Looping in a spiral

Question

A friend was in need of an algorithm that would let him loop through the elements of an NxM matrix (N and M are odd). I came up with a solution, but I wanted to see if my fe

Answer 1

Here's c#, linq'ish.

public static class SpiralCoords
{
  public static IEnumerable> GenerateOutTo(int radius)
  {
    //TODO trap negative radius.  0 is ok.

    foreach(int r in Enumerable.Range(0, radius + 1))
    {
      foreach(Tuple coord in GenerateRing(r))
      {
        yield return coord;
      }
    }
  }

  public static IEnumerable> GenerateRing(int radius)
  {
    //TODO trap negative radius.  0 is ok.

    Tuple currentPoint = Tuple.Create(radius, 0);
    yield return Tuple.Create(currentPoint.Item1, currentPoint.Item2);

    //move up while we can
    while (currentPoint.Item2 < radius)
    {
      currentPoint.Item2 += 1;
      yield return Tuple.Create(currentPoint.Item1, currentPoint.Item2);
    }
    //move left while we can
    while (-radius < currentPoint.Item1)
    {
      currentPoint.Item1 -=1;
      yield return Tuple.Create(currentPoint.Item1, currentPoint.Item2);    
    }
    //move down while we can
    while (-radius < currentPoint.Item2)
    {
      currentPoint.Item2 -= 1;
      yield return Tuple.Create(currentPoint.Item1, currentPoint.Item2);
    }
    //move right while we can
    while (currentPoint.Item1 < radius)
    {
      currentPoint.Item1 +=1;
      yield return Tuple.Create(currentPoint.Item1, currentPoint.Item2);    
    }
    //move up while we can
    while (currentPoint.Item2 < -1)
    {
      currentPoint.Item2 += 1;
      yield return Tuple.Create(currentPoint.Item1, currentPoint.Item2);
    }
  }

}

The question's first example (3x3) would be:

var coords = SpiralCoords.GenerateOutTo(1);

The question's second example (5x3) would be:

var coords = SpiralCoords.GenerateOutTo(2).Where(x => abs(x.Item2) < 2);