Re: [math-fun] Josephus = Fractal
I would say that this description could be written *far* more clearly. (And I don't see the problem as related to facility with English.) For example, are eliminated people included in the process of counting, or not? --Dan << The classical version of the Josephus problem counts off uneliminated people arranged in a circle and eliminates the person. The process continues until only one person is left. In this variant of the Josephus problem, two people are to be eliminated at each stage, but the two processes of elimination go in opposite directions. Suppose that there are people and every person is eliminated. The first process starts at the first person and eliminates the people at positions , , …. The second process starts with the person and eliminates the people at positions , , …. Suppose that the first process comes first and the second process second at each stage. Denote the position of the survivor by . The graph of the list is very interesting.
-----Forwarded Message-----
From: Ed Pegg Jr <ed@mathpuzzle.com> Sent: Aug 11, 2008 5:52 PM To: math-fun <math-fun@mailman.xmission.com> Subject: [math-fun] Josephus = Fractal
Student Daisuke Minematsu and his classmates have noticed that the Josephus problem has fractal behavior.
http://demonstrations.wolfram.com/TheJosephusProblemInBothDirections/
Here's a specific image: http://demonstrations.wolfram.com/TheJosephusProblemInBothDirections/HTMLIma...
Seems like a great result for a classic problem. It's new to me. Is it actually new?
--Ed Pegg Jr
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Dan Asimov