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.
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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.
Some do, some teach, the rest look it up.