The New Sudoku Players' Forum

[ronk]

The current file will be replaced in the next days by a file without the 329 non minimal puzzles I added by mistake. I keep in the file 2 previous non minimal puzzles included for historical reasons.

Using the filter "minimal&&uniq(%#.c)" with gsf's sudoku.exe, there are 220470 puzzles remaining. Apparently 5107 of the 225577 puzzles in the file have a problem, not 329.

[edit: Using the filter "valid&&minimal&&uniq(%#.c)" with gsf's sudoku.exe, there are 196534 puzzles remaining. Apparently at least 29043 of the 225577 puzzles in the file have a problem, not 329.]

[champagne]

It will be very interesting after the update of the data base to get examples of such puzzles

[dobrichev]

According to my solver, in the collection there are 6053 puzzles with redundant clues, 23936 with multiple solutions, and 5928 of them are both invalid and non-minimal. There are no duplicates. Most probably this garbage has some potential too.
Exemplars:

Code: Select all

1.......2.3.4...5...6...7...9.58........64......3.9.8...7...6...4...5.9.2.......1   22   redundant clue 8 at 31
98.7..6..75.4.........93.7.8...7..3..3.2.......7...1.4.7..5..8...6...2.....1.....   24   redundant clue 7 at 9, multiple solutions
98.7..6..75.....8.....6.....4...35.....58..6.......2.1...69..5......43..........2   22   338 solutions

Champagne,
Please investigate what happened and try to resurrect the lost puzzles as much as possible. It wold be a shame to lose your valuable work due to some stupid bug or wrong operation.
I know the risk of publishing intermediate results

Here is one not in your list

Code: Select all

........1.....2.3....14.5....3.....6.5.7..4..8.......2..6..8..5.4.51.....9..7.1.. skfr=11.8/11.8/2.6

Cheers,
MD

[champagne]

[Please investigate what happened and try to resurrect the lost puzzles as much as possible. It would be a shame to lose your valuable work due to some stupid bug or wrong operation.

I know what wrong manipulation was the source of trouble for multiple solution. No possible recovery, but it does not affect previous versions of the data base
Other non minimal are seeds I created for 'n' out of 'n-1' clues. They should not have been entered

I'll update the data base to-day. Many thanks

Here is one not in your list

Code: Select all

........1.....2.3....14.5....3.....6.5.7..4..8.......2..6..8..5.4.51.....9..7.1.. skfr=11.8/11.8/2.6

[/quote]

will be added later

[champagne]

I made disappointing trials to find new ideas for the list of puzzles qualified as "nothing special" in the data base.

Studying one of these puzzles, I came back to an old topic already discussed in that thread, the "partial exocet".

I modified my code to extract such puzzles and got pretty good results.

This puzzle (number 4946) is a good example of what can be done

....5..8....7.9..2..92...5..1.6.....3...2.8....5.9...7.4....3..6....4.....2.7...8;11.10;1.20;1.20;4946;elev;1945

It has a pattern very close to a double exocet. It is easy to show that it works like a double exocet.
The puzzle is immediately reduced to a skfr rating 9.00, and it is very easy to crack it excluding the invalid pairs.

I have to check again the results, but the number of puzzles having a partial exocet seems very important, and the ratio of double exocets seems to be in line with what has been seen for exocets.

[champagne]

A complete solution for the puzzle 4946 using the findings of my new solver

Code: Select all

....5..8....7.9..2..92...5..1.6.....3...2.8....5.9...7.4....3..6....4.....2.7...8
;4946;elev;1945;11.10;1.20;1.20


Code: Select all

1247  2367  13467 |134   5     136  |14679 8     13469
1458  3568  13468 |7     13468 9    |146   1346  2     
1478  3678  9     |2     13468 1368 |1467  5     1346 
------------------------------------------------------
24789 1     478   |6     348   3578 |2459  2349  3459 
3     679   467   |145   2     157  |8     1469  14569
248   268   5     |1348  9     138  |1246  12346 7     
------------------------------------------------------
15789 4     178   |1589  168   2    |3     1679  1569 
6     35789 1378  |13589 138   4    |1259  1279  159   
159   359   2     |1359  7     1356 |14569 1469  8 

In that puzzle we have a double exocet potential pattern

r1c4r1c6 r2c3 r3c9
r2c7r2c8 r1c3 r3c5

but the control fails for the digit 1.

Having 3 digits ok, it’s enough to show that 1r1c46 + 1r2c78 is not valid to apply the double exocet eliminations.

This is not to hard, but relatively long in my path.

Code: Select all

1r1c46 + 1r2c78
  => 1r3c1;1r78c35(XWing) =>14r9c78 =><4>r2c78 (UR); 6r9c6
  => 57r45c6;4r5c4;13r1c46;8r3c6;8r6c4(UR);3r4c5 ;3r3c9 ;16r2c78
  => 4r2c5 6r3c5 7r3c2 4r3c7 9r1c9 5r8c9 4r4c9
     5r4c7 7r4c6 8r4c3 7r5c3 46locked in r1c3 not valid


We can now apply the double exocet eliminations

r1c9=9 r1c7=7 r2c5=8

and we come quickly in that position

Code: Select all

124   236   1346  |134   5     136  |7     8     9
145   356   1346  |7     8     9    |146   1346  2     
78    78    9     |2     1346  136  |146   5     1346 
------------------------------------------------------
24789 1     478   |6     34    3578 |2459  2349  345   
3     679   467   |145   2     157  |8     1469  1456 
248   268   5     |134   9     138  |1246  12346 7     
------------------------------------------------------
15789 4     178   |1589  16    2    |3     1679  156   
6     35789 1378  |13589 13    4    |1259  1279  15   
159   359   2     |1359  7     1356 |14569 1469  8 

Skipping a possible XYZWing, we look at the ‘4’ floor

Code: Select all

4.4 4.. ...
4.4 ... 44.
... .4. 4.4

4.4 .4. 444
..4 4.. .44
4.. 4.. 44.

.g. ... ...
... ..g ...
... ... 44.


Easy here to show <4>r4c7 and <4>r4c8 (we need that one for the rank 0 logic)
Then a rank 0 logic appears

Code: Select all

18 Truths = {1R1269 3R1269 4R1269 6R1269 3N67 }
18 Links = {1c167 3c26 4c17 6c267 1n34 2n38 6n48 9n48 }


And the rest is easy.

This is one example of a rank 0 logic appearing after the start.
As the puzzle at that point has a rating of 9.0 (skfr), most players would not look for it

[David P Bird]

Champagne, from what I make out, you started from the assumption that (1)r1c46 and (1)r2c78 were both true, and used a forcing net to prove that this would (eventually) produce a contradiction and is therefore false.

As a result at least one of the two possible Excocets will reduce to the base digits (346), but that doesn't prove that both of them will, which is what you seem to assume.

Have I missed something?

[pjb]

Here is another (more straightforward) puzzle in which I don't detect a multifish initially, but after the eliminations brought about by it's double exocet, there is a simple multifish:

Code: Select all

...4....87.......9..3.9.5....9.6...51....23...7....6....6.8..5......1....2...4... Coly150

After the double exocet the state is:

Code: Select all

6      9      125    | 4      12357  357    | 127    1237   8     
7      1458   12458  | 12358  1235   6      | 124    1234   9     
248    148    3      | 1278   9      78     | 5      1247   6     
---------------------+----------------------+---------------------
234    34     9      | 137    6      37     | 8      1247   5     
1      6      458    | 5789   457    2      | 3      479    47     
23458  7      2458   | 13589  1345   3589   | 6      1249   124   
---------------------+----------------------+---------------------
349    134    6      | 2379   8      379    | 1247   5      1247   
34589  3458   4578   | 235679 2357   1      | 2479   68     2347   
3589   2      1578   | 35679  357    4      | 179    68     137

and then my solver detects the following:

Code: Select all

12 Truths = {1247R2, 1247R3, 1247R6}
12 Links = {24c0, 14c1, 127c3, 7c5,  2n7, 3n7, 6n6, 6n8}
14 Eliminations:  05<>7, 11<>1, 11<>4, 13<>1, 13<>2, 43<>7, 50<>2, 50<>4, 53<>1, 70<>4, 71<>4, 73<>2, 73<>7, 83<>7

which renders the puzzle easily solved.

Phil

[champagne]

Champagne, from what I make out, you started from the assumption that (1)r1c46 and (1)r2c78 were both true, and used a forcing net to prove that this would (eventually) produce a contradiction and is therefore false.

As a result at least one of the two possible Excocets will reduce to the base digits (346), but that doesn't prove that both of them will, which is what you seem to assume.

Have I missed something?

Hi David,

I made it short, but the detailed logic is the following.

1) The Double Exocet property is established for 3 of the 4 digits '3' '4' '6'. It does not work for digit '1'.

2) from the double exocet property if any of the exocet is solved by 2 digits among 346, the other one can not contain these digits and will be solved by '1' plus the fourth digit.

3) As a consequence, except if '1' is solution in both exocet, the double exocet will contain the four digits, giving the way for the classical eliminations.

As you say, the proof that '1' is not solution for both exocets is done here by what serate qualify as a "dynamic plus expansion leading to a contradiction" with '1' true in both exocets.
I just used URs in the expansion, what is not done in serate.

There is likely a shorter way to establish it.

[champagne]

Here is another (more straightforward) puzzle in which I don't detect a multifish initially, but after the eliminations brought about by it's double exocet, there is a simple multifish:

Code: Select all

...4....87.......9..3.9.5....9.6...51....23...7....6....6.8..5......1....2...4... Coly150

...
which renders the puzzle easily solved.

Phil

Hi Phil,

Nice example. In that case (and generally speaking, in all puzzles having an exocet or a partial exocet, which is the huge majority of hardest puzzles), there is a logic to restart the search of rank 0 logic after the start. We know the 4 digits to use.

The design of the rank 0 logic search in my code has to be changed to go in that direction. In my example, I got it because I solved separately the puzzle after assignment of the 3 cells solved by the double exocet.

[Leren]

Champagne wrote: A complete solution for the puzzle 4946 using the findings of my new solver

Hi Champagne, I've coded your Almost Exocet technique and it works for puzzle 4946 as you described.

Could you provide some more examples of puzzles where this applies ? I'd like to do some more testing of this technique.

Leren

[champagne]

Champagne wrote: A complete solution for the puzzle 4946 using the findings of my new solver

Hi Champagne, I've coded your Almost Exocet technique and it works for puzzle 4946 as you described.

Could you provide some more examples of puzzles where this applies ? I'd like to do some more testing of this technique.

Leren

Hi leren,

I on my side changed my code and nearly finished to test it to be in a position to catch and work out your rank1 logic.

But I leave home in the next hours for 2 months and I was preparing the last transfers to keep on my small PC the key data.
I have a revised code to extract partial exocets from the file of hardest, but I have to check the results.
I intended to add such a file in the next update of the data base of hardest in march.

This should be possible from my small PC, then I'll answer within 3 or 4 days.
I try to run a test before departure on the partial exocet issue.

[champagne]

I checked quickly the test results.

I used the command (for the file 02gr12_12)
skmpp -i02gr12_12 -c"142" "// test for partial exocet "

And I got many examples with the following output on 02gr12_12_file2


.....1..2....2.34...2....5.....1.4.5.6...3.7.8..9..6....5.3...443...7...9..8.....; ;1;match type;13;r1c7 r1c8 r2c3 r3c5 6789;;;;68;;
.......12....345....51.......3.56....1.7.....64.....8..5..1.4..3.......14..9.7...;D;2;match type;1212;r1c1 r3c1 r4c2 r7c3 2789;r5c3 r6c3 r2c2 r7c1 2789;;;;


First puzzle should have a partial exocet digits 6789 valid for digits 68

Second puzzle has a double partial exocet, but the output for valid digits is not clear. I have to check the code.

I suggest you compile the code than you can find here project skmpp

I can also in the next days prepare a win32.exe download for the program

[Leren]

Champagne wrote : .......12....345....51.......3.56....1.7.....64.....8..5..1.4..3.......14..9.7...;D;2;match type;1212;r1c1 r3c1 r4c2 r7c3 2789;r5c3 r6c3 r2c2 r7c1 2789;;;;

Hi Champagne, thanks for your response. I ran that second puzzle and find that checks fail for digits 28, OK for 79 for both Exocets.

I'm not sure how to proceed from here - does your Almost Exocet method apply for only 2 digits OK?

Leren

[champagne]

Champagne wrote : .......12....345....51.......3.56....1.7.....64.....8..5..1.4..3.......14..9.7...;D;2;match type;1212;r1c1 r3c1 r4c2 r7c3 2789;r5c3 r6c3 r2c2 r7c1 2789;;;;

Hi Champagne, thanks for your response. I ran that second puzzle and find that checks fail for digits 28, OK for 79 for both Exocets.

I'm not sure how to proceed from here - does your Almost Exocet method apply for only 2 digits OK?

Leren

well!! this is a new field. I found an answer with 3 digits matching out of 4 in the case of a double exocet.

With 2 digits matching as here, I guess something else should establish the proof that it is working as a classical double exocet.

In the same way, a single partial exocet will be easier to use in the solver if several digits are matching, especially with the "abi loop" pattern.


At that point, I can't say more, but in my old solver, I processed partial exocets (without using the "abi loop" property).
For many of the "hardest" puzzles, partial exocets had a significant contribution to the solution.

If I wanted to work in that field now, I would likely start, as you did, with double exocet partial patterns and then with single exocets with several "abi loop" eliminations