我如何确定两条直线是否相交，如果相交，在x,y点处?
当前回答
我认为这个问题有一个更简单的解决方案。今天我想到了另一个想法，看起来效果不错(至少在2D中)。你所要做的就是计算两条直线的交点，然后检查计算的交点是否在两条线段的边界框内。如果是，两条线段相交。就是这样。
编辑:
这就是我如何计算交集(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BoundingBox类(为了回答的目的而简化):
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));
}
}
其他回答
有一个很好的方法来解决这个问题就是用向量叉乘。定义二维向量叉乘v × w为vx wy−vy wx。
假设这两条线段从p到p + r，从q到q + s。那么第一行上的任意点都可以表示为p + t r(对于标量参数t)，第二行上的任意点可以表示为q + u s(对于标量参数u)。
如果t和u满足以下条件，两条直线相交:
P + t r = q + u s
两边叉乘s，得到
(p + 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)
同样地，我们可以解出u:
(p + 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，可以方便地将其重写为以下形式(记住s × r =−r × s):
U = q−p × r / (r × s)
现在有四种情况:
If r × s = 0 and (q − p) × r = 0, then the two lines are collinear. In this case, express the endpoints of the second segment (q and q + s) in terms of the equation of the first line segment (p + t r): t0 = (q − p) · r / (r · r) t1 = (q + s − p) · r / (r · r) = t0 + s · r / (r · r) If the interval between t0 and t1 intersects the interval [0, 1] then the line segments are collinear and overlapping; otherwise they are collinear and disjoint. Note that if s and r point in opposite directions, then s · r < 0 and so the interval to be checked is [t1, t0] rather than [t0, t1]. If r × s = 0 and (q − p) × r ≠ 0, then the two lines are parallel and non-intersecting. If r × s ≠ 0 and 0 ≤ t ≤ 1 and 0 ≤ u ≤ 1, the two line segments meet at the point p + t r = q + u s. Otherwise, the two line segments are not parallel but do not intersect.
来源:该方法是3D线相交算法的2维专门化，来自Ronald Goldman发表在Graphics Gems，第304页的文章“三条线在三维空间中的相交”。在三维空间中，通常的情况是直线是倾斜的(既不平行也不相交)，在这种情况下，该方法给出了两条直线最接近的点。
这是基于Gareth Ree的回答。它还返回线段重叠的情况。用c++编写的V是一个简单的向量类。其中二维中两个向量的外积返回一个标量。通过了学校自动测试系统的测试。
//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;
}
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;
}
}
找到两条线段的正确交点是一项具有大量边缘情况的非简单任务。下面是一个用Java编写的、有效的、经过测试的解决方案。
本质上，在求两条线段的交点时，有三种情况会发生:
线段不相交 有一个唯一的交点 交点是另一段
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，我在这个实现中允许线段作为点。
代码是从我的github回购
/**
* 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)};
}
}
下面是一个简单的用法示例:
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);
}
我已经尝试实现上述Jason所描述的算法;不幸的是，虽然在调试数学工作，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) h=。5然而，通过检查可以清楚地看到，这些部分彼此一点也不接近。
将其绘制成图可以清楚地看出，0 < h < 1条件仅表明如果存在截距点，则截距点将位于CD上，而不告诉我们该点是否位于AB上。 为了确保有一个交叉点，你必须对变量g进行对称计算，拦截的要求是: 0 < g < 1 AND 0 < h < 1
