XY-Chain

An XY-Chain is the XY-Wing stretched out into a full chain. It runs through a series of bi-value cells — cells with exactly two candidates — linked so that a single digit is forced at both ends. Any cell that sees both ends then loses that digit.

How the chain works

Every cell in the chain has two candidates, and each shares one candidate with the next cell in line. Follow the logic: assume the first cell takes one of its digits; that forces the next cell to its other digit, which forces the next, and so on down the chain. The clever part is arranging it so that both ends carry the same digit Z — meaning whichever way the chain falls, one end is a Z. And if you assume the first cell takes its other value, the same digit Z appears at the near end directly. Either way, one of the two ends is Z.

The elimination

Since one end must be Z, any cell that can see both ends of the chain cannot be Z. Delete it there. It's exactly the XY-Wing conclusion — a shared digit forced out — but the "wing" is now a chain of any length, letting it reach right across the grid.

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An XY-Chain of bi-value cells: r1c1{3,5} – r1c5{5,7} – r5c5{7,8} – r5c9{8,3}, each sharing a digit with the next. Both ends (amber) hold 3, so one of them is a 3 — and r5c1, seeing both ends, loses its 3.

Step by step

  1. Start at a bi-value cell and note its two candidates.
  2. Link to another bi-value cell it sees that shares one candidate.
  3. Keep going, each cell sharing a digit with the next, until both ends of the chain force the same digit Z.
  4. Eliminate Z from any cell seeing both ends.

Where it fits

XY-Chains move between digits as they go, where an X-Chain sticks to one. Both are among the most useful advanced techniques a manual solver can learn, and they stop just short of the fully general forcing chains.

Common mistakes

  • Non-bi-value cells. Every cell in an XY-Chain must have exactly two candidates.
  • Ends forcing different digits. The chain only works if both ends force the same digit Z.
  • Broken links. Each cell must see the next and share a candidate with it.

Which Sudoku types is this best for?

The same logic applies across variants, but it pays off more in some than others.