From: "Alexandru E. Ungur" <alexandru@...>
Date: 2006-03-06T18:01:07+09:00
Subject: Re: [SOLUTION] The Golden Fibbonacci Ratio (#69)

Hi all,

Nothing fancy here either, I just tried to solve this in the simplest way 
I could (first). Then I tried 'the Ruby way' (at least how I see it after
only 10 hours of playing with Ruby...).

The output uses a two characters to display one cell of data, that is
"##" not "#". Since a single character is 8x16 bits, two chars together
make a pretty good square, and thus the proportions in the final output
are more close to the reality. The blue rectangle is the current
rectangle, while the white square is the part that gets cut from it.

Needs ANSI capable terminal, don't know if it would work or not on
Windows...


--- cut here ---
cell, blank, clear = "\033[34;1m##", "\033[37;1m##", "\033[30;0m"

next_rect = lambda { |a,b| [[a,b].max, [a,b].min + [a,b].max] }
rect = next_rect

res = [1, 1]
(1..6).each do
  p res
  side = ''
  res[0].times { side = side + cell }
  res[1].times { side = side + blank }
  res[1].times { puts side }
  puts clear
  res = rect.call(res[0], res[1])
end
--- cut here ---



the OO solution, using recursion

--- cut here ---
class GoldenRectangles
  def initialize
    @cell, @blank, @clear = "\033[34;1m##", "\033[37;1m##", "\033[30;0m"
  end

  def next_rectangle(a, b)
    [[a,b].max, [a,b].min + [a,b].max]
  end

  def show_rectangles(rect, count)
    if count > 0
      p rect
      side = ''
      rect[0].times { side = side + @cell }
      rect[1].times { side = side + @blank }
      rect[1].times { puts side }
      puts @clear
      rect = next_rectangle(rect[0], rect[1])
      show_rectangles(rect, count - 1)
    end
  end
end

GoldenRectangles.new.show_rectangles([1,1], 5)
--- cut here ---


Have a nice day all,
Alex