Three cells, two candidates each, one elimination

The short version

An XY-Wing is three cells. One is the pivot, holding exactly two candidates. The other two are pincers, each holding exactly two candidates, each sharing one with the pivot, and sharing the other one with each other.

Whatever the pivot turns out to be, one pincer is forced to the shared value. So any cell that can see both pincers cannot hold that value.

The shape

Name the candidates so the structure is visible. Let the pivot be {X, Y}, one pincer {X, Z}, the other {Y, Z}.

• Pivot: {X, Y}
• Pincer A: {X, Z} — sees the pivot
• Pincer B: {Y, Z} — sees the pivot
• Both pincers share Z.
• Any cell that sees both pincers loses Z.

The digits X and Y are the pivot's two options. Z is the digit that appears in both pincers and in neither slot of the pivot. That asymmetry is the whole pattern: Z is the only digit that both pincers can hold, and neither pincer can hold it alongside the other's X/Y.

If that reads like a lot of conditions, it is — which is why XY-Wings are missed. But every condition is checkable, and the check takes about fifteen seconds per pivot candidate.

Why it works

The pivot has two values. Only two cases exist.

• Case 1: the pivot is X. Then pincer A, which holds {X, Z}, cannot be X — it shares a unit with the pivot. So pincer A is Z.
• Case 2: the pivot is Y. Then pincer B, which holds {Y, Z}, cannot be Y. So pincer B is Z.

Either way, some pincer holds Z. You never learn which one, and you do not need to. What matters is that at least one of them does, so Z is present in the pincer region no matter what.

Now take a cell that sees both pincers. If it held Z, both pincers would be blocked from Z — but one of them must be Z. Contradiction. So that cell cannot be Z.

This is the first technique on this site whose conclusion is a guarantee about a disjunction rather than a single cell. That is what makes it feel different: you are reasoning about two possibilities without resolving either.

A real board

Three cells, and the candidate lists are unusually informative:

• R1C5 holds {5, 6} — the pivot.
• R2C5 holds {6, 7} — a pincer, sharing 6 with the pivot.
• R3C6 holds {5, 7} — the other pincer, sharing 5 with the pivot.
• Both pincers hold 7, and neither holds it alongside the pivot's options.

Run the two cases:

• If R1C5 is 5, then R3C6 — which can only be 5 or 7, and shares a box with the pivot — cannot be 5. So R3C6 is 7.
• If R1C5 is 6, then R2C5 — which can only be 6 or 7, and shares a box with the pivot — cannot be 6. So R2C5 is 7.

Either way, a 7 lands in one of the two pincers. Now look at R2C6, which reads {4, 6, 7, 8}. It shares a row with R2C5 and a column with R3C6, so it sees both pincers. If it were 7, neither pincer could be — but one must be. So R2C6 loses its 7.

One elimination from three cells and a two-case argument. That is the entire yield, and on a stalled board it is often decisive.

How to scan for one

XY-Wings are found by hunting pivots, and the hunt is mechanical.

1. Collect every two-candidate cell. These are the only possible pivots, and there are usually only a handful on a hard board.
2. For each pivot {X, Y}, look for its pincers. Find a cell in a shared unit holding {X, Z} for some Z, and another holding {Y, Z} for the same Z.
3. Check both pincers see the pivot. One shared unit each is enough — row, column or box.
4. Find the Z-cells that see both pincers. For every cell containing Z, ask whether it shares a unit with both pincers. Those are your eliminations.
5. Verify before deleting. The most common false positive is a Z candidate that sees only one pincer. One pincer alone proves nothing.

A useful prioritisation: pivots with candidates that are scarce elsewhere on the board produce more eliminations, because more cells carry Z. If the pivot is {2, 8} and 8 is nearly finished, the wing will usually fire.

The traps

Pincers that do not see the pivot. Two perfect {X, Z} and {Y, Z} cells across the board from the pivot form nothing. The forcing only works through a shared unit.

Mismatched Z. Pivot {5, 6}, pincers {5, 7} and {6, 8}. There is no common Z, so there is no wing — just three cells with two candidates each.

Eliminating from cells that see one pincer. Deleting Z from a cell that only sees pincer A is not justified: pincer B might be the one holding Z, in which case the cell is free. The elimination requires sight of both.

Confusing it with XYZ-Wing. If the pivot has three candidates, it is a different pattern with a narrower elimination — XYZ-Wing, not XY-Wing. The extra candidate changes which cells are affected.

Where to use it

XY-Wings live in the band where pairs, triples and box-line reduction have all come up empty but the puzzle is not yet a guessing exercise. They are common enough in published hard puzzles that most solvers eventually need them, and rare enough in easy ones that learning them early is wasted effort.

They are also the gateway to the whole wing family. Once the two-case argument is comfortable, XYZ-Wing and skyscrapers are variations on the same move rather than new ideas.