... 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 'one-size-fits-all' strategy and so allow many more cases to be found.
The strategy for Bent Quads (WXYZ-wing) 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 inter-linked 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 Z-digits in the alliance and the spy cells. The spy cell can see all the other W-digits in the alliance. So if the Z-digit is given as a solution in the rogue cell, all the Z-digits are eliminated from the quad, and the W-digit is now placed in the spy cell removing the W-digits 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 Z-digit
these are . . .
XY-XY-XY these cells must all be in the same house { 1 }
XY-XY-X, XY-XY-Y, XY-X-XY, XY-Y-XY, X-XY-XY, Y-XY-XY
same house { 6 }
the next 3 cases follow a similar pattern
XY-X-Y, etc the cell with XY must be a pivot { 6 }
XY-X-X, etc the cells X-X must be in the same house { 6 }
X-X-Y, etc the cells X-X must be in the same house { 6 }
and the last 2 are . . .
X-X-X, Y-Y-Y impossible! it means that the 4th digit
is missing { 2 }
If any of the above conditions are met then the Z-digit can be eliminated from the rogue cell.
This idea could be extended to form any N-wing format, but the logistics are astronomical.
The number of combinations of the remaining (N-2) digits to fit in
the (N-1) alliance cells is given by the formula
C = {[2^(N-2)]-1}^{N-1}
So the above 4-wing has 27 (3^3), a 5-wing has 2401 (7^4) and the 6-wing 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 4-wing 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.