From: Brian Candler <b.candler@...>
Date: 2009-09-02T18:14:30+09:00
Subject: Re: Nice algorithm for 'spreading' indexes across an array?

Max Williams 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?

# I'd add 0.01 to give some margin for rounding errors

class Array
  def spread(n)
    interval = (size - 0.99) / (n - 1)
    res = []
    n.times do |i|
      res << self[i * interval]
      # or: res << self[i * interval + 0.5]
    end
    res
  end
end

a = (1..12).to_a
puts a.spread(3)
puts a.spread(4)
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