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 coloin
 Joined: 05 May 2005  Posts: 97  :   Items 

Posted: Thu May 05, 2005 8:53 pm Post subject: How many false/true Sudoku grids are out there ? 


I am no an expert math and I am quite prepared to be corrected but I reckon there are:
9! to the power of 9 sudoku grids  each with 19 in each 3by3 square
= 362880 to the power of 9
This may be devided by 4 because of 4 similar rotational forms .
Now how many true sudoko grids are there ?????????
Good luck 

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 Animator
 Joined: 26 Apr 2005  Posts: 18  :   Items 


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 coloin
 Joined: 05 May 2005  Posts: 97  :   Items 

Posted: Thu May 05, 2005 11:59 pm Post subject: how many.... 


Thanks
I see people are on the case !
Any other forums out there!!!!! 

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 upsidedownface
 Joined: 27 May 2005  Posts: 5  :  Location: Leeds  Items 

Posted: Fri May 27, 2005 9:01 pm Post subject: 


I think there are only 32 different 2x2 Sudoku squares.
There is only one first row, even though there look like lots. It is a trivial transformation of the symbols 1 2 3 4 into any other first line,by changing the symbol in the first cell on the first line into 1, the second into 2 etc.
Then you can put the next 1 into any of 8 cells, i.e cells 3,4 on row 2,cells 2,3,4 on row 3 or row 4.
The second 2 can then go into any of 7 cells, the same as the second 1 minus the cell already taken by the second 1.
The second 3 can go into either 8 or 7 or 6 cells, depending on whether the second 1 and 2 are on row 2 or not.
The 32 comes from there being 2 choices 5 times over of where in a box the next number can be put, i.e 2 to the 5th power is 32. 

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 upsidedownface
 Joined: 27 May 2005  Posts: 5  :  Location: Leeds  Items 

Posted: Sat May 28, 2005 2:29 pm Post subject: 


I overestimated the number of 2box x 2box solutions. I think here are 12 after trying to list them all!
There is only 1 first row because simple number substitution translates any into any other first row.
There are four possible second rows because the second 1 can go in two positions and the second 2 then has two possible places, but the positions of the second 3 and 4 then have no choice.
The third row also can have 1 in either of two places and 2 in the two remaining places. But the positions of the 3 and 4 on the third row are determined.
The fourth row has to have the remaining number in each column in the single remaining cell.
This makes 1x4x4x1 different squares = 16, BUT four of the possibilities break the Sudoku rules as there is no possibility for placing the third 4.
Here (I hope) is a copy of my listing!
1234 1234 1234 1234  1234 1234 1234 1234  1234 1234 1234 1234  1234 1234 1234 1234
3412 3412 3412 3412  3421 3421 3421 3421  4312 4312 4312 4312  4321 4321 4321 4321
2143 4123 4321 2341  4312 231. 2143 .132  2143 312. 2.31 3421  2143 3142 2413 3412
1234 1234 1234 1234  1234 1234  1234 1234  1234 1234 1234 1234
3412 3412 3412 3412  3421 3421  4312 4312  4321 4321 4321 4321
2143 4123 4321 2341  4312 2143  2143 3421  2143 3142 2413 3412
#@~" #@~" #@~" #@~"  #@~" #@~"  #@~" #@~"  #@~" #@~" #@~" #@~"
~"#@ ~"#@ ~"#@ ~"#@  ~"@# ~"@#  "~#@ "~#@  "~@# "~@# "~@# "~@#
@#"~ "#@~ "~@# @~"#  "~#@ @#"~  @#"~ ~"@#  @#"~ ~#"@ @"#~ ~"#@ 

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