Friday 25 December 2015

Friday Puzzles #288

That moment when you realise Christmas Day is also a Friday!

Nothing too taxing today as this was rather a last minute affair, but watch your step!  Each of the four shaded regions follows a different rule:

  • odd
  • even
  • fortress
  • renban group
Merry Christmas to all my dearest readers, and as ever, enjoy!

EDIT: I'd managed to leave a couple of clues out.  This should make things a little easier!

    #328 Mini-Matchmaker Sudoku – rated medium
All puzzles © Tom Collyer 2009-15.

Tuesday 15 December 2015

2015 WSC and Group Theory

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.


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!

Friday 11 December 2015

Friday Puzzles #287

Greetings all!  I've been a bit busy recently - not least because I have taken over the role of Competition Director of the WPF's Sudoku GP.  Exciting times indeed.

I have been posting for the Daily League once a fortnight for a while, as has Bastien on a Wednesday.  I think this means now at least we are posting on different weeks.  By-the-by, if you fancy yourself as an aspiring Sudoku author, then please do contribute!

Anyhow, only time for a classic this week.  Enjoy!
    #327 Sudoku – rated easy
All puzzles © Tom Collyer 2009-15.

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