This is the first time I am posting to this forum, I hope I'm in the right place.

The method I'm about to present is a generalization of ALS-Chain which is much rarer but could be spotted much more easily (like XY-Wing -> XYZ-Wing).

I looked and couldn't find it in popular websites of sudoku solving techniques so I'm posting it here.

First a definition:

An almost locked subset (ALSS) of a set A of N(>1) cells is a set of appearances of candidates in A such that at most 1 cell of A can contain a candidate NOT from the subset.

Observations:

1. If we take an ALS (which is not a bi-value cell) and choose all candidates but one they from an ALSS.

2. The key property of ALSSes: If we remove all appearances of a candidate of the ALSS then a subset of A of size N-1 forms a locked set on the remaining candidates of the ALSS (hence the name).

The method - ALSS-Chain (Loosely speaking, take the definition of singely linked ALS-Chain and every time you seethe word "ALS" replace it with "ALSS", that is actually exactly what I did):

ALSS-Chains are a series of ALSSes connected by restricted common candidates (RCCs from now on). The first and the last ALSS must contain a common candidate Z which is not the first nor last RCCS ((not Z=first RCC) and (not RCC=last RCC)), Z is eliminated from all cells that see all appearances of Z in both ends of the ALSS-Chain. Some restrictions are put on the RCCs: No two adjacent RCCs may be the same.

Proof:

Assume one end of the chain contains the common digit, then the elimination is immediate.

Otherwise, we just removed the common candidate from the ALSS in an end of the chain, so by observation 2 some subset of it is locked and contains the first RCC of the chain (since it's different from the common digit) so the same thing happens to the next ALSS.

We can continue all the way to the last ALSS, all of them (including the last and excluding the first) contain a locked subset of their size - 1 of all candidates but the previous RCC (we can view Z as a "before first" RCC and include the first in the last sentance).

If a cell can see all apperances of the common candidate in the last ALSS, it sees all of it's apperances in it's locked subset (there are appearances because (not Z = last RCC) so we didn't remove Z), which exists by observation 2; and therefore the common candidate can be eliminated from it.

Disscussion about finding ALSSes:

Note that an ALSS which is not an ALS is quite rare, in fact from the methods I know of it can appear in exactly two forms:

1. There is an ALSS to avoid a deadly pattern (different solutions to the sudoku).

2. The ALSS comes from two colors of 3D-Medusa (also in this case all ALSS-XZ eliminations are already made as part of the normal 3D-Medusa eliminations so those chains are of length > 1).

I actually came across both of these in Diabolical puzzles (unfortunately I didn't record it at the time) and only later generalized the pattern.

How I think one searches for it: You don't!

You can integrate it into other solving techniques as a "last resort":

When you search for a unique rectangle, unique rectangle type 2 is a locked subset. If you only find an ALSS, only then you look for a chain of ALS(S)'s that contains it.

When you do a 3D-coloring and find N cells in the same house with N colored candidates such that every color is spread across N-1 cells of this set it's an ALSS, so now you search for others to complete a chain.

Examples (click on the mini-images for a link to a normal size image):

1. A simple example with a deadly pattern:

- Code: Select all

+---------------------------------------------------------------------------+

| - - - | - - - | - - - |

| - - - | - - - | - - - |

| - - - | - - - | - - - |

|----------------------------+--------------------+-------------------------|

| 4- ab12347 4- | - - - | - ab1247 a47 |

| c3456 c3456 - | - c56 - | - - - |

| - b12 - | - - - | - b12 - |

|----------------------------+--------------------+-------------------------|

| - - - | - - - | - - - |

| - - - | - - - | - - - |

| - - - | - - - | - - - |

+---------------------------------------------------------------------------+

Anything in the grid that's marked with '-' can be anything.

The cells marked with b (blue candidates in the pic) are what is a potential deadly pattern, what's stopping it are the other candidates in those cell, taht together with the candidates in R4C9 (all marked with a or yellow candidates in the pic) form an ALSS.

The cells marked with c (yellow cells) form an ALS which is an ALSS by throwing away the 5.

The RCC is 3 and 4 is another common candidate so it can be eliminated from R4C1 and R4C3.

This rule is ALSS-XZ

2. A complicated example with 3D-Medusa:

I'm not adding code here since I couldn't explain add the 3D-Medusa markings inside the grid without making it unreadable, so you'll have to click on the image, sorry.

The colors (of candidates) obtained via the 3D-Medusa are yellow and green.

The first ALSS is a subset of the red squares that are colored = {12378}.

It is an ALSS because we have 4 candidates of both colors in the red squares => if 2 cells are not from the ALSS then both colors are wrong which is a contradiction to their definition (using inference of 3D-Medusa).

The second ALSS is the ALS colored in blue (minus 4) and the RCC is 1.

The third ALSS is the ALS colored in yellow (minus 8) and the RCC is 5.

R2C2 can see both ends of the ALSS chain so 3 can be eliminated from it.

Note:

You don't have to know which candidate to remove from an ALS to make it an ALSS that fits some ALSS-chain, the point is there is always one (because for the chain we need the previous RCC[common candidate for first] and the next RCC [common candidate for last] and the ALS contains at least 2 cells and 3 candidates, so one of them is neither and it doesn't affect the chain).

I don't think normal human beings (that know ALS-chains ) can find an ALSS chain with no ALSes in it (solvers probably can, but I assume this situation is super-rare).

I'd love your comments, suggestions and critique,

Omer Rosler