The New Sudoku Players' Forum

[denis_berthier]

But what proves that A(S) is a solution ?
This is precisely what we want to prove !

Probably, you're forgetting that an automorphism of S is first of all a Sudoku isomorphism.
If S is a solution, then A(S) satisfies the Sudoku axioms (they are invariant by any Sudoku isomorphism, by definition). In case P is invariant under A, A(S) also has the right clues for being a solution of P. qed.

[DEFISE]

OK Denis, it's more subtle than I thought !

[denis_berthier]

.
While I'm at it, let me prove the general version:
If P1, P2, ... Pn is an invariant set of puzzles wrt a Sudoku isomorphism A, the set made of all the solutions for any of these puzzles is also invariant under A.
Proof:
Let S be a solution for one of the puzzles, say Pi. We must show that A(S) is a solution for some of the Pk's.
A being a Sudoku isomorphism, A(S) is a complete Sudoku grid. By hypothesis, A(Pi) is equal to some Pk.
But, by definition, A(Pi) is also equal to A(S) restricted to the subset Pi of S.
Conclusion: A(S) is a solution of Pk. qed.