The Assassin Problem 3
Having a look at some of your solutions to The Assassin Problem (original video here http://www.youtube.com/watch?v=crkeHTaVOyY ) My solution video here http://www.youtube.com/watch?v=7d6E2oJ4-u0 Matt Parker's video solution here http://www.youtube.com/watch?v=kDl_MXWmlqs The vast majority of solutions, including all solutions in this video (except mine), are the same underneath. Mathematically speaking we say they are 'isomorphic', in that they may look different on the surface but they possess the same mathematical structure underneath. The majority involved assigning each player a number (either physically by position, or by assigning a number to each name) and then performing x+1 mod 8 (adding one to each number, where the numbers wrap round back to 1). Essentially, this is what an 8-cycle is, so they all answer the question given. However, how they perform this x+1 operation, practically and secretly, is what makes them different. Some solutions found ingenious ways to make use of a different shift, like x+2, x+3, x+4 etc (In fact this is the solution the original poster, Tom, had). However, if the shift divides the number of players then this will not make one full cycle, but instead will break up the players into subsets. So in our main example, since 2 and 4 both divide 8, these shifts do not make one complete cycle. My solution works on a completely different mathematical principle, which I did on purpose - I am meant to be the expert afterall. It uses an idea called conjugation; a permuation sandwiched between a function and its inverse will return a permuation of the same cycle type. The shuffling of the names and looking to the next name down in the pack is the x+1 idea again, but that is sandwiched between the idea of making pairs, with pairs being self-inverse. Giving each name a number is a bijection (a labelling). Which number kills which other number is the structure, with most answers having an x+1 structure, with the exceptions mentioned above.