The Josephus problem

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The Josephus problem

Legend has it that Josephus wouldn't have lived to become famous without his mathematical talents. During the Jewish-Roman war, he was among a band of 41 Jerish rebels trapped in a cave by Romans. Preferring suicide to capture, the rebels decide to form a circle, and proceeding around it, to kill every third remaining person until no one was left. But Josephus, among with an unindicted co-conspirator, wanted none of this suicide nonsense; so he quickly calculated where he and his friend should stand in the vicious circle.

In our variation, we start with $n$ people numbered 1 to $n$ around a circle, and we eliminate every $second$ remaining person until ONLY one survives. For example, here's the starting configuration for $n$ =10.

The elimination order is 2, 4, 6, 8, 10, 3, 7, 1, 9, so 5 survives. The problem: determine the survivor's number $J(n)$ .

Suppose that we have $2n$ people originally. After the first goround, we are left with 1, 3, 5, 7, ... , $2n-1$ . and 3 will be the next to go. this is just like starting out with $n$ people, except that each person's number has been doubled and decreased by 1. That is,

$J(2n)=2J(n)-1,n geq 1$

But what about the odd case? with $2n+1$ people, it turns out that person number 1 is wiped out just after person number $2n$ , and we are left with 3, 5, 7, 9, ..., $2n+1$ .

Again we almost have teach original situation with $n$ people, But this time their number is doublede and increased by 1. Thus

$J(2n+1)=2J(n)+1,n geq 1$

Noticing that $J(1)=1$ .

Now we seek a closed form, because that will be even quicker and more informative. After all, that is a matter of life or death.

Our recurrence makes it possible to build a table of small values very quickly. Perhaps we'll be able to splot a pattern and guess the answer.

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It seems we can group by power of 2. $J(n)$ is always 1 at the beginning of group and it increases by 2 within a group. So if we write $n$ in the form $n=2^m+l$ , where $2^m$ is the largest power of 2 not exceeding $n$ and where $l$ is what's left, the solution to our recurrence seems to be

$J(2^m+l)=2l+1,(m geq 0,0 leq l leq 2^m)$

We must now prove this function.The induction step has two parts, depending on whether $l$ is odd or even. If $m>0$ and $2^m+l=2n$ , then $l$ is even and

$J(2^m+l)=2J(2^{m-1}+l/2)-1=2(2l/2+1)-1=2l+1$

A similar proof works in the odd case, when $2^m+l=2n+1$ . We might also note that $J(2n+1)-J(2n)=2$ .

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原文地址：https://www.cnblogs.com/taokongcn/p/4034352.html