How many false/true Sudoku grids are out there ?

coloin

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 1-9 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 Shocked

Animator · Posted: Thu May 05, 2005 9:21 pm Post subject:

You might want to take a look at this thread:

http://www.sudoku.com/forums/viewtopic.php?t=44&postdays=0&postorder=asc&start=0

It's on another forum, but who cares?

coloin · 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!!!!!

upsidedownface · 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.

upsidedownface · 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

#@~" #@~" #@~" #@~" | #@~" #@~" | #@~" #@~" | #@~" #@~" #@~" #@~"
~"#@ ~"#@ ~"#@ ~"#@ | ~"@# ~"@# | "~#@ "~#@ | "~@# "~@# "~@# "~@#
@#"~ "#@~ "~@# @~"# | "~#@ @#"~ | @#"~ ~"@# | @#"~ ~#"@ @"#~ ~"#@