我需要一个精确和数值稳定的测试,用于2D中两条线段相交。有一种可能的解决方案检测4个位置,请参阅下面的代码。线段相交,数值稳定测试
getInters (double x1, double y1, double x2, double y2, double x3, double y3, double x4, double y4, double & x_int, double & y_int )
{
3: Intersect in two end points,
2: Intersect in one end point,
1: Intersect (but not in end points)
0: Do not intersect
unsigned short code = 2;
//Initialize intersections
x_int = 0, y_int = 0;
//Compute denominator
double denom = x1 * (y4 - y3) + x2 * (y3 - y4) + x4 * (y2 - y1) + x3 * (y1 - y2) ;
//Segments are parallel
if (fabs (denom) < eps)
{
//Will be solved later
}
//Compute numerators
double numer1 = x1 * (y4 - y3) + x3 * (y1 - y4) + x4 * (y3 - y1);
double numer2 = - (x1 * (y3 - y2) + x2 * (y1 - y3) + x3 * (y2 - y1));
//Compute parameters s,t
double s = numer1/denom;
double t = numer2/denom;
//Both segments intersect in 2 end points: numerically more accurate than using s, t
if ((fabs (numer1) < eps) && (fabs (numer2) < eps) ||
(fabs (numer1) < eps) && (fabs (numer2 - denom) < eps) ||
(fabs (numer1 - denom) < eps) && (fabs (numer2) < eps) ||
(fabs (numer1 - denom) < eps) && (fabs (numer2 - denom) < eps))
{
code = 3;
}
//Segments do not intersect: do not compute any intersection
else if ((s < 0.0) || (s > 1) ||
(t < 0.0) || (t > 1))
{
return 0;
}
//Segments intersect, but not in end points
else if ((s > 0) && (s < 1) && (t > 0) && (t < 1))
{
code = 1;
}
//Compute intersection
x_int = x1 + s * (x2 - x1);
y_int = y1 + s * (y2 - y1);
//Segments intersect in one end point
return code;
}
我不知道是否所有建议的条件设计得当(以避免圆度误差)。
使用参数s,t进行测试或仅将其用于计算交叉点有意义吗?
恐怕位置2(段一个终点相交)可能无法正确检测(最后剩余的情况下,没有任何条件)...
想法:第一次检查退化病例(并行,事件或不相交)。第二个计算交点。第三次检查交叉点是否位于任一段上,如果是,则在哪里。如果你有能力使用理性而不是实质,你甚至可以得到确切的答案。 – 2012-01-08 18:26:45