I think a more generalised proof of rule 1 would go something like this:
If 2 ALS have a common restricted digit that means you only have 1 more digits than cells to work with. If you were now to place a digit that completely removes a digit that is not commonly restricted from both ALS you effectively remove 2 digits, leaving 1 fewer digits than cells to work with (which is obviously wrong).

Also concerning Robert's comments I think his last comment is a correction for his rule 1 and not his rule 2 as the eliminations he describes do not work on sets with any "excess"


All in all I think this can be generalised into having any number of sets that each contain some number N of digits in some lower number C of cells and then some of them are restricted.
Now as soon as you look at 3 sets there will be different types of restrictions occuring as a digit could be able to appear only once across all 3 sets or it could be able to only appear once in 2 of the sets but could still appear in the third set in addition.
In other words in the first case placing it in one of the sets eliminates it from 2 other sets and in the second case from 1 other set.
Thus I call them 2-restricted or 1-restricted respectively.

If no digit is restricted you just add all N and all C of the sets you are looking at and end up with N digits (digits may repeat here) to put into C cells, in other words you have N - C degrees of freedom.

Now if digits are restricted commons that effectively reduces your N and it does so by exactly the strength as I have defined it, so a 1-restricted digit reduces N by 1, a 2-restricted by 2 etc.
(This is due to it appearing in N as often as you include sets it appears in, but for the sets it is restricted in it should've appeared only once)
C stays the same, so your degrees of freedom also go down by that amount.

If now a candidate would completely remove more digits than the degree of freedom it can be eliminated.
(I hope this is obvious as it would result in having to place a number of digits in a larger amount of cells inevitably leaving at least one empty.)
You have to be very careful though, let's say you are looking at the sets 1,2,8 and 3,4,8 where 8 is not a restricted digit removing 8 from both sets counts as 2 eliminations while removing it from one set only counts as 1 elimination. If you only remove it from cells within a set but it still can go somewhere in the set that is no elimination at all.
However if 8 were a restricted common in both sets eliminating it from just one of the sets would not amount to eliminating one digit, instead it has to be eliminated from all sets it is restricted in at once to count as a single elimination.

Let us look at the example for the double linked rule in this article to showcase this:
We have two sets: 2,4,7,9 and 1,2,4.
Thus our N is 7.
The sets have 3 and 2 cells respectively adding to C = 5.
We have 2 digits that are 1-restricted, so our N decreases by 2.
We can see that N - C = 0, so we have no degree of freedom.
Thus any cell that removes even just one of the digits from one of the sets can be eliminated.
We can also see that only the 2 and 4s get removed that remove the restricted common digits from both sets.

Of course all this also works with the complementary aka hidden sets, though thinking about that makes my head hurt.
I also do not know how many eliminations this generalisation even opens up.
Furthermore for a human solver I do not think this strategy is very feasible (Not that I myself am any good at spotting regular ALS eliminations), but I think the best strategy to search for these might just be picking one ALS you can see and then expanding from that by adding other cells / sets of cells that are linked to the original set in a way that the degree of freedom decreases or stays the same to increase the number of eliminations (Though with this approach there will be a LOT of links to pay attention to in the end).

What StrmCkr proposes here goes in the same direction as what he's mentioning are 3 ALS, which are triply linked.
3 ALS means 3 sets which one more digit than cells, so a degree of freedom of 3
Triply linked in this case means 3 1-restricted commons, so a reduction of the degree of freedom by 3 to 0.
Therefore you can apply the eliminations as discussed above.
The same will obviously work with 4 ALS and 4 links

I don't think the example they make in their comment is a good one cuz that can also just be done with a double linked ALS as they say, but the last example of their forum post has a better example as it contains 3 sets that are distinctly separate and at least your solver fails to solve the puzzle without guessing, but can do it if you put in the eliminations provided in that example (though the puzzle remains absolutely brutal even after that).

Hello Andrew

I have several examples of two overlapping ALS where one cell can be described as existing is both sets. If one accepts that this is allowed then eliminations are possible according to the ALS xz rules. The link to the diagram is below (sorry didn't know how to paste one in)




Set A {1,3,5,6} is the two green cells plus the yellow cell. n = 4 so n-1 = 3 cells
Set B {3,5} ids the yellow cell

The overlapping cell is the yellow one.

Restricted common x = 3 which will only end up in A or B
Common Candidate z = 5 which is eliminated and shown as the red cell.

Is this legal ? I have several other of these which worked perfectly for solutions to difficult puzzles. However....

Hi Robert

I did encounter ALS-XZ Rule 2 when using the solver and it seems that your assumption is actually correct. The solver's implementation seems similar to this here: http://hodoku.sourceforge.net/en/tech_als.php#axz2

However the solver uses the words 'unique to one ALS' which I think is wrong, as they need not be.

Bugger, I did not describe my idea for an extension to this technique properly.

My proposed "Rule 2" is correct as far as it goes, but it is incomplete. If there are two restricted digits in the ALS case, or a number of restricted digits equal to the "excess" (definition in my earlier comment) if we allow AALS or even more "A"s, then all restricted digits must occur in the one set or the other. Therefore, in any cell outside the two sets, if one of the restricted digits occurs and can "see" all occurrences within the two sets, it can be eliminated. So there are two types of eliminations - restricted digits need to be able to see all occurrences in both sets, but unrestricted digits only have to see all occurrences in one of the two sets. Unrestricted digits in this case do not have to occur in both sets; however, if they do, some of them may be eliminated, if they can "see" all occurrences of the same digit in the other set.

Follow-up to my earlier comment regarding my extension of this strategy.

In my database of 341 "advanced" puzzles, I run the ALS strategy (with my version of Rule 2, which may or may not be the same as David Bird's) together with basic techniques, and get 66 puzzles fully solved.

My extended version of the strategy only finds eliminations based on ALSs, AALSs, and AAALSs. Furthermore, although my solver does find some eliminations involving AAALSs, it finds the same eliminations when restricted to just ALSs and AALSs.

So I don't know whether there is some general wisdom here that there is no need to check for anything more than an AAALS, or if that is just a specific property of my sample of 341 puzzles.

So the documentation on Rule 2 is still on the to-do list, but I have speculated in a comment last week about what it might be. I repeat my speculation here, and also offer an extension to this method.

First, my speculation about Rule 2 - if there are two restricted digits in the pair of ALSs, then any candidate in any cell that can see all of the same digit in just *one* of the ALSs can be eliminated. This might possibly include occurrences of that digit in the other ALS (so the same value occurring in both ALSs, but it is not restricted, and therefore possible for the same digit to appear in the solved puzzle in both ALSs). My own solver finds some instances of this occurring in my sample of "advanced" puzzles (now up to 341 puzzles), and can solve 66 of those puzzles using the basic techniques, the ALS Rule 1 above, and my proposed version of the ALS Rule 2. Using only Rule 1, I get 65 puzzles solved instead of 66, and possibly some additional eliminations in puzzles that are partially solved (I would have to check).

Rule 1 is a special case of AICs with ALSs, but with the size limitation in the implementation of ALSs that appear in AICs, it is possible that the solver might find some eliminations using this strategy, that is doesn't find using AIC with ALSs. My version of Rule 2 is also a special case of AICs with ALSs.

Now, for my extension. We can use not just ALSs, but AALSs, AAALSs, etc. So here is how it works. Find two groups of cells that are not locked sets. (If they include locked sets as subsets, these should be thrown away - the strategy is still valid, but other strategies can be applied.) So they might be ALSs, AALSs, AAALSs, or maybe even more Almosts. Call the "excess" the combined number of "almosts" in the two sets. For example, if one set is an ALS, and another is an AALS, the "excess" is three. If there are two AALSs, the "excess" is four. If there are two ALSs, the "excess" is two. (I think this strategy probably works with locked sets as well, but in that case, easier strategies can be used instead.)

The number of restricted digits (same definition used for ALSs) cannot be more than the "excess". If it is, the puzzle has no solution (or we have made a logical error while trying to solve it).

Rule 1: if the number of restricted digits is equal to the "excess", then any candidate in any cell that can "see" all occurrences of an unrestricted digit in *either* (not both) of the sets can be eliminated. This might possibly include candidates in one of the two sets, if it can see all occurrences of an unrestricted digit in the other set.

Rule 2: If the number of restricted digits is equal to the "excess" minus one, then any candidate in any cell that can "see" all occurrences of an unrestricted digit in *both* of the sets (and the unrestricted digit must occur in both sets) can be eliminated.

I have implemented this in my own solver. Once I had the ALS code running smoothly, it probably took me five minutes to make the extension (definitely less than ten). It adds a lot of processing time, because *every* set of cells within a unit that is not a locked set, is an ALS, an AALS, an AAALS, an AAAALS, an AAAAALS, an AAAAAALS, an AAAAAAALS, or an AAAAAAAALS. In my database of advanced puzzles, it does not solve any additional puzzles (when combined with basic techniques) than the 66 solved using ALS Rule 1 and my version of Rule 2 (which may or may not match David Bird's Rule 2). However, it does find some additional eliminations, bring some of the partially solved puzzles a little closer to solution.

When I combine it with all the other strategies I have implemented, although a number of my "super-ALS" pairs are found, and some eliminations made, all the same eliminations would have been made by other strategies anyway. So it is possible this extended strategy is a special case of some other strategies (just as Rule 1 of the ALS strategy, and my version of Rule 2, are special cases of AIC with ALSs).

So no docs on the extension mentioned, but if I may speculate -

Is it that, if there are two restricted digits, then any candidate that can see all instances of a non-restricted digit in *either* (not necessarily both) ALS can be eliminated?

Almost Locked Set Example 3: cannot 7 also be removed from D5?

In my last comment, I put some of the content in angle brackets, which apparently made them invisible (I guess they are interpreted as formatting commands).

The missing parts simply state that the links in my AIC are weak-strong-weak-strong-weak.

I've been implementing my own solver, really just for laughs, and I've gotten as far as AIC with Groups. Still need to do AIC with ALSs, which is next on my to-do list. But AIC seems to be an extremely powerful technique. Mesusa (except for Rule 6), Digit Forcing, Nishio Forcing, and some other strategies are all special cases.

Which brings me to this one. All of the examples above look like they could be considered examples of AICs with ALSs, if we define a weak link between certain candidates in the ALSs. In these examples, the resulting AIC has five links, and the candidate to be eliminated as between two weak links. If this candidate is "on", then the same candidate must be "off" in both the ALSs. But because they are ALSs, if one of the candidates is "off", all the other candidates must be "on" (strong link), although if a candidate appears several times within the ALS, we won't know which one is "on". But for the restricted common, there is a weak link between the two ALSs - "on" in one means "off" in the other. So we have an AIC. In the last example shown,

Candidate 7 in G7

Candidate 7 in ALS {G2,G5,G6}

Candidate 1 in ALS {G2,G5,G6}

Candidate 1 in ALS {A2,A7}

Candidate 7 in ALS {A2,A7}

Candidate 7 in G7

So it is an AIC, Candidate 7 in G7 can be eliminated by Rule 3 (and Rule 1 and Rule 2 are really just special cases of Rule 3 anyway).

So I think by defining the notion of a "weak link" between ALSs (which is done using the restricted common), we could build more complex inference chains than the five-step ones shown above.

Whether this is computationally feasible, let alone human-feasible, I do not know at this point.

For rule 1, you state that:

"any other candidate found in both ALSs but not a restricted common; that, a candidate that still appears in both ALSs and is not exclusive to one or the other."

but, what if there are two restricted commons?

Here is an example of a 1 and 4 cell ALS:

https://www.sudokuwiki.org/sudoku.htm?bd=360020009920000050870009203490750000010000090030918002259400831080090020040080965

The solver claims that candidates 1,4 can be removed from A8; which contradicts the above statement.

Presently missing an explanation of the doubly linked ALS with two RCCs which becomes cannibalistic and is included in the solver as ALS XZ-Rule 2. See this link for more info: http://hodoku.sourceforge.net/en/tech_als.php I think you refer to David Bird about it.

Where are the docs you often refer to?
I was looking at the Almost Locked Sets web page, at the bottom you have a title, ALS-XZ Rule 2, with a thanks to Davis Bird and that Docs were to come. Where can they be found, or do they not exist yet?

Thanks in advance.

I have found it quite useful to extend this method with XY-Chains. XY-Chains find weak links between bi-value cells but you can use ALSs in place of XY cells (XY cells are just a simple type of ALS).

ALS XZs are just a two step chain and a "ALS Chain" is just continuing this logic.

These may seem hard to find at first, but if you are already looking for XY-Chains I have found it quite easy to include 2 and 3 cell ALSs in the chaining logic.

Of course all of these are examples of AICs but I find those to be quite hard to find in general.

I found a very interesting example for this strategy. It involves in total 9 cells and they are all in only one unit. I found it in your "18 elimination Jelly-Fish" example (..........7..3.92..19.2563...4...21...........57.9.46..9514.37.7......4..4236759.)

It all happens in column 9. The first ALS is a 6-cell set {A9, B9, C9, D9, E9, J9} and the second is the single cell set {F9}. The restricted common is 3. This results in the elimination of candidate 8 in cells F9 and H9. The fact that both sets can be in one unit is not so evident, that's why it find it so interesting.

I also find it interesting for the fact that it illustrates that there can be more than one restricted common in each set. The first ALS has a 3 in both D9 and E9. More explicitly: if all candidates for X in set 1 see all candidates for X in set 2 then X is a restricted common for both sets.

LOL
Still interested to know whether a restricted candidate is really necessary for an ALS to work

A simple explanation, I finally got it. Wonderful site.

There is a second way a set can be almost locked, namely, when N candidates appear in N + 1 cells within a unit. I call this a Hidden ALS (or HALS), as knocking out one of the cells reduces to a hidden pair/triple/quad/&c. Likewise, the ALS described on this page would be better called a NALS (Naked ALS).

I wish to show my appreciation to the witrer just for bailing me out of this particular issue. Just after browsing throughout the world-wide-web and coming across opinions which were not productive, I was thinking my life was gone. Being alive without the solutions to the problems you've resolved all through your entire report is a critical case, as well as ones that might have in a wrong way damaged my entire career if I had not discovered your web site. Your main expertise and kindness in handling all the pieces was very useful. I am not sure what I would have done if I hadn't come upon such a thing like this. I can now look forward to my future. Thanks so much for your high quality and amazing guide. I won't think twice to recommend your blog to any person who wants and needs assistance on this issue.

Ian,

If I understand correctly, adding G5 to the brown ALS does no good. You can only use this technique to make eliminations on numbers common to both the ALS.

In Example 1, adding G5 to the brown ALS picks up another candidate, the 9, but it will not lead to any more eliminations because the yellow ALS does not have a 9. If the yellow ALS happened to have a 9, then it could help.

Essentially, the relationships between the common elements of the two ALS means that the unrestricted common candidate (in Example 1, it's the 8) has to be the correct solution for a cell in at least one of the two ALS. So at least one of the yellow or brown cells that contain 8 as a candidate must actually be a 8 in the final answer.

It logically follows that 8 can be removed as a candidate in any cell that can see ALL of the 8s in the yellow and brown squares.

Hi Andrew!
Example 1: Could you not also have added G5 to the brown ALS and made it an almost locked triplet over 4 candidates?

If yes - is there a reason why you didn't?
If no - please could you explain why not?

Many thanks! Your site is fascinating.
Kind regards,
Ian

Ok, I've found out the answers myself...

Answer 1 = No, it's not true. A bi-value cell ALS can share the same box with one or more cells of the other ALS

Answer 2= Yes, they can.

I have two questions:

If one of the ALS is a bi-value cell, then the other ALS cannot have a cell in the same box of the bi-value cell, is that true?

If none of the ALS is a bi-value cell, and they both are in a row (or both in a column), can both these rows (or columns) appear in the same band (or stack)?

Actually, sorry. I was wrong on this one. It does present a contradiction. I stared at this for 15 minutes trying to see it. Wrote the comment, then looked again and saw it immediately. Looking forward to getting this on the iPad. :)

Andrew must be busy on the mobile solver since I have seen no answer. But I think Grandad is right. If you load the example, you will not even get to ALS as an example for those cells unless you manually remove the 2. However, even in that case, the proof doesn't seem to work. If you make F5 a 3, it does not present a contradiction in the two ALSs. It may present a contradiction elsewhere. But this type of example trips me up. I bought the book and I love it. But I often find that I can't prove a path I follow using its rules. I usually import the puzzle that I can't prove and watch what the solver does. Sometimes that helps, sometimes not.

I am almost certainly not the first to ask but - in the 2nd example, surely F1 should have possibles 1, 2 and 3 so isn't an ALS ?

Hi Andrew,

Excellent site ;-)
I'm just getting to grips with some of the more difficult strategies. I had a puzzle loaded in the solver and it came up with an ALS that I fail to understand :( The starting point for the puzzle is

..7...32.1..2.7..6.26.5.71..8...2....19.6.5...6.7......91.2..475..17.....78......

If you just keep clicking 'Take Step' until it comes to the first ALS, I can't see how 8 is restricted as 6 can also be seen by both ALSs.

What am I missing in my understanding of the ALS technique?

Many thanks and best wishes,

Terry

In the last example the {1,2,6,7,8} in row G is not an ALS because it has 5 candidates in 3 cells