From: John W Higgins <wishdev@...>
Date: 2009-09-02T02:11:36+09:00
Subject: Re: Nice algorithm for 'spreading' indexes across an array?

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Morning Max,

On Tue, Sep 1, 2009 at 9:41 AM, Max Williams <toastkid.williams@gmail.com>wrote:

> Little ruby algorithm puzzle...
>
> I have a situation where i have an array of 12 items.  If someone
> chooses to have n of them (where n can be between 3 and 12) then i want
> to always include the first and last, and then 'spread' the others out
> as evenly as possible between the rest.
>
> So, lets say for the sake of argument that the array holds the numbers 1
> to 12.
>
> >> arr = (1..12).to_a
> => [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]
>
> I would get results back like this
>
> arr.spread(3)
> => [1,6,12] (or [1,7,12], either is fine)
>
> arr.spread(4)
> => [1, 5, 9, 12]  (or [1,4,8,12] or [1, 5, 8, 12])
>
> It feels like there should be a simple solution for this but i can't
> think of a nice way.  Anyone?
>
> thanks
> max
> --
> Posted via http://ww <http://www.ruby-forum.com/>RijndaelManaged<http://www.ruby-forum.com/>
> w.ruby-forum.com/ <http://www.ruby-forum.com/>.
>
>
So here is my silly attempt - simple but probably not the best performing
answer you will get here.

class Array
  def spread(members)
    ret = Array.new
    #We want sections that total 1 less then the number of members we want
back - for example if we
    #want 3 members in the spread we want 2 equal groups. We use .to_f
because we care about fractions here.
    dist = count.to_f/(members-1)
    #Now we simply run thru the array and take the members at each spread
point.
    (members-1).times{ |slot|
      ret << at((slot*dist).ceil)
    }
    #Add the last member which is always the last member of the array
    ret << at(-1)
  end
end

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