For me the proof given is not a problem, but...What says the UR1.1 proof as far as I understand it:
A_ in a solution grid, assume that we find a pattern- Code: Select all
ab
ba of 2 digits a,b
then exchanging ‘a’ and ‘b’, we get another valid solution grid.
Is it a fact or a corollary of the basic rules, no matter, we have here the smallest unavoidable set.
In a solution grid, any unavoidable set (we have billions of them) must hit a clue to have a sudoku (not a rubbish sudoku).
B_ take now a puzzle supposed to be a sudoku or a rubbish sudoku. You get during the solving process the pattern
- Code: Select all
1 2
2 13
none of the four cells is a givenTheorem, with such a pattern the solution is ‘3’note: if the puzzle has no valid solution, whatever you take will not change the final conclusion.
I accept as hidden condition that the set of rules used to reach this point has no uniqueness assumption.- Code: Select all
proof
If the cell having the pm13 is filled with ‘1’
Applying the corollary A_, we have a second solution
2 1
1 2
but this in contradiction with the solved cells,
so ‘1’ is not valid
No problem in the logic IMO, but I have doubts that this situation will show up, so it could be a theorem of no interest.
And the proof is valid for a rubbish puzzle as well, again, the challenge is to reach the pattern during the solving process.