The New Sudoku Players' Forum

by **Myth Jellies** » Thu Sep 21, 2006 7:06 am

I've mentioned this in a couple of other threads. Maybe it is time to give this idea its own post.

Most puzzles have an abundance of Almost Locked Sets to consider. Many of these ALS's share overlapped cells. It turns out that you can consider two overlapping ALS's as one large structure, and the digits inside this structure have a potentially useful interrelationship.

You form a Combined Overlapping ALS structure by taking the union of 2 different ALS's that share at least one cell.

The useful relationship is as follows:

Combined Overlapping ALS Rule

For the cells that make up the entire combined ALS structure, either all the digits found in the overlap region are found in the structure, or all the digits not found in the overlap region are found in the structure, or both.

Lets show how you can use this information, first with a hypothetical example.

Code: Select all: *----------------------------------------------------------------------* | . . . | . C12347 . | . D67 . | | . . . | A125 . . | . . . | | . . . | A125 . . | . . . | +----------------------+------------------------+----------------------+ | . . . | . B345 . | . . . | | . . . | . B345 . | . . . | | . . . |AB56 . . | . -6* . | +----------------------+------------------------+----------------------+ | . . . | . . . | . . . | | . . . | . . . | . . . | | . . . | . . . | . . . | *----------------------------------------------------------------------*

Overlapping ALS A: r236c4 containing digits 1256.
Overlapping ALS B: r6c4|r45c5 containing digits 3456.
Combined ALS: r236c4|r45c5 (union of ALS A and B cells)
Overlap Cells: r6c4 (intersection of ALS A and B cells)
Digits in Overlap Cells: 5&6
Digits not in Overlap Cells: 1&2&3&4
CoALS relationship: ((5&6) = (1&2&3&4))r236c4|r45c5 a strong internal link.

Thus you can form this AIC...

((5&6) = (1&2&3&4))r236c4|r45c5 - ((1v2v3v4) = 7)r1c5 - (7=6)r1c8 => r6c8 <> 6

Here are some real world examples. First a simple one:

Code: Select all: +-------------------+-----------------+---------------------+ | 5 3468 2 | 367 78 9 | 146 347 1346 | | -3678 3469 3679 | 2367 1 368 | 24679 5 23469 | | 1 369 3679 | 4 25 356 | 8 2379 369 | +-------------------+-----------------+---------------------+ | 36AB 356 8 | 1 259 4 | 69B 29 7 | | 9 7 4 | 23 6 35 | 12 8 125 | | 2 1 56 | 8 59 7 | 3 49 4569 | +-------------------+-----------------+---------------------+ | 367A 3569 1 | 569 4 2 | 79C 379 8 | | 78 259 5679 | 5679 3 68 | 2479 1 249 | | 4 2389 379 | 79 78 1 | 5 6 239 | +-------------------+-----------------+---------------------+

A and B denote the overlapping ALSs
Overlap digits are 3&6, non-overlap digits are 7&9
Thus have ((3&6) = (7&9))r4c17|r7c1

One can either generate an AIC using r7c7 to show r4c17|r7c1 contains 3&6, or one can just observe that since all the 7s and 9s in the combined ALS see the same 79 cell in r7c7, the (7&9) premise must be false and the combined ALS must contain 3&6. Either way you come up with it, any candidate three seeing all the threes in the combined ALS can be removed.

Code: Select all: *------------------------------------------------------------------------------* | 356 4 1356 | 689 1569 1689 | 578 2 567 | | 9 578 567 | 2468 2456 68 | 1 4568 3 | | 56 18 2 | 3 14 7 | 9 48 56 | |--------------------------+-------------------------+-------------------------| | 23567 23579 3567 | 1 23689 4 | 23578 3568 2567 | | 8 123579 13567 | 269 2369 369 | 2357 1356 4 | | 2346 123 1346 | 7 2368 5 | 238 9 126 | |--------------------------+-------------------------+-------------------------| |AB347 AB37 8 | 5 -13479 2 | 6 A13 A19 | | 1 -2357 9 | 46 3467 36 | 2345 35 8 | | B2345 6 B345 | 489 1349 1389 | 2345 7 C1259 | *------------------------------------------------------------------------------*

A and B denote the overlapping ALSs
Overlap digits are 3&4&7, non-overlap digits are 1&2&5&9
Thus have ((3&4&7) = (1&2&5&9))r7c1289|r9c13
Since all the 1&2&5&9 candidates in the combined ALS see the r9c9 1259 cell, that cannot be true, therefore r7c1289|r9c13 contains 3&4&7.
We can remove the sevens from r7c5 and r8c2 because they see all the sevens in the combined ALS.

Code: Select all: +-----------------+-------------------+--------------------+ | 4 78 39 | 368 1 3689 | 23679 2378 5 | | 6 15 15 | 2 389 7 | 39 38 4 | | 2 78 39 | 3468 3489 5 | 1 378 67 | +-----------------+-------------------+--------------------+ | 357 4 18 | 9 2378A 238AB| 357 6 127 | | 37 16- 2 | 1367 5 36AB| 8 4 9 | | 357 9 168 | 16 278A 4 | 2357 12357 127 | +-----------------+-------------------+--------------------+ | 9 2356 7 | 348 2348 238B | 2456 125 126 | | 8 26D 46 | 5 279C 1 | 2467 279 3 | | 1 235 45 | 347 6 239B | 2457 2579 8 | +-----------------+-------------------+--------------------+

A and B denote the overlapping ALSs
Overlap digits are 2&3&6&8, non-overlap digits are 7&9
Thus have
((2&3&6&8) = (7&9))r46c5|r4579c6 - ((7or9) = 2)r8c5 - (2=6)r8c2 => r5c6 <> 6

Proof of the Combined Overlapping ALS (CoALS) rule

Assume two overlapping ALS's.

A digit found in the overlap must be part of both ALS's

If an overlap digit is not present in the combined ALS, then all the remaining digits in both ALS's must be true. This means that all non-overlap digits must be present.

If a non-overlap digit is not present in the combined ALS, then for at least one ALS, all the remaining digits must be present. This has to include all overlap digits, therefore all overlap digits must be present.

Therefore we can state that either all the overlap digits can be found in the combined ALS, or all the non-overlap digits can be found in the ALS.

This translates directly into a strong link between the ANDed overlap digits and the ANDed non-overlap digits over the entirety of the combined ALS cells.

Thanks to ronk and Mike Barker for discussions related to overlapping ALS which helped uncover this relationship.

by **tarek** » Thu Sep 21, 2006 8:30 am

nice work on this myth,

I haven't put it into practice, but it looks to me that this method would be difficult to spot for a manual solver because you should take into consideration spotting the overlap...

I'm wondering if -because all of the logic is a subset of COUNTING- that in this specific case, looking at it as one large ALS would be easier.

Having different digits makes this one big ALS easier to spot, it becomes more difficult when you have a digit in one section of this ALS that can't see the same digit in another section of this big angulating ALS....

tarek

by **Havard** » Mon Sep 25, 2006 12:29 am

Hi Myth, thanks for a nice post!

I have a slightly different view on what is going on in some of your posts though. I have been working quite a bit with locked sets that has "freedom bigger then 1", in other words AALS and even AAALS etc.
I have long thought to present some of these results in this forum, and hope to do so soon.

With an ALS you can do the following reasoning:
"if you can connect two other (chains of) ALS's to one ALS, and the ends of the two ALS's has a candidate in common, any other of that same candidate that both sets can see can be eliminated"
in other words in you have an ALS like this:
[12]

and are able to connect two other ALS's to this one:
[17]-1-[12]-2-[27]

you can now eliminate any 7 that both [17] and [27] can see. (simple XY wing)

now with an AALS the same holds true, except you now need THREE bindings going out of the AALS(yes, this means that you can have one single and one double binding). In it's simplest form this means:

Code: Select all: [17]-1-[123]-2-[27] | 3 | [37]

now any 7 that can see both [17], [27] and [37] can be eliminated!

now of course you can also have long chains of ALS's going out of the AALS, like this:

Code: Select all: [17]-1-[123]-2-[25]-5-[59]-9-[79] | 3 | [37]

now any 7 that can see both [17], [27] and [79] can be eliminated!

now the relevance to overlapping here, is that any of these three chains may be treated as a "separate" chain, and hence can overlap eachother but I think that this overlapping can only happen at the very end of the chains. I have at least only (so far) had any luck with letting chains of ALS's overlap at the endpoints. If we look at your last example in this light:

Code: Select all: +-----------------+-------------------+--------------------+ | | | | | | | | | | | | +-----------------+-------------------+--------------------+ | | 2378A 238AB| | | 16- | 36AB| | | | 278A | | +-----------------+-------------------+--------------------+ | | 238B | | | 26D | 279C | | | | 239B | | +-----------------+-------------------+--------------------+

you would get:

Code: Select all: [26]-2-[279]-9-[238][238][239][36] | 7 | [2378] [278] [238] [36]

all three "endsets" have a 6 in common, and all these 6es can "see" the 6 in r2c5, and hence it can be eliminated. The fact that several cells in the two "end" ALSs overlap does not matter, since chain-ends are allowed to.

So my question is, can you show me an example of ALS-overlap where the overlap does not happen in the "end" of a chain?

(added:)
with your hypothetical example, the logic would go like this:

your C(12347) is an AlmostAlmostAlmostAlmost locked set.

Then you will need 5 bindings going out of it if you want to use the "exclude any candidate that can see all occurences of a non-linked candidate in all of the end-sets"-rule...
in this case:

Code: Select all: *----------------------------------------------------------------------* | . . . | . C12347 . | . D67 . | | . . . | A125 . . | . . . | | . . . | A125 . . | . . . | +----------------------+------------------------+----------------------+ | . . . | . B345 . | . . . | | . . . | . B345 . | . . . | | . . . |AB56 . . | . -6* . | +----------------------+------------------------+----------------------+ | . . . | . . . | . . . | | . . . | . . . | . . . | | . . . | . . . | . . . | *----------------------------------------------------------------------*

Code: Select all: [56]-5-[125][125]=1,2=[12347]=3,4=[345][345]-5-[56] | 7 | [67]

any 6 that can see both [56], [56] and [67] can be eliminated. The fact that [56] is the same cell does not matter, since end-sets can overlap!

Havard

by **Myth Jellies** » Tue Sep 26, 2006 6:13 am

Havard wrote:I have a slightly different view on what is going on in some of your posts though. I have been working quite a bit with locked sets that has "freedom bigger then 1", in other words AALS and even AAALS etc.
I have long thought to present some of these results in this forum, and hope to do so soon.

Absolutely right. Your A*LS approach is spot on, and is basically equivalent to what Mike Barker is doing with his Death Blossom technique. It's also been tossed around a little bit in this thread. My goal in presenting the CoALS alternative is twofold:

1) You can take a branching, or multi-threaded, network and show it in a single-threaded AIC chain or deduction. I like single-threaded logic when I can get it

.

2) The CoALS relationship is actually pretty prevalent & might be easier to spot and analyze for potential usefulness than running a bunch of chains off of every number in every cell. There are tricks you can apply to the overlapping ALS's (each ALS has to have a non-overlap digit unique to itself, for example) which will allow you to quickly weed out non-productive combinations.

Havard wrote:So my question is, can you show me an example of ALS-overlap where the overlap does not happen in the "end" of a chain?

(added:)
with your hypothetical example, the logic would go like this:

your C(12347) is an AlmostAlmostAlmostAlmost locked set.
Then you will need 5 bindings going out of it if you want to use the "exclude any candidate that can see all occurences of a non-linked candidate in all of the end-sets"-rule...
in this case:

Code: Select all
*----------------------------------------------------------------------* | . . . | . C12347 . | . D67 . | | . . . | A125 . . | . . . | | . . . | A125 . . | . . . | +----------------------+------------------------+----------------------+ | . . . | . B345 . | . . . | | . . . | . B345 . | . . . | | . . . |AB56 . . | . -6* . | +----------------------+------------------------+----------------------+ | . . . | . . . | . . . | | . . . | . . . | . . . | | . . . | . . . | . . . | *----------------------------------------------------------------------*

Code: Select all
[56]-5-[125][125]=1,2=[12347]=3,4=[345][345]-5-[56] | 7 | [67]

any 6 that can see both [56], [56] and [67] can be eliminated. The fact that [56] is the same cell does not matter, since end-sets can overlap!

Would something like this count as an example where the overlap is not at the endpoints?

Code: Select all: *----------------------------------------------------------------------* | . . . | . C12347 . | . D67 . | | . . . | A125 . . | . . . | | . . . | A125 . . | . . . | +----------------------+------------------------+----------------------+ | . . . | . B345 . | . . . | | . . . | . B345 . | . . . | | 68 . . |AB56 . . | . . . | +----------------------+------------------------+----------------------+ | . . . | . . . | . . . | | . . . | . . . | . . . | | 68 . . | . . . | . -6* . | *----------------------------------------------------------------------*

by **Havard** » Tue Sep 26, 2006 3:02 pm

Myth Jellies wrote:Would something like this count as an example where the overlap is not at the endpoints?

Code: Select all
*----------------------------------------------------------------------* | . . . | . C12347 . | . D67 . | | . . . | A125 . . | . . . | | . . . | A125 . . | . . . | +----------------------+------------------------+----------------------+ | . . . | . B345 . | . . . | | . . . | . B345 . | . . . | | 68 . . |AB56 . . | . . . | +----------------------+------------------------+----------------------+ | . . . | . . . | . . . | | . . . | . . . | . . . | | 68 . . | . . . | . -6* . | *----------------------------------------------------------------------*

ok, so in this example, you would (to prevent going insane) still has to think of it as three different chains, but where two of the chains share three nodes (all at the end). Now can you make me an example where there are three different endpoints for three different chains, but at some stage some of the three chains shares a few cells?

Havard

by **Myth Jellies** » Thu Sep 28, 2006 6:27 am

Well, I had a chain branch, and then come together again. I could make it split again, but I don't think that is what you are after. Perhaps something more like this...

Code: Select all: A123 . . | . B1457 . | . . -9* . . . | . . . | . . -9* . . . | . . . | . . -9* --------------+----------------+---------------- D257 . . | . BDG457 . |D1457 . D1457 . . . | . BG457 . | . . E19 F36 . . |G568 BG567 . | . . H89 --------------+----------------+---------------- . . . | . . . | . . -9* . . . | . . . | . . -9* . . . | . B3456 . | . . C39 (1=4&5&6&7&3)B - (3=9)C | A(123)-(3=6)F - (6=4&5&7&8)G - (8=9)H | (2=4&5&7&1)D - (1=9)E

Cell sets B, D, and G represent overlapping ALS's. Chain-network endpoints are the 9s in r5c9, r6c9, and r9c9; which eliminate all other nines in that column.

by **ronk** » Thu Sep 28, 2006 1:18 pm

Myth Jellies wrote:
Code: Select all
*-------------------------------------------------------------* | . . . | . C12347 . | . D67 . | | . . . | A125 . . | . . . | | . . . | A125 . . | . . . | +-------------------+---------------------+-------------------+ | . . . | . B345 . | . . . | | . . . | . B345 . | . . . | | . . . | X56 . . | . -6* . | +-------------------+---------------------+-------------------+ | . . . | . . . | . . . | | . . . | . . . | . . . | | . . . | . . . | . . . | *-------------------------------------------------------------*

[edit: label X was your label AB]

Have you considered combining (relabeled) sets A, B and C?

If found this Bob Hanson style diagram helpful.

Code: Select all: 1 . . . . . . . 1 || \ || 2 . . . . 2---A---5 . || // . . 6---D---7 . . . 7====C *5 . . . . . 5---X---6 . . || \\ . . . . || 3 . . . . 3---B---5 . . || / *7 4 . . . . . . . 4 (1 cell AAAALS) -, |, / and \ denote ALS set =, ||, // and \\ denote AALS, AAALS, AAALS etc. sets ... denotes weak link *X denotes exclusion when chain is part of continuous loop

When the 7 is excluded from AAAALS set C, we don't know which of digits 1, 2, 3, or 4 it will contain. But we do know at least one of set A and B will contain a 5. If the combined sets A, B, and C become part of a continuous loop, other 5s that see all of the 5s in the combined sets may be excluded. Ditto for 7s. But no exclusions of the other four digits (1,2,3,4) may be made. So why not just "hide" those digits within a union of sets?

Code: Select all: 6---D---7 . . . 7---ABC---5 . . . 5---X---6 . . . . . . . . . . . . *7 *5

Is there a hitch in this get-along? One I guess. The peer cells for digit 5 in the combined sets is the intersection of sets A and B, the same as for overlapped sets A and B.

by **Havard** » Fri Sep 29, 2006 1:21 pm

Myth Jellies wrote:Well, I had a chain branch, and then come together again. I could make it split again, but I don't think that is what you are after. Perhaps something more like this...

Code: Select all
A123 . . | . B1457 . | . . -9* . . . | . . . | . . -9* . . . | . . . | . . -9* --------------+----------------+---------------- D257 . . | . BDG457 . |D1457 . D1457 . . . | . BG457 . | . . E19 F36 . . |G568 BG567 . | . . H89 --------------+----------------+---------------- . . . | . . . | . . -9* . . . | . . . | . . -9* . . . | . B3456 . | . . C39 (1=4&5&6&7&3)B - (3=9)C | A(123)-(3=6)F - (6=4&5&7&8)G - (8=9)H | (2=4&5&7&1)D - (1=9)E

Cell sets B, D, and G represent overlapping ALS's. Chain-network endpoints are the 9s in r5c9, r6c9, and r9c9; which eliminate all other nines in that column.

Hi Myth! I have spent quite some time on this one now, and I can't find anything to object to... This does not mean that I distrust your wise words, but the fact that I have been working with ALS-chains for quite some time now, and always when I have tried to make the chains overlap, I have gotten errors, and has then concluded that overlapping is bad.

However, in your example the overlapping is working like a charm, so I guess my question is wheter there are some exceptions to the overlap-rules, and how these can be formulated? When I get some time I will try and make my program allow overlapping, and then report back any errors I get here!

Thanks for your great example by the way!

Havard

by **Mike Barker** » Fri Sep 29, 2006 8:01 pm

Myth, actually what you have here is an excellent example of a Kraken Blossom which is known to have excellent overlap properties since each chain is defined independently (see a solution to U#23). You've already made a great contribution just by showing that ALS can overlap at the end of a nice loop. The question is, as Havard stated, what are the conditions for which ALS can overlap in the middle of a nice loop? Maybe the Kraken Blossom paradigm is the right way to understand the answer.

by **Mike Barker** » Fri Oct 20, 2006 11:57 pm

So far I've had success with overlapping ALS when they are

here

I'm still working on understanding use of CoALS within a nice loop/chain since clearly not all ALS/bivalues within a chain may be able to overlap. I have, however, made some progress understanding overlap between strong links in a nice loop/chain. This may not be the best place to post this, but I'd like to keep posts regarding overlapping nice loop nodes in a common place.

The standard nice loop architecture allows for the end of one strong link to be the same node as the start of the next as long as the labels are different and the nodes are a single cell. The can be expanded. Consider Rod Hagglund's solution to Unsolvable#1 which can be formulated as a Kraken cell. What is interesting is that one chain consists of two strong links which overlap: R3C4-1-R3C89=1=R2C9=4=R3c9-4-. The generalization is pretty straight forward:

Note that even though this can result in one node with more than two links, the loop remains a double implication loop since we are not forking at the common node.

Code: Select all: Kraken Cell (R3C4-4-, R3C4-1-R3C89=1=R2C9=4=R3c9-4-, R3C4-3-R58C4-6-R5C3-2-R12C3|R1C2-4-): R3C4=134 => R3C2<>4 +------------------+---------------------+------------------+ | 1 46@ 2456@ | 9 24568 7 | 258 258 3 | | 36 8 2456@ | 1346 23456 1235 | 9 7 124# | | 37 2-4 9 | 134* 23458 12358 | 6 1258# 1248# | +------------------+---------------------+------------------+ | 5 3 7 | 2 168 9 | 4 168 168 | | 4 1 26@ | 367@ 3678 38 | 28 9 5 | | 69 269 8 | 5 16 4 | 3 126 7 | +------------------+---------------------+------------------+ | 689 469 3 | 14 2459 125 | 7 2568 268 | | 6789 5 16 | 37@ 2379 23 | 18 4 268 | | 2 47 14 | 8 457 6 | 15 3 9 | +------------------+---------------------+------------------+

Here's an example with a grouped nice loop:

Code: Select all: Common 4-element Grouped Nice Loop: r8c7=2=r8c6=9=r56c6-9-ALS:r45c5|r4c4-8-r6c6=8=r6c7~8~r8c7 => r8c7<>8 +-----------------+--------------------+-------------------+ | 8 47 13 | 3579 379 6 | 123 124 345 | | 47 5 36 | 378 2 1 | 9 48 3468 | | 9 2 136 | 358 4 58 | 13568 7 3568 | +-----------------+--------------------+-------------------+ | 345 349 2 | 138*# 1389*# 7 | 158 6 4589 | | 6 1349 8 | 259 139*# 259*@ | 7 149 459 | | 15 19 7 | 4 6 589*@ | 158* 3 2 | +-----------------+--------------------+-------------------+ | 137 1378 9 | 128 5 4 | 2368 28 3678 | | 137 1378 4 | 6 189 29* | 23-8* 5 3789 | | 2 6 5 | 789 789 3 | 4 89 1 | +-----------------+--------------------+-------------------+

by **Myth Jellies** » Sat Oct 21, 2006 5:39 pm

Mike Barker wrote:I'm still working on understanding use of CoALS within a nice loop/chain since clearly not all ALS/bivalues within a chain may be able to overlap.

The CoALS rule always holds, but I don't think it will be useful above and beyond a regular ALS unless the two ALS's being combined have the following properties.

Mike Barker wrote:The standard nice loop architecture allows for the end of one strong link to be the same node as the start of the next as long as the labels are different and the nodes are a single cell. The can be expanded. Consider Rod Hagglund's solution to Unsolvable#1 which can be formulated as a Kraken cell. What is interesting is that one chain consists of two strong links which overlap: R3C4-1-R3C89=1=R2C9=4=R3c9-4-. The generalization is pretty straight forward:
1) the starting node of one stong link and the ending node of a second strong link can share a cell in a nice loop if the two nodes have different labels
2) the starting (or ending) nodes of two strong links can share a cell in a nice loop if the two nodes have the same label (this may not be very useful since the loop created at the common node is not required for the nice loop)
3) this can be expanded to grouped strong link nodes with possibly multiple overlaps as long as the first two rules are followed at the common nodes. Note this applies only to overlapping of non-contiguous nodes in a nice loop, not direct links which require the node be a single cell.
Note that even though this can result in one node with more than two links, the loop remains a double implication loop since we are not forking at the common node.

Code: Select all
Kraken Cell (R3C4-4-, R3C4-1-R3C89=1=R2C9=4=R3c9-4-, R3C4-3-R58C4-6-R5C3-2-R12C3|R1C2-4-): R3C4=134 => R3C2<>4 +------------------+---------------------+------------------+ | 1 46@ 2456@ | 9 24568 7 | 258 258 3 | | 36 8 2456@ | 1346 23456 1235 | 9 7 124# | | 37 2-4 9 | 134* 23458 12358 | 6 1258# 1248# | +------------------+---------------------+------------------+ | 5 3 7 | 2 168 9 | 4 168 168 | | 4 1 26@ | 367@ 3678 38 | 28 9 5 | | 69 269 8 | 5 16 4 | 3 126 7 | +------------------+---------------------+------------------+ | 689 469 3 | 14 2459 125 | 7 2568 268 | | 6789 5 16 | 37@ 2379 23 | 18 4 268 | | 2 47 14 | 8 457 6 | 15 3 9 | +------------------+---------------------+------------------+

Here's an example with a grouped nice loop:
Code: Select all
Common 4-element Grouped Nice Loop: r8c7=2=r8c6=9=r56c6-9-ALS:r45c5|r4c4-8-r6c6=8=r6c7~8~r8c7 => r8c7<>8 +-----------------+--------------------+-------------------+ | 8 47 13 | 3579 379 6 | 123 124 345 | | 47 5 36 | 378 2 1 | 9 48 3468 | | 9 2 136 | 358 4 58 | 13568 7 3568 | +-----------------+--------------------+-------------------+ | 345 349 2 | 138*# 1389*# 7 | 158 6 4589 | | 6 1349 8 | 259 139*# 259*@ | 7 149 459 | | 15 19 7 | 4 6 589*@ | 158* 3 2 | +-----------------+--------------------+-------------------+ | 137 1378 9 | 128 5 4 | 2368 28 3678 | | 137 1378 4 | 6 189 29* | 23-8* 5 3789 | | 2 6 5 | 789 789 3 | 4 89 1 | +-----------------+--------------------+-------------------+

Here is the AICs for the one's pathway in the first example:
(3or4=1)r3c4 - (1)r3c89 = (1-4)r2c9 = (4)r3c9

Here is the AIC for the second example:
(2)r8c7 = (2-9)r8c6 = (9)r56c6 - (9=1&3&8)r45c5|r4c4 - (8)r6c6 = (8)r6c7 => r8c7 <> 8

I'm not sure that I would characterize these two examples as overlaps. It is true that they pass through the same cell more than once, but when they pass through, they use different candidates. On a candidate-to-candidate linking basis, no candidate gets used or is needed in the chain more than once.

by **Mike Barker** » Sun Nov 26, 2006 7:44 pm

I posted the next installment in the study of common cells within nodes of basic and grouped nice loops in a separate thread because the study is different than COALS. That doesn't mean there isn't overlap

between the two topics. In checking out some of my latest code changes and looking at Carcul's observation and Steve's question I came across a couple of interesting COALS:

Code: Select all: Overlap 2-element Grouped Nice Loop: ALS:r6c459-7-ALS:r6c89|r5c79~16~ => r6c7<>1,r6c7<>6 +---------------------+------------------+---------------------+ | 2 13469 1349 | 139 139 8 | 1467 1467 5 | | 157 1478 1458 | 125 125 6 | 9 3 1248 | | 1356 13689 13589 | 4 12359 7 | 126 126 1268 | +---------------------+------------------+---------------------+ | 13 2 6 | 179 19 5 | 1347 8 14 | | 15 18 7 | 168 4 3 | 1256@ 9 126@ | | 9 1348 13458 | 1678* 168* 2 | 357-16 157@ 16*@ | +---------------------+------------------+---------------------+ | 8 369 239 | 2356 2356 1 | 2456 2456 7 | | 4 5 12 | 26 7 9 | 8 126 3 | | 1367 1367 123 | 23568 3568 4 | 1256 1256 9 | +---------------------+------------------+---------------------+ Overlap 2-element Grouped Nice Loop: ALS:r9c4|r78c5-7-ALS:r346789c4~5~ => r5c4<>5 +-------------------+---------------------+-------------------+ | 28 2348 2348 | 1 5 46 | 7 2368 9 | | 1 6 23458 | 9 478 47 | 358 238 258 | | 58 7 9 | 368@ 2 36 | 1 4 568 | +-------------------+---------------------+-------------------+ | 4 5 238 | 238@ 389 1 | 68 689 7 | | 6 28 1 | 278-5 789 257 | 4 589 3 | | 9 38 7 | 345@ 6 345 | 2 58 1 | +-------------------+---------------------+-------------------+ | 3 9 245 | 24567@ 47* 8 | 56 1 2456 | | 7 1 458 | 23456@ 34* 23456 | 9 2368 24568 | | 258 248 6 | 345*@ 1 9 | 38 7 2458 | +-------------------+---------------------+-------------------+ Overlap 2-element Grouped Nice Loop: ALS:r1c2348-6-ALS:r1c45|r3c46|r2c4~3~ => r3c5<>3 +-----------------+----------------------+--------------------+ | 34 789* 78* | 3789*@ 34679@ 2 | 1 678* 5 | | 34 2789 6 | 3789@ 1 5 | 2348 278 237 | | 5 1278 1278 | 378@ 467-3 347@ | 23468 9 2367 | +-----------------+----------------------+--------------------+ | 2 1357 9 | 6 357 8 | 35 4 13 | | 16 4567 3457 | 12379 23579 1379 | 2359 126 8 | | 8 356 35 | 4 2359 19 | 7 126 2369 | +-----------------+----------------------+--------------------+ | 7 45 45 | 129 8 6 | 29 3 129 | | 16 2368 238 | 5 2349 349 | 2689 1278 2679 | | 9 1368 1238 | 1237 237 137 | 268 5 4 | +-----------------+----------------------+--------------------+

by **Havard** » Sun Nov 26, 2006 10:27 pm

I have come to that as long as the overlapping ALS cells does not contain the "restricted common" of the two sets, they are legal. Is this the rule you use too Mike? I am having a bit of trouble proving why it is like this, but all the examples I have tried out point to this. Take your first example. If you add a "7" (the restricted common) in the overlap cell (r6c9) the eliminations are no longer valid. This would certainly add a lot to the power of ALS!

Havard

by **Mike Barker** » Sun Nov 26, 2006 10:52 pm

For the case of the above examples, that is sufficient. In the case of longer grouped nice loops I'm claiming in the common cell thread, that it is also possible for ALS to share cells if the "input" link label to each ALS is the same and the ALS are internal to the loop. If the ALS are at the start and end of the loop (which is which is arbitrary) and the loop is at least 3 nodes long, then there are no restrictions on which cells can be shared. Unfortunately I don't have mathematical proofs of any of the statements either.

by **Myth Jellies** » Mon Nov 27, 2006 7:21 am

Mike Barker wrote:I posted the next installment in the study of common cells within nodes of basic and grouped nice loops in a separate thread because the study is different than COALS. That doesn't mean there isn't overlap between the two topics. In checking out some of my latest code changes and looking at Carcul's observation and Steve's question I came across a couple of interesting COALS:

Code: Select all: Overlap 2-element Grouped Nice Loop: ALS:r6c459-7-ALS:r6c89|r5c79~16~ => r6c7<>1,r6c7<>6 +---------------------+------------------+---------------------+ | 2 13469 1349 | 139 139 8 | 1467 1467 5 | | 157 1478 1458 | 125 125 6 | 9 3 1248 | | 1356 13689 13589 | 4 12359 7 | 126 126 1268 | +---------------------+------------------+---------------------+ | 13 2 6 | 179 19 5 | 1347 8 14 | | 15 18 7 | 168 4 3 | 1256@ 9 126@ | | 9 1348 13458 | 1678* 168* 2 | 357-16 157@ 16*@ | +---------------------+------------------+---------------------+ | 8 369 239 | 2356 2356 1 | 2456 2456 7 | | 4 5 12 | 26 7 9 | 8 126 3 | | 1367 1367 123 | 23568 3568 4 | 1256 1256 9 | +---------------------+------------------+---------------------+

My AIC representation for this would be...

(1&6&8=7)r6c459 - (7=2&5&1&6)r5c79|r6c89

Thus all 1's and 6's which see all of the 1's and 6's in the two ALS's can be removed. It is easy to see that if you tried to include r6c8 in the *ALS (in order to eliminate the 5 from r6c7 as well) then the weak link between 7's in the two ALS's would no longer be valid--the shared 7 would allow both ALS's to contain a 7 and the weak link disappears.

Even though there is plenty of overlap here, this does not use the CoALS relationship. Here, CoALS gives you the following...

(1&6 = 2&5&7&8)r5c79|r6c4589

To use this, you would need something like a 28 cell in r5c456. That would be enough to show (2&5&7&8) cannot be true, therefore (1&6) must be true, and then you could make the same reduction. The remaining reductions do not use CoALS either. They are just two-ALS-element chains with significant overlap and a weak link between them.

Code: Select all: Overlap 2-element Grouped Nice Loop: ALS:r9c4|r78c5-7-ALS:r346789c4~5~ => r5c4<>5 +-------------------+---------------------+-------------------+ | 28 2348 2348 | 1 5 46 | 7 2368 9 | | 1 6 23458 | 9 478 47 | 358 238 258 | | 58 7 9 | 368@ 2 36 | 1 4 568 | +-------------------+---------------------+-------------------+ | 4 5 238 | 238@ 389 1 | 68 689 7 | | 6 28 1 | 278-5 789 257 | 4 589 3 | | 9 38 7 | 345@ 6 345 | 2 58 1 | +-------------------+---------------------+-------------------+ | 3 9 245 | 24567@ 47* 8 | 56 1 2456 | | 7 1 458 | 23456@ 34* 23456 | 9 2368 24568 | | 258 248 6 | 345*@ 1 9 | 38 7 2458 | +-------------------+---------------------+-------------------+

AIC rep

(5&2&3&4&6&8 = 7)r346789c4 - (7 = 3&4&5)r78c5|r9c4 => r5c4 <> 5

another way to do this is

(5&2&3&4&6&8 = 7)r346789c4 - (7 = 3&4)r78c5 - (3v4 = 5)r9c4

Code: Select all: Overlap 2-element Grouped Nice Loop: ALS:r1c2348-6-ALS:r1c45|r3c46|r2c4~3~ => r3c5<>3 +-----------------+----------------------+--------------------+ | 34 789* 78* | 3789*@ 34679@ 2 | 1 678* 5 | | 34 2789 6 | 3789@ 1 5 | 2348 278 237 | | 5 1278 1278 | 378@ 467-3 347@ | 23468 9 2367 | +-----------------+----------------------+--------------------+ | 2 1357 9 | 6 357 8 | 35 4 13 | | 16 4567 3457 | 12379 23579 1379 | 2359 126 8 | | 8 356 35 | 4 2359 19 | 7 126 2369 | +-----------------+----------------------+--------------------+ | 7 45 45 | 129 8 6 | 29 3 129 | | 16 2368 238 | 5 2349 349 | 2689 1278 2679 | | 9 1368 1238 | 1237 237 137 | 268 5 4 | +-----------------+----------------------+--------------------+

AIC is

(3&7&8&9 = 6)r1c2348 - (6 = 4&7&8&9&3)r123c4|r1c5|r3c6 => r3c5 <> 3

So even though these do not use CoALS, They likely are quite useful. The AIC representation is actually all the proof you need that you just need to avoid using overlap digits as your weak link (restricted common digit?) between the two ALS's

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