我需要一个基本函数来求点到线段的最短距离。你可以随意用任何你想要的语言来编写解决方案;我可以把它翻译成什么我正在使用(Javascript)。
编辑:线段由两个端点定义。线段AB由两点A (x1,y1)和B (x2,y2)定义。我要求的是这条线段到点C (x3,y3)的距离。我的几何技能生疏了,所以我看到的例子让我很困惑,我很遗憾地承认。
我需要一个基本函数来求点到线段的最短距离。你可以随意用任何你想要的语言来编写解决方案;我可以把它翻译成什么我正在使用(Javascript)。
编辑:线段由两个端点定义。线段AB由两点A (x1,y1)和B (x2,y2)定义。我要求的是这条线段到点C (x3,y3)的距离。我的几何技能生疏了,所以我看到的例子让我很困惑,我很遗憾地承认。
当前回答
这里它使用Swift
/* Distance from a point (p1) to line l1 l2 */
func distanceFromPoint(p: CGPoint, toLineSegment l1: CGPoint, and l2: CGPoint) -> CGFloat {
let A = p.x - l1.x
let B = p.y - l1.y
let C = l2.x - l1.x
let D = l2.y - l1.y
let dot = A * C + B * D
let len_sq = C * C + D * D
let param = dot / len_sq
var xx, yy: CGFloat
if param < 0 || (l1.x == l2.x && l1.y == l2.y) {
xx = l1.x
yy = l1.y
} else if param > 1 {
xx = l2.x
yy = l2.y
} else {
xx = l1.x + param * C
yy = l1.y + param * D
}
let dx = p.x - xx
let dy = p.y - yy
return sqrt(dx * dx + dy * dy)
}
其他回答
该算法基于求出指定直线与包含指定点的正交直线的交点,并计算其距离。在线段的情况下,我们必须检查交点是否在线段的点之间,如果不是这样,则最小距离是指定点与线段的一个端点之间的距离。这是一个c#实现。
Double Distance(Point a, Point b)
{
double xdiff = a.X - b.X, ydiff = a.Y - b.Y;
return Math.Sqrt((long)xdiff * xdiff + (long)ydiff * ydiff);
}
Boolean IsBetween(double x, double a, double b)
{
return ((a <= b && x >= a && x <= b) || (a > b && x <= a && x >= b));
}
Double GetDistance(Point pt, Point pt1, Point pt2, out Point intersection)
{
Double a, x, y, R;
if (pt1.X != pt2.X) {
a = (double)(pt2.Y - pt1.Y) / (pt2.X - pt1.X);
x = (a * (pt.Y - pt1.Y) + a * a * pt1.X + pt.X) / (a * a + 1);
y = a * x + pt1.Y - a * pt1.X; }
else { x = pt1.X; y = pt.Y; }
if (IsBetween(x, pt1.X, pt2.X) && IsBetween(y, pt1.Y, pt2.Y)) {
intersection = new Point((int)x, (int)y);
R = Distance(intersection, pt); }
else {
double d1 = Distance(pt, pt1), d2 = Distance(pt, pt2);
if (d1 < d2) { intersection = pt1; R = d1; }
else { intersection = pt2; R = d2; }}
return R;
}
c#版本
public static FP DistanceToLineSegment(FPVector3 a, FPVector3 b, FPVector3 point)
{
var d = b - a;
var s = d.SqrMagnitude;
var ds = d / s;
var lambda = FPVector3.Dot(point - a, ds);
var p = FPMath.Clamp01(lambda) * d;
return (a + p - point).Magnitude;
}
在数学
它使用线段的参数描述,并将点投影到线段定义的直线中。当参数在线段内从0到1时,如果投影在这个范围之外,我们计算到相应端点的距离,而不是法线到线段的直线。
Clear["Global`*"];
distance[{start_, end_}, pt_] :=
Module[{param},
param = ((pt - start).(end - start))/Norm[end - start]^2; (*parameter. the "."
here means vector product*)
Which[
param < 0, EuclideanDistance[start, pt], (*If outside bounds*)
param > 1, EuclideanDistance[end, pt],
True, EuclideanDistance[pt, start + param (end - start)] (*Normal distance*)
]
];
策划的结果:
Plot3D[distance[{{0, 0}, {1, 0}}, {xp, yp}], {xp, -1, 2}, {yp, -1, 2}]
画出比截断距离更近的点:
等高线图:
本想在GLSL中这样做,但如果可能的话,最好避免所有这些条件。使用clamp()可以避免两种端点情况:
// find closest point to P on line segment AB:
vec3 closest_point_on_line_segment(in vec3 P, in vec3 A, in vec3 B) {
vec3 AP = P - A, AB = B - A;
float l = dot(AB, AB);
if (l <= 0.0000001) return A; // A and B are practically the same
return AP - AB*clamp(dot(AP, AB)/l, 0.0, 1.0); // do the projection
}
如果您可以确定A和B彼此不会非常接近,则可以简化为删除If()。事实上,即使A和B是相同的,我的GPU仍然给出了这个无条件版本的正确结果(但这是使用pre-OpenGL 4.1,其中GLSL除零是未定义的):
// find closest point to P on line segment AB:
vec3 closest_point_on_line_segment(in vec3 P, in vec3 A, in vec3 B) {
vec3 AP = P - A, AB = B - A;
return AP - AB*clamp(dot(AP, AB)/dot(AB, AB), 0.0, 1.0);
}
计算距离是很简单的——GLSL提供了一个distance()函数,你可以在这个最近的点和P。
灵感来自Iñigo Quilez的胶囊距离函数代码
这是一个为有限线段而做的实现,而不是像这里的大多数其他函数那样的无限线(这就是为什么我做这个)。
Paul Bourke的理论实施。
Python:
def dist(x1, y1, x2, y2, x3, y3): # x3,y3 is the point
px = x2-x1
py = y2-y1
norm = px*px + py*py
u = ((x3 - x1) * px + (y3 - y1) * py) / float(norm)
if u > 1:
u = 1
elif u < 0:
u = 0
x = x1 + u * px
y = y1 + u * py
dx = x - x3
dy = y - y3
# Note: If the actual distance does not matter,
# if you only want to compare what this function
# returns to other results of this function, you
# can just return the squared distance instead
# (i.e. remove the sqrt) to gain a little performance
dist = (dx*dx + dy*dy)**.5
return dist
AS3:
public static function segmentDistToPoint(segA:Point, segB:Point, p:Point):Number
{
var p2:Point = new Point(segB.x - segA.x, segB.y - segA.y);
var something:Number = p2.x*p2.x + p2.y*p2.y;
var u:Number = ((p.x - segA.x) * p2.x + (p.y - segA.y) * p2.y) / something;
if (u > 1)
u = 1;
else if (u < 0)
u = 0;
var x:Number = segA.x + u * p2.x;
var y:Number = segA.y + u * p2.y;
var dx:Number = x - p.x;
var dy:Number = y - p.y;
var dist:Number = Math.sqrt(dx*dx + dy*dy);
return dist;
}
Java
private double shortestDistance(float x1,float y1,float x2,float y2,float x3,float y3)
{
float px=x2-x1;
float py=y2-y1;
float temp=(px*px)+(py*py);
float u=((x3 - x1) * px + (y3 - y1) * py) / (temp);
if(u>1){
u=1;
}
else if(u<0){
u=0;
}
float x = x1 + u * px;
float y = y1 + u * py;
float dx = x - x3;
float dy = y - y3;
double dist = Math.sqrt(dx*dx + dy*dy);
return dist;
}