Question

I had the following code of bresenham’s algorithm which represents adapted to Scala Java code.

def bresenham(x0: Int, y0: Int, x1: Int, y1: Int) = {
    import scala.math.abs

    val dx = abs(x1 - x0)
    val dy = abs(y1 - y0)

    val sx = if (x0 < x1) 1 else -1
    val sy = if (y0 < y1) 1 else -1

    new Iterator[(Int, Int)] {
      var (x, y) = (x0, y0)
      var err = dx - dy

      def next = {
        val omitted = (x, y)
        val e2 = 2 * err
        if (e2 > -dy) {
          err -= dy
          x += sx
        }
        if (e2 < dx) {
          err += dx
          y += sy
        }
        omitted
      }

      def hasNext = (x <= x1 && y <= y1)
    }
  }

For almost all lines all goes fine, but when I’m trying to compute vertical lines from top to bottom (i.e. (0,3) -> (0,0) ) I’m getting nothing.
I feel stupid about myself, because problem is not so hard and lies in hasNext which says nope for a case above).
I’ve dealt with that by swapping points, but that’s obviously a bad solution.
Can anybody help me to generalize algorithm?

Answer 1

In the failure case, you’re trying to go from y0 = 3 to y1 = 0. So the step will be negative, sy = -1. The condition for continuing in hasNext should then depend on y >= y1 rather than what you wrote (y <= y1).

hasNext must be generalized to handle either direction. A clever way would be,

def hasNext = (sx*x <= sx*x1 && sy*y <= sy*y1)

which works because sx and sy are nonzero, and their sign determines the direction of the steps.