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
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);
}
其他回答
上面有很多解决方案，但我认为下面的解决方案很简单，很容易理解。
矢量AB和矢量CD相交当且仅当
端点a和b在线段CD的两边。 端点c和d在线段AB的对边。
更具体地说，a和b在线段CD的对面当且仅当两个三元组中有一个是逆时针顺序的。
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;
这里的CCW代表逆时针，根据点的方向返回真/假。
来源:http://compgeom.cs.uiuc.edu/~jeffe/teaching/373/notes/x06-sweepline.pdf 第二页
iMalc回答的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
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;
}
}
}
};
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
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);
}
