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Alternating Inference Chains

ExtremeClassic + KillerStep 34 of 384 min read

Alternating Inference Chains (AIC) are the general form of every chain technique in Sudoku, and the core of AIC sudoku solving. Instead of following one digit, you hop between digits and cells, alternating two kinds of link, and read off what must be true at the ends. Once you can build them, X-Wings, wings and single-digit cycles all become special cases.

What the pattern is

A chain is a sequence of candidates, each one a digit in a specific cell. Two link types connect them:

LinkMeaningWhere it comes from
Strong linkif one is false, the other is truea digit with only two places in a unit, or a cell with only two candidates
Weak linkthey cannot both be truetwo candidates of one digit in a shared unit, or two candidates in one cell

An Alternating Inference Chain starts with a strong link, ends with a strong link, and alternates strong, weak, strong, weak all the way through. That structure guarantees one thing: at least one of the two end candidates is true. You never learn which, and you do not need to.

When to look for it

Reach for chains once singles, locked sets, fish and wings are exhausted and the grid is full of bivalue cells. Chains need complete, trustworthy pencil marks; a single stale candidate invents links that do not exist. Start from a bivalue cell or from a digit that has exactly two places in some unit, because those are the strong links that seed everything.

How to apply it, step by step

  1. Pick a starting candidate and a strong link leaving it.
  2. From the candidate you land on, take a weak link: same cell, different digit, or same digit in the same unit.
  3. From there take another strong link. Keep alternating.
  4. Stop on a strong link, so that at least one end is true.
  5. Eliminate any candidate that both ends rule out. With two ends of the same digit, that is every cell seeing both ends. With different digits, look at the end cells themselves (see the example).

If instead the chain closes on itself with the alternation intact, you have a continuous loop: every weak link in it becomes exclusive, so all other candidates of that digit in the shared unit go, and any extra candidate in a looped cell goes too. If the alternation breaks at one node, the chain is discontinuous: a node with two weak links pointing at it is false, and a node with two strong links pointing at it is true.

A worked example

Digits 3, 8 and 9 across boxes 5 and 8. The chain has three strong and two weak links, and two of its hops stay inside one cell (H6 and F5).

  1. Easier techniques are used up. Look at the 3s, 8s and 9s in boxes 5 and 8.

  2. Column 6 has only two places for 9: F6 and H6. If F6 is not 9, H6 is. That is a strong link.

  3. If H6 is 9, it cannot also be 8: a weak link inside the cell.

  4. Box 8 has only two places for 8: H6 and H5. If H6 is not 8, H5 is: another strong link.

  5. H5 and F5 share column 5, so they cannot both be 8 (weak link).

    F5 holds only 3 and 8, so if it is not 8 it is 3 (strong link). The chain ends there.

  6. Suppose F6 is not 9. Then H6 is 9, so not 8; H5 is 8; F5 is not 8, so F5 is 3. One end of the chain is always true: F6 is 9, or F5 is 3.

    If F6 is 9 it is not 3. If F5 is 3, F6 cannot be 3 because they share row F.

  7. 3 removed from F6 because either end of the chain rules it out.

Easier techniques are used up. Look at the 3s, 8s and 9s in boxes 5 and 8.

Written out: F6:9 = H6:9 - H6:8 = H5:8 - F5:8 = F5:3

  • Column 6 has 9 only in F6 and H6 (strong).
  • H6 is , so 9 and 8 in H6 exclude each other (weak).
  • Box 8 has 8 only in H6 and H5 (strong).
  • H5 and F5 share column 5 (weak).
  • F5 is (strong).

So F6 is 9 or F5 is 3. The ends carry different digits here, and the elimination sits in an end cell. If F6 is 9, it is not 3. If F5 is 3, F6 sees F5 in row F, so it is not 3 either. F6 cannot be 3 in either case.

When both ends hold the same digit, the rule is simpler: every cell that sees both ends loses that digit.

Common mistakes

  • Using a weak link where the pattern only supports "not both true" and then reading it as a strong link. A cell with three candidates gives weak links, not strong ones.
  • Ending on a weak link. The final link must be strong or the last node is not forced.
  • Eliminating on cells that see just one end.
  • Chaining from pencil marks you have not refreshed since your last placement.

Where you meet it

Alternating Inference Chains take over where X-Cycles stop, because they are free to change digit at every cell. Expect to need them in Master puzzles and to lean on them constantly at Legendary. Build short ones first: a five-node chain solves far more positions than most solvers expect.

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