Reset loop, or return to start of loop, in Ruby? (Finding a number's factors)

Question

I am trying to find the factors of a number N. I want to be able to iterate over my prime numbers array, and when a condition is met, go back to start of the iteration, rather than just continuing on.

arrFactors = []    
N = 150
[2,3,5,7,11].each do |a|
  if N % a == 0
    arrFactors.push(a)
    N = N/a
    break if N == 1
    <return to start of the collection [2,3,5,7,11]>
  end
end

How can I do this in Ruby 1.8.6? (I notice there is a "cycle" enumerator in 1.8.7 that maybe relavent?)

P.S
I know the below algorithm is not perfect yet (I think I need to provide for N = 1). I also know Ruby provides a factoring method, but I don't want to use 'cause I'm learning here.

Answer 1

you could use something like that:

[2,3,5,7,11].each do |a|

  while N % a == 0
    arrFactors.push(a)
    N = N/a
  end

  if N == 1 then
    break
  end

end

This would remove the need to reset the outer loop.

Answer 2

Rather than use the Array.each method, you could look into using the Array.select method. This method takes a block, and will return a new array with only the items that match your logic. If your block returns true, then that item will be included in the result set. The following code uses the select method and does what you're looking for:

[2,3,5,7,11].select do |a|
  if N % a == 0
    N = N/a
    break if N == 1
    return true
  end
  return false
end

The documentation for the select method can be found here.

Answer 3

"redo" resets the loop current iteration of the loop. "retry" would start the loop all over. (Oh, and "N" is a constant; should be "n".)

Answer 4

You can try this

n = 150
factors = [2,3,5,7,11]
arrFactors = []
factor_index = 0
  while ((factors.size > factor_index + 1) || n > 1) do
    factors.each_with_index { |f, i|      
      factor_index = i
      if (n % f) == 0
        arrFactors.push(f)
        n = n / f 
        break
      end
    }
  end
puts n
puts arrFactors