fighting again puzzles having the SK loop

Everything about Sudoku that doesn't fit in one of the other sections

fighting again puzzles having the SK loop

Postby champagne » Wed May 14, 2008 5:42 am

A significant part of the hardest puzzles has the Kurtzhals loop.
In JPF’s list at the time I picked it up, five of them had the highest rating 11.4.

The two at the top of the list

Code: Select all
11.4 # 500000009020100070008000300040600000000050000000207010003000800060004020900000005 # m_b_metcalf
11.4 # 500000009020100070008000300040002000000050000000706010003000800060004020900000005 # StrmCkr

Can be solved in a very similar way and should give a good example to figure out how to kill such monsters.
The solution of my solver can be seen in my set of examples.
I will develop here a tentative synthesis of what the solver did.

Normally a preliminary reading of the full tagging method is requested although most of the finding could be understood with a limited knowledge of the method.
This can be found on my website

http://pagesperso-orange.fr/gpenet/index_fichiers/AWHO.htm

I also experiment in that thread mixing of text and images, so I will likely have to do several edits before reaching what I want to do. Text can be copyed using this link
http://pagesperso-orange.fr/gpenet/UX/Sample5metc/metc_fichiers/V1_01_x1.htm

Image

to be continued soon
Last edited by champagne on Thu May 22, 2008 9:50 am, edited 1 time in total.
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Postby champagne » Wed May 14, 2008 5:57 pm

metcalf analysis phase 2 (can be copyed using the link)
http://pagesperso-orange.fr/gpenet/UX/Sample5metc/metc_fichiers/V1_01_x2.htm

Image
Last edited by champagne on Thu May 22, 2008 9:55 am, edited 1 time in total.
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Postby champagne » Thu May 15, 2008 9:07 am

metcalf analysis phase 3 (can be copyed using the link)
http://pagesperso-orange.fr/gpenet/UX/Sample5metc/metc_fichiers/V1_01_x3.htm
Image
Last edited by champagne on Thu May 22, 2008 9:53 am, edited 1 time in total.
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Re: fighting again puzzles having the SK loop

Postby ronk » Thu May 22, 2008 1:11 pm

champagne wrote:I also experiment in that thread mixing of text and images, so I will likely have to do several edits before reaching what I want to do.

Retyping text that is embedded in images takes more effort than I care to expend ... so this question will have to do.

Would you please put your "virus pattern" viewpoint for the K-loop (in your first post above) into terms of an inference stream (of strong and weak inferences):?: I see no strong inferences identified.
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Re: fighting again puzzles having the SK loop

Postby champagne » Thu May 22, 2008 2:13 pm

ronk wrote:
champagne wrote:I also experiment in that thread mixing of text and images, so I will likely have to do several edits before reaching what I want to do.

Retyping text that is embedded in images takes more effort than I care to expend ... so this question will have to do.



I feel the most important is to produce a reliable post easy to understand. But you are right, must be easy to copy it.

My images are in fact .gif files produced out of .htm files. I will take care that the .htm files are on my web site.

You normally reach them changing the extension of the file. I edited the three posts to give the link to the three pages.

Tell me if it solves your problem. (it could be that the .htm file has been updated and not the image).



ronk wrote:
Would you please put your "virus pattern" viewpoint for the K-loop (in your first post above) into terms of an inference stream (of strong and weak inferences):?: I see no strong inferences identified.


This is part of my level 4 process. You will find existing comments in the places here below.

I start preparing a post sticking to the metcalf situation and free from any tagging (tagging is not necessary in that process).

http://pagesperso-orange.fr/gpenet/UM/UM00_fichiers/UM41.htm

http://pagesperso-orange.fr/gpenet/UX/sample1/s1_fichiers/S1D1.htm


edit one, adding comments on metcalf sk loop

http://pagesperso-orange.fr/gpenet/UX/Sample5metc/metc_fichiers/V2_01_sk.htm
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Postby StrmCkr » Fri May 23, 2008 10:06 am

wow..
people are still trying to solve my puzzle same as me.

your first post is what i found as well
where paird numbers

4&6, 1&7, 3&9, 8&5
couldn't be found in columns and rows for all the puzzle.

as well as the fact specific paired numbers cannot exsits on a diagonal either for box 1 and 9 as it leaves a zero solution in the sk loop..
1&7, 4&6 are the pairs. see my link for further detail.

my first move is the sk loop not detailing my self. Steve has confirmed its correct on the link below for eliminations.

Code: Select all
 *-----------------------------------------------------------------------------*

 | 5       137     1467    | 23468   234678  378     | 1246    468     9       |

 | 346     2       469     | 1       34689   3589    | 456     7       468     |

 | 1467    179     8       | 24569   24679   579     | 3       456     1246    |

 |-------------------------+-------------------------+-------------------------|

 | 13678   4       15679   | 389     1389    2       | 5679    35689   3678    |

 | 123678  13789   12679   | 3489    5       1389    | 24679   34689   234678  |

 | 238     3589    259     | 7       3489    6       | 2459    1       2348    |

 |-------------------------+-------------------------+-------------------------|

 | 1247    157     3       | 2569    12679   1579    | 8       469     1467    |

 | 178     6       157     | 3589    13789   4       | 179     2       137     |

 | 9       178     1247    | 2368    123678  1378    | 1467    346     5       |

 *-----------------------------------------------------------------------------*


i know the reductions lead to this grid and from here i can do a intresting move

Code: Select all
 *-----------------------------------------------------------------------------*

 | 5       137     1467    | 23468   234678  378     | 12      468     9       |

 | 346     2       469     | 1       389@     3589    | 456     7       468     |

 | 1467    179     8       | 24569   24679   579     | 3       456     12      |

 |-------------------------+-------------------------+-------------------------|

 | 13678   4       15679   | 389     1389@    2       | 5679    3589    3678    |

 | 123687  389@      12679   | 3489@    5       1389@    | 24679   389@    34678  |

 | 238     3589    259     | 7       3489@    6       | 2459    1       2348    |

 |-------------------------+-------------------------+-------------------------|

 | 24      157     3       | 2569    12679   1579    | 8       469     1467    |

 | 178     6       157     | 3589      389@   4       | 179     2       137     |

 | 9       178     24      | 2368    123678  1378    | 1467    346     5       |

 *-----------------------------------------------------------------------------*


({R4C5,R5C6}389|(1 = 4)|389{R5C4,R6C5})=389(R2C58)(R5C38) - - loop

=>r5c19<>38,r5c37<>9, r19c5<>38, r37c5<>9

leaves me here:(

and im into combatronics with matrixs to solve past this point
{ but im am attempting to develope a dual placement theory for eliminations.)

i found a few moves that invloves the both halfs of 1/2 or 4/2 relation to strat in combination with the paired contradictions above:
where placing 6+n: where n =1 or 7 only in box 1 at specific points R1C3,R3C1
where all 4 infernances(1/2 both views of subsets 1/7.) leaves muti solutions in the sk loop, or no solution. eliminating 6 from R1C3 and R3C1.

the thread is here for reference...

http://www.sudoku.org.uk/SudokuThread.asp?fid=4&sid=10183&p1=1&p2=11

Code: Select all
 *--------------------------------------------------------------------*

 | 5      137    1467   | 23468  2467   378    | 12     468    9      |

 | 346    2      469    | 1      389    3589   | 456    7      468    |

 | 1467   179    8      | 24569  2467   579    | 3      456    12     |

 |----------------------+----------------------+----------------------|

 | 13678  4      15679  | 389    1389   2      | 5679   3589   3678   |

 | 1267   389    1267   | 3489   5      1389   | 2467   389    2467   |

 | 238    3589   259    | 7      3489   6      | 2459   1      2348   |

 |----------------------+----------------------+----------------------|

 | 24     157    3      | 2569   1267   1579   | 8      469    1467   |

 | 178    6      157    | 3589   389    4      | 179    2      137    |

 | 9      178    24     | 2368   1267   1378   | 1467   346    5      |

 *--------------------------------------------------------------------*
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Postby ronk » Fri May 23, 2008 11:45 am

champagne, what you're calling "virus pattern" seems to be what everyone else calls an AALS. As of now I see no utility in yet another term, so I'll stick with AALS.

Your continuous loop of AALSs is a somewhat different point-of-view (POV) than Steve Kurzhal's original hidden pair loop. Steve's AIC expression used paired strong inferences in r2, c8, r8 and c2 ... and exclusively used K-loop digits (5,8,3,9 in this puzzle).

However, I think your POV is better. For sure, the loop is shorter. It also means K-loops can exist in puzzles without the paired strong inferences noted above.

My interpretation of the K-loop deduction using only AALSs:

Code: Select all
 5      A137     1467    | 3478    234678  2368    | 12     I468     9
L346     2      L469     | 1      K389-46 K3589-6  |J456     7      J468
 1467   A179     8       | 4579    24679   2569    | 3      I456     12
-------------------------+-------------------------+------------------------
 12378   4       12579   | 6       1389    1389    | 2579   H3589    2378
 123678 B389-17  12679   | 3489    5       1389    | 24679  H389-46  234678
 368    B3589    569     | 2       3489    7       | 4569    1       3468
-------------------------+-------------------------+------------------------
 24     C157     3       | 579     12679   12569   | 8      G469     1467
D178     6      D157     |E3589-7 E389-17  4       |F179     2      F137
 9      C178     24      | 378     123678  12368   | 1467   G346     5
   
  A          C        D          F        G          I        J          L        A
r13c2 -17- r79c2-58-r8c13 -17- r8c79-39-r79c8 -46- r13c8-58-r2c79 -46- r2c13-39-r13c2 - continuous
 
 ==> r5c2<>17, r8c4<>7 r8c5<>17, r5c8<>46, r2c6<>6 and r2c5<>46

Nodes A, C, D, F, G, I, J and L are all AALSs, in this case, two cells with four candidates. The notation r13c2 -17- r79c2 means the two AALSs share two digits. R13c2 might ultimately contain both, none, or only one of the digits. The continuous property of the loop means that all weak inferences become conjugate inferences.

Steve's loop would have used nodes B, E, H and K as well.

[edit: Relabeled the K-loop in the CCW direction (to be consistent with my other posts elsewhere)]
Last edited by ronk on Sat Jun 14, 2008 10:42 am, edited 2 times in total.
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Postby champagne » Fri May 23, 2008 1:18 pm

StrmCkr wrote:wow..
people are still trying to solve my puzzle same as me.

your first post is what i found as well . . .



I agree that we reach quickly your last map with your puzzle

5.......9.2.1...7...8...3...4...2.......5.......7.6.1...3...8...6...4.2.9.......5


At that point you have normally shown that (3&9)r2c13 and (4&6)r2c13 are not valid.
Your last situation is tagged as follow by the solver showing clearly the strong inferences on isolated candidates within the

SK loop:
3r2c1 = 9r2c3 . . .


Code: Select all
5     13c7  1467   |23468  2467  378      |1d2D  468A  9   
3C46  2     469c   |1      389   35a89    |45A6  7     468a
1467  179C  8      |245q69 2467  579      |3     45a6  1D2d
-----------------------------------------------------------
13678 4     15p679 |389    1e389 2        |5679  35A89 3678
1267  389   1267   |3Î4e89 5     1E3Ë8Ì9Í |2467  389   2467
238   35b89 259    |7      34E89 6        |245p9 1     2348
-----------------------------------------------------------
2f4F  15B7  3      |2569   1267  15q79    |8     469g  1467
178B  6     15b7   |35B89  389   4        |179G  2     13g7
9     178b  2F4f   |2368   1267  1378     |1467  3G46  5   

As I wrote, next step for my solver is very similar to metcalf path, establishing
1r3c2==1r7c9 4r2c3==4r9c7 1r1c2==1r9c7 4r2c1==4r7c9 1r8c9==1r3c1 4r9c8==4r1c3 1r8c7==1r1c3 4r7c8==4r3c1

Again, I think we can limit easily boxes 1;9 to 8 scenarios, but I have to check it.

What is for sure is that the clearing sequence of the solver is more focused on the SK loop that in Metcalf.

It starts that way :
#4r3c1 #4r7c8 #7r8c9 #6r1c8 #4r2c7 #6r2c1 #7r7c2 #1r8c1 #1r3c2 #1r7c9 #7r1c3 #6r9c7.

I will study an improvement of the solver proposal based on the principles developped here and I will come back.

From what I have seen anlysing the 8 scenarios in Metcalf, some Ac2 must be studied to come to something reasonable.
(eg: r26c1 and r37c6). Others as here r7c2r8c1 can add strong inferences within the sk loop.
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Postby champagne » Fri May 23, 2008 1:32 pm

ronk wrote:champagne, what you're calling "virus pattern" seems to be what everyone else calls an AALS. As of now I see no utility in yet another term, so I'll stick with AALS.


these patterns are ALLS/AAHS belonging to wo units.

In my context, they are AAHS/AC2. We already raise that point and I said why I kept my name AC2 reminding the fact that they are seen as "almost cells" with freedom 2. So for me a virus pattern is an AC2 belonging to two units. I chose the name "virus pattern" for such AC2 because they can (if chained) be the source of weak inferences expansion for super candidates. (explained in my site)

ronk wrote:My interpretation of the K-loop deduction using all AALSs: . . .


I had in mind your way to explain it. I studied your post in October 2007. Nothing to say, I fully agree.
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Postby champagne » Sat May 24, 2008 5:08 pm

Hi strmckr

As promissed the way I would continue with your puzzle.
This is significantly different and shorter than the solver proposal.
I'll continue as soon as possible.

EDIT split in two pages after correction of an error
next post will be focused on clearing of scenarios

http://pagesperso-orange.fr/gpenet/UX/strmckr_fichiers/V1_01.htm
http://pagesperso-orange.fr/gpenet/UX/strmckr_fichiers/V1_02.htm
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Last edited by champagne on Sun May 25, 2008 4:03 am, edited 1 time in total.
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Postby StrmCkr » Sun May 25, 2008 4:56 am

if i am reading that correct it seems i was going in the same direction.

u are using the avoided sets of pairs that lead to a contradiction in the sk loop. to reduce one of the corners to only being one of 8 possible senario.

from this i mean
box 1 cannot have paid in a row or column.
1-7,3-9,6-4

then from corner to corner (r1c3 & r3C1) pairs 4-6,1-7 cannot be combined.

from here i use the sk loop to look for genereated errors by matching sets

I start with placing
the 1 or 2
or
4 or 2 found in box 3 and 7 { limits the senarios to 2 and 2 possible when u consider the anti parings}

using all 4 possible senarios but as sets of 2 diffrent functions.(the 1&2 as one and 4&2 as the second)

in conjucution

i test place 6 in: R1C3 & R3C1 with all its legal pairing ithe opposit corner R1C3 & R3C1 which can only 1 or 7 - and look for a contradiction (3 legal moves tests)

i test place 4 in: R1C3 & R3C1 with all its legal pairing ithe opposit corner R1C3 & R3C1 which can only 1 or 7 - and look for a contradiction.(3 tests)

i test place 7 in: R1C3 & R3C1 with all its legal pairing ithe opposit and look for contradictions. (3 legal tests)

corner R1C3 & R3C1 which can only 4 or 6

and look for a contradiction in all of each test numbers senarios.

the results are R1c3 & R3C1 cannot = 6, 1 and 7 from R1C3,

(all 4 senarios produce a contradiction} contradiction is in the sk loop.

u can repeat this proccess on box 9
using its properties of
cannot have paid in a row or column.
1-7,3-9,6-4
then from corner to corner (r8c9 & r9C7) pairs 4-6,1-7 cannot be combined.

in conjucution
i test place 6 in: R1C3 & R3C1 with all its legal pairing ithe opposit corner R1C3 & R3C1 which can only 1 or 7 - and look for a contradiction (3 legal moves tests)

i test place 4 in: R1C3 & R3C1 with all its legal pairing ithe opposit corner R1C3 & R3C1 which can only 1 or 7 - and look for a contradiction.(3 tests)

i test place 7 in: R1C3 & R3C1 with all its legal pairing ithe opposit and look for contradictions. (3 legal tests)

results are r8c9 & r9C7 cannot be 4,1. and 1 from r8C9.

then from here i use box line reduction to complete the puzzle as singles.

i need some one to check if i am doing this right or not...

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Postby champagne » Sun May 25, 2008 4:18 pm

Next step in strmckr, the clearing of scenarios.
Here three are cleared, I will edit the web file ASAP.

Comments to strmckr will come separatly

http://pagesperso-orange.fr/gpenet/UX/strmckr_fichiers/V1_03.htm

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Postby champagne » Mon May 26, 2008 6:35 am

Hi StrmCkr,

StrmCkr wrote:if i am reading that correct it seems i was going in the same direction.
u are using the avoided sets of pairs that lead to a contradiction in the sk loop. to reduce one of the corners to only being one of 8 possible scenarios.


What has clearly shown my solver is that if you can not solve a puzzle clearing candidates thru AICs nets. Using “avoided sets of pairs” you can crack all known puzzles, but there is an exponential growth of possibilities each time a new field is open.

The question is then to find a feasible process for a player. The general strategy followed in full tagging in to “reduce the count”. The SK loop when it exists has the best chances to offer logical clearings. When you have reached the point to say “I have only 8 scenarios. . .” you already made a lot of work.

StrmCkr wrote:I start with placing the 1 or 2 or 4 or 2 found in box 3 and 7 { limits the scenarios to 2 and 2 possible when u consider the anti parings}
using all 4 possible scenarios but as sets of 2 different functions.(the 1&2 as one and 4&2 as the second)
in conjunction


As you could see, in the solution I published, these cells r1c7 r3c9 r7c1 r9c3 play a key role to establish equivalences (and reduce the count).
Later on, their effect is hidden in the list of the 8 scenarios.


StrmCkr wrote:I test place 6 in: R1C3 & R3C1 with all its legal pairing in the opposite corner R1C3 & R3C1 which can only 1 or 7 - and look for a contradiction (3 legal moves tests)
I test place 4 in: R1C3 & R3C1 with all its legal pairing in the opposite corner R1C3 & R3C1 which can only 1 or 7 - and look for a contradiction.(3 tests)
I test place 7 in: R1C3 & R3C1 with all its legal pairing in the opposite and look for contradictions. (3 legal tests)


This is a very common way for players to face hardest puzzles . The more I work on solution, the more I see that a solver building a net of AICs as mine is doing something equivalent. The main difference is that the solver will limit itself to what can be achieved with strong an weak inferences coming out of an identified set of stuff and constraints.

Again, the number of possibilities to examine is in general very high. In the SK loop, we have tools and good reasons to identify and test a limited number of scenarios.



StrmCkr wrote:corner R1C3 & R3C1 which can only 4 or 6 and look for a contradiction in all of each test numbers scenarios.
the results are R1c3 & R3C1 cannot = 6, 1 and 7 from R1C3, (all 4 scenarios produce a contradiction} contradiction is in the sk loop.
u can repeat this process on box 9 using its properties of
cannot have paid in a row or column. 1-7,3-9,6-4
then from corner to corner (r8c9 & r9C7) pairs 4-6,1-7 cannot be combined.


I know it is a hard job, but showing how you did it would be interesting. Difficult to comment such a general statement.

The way I am working to prepare my solution is very specific. I try to work out a solution the solver could produce in the future. I am very keen to select a path complying with the full tagging process. There is plenty of other paths to solve this puzzle.
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Postby champagne » Wed May 28, 2008 9:09 am

I first intended to solve metcalf puzzle thru scenarios analysis. Finally, to meet STMCKR request, I finished the task on his puzzle.

As expected (may be with more success than expected by me), it could be achieved thru very short sequences.

Use of two colors (blue and red) has been enough to “see” the ‘groups’ in the sk loop. Combined to letters for isolated candidates, it gives, to my view, a quick and precise synthesis of where we are.

Puzzles having the SK loop can be split in two categories:

One group, including Easter Monster can be solved without entering level 4 (strong an weak inferences coming out of AC/AHS ALS addition are providing enough logical links to solve it). Puzzles of that group would be solved in an easier way if equivalences in the SK loop are searched in priority (including, if necessary use of AC). I’ll later on search for such a short version of Easter Monster solution.

Others are requesting more than ALS AC/AHS. Metcalf and strmckr puzzles are examples of that group. AC2 analysis can bring missing logical inferences. What is for sure is that combining layers to come to an extended scenario will not “create” in itself new inferences. It is just a way to express a complex AICs net in a simpler one.
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Postby StrmCkr » Fri May 30, 2008 8:21 am

when i learn how to post stuff more clearly i'll attempt to indicate how i came to the deductions u have reached. (

as some what vagly expressed above bye my self..)

i am using layers as well to indicate the invalid sequences of senarios as well..

to simplify the number of senarios
i started by using either of the bivavle sites setting it as 1 or 2, or 4 and 2.
leaves (2 sets of 2 choices.) wiht box 1 & 9 as having

3 senairos with only 2/1 potentially legal sub maps themselfs. most express contradiction chains quickly in the sk loop)

from this u can drop the comparing both layers down to only 2 versions

where R1c3 & R3C1 can be 6&1 or 7&4.
same goes for R7C9 & R9C7

most are only vaild combinations or arangments by the inital pair of 1/2 or 4/2.
u can then eliminate the invalid 4/2 or 1/2 combinations that don't map as well.

and only compare the 16 - 74 placemetns on a few submaps of
combinations of 42 & 1/2 alignments(that work).
only 1 is a valid map.

im sure thats not much help... but its how i did it.
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