2013-05-01 14:16:01 -04:00
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USING: kernel locals math math.ranges sequences ;
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IN: rosetta-code.josephus-problem
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! http://rosettacode.org/wiki/Josephus_problem
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! Problem: Josephus problem is a math puzzle with a grim
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! description: n prisoners are standing on a circle, sequentially
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! numbered from 0 to n − 1. An executioner walks along the circle,
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! starting from prisoner 0, removing every k-th prisoner and
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! killing him. As the process goes on, the circle becomes smaller
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! and smaller, until only one prisoner remains, who is then freed.
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! For example, if there are n = 5 prisoners and k = 2, the order
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! the prisoners are killed in (let's call it the "killing
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! sequence") will be 1, 3, 0, and 4, and the survivor will be #2.
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! Task: Given any n,k > 0, find out which prisoner will be the
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! final survivor. In one such incident, there were 41 prisoners
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! and every 3rd prisoner was being killed (k = 3). Among them was
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! ! a clever chap name Josephus who worked out the problem, stood at
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! the surviving position, and lived on to tell the tale. Which
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! number was he?
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! Extra: The captors may be especially kind and let m survivors
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! free, and Josephus might just have m − 1 friends to save.
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! Provide a way to calculate which prisoner is at any given
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! position on the killing sequence.
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! Notes:
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! 1. You can always play the executioner and follow the
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! procedure exactly as described, walking around the circle,
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! counting (and cutting off) heads along the way. This would yield
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! the complete killing sequence and answer the above questions,
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! with a complexity of probably O(kn). However, individually it
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! takes no more than O(m) to find out which prisoner is the m-th
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! to die.
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! 2. If it's more convenient, you can number prisoners from 1 to
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! n instead. If you choose to do so, please state it clearly.
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! 3. An alternative description has the people committing
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! assisted suicide instead of being executed, and the last person
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! simply walks away. These details are not relevant, at least not
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! mathematically.
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2013-05-01 14:25:40 -04:00
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:: josephus-k ( n k -- m )
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2013-05-01 14:16:01 -04:00
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n [1,b] 0 [ [ k + ] dip mod ] reduce ;
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2013-05-01 14:25:40 -04:00
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:: josephus-2 ( n -- m ) ! faster for k=2
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n n log2 2^ - 2 * ;
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:: josephus ( n k -- m )
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k 2 = [ n josephus-2 ] [ n k josephus-k ] if ;
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