Algorithm for iterating over an outward spiral on a discrete 2D grid from the origin

Question

For example, here is the shape of intended spiral (and each step of the iteration)

          y
          |
          |
   16 15 14 13 12
   17  4  3  2 11
--         


        

Answer 1

I've done pretty much the same thin as a training exercise, with some differences in the output and the spiral orientation, and with an extra requirement, that the functions spatial complexity has to be O(1).

After think for a while I came to the idea that by knowing where does the spiral start and the position I was calculating the value for, I could simplify the problem by subtracting all the complete "circles" of the spiral, and then just calculate a simpler value.

Here is my implementation of that algorithm in ruby:

def print_spiral(n)
  (0...n).each do |y|
    (0...n).each do |x|
      printf("%02d ", get_value(x, y, n))
    end
    print "\n"
  end
end


def distance_to_border(x, y, n)
  [x, y, n - 1 - x, n - 1 - y].min
end

def get_value(x, y, n)
  dist = distance_to_border(x, y, n)
  initial = n * n - 1

  (0...dist).each do |i|
    initial -= 2 * (n - 2 * i) + 2 * (n - 2 * i - 2)
  end        

  x -= dist
  y -= dist
  n -= dist * 2

  if y == 0 then
    initial - x # If we are in the upper row
  elsif y == n - 1 then
    initial - n - (n - 2) - ((n - 1) - x) # If we are in the lower row
  elsif x == n - 1 then
    initial - n - y + 1# If we are in the right column
  else
    initial - 2 * n - (n - 2) - ((n - 1) - y - 1) # If we are in the left column
  end
end

print_spiral 5

This is not exactly the thing you asked for, but I believe it'll help you to think your problem