brad89 Posted December 21, 2005 Share Posted December 21, 2005 Is there any type of equation that would represent the total possible amount of sudoku games there are possible to make? Link to comment Share on other sites More sharing options...

cosine Posted December 21, 2005 Share Posted December 21, 2005 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 [math]9(9!)[/math]. Link to comment Share on other sites More sharing options...

ecoli Posted December 21, 2005 Share Posted December 21, 2005 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. Link to comment Share on other sites More sharing options...

5614 Posted December 21, 2005 Share Posted December 21, 2005 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. Link to comment Share on other sites More sharing options...

ecoli Posted December 21, 2005 Share Posted December 21, 2005 According to these guys there are 6,670,903,752,021,072,936,960 Sudoku grids. http://www.shef.ac.uk/~pm1afj/sudoku/ Here's a forum that's debating the subject. http://www.sudoku.org.uk/discus/messages/29/34.html?1130280356 Link to comment Share on other sites More sharing options...

Sisyphus Posted December 21, 2005 Share Posted December 21, 2005 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. Link to comment Share on other sites More sharing options...

Morbid Steve Posted December 29, 2005 Share Posted December 29, 2005 That's it? I must already have done most of them. Yo tambien! Link to comment Share on other sites More sharing options...

swansont Posted December 29, 2005 Share Posted December 29, 2005 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... Link to comment Share on other sites More sharing options...

ecoli Posted December 29, 2005 Share Posted December 29, 2005 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?} Link to comment Share on other sites More sharing options...

Lyssia Posted December 29, 2005 Share Posted December 29, 2005 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. Link to comment Share on other sites More sharing options...

Sisyphus Posted December 29, 2005 Share Posted December 29, 2005 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.} Link to comment Share on other sites More sharing options...

cosine Posted December 29, 2005 Share Posted December 29, 2005 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?} Link to comment Share on other sites More sharing options...

swansont Posted December 30, 2005 Share Posted December 30, 2005 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. Link to comment Share on other sites More sharing options...

BigMoosie Posted December 30, 2005 Share Posted December 30, 2005 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 Link to comment Share on other sites More sharing options...

cosine Posted December 30, 2005 Share Posted December 30, 2005 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! Link to comment Share on other sites More sharing options...

brad89 Posted December 30, 2005 Author Share Posted December 30, 2005 Wow. I never would have thought it possible to find out how many boards could be made. Link to comment Share on other sites More sharing options...

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