The Illusion of Fata Morgana

Advanced methods and approaches for solving Sudoku puzzles

Postby Allan Barker » Tue Oct 28, 2008 7:52 am

ronk wrote:Color me skeptical. I don't think it's possible for a "minimal rank 29" constraint set to yield that many eliminations ... and there's a strong likelihood that it's masquerading for a yet-to-be-defined "complementary" structure with much lower rank -- maybe even rank 0.

You are for sure correct. I mentioned above, rather briefly, that this pattern is probably 3 interwoven loops. They would mostly share the same sets but have many seperate linksets, thus the 49 linksets. When I get time I will trim the logic to see what comes out. As is, the pattern contains all the R, C, and N sets from the three layers as per Champagne's output, thus some linksets are probably not required at all. I won't show another picture but, fully assembled, all linksets in layer 5 are rank 0, and about half the linksets in layers 1 and 7.

The relation between rank and cover sets can vary, e.g., in FM, adding additional linksets to sets that are already covered resulted in additional eliminations. I recently found a very simple example of this, I will put that on my website. One might say that the elimination is becasue of additional constraints, and not related to rank. However, as far as I have seen, the rank 'rules' correctly account for all such eliminations.

Rank can also vary inside a structure and there is a global rank for any group of cover sets that covers all sets. Becasue this variability, I refer to an 'overall' or 'raw' rank as simply the number of linksets - number of sets within any logic, 29 in this case. The actual rank(s) relating to cover sets and eliminatons must be specified more precisely. I'll change my output to say 'raw' rank.

ronk wrote:That's my WAG for this month.:)

Today is the 28th.:(
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Postby Allan Barker » Tue Oct 28, 2008 8:34 am

champagne wrote:Again, what I am doing is just testing all permutations for a given group of sets. I have no guaranty that this fit with a sets/linksets construction.
Nothing to object if you need additional nodes to conclude in a sets/linksets view.


OK, I think I understand, but I would still say (for example) the group of sets you give for TR does not contain all the constraints required to make eliminations. More sets must be added. (I do not need a set/linkset model to test, I can test using only sets). These sets give me 3876 permutations.

I understand you are now testing, but I enjoy testing your testing!

Allan
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Postby champagne » Tue Oct 28, 2008 10:28 am

Allan Barker wrote:but I would still say (for example) the group of sets you give for TR does not contain all the constraints required to make eliminations. More sets must be added.


As you insists, something must be true.

It's a fact that in my process, I apply known floor to unknown floors to clear candidates in cells already covered.

This is for sure equivalent to using "Nodes" sets of my HC: line.

So I woud say:

. possibility to eliminate is granted,
. some nodes are implicit in my process.

t's not a problem in that phase dedicated to selection of favourable patterns, I have to pay attention to use them explicitely in the final process.

champagne
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Postby Allan Barker » Tue Oct 28, 2008 4:50 pm

champagne wrote:So I woud say:

. possibility to eliminate is granted,
. some nodes are implicit in my process.

t's not a problem in that phase dedicated to selection of favourable patterns, I have to pay attention to use them explicitely in the final process.

OK, very clear.

Congratualtions, I think you have proved that applying permutations to selected floors works very well for these kinds of layered puzzles. Symetery is not needed, a puzzle can be morphed and have the same results.

By adding a couple more sets I now get 19 eliminations from your floors, as below. There are no linksets.
Code: Select all
ENG 101 Nodes, Raw Rank = -73 (linksets - sets)
     73 Sets = {1r1245678 5r1245689 7r2345678 1c2345678 5c1245689 7c2345678 2n5 4n37 5n268 6n268 8n4 1b2345678 5b1245689 7b1245689}
      0 Links = {}
      r1c1<>5, r1c4<>5, r1c6<>5, r2c5<>3, r2c5<>9, r2c6<>5, r3c3<>7, r3c6<>7, r4c4<>5, r4c5<>5,
      r5c1<>5, r5c5<>5, r5c9<>5, r6c5<>5, r6c9<>5, r7c7<>7, r8c4<>6, r8c6<>5, r9c5<>5
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Postby champagne » Tue Oct 28, 2008 11:50 pm

Allan Barker wrote:OK, very clear.

Congratualtions, I think you have proved that applying permutations to selected floors works very well for these kinds of layered puzzles. Symetery is not needed, a puzzle can be morphed and have the same results.



May be I can comment on why I explored that field.

The floors aproach can work on any puzzle at any moment. It can appear as "road roller" to focus on eliminations you cand find thru simple AIC's or AIC's nets.

On hardest puzzles; the most difficult is normally at the start. Many examples we have seen show that in such cases the solution comes out of a limited number of floors.

For a human player, clues are cells having candidates limited to these flors. (eg: 136 in Fata Morgana). The "flavouring" sense of strmckr.

I could have tried to use directly these clues. I wanted to have a kind of exhaustive search of active combinations to find out which clues had to be considered.

As all exhaustive search, you have to consider many possibilities to finally come to evidence (eg: only floors 136 active in Fata Morgana up to 3 floors, what strmckr had seen for long). So I looked for an efficient process to locate "evidence" and check we did not miss something else.

For me, the findings of my second phase are as important as the focus on the active combinations of layers. "What sets must enter an active SLG (sets/linksets group) for that combination "

These compulsory sets should be the basis for a human player to try to build a SLG. If we can qualify them, it will help.


BTW, I extended my solver to the processing of patterns as 136 136 in row 5 of Fata Morgana, I am not at all convinced that this will be enough to solve the puzzle thru AIC's nets. But I will open a specific thread in due time to discuss that question.

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Postby ttt » Wed Oct 29, 2008 12:34 am

Hi All,
Finally, I finished it by my way for Allan’s first move of Fata Morgana, using 2D diagram with 3 sides…:D

My way based on the table:
Code: Select all
                 column 2     column 5     column 8   
                     
Candidate(1) row   4 5 9       9 46 1      1 5 6
                                                 
Candidate(3) row   4 5 2       2 46 9      9 5 6

Candidate(6) row   4 5 1       1 46 8      8 5 6

Present as diagram:
Code: Select all
Loop: => r4c2<>24
                                  ---------------------
                                 |                     |
(1)r9c2----(1)r9c5     (1)r6c8---      (6)r4c2         |
 ||         ||          ||              ||             |
(1)r4c2    (1)r1c5-----(1)r1c8         (6)r1c2-----    |
 ||         ||          ||              ||         |   |
(1)r5c2    (1)r46c5    (1)r5c8      ---(6)r5c2     |   |
 |          |           |          |               |   |
  ----------------------           |   (6)r1c5-----    |
            |                      |    ||             |
     (13)=(16)=(36)--------------------(6)r46c5        |
          r5c46                    |    ||             |
            |                      |   (6)r8c5----     | 
  ----------------------           |              |    |
 |          |           |          |              |    |
(3)r5c2    (3)r46c5    (3)r5c8      ---(6)r5c8    |    |
 ||         ||          ||              ||        |    |
(3)r4c2    (3)r9c5-----(3)r9c8         (6)r8c8----     |
 ||         ||          ||              ||             |
(3)r2c2----(3)r2c5     (3)r6c8---------(6)r6c8         |
                                 |                     |
                                  ---------------------

Studying Allan’s first move for Fata Morgana & Tunsten Rod at first seems the same way based on floors (136)r5c46 for FM & (157)r56c6 for TR, but they are difference and that is very interesting… There are many things can study & extent from Allan’s great idea.

Allan Barker wrote:...because I don't think humans think to much about rank (takes HCMC vodka ??)

Yes, when you come HCMC, I’ll invite you…:D

PS
Allan & champagne: my brain can’t follow all your post quickly then I need times to sip them…:D

Thanks to all,
ttt
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