From: "Jesús Gabriel y Galán" <jgabrielygalan@...>
Date: 2009-09-02T02:44:04+09:00
Subject: Re: Nice algorithm for 'spreading' indexes across an array?

On Tue, Sep 1, 2009 at 6:41 PM, 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?

Here's my try. It fails when the number you are asking for is more
than half the size, but anyway, it might give you an idea:

class Array
        def spread how_many
                result = []
                each_slice(size / (how_many - 1)) { |a, *_| result << a}
                result << self[-1]
                result
        end
end

irb(main):004:0> load 'spread.rb'
=> true
irb(main):005:0> (1..12).to_a.spread 3
=> [1, 7, 12]
irb(main):006:0> (1..12).to_a.spread 4
=> [1, 5, 9, 12]
irb(main):007:0> (1..12).to_a.spread 12
=> [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 12]
irb(main):008:0> (1..12).to_a.spread 11
=> [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 12]
irb(main):019:0> (1..12).to_a.spread 8
=> [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 12]

Jesus.