C和Objective-C

基于Gareth Rees的回答

const AGKLine AGKLineZero = (AGKLine){(CGPoint){0.0, 0.0}, (CGPoint){0.0, 0.0}};

AGKLine AGKLineMake(CGPoint start, CGPoint end)
{
    return (AGKLine){start, end};
}

double AGKLineLength(AGKLine l)
{
    return CGPointLengthBetween_AGK(l.start, l.end);
}

BOOL AGKLineIntersection(AGKLine l1, AGKLine l2, CGPoint *out_pointOfIntersection)
{
    // http://stackoverflow.com/a/565282/202451

    CGPoint p = l1.start;
    CGPoint q = l2.start;
    CGPoint r = CGPointSubtract_AGK(l1.end, l1.start);
    CGPoint s = CGPointSubtract_AGK(l2.end, l2.start);
    
    double s_r_crossProduct = CGPointCrossProductZComponent_AGK(r, s);
    double t = CGPointCrossProductZComponent_AGK(CGPointSubtract_AGK(q, p), s) / s_r_crossProduct;
    double u = CGPointCrossProductZComponent_AGK(CGPointSubtract_AGK(q, p), r) / s_r_crossProduct;
    
    if(t < 0 || t > 1.0 || u < 0 || u > 1.0)
    {
        if(out_pointOfIntersection != NULL)
        {
            *out_pointOfIntersection = CGPointZero;
        }
        return NO;
    }
    else
    {
        if(out_pointOfIntersection != NULL)
        {
            CGPoint i = CGPointAdd_AGK(p, CGPointMultiply_AGK(r, t));
            *out_pointOfIntersection = i;
        }
        return YES;
    }
}

CGFloat CGPointCrossProductZComponent_AGK(CGPoint v1, CGPoint v2)
{
    return v1.x * v2.y - v1.y * v2.x;
}

CGPoint CGPointSubtract_AGK(CGPoint p1, CGPoint p2)
{
    return (CGPoint){p1.x - p2.x, p1.y - p2.y};
}

CGPoint CGPointAdd_AGK(CGPoint p1, CGPoint p2)
{
    return (CGPoint){p1.x + p2.x, p1.y + p2.y};
}

CGFloat CGPointCrossProductZComponent_AGK(CGPoint v1, CGPoint v2)
{
    return v1.x * v2.y - v1.y * v2.x;
}

CGPoint CGPointMultiply_AGK(CGPoint p1, CGFloat factor)
{
    return (CGPoint){p1.x * factor, p1.y * factor};
}

许多函数和结构都是私有的，但是你应该很容易就能知道发生了什么。 这是公开的在这个回购https://github.com/hfossli/AGGeometryKit/

基于@Gareth Rees的回答，Python版本:

import numpy as np

def np_perp( a ) :
    b = np.empty_like(a)
    b[0] = a[1]
    b[1] = -a[0]
    return b

def np_cross_product(a, b):
    return np.dot(a, np_perp(b))

def np_seg_intersect(a, b, considerCollinearOverlapAsIntersect = False):
    # https://stackoverflow.com/questions/563198/how-do-you-detect-where-two-line-segments-intersect/565282#565282
    # http://www.codeproject.com/Tips/862988/Find-the-intersection-point-of-two-line-segments
    r = a[1] - a[0]
    s = b[1] - b[0]
    v = b[0] - a[0]
    num = np_cross_product(v, r)
    denom = np_cross_product(r, s)
    # If r x s = 0 and (q - p) x r = 0, then the two lines are collinear.
    if np.isclose(denom, 0) and np.isclose(num, 0):
        # 1. If either  0 <= (q - p) * r <= r * r or 0 <= (p - q) * s <= * s
        # then the two lines are overlapping,
        if(considerCollinearOverlapAsIntersect):
            vDotR = np.dot(v, r)
            aDotS = np.dot(-v, s)
            if (0 <= vDotR  and vDotR <= np.dot(r,r)) or (0 <= aDotS  and aDotS <= np.dot(s,s)):
                return True
        # 2. If neither 0 <= (q - p) * r = r * r nor 0 <= (p - q) * s <= s * s
        # then the two lines are collinear but disjoint.
        # No need to implement this expression, as it follows from the expression above.
        return None
    if np.isclose(denom, 0) and not np.isclose(num, 0):
        # Parallel and non intersecting
        return None
    u = num / denom
    t = np_cross_product(v, s) / denom
    if u >= 0 and u <= 1 and t >= 0 and t <= 1:
        res = b[0] + (s*u)
        return res
    # Otherwise, the two line segments are not parallel but do not intersect.
    return None

曾经在这里被接受的答案是不正确的(它已经被不接受了，所以万岁!)它不能正确地消除所有非交点。简单地说，它可能有效，但也可能失败，特别是在0和1被认为对h有效的情况下。

考虑以下情况:

直线(4,1)-(5,1)和(0,0)-(0,2)

这两条垂线显然不重叠。

= (4,1) B =(5、1) C = (0, 0) D = (0, 2) E = (1) - (4,1) = (1,0) F = (0, 2) - (0, 0) = (0, 2) P = (0, 1) h =((4,1) -(0, 0))点(0,1)/((0,2)点(0,1))= 0

根据上面的答案，这两条线段在端点处相遇(值为0和1)。该端点为:

(0, 0) + (0, 2) * 0 = (0, 0)

So, apparently the two line segments meet at (0,0), which is on line CD, but not on line AB. So what is going wrong? The answer is that the values of 0 and 1 are not valid and only sometimes HAPPEN to correctly predict endpoint intersection. When the extension of one line (but not the other) would meet the line segment, the algorithm predicts an intersection of line segments, but this is not correct. I imagine that by testing starting with AB vs CD and then also testing with CD vs AB, this problem would be eliminated. Only if both fall between 0 and 1 inclusively can they be said to intersect.

如果你必须预测端点，我建议使用向量叉乘法。

-Dan

这对我来说很有效。从这里拍的。

 // calculates intersection and checks for parallel lines.  
 // also checks that the intersection point is actually on  
 // the line segment p1-p2  
 Point findIntersection(Point p1,Point p2,  
   Point p3,Point p4) {  
   float xD1,yD1,xD2,yD2,xD3,yD3;  
   float dot,deg,len1,len2;  
   float segmentLen1,segmentLen2;  
   float ua,ub,div;  

   // calculate differences  
   xD1=p2.x-p1.x;  
   xD2=p4.x-p3.x;  
   yD1=p2.y-p1.y;  
   yD2=p4.y-p3.y;  
   xD3=p1.x-p3.x;  
   yD3=p1.y-p3.y;    

   // calculate the lengths of the two lines  
   len1=sqrt(xD1*xD1+yD1*yD1);  
   len2=sqrt(xD2*xD2+yD2*yD2);  

   // calculate angle between the two lines.  
   dot=(xD1*xD2+yD1*yD2); // dot product  
   deg=dot/(len1*len2);  

   // if abs(angle)==1 then the lines are parallell,  
   // so no intersection is possible  
   if(abs(deg)==1) return null;  

   // find intersection Pt between two lines  
   Point pt=new Point(0,0);  
   div=yD2*xD1-xD2*yD1;  
   ua=(xD2*yD3-yD2*xD3)/div;  
   ub=(xD1*yD3-yD1*xD3)/div;  
   pt.x=p1.x+ua*xD1;  
   pt.y=p1.y+ua*yD1;  

   // calculate the combined length of the two segments  
   // between Pt-p1 and Pt-p2  
   xD1=pt.x-p1.x;  
   xD2=pt.x-p2.x;  
   yD1=pt.y-p1.y;  
   yD2=pt.y-p2.y;  
   segmentLen1=sqrt(xD1*xD1+yD1*yD1)+sqrt(xD2*xD2+yD2*yD2);  

   // calculate the combined length of the two segments  
   // between Pt-p3 and Pt-p4  
   xD1=pt.x-p3.x;  
   xD2=pt.x-p4.x;  
   yD1=pt.y-p3.y;  
   yD2=pt.y-p4.y;  
   segmentLen2=sqrt(xD1*xD1+yD1*yD1)+sqrt(xD2*xD2+yD2*yD2);  

   // if the lengths of both sets of segments are the same as  
   // the lenghts of the two lines the point is actually  
   // on the line segment.  

   // if the point isn’t on the line, return null  
   if(abs(len1-segmentLen1)>0.01 || abs(len2-segmentLen2)>0.01)  
     return null;  

   // return the valid intersection  
   return pt;  
 }  

 class Point{  
   float x,y;  
   Point(float x, float y){  
     this.x = x;  
     this.y = y;  
   }  

   void set(float x, float y){  
     this.x = x;  
     this.y = y;  
   }  
 }  

人们似乎对Gavin的答案很感兴趣，cortijon在评论中提出了一个javascript版本，iMalc提供了一个计算量略少的版本。一些人指出了各种代码建议的缺点，另一些人则评论了一些代码建议的效率。

iMalc通过Gavin的答案提供的算法是我目前在一个javascript项目中使用的算法，我只是想在这里提供一个清理过的版本，如果它可以帮助到任何人的话。

// Some variables for reuse, others may do this differently
var p0x, p1x, p2x, p3x, ix,
    p0y, p1y, p2y, p3y, iy,
    collisionDetected;

// do stuff, call other functions, set endpoints...

// note: for my purpose I use |t| < |d| as opposed to
// |t| <= |d| which is equivalent to 0 <= t < 1 rather than
// 0 <= t <= 1 as in Gavin's answer - results may vary

var lineSegmentIntersection = function(){
    var d, dx1, dx2, dx3, dy1, dy2, dy3, s, t;

    dx1 = p1x - p0x;      dy1 = p1y - p0y;
    dx2 = p3x - p2x;      dy2 = p3y - p2y;
    dx3 = p0x - p2x;      dy3 = p0y - p2y;

    collisionDetected = 0;

    d = dx1 * dy2 - dx2 * dy1;

    if(d !== 0){
        s = dx1 * dy3 - dx3 * dy1;
        if((s <= 0 && d < 0 && s >= d) || (s >= 0 && d > 0 && s <= d)){
            t = dx2 * dy3 - dx3 * dy2;
            if((t <= 0 && d < 0 && t > d) || (t >= 0 && d > 0 && t < d)){
                t = t / d;
                collisionDetected = 1;
                ix = p0x + t * dx1;
                iy = p0y + t * dy1;
            }
        }
    }
};

我从《多视图几何》这本书里读到了这些算法

以下文本使用

'作为转置符号

*作为点积

当用作算子时，X作为叉乘

1. 线的定义

点x_vec = (x, y)'在直线ax + by + c = 0上

标记L = (a, b, c)'，点为(x, y, 1)'为齐次坐标

直线方程可以写成

(x, y, 1)(a, b, c)' = 0或x' * L = 0

2. 直线交点

我们有两条直线L1=(a1, b1, c1)'， L2=(a2, b2, c2)'

假设x是一个点，一个向量，x = L1 x L2 (L1叉乘L2)。

注意，x始终是一个二维点，如果你对(L1xL2)是一个三元素向量，x是一个二维坐标感到困惑，请阅读齐次坐标。

根据三重积，我们知道

L1 * (L1 x L2) = 0, L2 * (L1 x L2) = 0，因为L1,L2共平面

我们用向量x代替L1*x，那么L1*x=0, L2*x=0，这意味着x在L1和L2上，x是交点。

注意，这里x是齐次坐标，如果x的最后一个元素是零，这意味着L1和L2是平行的。