The New Sudoku Players' Forum

[champagne]

champagne your results are worrying!

Puzzle No1 in your list of 42 should have been recognised with (2345)Exocet:r78c8,r1c9,r5c7 => r1c9 <> 1
It has empty cells in different columns as usual at r1c7 & r5c9, so it should have been found.


DPB

98.7..6..5..........4.3....3....5.8..7.9....6..2.4.....9.1...6......29.7......1..;;220;GP;H28;2BN H8H9

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9     8     13    |7     125   14     |6     12345  12345  xx
             -            --    -
5     1236  1367  |2468  12689 14689  |23478 123479 123489
1267  126   4     |2568  3     1689   |2578  12579  12589 
----------------------------------------------------------
3     146   9     |26    1267  5      |247   8      124   
148   7     158   |9     128   138    |2345  12345  6      xx
 -           -            -     -
168   156   2     |368   4     13678  |357   13579  1359   
----------------------------------------------------------
2478  9     3578  |1     578   3478   |23458 6      23458 
--          --           -     --
1468  13456 13568 |34568 568   2      |9     345    7     
24678 23456 35678 |34568 56789 346789 |1     2345   23458 

                                               xx
r89c8 match r1c9r5c7

The problem in that algorithm comes from 2 r7c1
I see no evidence just looking at rows 1;5;7 that 2r7c1 is not valid if the target is not filled

I am convinced it's true (my old solver found it), but it likely requires and extended view.

champagne

[ronk]

champagne your results are worrying!

Puzzle No1 in your list of 42 should have been recognised with (2345)Exocet:r78c8,r1c9,r5c7 => r1c9 <> 1
It has empty cells in different columns as usual at r1c7 & r5c9, so it should have been found.

Not if champagne took your "swordfish" requirement literally. This is a degenerate case with only an "x-wing" for digit <2>.

[champagne]

champagne your results are worrying!

Puzzle No1 in your list of 42 should have been recognised with (2345)Exocet:r78c8,r1c9,r5c7 => r1c9 <> 1
It has empty cells in different columns as usual at r1c7 & r5c9, so it should have been found.

Not if champagne took your "swordfish" requirement literally. This is a degenerate case with only an "x-wing" for digit <2>.

I did not use David wording in the last step. I just tried to stay within the generic condition to use that simplified process

If the target is empty, then, solving the "perpendicular rows" of the target, the "perpendicular row" of the last cell in the box containing the base must have no solution.

BTW, if I am right, the classical XSUDO construction associated to an exocet should not work.

champagne

[ronk]

If the target is empty, then, solving the "perpendicular rows" of the target, the "perpendicular row" of the last cell in the box containing the base must have no solution.

BTW, if I am right, the classical XSUDO construction associated to an exocet should not work.

Don't understand that 1st sentence, but for GP-H28 after one single:

____

Code: Select all

     13 Truths = {2345R1 2345R5 345R7 89N8}
     16 Links = {4c1 35c3 25c5 34c6 2345c8 5n7 1n9 345b9}
     5 Eliminations --> r2c58<>2, r1c9<>1, r3c8<>2, r4c5<>2

Because of a clue at r1c1, the 2R7 truth is not required, and r8c89=2 implies 2r1c9 = 2r5c7 due to a mere skyscraper.

[David P Bird]

I see no evidence just looking at rows 1;5;7 that 2r7c1 is not valid if the target is not filled

I don't see what your concern is. Only the digits that turn out to be true in the target cells will make Swordfish (degenerate or not). If (2) is false in the base cells and both target cells the Exocet pattern doesn't require (2)r7c1 to be false!

Not if champagne took your "swordfish" requirement literally. This is a degenerate case with only an "x-wing" for digit <2>.

I must be thick tonight as I don't grasp where the problem lies. I can't see that the degenerate case makes any difference. However I'd be happy for you to revise my wording to cover your concerns.

[daj95376]

Well, I'm lost. [Edit: Never mind.]

[ronk]

I'd be happy for you to revise my wording to cover your concerns.

I'm just pointing out that where ...

Each base digit restricted to 2 rows in these cells

... it's not clear that there might be only 1 row. Furthermore, I think the base cells might contain only "ab" and "bc", or even "ab" and "ab".

[David P Bird]

I'd be happy for you to revise my wording to cover your concerns.

I'm just pointing out that where ...

Each base digit restricted to 2 rows in these cells

... it's not clear that there might be only 1 row.

No that can't happen because we know the two target cells will contain different base digits.

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*-------*-------*-------*
| x y . | . . . | . . . | 
| . . . | / . . | / . . |   
| . . . | x . . | y . . | 
*-------*-------*-------* 
| . . S | S . . | S . . | 
| . . S | S . . | S . . |
| . . S | S . . | S . . |   
*-------*-------*-------*  S = Swordfish Cells
| . . S | S . . | S . . |      Each base digit restricted to 2 rows in these cells
| . . S | S . . | S . . |     
| . . S | S . . | S . . |  . = Any combination of candidates
*-------*-------*-------*

Say x & y are the true base digits. Each of them must be true in two of the S cells, so they must be on two different rows.

Furthermore, I think the base cells might contain only "ab" and "bc", or even "ab" and "ab".

(ab) and (bc) in the two target cells is still an Exocet and is OK, while I think (ab) and (ab) is just a trivial case that won’t provide any new exclusions. To screen it out the base cell criteria could be changed to read "Base Cells restricted to candidates from [abc] or [abcd], with at least 3 different candidates in the two cells".

[David P Bird]

This is in champagnes "nothing special" puzzle set
.2.4.......7.8...6.....3.5...9.6...1.....23.....5...4...1...8..6...1...797.......;243;elev;258;11.40;11.40;11.30;884

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┌───────────────────┬───────────────────┬───────────────────┐ 3Fish Candidates
│ 35    2     35 %  │ 4     5  %  5     │ F     3     3  %  │   35   
│ 345   345   F     │ 2           5     │ 24    23    F     │   -
│ 4     4     4  %  │ 2     2  %  3     │ 24    5     24 %  │   24   
├───────────────────┼───────────────────┼───────────────────┤
│ 2345  345   F     │ 3     F     4     │ 25    2     F     │   -   
│ 45    45    45 %  │ F     4  %  2     │ 3     F     5  %  │   45
│ 23    3     23 %  │ 5     3  %  F     │ 2     4     2  %  │   23
├───────────────────┼───────────────────┼───────────────────┤
│ 2345# 345#  F     │ 23    2345  45    │       23    2345  │         
│ F     345   2345  │ 23    \   # 45    │ 245   23    \   # │
│ F     F     2345  │ 23    2345# 45    │ 245   23    2345# │
└───────────────────┴───────────────────┴───────────────────┘
               ^             ^                         ^

Stripped of the non-member digits, it is clearer to see that it has a fully qualified Exocet
(2345)Exocet:r7c12,r9c5,r9c9 but unfortunately there are no exclusions in either of the target cells.
However it proves that this cell distribution with the two target cells on the same row is possible.

[ronk]

No that can't happen because we know the two target cells will contain different base digits.

I'm talking about the rows appearing as cover sets in the deductive logic. Did you even look at my graphic (where it's columns instead of rows) for GP-H28?

[ronk]

Did I misunderstand something?

It seems a lot is being lost in translation here, even between the King's English and American English, so let me interject with a Midwest USA translation.

Whatever change champagne made, presumably to implement David P Bird's suggestion, caused a loss of 42 of 11,454 exocets which were detected before the change. However, that change also caused the detection of 2973 exocets that weren't previously detected. If they're all valid, that's a huge improvement, and now we're just kicking around reasons why those 42 should or should not have been lost.

As to another proposed change, I don't believe champagne has yet allowed the target cells to see each other.

[champagne]

If the target is empty, then, solving the "perpendicular rows" of the target, the "perpendicular row" of the last cell in the box containing the base must have no solution.

BTW, if I am right, the classical XSUDO construction associated to an exocet should not work.

Don't understand that 1st sentence, but for GP-H28 after one single:

____

Code: Select all

     13 Truths = {2345R1 2345R5 345R7 89N8}
     16 Links = {4c1 35c3 25c5 34c6 2345c8 5n7 1n9 345b9}
     5 Eliminations --> r2c58<>2, r1c9<>1, r3c8<>2, r4c5<>2

Because of a clue at r1c1, the 2R7 truth is not required, and r8c89=2 implies 2r1c9 = 2r5c7 due to a mere skyscraper.

the diagram helped me to catch the problem
in fact, in that situation, 2 in the base forces 2 in the target through a bi value created in the rows 1 and 5.
No need to refer to the row 7;

I added one line to the code and the list is now reduced to 5 puzzles

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...4....94....923..8..2...4..6..3...8..59...2.......7.3..9....5..8..21...1...5...;1952;11.20;11.20;10.40;elev;1806;24
98.7..6..5..........4.8..5.3.9.....4.6.8..2......3.....1...6......1...26....781..;14808;10.90;10.90;2.60;GP;kz1a;23
98.7..6..7......9...6.5.....4......3..78..26.......5....96...52..85...7......1...;17207;11.00;11.00;10.60;GP;Kz1 b;23
98.7..6..7..6..5......4..3.6.85..9...2.........1......1...........1...96....571..;9554;11.00;11.00;2.60;GP;cy4;22
98.7..6..7..65.4.......3.2.6.84..9...5...........6....1...........1...96....741..;15122;11.00;1.20;1.20;GP;kz1a;23


champagne

[David P Bird]

No that can't happen because we know the two target cells will contain different base digits.

I'm talking about the rows appearing as cover sets in the deductive logic. Did you even look at my graphic (where it's columns instead of rows) for GP-H28?

You're right I didn't look at your graphic. I clearly relied on the cover sets used in Blues theorem, but you were referring to those in your proof so we were talking about different things.

Now if your collection of cover sets is not case-dependent and can be applied to every hit the latest search pattern will find, then we have a second way of proving the Exocet eliminations theorem. This would be useful as it might show some additional cases where these eliminations are still valid.

champagne, I've checked your 5 puzzles and they have all been properly excluded as they fail the partial Swordfish condition, so well done.

This puts them into the category where the predicted eliminations are good but are not covered by the current theorem and may just be the result of chance (which must happen sometimes). As 5 out of 11454 is very low, I would be happy to assume that.

DPB

[champagne]

champagne, I've checked your 5 puzzles and they have all been properly excluded as they fail the partial Swordfish condition, so well done.

This puts them into the category where the predicted eliminations are good but are not covered by the current theorem and may just be the result of chance (which must happen sometimes). As 5 out of 11454 is very low, I would be happy to assume that.

DPB

let's take it on the positive side with fresh statistics

The specific pattern you have described is seen in 24580 puzzles out of 30580 in the current status of my data base of "potential hardest"
this is 80% of the total (some should think about that ratio)

I run the test on a file of 1.5 million puzzles rating in the range 9.5 to 11.0 (skfr rating) but most of them on the high side

I got a ratio of 79% puzzles having the searched pattern. So with lower ratings, we can still expect to find it with a very high frequency.

On the other way, 4 of the 5 puzzles in the last list have a double exocet.
So solving them with the help of that pattern should not be to hard.

champagne

[David P Bird]

champage Those results are fantastic!

They may be biased though because they are all from the 'hardest' category where a lot of the same design tricks have often been used. I would expect lower figures for the general population.

As I see it, the 5 puzzles match your bi-bi definition, but don't match the Exocet definition. If your code made the Exocet eliminations as soon as the pattern was found there should be a saving in execution time. How big that might be will depend on how often you need to continue with a brute force template analysis for the bi-bi digits anyway.

DPB