The New Sudoku Players' Forum

[StrmCkr]

removed
updated for more clarity in the next post.

[SpAce]

I think a DL-ALS-XZ can eliminate all 9, because DL-ALS-XZ is also a Rank0 logic, not the same SL-ALS-XZ is a Rank1 logic.Rearrange coversets without affecting the final rank, so these combinations can form effective elimination。

That's only true if viewed as set logic. I have no problem with that, and we've already agreed that getting all eliminations requires three different combos of cover sets. Who says that ALS-XZ implies set logic, though? In XSudo such a POV is expected, but not in a solver that doesn't use set logic for everything (at least not openly). Patterns like DDS and MSLS (and fishes obviously) imply set logic directly, but I don't think it's true about ALS-XZ or many other techniques.

My problem is exactly that neither kind of ALS-XZ implies set logic necessarily, and it's probably not how most people see it. It's a simple interaction between two ALSs, which locks certain digits into two groups of cells allowing eliminations. With that POV it's impossible to get both -9 r19c1 and -9 r6c46 at once without sequential logic, i.e. taking the cannibal eliminations first. I strongly think that a single move should only list eliminations that are available in parallel using the expected logic of the move (and not some other logic like set logic, even if it produces more eliminations for the given pattern).

As me and eleven have said the eliminations are valid under als eliminatiins rules.

No. Only you and yzfwsf have claimed that, but eleven never did. His technique has nothing to do with ALSs. It looks at the full pattern as a whole, just like the set logic POV, without breaking it into individual ALSs. That allows getting all the eliminations at once, just like the set logic POV. On the other hand, when viewed as an interaction between two ALSs it's not possible, because it doesn't allow locking the digits in such a way that we'd get all eliminations simultaneously.

As an aic nope since it cannt really do internal parts of a chain.

Glad we agree about the AIC POV at least. However, I don't think it works any better even if we forget AICs and just look at the ALS-XZ logic.

First, if we break the pattern into these two ALSs with the dual RC(1,6):

Code: Select all

ALS A: (1 6 9)r56c1
        | |
ALS B: (1 6 925)r4c2346

=>

(1'9)r56c1 & (6'925)r4c2346

OR

(6'9)r56c1 & (1'925)r4c2346

What we see is that in both cases the 9 gets locked into both sets. Note that they're not one single set (like an MSLS), but two, and thus they must be treated separately. When looked at in parallel, the 9s in the larger LS are locked into r4 only but not in b5. Thus it only gives us the r4 eliminations, but not r6c46. To get them too you have to start with the smaller LS to get the cannibal eliminations (r4c23) first, and only then the 9s get locked into r4c46 in the larger LS eliminating r6c46. That's sequential logic which should not be assumed in a single move.

(You seem to be combining the two locked sets but I don't think that's warranted by the ALS-XZ logic. It's set logic or MSLS territory.)

The same thing happens if we use the other possible pair of ALSs with the dual RC(2,5):

Code: Select all

ALS A: (2 5 9)r4c46
        | |
ALS B: (2 5 916)b4p2347

=>

(2'9)r4c46 & (5'916)b4p2347

OR

(5'9)r4c46 & (2'916)b4p2347

Again the 9 gets locked into both sets, but now the larger LS has the 9s locked in b4 only and can't eliminate 9r19c1 on its own. Once again you need sequential logic, starting with the smaller LS in r4c46 and eliminating 9r4c23 first. Only then are the 9s locked in r56c1 in the larger LS, allowing the r19c1 eliminations.

Thus, to get all of the eliminations at once I would accept generic set logic, or a cannibalistic MSLS, or eleven's logic. They all work because they look at the full pattern as a whole. I don't accept them with DL-ALS-XZ because it has two separate locked sets which don't produce all the eliminations without sequential logic. Also, I still don't accept that there's a Sue de Coq here (cannibalistic or not), but I'm done arguing about that.

I've now made my case in as many ways as I can and already repeated myself enough. At this point it's probably better that we agree to disagree, as I'm not willing to argue this point any further. It has no practical consequence either because all the eliminations are caught one way or the other anyway. The main reason why I'm arguing this point of view is that it's MUCH simpler to understand yet the end result is exactly the same.

There's no real benefit in adding complexity by jamming extra eliminations into a move which doesn't natively justify them (or it's at least very questionable). It just makes the readers scratch their heads trying to figure out what's going on. As in programming, it's better that a method/function/procedure does only what it obviously implies instead of introducing unexpected side effects.

[StrmCkr]

You seem to be combining the two locked sets but I don't think that's warranted by the ALS-XZ logic. It's set logic or MSLS territory.

Almost Locked sets is set logic combining sector{s} between 2 sets is the RC -- how do you not see it as that? as sectors hold cells, cells on their own
do not define anything unless they share a commonality between them.

A side comment:
ALS search engine are based on fish size 1/1 {base/cover} to id the RC sector that are locked.

MSLS is also set-logic using multiple sectors housing n digits in n cells where N cells in N sectors are locked for N digits.
the difference is MSLS use MsHS to identify a msls or vise versa but both are defined by RC's in sectors.

MSls, MsHs are also based on fish size 4 /{4 -7} {base cover} for there search function to id the RC sectors that are locked.

the real difference is complexity by size Incrementation.

the rules between the two do not change
elimination code between the two increased by fish size due to increase in sectors size to compare in the elimination execution.

now back to:
Als - xz :
which gives us the basic eliminations: with 1 RC
Z digit shared between both sets is removed as that's the only digit we can isolate.

Compact eliminations are with 2 RC -[ multi-sectors{2} or singular ]
set A for its candidates - RC candidates is locked to A cells eliminate all peers per candidate
set B for its candidates - RC candidates is locked To B cells eliminate all peers per candidate
Shared candidate of A & B are eliminated from peers cells visibale to all candidates of A & B Cells
peers of RC' cells are Eliminated For RC candidate

up to here you agree with:
which is fine, we can agree to disagree when the eliminations that are caught post cleanup or via 2 different als-xz as the outcome is the same, albeit with +1 steps.

its these you don't agree with is the digestion of how the sets lock which uses the rc cells, and sets themsleves:

when the RC cell{s} of A or B contain the shared non restricted Common digits and is peers of all copies of the nRC of both sets
then the RC are placed/locked & nRC candidates are removed from 1 set leaving n-1 digits for n cells in the opposite set
there by:
the RC Cells of said set cannot contain nRC
set A, B cells out side RC contain the nRC & peers of these cells cannot contain the NRC

in the example in question:
if RC contains 9 at either R4c2 OR r4c3 for set A THEN SET b is reduced to 6 & 1 { 2 digits in 2 cells} = reduces set A to candidates 2,5 (2 digits in 3 cells)
=> R4c2, r4c3 <> 9 which means the remaining cells of each set A & B contain 9 , and their peers can be eliminated for 9.

The main reason why I'm arguing this point of view is that it's MUCH simpler to understand yet the end result is exactly the same.

Fair enough: added complexity for little/no gain.

in this configuration there is no ad-hock deduction or post ratification.
- is this intuitive: probably not.

- does it gain any advantage:
less steps in resolution counts, but with added complexity to deduction step
noting the "Extra" eliminations" are replicated with a blr/naked subsets to pick up the missed eliminations.

Capture1.PNG (6.65 KiB) Viewed 654 times

Capture2.PNG (7.03 KiB) Viewed 654 times

More complicated versions exsist and i can post an example where 1 nrc cell places a rc in 2nd set and locks the first s a n set leaving the 2nd as n-1.( happens with 2 sectors rcs are used) which means 1 rc is a Canabolistic removal.
The few examples I've found are the direct result of an internal smaller als xz.
If your intrested ill post it.