How to resolve the algorithm Sutherland-Hodgman polygon clipping step by step in the 11l programming language

Published on 12 May 2024 09:40 PM
#11l

How to resolve the algorithm Sutherland-Hodgman polygon clipping step by step in the 11l programming language

Table of Contents

Problem Statement

The   Sutherland-Hodgman clipping algorithm   finds the polygon that is the intersection between an arbitrary polygon (the “subject polygon”) and a convex polygon (the “clip polygon”). It is used in computer graphics (especially 2D graphics) to reduce the complexity of a scene being displayed by eliminating parts of a polygon that do not need to be displayed.

Take the closed polygon defined by the points: and clip it by the rectangle defined by the points: Print the sequence of points that define the resulting clipped polygon.

Display all three polygons on a graphical surface, using a different color for each polygon and filling the resulting polygon. (When displaying you may use either a north-west or a south-west origin, whichever is more convenient for your display mechanism.)

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Sutherland-Hodgman polygon clipping step by step in the 11l programming language

Source code in the 11l programming language

F clip(subjectPolygon, clipPolygon)
   F inside(p, cp1, cp2)
      R (cp2.x - cp1.x) * (p.y - cp1.y) > (cp2.y - cp1.y) * (p.x - cp1.x)

   F computeIntersection(s, e, cp1, cp2)
      V dc = cp1 - cp2
      V dp = s - e
      V n1 = cp1.x * cp2.y - cp1.y * cp2.x
      V n2 = s.x * e.y - s.y * e.x
      V n3 = 1.0 / (dc.x * dp.y - dc.y * dp.x)
      R ((n1 * dp.x - n2 * dc.x) * n3, (n1 * dp.y - n2 * dc.y) * n3)

   V outputList = subjectPolygon
   V cp1 = clipPolygon.last

   L(clipVertex) clipPolygon
      V cp2 = clipVertex
      V inputList = outputList
      outputList.clear()
      V s = inputList.last

      L(subjectVertex) inputList
         V e = subjectVertex
         I inside(e, cp1, cp2)
            I !inside(s, cp1, cp2)
               outputList.append(computeIntersection(s, e, cp1, cp2))
            outputList.append(e)
         E I inside(s, cp1, cp2)
            outputList.append(computeIntersection(s, e, cp1, cp2))
         s = e
      cp1 = cp2
   R (outputList)

V subjectp = [(50.0, 150.0), (200.0, 50.0), (350.0, 150.0), (350.0, 300.0), (250.0, 300.0), (200.0, 250.0), (150.0, 350.0), (100.0, 250.0), (100.0, 200.0)]
V clipp = [(100.0, 100.0), (300.0, 100.0), (300.0, 300.0), (100.0, 300.0)]
print_elements(clip(subjectp, clipp), sep' "\n")

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