IBM Research | Ponder This | May 2012 solutions
Skip to main content

May 2012

<<April May June>>


There are many solutions, some of which have quite a few interesting symmetries, like Joël Bleuse's solution (note the place of the 7-s):

  1    8    3    2    5    9    7  10    4    6
  2    5  10    9    7    1    6    4    8    3
  5    3    8    2    6    1    7  10    9    4
  4    9  10    7    1    6    2    8    3    5
  3    8    4    6    1    7    9  10    5    2
  6    4  10    7    9    5    2    3    8    1

Matej Kollár even wrote a nice tool http://www.fi.muni.cz/~xkollar2/symmetry-helper-1/ and used it to find a nice symmetric solution.
We think that for any even n>3, there is a solution for n permutations of 2n-2 elements, but we don't have a proof for it.


If you have any problems you think we might enjoy, please send them in. All replies should be sent to: ponder@il.ibm.com