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
GLSL版:
// line (a -> b ) point p[enter image description here][1]
float distanceToLine(vec2 a, vec2 b, vec2 p) {
float aside = dot((p - a),(b - a));
if(aside< 0.0) return length(p-a);
float bside = dot((p - b),(a - b));
if(bside< 0.0) return length(p-b);
vec2 pointOnLine = (bside*a + aside*b)/pow(length(a-b),2.0);
return length(p - pointOnLine);
}
其他回答
%Matlab solution by Tim from Cody
function ans=distP2S(x0,y0,x1,y1,x2,y2)
% Point is x0,y0
z=complex(x0-x1,y0-y1);
complex(x2-x1,y2-y1);
abs(z-ans*min(1,max(0,real(z/ans))));
只是遇到了这个,我想我应该添加一个Lua实现。它假设点以表{x=xVal, y=yVal}给出,直线或线段由包含两个点的表给出(见下面的例子):
function distance( P1, P2 )
return math.sqrt((P1.x-P2.x)^2 + (P1.y-P2.y)^2)
end
-- Returns false if the point lies beyond the reaches of the segment
function distPointToSegment( line, P )
if line[1].x == line[2].x and line[1].y == line[2].y then
print("Error: Not a line!")
return false
end
local d = distance( line[1], line[2] )
local t = ((P.x - line[1].x)*(line[2].x - line[1].x) + (P.y - line[1].y)*(line[2].y - line[1].y))/(d^2)
local projection = {}
projection.x = line[1].x + t*(line[2].x-line[1].x)
projection.y = line[1].y + t*(line[2].y-line[1].y)
if t >= 0 and t <= 1 then -- within line segment?
return distance( projection, {x=P.x, y=P.y} )
else
return false
end
end
-- Returns value even if point is further down the line (outside segment)
function distPointToLine( line, P )
if line[1].x == line[2].x and line[1].y == line[2].y then
print("Error: Not a line!")
return false
end
local d = distance( line[1], line[2] )
local t = ((P.x - line[1].x)*(line[2].x - line[1].x) + (P.y - line[1].y)*(line[2].y - line[1].y))/(d^2)
local projection = {}
projection.x = line[1].x + t*(line[2].x-line[1].x)
projection.y = line[1].y + t*(line[2].y-line[1].y)
return distance( projection, {x=P.x, y=P.y} )
end
使用示例:
local P1 = {x = 0, y = 0}
local P2 = {x = 10, y = 10}
local line = { P1, P2 }
local P3 = {x = 7, y = 15}
print(distPointToLine( line, P3 )) -- prints 5.6568542494924
print(distPointToSegment( line, P3 )) -- prints false
特征c++版本的3D线段和点
// Return minimum distance between line segment: head--->tail and point
double MinimumDistance(Eigen::Vector3d head, Eigen::Vector3d tail,Eigen::Vector3d point)
{
double l2 = std::pow((head - tail).norm(),2);
if(l2 ==0.0) return (head - point).norm();// head == tail case
// Consider the line extending the segment, parameterized as head + t (tail - point).
// We find projection of point onto the line.
// It falls where t = [(point-head) . (tail-head)] / |tail-head|^2
// We clamp t from [0,1] to handle points outside the segment head--->tail.
double t = max(0,min(1,(point-head).dot(tail-head)/l2));
Eigen::Vector3d projection = head + t*(tail-head);
return (point - projection).norm();
}
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)
}
这里是与c++答案相同的东西,但移植到pascal。点参数的顺序已经改变,以适应我的代码,但还是一样的东西。
function Dot(const p1, p2: PointF): double;
begin
Result := p1.x * p2.x + p1.y * p2.y;
end;
function SubPoint(const p1, p2: PointF): PointF;
begin
result.x := p1.x - p2.x;
result.y := p1.y - p2.y;
end;
function ShortestDistance2(const p,v,w : PointF) : double;
var
l2,t : double;
projection,tt: PointF;
begin
// Return minimum distance between line segment vw and point p
//l2 := length_squared(v, w); // i.e. |w-v|^2 - avoid a sqrt
l2 := Distance(v,w);
l2 := MPower(l2,2);
if (l2 = 0.0) then begin
result:= Distance(p, v); // v == w case
exit;
end;
// 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
t := Dot(SubPoint(p,v),SubPoint(w,v)) / l2;
if (t < 0.0) then begin
result := Distance(p, v); // Beyond the 'v' end of the segment
exit;
end
else if (t > 1.0) then begin
result := Distance(p, w); // Beyond the 'w' end of the segment
exit;
end;
//projection := v + t * (w - v); // Projection falls on the segment
tt.x := v.x + t * (w.x - v.x);
tt.y := v.y + t * (w.y - v.y);
result := Distance(p, tt);
end;