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.



Task

Take the closed polygon defined by the points:

 



[
(
50
,
150
)
,
(
200
,
50
)
,
(
350
,
150
)
,
(
350
,
300
)
,
(
250
,
300
)
,
(
200
,
250
)
,
(
150
,
350
)
,
(
100
,
250
)
,
(
100
,
200
)
]


{\displaystyle [(50,150),(200,50),(350,150),(350,300),(250,300),(200,250),(150,350),(100,250),(100,200)]}


and clip it by the rectangle defined by the points:

 



[
(
100
,
100
)
,
(
300
,
100
)
,
(
300
,
300
)
,
(
100
,
300
)
]


{\displaystyle [(100,100),(300,100),(300,300),(100,300)]}


Print the sequence of points that define the resulting clipped polygon.



Extra credit

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.)




 
def clip(subjectPolygon, clipPolygon):
   def inside(p):
      return(cp2[0]-cp1[0])*(p[1]-cp1[1]) > (cp2[1]-cp1[1])*(p[0]-cp1[0])
 
   def computeIntersection():
      dc = [ cp1[0] - cp2[0], cp1[1] - cp2[1] ]
      dp = [ s[0] - e[0], s[1] - e[1] ]
      n1 = cp1[0] * cp2[1] - cp1[1] * cp2[0]
      n2 = s[0] * e[1] - s[1] * e[0] 
      n3 = 1.0 / (dc[0] * dp[1] - dc[1] * dp[0])
      return [(n1*dp[0] - n2*dc[0]) * n3, (n1*dp[1] - n2*dc[1]) * n3]
 
   outputList = subjectPolygon
   cp1 = clipPolygon[-1]
 
   for clipVertex in clipPolygon:
      cp2 = clipVertex
      inputList = outputList
      outputList = []
      s = inputList[-1]
 
      for subjectVertex in inputList:
         e = subjectVertex
         if inside(e):
            if not inside(s):
               outputList.append(computeIntersection())
            outputList.append(e)
         elif inside(s):
            outputList.append(computeIntersection())
         s = e
      cp1 = cp2
   return(outputList)
 