# The Josephus Problem in Both Directions

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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.

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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.

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Contributed by: Hiroshi Matsui, Toshiyuki Yamauchi, Daisuke Minematsu, Soh Tatsumi, Masakazu Naito, Takafumi Inoue and Ryohei Miyadera (March 2011)
Open content licensed under CC BY-NC-SA

## Details

If you look at the graph of the list , you will discover a very interesting fact: the graph is self-similar, a fractal. To prove that the graph is indeed a fractal, you need to find recursive relations for . The authors are going to publish their research about the self-similarity of this graph in Kokyuroku at the Research Institute of Mathematical Science at Kyoto University. As to the other variants of the Josephus Problem, see S. Hashiba, D. Minematsu, and R. Miyadera, "How High School Students Can Discover Original Ideas of Mathematics Using Mathematica", Mathematica in Education and Research, 11(3), 2006.

In this problem every person is eliminated. By changing the number you can make an interesting animation.

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