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
以下是Grumdrig解决方案的一个更完整的说明。这个版本还返回最近的点本身。
#include "stdio.h"
#include "math.h"
class Vec2
{
public:
float _x;
float _y;
Vec2()
{
_x = 0;
_y = 0;
}
Vec2( const float x, const float y )
{
_x = x;
_y = y;
}
Vec2 operator+( const Vec2 &v ) const
{
return Vec2( this->_x + v._x, this->_y + v._y );
}
Vec2 operator-( const Vec2 &v ) const
{
return Vec2( this->_x - v._x, this->_y - v._y );
}
Vec2 operator*( const float f ) const
{
return Vec2( this->_x * f, this->_y * f );
}
float DistanceToSquared( const Vec2 p ) const
{
const float dX = p._x - this->_x;
const float dY = p._y - this->_y;
return dX * dX + dY * dY;
}
float DistanceTo( const Vec2 p ) const
{
return sqrt( this->DistanceToSquared( p ) );
}
float DotProduct( const Vec2 p ) const
{
return this->_x * p._x + this->_y * p._y;
}
};
// return minimum distance between line segment vw and point p, and the closest point on the line segment, q
float DistanceFromLineSegmentToPoint( const Vec2 v, const Vec2 w, const Vec2 p, Vec2 * const q )
{
const float distSq = v.DistanceToSquared( w ); // i.e. |w-v|^2 ... avoid a sqrt
if ( distSq == 0.0 )
{
// v == w case
(*q) = v;
return v.DistanceTo( p );
}
// consider the line extending the segment, parameterized as v + t (w - v)
// we find projection of point p onto the line
// it falls where t = [(p-v) . (w-v)] / |w-v|^2
const float t = ( p - v ).DotProduct( w - v ) / distSq;
if ( t < 0.0 )
{
// beyond the v end of the segment
(*q) = v;
return v.DistanceTo( p );
}
else if ( t > 1.0 )
{
// beyond the w end of the segment
(*q) = w;
return w.DistanceTo( p );
}
// projection falls on the segment
const Vec2 projection = v + ( ( w - v ) * t );
(*q) = projection;
return p.DistanceTo( projection );
}
float DistanceFromLineSegmentToPoint( float segmentX1, float segmentY1, float segmentX2, float segmentY2, float pX, float pY, float *qX, float *qY )
{
Vec2 q;
float distance = DistanceFromLineSegmentToPoint( Vec2( segmentX1, segmentY1 ), Vec2( segmentX2, segmentY2 ), Vec2( pX, pY ), &q );
(*qX) = q._x;
(*qY) = q._y;
return distance;
}
void TestDistanceFromLineSegmentToPoint( float segmentX1, float segmentY1, float segmentX2, float segmentY2, float pX, float pY )
{
float qX;
float qY;
float d = DistanceFromLineSegmentToPoint( segmentX1, segmentY1, segmentX2, segmentY2, pX, pY, &qX, &qY );
printf( "line segment = ( ( %f, %f ), ( %f, %f ) ), p = ( %f, %f ), distance = %f, q = ( %f, %f )\n",
segmentX1, segmentY1, segmentX2, segmentY2, pX, pY, d, qX, qY );
}
void TestDistanceFromLineSegmentToPoint()
{
TestDistanceFromLineSegmentToPoint( 0, 0, 1, 1, 1, 0 );
TestDistanceFromLineSegmentToPoint( 0, 0, 20, 10, 5, 4 );
TestDistanceFromLineSegmentToPoint( 0, 0, 20, 10, 30, 15 );
TestDistanceFromLineSegmentToPoint( 0, 0, 20, 10, -30, 15 );
TestDistanceFromLineSegmentToPoint( 0, 0, 10, 0, 5, 1 );
TestDistanceFromLineSegmentToPoint( 0, 0, 0, 10, 1, 5 );
}
其他回答
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:
基于Joshua Javascript的AutoHotkeys版本:
plDist(x, y, x1, y1, x2, y2) {
A:= x - x1
B:= y - y1
C:= x2 - x1
D:= y2 - y1
dot:= A*C + B*D
sqLen:= C*C + D*D
param:= dot / sqLen
if (param < 0 || ((x1 = x2) && (y1 = y2))) {
xx:= x1
yy:= y1
} else if (param > 1) {
xx:= x2
yy:= y2
} else {
xx:= x1 + param*C
yy:= y1 + param*D
}
dx:= x - xx
dy:= y - yy
return sqrt(dx*dx + dy*dy)
}
忍不住用python来编码:)
from math import sqrt, fabs
def pdis(a, b, c):
t = b[0]-a[0], b[1]-a[1] # Vector ab
dd = sqrt(t[0]**2+t[1]**2) # Length of ab
t = t[0]/dd, t[1]/dd # unit vector of ab
n = -t[1], t[0] # normal unit vector to ab
ac = c[0]-a[0], c[1]-a[1] # vector ac
return fabs(ac[0]*n[0]+ac[1]*n[1]) # Projection of ac to n (the minimum distance)
print pdis((1,1), (2,2), (2,0)) # Example (answer is 1.414)
fortran也是一样:)
real function pdis(a, b, c)
real, dimension(0:1), intent(in) :: a, b, c
real, dimension(0:1) :: t, n, ac
real :: dd
t = b - a ! Vector ab
dd = sqrt(t(0)**2+t(1)**2) ! Length of ab
t = t/dd ! unit vector of ab
n = (/-t(1), t(0)/) ! normal unit vector to ab
ac = c - a ! vector ac
pdis = abs(ac(0)*n(0)+ac(1)*n(1)) ! Projection of ac to n (the minimum distance)
end function pdis
program test
print *, pdis((/1.0,1.0/), (/2.0,2.0/), (/2.0,0.0/)) ! Example (answer is 1.414)
end program test
我需要一个Godot (GDscript)的实现,所以我写了一个基于grumdrig接受的答案:
func minimum_distance(v: Vector2, w: Vector2, p: Vector2):
# Return minimum distance between line segment vw and point p
var l2: float = (v - w).length_squared() # i.e. |w-v|^2 - avoid a sqrt
if l2 == 0.0:
return p.distance_to(v) # v == w case
# Consider the line extending the segment, parameterized as v + t (w - v).
# We find projection of point p onto the line.
# It falls where t = [(p-v) . (w-v)] / |w-v|^2
# We clamp t from [0,1] to handle points outside the segment vw.
var t: float = max(0, min(1, (p - v).dot(w - v) / l2))
var projection: Vector2 = v + t * (w - v) # Projection falls on the segment
return p.distance_to(projection)