# WXYZ-Wing (4-Y-Wing) and higher

Part of a family: Y-Wing-Family

Brutal

## Description

Next, we will not only describe the WXYZ-Wing technique, but we will also show how all those different Y-Wing techniques are basically the same.

All the Y-Wing techniques differ in just two things:

• the number of candidates involved
• whether it is restricted or unrestricted (is the elimination candidate also present in the pivot cells)

### Number of candidates

Of course, the more candidates are involved, the more difficult is it to spot a Y-Wing configuration.

Candidates Name Official Name
3 3-Y-Wing Y-Wing / XYZ-Wing
4 4-Y-Wing WXYZ-Wing
5 5-Y-Wing VWXYZ-Wing
...
9 9-Y-Wing RSTUVWXYZ-Wing

Yes, if we continue the official naming scheme, it gets a bit ridiculous. That is why I will rather say 5-Y-Wing instead of VWXYZ-Wing.

### Restricted vs Unrestricted

4-Y-Wings and higher ones include both the restricted version as well as the unrestricted one. For 3-Y-Wings we bothered to give them separate names (Y-Wing / XYZ-Wing) because the Y-Wing can be spotted easier. 4-Y-Wings are all very tough to spot as is, so we do not bother.

### General definition of n-Y-Wings

A general n-Y-Wing can be described thusly:
• There must be a pivot cell group ()
• There must be two wing cell groups ( and )
• Every wing cell must see all pivot cells.
• The pivot cells and wing cells cannot contain any candidates that are not from the same pool of n candidates.
• There are exactly n cells (pivot cells and wing cells) involved.
• The elimination candidate may be in all wing cells.
• The wing cells' candidates must be disjoint, except for the elimination candidate. (The two wings may only share a single candidate, which is the elimination candidate.)
The result of such an n-Y-Wing is this:
• If the elimination candidate is in one of the pivot cells, we can eliminate that candidate from all cells that see all cells involved.
• If the elimination candidate is not in any of the pivot cells, we can eliminate that candidate from all cells that see all wing cells.

### Abstract Examples

Let us take a look at some abstract examples.

the pivot cell(s)
wing 1
wing 2
the candidates we can eliminate

• There is a pivot cell, which is seen by both wings.
• The wing cells only share exactly one candidate (3).
• There are exactly three cells and those cells use only three different candidates (1, 2, 3) overall.

Since (3) is not inside the pivot cell, we can eliminate (3) from all cells that see all wing cells. ()

This one is a WXYZ-Wing (4-Y-Wing).

• There are pivot cells, which are seen by both wings.
• The wing cells only share exactly one candidate (4).
• Four cells and four candidates overall.

Since it is not restricted (4 is inside the pivot cell), we can only eliminate (4) from all cells that see all cells, which is just the one cell.

Here is another WXYZ-Wing (4-Y-Wing).

This time, there is only one pivot cell, but two cells in a wing. It doesn't matter where the fourth cell is, as long as the above rules are satisfied.

We can eliminate from both cells.

This one is restricted, so we can eliminate the 4 from some more cells.

Here is another one.

The cells of the wing groups do not need to hold all candidates from the group. (See the cell on the left.)

Even the elimination candidate (4) does not have to be there. (See the cell on the right.)

Here is an 8-Y-Wing.

Eight candidates. Eight cells.

See how all rules still apply?

Here, you can also see, that the pivot cells do not need to have all candidates in them.

## Why does it work?

### Restricted version

Imagine any of the cells would be a 7.

Then all 7s would be eliminated.

Now all six remaining candidates (1,2,3,4,5,6) need to be distributed over the seven involved cells (pivot + wings).

This is impossible for all Y-Wing configurations. But why?

The pivot cells must definitely be any of the six remaining candidates.

Regardless of which candidates are correct for the pivot cells, we can eliminate those two candidates from all wing cells, since the pivot cells see all wing cells.

Let us say, for this example, that the pivot cells' solutions are 1 and 2.

Those two candidates would be eliminated from the pivot cells and now we have four candidates left (3,4,5,6) for five wing cells.

The wing cells don't see each other, so why would that be a problem?

Usually, when the cells don't see each other, it isn't a problem to distribute four candidates over five cells because we can simply repeat a candidate.

But here the cells that don't see each other have no candidate in common, they are completely disjoint (as we required them to be).

Since we now know that one cell will be without a solution, we know that there cannot be a 7 in any of the cells.

### Unrestricted version

If we make the same assumption that the one cell is a 7, but there are any 7s in the pivot cells, then we cannot derive a contradiction.

Even after elimination, we would still have seven candidates for seven cells, which would not be a problem. So we cannot eliminate this assumed 7.

To eliminate all 7s, we would need to put a 7 in one of the cells that see all other cells.

So if the elimination candidate is in one of the pivot cells, we can only eliminate it from this cell.

## Examples from real Sudokus

the pivot cell(s)
wing 1
wing 2
the candidates we can eliminate