The New Sudoku Players' Forum

Website · by **denis_berthier** » Tue Jun 04, 2013 6:32 am

champagne wrote: I found only 75 puzzles having the exocet pattern in the rand 8.6 file

Taking these data as they are:

Code: Select all: Distribution of clues in the 75 list with exocet: nb-clues nb-instances 19 0 20 0 21 0 22 3 23 14 24 19 25 24 26 12 27 2 28 0 29 1 30 0 31 0 32 0 33 0 34 0 35 0

Code: Select all: Distribution of clues in the full list of 282588 puzzles nb-clues nb-instances 19 0 20 2 21 283 22 5814 23 36563 24 87547 25 91557 26 46586 27 12283 28 1794 29 149 30 10 31 0 32 0 33 0 34 0 35 0

Code: Select all: Proportion of puzzles having an exocet: nb-clues proportion 19 0 20 0 21 0 22 3/5814 23 14/36563 24 19/87547 25 24/91557 26 12/46586 27 2/12283 28 0 29 1/149 30 0 31 0 32 0 33 0 34 0 35 0

Results recalled from here: http://forum.enjoysudoku.com/the-real-distribution-of-minimal-puzzles-t30127-8.html

Code: Select all: Weights: #clues %puzzles 20 0.0 21 0.000034 22 0.0034 23 0.149 24 2.28 25 13.42 26 31.94 27 32.74 28 15.48 29 3.56 30 0.41 31 0.022

Application of the formula in the above-mentioned post

Code: Select all: Unbiased proportion of puzzles having an exocet nb-clues weighted-proportion * 100 22 3/5814 * 0.0034 = 1.75438596491228e-06 23 14/36563 * 0.149 = 5.70522112518119e-05 24 19/87547 * 2.28 = 0.000494819925297269 25 24/91557 * 13.42 = 0.00351780857826272 26 12/46586 * 31.94 = 0.00822736444425364 27 2/12283 * 32.74 = 0.00533094520882521 28 0 = 0 29 1/149 * 0.022 = 0.000147651006711409

Estimated unbiased proportion of puzzles having an exocet in the SER > 8.6 area
(+ 1.75438596491228e-06 5.70522112518119e-05 0.000494819925297269 0.00351780857826272 0.00822736444425364 0.00533094520882521 0.000147651006711409)

= 0.017777395760567%
= 0.018%

This is a small proportion (less than 2 in 10,000), although this is more than I expected. The reason is, there are more cases than expected with more than 24 clues.

by **blue** » Tue Jun 04, 2013 7:23 am

Hi Denis & champagne,

We may be getting different results because we don't apply the same list of "simple techniques" before the search -- champagne uses more then I do.

I mentioned this in the other thread -- if sometimes champagne's puzzles end up with a filled base or target cell in one of the JE's that I find, then (of course) he wouldn't find it.

For champagne, these are the 129 that I find in the grey area:

Hidden Text: Show

For Denis, here are the statistics by clue count for JE3/JE4 and JE3/4 with eliminations:

Code: Select all: Green zone: Puzzles JE3 JE4 JE3 JE4 20 8 21 667 22 11495 9 7 5 5 23 71975 83 29 22 7 24 175668 167 66 61 22 25 190921 215 43 71 21 26 100264 104 30 42 13 27 27129 32 9 9 7 28 3985 2 2 29 307 30 19 31 7 32 2 33 3 Grey zone: Puzzles JE3 JE4 JE3 JE4 20 2 21 283 22 5814 3 1 1 1 23 36563 17 5 6 4 24 87547 29 6 10 2 25 91557 39 10 16 6 26 46586 14 2 5 27 12283 2 1 28 1794 29 149 30 10

denis_berthier wrote:
blue wrote:One more note: For both puzzle lists, there were puzzles containing 100-200 exocets (of general type).
This didn't happen for champagne's "potential hardest" list.
For the grey zone puzzles, it was rare: nothing with >= 100 exocets, in the "filtered" tables, 8 with >= 100 in the "all" tables.
For the green zone, it happens often enough to effect the "exocets per 'puzzle with an exocet'" ratio -- 39 and 475 puzzles, respectively.

After elimination of the candidates in direct contradiction with the clues, there remain in general between 200 and 250 candidates. I probably don't understand what "exocet of general type" means, but whatever it means, it seems very strange that there can be almost as many instances of a "pattern" as there are candidates. Obviously, they cannot have independent eliminations. Does your "exocet of general type" rely on indirect contradictions (with more or less unrestricted length) between candidates?

By "exocet of general type", I meant exocets the way champagne defined them (including JE).
I use single digit templates in the search, requiring that every viable template for a base digit, either doesn't contain a base cell, or (does, and also) does contain a target cell (at least one).
[ I think that corresponds with champagne's definition. ]

The puzzles with a large number number of exocets, seem to be solvable using single digit templates.
After applying the "simple solving techniques", they seem to have a small number of empty cells, and a small number of candidates and digits with candidates.

Here is an example with 220 exocets with (standard) eliminations.

Code: Select all: 5...6.......7...4..392.1.5.2.1..6.......2.4........2.6.5.....37..6.17.....2.3.... ED=7.6/1.2/1.2 +-------------+-----------------+-------------+ | 5 4 7 | 389 6 389 | 39 1 2 | | 1 2 8 | 7 59 359 | 6 4 39 | | 6 3 9 | 2 4 1 | 7 5 8 | +-------------+-----------------+-------------+ | 2 789 1 | 4 789 6 | 35 89 35 | | 89 6 35 | 389 2 3589 | 4 7 1 | | 4 789 35 | 1 5789 3589 | 2 89 6 | +-------------+-----------------+-------------+ | 89 5 4 | 6 89 2 | 1 3 7 | | 3 89 6 | 589 1 7 | 589 2 4 | | 7 1 2 | 589 3 4 | 589 6 59 | +-------------+-----------------+-------------+ 73 "live" candidates

The 220 exocets (with types):

Hidden Text: Show

126 of them are 2-digit exocets.

Regards,
Blue.

Website · by **denis_berthier** » Tue Jun 04, 2013 7:57 am

blue wrote:Here is an example with 220 exocets with (standard) eliminations.
Code: Select all
5...6.......7...4..392.1.5.2.1..6.......2.4........2.6.5.....37..6.17.....2.3.... ED=7.6/1.2/1.2

[...]
The puzzles with a large number number of exocets, seem to be solvable using single digit templates.

Starting from your PM:

Code: Select all: whip[2]: c1n8{r5 r7} - b8n8{r7c5 .} ==> r5c4 <> 8 whip[2]: r5n8{c1 c6} - c5n8{r6 .} ==> r7c1 <> 8 singles to the end

This illustrates what you say about 1-digit templates.

Generally speaking, I don't think exocets will have much interest in the "green zone". Of course, all depends on what rules one chooses to use before them.

A short comment about the stats with/without eliminations: they give a rough idea of how useful it is to make a difference between useful instances and useless ones.

by **champagne** » Tue Jun 04, 2013 8:15 am

blue wrote:For champagne, these are the 129 that I find in the grey area:

I tried a quick match "by eyes", but it seems that the appearance of JE's is very sensitive to the level of moves at the start.

each of the 2 lists has puzzles not in the other one.

this requires some investigations to do later

I work currently on a similar task which is to filter trivial puzzles in the sample I made in the vicinity of the hardest.

by **eleven** » Tue Jun 04, 2013 8:28 am

Thanks for the analysis, champagne, blue and Denis.
Your results are much better than i had expected to get out of my puzzle sets.
I am short of time now, but as soon as i can, i will have a closer look at the details of blues' stats.

by **blue** » Tue Jun 04, 2013 8:40 am

Hi Denis,

denis_berthier wrote:
Code: Select all
Weights: #clues %puzzles 20 0.0 21 0.000034 22 0.0034 23 0.149 24 2.28 25 13.42 26 31.94 27 32.74 28 15.48 29 3.56 30 0.41 31 0.022

Application of the formula in the above-mentioned post

Code: Select all
Unbiased proportion of puzzles having an exocet nb-clues weighted-proportion * 100 22 3/5814 * 0.0034 = 1.75438596491228e-06 23 14/36563 * 0.149 = 5.70522112518119e-05 24 19/87547 * 2.28 = 0.000494819925297269 25 24/91557 * 13.42 = 0.00351780857826272 26 12/46586 * 31.94 = 0.00822736444425364 27 2/12283 * 32.74 = 0.00533094520882521 28 0 = 0 29 1/149 * 0.022 = 0.000147651006711409

It seems like you should be using a different set of weights, specifically for grey zone puzzles.
Weights from a table built like this:

Code: Select all: Weights: #clues %puzzles Prob(in grey zone) product %grey_zone_puzzles 20 0.0 P20 21 0.000034 P21 22 0.0034 P22 23 0.149 P23 24 2.28 P24 25 13.42 P25 X25=13.42*P25 100*X25/(X20+X21+...+X31) 26 31.94 P26 (etc.) (etc.) 27 32.74 P27 28 15.48 P28 29 3.56 P29 30 0.41 P30 31 0.022 P31

I may be wrong, but it doesn't seem like the probabilities (P20-P31) would be the same (or even roughly the same) for each clue count. You probably have the right kind of data to compute the values -- for each clue count, a truly random selection of puzzles.
[ "Probabliity", maybe isn't the right word. "Percentage" or "fraction" would be better. It's the same number in any case. ]

Regards,
Blue.

Website · by **denis_berthier** » Tue Jun 04, 2013 9:23 am

Hi Blue,

blue wrote:It seems like you should be using a different set of weights, specifically for grey zone puzzles.

In theory, you are right.
However (and this is one thing I should have recalled in the other thread), the correlation coefficient between SER (or W rating) and #clues is very small (0.12), so I don't think the weights depend much on the SER (at least for non-extreme numbers of clues and non-extreme SER - but this is precisely what we have). In any case, it can't change the order of magnitude of the result.
Considering that there are many other causes of uncertainty or debate about what should be taken into account (exact definitions, rules applied before, my "grey zone" vs champagne's broader "grey area", ...), I think more detailed calculations are not necessary.

by **JC Van Hay** » Tue Jun 04, 2013 10:07 am

FWIW, here is a solution of a puzzle studied by Leren ...

98.7.......7.65.........7..4...3..2..1......9..95..8..1......4...59..6.......2..3; ;2;0;match type;50;r1c5 r1c6 r3c3 r2c7 1234;r2c1 r2c2 r3c4 r1c7 234;;;4;;

#1. SER=9.7; UP22; r4c6=9=r3c5; LC(5B1) :=> -5r3c89; UP24
#2. ALS(234)r2c12 :

Code: Select all: +------------------------+-----------------------+-----------------------+ | 9 8 146(23) | 7 124 134 | 145(23) 1356 12456 | | (23) 4-23 7 | 148(23) 6 5 | 149(23) 1389 1248 | | 2356 23456 146(23) | 148(23) 9 1348 | 7 1368 12468 | +------------------------+-----------------------+-----------------------+ | 4 567 68 | 168 3 9 | 15 2 1567 | | 235678 1 68(23) | 468(2) 2478 4678 | 45(3) 3567 9 | | 2367 2367 9 | 5 1247 1467 | 8 1367 1467 | +------------------------+-----------------------+-----------------------+ | 1 23679 68(23) | 68(3) 578 3678 | 59(2) 4 2578 | | 2378 2347 5 | 9 1478 13478 | 6 178 1278 | | 678 4679 468 | 1468 14578 2 | 159 15789 3 | +------------------------+-----------------------+-----------------------+

a. r2c12=23->HP(23)r57c3; FSF(2C437)+FSF(3C734) has no solution :=> r2c2=4; r9c3=4; UP25
b. r2c12=24->FSF(2C437) :=> r3c4=2; Contradiction(Singles) :=> r2c12=34
c. FSF(3C437)+XW(4C47) :=> r1c56=12; UP81

by **David P Bird** » Tue Jun 04, 2013 10:37 am

Here's my current draft for specifying the eliminations from a Double Exocet with 4 targets. In Lerens example I made such a mess of, the unconditional eliminations would kill the puzzle in one shot. It looks like a good one to use as an example.

Double JEs

Depending on the positions of the base cells and their corresponding targets there may be either 3 or 4 target cells.

When there are 4 target cells

Code: Select all: *-------------*-------------*-------------* ab" = base digits r1c12 ab' = targets | ab" ab" \ | cdx cd' cdx | * * * | cd" = base digits r3c89 cd' = targets | cdx cdx cd' | * \ * | ab' abx abx | x = any other digits | * * * | abx ab' abx | \ cd" cd" | \ = companion cells *-------------*-------------*-------------* * = abcd can be eliminated

This is how the two pairs of base digits are forced to repeat.

As with the single JE, the target cells (ab)r2c6,r34 and (cd)r1c5,r2c3 must contain different digits.

1. Unconditional Eliminations
a) Any base digit in the 6 cells seen by both base pairs and 2 two cells seen by all 4 targets
b) Any non-base digit in the 4 target cells
c) Fin cells for the partial fish for each of the 4 base digits

2. Conditional Eliminations
a) If one of the target cells is known, the same digit can be eliminated from the other 3 target cells.
b) If any of the cells containing 'x' digits are known, the non-base digits in the same mini-line can be eliminated.
c) If one of the 'x' digits is locked in a mini-line the other 'x' digits in the mini-line can be eliminated

3. Pattern Inferences
a) Instances of the same digit in each pair of target cells are weakly linked
b) The non base candidates in the mini-lines containing a target cell form a weak inference set
c) A digit confined to a mini-line containing a target cell will make a hidden triple with the true base digits in that target.

4 Proof
1) The digits occupying the two pairs of base cells must also occupy a target cell to bring the number of truths in the partial fish columns to 3.
2) But if a digit is true on both pairs of base cells, it can't be true in any of the target cells as each base pair see the target cells for the other.
3) Consequently no digit can be common to both base pairs and, together, they must hold 4 different digits.
4) From 1) this in turn will require the 4 target cells to hold one instance of each of the base digits.

by **champagne** » Tue Jun 04, 2013 5:28 pm

to go in the direction of a better efficiency of JEs found in the green area, I added three filters

1) no analysis if the number of unknown go below 50
2) no JE accepted if one of the digits of the base is known in the band
3) no JE without elimination if it is a single JE (target 2 cells)

I did not retain David's request for point 3 (no JE without elimination) because it killed some double exocet patterns.
The new run cleared 87% of the JE found in the first run in the green area.

I have now 0.5% of selected puzzles against 3.78% and I reached in the green area a volume authorizing the uploading of the full lot of puzzles eligible to the JE in that area.

I am checking the status for other kind of exocets (excluding the last findings of blues) to decide of the best way to share that.

More filters can be studied on the reduced file

by **Leren** » Tue Jun 04, 2013 10:00 pm

An improved solution for puzzle 98.7.......7.65.........7..4...3..2..1......9..95..8..1......4...59..6.......2..3 is as follows:

Code: Select all: *--------------------------------------------------------------------------------* | 9 8 12346 | 7 124 134 | 12345 1356 12456 | | 23 234 7 | 12348 6 5 | 234-19 1389 1248 | | 235-6 2345-6 1234-6 | 12348 9 1348 | 7 1368 12468 | |--------------------------+--------------------------+--------------------------|

Exocet 1 : r1c5 r1c6 r3c3 r2c7 1234;

Eliminate non-base candidates in Targets => - 9 r2c7, - 6 r3c3;

Tertiary Equivalence r2c7==r3c12(5), r3c3 => - 6 r3c12, - 1 r2c7; lclste

Leren

by **David P Bird** » Wed Jun 05, 2013 10:14 pm

While it's gone quiet here...

I've been exploring the possible configurations of the target cells in different types of single and double JEs with paired target cells in different boxes. The paired targets will either be collinear (one the same line) or diagonal (on different lines). We can therefore categorise single and double JEs by the number of target cells they have and whether these are diagonal (D) or collinear (C) for each pair of base cells.

Code: Select all: *-------*-------*-------* *-------*-------*-------* | B B . | . . . | . . . | | B B . | . . . | . . . | B, b = base cells | . . . | . \ . | T . . | | . . . | . T . | T . . | T, t = target cells | . . . | . T . | \ . . | | . . . | . \ . | \ . . | \ = companion cells *-------*-------*-------* *-------*-------*-------* JE2D JE2C *-------*-------*-------* *-------*-------*-------* *-------*-------*-------* | B B t | . \ . | . . . | | B B \ | . \ . | . . . | | B B \ | . \ . | . . . | | . . \ | . T'. | \ . . | | . . t | . T'. | \ . . | | . . t | . T'. | T . . | | . . . | . \ . | T b b | | . . . | . \ . | T b b | | . . . | . \ . | \ b b | *-------*-------*-------* *-------*-------*-------* *-------*-------*-------* JE3DD JE3DC JE3CC *-------*-------*-------* *-------*-------*-------* *-------*-------*-------* | B B \ | . t . | . . . | | B B t | . t . | . . . | | B B t | . t . | . . . | | . . t | . \ . | T . . | | . . \ | . \ . | T . . | | . . \ | . \ . | \ . . | | . . . | . T . | \ b b | | . . . | . T . | \ b b | | . . . | . T . | T b b | *-------*-------*-------* *-------*-------*-------* *-------*-------*-------* JE4DD JE4DC JE4CC *-------*-------*-------* *-------*-------*-------* *-------*-------*-------* | B B t | . . \ | . . . | | B B \ | . . \ | . . . | | B B \ | . . \ | . . . | | . . \ | \ . t | T . . | | . . t | \ . t | T . . | | . . t | T . t | T . . | | . . . | T . . | \ b b | | . . . | T . . | \ b b | | . . . | \ . . | \ b b | *-------*-------*-------* *-------*-------*-------* *-------*-------*-------* JE4DD JE4DC JE4CC *-------*-------*-------* *-------*-------*-------* *-------*-------*-------* | B B t | . . \ | . . . | | B B t | . . t | . . . | | B B t | . . t | . . . | | . . \ | T . t | \ . . | | . . \ | T . \ | \ . . | | . . \ | \ . \ | \ . . | | . . . | \ . . | T b b | | . . . | \ . . | T b b | | . . . | T . . | T b b | *-------*-------*-------* *-------*-------*-------* *-------*-------*-------* JE4DD JE4DC JE4CC

For JE+ or Twin JExocet patterns the D or C suffix won't be known as there is uncertainty which object cell in a pair will be the target and which will be the companion. In these cases a T suffix should be used.

I'm using these grids to check what inferences that will be available under different conditions. In the case of the Twin JExocets some inferences won't be available until the identity of the target cell is established which is easy enough to describe. So far I haven't looked at cases where the two targets associate with a base pair are in the same box because I'd like to see if they survive the screening regime Champagne's now using. I have an intuition that they may be rather trivial to solve and might not get through.

I suggest that this categorisation system should be adopted in place of the one based on the number of digits in the base sets which is proving troublesome. It allows the available inferences to be identified quite neatly and should go a long way to satisfying Denis' requirements.

by **champagne** » Thu Jun 06, 2013 12:13 pm

David P Bird wrote:While it's gone quiet here...

I've been exploring the possible configurations of the target cells in different types of single and double JEs with paired target cells in different boxes.
Code: Select all
*-------*-------*-------* *-------*-------*-------* *-------*-------*-------* | B B \ | . t . | . . . | | B B t | . t . | . . . | | B B t | . t . | . . . | | . . t | . \ . | T . . | | . . \ | . \ . | T . . | | . . \ | . \ . | \ . . | | . . . | . T . | \ b b | | . . . | . T . | \ b b | | . . . | . T . | T b b | *-------*-------*-------* *-------*-------*-------* *-------*-------*-------* JE4DD JE4DC JE4CC ...

Hi David,

as often, we have slightly different views on the same facts.

I have to revise seriously my code establishing the double exocet pattern, so I studied yours patterns.

To have a wording very close to what I can put in code, I would write that in the following way.

A) JE4 + JE 4 (same four digits, same band, 2 different boxes for the base)

- if one base see the 2 targets of the other JE, then the double exocet is established
this cover all your patterns except JE4CC and JE4DD

-if all targets share the same unit, then the double exocet is established (JE4CC)
-if each base see a target of the other exocet and the 2 other targets share the same unit then the double exocet is established (JE4DD)

B) JE3 + JE3

here, none base can see the 2 other targets, the puzzle would have no solution
and for the same reason, the four targets can not share the same unit.
At the end, I strongly believe that one target must be common to both exocets. That target ins in the thir line and in the third row as in you diagrams.

- it is then enough that one base sees one target of the other exocet to have the double exocet established.

In that case, the 2 exocets share one digit and have one of the 2 others digits as second digit in base
All your diagrams fill that condition.

I see a small window with the 3 target in a mini row but I don't know if it is realistic.

C) JE4 + JE3

we have seen that in a former post.
The rules to apply to the 3 digits base are the same as for a 2*JE4 ;
I'll study several cases to have the relevant rules

by **David P Bird** » Thu Jun 06, 2013 1:24 pm

Hi Champagne,

With these double JEs there will be four base cells with either 3 or 4 target cells between them.
When there are three target cells we know that one digit will be true in both base sets so there are three true digits in total and the common one will occupy the shared target cell.
When there are four target cells, there will be four true digits.

In my view it is far better to identify the number of true digits in the 4 base cells than the number of candidates in the two base sets.

A general proof that no two target cells can hold the same digit for the JE3 and JE4 patterns:
1) It has already been proved that the target cells belonging to a base set must hold different digits
2) The target cells in the same boxes as a base set must hold a true base digit in the other base set.
3) The second true member in each base set must therefore be true in a target cell in the third box.
4) If there is a single target cell in this box, it must hold a digit that is common to both base sets, otherwise two targets must hold different digits that are true in their different base sets.
5) Each target cell must therefore hold a different digit, and together these cells will hold the same combination of digits as the full set of base cells.

This means that as soon as a digit is known to be true in one or other of the base sets it can be eliminated from the cells in sight of all the base cells or all the target cells, and can also be eliminated from the fin cells for its partial fish.

So, I think the way you are accepting double JEs even if there are no eliminations in the target cells is therefore right - these other eliminations might still exist.

I too was interested in the target cells for set A that could be seen by the base cells for Set B. This is why I had to show three different ways each of the JE4 patterns could have their targets positioned. From that I produced the general proof, which shows that it's not important!

Your points don't seem to agree with my findings, but maybe that’s because of language issues. If you work though my proof and then go back to your findings you may see what I mean.

David

by **champagne** » Thu Jun 06, 2013 2:08 pm

David P Bird wrote:
A general proof that no two target cells can hold the same digit for the JE3 and JE4 patterns:
1) It has already been proved that the target cells belonging to a base set must hold different digits
2) The target cells in the same boxes as a base set must hold a true base digit in the other base set.
3) The second true member in each base set must therefore be true in a target cell in the third box.
4) If there is a single target cell in this box, it must hold a digit that is common to both base sets, otherwise two targets must hold different digits that are true in their different base sets.
5) Each target cell must therefore hold a different digit, and together these cells will hold the same combination of digits as the full set of base cells.

The proof works well for JEs, and this is why I had no problem in the solving part of my program with JE's seen as double, although I made no checking.
The point 3) of your proof is not granted for exocets "not JEs" in the same band and that was part of my concern.

BTW, there is no contradiction between having a list of patterns or a list of rules. But I accept your point that with JEs it always work.

What about twin JEs?? I did not yet look at that pattern. Did you make a similar analysis

(Normally, I can apply the same rules but...)

meantime I thought a little more about JE3 + JE4. It works exactly as 2*JE4.

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