From: Harry Kakueki <list.push@...>
Date: 2009-09-02T18:20:51+09:00
Subject: Re: Nice algorithm for 'spreading' indexes across an array?

On Wed, Sep 2, 2009 at 5:30 PM, Max Williams<toastkid.williams@gmail.com> wrote:
> Thanks, everyone.  Harry, i like your solution but i thought of a way to
> tweak it in a way which seems more readable, to me at least:
>
> class Array
>  def spread(num)
>    discarded = []
>    (1..(size-num)).each{|u| discarded << self[size*u/(size-num+1)]}
>    self - discarded
>  end
> end
>
> The clever bit is of course line 4.  I'm still trying to work out why
> that works :)
>

That is not exactly the same thing.
But it depends on your specs.

class Array
  def spread(num)
    res,idue = [],[]
    (1..(size-num)).each{|u| idue << size*u/(size-num+1)}
    (0...size).each{|y| res << self[y] if idue.include?(y) == false}
    res
  end

  def spread2(num)
    discarded = []
    (1..(size-num)).each{|u| discarded << self[size*u/(size-num+1)]}
    self - discarded
  end

end

arr = [1,2,3,4,5,6,7,8,9,9,10,11,12]
(3..13).each{|t| p arr.spread(t)}
puts
(3..13).each{|t| p arr.spread2(t)}

############output

#=> [1, 7, 12]
#=> [1, 5, 9, 12]
#=> [1, 4, 7, 9, 12]
#=> [1, 3, 6, 8, 10, 12]
#=> [1, 3, 5, 7, 9, 10, 12]
#=> [1, 2, 4, 6, 8, 9, 11, 12]
#=> [1, 2, 4, 5, 7, 9, 9, 11, 12]
#=> [1, 2, 3, 5, 6, 8, 9, 10, 11, 12]
#=> [1, 2, 3, 4, 6, 7, 8, 9, 10, 11, 12]
#=> [1, 2, 3, 4, 5, 6, 8, 9, 9, 10, 11, 12]
#=> [1, 2, 3, 4, 5, 6, 7, 8, 9, 9, 10, 11, 12]

#=> [1, 7, 12]
#=> [1, 5, 12]
#=> [1, 4, 7, 12]
#=> [1, 3, 6, 8, 10, 12]
#=> [1, 3, 5, 7, 10, 12]
#=> [1, 2, 4, 6, 8, 11, 12]
#=> [1, 2, 4, 5, 7, 9, 9, 11, 12]
#=> [1, 2, 3, 5, 6, 8, 10, 11, 12]
#=> [1, 2, 3, 4, 6, 7, 8, 10, 11, 12]
#=> [1, 2, 3, 4, 5, 6, 8, 9, 9, 10, 11, 12]
#=> [1, 2, 3, 4, 5, 6, 7, 8, 9, 9, 10, 11, 12]


Harry

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