not much hope to find more fruit on the 9*9-tree

but maybe the method is useful for 16*16

Presumably it had been done for Latin Squares ?!

rookeries = k-plexes

not much hope to find more fruit on the 9*9-tree

but maybe the method is useful for 16*16

Presumably it had been done for Latin Squares ?!

rookeries = k-plexes

but maybe the method is useful for 16*16

Presumably it had been done for Latin Squares ?!

rookeries = k-plexes

- dukuso
**Posts:**479**Joined:**25 June 2005

dukuso wrote:not much hope to find more fruit on the 9*9-tree

but maybe the method is useful for 16*16

Presumably it had been done for Latin Squares ?!

rookeries = k-plexes

well I do have one more hope for the 9*9 17s:

that we can prove that all have been found

- gsf
- 2014 Supporter
**Posts:**7306**Joined:**21 September 2005**Location:**NJ USA

Nice idea.

But as i understand it, this method would not find all 17's (only those, where 10 givens are from 3 dgits).

But as i understand it, this method would not find all 17's (only those, where 10 givens are from 3 dgits).

Last edited by eleven on Tue Jun 01, 2010 4:19 pm, edited 1 time in total.

- eleven
**Posts:**1948**Joined:**10 February 2008

eleven wrote:Nice idea.

But as i understand it, this method would not find all 17's (not those with less than 5 digits, which only have 1 given).

I figure it would be a divide and conquer situation

where we could assert: all 17s of this type have been counted up to isomorphism, now on to the next type ...

- gsf
- 2014 Supporter
**Posts:**7306**Joined:**21 September 2005**Location:**NJ USA

Ok, the last type would be 2 digits, which solve a 3-rookery, and placing 15 out of 6 numbers in the 6-rookery.

There is much between ...

There is much between ...

- eleven
**Posts:**1948**Joined:**10 February 2008

eleven wrote:Ok, the last type would be 2 digits, which solve a 3-rookery, and placing 15 out of 6 numbers in the 6-rookery.

There is much between ...

Not really. Once you've done 7 clues in a 6-rookery, then repeat it for 8, 9 10 and 11 clues in a 6-rookery and I think we have them all. For the later searches you can limit the search to puzzles with specific digit distributions, for example the last stage only needs to look for puzzles with the distribution 222222221.

RW

- RW
- 2010 Supporter
**Posts:**1000**Joined:**16 March 2006

You are right with the clue numbers "between", but the cases with more clues in the 6-rookery would need a multiple of the computation time.

Though dukuso and gsf always had surprising fast implementations, i dont think, that all could be calculated in feasible time.

Though dukuso and gsf always had surprising fast implementations, i dont think, that all could be calculated in feasible time.

- eleven
**Posts:**1948**Joined:**10 February 2008

Sure, it would take an enormous amount of time. But if gsf sees it plausible to find all puzzles with 7 clues in a 6-rookery, I say go for it! Perhaps that inspires some new ideas that make the rest of the puzzles easier to find...

If we believe these numbers are true for all 17s, specifically if we believe that a 5-rookery cannot be solved in 4 clues (which I seem to remember is proven already), then the 7 clues in the 6-rookery must have one of the following digit distributions:

011122

111112

If we also assume that a 7-rookery cannot be solved in less than 10 clues, then the remaining 3 digits must have the distribution 334. This should narrow the search quite a bit.

RW

dukuso wrote:checking gordon's 17s, the minimum numbers of clues required to uniquely

solve a 1,2,..,9 rookery are 0,1,2,3,5,7,10,13,17

If we believe these numbers are true for all 17s, specifically if we believe that a 5-rookery cannot be solved in 4 clues (which I seem to remember is proven already), then the 7 clues in the 6-rookery must have one of the following digit distributions:

011122

111112

If we also assume that a 7-rookery cannot be solved in less than 10 clues, then the remaining 3 digits must have the distribution 334. This should narrow the search quite a bit.

RW

- RW
- 2010 Supporter
**Posts:**1000**Joined:**16 March 2006

Hi,

I traversed the neighbours of the SF grid (the solution grids which differ from it by only one UA set), then checked for a grids having known 17's. Result is posted here.

MD

I traversed the neighbours of the SF grid (the solution grids which differ from it by only one UA set), then checked for a grids having known 17's. Result is posted here.

MD

- dobrichev
- 2016 Supporter
**Posts:**1624**Joined:**24 May 2010

do we have a list of all sudokugrids (upto isomorphism, S-classes) with nontrivial

automorphism-group ?

automorphism-group ?

- dukuso
**Posts:**479**Joined:**25 June 2005

dukuso wrote:do we have a list of all sudokugrids (upto isomorphism, S-classes) with nontrivial

automorphism-group ?

this reminds me that I promised to show you some of the sudz-specific options in my solver

still planning to do that ...

these two data files are posted

300-416.sudz

a2.sudz

300-416.sudz are the essentially different grids from bands 300 through 416 inclusive (2,097,068 grids in lexicographic order)

it was generated by

- Code: Select all
`sudoku -gb300,416 -f%#ec > 300-416.sudz`

a2.sudz is the list of all essentially different grids with #automorphisms > 1 (560,151 grids in lexicographic order)

it was generated in a directory containing all .sudz files by

- Code: Select all
`sudoku -e '(%#An)>1' -f%#ec *.sudz > a2.sudz`

to list all essentially different grids with #automorphisms > 1 with band-index and #automorphisms comment

- Code: Select all
`sudoku -f'%v # %03#Bn %#An' a2.sudz`

as a check, these two commands should produce the same output (5,203 lines)

- Code: Select all
`sudoku -f'%v # %03#Bn %#An' -e '(%#An)>1' 300-416.sudz`

sudoku -f'%v # %03#Bn %#An' a2.sudz | grep '# [34]'

Last edited by gsf on Thu Jun 03, 2010 9:04 pm, edited 1 time in total.

- gsf
- 2014 Supporter
**Posts:**7306**Joined:**21 September 2005**Location:**NJ USA

OK, thanks. 1.9MB only , now decompress ...(how ?)

and then determine the six 416-gangsters in each and the G-class(es) of each

and then determine the six 416-gangsters in each and the G-class(es) of each

- dukuso
**Posts:**479**Joined:**25 June 2005

dukuso wrote:OK, thanks. 1.9MB only , now decompress ...(how ?)

and then determine the six 416-gangsters in each and the G-class(es) of each

decompress using my solver

example command in previous post

this command is similar but also lists the sorted band signature

- Code: Select all
`sudoku -f'%v # %03#Bn %3#An %#bc' a2.sudz`

use %#Bc for the unsorted signature

don't have G-class code in my solver

- gsf
- 2014 Supporter
**Posts:**7306**Joined:**21 September 2005**Location:**NJ USA

sudoku: %O3#Bn: cannot read

sudoku: %3#An: cannot read

sudoku: %#bc': cannot read

sudoku: %3#An: cannot read

sudoku: %#bc': cannot read

- dukuso
**Posts:**479**Joined:**25 June 2005

how many of the ~5.5e9 S-classes have 6(5,..,1) different 416-gangsters in the 6 chutes ?

(~95% of random G-class members)

how many of the ~5.5e9 S-classes have 6(5,..,1) different 44-gangsters in the 6 chutes ?

(~50% of random G-class members)

(~95% of random G-class members)

how many of the ~5.5e9 S-classes have 6(5,..,1) different 44-gangsters in the 6 chutes ?

(~50% of random G-class members)

Last edited by dukuso on Thu Jun 03, 2010 5:15 pm, edited 2 times in total.

- dukuso
**Posts:**479**Joined:**25 June 2005