# Iterate through array twice

I have array of horizontal segments and I need to find which of them
may be sides of a box, so I implemented this method:

def find_square_sides(horiz_segments)
squares = Array.new
horiz_segments.each_with_index do |side, i|
horiz_segments[(i+1)..(horiz_segments.size - 1)].each do |
condidate>
if(## here goes long-long check ##)
squares << Square.new(side, condidate)
end
end
end
squares
end

Knowing the beauty of Ruby, I hope that there is a better solution. I
guess that double iterating isn't the best.

Unfortunately the problem you are trying to solve is O(n*n) and there is nothing that will make this go away - at least not with the knowledge we have so far. It may be that elements to compare to can be picked more efficiently but if you have to compare each with each other your solution is probably as good as it gets.

Kind regards

robert

PS: The only small improvement that came to mind was to use -1 as ending index for the range of the second iteration.

···

On 16.02.2008 15:46, Ruhe wrote:

I have array of horizontal segments and I need to find which of them
may be sides of a box, so I implemented this method:

def find_square_sides(horiz_segments)
squares = Array.new
horiz_segments.each_with_index do |side, i|
horiz_segments[(i+1)..(horiz_segments.size - 1)].each do |
condidate>
if(## here goes long-long check ##)
squares << Square.new(side, condidate)
end
end
end
squares
end

Knowing the beauty of Ruby, I hope that there is a better solution. I
guess that double iterating isn't the best.

With Ruby 1.9:

def find_square_sides(horiz_segments)
squares = Array.new
horiz_segments.combination(2) do |side, candidate|
if (...whatever...)
squares << Square.new(side, candidate)
end
end
end

···

On Feb 16, 2008 6:49 AM, Ruhe <nocturneer@gmail.com> wrote:

I have array of horizontal segments and I need to find which of them
may be sides of a box, so I implemented this method:

def find_square_sides(horiz_segments)
squares = Array.new
horiz_segments.each_with_index do |side, i|
horiz_segments[(i+1)..(horiz_segments.size - 1)].each do |
condidate>
if(## here goes long-long check ##)
squares << Square.new(side, condidate)
end
end
end
squares
end

Knowing the beauty of Ruby, I hope that there is a better solution. I
guess that double iterating isn't the best.

Yes, Ruby is the most beautiful language!

···

On 16 фев, 21:20, Christopher Dicely <cmdic...@gmail.com> wrote:

With Ruby 1.9:

def find_square_sides(horiz_segments)
squares = Array.new
horiz_segments.combination(2) do |side, candidate|
if (...whatever...)
squares << Square.new(side, candidate)
end
end
end

On Feb 16, 2008 6:49 AM, Ruhe <nocturn...@gmail.com> wrote: