An AIC, short for alternating inference chain, joins candidates with strong and weak links in turn. A strong link can run between two cells or between the two candidates of one cell, so the digit can change along the way. When the chain starts and ends on a strong link and both ends carry the same digit, one end holds it, and a cell that sees both ends loses it.
How to spot it
Look for strong links first. There are two kinds. A row, column or box with only two places for a digit gives a strong link between two cells. A cell with only two candidates gives one inside the cell.
Pick the digit you hope to remove and a cell that holds it as the first end. Assume that cell is not the digit and follow a strong link to the candidate that must then be true. From there take a weak link. Another cell in the same house loses that digit, or the same cell loses its other candidates. Then a strong link again, and keep alternating.
Stop when a strong link lands on the digit you started with, in another cell. That cell is the second end, and any cell that sees both ends loses that digit.
Assume the first end is not the 2
Row six, column five has a 2 and a 4. Column five has only two places for a 2, this cell and row five, column five. So if row six, column five is not the 2, row five, column five is. The solid line between the two cells stands for that.
Switch digits inside the cell
Row five, column five holds one digit, so if it is the 2, it is not the 4. Row five has only two places for a 4, and the other one is row five, column eight. So that cell is the 4. The digit has changed from 2 to 4 without leaving the chain.
Go down column eight and switch back
Row seven, column eight shares column eight with that cell, so it is not the 4. This link is weak, since column eight has a third 4 at row four. Row seven, column eight has only a 4 and a 2, so it is the 2.
Land on a 2 again
Row seven, column two shares row seven, so it is not the 2. It still has a 5 and a 9, so the chain doesn't stop there. It uses box seven, which has only two places for a 2. The other is row eight, column three, so that cell is the 2. Put together, if row six, column five is not the 2, row eight, column three is.
Strike the 2 that sees both ends
Row six, column three shares row six with the first end and column three with the second. One of the two ends holds the 2, and either way this cell sits in a house with it. So the 2 comes out of row six, column three, which keeps a 1 and a 4.
Why one end holds the 2
Now start at row eight, column three instead. If row eight, column three is not the 2, box seven puts the 2 at row seven, column two. Row seven takes it away from row seven, column eight, so that cell is the 4. Column eight takes the 4 away from row five, column eight, and row five puts it at row five, column five. That cell is then not the 2, and column five puts the 2 at row six, column five. Whichever end you start from, if that end is not the 2, the far end is.
In the solution both ends are the 2, though the chain proves only that one of them is.
The two-ends rule
The two ends must carry the same digit, and the link at each end must be strong. A strong link says one of its two candidates is true. A weak link says only that they can't both be true, so a chain that ends on one only rules the digit out of its last cell. It doesn't place the digit at either end.
The cell you strike must also see both ends. Row six, column three shares row six with one end and column three with the other. A 2 that sees only one end, such as row six, column nine, stays.
How it differs from short AICs, X-chains and XY-chains
A short AIC is four cells and three links on one digit. An X-chain is longer but still stays on one digit, with every link between two cells. This chain changes digit twice, from 2 to 4 inside row five, column five and from 4 back to 2 inside row seven, column eight.
An XY-chain changes digit too, but every cell in it has two candidates and every strong link sits inside a cell. Here row five, column eight has a 1, a 2 and a 4, and row seven, column two has a 2, a 5 and a 9. The chain passes through each of them on a single digit, using a house with two places for it. Read it as an X-chain and it breaks at row five, column five, where the 2 turns into a 4. Read it as an XY-chain and it breaks at row five, column eight, the first cell with three candidates. On this grid no X-chain and no XY-chain removes anything.
A link inside a cell with three or more candidates is weak only. If the cell is one digit it is none of the others, but ruling one out doesn't decide it. This chain doesn't need such a link. The second chain at the end of this page does.
Box seven has only two places for a 2. Rule it out of row seven, column two and it must be at row eight, column three.
Stop at row seven, column two and all you know is that it is not the 2. Nothing says the 2 must sit at that end, so nothing can be struck.
A second route between the same ends
From row seven, column eight the 2 can also go to row eight, column nine, which shares box nine with it and has a 1, a 2 and an 8. Row eight has only two places for a 2, so the chain lands on row eight, column three again. The ends and the strike don't change. The in-game hint shows this route.
Counting chains of up to eight candidates without cell groups, this grid has four, and every one strikes the same 2. A path that differs from the hint's can still be right, as long as every link checks out.
Alternate strong and weak links, and start and end on a strong link. A strong link can join two cells or the two candidates of one cell. When both ends carry the same digit, strike it from every cell that sees both ends.
Where this sits on the path
An AIC never places a digit by itself. Here the strike leaves row six, column three with a 1 and a 4, and no single yet. That cell now has two candidates, so it can carry a strong link. A second AIC runs through it and takes the 4 out of row six, column five. It enters row five, column eight on the 4 and leaves on the 1, a weak link inside a cell with three candidates.
AICs belong to the chapter on chains and coloring, after XY-chains. On this grid nothing earlier on the path removes a candidate, X-chains, short AICs and XY-chains included.
AICs with cell groups come next.
On a different grid, this chain runs from row one, column six to row four, column five, and both ends carry a 4. Which 4 does it remove?
Find a 4 that shares a house with each end.
Recap
- Look for
- A path of candidates joined by strong and weak links in turn, where each strong link comes from a house with two places for a digit or a cell with two candidates, and both ends carry the same digit.
- Scope
- The candidates in the chain, and any cell holding the end digit that sees both ends.
- Result
- Every cell that sees both ends loses the end digit.
- Tell-tale
- Houses with two places for a digit next to cells with two candidates, when no X-chain or XY-chain works on its own.
- If it fails
- Check that both end links are strong and both ends carry the same digit. If no cell sees both ends, start from another candidate.