From: Sebastian Hungerecker <sepp2k@...>
Date: 2008-09-26T17:48:26+09:00
Subject: Re: [SUMMARY] One-Liners Mashup (#177)

Matthew Moss wrote:
> The problem:
>
>     Write a function per(n) which returns the periodicity of 1/n, i.e.
>       per(3) => 1
>       per(4) => 0
>       per(7) => 6
>       per(11) => 2
>
> The "Ummm... yeah... how do these work?" solution:
>
>     def per(n, b=10)
>       i=1;x=b;h={};loop {x=x%n*b;break 0 if x==0;h[x]?(break
> i-h[x]):h[x]=i; i+=1}
>     end

It works like this:
http://img243.imageshack.us/img243/9296/perwt4.png
It simply divides until the remainder repeats itself (storing the position 
where each remainder occured in a hash) and then returns the difference 
between the first position where the remainder occured (i.e. the start of the 
period) and the current position.

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