I have been working a lot with ALSs for a while now, and I have come to realize that the definintion I use for finding an ALS might not be the best way...

Some of this might better belong in the programmers forum, but I hope that some of this might be general enough to apply to manual solvers as well...

In short, the definition I have used for an ALS is: "n cells with exactly n+1 different candidates". Hence our beloved bivaluecell are the smallest known ALS.

However, even though all sets found with this definition is arguably an ALS, I have found that it produces a lot of so called "useless" ALSs, or ALS's that is not suited to be used in ALS related eliminations...

This is just screaming for an example, so let's start with this first argument:

"many ALSs that are found with the "n cells, n+1 candidates"-argument are actually several ALSs.":

The simplest example is if you have two cells: [13] and [14].

These two seen under one fulfills the "n cells, n+1 candidates"-argument because we have 2 cells and (1,3,4)=3 different candidates. However, this is also two different ALS's [13] and [14], and it turns out that any elimination that can be done using [13] and [14] as one ALS, can also be done with the short ALS chain: [13]-1-[14].

The removal of such superflous ALSs from a recursive ALS-chain-builder will speed it up conciderably without any loss of potential eliminations.

A more complex example would be the three cells: [123] [12] [14]

All these cells combined could be concidered an ALS. Both [12] and [14] can be concidered separate ALSs, but [123] can not, so here we can not say that this is one ALS concisting only of smaller ALSs. However, [123] and [12] can be concidered one ALS when combined and hence the [14] is not needed. And again, any elimination that can be done with the ALS: [123][12][14] can also be done with the little chain [123][12]-1-[14]

So that is my first objection... The second one is even more obvious and happens when a Locked Set finds it's way into an ALS.

"if an ALS contains a Locked Set, the cells that contain the locked set are useless to the ALS."

This might be stating obvious, but following the same rule, this has to be concidered an ALS:

[12] [12] [45]

We have three cells, and exactly 4 different candidates, and yet what have happend here is that a bivaluecell has clinged on to a Locked Set which it has absolutely no relation to whatsoever! You see my point? This is an absurd ALS, and yet it fits the definition perfectly?!?

What I want is for some of the great brains at this forum to help me come up with a new definition for what I call the "useful" ALSs...

Thanks!

Havard