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Sudoku


brad89

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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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