An ALS chain links three or more almost locked sets in sequence. Each set is joined to the next by a restricted common candidate, with a different digit at each joint. If the two end sets share a digit that no joint uses, at least one of them holds it, so any cell that sees every copy of it in both end sets loses it.
From two sets to a chain
Two almost locked sets joined by a restricted digit make the restricted common candidate move. At least one of the two sets goes without that digit and locks.
An ALS chain keeps going. A set that loses a joint's digit locks on the rest, so it must place every other digit it holds. If one of those is restricted with a third set, that third set loses it and locks in turn. Each new set adds one joint.
Why each joint needs its own digit
The middle set has just lost the first joint's digit. It can only pass on a digit it still has to place, so the next joint must use a different one.
How to spot one
Once the pencil marks are complete and the singles are used up, start from a small set. Find a nearby set that shares a digit with it, where every copy of that digit in one set sees every copy in the other. From there, repeat with a new digit. Stop when the last set shares a digit with the first that no joint uses.
Cells with two candidates are quick to check, since each holds a digit in one cell only. Sets of three or four cells are easier to miss, yet one can be what makes the chain work, as the middle set does here.
Find a first set
Row five, columns three and four hold 5 and 7, and 5, 6 and 7. Two cells, three digits: an almost locked set.
Add a second set
Row nine, columns three to five hold 5, 6, 7 and 9 between them. Three cells, four digits. It shares no cell with the first set.
Join them on the 6
The first set holds its 6 in one cell only, and so does the second. Both cells are in column four, so at most one set gets the 6. That makes the 6 the first joint.
Add a third set on the 5
Row eight, columns one and two hold 5, 6 and 7. The second set's only 5 and this set's only 5 share box seven, so the 5 is the second joint. It differs from the first.
Find a digit both ends hold
The first and last sets both hold 5, 6 and 7. The 5 and the 6 are joint digits, so the 7 is the one to use.
Look for a cell that sees all three 7s
Row five, column one does: it sees two of the 7s along row five and the third down column one. It is the only cell with a 7 that sees all three.
Strike the 7
At least one end set holds a 7, and row five, column one sees every 7 either set could hold. So this cell can't be the 7, and the 6 is all it has left.
Why it works
Suppose the first set has no 7. The chain then falls like a row of dominoes, and the last set ends up with a 7.
So whatever happens, the first set or the last one holds a 7. A cell that sees every 7 in both can't be the 7.
The chain also runs the other way. Start from a last set with no 7 and the dominoes fall back toward the first set, which then holds a 7.
In general: link the sets with restricted digits, a new digit at each joint. Any digit both end sets hold, other than a joint's digit, goes from every cell that sees all its copies in both end sets. Longer chains work the same way, one joint per extra set. The game looks for chains of up to four sets.
The first two sets share the 7 as well. Their 7s sit in row five and row nine, and no cell sees them all. On their own, the two sets remove nothing.
With the third set on the end, the 7s to watch are in row five and column one. One cell sees them all.
How it relates to other moves
With only two sets, this is the restricted common candidate move. On this grid no pair of sets removes the 7. It takes the third.
Make every set a single cell and the chain is an XY-chain. On that article's grid, the game's ALS-chain check finds the same four cells as four one-cell sets and strikes the same 1. Three single cells make a Y-wing. Run on the Y-wing article's grid, the same check takes the wing, the pivot and the other wing as three one-cell sets and removes the Y-wing's 5.
Here the middle set has three cells. On the starting grid, none of the game's wing, chain or cycle checks removes this 7.
Common slips
Striking a joint's digit. The end sets here also share the 5 and the 6. Those carry the joints, so they say nothing about the ends, and the rule leaves them alone.
Using the same digit at two joints in a row. The middle set has already lost that digit, so it can't pass the same digit on.
Calling a digit restricted when it is not. Every copy in one set must see every copy in the next. The first two sets both hold 7s, but row five, column three and row nine, column four share no row, column or box, so the 7 can't be a joint there.
Checking only one end. The cell you strike must see every copy of the digit in both end sets.
What the in-game hint shows
The in-game hint may show a different or simpler step first. On this grid it opens with a pointing pair: in box four, the 7s sit only in row five, so row five, column four loses its 7. The first set still holds 5, 6 and 7.
Another pointing pair and a naked pair follow, then a 3D Medusa step that strikes this 7 along with others.
On the starting grid, no step the hint checks before Nishio strikes that 7 on its own. Nishio does. If row five, column one is the 7, row five, column four must be the 6 and row nine, column three the 7, and then row nine, columns four and five both need the 9. The game checks Nishio before this move, so the hint seldom shows it. A different route can still be right.
Chain almost locked sets through restricted digits, never the same digit at two joints in a row. Then look at the two ends: a digit they both hold, other than a joint's digit, can't sit in any cell that sees every copy of it at both ends.
Where this sits on the path
This is the fifth technique of the last chapter, after aligned triple exclusion. It builds on the restricted common candidate. Sue de Coq comes next, and death blossoms close the chapter.
On a different grid, three almost locked sets are outlined. In order: row two, column five, then row three, columns three and four, then row nine, columns three and four. Which candidate does this chain remove?
Find the digit at each joint first. The digit to strike is in both end sets and is no joint's digit.
الخلاصة
- ابحث عن
- Three or more almost locked sets in sequence, each joined to the next by a restricted digit, a different one at each joint.
- النطاق
- The chain's own cells, plus any cell that sees both end sets.
- النتيجة
- Outside the chain, a cell loses a digit both end sets hold, other than a joint's digit, when it sees each copy of it in the first and last sets.
- العلامة
- A set that shares one restricted digit with a neighbour on each side, a different digit each time.
- إذا لم تنجح
- If the end sets share no digit beyond the joints' digits, try another end set.