Comments  Talk
... by: Fred Kong
Dear Andrew, I am very thankful to have learned about your WXYZWings theory. With this skill I have solved quite a few very toughed puzzles in my notcompleted storage. However, I met an exception. That is Puzzle 125 in your book: Extreme Sudoku For Dummies. Please take a look at where I cannot go any further:
 6 35 8  579 4579 4579  1 347 2   1 4 7  3 6 2  9 58 58   239 25 29  8 457 1  46 467 37  ++  4 28 1  256 3 56  7 9 568   78 9 3  1567 14578 4567  458 2 568   278 6 5  279 4789 479  3 48 1  ++  289 1 269  4 59 3  568 5678 79   39 7 4  56 2 8  56 1 39   5 38 69  1679 179 679  2 38 4 
I found there is a WXYZWing set as C1, G1, H1 and I2. I thought the Z should be 3, and I could eliminate the 3 in A2 cell. But I could not go to the right end. According to the Answer page, A2 cell is 3.
Please tell me why I cannot use WXYZWing as the above.
Thanks.
Fred Kong
... by: strmckr
Andrew Stuart writes:
Correct and I've put your link in the text. I have now completed the expansion of the search algorithm to get these cases.
thanks for the update, as for the remark on more then 10 cases, true there is a large list of positional locations for the 4 cells{ i originally mapped out 20 ish cases}, however after rotation, reflection, relabeling, band swaping ect is applied the 1  4 types I've identified and cross checked via programing should be the minimal exemplars. if you do have any examples cases that don't match my 4 primary types and hold additional eliminations feel free to email me as id love to examine them.
strmckr
... by: Gcd
 4 5678 1679  789 789 3  2 569 1568   159 58 19  6 2489 2489  1458 7 3   2379 23678 3679  4789 5 1  48 469 68  ++  37 9 2  1378 1378 58  6 35 4   6 1 4  39 239 259  35 8 7   357 357 8  347 347 6  9 1 2  ++  12379 237 1379  5 6 489  13478 34 18   8 4 1369  139 139 7  135 2 156   137 367 5  2 1348 48  13478 346 9 
Your Solver shows this wxyz uses H4, H5,G6 G9 3,9,8,1 With 1 being excluded from H7. This is from your Sunday daily puzzle of a Sudoku from early September.\
But if you change to G9 to G8, you now have 1,3,8,4 and the 3 in H7 now see's 3'3 all of the wing 3's yet it cannot be excluded. Why not? Where have I made my error?
... by: strmckr
i have covered more extensively the idea of bent quads on numerous forum boards for extending this strategy. prior to 2010
http://forum.enjoysudoku.com/wxyzwingst30012.html
however it can be noted that most if not all wxyz  wings are specific simplified cases of alsxz rule.
Andrew Stuart writes: Correct and I've put your link in the text. I have now completed the expansion of the search algorithm to get these cases.
... by: SudoNova
On various Sudoku sites on the internet I have found about a dozen different cases for the use of this technique. I would like to propose the following 'onesizefitsall' strategy and so allow many more cases to be found.
The strategy for Bent Quads (WXYZwing) can be summarised like this. If there are 4 cells having 4 digits each (W, X, Y, Z) distributed within them such that any 2 cells aren't a pair and any 3 cells aren't a triple. The cells aren't all in the same house but they are interlinked somehow.
One of the cells in the quad has digits WZ (I call this a 'spy' cell), and the other 3 cells of the quad form an 'alliance'. Outside the quad is a 'rogue' cell with digit Z among its candidates and if the rogue declares war, the spy acts as a 5th columnist  unless the alliance can prevent this.
The rogue cell can see all the Zdigits in the alliance and the spy cells. The spy cell can see all the other Wdigits in the alliance. So if the Zdigit is given as a solution in the rogue cell, all the Zdigits are eliminated from the quad, and the Wdigit is now placed in the spy cell removing the Wdigits from the alliance cells.
This leaves the X and Y digits distributed in the 3 remaining cells of the quad
There are { 27 } combinations in which X, Y or XY can be distributed in 3 cells so that the allies can then defeat the rogue Zdigit
these are . . .
XYXYXY these cells must all be in the same house { 1 }
XYXYX, XYXYY, XYXXY, XYYXY, XXYXY, YXYXY same house { 6 }
the next 3 cases follow a similar pattern
XYXY, etc the cell with XY must be a pivot { 6 }
XYXX, etc the cells XX must be in the same house { 6 }
XXY, etc the cells XX must be in the same house { 6 }
and the last 2 are . . .
XXX, YYY impossible! it means that the 4th digit is missing { 2 }
If any of the above conditions are met then the Zdigit can be eliminated from the rogue cell.
This idea could be extended to form any Nwing format, but the logistics are astronomical.
The number of combinations of the remaining (N2) digits to fit in the (N1) alliance cells is given by the formula
C = {[2^(N2)]1}^{N1}
So the above 4wing has 27 (3^3), a 5wing has 2401 (7^4) and the 6wing has 759375 (15^5).
Also, we could remove the restriction that the spy cell only has W and Z in its candidates and have any combination with X or Y included, but using a modified version of the above formula means that the 4wing now has 2401 combinations of W, X or Y to consider instead of just 27 combinations of X or Y.
Andrew Stuart writes: This is sound and could form the basis of a more generalised theory.
... by: Mats Anderbok
Why don't you have a WXYwing? With 3 candidates in the hinge, I think it fits in perfectly between XYwing and WXYZwing. Now it isn't caught until APE.
... by: Nassir M
also you could remove the 7s from c1 and c6 pls 4 from c1 and also to remove the 5 from A3 because the 5 is going to be either in B3 or C3. Thanks
... by: rich schrader
Isn't this strategy another way of interpreting an ALS?
... by: Jeff Sanborn
What if the cell containing 4/9 were located at Row 1, Column 2 (instead of Row 3, Column 8).
Must there be one cell within the box and two cells in the row, or is it just a total of three cells that is important?
Furthermore, could you have a VWXYZwing if you had a hinge cell with five values and four outlying cells?
Andrew Stuart writes: All the yellow cells must see the hinge  the brown cells. Yes you could have a 5cell VWXYZwing although I've not searched for one.
... by: LokiMomus
Is there a more generic strategy for this wxyz already mentioned? Couldnt find it, though I admit I havent read all strategies thoroughly. I don't have an example sudoku for this, but consider the given example and lets say row 3, column 5 has 379 as available options instead and i include an extra cell into the strategy, row 3, column 6, which has 379 as available options. Row 3, column 3 has instead the values 34579.
The theory still holds. Filling in the 9 in the yellow box will result in an invalid situation, so i can scratch the 9 in the yellow box.
In other words, the more generic requirement should be that there can be n cells of the same cellgroup as long as they have n+1 distinct symbols. (the +1 is for the symbol (9 in this case) that we are considering). The other rules stay intact.
The substrategy of wxyzwing would be with n=1.
My question to you is: Is this strategy already mentioned on the site? If not, can it be solved by using multiple others?
Note: This strategy can be even made more generic if there are more than 3 dimensions (special sudokus). In a traditional sudoku there are 3 dimensions (a cell is part of a horizontal cellgroup, a vertical cellgroup and a 3x3 cellgroup). A 4dimensional sudoku would for example be a jigsaw sudoku on top of a traditional one.

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