4 ms·
The exponential number of symmetries present in sudoku problems, means that once you've found one valid instance, you've actually found up to 9! * 3!^4 * 8 whic
by lifebeyondfife 2y ago
The exponential number of symmetries present in sudoku problems, means that once you've found one valid instance, you've actually found up to 9! * 3!^4 * 8 which are exactly the same.
The numbers themselves are all interchangeable, so you have 9! combinations: 362,880.
Columns 1-to-3 are all interchangeable, as are 4-to-6, and 7-to-9. On top of this, these blocks of columns (1-to-3, 4-to-6, 7-to-9) are all interchangeable. Read about wreath products in group theory to know more. Each of the above symmetries are 3!, combined to yield 3! * 3! = 36 combinations. As well as the columns though, the rows have the same property, so those can be combined too: 36 * 36 =
1,296.
Finally, there are the symmetries of a square. Combining all rotations and flips yields a further 8.
In total, sudoku has 3,762,339,840 symmetries. Owing to the starting state of the sudoku puzzle being incomplete, the orbit of the set of points (more group theory) will be smaller than 3 billion, but it provides an efficient method of recreating many more puzzles with the same property. In this case, human complexity.
- xjm 2y agoI counted one trillion or 9! * 3!^8 * 2 : the 8 because you have can choose 3 independent permutations of columns inside column blocks + 1 permutation of column blocks, plus same for rows. Then only one rotation should be counted, because flips are included in col/row permutations. I think wreath products relate to the second sentence; see this page, which mentions the same result: https://en.wikipedia.org/wiki/Mathematics_of_Sudoku#The_sudoku_symmetry_group https://en.wikipedia.org/wiki/Mathematics_of_Sudoku#The_sudo...
- lifebeyondfife 2y agoYou're correct, the horizontal and vertical flips for the square, are already accounted for in the wreath product. And I miscounted the products themselves. Up to 1.2*10^12 symmetries.