The New Sudoku Players' Forum

by **champagne** » Sat Jan 22, 2022 7:21 am

coloin wrote:Jus wondering what progress if any has been made !
Generating ever reducing yields with random search [it seems it needs to be better than a +2 to find some 18C] has made it difficult to see how it could ever be completed.

However... ... been reading through this thread and maybe some /many solution grids can be excluded from the search

On page 4
blue wrote:There are 983,959,110 ED double bands.
The 44 classes that you have in mind, can be partitioned into 913,393 smaller classes too

The proposal that I have ... is that we have this very incomplete table
Double bands have got 54 clues and they have been shown to be only 4 which complete in 7 clues ....[code]

Hi coloin,

Just in case this could help

In the 17 search, we are considering ED bands 12 "minimal against the diagonal view"

This reduce the number of minimal in my case to 610 163 364 but using the 983,959,110 ED double bands would not be a problem

It happens that trying to produce an efficient code for the 18 search for the "mathimagics" project, I have somehow to produce all low count double band valid puzzles.

If this is very important for you, this code can easily (as soon as the validation tests will be done) be modified to answer to your point.

My rough estimate is that the process would take less than one second per double band (may be around 10 double bands per second in the best case).

by **champagne** » Sat Jan 22, 2022 3:38 pm

coloin wrote:
blue did find a 12C in the MCgrid ... so indeed he has the methods !!
blue wrote:The MC grid has 6+6+27's:
Code: Select all
12...6......7....3.......5....5...9..64......8....1...312645978645978312978312645

which was a shame !

As I follow blues findings, I can see this one among a long list of potential 12 hitting all stored uas(and in a quick overview of the list 12 seems to be the minimum)
I did not make the final check of the validity of these 12. I can do it if you are interested and put them in the appropriate format (sub puzzle 54 cells)

EDIT here the first of a long list
..3.5....4....9.2....1.........6...7...8......9...15..312645978645978312978312645

by **champagne** » Sat Jan 22, 2022 4:21 pm

coloin wrote:This is the one to look at I think
Code: Select all
The first grid without a 21 [Grid#1] 123456789456789123789123456214897365365214897897365214541632978632978541978541632

Has it got a DB in 12 ?

As the MC grid, this one has 3 times band 1 (in 1-416)
I get similar result with the first in my list

....5...9..67...2.......4..2.......5....148...9.......541632978632978541978541632 (typo edit)

some more
...45......6....2.7....3...2....7.6...5.1.8...........541632978632978541978541632
....56.........1237........2....7.6...5.1.8...........541632978632978541978541632

by **champagne** » Sat Jan 22, 2022 4:45 pm

coloin wrote:Thanks for doing that ! ....
But it has scuppered it somewhat
Actually ........ you need to test all 6 DB

hi coloin,
don't forget my poor knowledge of your language.
I can misunderstand your last sentence.
If my code is correct, this is part of 1392 possible 12 :roll:

by **champagne** » Sat Jan 22, 2022 5:01 pm

coloin wrote:
Actually ........ you need to test all 6 DB

Assuming I understand your point, some comments on the process.

This is part of the 18 clue search.
the first part of the process is to produce all valid 2 bands sub puzzles.

To do this, a long list of UAs is first established, trying to select uas of value (here 294 uas for bands 1+2)
then this list of uas is expanded up to 12 clues (for a 6+6+6 18 split of clues)
in most cases, the potential sub puzzles produced here are valid.

and in this case, the entire search takes less than half a second

by **champagne** » Sat Jan 22, 2022 5:14 pm

coloin wrote:Actually I see that this grid is highly automorphic and is referred to as blue1, so it may well be that you have shown that all the DB have a 12-completion.
Your language is pretty good ... Thanks for your help

another remark

6 of the 416 bands have a minimum of 6 clues in band. (1;224;237;254;307;413)
Any pair of these bands has a minimum number of clues of 12.
the best chance to pass 12 clues as minimum for 2 bands is with such a pair.

note : the lowest number of cases to consider is with the pair 224;237 (162 x 606 possibilities). likely the best chance to pass 12

by **champagne** » Sat Jan 22, 2022 6:49 pm

coloin wrote:
Previously published details on the 4 grids
224,224,224,254,254,413 0, 0, 0, 0, 0, 0 # 6
224,224,224,225,225,414 0, 0, 0, 90, 90, 324 # 2
123456789457189326689327154216534897745891632938672541361245978574918263892763415 : blue3 puz20=0 puz21= 3894
123456789457189326689327154216534897745891632938672541392765418574918263861243975 : blue4 puz20=0 puz21=19682

Sorry first for all the trouble in the band Id .; Working on index, I have from time to time to adjust the number by +-1. I'll did it badly. I'll adjust the posts.

I was interested in checking the band "224" having the lowest count of "6 clues".

162 *162 remains a big number. I have less valids out of these 162*162 , but still 576 about 2/100 of the raw count

by **champagne** » Sun Jan 23, 2022 10:06 am

coloin wrote:Yes they are all big numbers .....
I an getting pessimistic to find a DB13 now, although it wouldn't be feasible to check all the potential DB.....
eg all 15 band pairs from (1;224;237;254;307;413 )

Yes, in theory it's not to hard, assuming that the 13 is not in another pair (a much lower probability)
I have the code to produce all ED pairs, just missing time and cores to do it.
The number of pairs containing only the six bands is not that big,

BTW, the 900 million pairs have a minimum number of clues in the range 8-12. (forgetting the case 7) and likely a small number of pairs requiring 12 clues.
the pic must be in 9-10 with hundred millions of pairs

by **champagne** » Sun Jan 23, 2022 10:27 pm

coloin wrote:I think you have a good appreciation of the task - there is almost certainly DB which require at least 8 [4+4] clues but actually need 9.

983 Million DB ... how many pairs both with 6 clues needed.....

416 x 415 / 2 combinations = 86,320 pairs each equally likely perhaps [?] and we have 15 pairs both with 6 clues required

983 M * 15 / 86,300 = 170,000 pairs

Hi coloin,

The rough count is likely correct.
As I wrote in another post, if we just consider ED pairs, the total is closer to 620 millions pairs.

Anyway, I have the code to produce any of the options. I have just to clean the inline producer of the bands for the search 17.

this could be done in "say" one week if I put it on the top of the to do file.

Champagne

by **champagne** » Sun Jan 23, 2022 10:34 pm

coloin wrote:Addendum ... maybe its just not possible to put 13 UAs in 54 clues [cf the above thread] . Certainly not possible to put 13 disjoint U4s .

13 disjoint uas minimum 4x13=52 surely not possible.

13 UAs in 54 clues: I work with hundreds sometimes more than 1000 uas depending on the pair, this with a strict limitation in the UAs size (around 20 clues at maximum). Could you tell more about this

Note : the band set of UAs to get all valid one band solutions can reach 81 UAs (from memory) this is for band 29. Generally, the set of UAs for one band is around 30

by **champagne** » Sun Jan 23, 2022 11:41 pm

coloin wrote:
champagne wrote:....
In the 17 search, we are considering ED bands 12 "minimal against the diagonal view"
This reduces the number of minimal in my case to 610 163 364 but using the 983,959,110 ED double bands would not be a problem....
"
Yes I didn't quite understand the .... " minimal against the diagonal view" .... but finally the penny dropped !!!
It must mean a diagonal reflection in a DB - which maintains equivalence - but the transformation is difficult to visualise !
I think that is a new one for most of us ... but well worth it !!!!

Sorry if I did not explain it in the right way.
And I am not so sure that we don't have to use the 983,959,110 ED double bands in your exercise.

In the 17 search, we want to avoid redundancy, so, to make it simple, we process the 5 472 730 538 ED solution grids.
To catch these ED solution grids, we don't process bands having a lower morph in the canonicalization process. 90° rotation or main diagonal symmetry must be used to cover all morphs
Thinking a little more about that, I think that you would like to have all pairs, not only pairs relevant in the catalog of ED solution grids.

by **champagne** » Mon Jan 24, 2022 6:56 am

hi coloin,
more about the bands requiring 6 clues.

In my 17 search, these bands have index 0 to 5. The reason is that the bands 1 are ordered to have the lowest count of valid 6 clues (as in mladen table) in the solution grid.

These six bands ordered 224, 237, 1, 413, 254, 307 have auto morphs respectively 5,1,107,35,5,5.
If as I think, the average auto morph is somewhere between 1 and 2, we should find less than average pairs.

to give an idea, in this area, a band 1 has more than 5Million bands 2 attached in the process with no auto morph.
band 1 (index2) has only 52 561
band 224 (index 0) 912 650 (5 auto morphs)
band 224(index 1) 2 680 806 (1 auto morph)

The number of pairs out of the 983,959,110 ED double bands could be below 100 000.

by **champagne** » Mon Jan 24, 2022 1:45 pm

coloin wrote:Ok ...thats good they all need 12 clues ... and each DB expands to give approx 40 ED solution grids ...

maybe in those 100K DB ... there will be those which have a "ganster44reverse" pattern which isnt solvable in 6 clues ......

there are only 44 "gangster44reverse equivalents" or DBgangsters perhaps is better

Hi coloin

if I got your point, this should be a 17 distribution 6+6+5

here is the list of known for the corresponding band 1 6 clues

.23.........1....6.9..7........9....5....2...8......71.4.....9.....3.2.....6.....;1;63;354;5
.23.........1....6.9..7........9....5...2....8......51.4.....9.....3.2.....5.....;1;137;330;5
1.......9...7...2..8...3..........7.5....1...6...9...8...............5.1.726.....;2;180;340;5
12...6...........3......45..9.5.......7.....1...34....3............97..2..5......;2;204;292;5
12..........78...3..9......5...4.9..6....1...........8.......1....8.2....7.....6.;2;206;209;5
1...5......6....2......34.......7.6.39...8...5..............3....126............8;2;231;410;5
1...5......6....2......34.........6734...8...9.......................3.8..126....;2;294;347;5
1...........8..6.2.9...7...2.8......6............34.7....21..5..4.....3..........;3;329;372;5
.234...........6.........512...1.........89...4.....3.............3...4.8.5..1...;4;31;370;5
12...........8.6.3...7.......8....7...4.3.........2.1......1...........8.46..5...;4;320;334;5
1....6.8...7.....3....2....26...1......9..5.7...........57......3.....6.....9....;5;23;52;5
12...6...............3..45..1......6..45...7.....9......5....9......1..27........;5;122;318;5
12...6...............3..45...45......6..9...2.......7......2..6..5....9.7........;5;160;318;5
..3.....9...1...2..8...7......5......126...........3.5....9.8..6......1......3...;5;172;343;5

here index 0-5 as first band
none of them has the second band with an index 0-5

by **champagne** » Mon Jan 24, 2022 5:34 pm

coloin wrote:maybe this explains .....

but maybe you knew this already !!!!!

Code: Select all
513624789984715326276389154425136897791458632638297541142563978857941263369872415 193 289 284 , 307 307 413

This grid solution cant have a 17

I don't know why

but as you got it, from the list above where we have no 5,5 I agree, it can not have one.

by **champagne** » Tue Jan 25, 2022 10:45 am

Hi Coloin,

more on this topic.

First of all, I can produce (reliable??) a list expected exhaustive of solution ED grids requiring a minimum of 12 clues just due to the fact that each band requires a minimum of 6 clues.
The best chance to have no valid sub puzzle with 12 clues in is this list where both bands have the minimum count of valid puzzles (band 224 index 0 count = 162)

Hidden Text: Show

in pairs with one or more band requiring less than 6 clues, the number of potential valid bands 1+2 increases sharply, and the chances for none of these to be a valid goes down in the same way.

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