Determining whether or not a polygon (2D) has its vertices ordered clockwise or counterclockwise

本文介绍了一种利用交叉乘积来判断2D多边形顶点顺序是否为顺时针或逆时针的方法,并通过实例展示了如何判断多边形是否为凸形或凹形。适用于计算机图形学和几何处理。

Determining whether or not a polygon (2D) has itsvertices ordered clockwise or counterclockwise

Written by Paul Bourke
March 1998

The following describes a method for determining whether or nota polygon has its vertices ordered clockwise or anticlockwise.As a consequence the test can also be used to determine whether ornot a polygon is concave or convex.A polygon will be assumed to be described by N vertices, ordered

(x 0,y 0), (x 1,y 1), (x 2,y 2), . . . (x n-1,y n-1)

A convenient definition of clockwise is based on considerations ofthe cross product between adjacent edges. If the crossproduct ispositive then it rises above the plane (z axis up out of the plane)and if negative then the cross product is into the plane.

cross product = ((x i - x i-1),(y i - y i-1)) x ((x i+1 - x i),(y i+1 - y i))

=(xi - xi-1) * (yi+1 - yi)- (yi - yi-1) * (xi+1 - xi)

If the polygon is known to be convex then one only has to considerthe cross product between any two adjacent edges. A positive cross product means we have a counterclockwise polygon. There are sometests that may need to be done if the polygons may not be "clean".In particular two vertices must not be coincident and two edges mustnot be colinear.

For the more general case where the polygons may be convex, it isnecessary to consider the sign of the cross product between adjacentedges as one moves around the polygon. If there are more positivecross products then the overall polygon is ordered counterclockwise.There are pathological cases to consider here as well, all the edgescannot be colinear, there must be at least 3 vertices, the polygonmust be simple, that is, it cannot intersect itself or have holes.

Test for concave/convex polygon

A similar argument to the above can be used to determine whether a polygon is concave or convex.For a convex polygon all the cross products of adjacent edges willbe the same sign, a concave polygon will have a mixture of crossproduct signs.

Source Code

Example and test program for testingwhether a polygon is ordered clockwise or counterclockwise.For MICROSOFT WINDOWS, contributed by G. Adam Stanislav.

C function by Paul Bourke

/*
   Return the clockwise status of a curve, clockwise or counterclockwise
   n vertices making up curve p
   return 0 for incomputables eg: colinear points
          CLOCKWISE == 1
          COUNTERCLOCKWISE == -1
   It is assumed that
   - the polygon is closed
   - the last point is not repeated.
   - the polygon is simple (does not intersect itself or have holes)
*/
int ClockWise(XY *p,int n)
{
   int i,j,k;
   int count = 0;
   double z;

   if (n < 3)
      return(0);

   for (i=0;i<n;i++) {
      j = (i + 1) % n;
      k = (i + 2) % n;
      z  = (p[j].x - p[i].x) * (p[k].y - p[j].y);
      z -= (p[j].y - p[i].y) * (p[k].x - p[j].x);
      if (z < 0)
         count--;
      else if (z > 0)
         count++;
   }
   if (count > 0)
      return(COUNTERCLOCKWISE);
   else if (count < 0)
      return(CLOCKWISE);
   else
      return(0);
}

Example and test program for testing whether a polygon is convex or concave.For MICROSOFT WINDOWS, contributed by G. Adam Stanislav.

/*
   Return whether a polygon in 2D is concave or convex
   return 0 for incomputables eg: colinear points
          CONVEX == 1
          CONCAVE == -1
   It is assumed that the polygon is simple
   (does not intersect itself or have holes)
*/
int Convex(XY *p,int n)
{
   int i,j,k;
   int flag = 0;
   double z;

   if (n < 3)
      return(0);

   for (i=0;i<n;i++) {
      j = (i + 1) % n;
      k = (i + 2) % n;
      z  = (p[j].x - p[i].x) * (p[k].y - p[j].y);
      z -= (p[j].y - p[i].y) * (p[k].x - p[j].x);
      if (z < 0)
         flag |= 1;
      else if (z > 0)
         flag |= 2;
      if (flag == 3)
         return(CONCAVE);
   }
   if (flag != 0)
      return(CONVEX);
   else
      return(0);
}


转自:http://debian.fmi.uni-sofia.bg/~sergei/cgsr/docs/clockwise.htm

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