I've yet to post any sort of championship report for the WSC/WPC held in Sofia in October. I might still get round to that soon! But the rationale for this post comes from finishing off the broken/unsolved puzzles from my booklets recently. I came across this gem of a puzzle.
This puzzle is actually taken from the instruction booklet - however it's not the puzzle itself which interests me, rather the phrasing of the instructions. The Twin Regions in question are effectively extra regions, which are related by a cyclic permutation - and here is the crucial bit - which might possibly be reversed in order.
Now given the cyclic nature of the constraint, it was reasonable to assume that if the twin regions were not quite parallel rows/columns, then they would at least both go from top edge to bottom or left edge to right. But if this is the case, then that cyclic permutation couldn't possibly be reversed!
And why is that, you ask?
Answer 1: Any orientation reversing symmetry of the regular nonagon is a relfection whose axis passes through a vertex and the midpoint of the opposite edge.
Answer 2: Lets label the cells in the 1st twin region ABCDEFGHI. Any order reversing cyclic permutation has a fixed point! This is bad for Sudoku because that means you will end up with two of the same number in a column.
And why is there a fixed point? Well, this is because any order-reversing cyclic permutation is in fact a reflection of the set X, and because X has 9 elements, when you start pairing off the elements of X that define the reflection, you will always be left with one left over. The fixed point!
Don't believe me about the reflection claim? Well let's take the easy case: the order reversing cyclic permutation.
ABCDEFGHI -> IHGFEDBCA
This pairs off (AI)(BH)(CG)(DF) and fixes E. It's a reflection in E.
Now, we can get any other order-reversing permutation of X by taking this one and applying a (non-trival) order-preserving permutation. This will move the fixed point from E, and rotate it round to another fixed point, which will then define another reflection.
Answer 3: The D_18 action on X has 9 distinct stabiliser subgroups!