Counting problem: possible sudoko tables?
Question:
I’m working on a sudoko solver (python). my method is using a game tree and explore possible permutations for each set of digits by DFS Algorithm.
in order to analyzing problem, i want to know what is the count of possible valid and invalid sudoko tables?
-> a 9*9 table that have 9 one, 9 two, … , 9 nine.
(this isn’t exact duplicate by this question)
my solution is:
1- First select 9 cells for 1s: (*)
(source: sitmo.com)
2- and like (1) for other digits (each time, 9 cells will be deleted from remaining available cells):
C(81-9,9) , C(81-9*2,9) …. =
(source: sitmo.com)
3- finally multiply the result by 9! (permutation of 1s-2s-3s…-9s in (*))
(source: sitmo.com)
this is not equal to accepted answer of this question but problems are equivalent. what did i do wrong?
Answers:
The number of valid Sudoku solution grids for the standard 9×9 grid was calculated by Bertram Felgenhauer and Frazer Jarvis in 2005 to be 6,670,903,752,021,072,936,960.
Mathematics of Sudoku |
source
I think problem with your solution is that deleting 9 cells each time from available cells does not necessarily create a valid grid. What I mean is just deleting 9 cells won’t suffice.
That is why 81! / (9!)^9 is much bigger number than actual valid solutions.
EDIT:
Permutations with repeated elements
Your solutions is almost correct if you want all the tables not just valid sudoku tables.
There is a formula:
(a+b+c+…)! / [a! b! c! ….]
Suppose there are 5 boys and 3 girls and we have 8 seats then number of different ways in which they can seat is
(5+3)! / (5! 3!)
Your problem is analogous to this one.
There are 9 1s , 9 2s … 9 9s.
and 81 places
so answer should be (9+9+…)! / (9!)^9
Now if you multiply again by 9! then this will add duplicate arrangements to the number by shuffling them.
According to this Wikipedia article (or this OEIS sequence), there are roughly 6.6 * 10^21 different sudoku squares.
What you did wrong was the last step: you shouldn’t multiply the answer by 9!. You have already counted all possible squares.
This doesn’t help you much when counting the possible Sudoku-tables. One other thing you could do is to count the tables where the “row-condition” holds: that is just (9!)^9, because you just choose one permutation of 1..9 for every row.
Still closer to the Sudoku-problem is counting Latin squares. Latin square has to satisfy both the “row-condition” and “column condition”. That is already a difficult problem and no closed form formula is known. Sudoku is a Latin square with the additional “subsquare-condition”.
I’m working on a sudoko solver (python). my method is using a game tree and explore possible permutations for each set of digits by DFS Algorithm.
in order to analyzing problem, i want to know what is the count of possible valid and invalid sudoko tables?
-> a 9*9 table that have 9 one, 9 two, … , 9 nine.
(this isn’t exact duplicate by this question)
my solution is:
1- First select 9 cells for 1s: (*)
(source: sitmo.com)
2- and like (1) for other digits (each time, 9 cells will be deleted from remaining available cells):
C(81-9,9) , C(81-9*2,9) …. =
(source: sitmo.com)
3- finally multiply the result by 9! (permutation of 1s-2s-3s…-9s in (*))
(source: sitmo.com)
this is not equal to accepted answer of this question but problems are equivalent. what did i do wrong?
The number of valid Sudoku solution grids for the standard 9×9 grid was calculated by Bertram Felgenhauer and Frazer Jarvis in 2005 to be 6,670,903,752,021,072,936,960.
Mathematics of Sudoku |
source
I think problem with your solution is that deleting 9 cells each time from available cells does not necessarily create a valid grid. What I mean is just deleting 9 cells won’t suffice.
That is why 81! / (9!)^9 is much bigger number than actual valid solutions.
EDIT:
Permutations with repeated elements
Your solutions is almost correct if you want all the tables not just valid sudoku tables.
There is a formula:
(a+b+c+…)! / [a! b! c! ….]
Suppose there are 5 boys and 3 girls and we have 8 seats then number of different ways in which they can seat is
(5+3)! / (5! 3!)
Your problem is analogous to this one.
There are 9 1s , 9 2s … 9 9s.
and 81 places
so answer should be (9+9+…)! / (9!)^9
Now if you multiply again by 9! then this will add duplicate arrangements to the number by shuffling them.
According to this Wikipedia article (or this OEIS sequence), there are roughly 6.6 * 10^21 different sudoku squares.
What you did wrong was the last step: you shouldn’t multiply the answer by 9!. You have already counted all possible squares.
This doesn’t help you much when counting the possible Sudoku-tables. One other thing you could do is to count the tables where the “row-condition” holds: that is just (9!)^9, because you just choose one permutation of 1..9 for every row.
Still closer to the Sudoku-problem is counting Latin squares. Latin square has to satisfy both the “row-condition” and “column condition”. That is already a difficult problem and no closed form formula is known. Sudoku is a Latin square with the additional “subsquare-condition”.