 2 rows, 2 columns, and 4 boxes. Wrong logic, but a coincidently correct deduction.
Thanks... My mistake...

Carcul wrote:
Udosuk wrote:After r8c9=7 why must r5c9=3?
Because due to an X-Wing, there is no "8" in r5c9.
But if we apply that x-wing, the essense of that move (from Finlip's position) would be the xy-wing deduction r9c7<>8 (r59c7+r58c9), and all the fuss about r7c8 would be obsolete...
udosuk

Posts: 2698
Joined: 17 July 2005

Udosuk wrote:But if we apply that x-wing, the essense of that move (from Finlip's position) would be the xy-wing deduction r9c7<>8 (r59c7+r58c9), and all the fuss about r7c8 would be obsolete...

Yes, you are absolutely right on that.

Carcul
Carcul

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Joined: 04 November 2005

udosuk wrote:Like Daj said the UR r59c37 (r5c3=7) solves it easily. It's just a moral issue whether we use it or not...

I was misunderstood. I was referring to the UR Type 2 in [r12c67] that was performed before the PM was listed. Once the PM has been reached, a 6-node Remote Pairs chain will lead to the puzzle being solved. If you want to skip the UR Type 2 and the RP, then the following works for the original puzzle.

Code: Select all
`  c1    -  78    Naked  Pair  c37   -  8     X-Wingr5c7    <> 3     XY-Wing  on [r8c9]r7c9    <> 3     XY-Wing  on [r8c9]  c9    -  7     Locked Candidate (2)r1c8    <> 6     XYZ-Wing on [r3c8]`
daj95376
2014 Supporter

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Since this was only a simple sudoku extreme, there are a number of other sovling methods. There may even be easier methods to msot. But the one I pointed out can easily be seen while looking for xy-wings which is my favourite solving methods. I haven't got the hang of the other methods not covered in simple sudoku. But after reading a few replies from you, I think this method may be helpful for a few more people.
Finlip

Posts: 49
Joined: 15 July 2005

Here is another example that I encountered today.
Code: Select all
` *-----------* |42.|3..|..5| |...|.7.|...| |.98|6..|.7.| |---+---+---| |.7.|..2|6.1| |.8.|...|.4.| |9.2|1..|.3.| |---+---+---| |.6.|..9|12.| |...|.4.|...| |7..|..6|.83| *-----------* *-----------* |427|3..|.65| |..6|.7.|.1.| |198|6..|37.| |---+---+---| |.7.|..2|6.1| |681|...|.4.| |9.2|16.|.3.| |---+---+---| |86.|..9|12.| |2..|.4.|..6| |7..|..6|.83| *-----------*  *--------------------------------------------------------------------* | 4      2      7      | 3      189    18     | 89     6      5      | | 35     35     6      | 2489   7      48     | 2489   1      2489   | | 1      9      8      | 6      25     45     | 3      7      24     | |----------------------+----------------------+----------------------| | 35     7      345    | 4589   589    2      | 6      59     1      | | 6      8      1      | 579    359    357    | 2579   4      279    | | 9      45     2      | 1      6      457    | 578    3      78     | |----------------------+----------------------+----------------------| | 8      6      345    | 57a    35     9      | 1      2      47a    | | 2      135    359    | 578    4      13578  | 579    59     6      | | 7      145    459    | 25*    125    6      | 459a   8      3      | *--------------------------------------------------------------------*`

The cells marked a would have formed an xy wing had r9c7 not had the extra 9.

If r9c7<>9, r9c4=2 by XY wing.
If r9c7=9, r1c7=8, r1c6=1, r1c5=9, r9c5=1, r9c4=2

Therefore r9c4=2
Finlip

Posts: 49
Joined: 15 July 2005

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