# Solving Sudoku - XY-Chain

This is a technique that allows you to make eliminations by following a chain of cells that have only two candidates each. The logic is similar to forcing chains, but while forcing chains make a deduction about a single cell, XY chains can make eliminations for multiple cells.

Consider the following Sudoku puzzle fragment:

 {1,3} * * *
 {3,7} {6,7}
 * * * {6,8} {1,8}

(The numbers in curly brackets { } are the only candidates for the cell.)

We can build up the implication chain:

if r1c1 = 3 then r1c5 = 7, so r2c6 = 6, so r2c8 = 8, so r3c9 = 1

So we know that r1c1 is 1, or if r1c1 is 3, then r3c9 must be 1. Either r1c1 and/or r3c9 must be 1. So any cells that share a unit with both of these cells can have 1 eliminated. These are the cells marked by the asterisks in the above example.

Note: we didn't make any deduction about the contents of r3c9 when r1c1 is 1, so this doesn't mean either r1c1 is 1 or r3c9 is 1 - it just means that at least one of the cells r1c1 and r3c9 is 1, and there's no reason why they can't both be 1

Consider the following Sudoku puzzle:

We can build up the implication chain:

if r1c2 = 5 then r6c2 = 4, and so r6c8 = 7, and so r6c1 = 1.

So we know that r1c2 is 1, or if r1c2 is 5, then r6c1 must be 1. So any cells that share a unit with both of these cells can have 1 eliminated. These are the blue cells in the above example.

(If r1c2 = 5, then after making the elimination above, there remains no cell in block one that can contain 1. This doesn't mean that the above logic is faulty, it just means that r1c2 must be 1.)

More generally, as long as the same number is "unused" at both ends of the chain, then this technique allows the elimination of that candidate for any cells that share a unit with both end cells.

These Sudoku puzzles can be solved using XY-chains: (What are .sdk files?)

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