Is there a quick way to determine if it is possible to rearrange a sudoku grid into an equivalent grid such that it creates an automorphism under some symmetry?
An automorphic grid under rotational symmetry (180º) is a grid such that all instances of a single digit X will map to all instances of a single digit Y after a rotation of 180º, for example.
Given an arbitrary sudoku grid A, an grid B is said to be equivalent if B can be obtained from a using the following operations:
1) Permuting rows in a horizontal band (A band is a part of the grid that encapsulates 3 rows and 3 boxes).
2) Permuting columns in a stack (A stack is a part of the grid that encapsulates 3 columns and 3 boxes)
3) Relabelling symbols (all 9s chance to 1s, for example)
4) Permuting bands
5) Permuting stacks
6) Reflecting or rotating the grid
If a board can have its givens rearranged such that they create an automorphism, it becomes easy to solve the puzzle via Gurth's symmetrical placement theorem.
Most grids are not automorphic as given, but they might be equivalent to an automorphic grid. This means one could solve the automorphic version, then undo any transformations to get the solution.
An automorphic grid under rotational symmetry (180º) is a grid such that all instances of a single digit X will map to all instances of a single digit Y after a rotation of 180º, for example.
Given an arbitrary sudoku grid A, an grid B is said to be equivalent if B can be obtained from a using the following operations:
1) Permuting rows in a horizontal band (A band is a part of the grid that encapsulates 3 rows and 3 boxes).
2) Permuting columns in a stack (A stack is a part of the grid that encapsulates 3 columns and 3 boxes)
3) Relabelling symbols (all 9s chance to 1s, for example)
4) Permuting bands
5) Permuting stacks
6) Reflecting or rotating the grid
If a board can have its givens rearranged such that they create an automorphism, it becomes easy to solve the puzzle via Gurth's symmetrical placement theorem.
Most grids are not automorphic as given, but they might be equivalent to an automorphic grid. This means one could solve the automorphic version, then undo any transformations to get the solution.
