The New Sudoku Players' Forum

[pjb]

ronk, thank you for the interesting example, with 3 deviations from standard SK loop: the 123's in row 7 linking boxes 7 & 8, the linking 479's in box 4, and the single linking 6 in box 7. Never-the-less, it still conforms with my results.

I have simply been making observations, and haven't as yet found the logical framework to explain it. The observations however are rather compelling as I can't find an exception having studied over a dozen examples of 'almost SK loops'. In all cases where there is a single deviation, ie 3 digits linking the chain in one of the 4 boxes, the following holds: the 3 digits end up in 3 of the 4 cells that are part of the chain in the box. In the other 3 boxes, the 2 linking digits end up in 2 of the 4 cells. Thus the usual eliminations in the 4 boxes of the loop can be made. The problem is that one of the digits of the pairs of digits linking the row or column is 'squeezed' out because the 3 digits have been accommodated, so one row or column does not end up with the two row or column linking digits in the 4 cells, thus in one row or column the usual eliminations cannot be made. At present I don't know how to predict which one without the hindsight of seeing the solution.

I think this has implications for manual solving as these loops are relatively easily seen.

pjb

[champagne]

That's slightly different from your explanation champagne, but if you exclude (69) in the rest of c7, why not do the same in b3 ?

It's quite common to go left to right in such chains and to do things when they are compulsory.
The logic clearly starts in columns 7.
In my chains posted below, we don't use the box 3

You'll also notice I didn't notate the large almost naked sets but used the smaller complementary almost hidden sets which are easier to follow and less error prone.

my current solver (the old one) does not recognise ALS 'ors', so I somehow follow you, but we have as many examples as one could ask of a chain nicer with ALS and the same for chains using ALS.

Here, the chain using ALS seems the shortest

When a JExocet exists I still like my method better as I believe that it's quicker.

This is more a matter of taste.

ABI loop does not use the exocet property, but when it is established, my preferred proof is the double contradiction chain directly derived form ABI 's logic

Code: Select all

 *----------------------*----------------------*----------------------*
 | 35678  34589  34679  | 4679   5689   4578   | 679 #  1      2      |
 | 15678  14589  4679   | 124679 125689 124578 | 679 #  56789  3      |
>| 15678  1589   2      | 3      15689  1578   | 4      56789  6789   | < 
 *----------------------*----------------------*----------------------*
>| 237    2349   1      | 8      236    234    | 23679  234679 5      | <
 | 235    6      349    | 124    7      12345  | 8      2349   149    |
 | 23578  23458  347    | 1246   12356  9      | 12367  23467  1467   |
 *----------------------*----------------------*----------------------*
>| 1236   123    8      | 5      1239   1237   | 123679 234679 14679  | <
 | 9      123    36     | 127    4      12378  | 5      23678  1678   |
 | 4      7      5      | 129    12389  6      | 1239   2389   189    |
 *----------------------*----------------------*----------------------*

proving <67> r12c7

Code: Select all

      6r8c3    7r6c3                             UR r12c37
                 7r4c1  6r4c5                   ALS r4c12567
                          6r6c4   7r8c4          UR r12c47
      6r7c1                       7r7c6          ALS r7c12567

    6r4c5 ^6r7c1                               twin exocet property

Code: Select all

6r7c1 - 6r4c5 = 7r4c1 - 7r6c3 = 6r8c3 - 6r7c1     <6>r7c1
6r4c5 - 6r7c1 = 7r7c6 - 7r8c4 = 6r6c4 - 6r4c5    <6>r4c5
contradiction, we must have 6r4c5 ^6r7c1

we really use here the minimum of data

BTW, with your twin Jexocet, we also know at the end that
r4c8 is reduced to 79
r7c89 is reduced to 479

[ronk]

(4) Must be true in either r7c8 or r7c9 so there will be one of two JExocets with common cross line candidates in the other stacks

I could be interesting to name that "twin Jexocet"

Exocet with twin targets would be more descriptive, and since these targets are "joined at the hip" (r4c8), conjoined or joined targets is even better IMO.

[edit: Thanks to coloin, replaced the politically-incorrect "Siamese" term.]

[ronk]

Platinum Blonde: partial pencilmarks after one hidden single r9c3=5 and assuming r12c7=69

Code: Select all

 .   .  *69+ |*69+ .   .   |*69  .    .   
 .   .  *69+ |*69+ .   .   |*69  .    .   
 .   .   /   | /   .   .   | .   .    .   
-------------+-------------+--------------
 /   9+  /   | /   6+  /   | /   69+  /   
 .   .   9+  | /   .   .   | .   .    .   
 .   .   /   | 6+  .   .   | .   .    .   
-------------+-------------+--------------
 /   6+  /   | /   9+  /   | /   469+ 469+
 .   .   6+  | /   .   .   | .   .    .   
 .   .   /   | 9+  .   .   | .   .    .   

11 SIS: r12c7, 69r47, 69c34, 4r7
2 AURs: (69)r12c37, (69)r12c47

The above is an impossible pattern, as is demonstrated below. The strong inference sets ("sis") shown are the strong sets used by champagne and David P Bird in a so-called "abi-loop", which is treated as a continuous loop. However, it's hard to imagine the existence of a continuous loop in an impossible pattern, when each uses exactly the same sis and AURs.

But let's pretend that r4c2 and r5c3 don't see each other. In this scenario, there are only 16 "valid" permutations of candidate placements in the sis. Note the contradictory r4c2=9 and r5c3=9 in all 16 permutations. That's hardly a recipe for a continuous loop.

So how do champagne and David P Bird come up with a continuous loop? I suspect constraints, cover constraints, in r1 and r2 are being ignored. Is doing so valid? Perhaps, but I'm not aware of a proof for the validity of such a practice.

Code: Select all

                                +-r4c2    +-r5c3                                     
                                v         v                                         
01: ...9..6....6...9............9.....6...9.........6.....6...9..4...................
02: ...9..6....6...9............9.....6...9.........6.....6......49..................
03: ......6....6...9............9.....6...9.........6.....6......49............9.....
04: ...9..6....6...9............9.....6...9.........6.........9..46..................
05: ...9..6........9............9.....6...9.........6.........9..46..6...............
06: ..6...9.....9..6............9.....6...9.........6.....6...9..4...................
07: ..6...9.....9..6............9.....6...9.........6.....6......49..................
08: ..6...9........6............9.....6...9.........6.....6......49............9.....
09: ..6...9.....9..6............9.....6...9.........6.........9..46..................
10: ......9.....9..6............9.....6...9.........6.........9..46..6...............
11: ...9..6....6...9............9.....6...9.........6.....6...9...4..................
12: ...9..6....6...9............9.....6...9.........6.....6......94..................
13: ......6....6...9............9.....6...9.........6.....6......94............9.....
14: ..6...9.....9..6............9.....6...9.........6.....6...9...4..................
15: ..6...9.....9..6............9.....6...9.........6.....6......94..................
16: ..6...9........6............9.....6...9.........6.....6......94............9.....

For reference, the first pernutation is illustrated below in grid format.

Code: Select all

01:
 . . . | 9 . . | 6 . .
 . . 6 | . . . | 9 . .
 . . . | . . . | . . .
-------+-------+-------
 . 9 . | . . . | . 6 .
 . . 9 | . . . | . . .
 . . . | 6 . . | . . .
-------+-------+-------
 6 . . | . 9 . | . 4 .
 . . . | . . . | . . .
 . . . | . . . | . . .

[champagne]

Platinum Blonde: partial pencilmarks after one hidden single r9c3=5 and assuming r12c7=69

But let's pretend that r4c2 and r5c3 don't see each other. In this scenario, there are only 16 "valid" permutations of candidate placements in the sis. Note the contradictory r4c2=9 and r5c3=9 in all 16 permutations. That's hardly a recipe for a continuous loop.

Code: Select all

01:
 . . . | 9 . . | 6 . .
 . . 6 | . . . | 9 . .
 . . . | . . . | . . .
-------+-------+-------
 . 9 . | . . . | . 6 .
 . . 9 | . . . | . . .
 . . . | 6 . . | . . .
-------+-------+-------
 6 . . | . 9 . | . 4 .
 . . . | . . . | . . .
 . . . | . . . | . . .

just one question before digging in that post

What makes that permutation with 2 "9" in box 4 valid

EDIT: the ABI loop logic says there is no valid permutation
the 16 permutations shown here have all 2 '9' in box 4
so we are still looking for a valid one to have a counter proof

[champagne]

I entered the ABI logic in XSUDO

No surprise, I got the answer illegal logic

here are the corresponding results
XSUDO did not find any valid perm in that situation

Code: Select all

+--------------------------+---------------------------+-------------------------+
| 35678    34589   347(69) | 47(69)    5689     4578   | (69)  1         2       |
| 15678    14589   47(69)  | 1247(69)  125689   124578 | (69)  56789     3       |
| 1578(6)  158(9)  2       | 3         158(69)  1578   | 4     578(69)   78(69)  |
+--------------------------+---------------------------+-------------------------+
| 237      234(9)  1       | 8         23(6)    234    | 237   2347(69)  5       |
| 235      6       34(9)   | 124       7        12345  | 8     2349      149     |
| 23578    23458   347     | 124(6)    12356    9      | 1237  23467     1467    |
+--------------------------+---------------------------+-------------------------+
| 123(6)   123     8       | 5         123(9)   1237   | 1237  237(469)  17(469) |
| 9        123     3(6)    | 127       4        12378  | 5     23678     1678    |
| 4        7       5       | 12(9)     12389    6      | 123   2389      189     |
+--------------------------+---------------------------+-------------------------+

plat [22,243] 36 Candidates,

13 Truths = {4R7 6R347 9R347 6C347 9C347}

16 Links = {6r12 9r2 6c15 9c25 47n8 7n9 6b357 9b348}

2 AURs = (96)R21C73, (96)R21C74

AUR points {aur 3r1c3 6r1c3 6r1c4 9r2c7 }

illegal logic

[David P Bird]

When a wrong conditional placement or elimination is made in an otherwise valid pencil mark grid it will eventually produce multiple contradictions any one of which can be used to show that it is impossible. Saying that the ABI logic is invalid because there is an alternative contradiction available is therefore ridiculous.

ABI loops identify that whichever combination of base digits are tested, there will be a potential contradiction in a particular place. Using that as the focus then brings consistency to the approach.

If the full proof of the logic must be used to eliminate any particular pair of candidates then I wouldn't find them acceptable as being both net-based and blatant trial & error. However champagne has said that this approach provides a quick incompatibility check for any pair of digits. In that case the ABI logic would just establish a theorem and it would be acceptable to use the resultant inferences. I have yet to check if identifying the qualifying ABI pattern requires following logical streams (as opposed to counting) or not, but it appears to be very similar to identifying the JExcocet pattern with a relaxation on the target cell conditions.

My alternative approach counting the links in a single digit chain between two nodes has very similar considerations to weigh up before we can conclude which one is more comprehensive.

[champagne]

Just to continue in the same way
In fact abi chains the two eliminations to come to the final result
r12c7 = 79.

This is the exact transcription in XSUDO of ABI double loop.
Again no surprise, we get exactly the same result

Code: Select all

+--------------------------+----------------------------+----------------------------------+
| 3568-7   3458-9  34(679) | 4(679)    568-9    458-7   | (79-6)     1           2         |
| 1568-7   1458-9  4(679)  | 124(679)  12568-9  12458-7 | (79-6)     568-79      3         |
| 158(67)  158(9)  2       | 3         158(69)  158(7)  | 4          58(6-79)    8(6-79)   |
+--------------------------+----------------------------+----------------------------------+
| 23(7)    234(9)  1       | 8         23(6)    234     | 23(6-79)   -234(79-6)  5         |
| 235      6       34(9)   | 124       7        12345   | 8          2349        149       |
| 2358-7   23458   34(7)   | 124(6)    12356    9       | 1236-7     23467       1467      |
+--------------------------+----------------------------+----------------------------------+
| 123(6)   123     8       | 5         123(9)   123(7)  | 123(6-79)  -23(479-6)  -1(479-6) |
| 9        123     3(6)    | 12(7)     4        1238-7  | 5          23678       1678      |
| 4        7       5       | 12(9)     1238-9   6       | 123-9      2389        189       |
+--------------------------+----------------------------+----------------------------------+

plat [22,251] 59 Candidates,

18 Truths = {679R3 679R4 4679R7 679C3 679C4 12N7}

27 Links = {679r1 679r2 6c157 7c167 9c257 47n8 7n9 6b357 7b348 9b348}

4 AURs = (76)R21C73, (96)R21C73, (76)R21C74, (96)R21C74

AUR points {aur 6r1c3 6r1c4 7r2c7 9r2c7 }

34 Eliminations --> r4c8<>2346, r2c168<>7, r2c258<>9, r467c7<>7, r479c7<>9, r7c8<>236,

r1c16<>7, r1c25<>9, r3c89<>7, r3c89<>9, r12c7<>6, r7c9<>16, r6c1<>7,

r8c6<>7, r9c5<>9,

EDIT :

IMO abi loop is easy to understand, a XSUDO logic of rank 9 is not.
I would not accept that construction without abi's chain explanation.

EDIT 2

I checked again the XSUDO diagram and suppressed the links unused in the logic.
I finally got the following

plat [22,251] 59 Candidates,
18 Truths = {679R3 679R4 4679R7 679C3 679C4 12N7}
21 Links = {6c157 7c167 9c257 47n8 7n9 6b357 7b348 9b348}

so a final rank 3 logic,
In fact, the AUR logic in XSUDO does not require the link between the AUR cells

[ronk]

When a wrong conditional placement or elimination is made in an otherwise valid pencil mark grid it will eventually produce multiple contradictions any one of which can be used to show that it is impossible. Saying that the ABI logic is invalid because there is an alternative contradiction available is therefore ridiculous.

ABI loops identify that whichever combination of base digits are tested, there will be a potential contradiction in a particular place. Using that as the focus then brings consistency to the approach.

To clarify, I'm saying the "abi-loop" is not a continuous loop because it results in a contradiction within itself, i.e., between the abi-loop members. Specifically in this instance, it results in "multiple conditional placements" of the same UR digit in UR cells within one row. IMO that's not just your old everyday "alternative contradiction."

[David P Bird]

To clarify, I'm saying the "abi-loop" is not a continuous loop because it results in a contradiction within itself, i.e., between the abi-loop members. Specifically in this instance, it results in "multiple conditional placements" of the same UR digit in UR cells within one row. IMO that's not just your old everyday "alternative contradiction."

Following the classic interpretation of AIC rules, all the inferences in the loop I gave <here> are stand-alone with no memory or anticipation of the rival truth states occurring elsewhere in the loop.

I take it that if weak and strong inference sets are used instead, another set of alternative contradictions will be produced. Either way the opening premise must be false.

What is of interest to me is whether or not a theorem can be composed using either approach that will produce inferences that can be used by a player whenever its associated pattern is recognised.

In the framework of existing patterns this would amount to registering a new type of Deadly Pattern that must be avoided and so allowing a similar notation style to be adopted.

As useful as XSudoku is, adopting its approach as the norm seems to have been counterproductive in this area.

[champagne]

What is of interest to me is whether or not a theorem can be composed using either approach that will produce inferences that can be used by a player whenever its associated pattern is recognised.

In the framework of existing patterns this would amount to registering a new type of Deadly Pattern that must be avoided and so allowing a similar notation style to be adopted.

David I follow you but I have a small practical remark.

we are working here mainly on my data base of potential hardest.

In that data base i have seen only 2 puzzles having the 3 digits exocet pattern. As shown in the case of Golden Nugget, the theorem would not be the same with 4 digits

reversely, in my data base of all published puzzles with high rating I collected before eleven and I started the search for rating over SE ED 11.0,
I have a huge majority of puzzles with an exocet 3 digits.

fata morgana, trompe l'oeil, platinum Blonde are in that data base.

I have also a data base of more than 4 millions puzzles rating (skfr ER) ER 10 or 10.2 and more

I can easily upload in my site the 51 000 puzzles already known with the 3 digits exocets.

I intend to run the test to find Platinum Blonde type exocets on the 4 millions puzzle file when the code (not started) is ready.

I guess you will soon ask to split the 3 digits file in "pure Jexocets" and others. I have to think about that

[champagne]

As useful as XSudoku is, adopting its approach as the norm seems to have been counter-productive in this area.

XSUDO is perfect to check if your logic is valid, but not so easy to use in complex diagrams.

Moreover, as I often stated, finding the rank 0 areas in an XSUDO not rank 0 logic is not a simple exercise.

So I agree with you, it's by far not the best tool to build a logical elimination

[ronk]

To clarify, I'm saying the "abi-loop" is not a continuous loop because it results in a contradiction within itself, i.e., between the abi-loop members. Specifically in this instance, it results in "multiple conditional placements" of the same UR digit in UR cells within one row. IMO that's not just your old everyday "alternative contradiction."

Following the classic interpretation of AIC rules, all the inferences in the loop I gave <here> are stand-alone with no memory or anticipation of the rival truth states occurring elsewhere in the loop.

I take it that if weak and strong inference sets are used instead, another set of alternative contradictions will be produced. Either way the opening premise must be false.

No surprise (in my best champagne imitation), that our difference lies in inference stream versus Boolean POVs. For champagne and abi, I'm not so sure.

I rarely use the Boolean POV, preferring inference stream and cover set POVs, as you must be aware by now. When someone provides a graphical tool (similar to Xsudo) for Booleans, I'll seriously consider changing. BTW Xsudo enables its user to specify one derived [strong] inference, but not more.

[champagne]

I processed my data base of seeds for the search of potential hardest.
That file contains a little more than 4 millions puzzles rating (skfr) 10 and more, mainly 10.2 and more.
I excluded from that base puzzles already processed as potential hardest, so the remaining file should not have puzzles with a SE rating over 11.2.

I change my code to extract and identify

Jexocets of 3 or 4 digits split in four lots

Code: Select all

lot 1 3 digits with one occurrence of each digit in the orthogonal rows/cols
      outside the band/stack holding the Jexocet
lot 0 3 digits other Jexocets
lot 5 4 digits same as lot 1
lot 4 4 digits same as lot 0

Pattern with potential twin exocet (1 + 2 cells or 2+2 cells)
So far, I did not check that the "2 cells" postions are an AHS
This is split again in four sub lots

Code: Select all

lot 3 3 digits similar to lot 1
lot 2 3 digits similar to lot 0
lot 7 4 digits similar to lot 1
lot 6 4 digits similar to lot 0

3.36 Millions of puzzles have been extracted out of the 4.3 millions.
It seems that the ratio of "exocet" is the same as for higher ratings

A huge majority of puzzles have a 4 digits exocet
Most of the exocets are not the ideal mode to apply the ABI loop
(one occurrence of each digit in the orthogonal lines)
But as we could see for Golden Nugget, use of ABI loop is partial in exocet with 4 digits
For exocets with 3 digits, about 50% have the adequate pattern to apply the ABI loop
Puzzles with a 3 digits twin type pattern seem to have in majority the adequate pattern to apply the ABI loop

Code: Select all

lot    number 
0   182343      
1   175836      

2   300      
3   4451      362 930 total 3 digits

4   2870053      
5   120083      

6   5099      
7   413      2 995 648 total 4 digits

24   18      
42   215      

   3358811

A small number of puzzles (lots 24 and 42) have given a double extraction

All this is delivered subject to deeper investigation.
I'll study in priority lots 3 and 7 where ABI loop should be valid if the AHS is established


EDIT :
I forgot to mention that in that code, I added a new constraint for Jexocets,
All given of the band/stack + digits of the base = 9 digits
With that constraint, I lost only one Jexocet in the file of potential hardest in the following puzzle
....5..8...67..1..7....3..42..9............1.96....2...3..4.....7...8.5...21..6..;554;elev;864

[ronk]

But as we could see for Golden Nugget, use of ABI loop is partial in exocet with 4 digits

What means "partial?"