The New Sudoku Players' Forum

[champagne]

Exploring the data base of potential hardest to find SK loops, including loops with given, I got this one


..5......67.5.....89....7....78..9.......4.3.....2...1....3184...89..6..........2;10.50;10.50;3.40;GP;13_11gexo;1347553;22;*
(* means SKFR rating)

The program is not yet correct, in this case missing the "6 bi local" in row 4, but the SK loop is "valid" 15 truths, 15 links.
This puzzle has also an exocet

Code: Select all

1234   1234  5     |123467 146789 236789 |1234 12689 34689
6      7     1234  |5      1489   2389   |1234 1289  3489 
8      9     1234  |12346  146    236    |7    1256  3456 
----------------------------------------------------------
12345  12345 7     |8      15     35     |9    256   456     
1259   12568 1269  |167    15679  4      |25   3     78   
3459   34568 3469  |367    2      35679  |45   78    1     
----------------------------------------------------------
2579   256   269   |267    3      1      |8    4     579   
123457 12345 8     |9      457    257    |6    157   357     
134579 13456 13469 |467    45678  5678   |135  1579  2     

exocet base r1c1 r1c2 target r2c7 r3c4
sk loop rows 48 columns 47 given 8r7c7

clearing the SK loop, we can eliminate
5r4c12 1r9c8 57r8c1 5r8c2 4r9c5 6r3c4 67r3c1

The exocet would clear here 6r3c4 also done by the SK loop
After the naked quad in row 1, we end here

Code: Select all

1234   1234  5     |1234   6789   6789   |1234 689   689
6      7     1234  |5      1489   2389   |1234t1289  3489 
8      9     1234  |1234t  146    236    |7    1256  3456 
----------------------------------------------------------
1234   1234  7     |8      15     35     |9    256   456     
1259   12568 1269  |167    15679  4      |25   3     78   
3459   34568 3469  |367    2      35679  |45   78    1     
----------------------------------------------------------
2579   256   269   |267    3      1      |8    4     579   
1234   1234  8     |9      457    257    |6    157   357       
134579 13456 13469 |467    5678  5678    |135  579   2     

In the exocet, we have an abi loop for the pairs 12 14 32 34
So the exocet solution is 13 or 24

We can not have 5r5c5 or 5r6c6 forcing 13r13c4 not possible with the exocet
In the same way, we can not have 57 r8c56, but this is not a direct elimination.

This put us here, but I didn't see here an easy step despite all "or" conditions coming out of the exocet as .

6^7 r79c4
5^7 r8c56 etc

Code: Select all

1234b  1234b 5     |1234   6789   6789   |1234 689   689
6      7     1234  |5      1489   2389   |1234t1289  3489 
8      9     1234  |1234t  146    236    |7    1256  3456 
----------------------------------------------------------
1234   1234  7     |8      15     35     |9    26    46       
1259   12568 1269  |167    1679   4      |25   3     78   
3459   34568 3469  |367    2      3679   |45   78    1     
----------------------------------------------------------
2579   256   269   |267    3      1      |8    4     579   
1234   1234  8     |9      457    257    |6    157   357     
134579 13456 13469 |467    5678  5678    |13   579   2     

may be a skill manual solver can find the missing easy move

BTW, I didn't get many sk loop with given, but the data base of hardest is not the best chance to find them.
Now back home, I'll finish tests on the code and explore my data base of seeds (ratings in the range 10.0 to 10.3 available) and the game data base

[SpAce]

exocet base r1c1 r1c2 target r2c7 r3c4
The exocet would clear here 6r3c4 also done by the SK loop

Hi champagne! Could you help me out with a bit off-topic question about exocets? I'd like to be able to express them as truths and links, and I've sort of figured out how. The devil is in the details, however. How exactly are the rank zones determined? As far as I understand, the base digit eliminations from the targets happen because those cells are Rank 0, but understanding why is not exactly easy. I think I kind of understand it for the 3-digit case like Fata Morgana which I see like this:

Alien 11-Fish (Mixed Rank [3]/1/0): {5N46 136C258} \ {136r5b5 16r1 3r2 6r8 13r9 / 4n2 6n8} => -24 r4c2

It has 11 truths and 14 links, thus the base rank is 3. However, since the base digits have double links (r5, b5), i.e. link triplets, and two of them are always occupied (because the base cells must have two different digits), it lowers the overall rank to 1. Also, one of them is always unoccupied which lowers the left-over digit's minor branch by one. All three paths connect at r4c2 and r6c8, which apparently puts them in Rank 0 zone. Or something like that. It's pretty much how I thought about it myself, and Allan himself seems to agree here. That doesn't mean I fully understand it.

Assuming that's about correct, how does the same work with the 4-digit case like your example? That one I see like this:

Alien 14-Fish (Mixed Rank [4]/2/0) {1N12 1234C3 1234C4 1234C7} \ {1234r1b1 12r5 34r6 2r7 134r9 / 2n7 3n4} => -6 r3c4

It has 14 truths and 18 links, thus base rank 4. Again the base digits have link triplets, and two of them are always occupied, so I guess it lowers the overall rank by 2 -- but it's then 2 (unlike the previous case). Is it now because there are two left-over links that are unoccupied, which gives us the further drop of 2 for the two target cells where all four paths connect?

Of course this is all pretty academic because it's easy to see in both cases that two base digits must occupy those target cells. I'm just interested because isn't this kind of how the Exocet property was originally found? Plus I kind of prefer the elegance of the original idea to the complexity of the zillion JExocet rules (while helpful, I think most of them are probably unnecessary -- i.e. easily derivable -- if one understands the fundamentals well enough).

[SpAce]

clearing the SK loop, we can eliminate
5r4c12 1r9c8 57r8c1 5r8c2 4r9c5 6r3c4 67r3c1

I think you missed: -1 r5c5, -3 r6c6 (and a typo in 67r3c1)? This is what I got:

Alien 15-Fish (Rank 0) {48N5689 4579N4 569N7} \ {56r4 57r8 67c4 5c7 13b5 24b6 24b8 13b9}
=> -67 r1c4, [-6 r3c4], -5 r4c12, -1 r5c5, -3 r6c6, -57 r8c1, -5 r8c2, -4 r9c5, -1 r9c8 (12 elims, or 11 if exocet first)

In the exocet, we have an abi loop for the pairs 12 14 32 34
So the exocet solution is 13 or 24

Agreed.

We can not have 5r5c5 or 5r6c6 forcing 13r13c4 not possible with the exocet
In the same way, we can not have 57 r8c56, but this is not a direct elimination.

Clever! I would have missed those. I guess they should be kind of standard things to check with exocets.

Anyway, with the missed SK-eliminations added I think we're here:

Code: Select all

.----------------------.------------------.------------------.
| 1234    1234   5     | 1234  6789  6789 | 1234  689   689  |
| 6       7      1234  | 5     1489  2389 | 1234  1289  3489 |
| 8       9      1234  | 1234  146   236  | 7     1256  3456 |
:----------------------+------------------+------------------:
| 1234    1234   7     | 8     15    35   | 9     26    46   |
| 1259    12568  1269  | 167   679   4    | 25    3     78   |
| 3459    34568  3469  | 367   2     679  | 45    78    1    |
:----------------------+------------------+------------------:
| 2579    256    269   | 267   3     1    | 8     4     579  |
| 1234    1234   8     | 9     457   257  | 6     157   357  |
| 134579  13456  13469 | 467   5678  5678 | 13    579   2    |
'----------------------'------------------'------------------'

I don't see any obvious openings, but my skills at these levels are quite limited anyway. Perhaps there are some more exocet inferences available, maybe with some chains or something. (Missing them, the exocet is pretty easy to crack open with simple T&E even manually. Finding out which base pair is correct requires easy techniques only, after which the puzzle is solvable normally (but requires several forcing chains). It's also easy to find out which way the digits are in the targets, which makes the rest trivial. I guess that's one way to use an exocet, though a pretty boring one.)

[champagne]

exocet base r1c1 r1c2 target r2c7 r3c4
The exocet would clear here 6r3c4 also done by the SK loop

Hi champagne! Could you help me out with a bit off-topic question about exocets? .....

Hi SpAce,

The first "Exocet" proof has clearly been Allan Barkers's Xsudo Truths/Links construction.

It has been done for "fata morgana" and "trompe l'oeil" puzzles at the same moment.

I long discussed this with Allan when he produced the solution. Although Allan had some clear views on how a rank >0 truths/links could have some areas of lower rank, he was not quite comfortable with the "complexity" of this rank 3 group.

This is why I looked for another explanation, and, somehow rewording what he wrote came to the exocet view: Each of the digits had to occupy one of the target, consequently, the digits occupying the base had to occupy the target.

The same rule works with four digits, so we did not try to explain why a rank 4 group could produce a rank 0 area.

It appeared later that nearly all known exocets were in fact JExocets or very close to Jexocets (same band as a minimum).

Such exocets have always a truths/links group having as truths the base, the target and crossing rows/columns and as links the columns/rows plus the base box sets, with triple points in the 2 cells of the base.

Note : For platinum blonde and similar exocets, the locked digit(s) has to be added as link. With platinum blonde, only one link is added, blue said somewhere that he found examples of 2 locked digits (one more cell in box needed), but I don't remember having seen an example of such a puzzle.

All this to say that if your expectation is in explanations on how you can cut a truths/links of rank >0 in sub areas of lower rank, I can't help you.

Reversely, there is room for discussion around the exocet property.

[champagne]

...

Anyway, with the missed SK-eliminations added I think we're here:

Code: Select all

.----------------------.------------------.------------------.
| 1234    1234   5     | 1234  6789  6789 | 1234  689   689  |
| 6       7      1234  | 5     1489  2389 | 1234  1289  3489 |
| 8       9      1234  | 1234  146   236  | 7     1256  3456 |
:----------------------+------------------+------------------:
| 1234    1234   7     | 8     15    35   | 9     26    46   |
| 1259    12568  1269  | 167   679   4    | 25    3     78   |
| 3459    34568  3469  | 367   2     679  | 45    78    1    |
:----------------------+------------------+------------------:
| 2579    256    269   | 267   3     1    | 8     4     579  |
| 1234    1234   8     | 9     457   257  | 6     157   357  |
| 134579  13456  13469 | 467   5678  5678 | 13    579   2    |
'----------------------'------------------'------------------'

I don't see any obvious openings, but my skills at these levels are quite limited anyway. Perhaps there are some more exocet inferences available, maybe with some chains or something. (Missing them, the exocet is pretty easy to crack open with simple T&E even manually. Finding out which base pair is correct requires easy techniques only, after which the puzzle is solvable normally (but requires several forcing chains). It's also easy to find out which way the digits are in the targets, which makes the rest trivial. I guess that's one way to use an exocet, though a pretty boring one.)

We have the same position.

When nothing easy comes out, I prefer, with an exocet, to work on the sub grids, here assuming 24 (false) in base and seeing what happens. Having appied the exocet properties with 24 in base, it is easy to clear 5 r8c5 and 5r8c6.

I have still to work on it, but it seems that the only "easy way" to kill r1c12 is to push to the 2 options in targets.

This would confirm that no easy move is available.

I'll give my final preferred solution later

[SpAce]

Ok, thanks a lot, champagne! Glad to know I'm not the only one with difficulties understanding the rank 0 explanation for JExocets. That part excluded, I think the truths/links construction helps to understand them better. With that, all the various S-Cell rules and stuff make more sense (and can be safely forgotten, I think).

I'll give my final preferred solution later

Looking forward to that!

[Ajò Dimonios]

Hi Phill. This is my first post in this forum. I read your document on "Sk and related Loop" carefully. I was especially interested in the topic "Almost SK-Loop". I thank you for the solver on the internet that I use frequently. I have a question to ask. It can be shown that an MSLS can be associated to each SK-Loop ?. Clearly the answer can be positive only if the MSLS model is implemented including also those MSLS containing 2 lines, 2 columns and 4 blocks. However, to date, for all those schemes in which the answer is positive, the regularity of the eliminations and insertions of the "Almost SK-Loop" method can be demonstrated through a method that I would call "Almost MSLS".

Code: Select all

Phill wrote

Almost SK (+1) Loops
This section is added out of interest, as trial and error is employed.

These are every bit as common as SK loops. They have 17 links rather than the 16 of normal SK loops. They differ by having a linking triple rather
than a linking pair in one of the boxes or rows/columns. The bidirectional logic is still there, but an absurdity is introduced by suggesting
3 numbers to be true in the 2 linking cells. The result of this is that in the final solution one is trying to fit 17 numbers into the 16 cells
of the SK pattern. The one that is squeezed out leads to just one of the eliminations being false, ie the candidate elimination must be true.
One can test this simply by in turn making each candidate elimination true and the rest false, and then testing to see if it creates a
contradiction. In all of the vast number of cases I've tested, this theory holds up. Thus this number can be assigned to the cell, and the
remaining candidate eliminations can be carried out. I would be delighted if someone could discover a theoretic framework to establish this
approach to be valid.

To explain myself better I use an example. The golden Nugget.

Code: Select all

+-----------------------+------------------------+--------------------+
| 25678  14568  124567  | 12568  124567  145678  | 1247   3     9     |
| 236789 134689 123467  | 123689 123467  1346789 | 1247   12467 5     |
| 235679 134569 1234567 | 123569 1234567 1345679 | 8      12467 1247  |
+-----------------------+------------------------+--------------------+
| 235    345    8       | 135    9       1357    | 123457 1247  6     |
| 3569   7      3456    | 13568  1356    2       | 13459  1489  1348  |
| 1      3569   2356    | 4      3567    35678   | 23579  2789  2378  |
+-----------------------+------------------------+--------------------+
| 367    136    9       | 1236   8       1346    | 12347  5     12347 |
| 3578   2      1357    | 1359   1345    13459   | 6      14789 13478 |
| 4      13568  1356    | 7      12356   13569   | 1239   1289  1238  |
+-----------------------+------------------------+--------------------+


The Almost (+1) SK loop produces this answer:
Almost (+1) SK loop detected (red cells): (35=24)r4c12 - (24=56)r56c3 - (56=17)r89c3 - (17=36)r7c12 - (36=124)r7c46 - (124=36)r89c5 - (36=17)r56c5 - (17=35)r4c46 - loop
No contradiction when 5 at r4c7 is true and others are all false
Eliminations (green cells): r4c7 <> 3, r7c7 <> 3, r7c9 <> 3, r1c3 <> 5, r1c3 <> 6, r2c3 <> 6, r1c5 <> 6, r2c5 <> 3, r2c5 <> 6, r5c4 <> 1, r6c6 <> 7, r8c1 <> 7, r9c2 <> 1, r8c4 <> 1, r8c6 <> 4

The Nugget has several possibilities of "Almost (+1) MSLS". In practice, I can optionally insert a candidate as true by creating an MSLS. What I noticed was that by inserting as one of the candidates that the Almost (+1) SK loop method selects as certain or eliminable, I always trigger at least one MSLS which determines the elimination of all the other candidates indicated by the Almost method (+1 ) SK loop. I report one at random by entering R2C5 = 3 which triggers the MSLS::

Base: 1247
16 cell Truths: r5689 c3589
16 links: 14r5, 27r6, 147r8, 12r9, 56c3, 6c5, 89c8, 38c9
17 eliminations: r5c4<>1, r5c7<>1, r5c7<>4, r6c6<>7, r6c7<>2, r6c7<>7, r8c1<>7, r8c4<>1, r8c6<>4, r9c2<>1, r9c7<>1, r9c7<>2, r1c3<>5, r1c3<>6, r2c3<>6, r1c5<>6, r7c9<>3, Naked triplets of 359 at r569c7 => -3 r47c7, -5 r4c7.

In the following example I report also the SK-loop that allows me to understand why a candidate is true and the rest are false.



By entering R8C6 = 4 I find the following SK-Loop which contains all the eliminations of Almost (+1) SK-Loop.

Naked quads of 1236 at r7c46, r89c5 => -1 r8c4, -3 r8c4, r9c6, -6 r9c6


Normal SK loop detected (red cells): (35=24)r4c12 - (24=56)r56c3 - (56=17)r89c3 - (17=36)r7c12 - (36=12)r7c46 - (12=36)r89c5 - (36=17)r56c5 - (17=35)r4c46 - loop
14 Eliminations (green cells): r4c7 <> 3, r4c7 <> 5, r7c7 <> 3, r7c9 <> 3, r1c3 <> 5, r1c3 <> 6, r2c3 <> 6, r1c5 <> 6, r2c5 <> 3, r2c5 <> 6, r5c4 <> 1, r6c6 <> 7, r8c1 <> 7, r9c2 <> 1

But also two MSLS:


1) Naked quads of 1236 at r7c46, r89c5 => -1 r8c4, -3 r8c4, r9c6, -6 r9c6

Base: 1247
16 cell Truths: r5689 c3589
16 links: 14r5, 27r6, 17r8, 12r9, 56c3, 36c5, 89c8, 38c9
17 eliminations: r5c4<>1, r5c7<>1, r5c7<>4, r6c6<>7, r6c7<>2, r6c7<>7, r8c1<>7, r9c2<>1, r9c7<>1, r9c7<>2, r1c3<>5, r1c3<>6, r2c3<>6, r1c5<>6, r2c5<>3, r2c5<>6, r7c9<>3,

and


2) Naked quads of 1236 at r7c46, r89c5 => -1 r8c4, -3 r8c4, r9c6, -6 r9c6

Base: 1247
16 cell Truths: r47 c35 b4578
16 links: 35r4, 36r7, 56c3, 36c5, 24b4, 17b5, 17b7, 12b8
14 eliminations: ): r4c7 <> 3, r4c7 <> 5, r7c7 <> 3, r7c9 <> 3, r1c3 <> 5, r1c3 <> 6, r2c3 <> 6, r1c5 <> 6, r2c5 <> 3, r2c5 <> 6, r5c4 <> 1, r6c6 <> 7, r8c1 <> 7, r9c2 <> 1.


. If we consider the -1 r8c4 of the naked quads, in the second MSLS we confirm the 16 eliminations + an insertion of the Almost (+1) SK loop, clearly reversing the R4C7 = 5 with R8C6 = 4.
In the second MSLS I found the reason why of all these candidates one must be true and all the others false. In fact, even if R8C6 = 4 were false (as it really is) in the last MSLS compulsory to avoid creating a contradiction, one of the eliminated group must be true. I would in fact in the opposite case (all the real eliminations including R8C6 ≠ 4) always:
16 cell Truths: r47 c35 b4578
but 17 links: 35r4, 36r7, 56c3, 36c5, 24b4, 17b5, 17b7, 124b8 with an obvious contradiction. In this case, in order to return the Links to 16, I must insert one of the eliminated candidates as true. This reasoning can clearly be done even starting from another Almost-MSLS obtained with a different candidate.


Ciao a tutti
Paolo

[numpl_npm]

Code: Select all

Normal SK loops:

. _ 8 _ . _ _ _ . _ 9 _ .
. _ 4 _ . _ _ _ . _ 3 _ .

| 1 x _ | _ _ _ | _ x 2 | A
| o 9 o | 4 _ _ | o 5 o | B 38
| _ x 6 | _ _ _ | 7 x _ | C

| _ 5 _ | 9 _ 3 | _ _ _ | D
| _ _ _ | _ 7 _ | _ _ _ | E
| _ _ _ | 8 5 _ | _ 4 _ | F

| 7 x _ | _ _ _ | 6 x _ | G
| o 3 o | _ _ 9 | o 8 o | H 45
| _ x 2 | _ _ _ | _ x 1 | J

(B13.AC2)(B79.AC8)
(H13.GJ2)(H79.GJ8)

U=[1267] N=[34589] (u in U) (n in N)

(counts of ox) = 16
(counts of n in ox) <= 8

(counts of u in B13.AC2) <= 2
(counts of u in B79.AC8) <= 2
(counts of u in H13.GJ2) <= 2
(counts of u in H79.GJ8) <= 2
(counts of u in ox) <= 8



[numpl_npm]

Code: Select all

Almost SK (+1) Loops - (Type B) Patterns game Jun 2010, joel64:

. _ 8 _ . _ _ _ . _ 4 _ .
. _ 7 _ . _ _ _ . _ 3 _ .

| _ x 1 | _ _ _ | 2 x _ | A
| o 3 o | _ _ 4 | o 5 o | B 78
| 2 x _ | _ _ _ | _ x 6 | C

| _ _ _ | 7 _ _ | _ 8 _ | D
| _ _ _ | _ 9 _ | 1 _ _ | E
| _ 5 _ | _ _ 3 | _ _ _ | F

| 9 x _ | _ 1 _ | _ x 2 | G
| o 4 o | 9 _ _ | o 7 o | H (two of 358)
| _ x 6 | _ _ _ | 9 x _ | J

(B13.AC2)(B79.AC8)
(H13.GJ2)(H79.GJ8)

U=[1269] N=[34578] (u in U) (n in N)

(nn)H56 -> 6H7 2H3 then to contradiction with basics
(uu)H56 -> 2J2 1D2 then to contradiction with basics

(counts of ox6) = 16
(counts of n in ox) <= 8

(counts of u in B13.AC2) <= 2
(counts of u in B79.AC8) <= 2
(counts of u in H13.GJ2) <= 2
(counts of u in H79.GJ8) <= 2
(counts of u in ox) <= 8

(un)H56