geometry - 你如何检测两条线段相交的位置？

Question

如何确定两条线是否相交，如果相交，在什么 x,y 点？

score 658 · Accepted Answer

有一个很好的方法来解决这个问题，它使用向量叉积。将二维向量叉积v × w定义为v _x w _y − v _y w _x。

假设两条线段从p到p + r和从q到q + s。那么第一行上的任何点都可以表示为p + t r（对于标量参数 t），第二行上的任何点都可以表示为q + u s（对于标量参数 u）。

两条相交的线段

如果我们可以找到t和u，那么两条线相交：

p + t r = q + u s

交点的公式

两边用s交叉，得到

( p + t r ) × s = ( q + u s ) × s

由于s × s = 0，这意味着

t ( r × s ) = ( q - p ) × s

因此，求解t：

t = ( q - p ) × s / ( r × s )

同样，我们可以求解u：

( p + t r ) × r = ( q + u s ) × r

u ( s × r ) = ( p - q ) × r

u = ( p - q ) × r / ( s × r )

为了减少计算步骤的数量，将其重写如下（记住s × r = − r × s）很方便：

u = ( q - p ) × r / ( r × s )

现在有四种情况：

如果r × s = 0 且 ( q − p) × r = 0，则两条线共线。

在这种情况下，用第一条线段的方程 ( p + t r )表示第二条线段的端点 ( q和q + s )：

t ₀ = ( q - p ) · r / ( r · r )

t ₁ = ( q + s - p ) · r / ( r · r ) = t ₀ + s · r / ( r · r )

如果t ₀和t ₁之间的区间与区间 [0, 1] 相交，则线段共线且重叠；否则它们是共线且不相交的。

请注意，如果s和r指向相反的方向，则s · r < 0，因此要检查的区间是 [ t ₁ , t ₀ ] 而不是 [ t ₀ , t ₁ ]。
如果r × s = 0 且 ( q − p) × r ≠ 0，则两条线平行且不相交。
如果r × s ≠ 0 且 0 ≤ t ≤ 1 且 0 ≤ u ≤ 1，则两条线段在点p + t r = q + u s处相交。
否则，两条线段不平行但不相交。

学分：此方法是 Ronald Goldman 发表于Graphics Gems第 304 页的文章“三空间中两条线的交点”中 3D 线相交算法的二维特化。在三维中，通常的情况是这些线是倾斜的（既不平行也不相交），在这种情况下，该方法给出了两条线最接近的点。

score 230 · Accepted Answer

FWIW，以下函数（在 C 中）都检测线交点并确定交点。它基于 Andre LeMothe 的“ Windows 游戏编程大师技巧”中的算法。它与其他答案中的某些算法（例如 Gareth 的）没有什么不同。LeMothe 然后使用克莱默规则（不要问我）来解方程本身。

我可以证明它在我微弱的小行星克隆中有效，并且似乎可以正确处理 Elemental、Dan 和 Wodzu 的其他答案中描述的边缘情况。它也可能比 KingNestor 发布的代码更快，因为它都是乘法和除法，没有平方根！

我想那里有一些被零除的可能性，尽管在我的情况下这不是问题。很容易修改以避免崩溃。

// Returns 1 if the lines intersect, otherwise 0. In addition, if the lines 
// intersect the intersection point may be stored in the floats i_x and i_y.
char get_line_intersection(float p0_x, float p0_y, float p1_x, float p1_y, 
    float p2_x, float p2_y, float p3_x, float p3_y, float *i_x, float *i_y)
{
    float s1_x, s1_y, s2_x, s2_y;
    s1_x = p1_x - p0_x;     s1_y = p1_y - p0_y;
    s2_x = p3_x - p2_x;     s2_y = p3_y - p2_y;

    float s, t;
    s = (-s1_y * (p0_x - p2_x) + s1_x * (p0_y - p2_y)) / (-s2_x * s1_y + s1_x * s2_y);
    t = ( s2_x * (p0_y - p2_y) - s2_y * (p0_x - p2_x)) / (-s2_x * s1_y + s1_x * s2_y);

    if (s >= 0 && s <= 1 && t >= 0 && t <= 1)
    {
        // Collision detected
        if (i_x != NULL)
            *i_x = p0_x + (t * s1_x);
        if (i_y != NULL)
            *i_y = p0_y + (t * s1_y);
        return 1;
    }

    return 0; // No collision
}

顺便说一句，我必须说，在 LeMothe 的书中，尽管他显然得到了正确的算法，但他展示的具体示例插入了错误的数字并进行了错误的计算。例如：

(4 * (4 - 1) + 12 * (7 - 1)) / (17 * 4 + 12 * 10)

= 844/0.88

= 0.44

这让我困惑了好几个小时。:(

score 63 · Accepted Answer

问题归结为这个问题：从 A 到 B 和从 C 到 D 的两条线是否相交？然后你可以问它四次（在直线和矩形的四个边之间）。

这是执行此操作的矢量数学。我假设从 A 到 B 的线是有问题的线，从 C 到 D 的线是矩形线之一。我的符号是Ax“A 的 x 坐标”和Cy“C 的 y 坐标”。并且“ *”表示点积，例如A*B = Ax*Bx + Ay*By。

E = B-A = ( Bx-Ax, By-Ay )
F = D-C = ( Dx-Cx, Dy-Cy ) 
P = ( -Ey, Ex )
h = ( (A-C) * P ) / ( F * P )

这个h数字是关键。如果在和h之间，则线相交，否则不相交。如果为零，您当然无法进行计算，但在这种情况下，线是平行的，因此仅在明显的情况下相交。01F*P

确切的交点是C + F*h。

更多乐趣：

如果h恰好是 0或1线在端点处接触。您可以将其视为“交叉点”，也可以不视为您认为合适的。

具体来说，h就是你必须乘以线的长度才能准确地接触另一条线。

因此，如果h<0，则表示矩形线在给定线的“后面”（“方向”为“从 A 到 B”），如果h>1矩形线在给定线的“前面”。

推导：

A 和 C 是指向行首的向量；E 和 F 是形成直线的 A 和 C 末端的向量。

对于平面上的任何两条不平行的线，必须恰好有一对标量g，并且h使得这个方程成立：

A + E*g = C + F*h

为什么？因为两条非平行线必须相交，这意味着您可以将两条线分别缩放一定量并相互接触。

（起初，这看起来像是一个有两个未知数的单一方程！ 但当你认为这是一个二维向量方程时，它并不是，这意味着这实际上是x和中的一对方程y。）

我们必须消除这些变量之一。一个简单的方法是将E术语设为零。为此，请使用一个向量对等式两边进行点积，该向量将与 E 点到零。我P在上面调用的那个向量，我对 E 做了明显的变换。

你现在有：

A*P = C*P + F*P*h
(A-C)*P = (F*P)*h
( (A-C)*P ) / (F*P) = h

score 46 · Accepted Answer

我试图实现上面 Jason 如此优雅地描述的算法；不幸的是，在调试数学时，我发现许多情况下它不起作用。

例如，考虑点 A(10,10) B(20,20) C(10,1) D(1,10) 给出 h=.5，但通过检查可以清楚地看出这些段在每个段附近都没有其他。

用图表可以清楚地看出，0 < h < 1 标准仅表明如果存在，截点将位于 CD 上，但没有说明该点是否位于 AB 上。为确保存在交叉点，您必须对变量 g 进行对称计算，截距要求为：0 < g < 1 AND 0 < h < 1

score 45 · Accepted Answer

这是对加文答案的改进。marcp 的解决方案也类似，但都没有推迟除法。

这实际上也是 Gareth Rees 答案的实际应用，因为叉积在 2D 中的等价物是 perp-dot-product，这是这段代码使用的三个。切换到 3D 并使用叉积，在最后插入 s 和 t，得到 3D 中线之间的两个最近点。无论如何，二维解决方案：

int get_line_intersection(float p0_x, float p0_y, float p1_x, float p1_y, 
    float p2_x, float p2_y, float p3_x, float p3_y, float *i_x, float *i_y)
{
    float s02_x, s02_y, s10_x, s10_y, s32_x, s32_y, s_numer, t_numer, denom, t;
    s10_x = p1_x - p0_x;
    s10_y = p1_y - p0_y;
    s32_x = p3_x - p2_x;
    s32_y = p3_y - p2_y;

    denom = s10_x * s32_y - s32_x * s10_y;
    if (denom == 0)
        return 0; // Collinear
    bool denomPositive = denom > 0;

    s02_x = p0_x - p2_x;
    s02_y = p0_y - p2_y;
    s_numer = s10_x * s02_y - s10_y * s02_x;
    if ((s_numer < 0) == denomPositive)
        return 0; // No collision

    t_numer = s32_x * s02_y - s32_y * s02_x;
    if ((t_numer < 0) == denomPositive)
        return 0; // No collision

    if (((s_numer > denom) == denomPositive) || ((t_numer > denom) == denomPositive))
        return 0; // No collision
    // Collision detected
    t = t_numer / denom;
    if (i_x != NULL)
        *i_x = p0_x + (t * s10_x);
    if (i_y != NULL)
        *i_y = p0_y + (t * s10_y);

    return 1;
}

基本上它将划分推迟到最后一刻，并将大部分测试移动到完成某些计算之前，从而增加了提前退出。最后，它还避免了线平行时发生的除以零的情况。

您可能还需要考虑使用 epsilon 测试而不是与零进行比较。非常接近平行的线会产生略微偏离的结果。这不是错误，而是浮点数学的限制。

score 40 · Accepted Answer

问题 C：如何检测两条线段是否相交？

我搜索了相同的主题，但我对答案不满意。所以我写了一篇文章，非常详细地解释了如何检查两条线段是否与大量图像相交。有完整的（和经过测试的）Java 代码。

这是文章，裁剪到最重要的部分：

检查线段 a 是否与线段 b 相交的算法如下所示：

在此处输入图像描述

什么是边界框？这是两个线段的两个边界框：

在此处输入图像描述

如果两个边界框都有交点，则移动线段 a 以使一个点位于 (0|0) 处。现在你有一条通过 a 定义的原点的线。现在以同样的方式移动线段 b 并检查线段 b 的新点是否在线 a 的不同侧。如果是这种情况，请反过来检查。如果也是这种情况，则线段相交。如果不是，它们不会相交。

问题 A：两条线段在哪里相交？

你知道两条线段 a 和 b 相交。如果您不知道，请使用我在“问题 C”中提供的工具进行检查。

现在您可以通过一些案例并通过 7 年级数学获得解决方案（参见代码和交互式示例）。

问题 B：如何检测两条线是否相交？

假设你的观点，A = (x1, y1)观点B = (x2, y2)，，C = (x_3, y_3)。D = (x_4, y_4)您的第一行由 AB 定义（使用 A ！= B），第二行由 CD 定义（使用 C ！= D）。

function doLinesIntersect(AB, CD) {
    if (x1 == x2) {
        return !(x3 == x4 && x1 != x3);
    } else if (x3 == x4) {
        return true;
    } else {
        // Both lines are not parallel to the y-axis
        m1 = (y1-y2)/(x1-x2);
        m2 = (y3-y4)/(x3-x4);
        return m1 != m2;
    }
}

问题 D：两条线在哪里相交？

如果它们完全相交，请检查问题 B。

线 a 和 b 由每条线的两个点定义。您基本上可以应用问题 A 中使用的相同逻辑。

score 21 · Accepted Answer

曾经在这里接受的答案是不正确的（此后一直未被接受，所以万岁！）。它不能正确消除所有非交叉点。微不足道，它可能看起来有效，但可能会失败，尤其是在 0 和 1 被认为对 h 有效的情况下。

考虑以下情况：

(4,1)-(5,1) 和 (0,0)-(0,2) 处的行

这些是明显不重叠的垂直线。

A=(4,1)
B=(5,1)
C=(0,0)
D=(0,2)
E=(5,1)-(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

根据上述答案，这两条线段在端点处相交（值 0 和 1）。该端点将是：

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

因此，显然两条线段在 (0,0) 处相交，在 CD 线上，但不在 AB 线上。那么出了什么问题呢？答案是 0 和 1 的值是无效的，只是有时会发生正确预测端点相交的情况。当一条线（而不是另一条）的延长线与线段相交时，算法会预测线段的交点，但这是不正确的。我想通过从 AB 与 CD 开始进行测试，然后再用 CD 与 AB 进行测试，这个问题将被消除。只有当两者都落在 0 和 1 之间时，才能说它们相交。

如果您必须预测端点，我建议使用向量叉积方法。

-担

score 14 · Accepted Answer

iMalc 答案的 Python 版本：

def find_intersection( p0, p1, p2, p3 ) :

    s10_x = p1[0] - p0[0]
    s10_y = p1[1] - p0[1]
    s32_x = p3[0] - p2[0]
    s32_y = p3[1] - p2[1]

    denom = s10_x * s32_y - s32_x * s10_y

    if denom == 0 : return None # collinear

    denom_is_positive = denom > 0

    s02_x = p0[0] - p2[0]
    s02_y = p0[1] - p2[1]

    s_numer = s10_x * s02_y - s10_y * s02_x

    if (s_numer < 0) == denom_is_positive : return None # no collision

    t_numer = s32_x * s02_y - s32_y * s02_x

    if (t_numer < 0) == denom_is_positive : return None # no collision

    if (s_numer > denom) == denom_is_positive or (t_numer > denom) == denom_is_positive : return None # no collision


    # collision detected

    t = t_numer / denom

    intersection_point = [ p0[0] + (t * s10_x), p0[1] + (t * s10_y) ]


    return intersection_point

score 11 · Accepted Answer

找到两条线段的正确交点是一项具有大量边缘情况的非平凡任务。这是一个文档齐全、工作良好且经过测试的 Java 解决方案。

本质上，在找到两条线段的交点时会发生三件事：

线段不相交
有一个独特的交点
交叉口是另一段

注意：在代码中，我假设 x1 = x2 和 y1 = y2 的线段 (x1, y1), (x2, y2) 是有效的线段。从数学上讲，线段由不同的点组成，但为了完整起见，我允许线段在此实现中成为点。

代码取自我的github 仓库

/**
 * This snippet finds the intersection of two line segments.
 * The intersection may either be empty, a single point or the
 * intersection is a subsegment there's an overlap.
 */

import static java.lang.Math.abs;
import static java.lang.Math.max;
import static java.lang.Math.min;

import java.util.ArrayList;
import java.util.List;

public class LineSegmentLineSegmentIntersection {

  // Small epsilon used for double value comparison.
  private static final double EPS = 1e-5;

  // 2D Point class.
  public static class Pt {
    double x, y;
    public Pt(double x, double y) {
      this.x = x; 
      this.y = y;
    }
    public boolean equals(Pt pt) {
      return abs(x - pt.x) < EPS && abs(y - pt.y) < EPS;
    }
  }

  // Finds the orientation of point 'c' relative to the line segment (a, b)
  // Returns  0 if all three points are collinear.
  // Returns -1 if 'c' is clockwise to segment (a, b), i.e right of line formed by the segment.
  // Returns +1 if 'c' is counter clockwise to segment (a, b), i.e left of line
  // formed by the segment.
  public static int orientation(Pt a, Pt b, Pt c) {
    double value = (b.y - a.y) * (c.x - b.x) - 
                   (b.x - a.x) * (c.y - b.y);
    if (abs(value) < EPS) return 0;
    return (value > 0) ? -1 : +1;
  }

  // Tests whether point 'c' is on the line segment (a, b).
  // Ensure first that point c is collinear to segment (a, b) and
  // then check whether c is within the rectangle formed by (a, b)
  public static boolean pointOnLine(Pt a, Pt b, Pt c) {
    return orientation(a, b, c) == 0 && 
           min(a.x, b.x) <= c.x && c.x <= max(a.x, b.x) && 
           min(a.y, b.y) <= c.y && c.y <= max(a.y, b.y);
  }

  // Determines whether two segments intersect.
  public static boolean segmentsIntersect(Pt p1, Pt p2, Pt p3, Pt p4) {

    // Get the orientation of points p3 and p4 in relation
    // to the line segment (p1, p2)
    int o1 = orientation(p1, p2, p3);
    int o2 = orientation(p1, p2, p4);
    int o3 = orientation(p3, p4, p1);
    int o4 = orientation(p3, p4, p2);

    // If the points p1, p2 are on opposite sides of the infinite
    // line formed by (p3, p4) and conversly p3, p4 are on opposite
    // sides of the infinite line formed by (p1, p2) then there is
    // an intersection.
    if (o1 != o2 && o3 != o4) return true;

    // Collinear special cases (perhaps these if checks can be simplified?)
    if (o1 == 0 && pointOnLine(p1, p2, p3)) return true;
    if (o2 == 0 && pointOnLine(p1, p2, p4)) return true;
    if (o3 == 0 && pointOnLine(p3, p4, p1)) return true;
    if (o4 == 0 && pointOnLine(p3, p4, p2)) return true;

    return false;
  }

  public static List<Pt> getCommonEndpoints(Pt p1, Pt p2, Pt p3, Pt p4) {

    List<Pt> points = new ArrayList<>();

    if (p1.equals(p3)) {
      points.add(p1);
      if (p2.equals(p4)) points.add(p2);

    } else if (p1.equals(p4)) {
      points.add(p1);
      if (p2.equals(p3)) points.add(p2);

    } else if (p2.equals(p3)) {
      points.add(p2);
      if (p1.equals(p4)) points.add(p1);

    } else if (p2.equals(p4)) {
      points.add(p2);
      if (p1.equals(p3)) points.add(p1);
    }

    return points;
  }

  // Finds the intersection point(s) of two line segments. Unlike regular line 
  // segments, segments which are points (x1 = x2 and y1 = y2) are allowed.
  public static Pt[] lineSegmentLineSegmentIntersection(Pt p1, Pt p2, Pt p3, Pt p4) {

    // No intersection.
    if (!segmentsIntersect(p1, p2, p3, p4)) return new Pt[]{};

    // Both segments are a single point.
    if (p1.equals(p2) && p2.equals(p3) && p3.equals(p4))
      return new Pt[]{p1};

    List<Pt> endpoints = getCommonEndpoints(p1, p2, p3, p4);
    int n = endpoints.size();

    // One of the line segments is an intersecting single point.
    // NOTE: checking only n == 1 is insufficient to return early
    // because the solution might be a sub segment.
    boolean singleton = p1.equals(p2) || p3.equals(p4);
    if (n == 1 && singleton) return new Pt[]{endpoints.get(0)};

    // Segments are equal.
    if (n == 2) return new Pt[]{endpoints.get(0), endpoints.get(1)};

    boolean collinearSegments = (orientation(p1, p2, p3) == 0) && 
                                (orientation(p1, p2, p4) == 0);

    // The intersection will be a sub-segment of the two
    // segments since they overlap each other.
    if (collinearSegments) {

      // Segment #2 is enclosed in segment #1
      if (pointOnLine(p1, p2, p3) && pointOnLine(p1, p2, p4))
        return new Pt[]{p3, p4};

      // Segment #1 is enclosed in segment #2
      if (pointOnLine(p3, p4, p1) && pointOnLine(p3, p4, p2))
        return new Pt[]{p1, p2};

      // The subsegment is part of segment #1 and part of segment #2.
      // Find the middle points which correspond to this segment.
      Pt midPoint1 = pointOnLine(p1, p2, p3) ? p3 : p4;
      Pt midPoint2 = pointOnLine(p3, p4, p1) ? p1 : p2;

      // There is actually only one middle point!
      if (midPoint1.equals(midPoint2)) return new Pt[]{midPoint1};

      return new Pt[]{midPoint1, midPoint2};
    }

    /* Beyond this point there is a unique intersection point. */

    // Segment #1 is a vertical line.
    if (abs(p1.x - p2.x) < EPS) {
      double m = (p4.y - p3.y) / (p4.x - p3.x);
      double b = p3.y - m * p3.x;
      return new Pt[]{new Pt(p1.x, m * p1.x + b)};
    }

    // Segment #2 is a vertical line.
    if (abs(p3.x - p4.x) < EPS) {
      double m = (p2.y - p1.y) / (p2.x - p1.x);
      double b = p1.y - m * p1.x;
      return new Pt[]{new Pt(p3.x, m * p3.x + b)};
    }

    double m1 = (p2.y - p1.y) / (p2.x - p1.x);
    double m2 = (p4.y - p3.y) / (p4.x - p3.x);
    double b1 = p1.y - m1 * p1.x;
    double b2 = p3.y - m2 * p3.x;
    double x = (b2 - b1) / (m1 - m2);
    double y = (m1 * b2 - m2 * b1) / (m1 - m2);

    return new Pt[]{new Pt(x, y)};
  }

}

这是一个简单的使用示例：

  public static void main(String[] args) {

    // Segment #1 is (p1, p2), segment #2 is (p3, p4)
    Pt p1, p2, p3, p4;

    p1 = new Pt(-2, 4); p2 = new Pt(3, 3);
    p3 = new Pt(0, 0);  p4 = new Pt(2, 4);
    Pt[] points = lineSegmentLineSegmentIntersection(p1, p2, p3, p4);
    Pt point = points[0];

    // Prints: (1.636, 3.273)
    System.out.printf("(%.3f, %.3f)\n", point.x, point.y);

    p1 = new Pt(-10, 0); p2 = new Pt(+10, 0);
    p3 = new Pt(-5, 0);  p4 = new Pt(+5, 0);
    points = lineSegmentLineSegmentIntersection(p1, p2, p3, p4);
    Pt point1 = points[0], point2 = points[1];

    // Prints: (-5.000, 0.000) (5.000, 0.000)
    System.out.printf("(%.3f, %.3f) (%.3f, %.3f)\n", point1.x, point1.y, point2.x, point2.y);
  }

score 8 · Accepted Answer

上面有很多解决方案，但我认为下面的解决方案非常简单易懂。

两个线段向量 AB 和向量 CD 相交当且仅当

端点 a 和 b 位于段 CD 的相对两侧。
端点 c 和 d 位于线段 AB 的相对侧。

更具体地说，当且仅当两个三元组 a、c、d 和 b、c、d 中的一个恰好是逆时针顺序时，a 和 b 位于段 CD 的相对侧。

Intersect(a, b, c, d)
 if CCW(a, c, d) == CCW(b, c, d)
    return false;
 else if CCW(a, b, c) == CCW(a, b, d)
    return false;
 else
    return true;

这里 CCW 表示逆时针方向，它根据点的方向返回真/假。

资料来源：http ://compgeom.cs.uiuc.edu/~jeffe/teaching/373/notes/x06-sweepline.pdf 第2页

score 8 · Accepted Answer

C 和 Objective-C

基于加雷斯里斯的回答

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，但你应该很容易知道发生了什么。这是在这个 repo 上公开的https://github.com/hfossli/AGGeometryKit/

score 8 · Accepted Answer

只是想提一下，可以在数字食谱系列中找到一个很好的解释和明确的解决方案。我有第 3 版，答案在第 1117 页，第 21.4 节。在 Marina Gavrilova Reliable Line Section Intersection Testing的论文中可以找到另一种具有不同命名法的解决方案。在我看来，她的解决方案要简单一些。

我的实现如下：

bool NuGeometry::IsBetween(const double& x0, const double& x, const double& x1){
   return (x >= x0) && (x <= x1);
}

bool NuGeometry::FindIntersection(const double& x0, const double& y0, 
     const double& x1, const double& y1,
     const double& a0, const double& b0, 
     const double& a1, const double& b1, 
     double& xy, double& ab) {
   // four endpoints are x0, y0 & x1,y1 & a0,b0 & a1,b1
   // returned values xy and ab are the fractional distance along xy and ab
   // and are only defined when the result is true

   bool partial = false;
   double denom = (b0 - b1) * (x0 - x1) - (y0 - y1) * (a0 - a1);
   if (denom == 0) {
      xy = -1;
      ab = -1;
   } else {
      xy = (a0 * (y1 - b1) + a1 * (b0 - y1) + x1 * (b1 - b0)) / denom;
      partial = NuGeometry::IsBetween(0, xy, 1);
      if (partial) {
         // no point calculating this unless xy is between 0 & 1
         ab = (y1 * (x0 - a1) + b1 * (x1 - x0) + y0 * (a1 - x1)) / denom; 
      }
   }
   if ( partial && NuGeometry::IsBetween(0, ab, 1)) {
      ab = 1-ab;
      xy = 1-xy;
      return true;
   }  else return false;
}

score 6 · Accepted Answer

这对我来说效果很好。取自这里。

 // 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;  
   }  
 }

score 6 · Accepted Answer

我尝试了其中一些答案，但它们对我不起作用（对不起，伙计们）；经过更多的网络搜索后，我发现了这个。

对他的代码稍作修改，我现在有了这个函数，它将返回交点，或者如果没有找到交点，它将返回-1，-1。

    Public Function intercetion(ByVal ax As Integer, ByVal ay As Integer, ByVal bx As Integer, ByVal by As Integer, ByVal cx As Integer, ByVal cy As Integer, ByVal dx As Integer, ByVal dy As Integer) As Point
    '//  Determines the intersection point of the line segment defined by points A and B
    '//  with the line segment defined by points C and D.
    '//
    '//  Returns YES if the intersection point was found, and stores that point in X,Y.
    '//  Returns NO if there is no determinable intersection point, in which case X,Y will
    '//  be unmodified.

    Dim distAB, theCos, theSin, newX, ABpos As Double

    '//  Fail if either line segment is zero-length.
    If ax = bx And ay = by Or cx = dx And cy = dy Then Return New Point(-1, -1)

    '//  Fail if the segments share an end-point.
    If ax = cx And ay = cy Or bx = cx And by = cy Or ax = dx And ay = dy Or bx = dx And by = dy Then Return New Point(-1, -1)

    '//  (1) Translate the system so that point A is on the origin.
    bx -= ax
    by -= ay
    cx -= ax
    cy -= ay
    dx -= ax
    dy -= ay

    '//  Discover the length of segment A-B.
    distAB = Math.Sqrt(bx * bx + by * by)

    '//  (2) Rotate the system so that point B is on the positive X axis.
    theCos = bx / distAB
    theSin = by / distAB
    newX = cx * theCos + cy * theSin
    cy = cy * theCos - cx * theSin
    cx = newX
    newX = dx * theCos + dy * theSin
    dy = dy * theCos - dx * theSin
    dx = newX

    '//  Fail if segment C-D doesn't cross line A-B.
    If cy < 0 And dy < 0 Or cy >= 0 And dy >= 0 Then Return New Point(-1, -1)

    '//  (3) Discover the position of the intersection point along line A-B.
    ABpos = dx + (cx - dx) * dy / (dy - cy)

    '//  Fail if segment C-D crosses line A-B outside of segment A-B.
    If ABpos < 0 Or ABpos > distAB Then Return New Point(-1, -1)

    '//  (4) Apply the discovered position to line A-B in the original coordinate system.
    '*X=Ax+ABpos*theCos
    '*Y=Ay+ABpos*theSin

    '//  Success.
    Return New Point(ax + ABpos * theCos, ay + ABpos * theSin)
End Function

score 6 · Accepted Answer

Gavin 的回答似乎引起了一些兴趣，cortijon 在评论中提出了一个 javascript 版本，而 iMalc 提供了一个计算量略少的版本。一些人指出了各种代码提案的缺点，而其他人则评论了一些代码提案的效率。

iMalc 通过 Gavin 的回答提供的算法是我目前在 javascript 项目中使用的算法，如果它可以帮助任何人，我只想在这里提供一个清理后的版本。

// 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;
            }
        }
    }
};

score 5 · Accepted Answer

我认为这个问题有一个更简单的解决方案。我今天想出了另一个想法，它似乎工作得很好（至少现在是 2D 的）。您所要做的就是计算两条线之间的交点，然后检查计算出的交点是否在两条线段的边界框内。如果是，则线段相交。而已。

编辑：

这就是我计算交集的方式（我不知道我在哪里找到了这个代码片段）

Point3D

来自

System.Windows.Media.Media3D

public static Point3D? Intersection(Point3D start1, Point3D end1, Point3D start2, Point3D end2) {

        double a1 = end1.Y - start1.Y;
        double b1 = start1.X - end1.X;
        double c1 = a1 * start1.X + b1 * start1.Y;

        double a2 = end2.Y - start2.Y;
        double b2 = start2.X - end2.X;
        double c2 = a2 * start2.X + b2 * start2.Y;

        double det = a1 * b2 - a2 * b1;
        if (det == 0) { // lines are parallel
            return null;
        }

        double x = (b2 * c1 - b1 * c2) / det;
        double y = (a1 * c2 - a2 * c1) / det;

        return new Point3D(x, y, 0.0);
    }

这是我的（为回答目的而简化）BoundingBox 类：

public class BoundingBox {
    private Point3D min = new Point3D();
    private Point3D max = new Point3D();

    public BoundingBox(Point3D point) {
        min = point;
        max = point;
    }

    public Point3D Min {
        get { return min; }
        set { min = value; }
    }

    public Point3D Max {
        get { return max; }
        set { max = value; }
    }

    public bool Contains(BoundingBox box) {
        bool contains =
            min.X <= box.min.X && max.X >= box.max.X &&
            min.Y <= box.min.Y && max.Y >= box.max.Y &&
            min.Z <= box.min.Z && max.Z >= box.max.Z;
        return contains;
    }

    public bool Contains(Point3D point) {
        return Contains(new BoundingBox(point));
    }

}

score 3 · Accepted Answer

此解决方案可能会有所帮助

public static float GetLineYIntesept(PointF p, float slope)
    {
        return p.Y - slope * p.X;
    }

    public static PointF FindIntersection(PointF line1Start, PointF line1End, PointF line2Start, PointF line2End)
    {

        float slope1 = (line1End.Y - line1Start.Y) / (line1End.X - line1Start.X);
        float slope2 = (line2End.Y - line2Start.Y) / (line2End.X - line2Start.X);

        float yinter1 = GetLineYIntesept(line1Start, slope1);
        float yinter2 = GetLineYIntesept(line2Start, slope2);

        if (slope1 == slope2 && yinter1 != yinter2)
            return PointF.Empty;

        float x = (yinter2 - yinter1) / (slope1 - slope2);

        float y = slope1 * x + yinter1;

        return new PointF(x, y);
    }

score 3 · Accepted Answer

我将 Kris 的上述答案移植到了 JavaScript。在尝试了许多不同的答案后，他提供了正确的观点。我以为我疯了，我没有得到我需要的分数。

function getLineLineCollision(p0, p1, p2, p3) {
    var s1, s2;
    s1 = {x: p1.x - p0.x, y: p1.y - p0.y};
    s2 = {x: p3.x - p2.x, y: p3.y - p2.y};

    var s10_x = p1.x - p0.x;
    var s10_y = p1.y - p0.y;
    var s32_x = p3.x - p2.x;
    var s32_y = p3.y - p2.y;

    var denom = s10_x * s32_y - s32_x * s10_y;

    if(denom == 0) {
        return false;
    }

    var denom_positive = denom > 0;

    var s02_x = p0.x - p2.x;
    var s02_y = p0.y - p2.y;

    var s_numer = s10_x * s02_y - s10_y * s02_x;

    if((s_numer < 0) == denom_positive) {
        return false;
    }

    var t_numer = s32_x * s02_y - s32_y * s02_x;

    if((t_numer < 0) == denom_positive) {
        return false;
    }

    if((s_numer > denom) == denom_positive || (t_numer > denom) == denom_positive) {
        return false;
    }

    var t = t_numer / denom;

    var p = {x: p0.x + (t * s10_x), y: p0.y + (t * s10_y)};
    return p;
}

score 2 · Accepted Answer

这基于 Gareth Ree 的回答。如果确实如此，它还会返回线段的重叠。V 用 C++ 编码，是一个简单的向量类。其中 2D 中两个向量的叉积返回单个标量。经我校自动测试系统测试通过。

//Required input point must be colinear with the line
bool on_segment(const V& p, const LineSegment& l)
{
    //If a point is on the line, the sum of the vectors formed by the point to the line endpoints must be equal
    V va = p - l.pa;
    V vb = p - l.pb;
    R ma = va.magnitude();
    R mb = vb.magnitude();
    R ml = (l.pb - l.pa).magnitude();
    R s = ma + mb;
    bool r = s <= ml + epsilon;
    return r;
}

//Compute using vector math
// Returns 0 points if the lines do not intersect or overlap
// Returns 1 point if the lines intersect
//  Returns 2 points if the lines overlap, contain the points where overlapping start starts and stop
std::vector<V> intersect(const LineSegment& la, const LineSegment& lb)
{
    std::vector<V> r;

    //http://stackoverflow.com/questions/563198/how-do-you-detect-where-two-line-segments-intersect
    V oa, ob, da, db; //Origin and direction vectors
    R sa, sb; //Scalar values
    oa = la.pa;
    da = la.pb - la.pa;
    ob = lb.pa;
    db = lb.pb - lb.pa;

    if (da.cross(db) == 0 && (ob - oa).cross(da) == 0) //If colinear
    {
        if (on_segment(lb.pa, la) && on_segment(lb.pb, la))
        {
            r.push_back(lb.pa);
            r.push_back(lb.pb);
            dprintf("colinear, overlapping\n");
            return r;
        }

        if (on_segment(la.pa, lb) && on_segment(la.pb, lb))
        {
            r.push_back(la.pa);
            r.push_back(la.pb);
            dprintf("colinear, overlapping\n");
            return r;
        }

        if (on_segment(la.pa, lb))
            r.push_back(la.pa);

        if (on_segment(la.pb, lb))
            r.push_back(la.pb);

        if (on_segment(lb.pa, la))
            r.push_back(lb.pa);

        if (on_segment(lb.pb, la))
            r.push_back(lb.pb);

        if (r.size() == 0)
            dprintf("colinear, non-overlapping\n");
        else
            dprintf("colinear, overlapping\n");

        return r;
    }

    if (da.cross(db) == 0 && (ob - oa).cross(da) != 0)
    {
        dprintf("parallel non-intersecting\n");
        return r;
    }

    //Math trick db cross db == 0, which is a single scalar in 2D.
    //Crossing both sides with vector db gives:
    sa = (ob - oa).cross(db) / da.cross(db);

    //Crossing both sides with vector da gives
    sb = (oa - ob).cross(da) / db.cross(da);

    if (0 <= sa && sa <= 1 && 0 <= sb && sb <= 1)
    {
        dprintf("intersecting\n");
        r.push_back(oa + da * sa);
        return r;
    }

    dprintf("non-intersecting, non-parallel, non-colinear, non-overlapping\n");
    return r;
}

score 2 · Accepted Answer

我尝试了很多方法，然后我决定自己写。所以这里是：

bool IsBetween (float x, float b1, float b2)
{
   return ( ((x >= (b1 - 0.1f)) && 
        (x <= (b2 + 0.1f))) || 
        ((x >= (b2 - 0.1f)) &&
        (x <= (b1 + 0.1f))));
}

bool IsSegmentsColliding(   POINTFLOAT lineA,
                POINTFLOAT lineB,
                POINTFLOAT line2A,
                POINTFLOAT line2B)
{
    float deltaX1 = lineB.x - lineA.x;
    float deltaX2 = line2B.x - line2A.x;
    float deltaY1 = lineB.y - lineA.y;
    float deltaY2 = line2B.y - line2A.y;

    if (abs(deltaX1) < 0.01f && 
        abs(deltaX2) < 0.01f) // Both are vertical lines
        return false;
    if (abs((deltaY1 / deltaX1) -
        (deltaY2 / deltaX2)) < 0.001f) // Two parallel line
        return false;

    float xCol = (  (   (deltaX1 * deltaX2) * 
                        (line2A.y - lineA.y)) - 
                    (line2A.x * deltaY2 * deltaX1) + 
                    (lineA.x * deltaY1 * deltaX2)) / 
                 ((deltaY1 * deltaX2) - (deltaY2 * deltaX1));
    float yCol = 0;
    if (deltaX1 < 0.01f) // L1 is a vertical line
        yCol = ((xCol * deltaY2) + 
                (line2A.y * deltaX2) - 
                (line2A.x * deltaY2)) / deltaX2;
    else // L1 is acceptable
        yCol = ((xCol * deltaY1) +
                (lineA.y * deltaX1) -
                (lineA.x * deltaY1)) / deltaX1;

    bool isCol =    IsBetween(xCol, lineA.x, lineB.x) &&
            IsBetween(yCol, lineA.y, lineB.y) &&
            IsBetween(xCol, line2A.x, line2B.x) &&
            IsBetween(yCol, line2A.y, line2B.y);
    return isCol;
}

基于这两个公式：（我从线方程和其他公式中简化了它们）

x 的公式

y 的公式

score 2 · Accepted Answer

这是 C# 中线段的基本实现，以及相应的交叉点检测代码。它需要一个名为的 2D 矢量/点结构Vector2f，但您可以将其替换为具有 X/Y 属性的任何其他类型。如果更适合您的需求，您也可以替换float为。double

此代码在我的 .NET 物理库Boing中使用。

public struct LineSegment2f
{
    public Vector2f From { get; }
    public Vector2f To { get; }

    public LineSegment2f(Vector2f @from, Vector2f to)
    {
        From = @from;
        To = to;
    }

    public Vector2f Delta => new Vector2f(To.X - From.X, To.Y - From.Y);

    /// <summary>
    /// Attempt to intersect two line segments.
    /// </summary>
    /// <remarks>
    /// Even if the line segments do not intersect, <paramref name="t"/> and <paramref name="u"/> will be set.
    /// If the lines are parallel, <paramref name="t"/> and <paramref name="u"/> are set to <see cref="float.NaN"/>.
    /// </remarks>
    /// <param name="other">The line to attempt intersection of this line with.</param>
    /// <param name="intersectionPoint">The point of intersection if within the line segments, or empty..</param>
    /// <param name="t">The distance along this line at which intersection would occur, or NaN if lines are collinear/parallel.</param>
    /// <param name="u">The distance along the other line at which intersection would occur, or NaN if lines are collinear/parallel.</param>
    /// <returns><c>true</c> if the line segments intersect, otherwise <c>false</c>.</returns>
    public bool TryIntersect(LineSegment2f other, out Vector2f intersectionPoint, out float t, out float u)
    {
        var p = From;
        var q = other.From;
        var r = Delta;
        var s = other.Delta;

        // t = (q − p) × s / (r × s)
        // u = (q − p) × r / (r × s)

        var denom = Fake2DCross(r, s);

        if (denom == 0)
        {
            // lines are collinear or parallel
            t = float.NaN;
            u = float.NaN;
            intersectionPoint = default(Vector2f);
            return false;
        }

        var tNumer = Fake2DCross(q - p, s);
        var uNumer = Fake2DCross(q - p, r);

        t = tNumer / denom;
        u = uNumer / denom;

        if (t < 0 || t > 1 || u < 0 || u > 1)
        {
            // line segments do not intersect within their ranges
            intersectionPoint = default(Vector2f);
            return false;
        }

        intersectionPoint = p + r * t;
        return true;
    }

    private static float Fake2DCross(Vector2f a, Vector2f b)
    {
        return a.X * b.Y - a.Y * b.X;
    }
}

score 1 · Accepted Answer

用于检查两个给定线段是否相交的 C++ 程序

#include <iostream>
using namespace std;

struct Point
{
    int x;
    int y;
};

// Given three colinear points p, q, r, the function checks if
// point q lies on line segment 'pr'
bool onSegment(Point p, Point q, Point r)
{
    if (q.x <= max(p.x, r.x) && q.x >= min(p.x, r.x) &&
        q.y <= max(p.y, r.y) && q.y >= min(p.y, r.y))
       return true;

    return false;
}

// To find orientation of ordered triplet (p, q, r).
// The function returns following values
// 0 --> p, q and r are colinear
// 1 --> Clockwise
// 2 --> Counterclockwise
int orientation(Point p, Point q, Point r)
{
    // See 10th slides from following link for derivation of the formula
    // http://www.dcs.gla.ac.uk/~pat/52233/slides/Geometry1x1.pdf
    int val = (q.y - p.y) * (r.x - q.x) -
              (q.x - p.x) * (r.y - q.y);

    if (val == 0) return 0;  // colinear

    return (val > 0)? 1: 2; // clock or counterclock wise
}

// The main function that returns true if line segment 'p1q1'
// and 'p2q2' intersect.
bool doIntersect(Point p1, Point q1, Point p2, Point q2)
{
    // Find the four orientations needed for general and
    // special cases
    int o1 = orientation(p1, q1, p2);
    int o2 = orientation(p1, q1, q2);
    int o3 = orientation(p2, q2, p1);
    int o4 = orientation(p2, q2, q1);

    // General case
    if (o1 != o2 && o3 != o4)
        return true;

    // Special Cases
    // p1, q1 and p2 are colinear and p2 lies on segment p1q1
    if (o1 == 0 && onSegment(p1, p2, q1)) return true;

    // p1, q1 and p2 are colinear and q2 lies on segment p1q1
    if (o2 == 0 && onSegment(p1, q2, q1)) return true;

    // p2, q2 and p1 are colinear and p1 lies on segment p2q2
    if (o3 == 0 && onSegment(p2, p1, q2)) return true;

     // p2, q2 and q1 are colinear and q1 lies on segment p2q2
    if (o4 == 0 && onSegment(p2, q1, q2)) return true;

    return false; // Doesn't fall in any of the above cases
}

// Driver program to test above functions
int main()
{
    struct Point p1 = {1, 1}, q1 = {10, 1};
    struct Point p2 = {1, 2}, q2 = {10, 2};

    doIntersect(p1, q1, p2, q2)? cout << "Yes\n": cout << "No\n";

    p1 = {10, 0}, q1 = {0, 10};
    p2 = {0, 0}, q2 = {10, 10};
    doIntersect(p1, q1, p2, q2)? cout << "Yes\n": cout << "No\n";

    p1 = {-5, -5}, q1 = {0, 0};
    p2 = {1, 1}, q2 = {10, 10};
    doIntersect(p1, q1, p2, q2)? cout << "Yes\n": cout << "No\n";

    return 0;
}

score 1 · Accepted Answer

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

score 0 · Accepted Answer

如果矩形的每一边都是一条线段，而用户绘制的部分是一条线段，那么您只需检查用户绘制的线段是否与四个边线段相交。考虑到每个段的起点和终点，这应该是一个相当简单的练习。

score 0 · Accepted Answer

基于 t3chb0t 的回答：

int intersezione_linee(int x1, int y1, int x2, int y2, int x3, int y3, int x4, int y4, int& p_x, int& p_y)
{
   //L1: estremi (x1,y1)(x2,y2) L2: estremi (x3,y3)(x3,y3)
   int d;
   d = (x1-x2)*(y3-y4) - (y1-y2)*(x3-x4);
   if(!d)
       return 0;
   p_x = ((x1*y2-y1*x2)*(x3-x4) - (x1-x2)*(x3*y4-y3*x4))/d;
   p_y = ((x1*y2-y1*x2)*(y3-y4) - (y1-y2)*(x3*y4-y3*x4))/d;
   return 1;
}

int in_bounding_box(int x1, int y1, int x2, int y2, int p_x, int p_y)
{
    return p_x>=x1 && p_x<=x2 && p_y>=y1 && p_y<=y2;

}

int intersezione_segmenti(int x1, int y1, int x2, int y2, int x3, int y3, int x4, int y4, int& p_x, int& p_y)
{
    if (!intersezione_linee(x1,y1,x2,y2,x3,y3,x4,y4,p_x,p_y))
        return 0;

    return in_bounding_box(x1,y1,x2,y2,p_x,p_y) && in_bounding_box(x3,y3,x4,y4,p_x,p_y);
}

score 0 · Accepted Answer

许多答案已将所有计算都包含在一个函数中。如果您需要计算直线斜率、y 轴截距或 x 轴截距以在代码中的其他位置使用，那么您将冗余地进行这些计算。我已经分离出各自的函数，使用了明显的变量名，并注释了我的代码以使其更容易理解。我需要知道线是否无限相交超出其端点，所以在 JavaScript 中：

http://jsfiddle.net/skibulk/evmqq00u/

var point_a = {x:0, y:10},
    point_b = {x:12, y:12},
    point_c = {x:10, y:0},
    point_d = {x:0, y:0},
    slope_ab = slope(point_a, point_b),
    slope_bc = slope(point_b, point_c),
    slope_cd = slope(point_c, point_d),
    slope_da = slope(point_d, point_a),
    yint_ab = y_intercept(point_a, slope_ab),
    yint_bc = y_intercept(point_b, slope_bc),
    yint_cd = y_intercept(point_c, slope_cd),
    yint_da = y_intercept(point_d, slope_da),
    xint_ab = x_intercept(point_a, slope_ab, yint_ab),
    xint_bc = x_intercept(point_b, slope_bc, yint_bc),
    xint_cd = x_intercept(point_c, slope_cd, yint_cd),
    xint_da = x_intercept(point_d, slope_da, yint_da),
    point_aa = intersect(slope_da, yint_da, xint_da, slope_ab, yint_ab, xint_ab),
    point_bb = intersect(slope_ab, yint_ab, xint_ab, slope_bc, yint_bc, xint_bc),
    point_cc = intersect(slope_bc, yint_bc, xint_bc, slope_cd, yint_cd, xint_cd),
    point_dd = intersect(slope_cd, yint_cd, xint_cd, slope_da, yint_da, xint_da);

console.log(point_a, point_b, point_c, point_d);
console.log(slope_ab, slope_bc, slope_cd, slope_da);
console.log(yint_ab, yint_bc, yint_cd, yint_da);
console.log(xint_ab, xint_bc, xint_cd, xint_da);
console.log(point_aa, point_bb, point_cc, point_dd);

function slope(point_a, point_b) {
  var i = (point_b.y - point_a.y) / (point_b.x - point_a.x);
  if (i === -Infinity) return Infinity;
  if (i === -0) return 0;
  return i;
}

function y_intercept(point, slope) {
    // Horizontal Line
    if (slope == 0) return point.y;
  // Vertical Line
    if (slope == Infinity)
  {
    // THE Y-Axis
    if (point.x == 0) return Infinity;
    // No Intercept
    return null;
  }
  // Angled Line
  return point.y - (slope * point.x);
}

function x_intercept(point, slope, yint) {
    // Vertical Line
    if (slope == Infinity) return point.x;
  // Horizontal Line
    if (slope == 0)
  {
    // THE X-Axis
    if (point.y == 0) return Infinity;
    // No Intercept
    return null;
  }
  // Angled Line
  return -yint / slope;
}

// Intersection of two infinite lines
function intersect(slope_a, yint_a, xint_a, slope_b, yint_b, xint_b) {
  if (slope_a == slope_b)
  {
    // Equal Lines
    if (yint_a == yint_b && xint_a == xint_b) return Infinity;
    // Parallel Lines
    return null;
  }
  // First Line Vertical
    if (slope_a == Infinity)
  {
    return {
        x: xint_a,
      y: (slope_b * xint_a) + yint_b
    };
  }
  // Second Line Vertical
    if (slope_b == Infinity)
  {
    return {
        x: xint_b,
      y: (slope_a * xint_b) + yint_a
    };
  }
  // Not Equal, Not Parallel, Not Vertical
  var i = (yint_b - yint_a) / (slope_a - slope_b);
  return {
    x: i,
    y: (slope_a * i) + yint_a
  };
}

score 0 · Accepted Answer

我从“多视图几何”一书中阅读了这些算法

以下文字使用

' 作为转置符号

* 作为点积

x 作为叉积，当用作运算符时

1.线定义

一个点 x_vec = (x, y)' 位于 ax + by + c = 0 线上

我们将 L = (a, b, c)' 表示为 (x, y, 1)' 的点为齐次坐标

线方程可以写成

(x, y, 1)(a, b, c)' = 0 或 x' * L = 0

2. 线的交点

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

假设 x 是一个点、一个向量，并且 x = L1 x L2（L1 叉积 L2）。

请注意，x 始终是 2D 点，如果您对 (L1xL2) 是三元素向量感到困惑，请阅读齐次坐标，而 x 是 2D 坐标。

根据三重产品，我们知道

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

我们将 (L1xL2) 替换为向量 x，然后我们有 L1*x=0，L2*x=0，这意味着 x 位于 L1 和 L2 上，x 是交点。

注意，这里x是齐次坐标，如果x的最后一个元素为零，则表示L1和L2平行。