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Is there any type of equation that would represent the total possible amount of sudoku games there are possible to make?

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Is there any type of equation that would represent the total possible amount of sudoku games there are possible to make?

I think you can make an upperbound of $9(9!)$.

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Is there any type of equation that would represent the total possible amount of sudoku games there are possible to make?

That is so strange... I was just thinking about that today.

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Yes you can and I know that someone did it (read it in the newspaper) however I can't remember any more than that, so I can't help much other than saying that it is possible and do-able.

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According to these guys there are 6' date='670,903,752,021,072,936,960 Sudoku grids.

[/quote']

That's it? I must already have done most of them.

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That's it? I must already have done most of them.

Yo tambien!

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I think you can make an upperbound of [math]9(9!)[/math'].

There are 9! ways of arranging the grids (before restrictions are applied), not 9. But 9!2 is much, much smaller than the number in ecoli's link...

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There are 9! ways of arranging the grids (before restrictions are applied), not 9. But 9!2[/sup'] is much, much smaller than the number in ecoli's link...

I'm not sure if there's any valiity to that link... anybody understand what they did and if it's correct?

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I've only read it a couple of times through, and not in any great detail, but it doesn't seem to make any glaring errors (not that I'm an expert of anything, however). I'm not sure if it's been submitted anywhere, so I wonder what kind of peer reviewing it would get.

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There are 9! ways of arranging the grids (before restrictions are applied), not 9. But 9!2[/sup'] is much, much smaller than the number in ecoli's link...

It wouldn't just be the number of ways of arranging the grid, I don't think. Because for each possible solution, you also have every single combination of starting numbers that lead uniquely to that solution. I have no idea how to go about calculating that, but it seems like it would be a pretty big number.

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There are 9! ways of arranging the grids (before restrictions are applied), not 9. But 9!2[/sup'] is much, much smaller than the number in ecoli's link...

Well I just got that number as a trivial upper bound this way:

Consider a line of 9 boxes. The first box has 9 possible entries, the second box has 8,... so this line has 9! possible permutations. Then there are 9 of these rows. So I figured that 9*9! would be an upperbound (since I didn't put any more restrictions on it). But that would leave an upperbound of 3265920 possibilities. Thats intuitively small. I apologize since I'm not really that much of a probability/combinatorics person yet (first class will be tomorrow). Can you tell me where the flaw in my logic is?

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Well I just got that number as a trivial upper bound this way:

Consider a line of 9 boxes. The first box has 9 possible entries' date=' the second box has 8,... so this line has 9! possible permutations. Then there are 9 of these rows. So I figured that 9*9! would be an upperbound (since I didn't put any more restrictions on it). But that would leave an upperbound of 3265920 possibilities. Thats intuitively small. I apologize since I'm not really that much of a probability/combinatorics person yet (first class will be tomorrow). Can you tell me where the flaw in my logic is?[/quote']

There are 9 rows, as you say. There are 9! ways of arranging those rows.

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Well I just got that number as a trivial upper bound this way:

Consider a line of 9 boxes. The first box has 9 possible entries' date=' the second box has 8,... so this line has 9! possible permutations. Then there are 9 of these rows. So I figured that 9*9! would be an upperbound (since I didn't put any more restrictions on it). But that would leave an upperbound of 3265920 possibilities. Thats intuitively small. I apologize since I'm not really that much of a probability/combinatorics person yet (first class will be tomorrow). Can you tell me where the flaw in my logic is?[/quote']

You are considering P(a or b or c...) for each row in which case the numbers are added giving 9(9!) but it is really P(a and b and c...) in which case they should be multiplied giving 9!^9 which is about 50 digits long.

EDIT: wolfram gives the same number as the link previously posted: http://mathworld.wolfram.com/Sudoku.html

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You are considering P(a or b or c...) for each row in which case the numbers are added giving 9(9!) but it is really P(a and b and c...) in which case they should be multiplied giving 9!^9 which is about 50 digits long.

EDIT: wolfram gives the same number as the link previously posted: http://mathworld.wolfram.com/Sudoku.html

Ohh, I see. Thanks!

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Wow. I never would have thought it possible to find out how many boards could be made.

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