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
下面是一个基本的c#线段实现,并有相应的交点检测代码。它需要一个名为Vector2f的2D向量/点结构,不过你可以用任何其他具有X/Y属性的类型替换它。如果更适合你的需要,你也可以用double替换float。
这段代码用于我的. net物理库Boing。
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;
}
}
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。
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
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;
}
}
}
};
只是想提一下,一个很好的解释和明确的解决方案可以在数字食谱系列中找到。我有这本书的第三版,答案在1117页21.4节。另一种不同命名的解决方案可以在玛丽娜·加夫里洛娃(Marina Gavrilova)的论文中找到。在我看来,她的解决办法要简单一些。
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;
}
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;
}
}