When n cells in the same row, column or box have between them n+1 candidate numbers, then this is called an
almost locked set (ALS). The simplest example is a single cell with two candidates. Almost locked sets provide one of the
most versatile and powerful methods for cracking middle range difficulty puzzles.
When there are two ALSs which each have a common number such that this number in ALS 1 can 'see' all instances of the same
number in ALS 2, and vice versa, then eliminations may be possible. This first number is often called the 'X' value. If a
second number occurs in both ALSs, then any instance of this number not in either ALS that can 'see' all the instances of
the second number in both ALSs, then this number can be eliminated. This second number is often called the 'Z' value. The logic
is that if the Z-value(s) in ALS1 is/are false, then ALS1 becomes a locked set. The X-value(s) in ALS1 is/are then true, and renders the
X-value(s) in ALS2 false, thus making ALS2 into a locked set also. The Z-value(s) in ALS2 must be true. Then any instance of Z outside
of both ALSs which can see the Z-value(s) in both ALSs can be eliminated.
A typical 1-4 ALS from Top95 #14 is shown below:
The X-value is 8 at r6c9 (ALS1) and r6c2 (ALs2). The Z-value is 3 at r6c9 (ALS1) and r45c2 (ALS2). The 3 at r6c3 can see all 3 Z-values
and can thus be eliminated. This is called the ALS-XZ rule, or alternatively ca be presented as the AIC (8=3) r6c9 - (3=8)r2c2568.
More complicated is the 4-5 ALS below that occurs in Top95 #11:
X=7, Z=3. If the 3 at r2c7 were true, then the 3 at r4c7 would be false. If the 3 at r2c7 were false, then the yellow ALS would reduce to a
naked set of 46789 inwhich the two 7s are confined to box 2 and one must be true. Therefor the 7s in box 3 in the red ALS must be false
and the red ALS reduces to a naked set of 1389. The 3 at r4c5 therefore must be true. So either the 3 at r2c7 or the 3 at r4c5 is true.
Therefore the 3 at r4c7 is false and can be eliminated. This could be represented as a chain (3=7)r2c14567 - (7=3)r1346c5.
There are numerous ALSs to be found in the Top95 set, eg #31 which may be solved with ALSs alone.
Alternatively,the following is a nice example of a 4-5 ALS which solves an otherwise hard puzzle in one hit:
This involves 3 linked ALSs with a Z-value(s) in the first and third ALS, with an X=value (X1) between ALS1 and ALS2, and a different
X-value (X2) between ALS2 and ALS3. It's a logical extension of the ALS-XZ rule. Here's an example, also from Top_95 #31 after a series of
X1 = 5, X2 = 9, Z = 4. The logic is that if all the 4s in r9c1345 are false, the 5 at r9c1 is true, the 5 at r5c1 is false and the 9 at
r5c1 is true. Thus the 9 in ALS3 is false and one of the 4s is true. The 4 at r9c2 which sees all 4 Zs in ALS1 and all 4 Zs in ALS3
can be eliminated. The corresponding AIC is (4=5)r9c1345 - (5=9)r5c1 - (9=4)r248c2 => -4 r9c2. Note that this process can be extended to
four or more ALSs.
When the X and Z in the first ALS see all the X and Z's in the second ALS and vice versa, then extra eliminations are possible. Returning to
Top95 #31, the first ALS found is a 1-4 double ALS with x=4 and Z=6. The 4 and 6's in the r9c1234 ALS are confined to r9c123, all visable to r8c2. The
ALS can be represented as a loop: (4=6)r8c2 - (6=4)r9c1234 - loop. Thus any 4 or 6 in Box 7, not part of the ALSs, can be eliminated (-46 r8c1).
Addditionally, if 4 is true in r8c2, the second ALS becomes a locked set of 5689. If the 6 is true, it becomes a locked set of 4589. So whichever is true,
any 5, 8 or 9 in a cell that sees all of the second ALS can be eliminated (-89 r9c5).
Two further great examples of double ALSs are the following:
7.8...3.....2.1...5.........4.....263...8.......1...9..9.6....4.8..7.5........... top1465#2 with an extra clue 8r8c2 (thanks 999_Springs)
....8......24.35...9.....2.3.......184.108.40.206.1...8...4...6..1..9...4...8......6.... Player's Forum, Puzzle "July 26, 2018"
...3.1..7..2...39........5....1...2...3..64..96..2..7.47.....1.6..8.5...2.9...... Mike Barker's collection via Tarek