Question

我需要对2D中的2个线段交叉进行精确且数值稳定的测试。检测4个位置有一种可能的解决方案，请参见下面的代码。

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;
 }

我不确定是否所有建议的条件都设计得恰当（以避免圆度误差）。

使用参数s，t进行测试或仅用于计算交叉点是否有意义？

我担心可能无法正确检测到位置2（在一个终点处相交的段）（没有任何条件的最后剩余情况）......

Answer 1

这似乎是一个非常常见的数学问题。有一个关于topcoder公式的好教程可以回答你的问题，很容易用你想要的任何编程语言实现基础：Line Intersection Tutorial

此致伊芙吉尼娅

Answer 2

if(fabs(denom) < eps){
    if((fabs(len(x2, y2, x3, y3) + len(x2, y2, x4, y4) - len(x3, y3, x4, y4)) < eps) || (fabs(len(x1, y1, x3, y3) + len(x1, y1, x4, y4) - len(x3, y3, x4, y4)) < eps) || (fabs(len(x3, y3, x1, y1) + len(x3, y3, x2, y2) - len(x1, y1, x2, y2)) < eps) || (fabs(len(x4, y4, x1, y1) + len(x4, y4, x2, y2) - len(x1, y1, x2, y2)) < eps)){
      return 1;
    }else{
      return 0;
    }
}

len = sqrt(sqr(c - a) + sqr(d - b))

线段交叉，数值稳定测试

2 个答案: