Angle of a given point on a Bezier curve? - actionscript-3

I created this class in actionscript, it returns a given point of the bezier. And what I am trying to achieve is to get the angle of the current point. I searched on the internet but I couldn't found much. How can I do this?
public static function quadraticBezierPoint(u:Number, anchor1:Point, anchor2:Point, control:Point):Point {
var uc:Number = 1 - u;
var posx:Number = Math.pow(uc, 2) * anchor1.x + 2 * uc * u * control.x + Math.pow(u, 2) * anchor2.x;
var posy:Number = Math.pow(uc, 2) * anchor1.y + 2 * uc * u * control.y + Math.pow(u, 2) * anchor2.y;
return new Point(posx, posy);
}

Given:
control points p0, p1, p2
time t
point B is the point on the quadratic bezier curve described by p0, p1, and p2 at time t.
q0 is the point on the linear bezier curve described by p0 and p1 at time t.
q1 is the point on the linear bezier curve described by p1 and p2 at time t.
The line segment between q0 and q1 is tangent to your quadratic bezier curve at point B.
Therefore, the angle of the bezier curve at time t is equal to the slope of the line segment between q0 and q1.
Wikipedia has a lovely gif demonstrating this. The black dot is point B, and the endpoints of the green line segment are q0 and q1.
The principle is identical for bezier curves of higher dimensions. To find the angle of a point on an N-degree bezier curve, find q0 and q1, which are the points on the N-1-degree bezier curves for control points [p0,p1,...,p(N-1)] and [p1, p2,...,pN]. The angle is equal to the slope of the q0-q1 line segment.
In pseudocode:
def bezierCurve(controlPoints, t):
if len(controlPoints) == 1:
return controlPoints[0]
else:
allControlPointsButTheLastOne = controlPoints[:-1]
allControlPointsButTheFirstOne = controlPoints[1:]
q0 = bezierCurve(allControlPointsButTheLatOne, t)
q1 = bezierCurve(allControlPointsButTheFirstOne, t)
return (1-t) * q0 + t * q1
def bezierAngle(controlPoints, t):
q0 = bezierCurve(controlPoints[:-1], t)
q1 = bezierCurve(controlPoints[1:], t)
return math.atan2(q1.y - q0.y, q1.x - q0.x)

After the explanation from Kevin I made a dynamic but simple solution:
public static function quadraticBezierAngle(u:Number, anchor1:Point, anchor2:Point, control:Point):Number {
var uc:Number = 1 - u;
var dx:Number = (uc * control.x + u * anchor2.x) - (uc * anchor1.x + u * control.x);
var dy:Number = (uc * control.y + u * anchor2.y) - (uc * anchor1.y + u * control.y);
return Math.atan2(dy, dx);
}

Related

convex square - matlab

Given an interactive input of 4 points marked clockwise with the mouse pointer, I need to check using Matlab whether the shape that was drawn is a quadrilateral convex or not. I saw some people that suggested a gift wrapping algorithm. My thought was just using tan, such that if I have an angle greater than 180 degrees, the shape is not a convex.
Can you suggest a better way to do it? I'd appreciate your reference to following code:
showImage(imageA)
hold on
% Initially, the list of points is empty.
xy = [];
n = 0;
% Loop, picking up the points.
disp('Please enter corners of place to insert image in clockwise order.')
for i = 1:4
[xi,yi] = ginput(1);
plot(xi,yi,'yo')
xy(:,i) = [xi;yi];
end
%check if this is a convex quadrillateral
a1 = ( xy(2,2) - xy(2,1) ) / ( xy(1,2) - xy(1,1) );
a2 = ( xy(2,3) - xy(2,2) ) / ( xy(1,3) - xy(1,2) );
a3 = ( xy(2,4) - xy(2,3) ) / ( xy(1,4) - xy(1,3) );
a4 = ( xy(2,1) - xy(2,4) ) / ( xy(1,1) - xy(1,4) );
tan1 = abs( atand( (a2-a1) /( 1+a1*a2) ) );
tan2 = abs( atand( (a3-a2) / (1+a3*a2) ) );
tan3 = abs( atand( (a4-a3) / (1+a4*a3) ) );
tan4 = abs( atand( (a1-a4) / (1+a1*a4) ) );
if ((tan1 > 180) | (tan2 > 180) | (tan3 > 180) | (tan4 > 180))
disp('this is not a convex quadrillateral!!')
end
Here is a very simple way to do it:
Take all combinations of 3 points (there are 4 in total).
Check if the fourth point is in the triangle defined by using those points as corners.
If any of the fourth points is in the triangle it is not convex, otherwise it is.
I think this will work for n point in general, if you are prepared to do n+1 checks.

Mathematic hidden collision

I'm working on a project that uses tile engine (self made), and my next task is to create an AI (besides other AIs that are done), this one is tricky because the AI should only spot the player if the player is in the AI's sight. Tried it with for cycles to after calculating the ranges (in tiles) [1 tile = 32*32].
Then I thought about creating an equation of a straight line. And here I am, puzzled in math.
Any idea how could I calculate if it's overlaps one of these "hidden" tiles?
NOTE that I want only use math!
TileInfo.tileData[la[floor(y / 32)][floor(x / 32)]];
//la -> array of tile positions, if it's >0 then there is a tile.
Say that the viewer is at position (x1,y1) and the target at (x2,y2). Now, I am assuming that there is a set of n contiguous tiles along along x and m along y. The lower, left corner of the first of these tiles is at position (x0,y0). The size of tiles are d along x and t along y. Now the math:
The line connecting viewer and target is
y = y1 + (y2 - y1) * (x - x1) / (x2 - x1)
The tiles corners are at points p1 = (x0,y0); p2 = (x0 + n * d, y0); p3 = (x0 + n * d, y0 + m * t); p4 = (x0, y0 + m * t). Now the job is to find if that line crosses any of the 4 segments connecting two consecutive corners. Let's take the segment between p1 and p2 (a horizontal line) defined by y = y0. If you set this into the line equation you can find the possible interception x which I named xi:
y0 = (y2 - y1) * (xi - x1) / (x2 - x1) + y1
You can invert this equation and find the possibx:
xi = x1 + (y0 - y1) * (x2 - x1) / (y2 - y1)
Now if xi > x0 and xi < x0 + n * d you have an interception for this segment. Otherwise you have a free line of sight.
Do the same for the other three segments whose straight lines are defined by p2 -> p3: x = x0 + n * d; p3 -> p4: y = y0 + m * d; and p4 -> p1: x = x0.
Note that when the segment is horizontal (y = const) you have to put this y in the line of sight straight line, calculate x and compare this x with the intercept. If the segment is vertical (x = const) then you have to put x in the straight line equation, calculate y and check if it falls in the interval or not.
A final remark is that you have to take particular care of cases where x1 = x2 or y1 = y2. This are vertical and horizontal line of sights and may lead to division by zero in the above equations. The solution: deal with these cases separately.

Intersection of parabolic curve and line segment

I have an equation for a parabolic curve intersecting a specified point, in my case where the user clicked on a graph.
// this would typically be mouse coords on the graph
var _target:Point = new Point(100, 50);
public static function plot(x:Number, target:Point):Number{
return (x * x) / target.x * (target.y / target.x);
}
This gives a graph such as this:
I also have a series of line segments defined by start and end coordinates:
startX:Number, startY:Number, endX:Number, endY:Number
I need to find if and where this curve intersects these segments (A):
If it's any help, startX is always < endX
I get the feeling there's a fairly straight forward way to do this, but I don't really know what to search for, nor am I very well versed in "proper" math, so actual code examples would be very much appreciated.
UPDATE:
I've got the intersection working, but my solution gives me the coordinate for the wrong side of the y-axis.
Replacing my target coords with A and B respectively, gives this equation for the plot:
(x * x) / A * (B/A)
// this simplifies down to:
(B * x * x) / (A * A)
// which i am the equating to the line's equation
(B * x * x) / (A * A) = m * x + b
// i run this through wolfram alpha (because i have no idea what i'm doing) and get:
(A * A * m - A * Math.sqrt(A * A * m * m + 4 * b * B)) / (2 * B)
This is a correct answer, but I want the second possible variation.
I've managed to correct this by multiplying m with -1 before the calculation and doing the same with the x value the last calculation returns, but that feels like a hack.
SOLUTION:
public static function intersectsSegment(targetX:Number, targetY:Number, startX:Number, startY:Number, endX:Number, endY:Number):Point {
// slope of the line
var m:Number = (endY - startY) / (endX - startX);
// where the line intersects the y-axis
var b:Number = startY - startX * m;
// solve the two variatons of the equation, we may need both
var ix1:Number = solve(targetX, targetY, m, b);
var ix2:Number = solveInverse(targetX, targetY, m, b);
var intersection1:Point;
var intersection2:Point;
// if the intersection is outside the line segment startX/endX it's discarded
if (ix1 > startX && ix1 < endX) intersection1 = new Point(ix1, plot(ix1, targetX, targetY));
if (ix2 > startX && ix2 < endX) intersection2 = new Point(ix2, plot(ix2, targetX, targetY));
// somewhat fiddly code to return the smallest set intersection
if (intersection1 && intersection2) {
// return the intersection with the smaller x value
return intersection1.x < intersection2.x ? intersection1 : intersection2;
} else if (intersection1) {
return intersection1;
}
// this effectively means that we return intersection2 or if that's unset, null
return intersection2;
}
private static function solve(A:Number, B:Number, m:Number, b:Number):Number {
return (m + Math.sqrt(4 * (B / (A * A)) * b + m * m)) / (2 * (B / (A * A)));
}
private static function solveInverse(A:Number, B:Number, m:Number, b:Number):Number {
return (m - Math.sqrt(4 * (B / (A * A)) * b + m * m)) / (2 * (B / (A * A)));
}
public static function plot(x:Number, targetX:Number, targetY:Number):Number{
return (targetY * x * x) / (targetX * targetX);
}
Or, more explicit yet.
If your parabolic curve is
y(x)= A x2+ B x + C (Eq 1)
and your line is
y(x) = m x + b (Eq 2)
The two possible solutions (+ and -) for x are
x = ((-B + m +- Sqrt[4 A b + B^2 - 4 A C - 2 B m + m^2])/(2 A)) (Eq 3)
You should check if your segment endpoints (in x) contains any of these two points. If they do, just replace the corresponding x in the y=m x + b equation to get the y coordinate for the intersection
Edit>
To get the last equation you just say that the "y" in eq 1 is equal to the "y" in eq 2 (because you are looking for an intersection!).
That gives you:
A x2+ B x + C = m x + b
and regrouping
A x2+ (B-m) x + (C-b) = 0
Which is a quadratic equation.
Equation 3 are just the two possible solutions for this quadratic.
Edit 2>
re-reading your code, it seems that your parabola is defined by
y(x) = A x2
where
A = (target.y / (target.x)2)
So in your case Eq 3 becomes simply
x = ((m +- Sqrt[4 A b + m^2])/(2 A)) (Eq 3b)
HTH!
Take the equation for the curve and put your line into y = mx +b form. Solve for x and then determine if X is between your your start and end points for you line segment.
Check out: http://mathcentral.uregina.ca/QQ/database/QQ.09.03/senthil1.html
Are you doing this often enough to desire a separate test to see if an intersection exists before actually computing the intersection point? If so, consider the fact that your parabola is a level set for the function f(x, y) = y - (B * x * x) / (A * A) -- specifically, the one for which f(x, y) = 0. Plug your two endpoints into f(x,y) -- if they have the same sign, they're on the same side of the parabola, while if they have different signs, they're on different sides of the parabola.
Now, you still might have a segment that intersects the parabola twice, and this test doesn't catch that. But something about the way you're defining the problem makes me feel that maybe that's OK for your application.
In other words, you need to calulate the equation for each line segment y = Ax + B compare it to curve equation y = Cx^2 + Dx + E so Ax + B - Cx^2 - Dx - E = 0 and see if there is a solution between startX and endX values.

Shortest distance between a point and a line segment

I need a basic function to find the shortest distance between a point and a line segment. Feel free to write the solution in any language you want; I can translate it into what I'm using (Javascript).
EDIT: My line segment is defined by two endpoints. So my line segment AB is defined by the two points A (x1,y1) and B (x2,y2). I'm trying to find the distance between this line segment and a point C (x3,y3). My geometry skills are rusty, so the examples I've seen are confusing, I'm sorry to admit.
Eli, the code you've settled on is incorrect. A point near the line on which the segment lies but far off one end of the segment would be incorrectly judged near the segment. Update: The incorrect answer mentioned is no longer the accepted one.
Here's some correct code, in C++. It presumes a class 2D-vector class vec2 {float x,y;}, essentially, with operators to add, subract, scale, etc, and a distance and dot product function (i.e. x1 x2 + y1 y2).
float minimum_distance(vec2 v, vec2 w, vec2 p) {
// Return minimum distance between line segment vw and point p
const float l2 = length_squared(v, w); // i.e. |w-v|^2 - avoid a sqrt
if (l2 == 0.0) return distance(p, 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.
const float t = max(0, min(1, dot(p - v, w - v) / l2));
const vec2 projection = v + t * (w - v); // Projection falls on the segment
return distance(p, projection);
}
EDIT: I needed a Javascript implementation, so here it is, with no dependencies (or comments, but it's a direct port of the above). Points are represented as objects with x and y attributes.
function sqr(x) { return x * x }
function dist2(v, w) { return sqr(v.x - w.x) + sqr(v.y - w.y) }
function distToSegmentSquared(p, v, w) {
var l2 = dist2(v, w);
if (l2 == 0) return dist2(p, v);
var t = ((p.x - v.x) * (w.x - v.x) + (p.y - v.y) * (w.y - v.y)) / l2;
t = Math.max(0, Math.min(1, t));
return dist2(p, { x: v.x + t * (w.x - v.x),
y: v.y + t * (w.y - v.y) });
}
function distToSegment(p, v, w) { return Math.sqrt(distToSegmentSquared(p, v, w)); }
EDIT 2: I needed a Java version, but more important, I needed it in 3d instead of 2d.
float dist_to_segment_squared(float px, float py, float pz, float lx1, float ly1, float lz1, float lx2, float ly2, float lz2) {
float line_dist = dist_sq(lx1, ly1, lz1, lx2, ly2, lz2);
if (line_dist == 0) return dist_sq(px, py, pz, lx1, ly1, lz1);
float t = ((px - lx1) * (lx2 - lx1) + (py - ly1) * (ly2 - ly1) + (pz - lz1) * (lz2 - lz1)) / line_dist;
t = constrain(t, 0, 1);
return dist_sq(px, py, pz, lx1 + t * (lx2 - lx1), ly1 + t * (ly2 - ly1), lz1 + t * (lz2 - lz1));
}
Here, in the function parameters, <px,py,pz> is the point in question and the line segment has the endpoints <lx1,ly1,lz1> and <lx2,ly2,lz2>. The function dist_sq (which is assumed to exist) finds the square of the distance between two points.
Here is the simplest complete code in Javascript.
x, y is your target point and x1, y1 to x2, y2 is your line segment.
UPDATED: fix for 0 length line problem from comments.
function pDistance(x, y, x1, y1, x2, y2) {
var A = x - x1;
var B = y - y1;
var C = x2 - x1;
var D = y2 - y1;
var dot = A * C + B * D;
var len_sq = C * C + D * D;
var param = -1;
if (len_sq != 0) //in case of 0 length line
param = dot / len_sq;
var xx, yy;
if (param < 0) {
xx = x1;
yy = y1;
}
else if (param > 1) {
xx = x2;
yy = y2;
}
else {
xx = x1 + param * C;
yy = y1 + param * D;
}
var dx = x - xx;
var dy = y - yy;
return Math.sqrt(dx * dx + dy * dy);
}
UPDATED: Kotlin version
fun getDistance(x: Double, y: Double, x1: Double, y1: Double, x2: Double, y2: Double): Double {
val a = x - x1
val b = y - y1
val c = x2 - x1
val d = y2 - y1
val lenSq = c * c + d * d
val param = if (lenSq != .0) { //in case of 0 length line
val dot = a * c + b * d
dot / lenSq
} else {
-1.0
}
val (xx, yy) = when {
param < 0 -> x1 to y1
param > 1 -> x2 to y2
else -> x1 + param * c to y1 + param * d
}
val dx = x - xx
val dy = y - yy
return hypot(dx, dy)
}
This is an implementation made for FINITE LINE SEGMENTS, not infinite lines like most other functions here seem to be (that's why I made this).
Implementation of theory by Paul Bourke.
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;
}
In my own question thread how to calculate shortest 2D distance between a point and a line segment in all cases in C, C# / .NET 2.0 or Java? I was asked to put a C# answer here when I find one: so here it is, modified from http://www.topcoder.com/tc?d1=tutorials&d2=geometry1&module=Static :
//Compute the dot product AB . BC
private double DotProduct(double[] pointA, double[] pointB, double[] pointC)
{
double[] AB = new double[2];
double[] BC = new double[2];
AB[0] = pointB[0] - pointA[0];
AB[1] = pointB[1] - pointA[1];
BC[0] = pointC[0] - pointB[0];
BC[1] = pointC[1] - pointB[1];
double dot = AB[0] * BC[0] + AB[1] * BC[1];
return dot;
}
//Compute the cross product AB x AC
private double CrossProduct(double[] pointA, double[] pointB, double[] pointC)
{
double[] AB = new double[2];
double[] AC = new double[2];
AB[0] = pointB[0] - pointA[0];
AB[1] = pointB[1] - pointA[1];
AC[0] = pointC[0] - pointA[0];
AC[1] = pointC[1] - pointA[1];
double cross = AB[0] * AC[1] - AB[1] * AC[0];
return cross;
}
//Compute the distance from A to B
double Distance(double[] pointA, double[] pointB)
{
double d1 = pointA[0] - pointB[0];
double d2 = pointA[1] - pointB[1];
return Math.Sqrt(d1 * d1 + d2 * d2);
}
//Compute the distance from AB to C
//if isSegment is true, AB is a segment, not a line.
double LineToPointDistance2D(double[] pointA, double[] pointB, double[] pointC,
bool isSegment)
{
double dist = CrossProduct(pointA, pointB, pointC) / Distance(pointA, pointB);
if (isSegment)
{
double dot1 = DotProduct(pointA, pointB, pointC);
if (dot1 > 0)
return Distance(pointB, pointC);
double dot2 = DotProduct(pointB, pointA, pointC);
if (dot2 > 0)
return Distance(pointA, pointC);
}
return Math.Abs(dist);
}
I'm #SO not to answer but ask questions so I hope I don't get million down votes for some reasons but constructing critic. I just wanted (and was encouraged) to share somebody else's ideas since the solutions in this thread are either with some exotic language (Fortran, Mathematica) or tagged as faulty by somebody. The only useful one (by Grumdrig) for me is written with C++ and nobody tagged it faulty. But it's missing the methods (dot etc.) that are called.
For anyone interested, here's a trivial conversion of Joshua's Javascript code to Objective-C:
- (double)distanceToPoint:(CGPoint)p fromLineSegmentBetween:(CGPoint)l1 and:(CGPoint)l2
{
double A = p.x - l1.x;
double B = p.y - l1.y;
double C = l2.x - l1.x;
double D = l2.y - l1.y;
double dot = A * C + B * D;
double len_sq = C * C + D * D;
double param = dot / len_sq;
double xx, yy;
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;
}
double dx = p.x - xx;
double dy = p.y - yy;
return sqrtf(dx * dx + dy * dy);
}
I needed this solution to work with MKMapPoint so I will share it in case someone else needs it. Just some minor change and this will return the distance in meters :
- (double)distanceToPoint:(MKMapPoint)p fromLineSegmentBetween:(MKMapPoint)l1 and:(MKMapPoint)l2
{
double A = p.x - l1.x;
double B = p.y - l1.y;
double C = l2.x - l1.x;
double D = l2.y - l1.y;
double dot = A * C + B * D;
double len_sq = C * C + D * D;
double param = dot / len_sq;
double xx, yy;
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;
}
return MKMetersBetweenMapPoints(p, MKMapPointMake(xx, yy));
}
In F#, the distance from the point c to the line segment between a and b is given by:
let pointToLineSegmentDistance (a: Vector, b: Vector) (c: Vector) =
let d = b - a
let s = d.Length
let lambda = (c - a) * d / s
let p = (lambda |> max 0.0 |> min s) * d / s
(a + p - c).Length
The vector d points from a to b along the line segment. The dot product of d/s with c-a gives the parameter of the point of closest approach between the infinite line and the point c. The min and max function are used to clamp this parameter to the range 0..s so that the point lies between a and b. Finally, the length of a+p-c is the distance from c to the closest point on the line segment.
Example use:
pointToLineSegmentDistance (Vector(0.0, 0.0), Vector(1.0, 0.0)) (Vector(-1.0, 1.0))
In Mathematica
It uses a parametric description of the segment, and projects the point into the line defined by the segment. As the parameter goes from 0 to 1 in the segment, if the projection is outside this bounds, we compute the distance to the corresponding enpoint, instead of the straight line normal to the segment.
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*)
]
];
Plotting result:
Plot3D[distance[{{0, 0}, {1, 0}}, {xp, yp}], {xp, -1, 2}, {yp, -1, 2}]
Plot those points nearer than a cutoff distance:
Contour Plot:
Hey, I just wrote this yesterday. It's in Actionscript 3.0, which is basically Javascript, though you might not have the same Point class.
//st = start of line segment
//b = the line segment (as in: st + b = end of line segment)
//pt = point to test
//Returns distance from point to line segment.
//Note: nearest point on the segment to the test point is right there if we ever need it
public static function linePointDist( st:Point, b:Point, pt:Point ):Number
{
var nearestPt:Point; //closest point on seqment to pt
var keyDot:Number = dot( b, pt.subtract( st ) ); //key dot product
var bLenSq:Number = dot( b, b ); //Segment length squared
if( keyDot <= 0 ) //pt is "behind" st, use st
{
nearestPt = st
}
else if( keyDot >= bLenSq ) //pt is "past" end of segment, use end (notice we are saving twin sqrts here cuz)
{
nearestPt = st.add(b);
}
else //pt is inside segment, reuse keyDot and bLenSq to get percent of seqment to move in to find closest point
{
var keyDotToPctOfB:Number = keyDot/bLenSq; //REM dot product comes squared
var partOfB:Point = new Point( b.x * keyDotToPctOfB, b.y * keyDotToPctOfB );
nearestPt = st.add(partOfB);
}
var dist:Number = (pt.subtract(nearestPt)).length;
return dist;
}
Also, there's a pretty complete and readable discussion of the problem here: notejot.com
One line solution using arctangents:
The idea is to move A to (0, 0) and rotate triangle clockwise to make C lay on X axis,
when this happen, By will be the distance.
a angle = Atan(Cy - Ay, Cx - Ax);
b angle = Atan(By - Ay, Bx - Ax);
AB length = Sqrt( (Bx - Ax)^2 + (By - Ay)^2 )
By = Sin ( bAngle - aAngle) * ABLength
C#
public double Distance(Point a, Point b, Point c)
{
// normalize points
Point cn = new Point(c.X - a.X, c.Y - a.Y);
Point bn = new Point(b.X - a.X, b.Y - a.Y);
double angle = Math.Atan2(bn.Y, bn.X) - Math.Atan2(cn.Y, cn.X);
double abLength = Math.Sqrt(bn.X*bn.X + bn.Y*bn.Y);
return Math.Sin(angle)*abLength;
}
One line C# (to be converted to SQL)
double distance = Math.Sin(Math.Atan2(b.Y - a.Y, b.X - a.X) - Math.Atan2(c.Y - a.Y, c.X - a.X)) * Math.Sqrt((b.X - a.X) * (b.X - a.X) + (b.Y - a.Y) * (b.Y - a.Y))
For the lazy, here's my Objective-C port of #Grumdrig's solution above:
CGFloat sqr(CGFloat x) { return x*x; }
CGFloat dist2(CGPoint v, CGPoint w) { return sqr(v.x - w.x) + sqr(v.y - w.y); }
CGFloat distanceToSegmentSquared(CGPoint p, CGPoint v, CGPoint w)
{
CGFloat l2 = dist2(v, w);
if (l2 == 0.0f) return dist2(p, v);
CGFloat t = ((p.x - v.x) * (w.x - v.x) + (p.y - v.y) * (w.y - v.y)) / l2;
if (t < 0.0f) return dist2(p, v);
if (t > 1.0f) return dist2(p, w);
return dist2(p, CGPointMake(v.x + t * (w.x - v.x), v.y + t * (w.y - v.y)));
}
CGFloat distanceToSegment(CGPoint point, CGPoint segmentPointV, CGPoint segmentPointW)
{
return sqrtf(distanceToSegmentSquared(point, segmentPointV, segmentPointW));
}
Couldn't resist coding it in python :)
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)
Ditto for fortran :)
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
Here is a more complete spelling out of Grumdrig's solution. This version also returns the closest point itself.
#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 );
}
Consider this modification to Grumdrig's answer above. Many times you'll find that floating point imprecision can cause problems. I'm using doubles in the version below, but you can easily change to floats. The important part is that it uses an epsilon to handle the "slop". In addition, you'll many times want to know WHERE the intersection happened, or if it happened at all. If the returned t is < 0.0 or > 1.0, no collision occurred. However, even if no collision occurred, many times you'll want to know where the closest point on the segment to P is, and thus I use qx and qy to return this location.
double PointSegmentDistanceSquared( double px, double py,
double p1x, double p1y,
double p2x, double p2y,
double& t,
double& qx, double& qy)
{
static const double kMinSegmentLenSquared = 0.00000001; // adjust to suit. If you use float, you'll probably want something like 0.000001f
static const double kEpsilon = 1.0E-14; // adjust to suit. If you use floats, you'll probably want something like 1E-7f
double dx = p2x - p1x;
double dy = p2y - p1y;
double dp1x = px - p1x;
double dp1y = py - p1y;
const double segLenSquared = (dx * dx) + (dy * dy);
if (segLenSquared >= -kMinSegmentLenSquared && segLenSquared <= kMinSegmentLenSquared)
{
// segment is a point.
qx = p1x;
qy = p1y;
t = 0.0;
return ((dp1x * dp1x) + (dp1y * dp1y));
}
else
{
// Project a line from p to the segment [p1,p2]. By considering the line
// extending the segment, parameterized as p1 + (t * (p2 - p1)),
// we find projection of point p onto the line.
// It falls where t = [(p - p1) . (p2 - p1)] / |p2 - p1|^2
t = ((dp1x * dx) + (dp1y * dy)) / segLenSquared;
if (t < kEpsilon)
{
// intersects at or to the "left" of first segment vertex (p1x, p1y). If t is approximately 0.0, then
// intersection is at p1. If t is less than that, then there is no intersection (i.e. p is not within
// the 'bounds' of the segment)
if (t > -kEpsilon)
{
// intersects at 1st segment vertex
t = 0.0;
}
// set our 'intersection' point to p1.
qx = p1x;
qy = p1y;
// Note: If you wanted the ACTUAL intersection point of where the projected lines would intersect if
// we were doing PointLineDistanceSquared, then qx would be (p1x + (t * dx)) and qy would be (p1y + (t * dy)).
}
else if (t > (1.0 - kEpsilon))
{
// intersects at or to the "right" of second segment vertex (p2x, p2y). If t is approximately 1.0, then
// intersection is at p2. If t is greater than that, then there is no intersection (i.e. p is not within
// the 'bounds' of the segment)
if (t < (1.0 + kEpsilon))
{
// intersects at 2nd segment vertex
t = 1.0;
}
// set our 'intersection' point to p2.
qx = p2x;
qy = p2y;
// Note: If you wanted the ACTUAL intersection point of where the projected lines would intersect if
// we were doing PointLineDistanceSquared, then qx would be (p1x + (t * dx)) and qy would be (p1y + (t * dy)).
}
else
{
// The projection of the point to the point on the segment that is perpendicular succeeded and the point
// is 'within' the bounds of the segment. Set the intersection point as that projected point.
qx = p1x + (t * dx);
qy = p1y + (t * dy);
}
// return the squared distance from p to the intersection point. Note that we return the squared distance
// as an optimization because many times you just need to compare relative distances and the squared values
// works fine for that. If you want the ACTUAL distance, just take the square root of this value.
double dpqx = px - qx;
double dpqy = py - qy;
return ((dpqx * dpqx) + (dpqy * dpqy));
}
}
I'm assuming you want to find the shortest distance between the point and a line segment; to do this, you need to find the line (lineA) which is perpendicular to your line segment (lineB) which goes through your point, determine the intersection between that line (lineA) and your line which goes through your line segment (lineB); if that point is between the two points of your line segment, then the distance is the distance between your point and the point you just found which is the intersection of lineA and lineB; if the point is not between the two points of your line segment, you need to get the distance between your point and the closer of two ends of the line segment; this can be done easily by taking the square distance (to avoid a square root) between the point and the two points of the line segment; whichever is closer, take the square root of that one.
Here it is using 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)
}
Grumdrig's C++/JavaScript implementation was very useful to me, so I have provided a Python direct port that I am using. The complete code is here.
class Point(object):
def __init__(self, x, y):
self.x = float(x)
self.y = float(y)
def square(x):
return x * x
def distance_squared(v, w):
return square(v.x - w.x) + square(v.y - w.y)
def distance_point_segment_squared(p, v, w):
# Segment length squared, |w-v|^2
d2 = distance_squared(v, w)
if d2 == 0:
# v == w, return distance to v
return distance_squared(p, v)
# 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 = ((p.x - v.x) * (w.x - v.x) + (p.y - v.y) * (w.y - v.y)) / d2;
if t < 0:
# Beyond v end of the segment
return distance_squared(p, v)
elif t > 1.0:
# Beyond w end of the segment
return distance_squared(p, w)
else:
# Projection falls on the segment.
proj = Point(v.x + t * (w.x - v.x), v.y + t * (w.y - v.y))
# print proj.x, proj.y
return distance_squared(p, proj)
And now my solution as well......
(Javascript)
It is very fast because I try to avoid any Math.pow functions.
As you can see, at the end of the function I have the distance of the line.
code is from the lib http://www.draw2d.org/graphiti/jsdoc/#!/example
/**
* Static util function to determine is a point(px,py) on the line(x1,y1,x2,y2)
* A simple hit test.
*
* #return {boolean}
* #static
* #private
* #param {Number} coronaWidth the accepted corona for the hit test
* #param {Number} X1 x coordinate of the start point of the line
* #param {Number} Y1 y coordinate of the start point of the line
* #param {Number} X2 x coordinate of the end point of the line
* #param {Number} Y2 y coordinate of the end point of the line
* #param {Number} px x coordinate of the point to test
* #param {Number} py y coordinate of the point to test
**/
graphiti.shape.basic.Line.hit= function( coronaWidth, X1, Y1, X2, Y2, px, py)
{
// Adjust vectors relative to X1,Y1
// X2,Y2 becomes relative vector from X1,Y1 to end of segment
X2 -= X1;
Y2 -= Y1;
// px,py becomes relative vector from X1,Y1 to test point
px -= X1;
py -= Y1;
var dotprod = px * X2 + py * Y2;
var projlenSq;
if (dotprod <= 0.0) {
// px,py is on the side of X1,Y1 away from X2,Y2
// distance to segment is length of px,py vector
// "length of its (clipped) projection" is now 0.0
projlenSq = 0.0;
} else {
// switch to backwards vectors relative to X2,Y2
// X2,Y2 are already the negative of X1,Y1=>X2,Y2
// to get px,py to be the negative of px,py=>X2,Y2
// the dot product of two negated vectors is the same
// as the dot product of the two normal vectors
px = X2 - px;
py = Y2 - py;
dotprod = px * X2 + py * Y2;
if (dotprod <= 0.0) {
// px,py is on the side of X2,Y2 away from X1,Y1
// distance to segment is length of (backwards) px,py vector
// "length of its (clipped) projection" is now 0.0
projlenSq = 0.0;
} else {
// px,py is between X1,Y1 and X2,Y2
// dotprod is the length of the px,py vector
// projected on the X2,Y2=>X1,Y1 vector times the
// length of the X2,Y2=>X1,Y1 vector
projlenSq = dotprod * dotprod / (X2 * X2 + Y2 * Y2);
}
}
// Distance to line is now the length of the relative point
// vector minus the length of its projection onto the line
// (which is zero if the projection falls outside the range
// of the line segment).
var lenSq = px * px + py * py - projlenSq;
if (lenSq < 0) {
lenSq = 0;
}
return Math.sqrt(lenSq)<coronaWidth;
};
C#
Adapted from #Grumdrig
public static double MinimumDistanceToLineSegment(this Point p,
Line line)
{
var v = line.StartPoint;
var w = line.EndPoint;
double lengthSquared = DistanceSquared(v, w);
if (lengthSquared == 0.0)
return Distance(p, v);
double t = Math.Max(0, Math.Min(1, DotProduct(p - v, w - v) / lengthSquared));
var projection = v + t * (w - v);
return Distance(p, projection);
}
public static double Distance(Point a, Point b)
{
return Math.Sqrt(DistanceSquared(a, b));
}
public static double DistanceSquared(Point a, Point b)
{
var d = a - b;
return DotProduct(d, d);
}
public static double DotProduct(Point a, Point b)
{
return (a.X * b.X) + (a.Y * b.Y);
}
Matlab code, with built-in "self test" if they call the function with no arguments:
function r = distPointToLineSegment( xy0, xy1, xyP )
% r = distPointToLineSegment( xy0, xy1, xyP )
if( nargin < 3 )
selfTest();
r=0;
else
vx = xy0(1)-xyP(1);
vy = xy0(2)-xyP(2);
ux = xy1(1)-xy0(1);
uy = xy1(2)-xy0(2);
lenSqr= (ux*ux+uy*uy);
detP= -vx*ux + -vy*uy;
if( detP < 0 )
r = norm(xy0-xyP,2);
elseif( detP > lenSqr )
r = norm(xy1-xyP,2);
else
r = abs(ux*vy-uy*vx)/sqrt(lenSqr);
end
end
function selfTest()
%#ok<*NASGU>
disp(['invalid args, distPointToLineSegment running (recursive) self-test...']);
ptA = [1;1]; ptB = [-1;-1];
ptC = [1/2;1/2]; % on the line
ptD = [-2;-1.5]; % too far from line segment
ptE = [1/2;0]; % should be same as perpendicular distance to line
ptF = [1.5;1.5]; % along the A-B but outside of the segment
distCtoAB = distPointToLineSegment(ptA,ptB,ptC)
distDtoAB = distPointToLineSegment(ptA,ptB,ptD)
distEtoAB = distPointToLineSegment(ptA,ptB,ptE)
distFtoAB = distPointToLineSegment(ptA,ptB,ptF)
figure(1); clf;
circle = #(x, y, r, c) rectangle('Position', [x-r, y-r, 2*r, 2*r], ...
'Curvature', [1 1], 'EdgeColor', c);
plot([ptA(1) ptB(1)],[ptA(2) ptB(2)],'r-x'); hold on;
plot(ptC(1),ptC(2),'b+'); circle(ptC(1),ptC(2), 0.5e-1, 'b');
plot(ptD(1),ptD(2),'g+'); circle(ptD(1),ptD(2), distDtoAB, 'g');
plot(ptE(1),ptE(2),'k+'); circle(ptE(1),ptE(2), distEtoAB, 'k');
plot(ptF(1),ptF(2),'m+'); circle(ptF(1),ptF(2), distFtoAB, 'm');
hold off;
axis([-3 3 -3 3]); axis equal;
end
end
coded in t-sql
the point is (#px, #py) and the line segment runs from (#ax, #ay) to (#bx, #by)
create function fn_sqr (#NumberToSquare decimal(18,10))
returns decimal(18,10)
as
begin
declare #Result decimal(18,10)
set #Result = #NumberToSquare * #NumberToSquare
return #Result
end
go
create function fn_Distance(#ax decimal (18,10) , #ay decimal (18,10), #bx decimal(18,10), #by decimal(18,10))
returns decimal(18,10)
as
begin
declare #Result decimal(18,10)
set #Result = (select dbo.fn_sqr(#ax - #bx) + dbo.fn_sqr(#ay - #by) )
return #Result
end
go
create function fn_DistanceToSegmentSquared(#px decimal(18,10), #py decimal(18,10), #ax decimal(18,10), #ay decimal(18,10), #bx decimal(18,10), #by decimal(18,10))
returns decimal(18,10)
as
begin
declare #l2 decimal(18,10)
set #l2 = (select dbo.fn_Distance(#ax, #ay, #bx, #by))
if #l2 = 0
return dbo.fn_Distance(#px, #py, #ax, #ay)
declare #t decimal(18,10)
set #t = ((#px - #ax) * (#bx - #ax) + (#py - #ay) * (#by - #ay)) / #l2
if (#t < 0)
return dbo.fn_Distance(#px, #py, #ax, #ay);
if (#t > 1)
return dbo.fn_Distance(#px, #py, #bx, #by);
return dbo.fn_Distance(#px, #py, #ax + #t * (#bx - #ax), #ay + #t * (#by - #ay))
end
go
create function fn_DistanceToSegment(#px decimal(18,10), #py decimal(18,10), #ax decimal(18,10), #ay decimal(18,10), #bx decimal(18,10), #by decimal(18,10))
returns decimal(18,10)
as
begin
return sqrt(dbo.fn_DistanceToSegmentSquared(#px, #py , #ax , #ay , #bx , #by ))
end
go
--example execution for distance from a point at (6,1) to line segment that runs from (4,2) to (2,1)
select dbo.fn_DistanceToSegment(6, 1, 4, 2, 2, 1)
--result = 2.2360679775
--example execution for distance from a point at (-3,-2) to line segment that runs from (0,-2) to (-2,1)
select dbo.fn_DistanceToSegment(-3, -2, 0, -2, -2, 1)
--result = 2.4961508830
--example execution for distance from a point at (0,-2) to line segment that runs from (0,-2) to (-2,1)
select dbo.fn_DistanceToSegment(0,-2, 0, -2, -2, 1)
--result = 0.0000000000
Looks like just about everyone else on StackOverflow has contributed an answer (23 answers so far), so here's my contribution for C#. This is mostly based on the answer by M. Katz, which in turn is based on the answer by Grumdrig.
public struct MyVector
{
private readonly double _x, _y;
// Constructor
public MyVector(double x, double y)
{
_x = x;
_y = y;
}
// Distance from this point to another point, squared
private double DistanceSquared(MyVector otherPoint)
{
double dx = otherPoint._x - this._x;
double dy = otherPoint._y - this._y;
return dx * dx + dy * dy;
}
// Find the distance from this point to a line segment (which is not the same as from this
// point to anywhere on an infinite line). Also returns the closest point.
public double DistanceToLineSegment(MyVector lineSegmentPoint1, MyVector lineSegmentPoint2,
out MyVector closestPoint)
{
return Math.Sqrt(DistanceToLineSegmentSquared(lineSegmentPoint1, lineSegmentPoint2,
out closestPoint));
}
// Same as above, but avoid using Sqrt(), saves a new nanoseconds in cases where you only want
// to compare several distances to find the smallest or largest, but don't need the distance
public double DistanceToLineSegmentSquared(MyVector lineSegmentPoint1,
MyVector lineSegmentPoint2, out MyVector closestPoint)
{
// Compute length of line segment (squared) and handle special case of coincident points
double segmentLengthSquared = lineSegmentPoint1.DistanceSquared(lineSegmentPoint2);
if (segmentLengthSquared < 1E-7f) // Arbitrary "close enough for government work" value
{
closestPoint = lineSegmentPoint1;
return this.DistanceSquared(closestPoint);
}
// Use the magic formula to compute the "projection" of this point on the infinite line
MyVector lineSegment = lineSegmentPoint2 - lineSegmentPoint1;
double t = (this - lineSegmentPoint1).DotProduct(lineSegment) / segmentLengthSquared;
// Handle the two cases where the projection is not on the line segment, and the case where
// the projection is on the segment
if (t <= 0)
closestPoint = lineSegmentPoint1;
else if (t >= 1)
closestPoint = lineSegmentPoint2;
else
closestPoint = lineSegmentPoint1 + (lineSegment * t);
return this.DistanceSquared(closestPoint);
}
public double DotProduct(MyVector otherVector)
{
return this._x * otherVector._x + this._y * otherVector._y;
}
public static MyVector operator +(MyVector leftVector, MyVector rightVector)
{
return new MyVector(leftVector._x + rightVector._x, leftVector._y + rightVector._y);
}
public static MyVector operator -(MyVector leftVector, MyVector rightVector)
{
return new MyVector(leftVector._x - rightVector._x, leftVector._y - rightVector._y);
}
public static MyVector operator *(MyVector aVector, double aScalar)
{
return new MyVector(aVector._x * aScalar, aVector._y * aScalar);
}
// Added using ReSharper due to CodeAnalysis nagging
public bool Equals(MyVector other)
{
return _x.Equals(other._x) && _y.Equals(other._y);
}
public override bool Equals(object obj)
{
if (ReferenceEquals(null, obj)) return false;
return obj is MyVector && Equals((MyVector) obj);
}
public override int GetHashCode()
{
unchecked
{
return (_x.GetHashCode()*397) ^ _y.GetHashCode();
}
}
public static bool operator ==(MyVector left, MyVector right)
{
return left.Equals(right);
}
public static bool operator !=(MyVector left, MyVector right)
{
return !left.Equals(right);
}
}
And here's a little test program.
public static class JustTesting
{
public static void Main()
{
Stopwatch stopwatch = new Stopwatch();
stopwatch.Start();
for (int i = 0; i < 10000000; i++)
{
TestIt(1, 0, 0, 0, 1, 1, 0.70710678118654757);
TestIt(5, 4, 0, 0, 20, 10, 1.3416407864998738);
TestIt(30, 15, 0, 0, 20, 10, 11.180339887498949);
TestIt(-30, 15, 0, 0, 20, 10, 33.541019662496844);
TestIt(5, 1, 0, 0, 10, 0, 1.0);
TestIt(1, 5, 0, 0, 0, 10, 1.0);
}
stopwatch.Stop();
TimeSpan timeSpan = stopwatch.Elapsed;
}
private static void TestIt(float aPointX, float aPointY,
float lineSegmentPoint1X, float lineSegmentPoint1Y,
float lineSegmentPoint2X, float lineSegmentPoint2Y,
double expectedAnswer)
{
// Katz
double d1 = DistanceFromPointToLineSegment(new MyVector(aPointX, aPointY),
new MyVector(lineSegmentPoint1X, lineSegmentPoint1Y),
new MyVector(lineSegmentPoint2X, lineSegmentPoint2Y));
Debug.Assert(d1 == expectedAnswer);
/*
// Katz using squared distance
double d2 = DistanceFromPointToLineSegmentSquared(new MyVector(aPointX, aPointY),
new MyVector(lineSegmentPoint1X, lineSegmentPoint1Y),
new MyVector(lineSegmentPoint2X, lineSegmentPoint2Y));
Debug.Assert(Math.Abs(d2 - expectedAnswer * expectedAnswer) < 1E-7f);
*/
/*
// Matti (optimized)
double d3 = FloatVector.DistanceToLineSegment(new PointF(aPointX, aPointY),
new PointF(lineSegmentPoint1X, lineSegmentPoint1Y),
new PointF(lineSegmentPoint2X, lineSegmentPoint2Y));
Debug.Assert(Math.Abs(d3 - expectedAnswer) < 1E-7f);
*/
}
private static double DistanceFromPointToLineSegment(MyVector aPoint,
MyVector lineSegmentPoint1, MyVector lineSegmentPoint2)
{
MyVector closestPoint; // Not used
return aPoint.DistanceToLineSegment(lineSegmentPoint1, lineSegmentPoint2,
out closestPoint);
}
private static double DistanceFromPointToLineSegmentSquared(MyVector aPoint,
MyVector lineSegmentPoint1, MyVector lineSegmentPoint2)
{
MyVector closestPoint; // Not used
return aPoint.DistanceToLineSegmentSquared(lineSegmentPoint1, lineSegmentPoint2,
out closestPoint);
}
}
As you can see, I tried to measure the difference between using the version that avoids the Sqrt() method and the normal version. My tests indicate you can maybe save about 2.5%, but I'm not even sure of that - the variations within the various test runs were of the same order of magnitude. I also tried measuring the version posted by Matti (plus an obvious optimization), and that version seems to be about 4% slower than the version based on Katz/Grumdrig code.
Edit: Incidentally, I've also tried measuring a method that finds the distance to an infinite line (not a line segment) using a cross product (and a Sqrt()), and it's about 32% faster.
Here is devnullicus's C++ version converted to C#. For my implementation I needed to know the point of intersection and found his solution to work well.
public static bool PointSegmentDistanceSquared(PointF point, PointF lineStart, PointF lineEnd, out double distance, out PointF intersectPoint)
{
const double kMinSegmentLenSquared = 0.00000001; // adjust to suit. If you use float, you'll probably want something like 0.000001f
const double kEpsilon = 1.0E-14; // adjust to suit. If you use floats, you'll probably want something like 1E-7f
double dX = lineEnd.X - lineStart.X;
double dY = lineEnd.Y - lineStart.Y;
double dp1X = point.X - lineStart.X;
double dp1Y = point.Y - lineStart.Y;
double segLenSquared = (dX * dX) + (dY * dY);
double t = 0.0;
if (segLenSquared >= -kMinSegmentLenSquared && segLenSquared <= kMinSegmentLenSquared)
{
// segment is a point.
intersectPoint = lineStart;
t = 0.0;
distance = ((dp1X * dp1X) + (dp1Y * dp1Y));
}
else
{
// Project a line from p to the segment [p1,p2]. By considering the line
// extending the segment, parameterized as p1 + (t * (p2 - p1)),
// we find projection of point p onto the line.
// It falls where t = [(p - p1) . (p2 - p1)] / |p2 - p1|^2
t = ((dp1X * dX) + (dp1Y * dY)) / segLenSquared;
if (t < kEpsilon)
{
// intersects at or to the "left" of first segment vertex (lineStart.X, lineStart.Y). If t is approximately 0.0, then
// intersection is at p1. If t is less than that, then there is no intersection (i.e. p is not within
// the 'bounds' of the segment)
if (t > -kEpsilon)
{
// intersects at 1st segment vertex
t = 0.0;
}
// set our 'intersection' point to p1.
intersectPoint = lineStart;
// Note: If you wanted the ACTUAL intersection point of where the projected lines would intersect if
// we were doing PointLineDistanceSquared, then intersectPoint.X would be (lineStart.X + (t * dx)) and intersectPoint.Y would be (lineStart.Y + (t * dy)).
}
else if (t > (1.0 - kEpsilon))
{
// intersects at or to the "right" of second segment vertex (lineEnd.X, lineEnd.Y). If t is approximately 1.0, then
// intersection is at p2. If t is greater than that, then there is no intersection (i.e. p is not within
// the 'bounds' of the segment)
if (t < (1.0 + kEpsilon))
{
// intersects at 2nd segment vertex
t = 1.0;
}
// set our 'intersection' point to p2.
intersectPoint = lineEnd;
// Note: If you wanted the ACTUAL intersection point of where the projected lines would intersect if
// we were doing PointLineDistanceSquared, then intersectPoint.X would be (lineStart.X + (t * dx)) and intersectPoint.Y would be (lineStart.Y + (t * dy)).
}
else
{
// The projection of the point to the point on the segment that is perpendicular succeeded and the point
// is 'within' the bounds of the segment. Set the intersection point as that projected point.
intersectPoint = new PointF((float)(lineStart.X + (t * dX)), (float)(lineStart.Y + (t * dY)));
}
// return the squared distance from p to the intersection point. Note that we return the squared distance
// as an optimization because many times you just need to compare relative distances and the squared values
// works fine for that. If you want the ACTUAL distance, just take the square root of this value.
double dpqX = point.X - intersectPoint.X;
double dpqY = point.Y - intersectPoint.Y;
distance = ((dpqX * dpqX) + (dpqY * dpqY));
}
return true;
}
A 2D and 3D solution
Consider a change of basis such that the line segment becomes (0, 0, 0)-(d, 0, 0) and the point (u, v, 0). The shortest distance occurs in that plane and is given by
u ≤ 0 -> d(A, C)
0 ≤ u ≤ d -> |v|
d ≤ u -> d(B, C)
(the distance to one of the endpoints or to the supporting line, depending on the projection to the line. The iso-distance locus is made of two half-circles and two line segments.)
In the above expression, d is the length of the segment AB, and u, v are respectivey the scalar product and (modulus of the) cross product of AB/d (unit vector in the direction of AB) and AC. Hence vectorially,
AB.AC ≤ 0 -> |AC|
0 ≤ AB.AC ≤ AB² -> |ABxAC|/|AB|
AB² ≤ AB.AC -> |BC|
see the Matlab GEOMETRY toolbox in the following website:
http://people.sc.fsu.edu/~jburkardt/m_src/geometry/geometry.html
ctrl+f and type "segment" to find line segment related functions. the functions "segment_point_dist_2d.m" and "segment_point_dist_3d.m" are what you need.
The GEOMETRY codes are available in a C version and a C++ version and a FORTRAN77 version and a FORTRAN90 version and a MATLAB version.
AutoHotkeys version based on Joshua's Javascript:
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)
}
the accepted answer does not work
(e.g. distance between 0,0 and (-10,2,10,2) should be 2).
here's code that works:
def dist2line2(x,y,line):
x1,y1,x2,y2=line
vx = x1 - x
vy = y1 - y
ux = x2-x1
uy = y2-y1
length = ux * ux + uy * uy
det = (-vx * ux) + (-vy * uy) #//if this is < 0 or > length then its outside the line segment
if det < 0:
return (x1 - x)**2 + (y1 - y)**2
if det > length:
return (x2 - x)**2 + (y2 - y)**2
det = ux * vy - uy * vx
return det**2 / length
def dist2line(x,y,line): return math.sqrt(dist2line2(x,y,line))
Didn't see a Java implementation here, so I translated the Javascript function from the accepted answer to Java code:
static double sqr(double x) {
return x * x;
}
static double dist2(DoublePoint v, DoublePoint w) {
return sqr(v.x - w.x) + sqr(v.y - w.y);
}
static double distToSegmentSquared(DoublePoint p, DoublePoint v, DoublePoint w) {
double l2 = dist2(v, w);
if (l2 == 0) return dist2(p, v);
double t = ((p.x - v.x) * (w.x - v.x) + (p.y - v.y) * (w.y - v.y)) / l2;
if (t < 0) return dist2(p, v);
if (t > 1) return dist2(p, w);
return dist2(p, new DoublePoint(
v.x + t * (w.x - v.x),
v.y + t * (w.y - v.y)
));
}
static double distToSegment(DoublePoint p, DoublePoint v, DoublePoint w) {
return Math.sqrt(distToSegmentSquared(p, v, w));
}
static class DoublePoint {
public double x;
public double y;
public DoublePoint(double x, double y) {
this.x = x;
this.y = y;
}
}
WPF version:
public class LineSegment
{
private readonly Vector _offset;
private readonly Vector _vector;
public LineSegment(Point start, Point end)
{
_offset = (Vector)start;
_vector = (Vector)(end - _offset);
}
public double DistanceTo(Point pt)
{
var v = (Vector)pt - _offset;
// first, find a projection point on the segment in parametric form (0..1)
var p = (v * _vector) / _vector.LengthSquared;
// and limit it so it lays inside the segment
p = Math.Min(Math.Max(p, 0), 1);
// now, find the distance from that point to our point
return (_vector * p - v).Length;
}
}
I've made an interactive Desmos graph to demonstrate how to achieve this:
https://www.desmos.com/calculator/kswrm8ddum
The red point is A, the green point is B, and the point C is blue.
You can drag the points in the graph to see the values change.
On the left, the value 's' is the parameter of the line segment (i.e. s = 0 means the point A, and s = 1 means the point B).
The value 'd' is the distance from the third point to the line through A and B.
EDIT:
Fun little insight: the coordinate (s, d) is the coordinate of the third point C in the coordinate system where AB is the unit x-axis, and the unit y-axis is perpendicular to AB.
Python Numpy implementation for 2D coordinate array:
import numpy as np
def dist2d(p1, p2, coords):
''' Distance from points to a finite line btwn p1 -> p2 '''
assert coords.ndim == 2 and coords.shape[1] == 2, 'coords is not 2 dim'
dp = p2 - p1
st = dp[0]**2 + dp[1]**2
u = ((coords[:, 0] - p1[0]) * dp[0] + (coords[:, 1] - p1[1]) * dp[1]) / st
u[u > 1.] = 1.
u[u < 0.] = 0.
dx = (p1[0] + u * dp[0]) - coords[:, 0]
dy = (p1[1] + u * dp[1]) - coords[:, 1]
return np.sqrt(dx**2 + dy**2)
# Usage:
p1 = np.array([0., 0.])
p2 = np.array([0., 10.])
# List of coordinates
coords = np.array(
[[0., 0.],
[5., 5.],
[10., 10.],
[20., 20.]
])
d = dist2d(p1, p2, coords)
# Single coordinate
coord = np.array([25., 25.])
d = dist2d(p1, p2, coord[np.newaxis, :])

Followup: Finding an accurate "distance" between colors

Original Question
I am looking for a function that attempts to quantify how "distant" (or distinct) two colors are. This question is really in two parts:
What color space best represents human vision?
What distance metric in that space best represents human vision (euclidean?)
Convert to La*b* (aka just plain "Lab", and you'll also see reference to "CIELAB"). A good quick measaure of color difference is
(L1-L2)^2 + (a1-a2)^2 + (b1-b2)^2
Color scientists have other more refined measures, which may not be worth the bother, depending on accuracy needed for what you're doing.
The a and b values represent opposing colors in a way similar to how cones work, and may be negative or positive. Neutral colors - white, grays are a=0,b=0. The L is brightness defined in a particular way, from zero (pure darkness) up to whatever.
Crude explanation :>> Given a color, our eyes distinguish between two broad ranges of wavelength - blue vs longer wavelengths. and then, thanks to a more recent genetic mutation, the longer wavelength cones bifurcated into two, distinguishing for us red vs. green.
By the way, it'll be great for your career to rise above your color caveman collegues who know of only "RGB" or "CMYK" which are great for devices but suck for serious perception work. I've worked for imaging scientists who didn't know a thing about this stuff!
For more fun reading on color difference theory, try:
http://white.stanford.edu/~brian/scielab/introduction.html and info
and links on color theory in general, websurf starting with http://www.efg2.com/Lab/Library/Color/ and
http://www.poynton.com/Poynton-color.html
More detail on Lab at http://en.kioskea.net/video/cie-lab.php3 I can't at this time find a non-ugly page that actually had the conversion formulas but I'm sure someone will edit this answer to include one.
as cmetric.htm link above failed for me, as well as many other implementations for color distance I found (after a very long jurney..) how to calculate the best color distance, and .. most scientifically accurate one: deltaE and from 2 RGB (!) values using OpenCV:
This required 3 color space conversions + some code conversion from javascript (http://svn.int64.org/viewvc/int64/colors/colors.js) to C++
And finally the code (seems to work right out of the box, hope no one finds a serious bug there ... but it seems fine after a number of tests)
#include <opencv2/core/core.hpp>
#include <opencv2/imgproc/imgproc.hpp>
#include <opencv2/highgui/highgui.hpp>
#include <opencv2/photo/photo.hpp>
#include <math.h>
using namespace cv;
using namespace std;
#define REF_X 95.047; // Observer= 2°, Illuminant= D65
#define REF_Y 100.000;
#define REF_Z 108.883;
void bgr2xyz( const Vec3b& BGR, Vec3d& XYZ );
void xyz2lab( const Vec3d& XYZ, Vec3d& Lab );
void lab2lch( const Vec3d& Lab, Vec3d& LCH );
double deltaE2000( const Vec3b& bgr1, const Vec3b& bgr2 );
double deltaE2000( const Vec3d& lch1, const Vec3d& lch2 );
void bgr2xyz( const Vec3b& BGR, Vec3d& XYZ )
{
double r = (double)BGR[2] / 255.0;
double g = (double)BGR[1] / 255.0;
double b = (double)BGR[0] / 255.0;
if( r > 0.04045 )
r = pow( ( r + 0.055 ) / 1.055, 2.4 );
else
r = r / 12.92;
if( g > 0.04045 )
g = pow( ( g + 0.055 ) / 1.055, 2.4 );
else
g = g / 12.92;
if( b > 0.04045 )
b = pow( ( b + 0.055 ) / 1.055, 2.4 );
else
b = b / 12.92;
r *= 100.0;
g *= 100.0;
b *= 100.0;
XYZ[0] = r * 0.4124 + g * 0.3576 + b * 0.1805;
XYZ[1] = r * 0.2126 + g * 0.7152 + b * 0.0722;
XYZ[2] = r * 0.0193 + g * 0.1192 + b * 0.9505;
}
void xyz2lab( const Vec3d& XYZ, Vec3d& Lab )
{
double x = XYZ[0] / REF_X;
double y = XYZ[1] / REF_X;
double z = XYZ[2] / REF_X;
if( x > 0.008856 )
x = pow( x , .3333333333 );
else
x = ( 7.787 * x ) + ( 16.0 / 116.0 );
if( y > 0.008856 )
y = pow( y , .3333333333 );
else
y = ( 7.787 * y ) + ( 16.0 / 116.0 );
if( z > 0.008856 )
z = pow( z , .3333333333 );
else
z = ( 7.787 * z ) + ( 16.0 / 116.0 );
Lab[0] = ( 116.0 * y ) - 16.0;
Lab[1] = 500.0 * ( x - y );
Lab[2] = 200.0 * ( y - z );
}
void lab2lch( const Vec3d& Lab, Vec3d& LCH )
{
LCH[0] = Lab[0];
LCH[1] = sqrt( ( Lab[1] * Lab[1] ) + ( Lab[2] * Lab[2] ) );
LCH[2] = atan2( Lab[2], Lab[1] );
}
double deltaE2000( const Vec3b& bgr1, const Vec3b& bgr2 )
{
Vec3d xyz1, xyz2, lab1, lab2, lch1, lch2;
bgr2xyz( bgr1, xyz1 );
bgr2xyz( bgr2, xyz2 );
xyz2lab( xyz1, lab1 );
xyz2lab( xyz2, lab2 );
lab2lch( lab1, lch1 );
lab2lch( lab2, lch2 );
return deltaE2000( lch1, lch2 );
}
double deltaE2000( const Vec3d& lch1, const Vec3d& lch2 )
{
double avg_L = ( lch1[0] + lch2[0] ) * 0.5;
double delta_L = lch2[0] - lch1[0];
double avg_C = ( lch1[1] + lch2[1] ) * 0.5;
double delta_C = lch1[1] - lch2[1];
double avg_H = ( lch1[2] + lch2[2] ) * 0.5;
if( fabs( lch1[2] - lch2[2] ) > CV_PI )
avg_H += CV_PI;
double delta_H = lch2[2] - lch1[2];
if( fabs( delta_H ) > CV_PI )
{
if( lch2[2] <= lch1[2] )
delta_H += CV_PI * 2.0;
else
delta_H -= CV_PI * 2.0;
}
delta_H = sqrt( lch1[1] * lch2[1] ) * sin( delta_H ) * 2.0;
double T = 1.0 -
0.17 * cos( avg_H - CV_PI / 6.0 ) +
0.24 * cos( avg_H * 2.0 ) +
0.32 * cos( avg_H * 3.0 + CV_PI / 30.0 ) -
0.20 * cos( avg_H * 4.0 - CV_PI * 7.0 / 20.0 );
double SL = avg_L - 50.0;
SL *= SL;
SL = SL * 0.015 / sqrt( SL + 20.0 ) + 1.0;
double SC = avg_C * 0.045 + 1.0;
double SH = avg_C * T * 0.015 + 1.0;
double delta_Theta = avg_H / 25.0 - CV_PI * 11.0 / 180.0;
delta_Theta = exp( delta_Theta * -delta_Theta ) * ( CV_PI / 6.0 );
double RT = pow( avg_C, 7.0 );
RT = sqrt( RT / ( RT + 6103515625.0 ) ) * sin( delta_Theta ) * -2.0; // 6103515625 = 25^7
delta_L /= SL;
delta_C /= SC;
delta_H /= SH;
return sqrt( delta_L * delta_L + delta_C * delta_C + delta_H * delta_H + RT * delta_C * delta_H );
}
Hope it helps someone :)
HSL and HSV are better for human color perception. According to Wikipedia:
It is sometimes preferable in working with art materials, digitized images, or other media, to use the HSV or HSL color model over alternative models such as RGB or CMYK, because of differences in the ways the models emulate how humans perceive color. RGB and CMYK are additive and subtractive models, respectively, modelling the way that primary color lights or pigments (respectively) combine to form new colors when mixed.
The easiest distance would of course be to just consider the colors as 3d vectors originating from the same origin, and taking the distance between their end points.
If you need to consider such factors that green is more prominent in judging intensity, you can weigh the values.
ImageMagic provides the following scales:
red: 0.3
green: 0.6
blue: 0.1
Of course, values like this would only be meaningful in relation to other values for other colors, not as something that would be meaningful to humans, so all you could use the values for would be similiarity ordering.
Well, as a first point of call, I'd say of the common metrics HSV (Hue, Saturation and Value) or HSL are better representative of how humans perceive colour than say RGB or CYMK. See HSL, HSV on Wikipedia.
I suppose naively I would plot the points in the HSL space for the two colours and calculate the magnitude of the difference vector. However this would mean that bright yellow and bright green would be considered just as different as green to dark green. But then many consider red and pink two different colours.
Moreover, difference vectors in the same direction in this parameter space are not equal. For instance, the human eye picks up green much better than other colours. A shift in hue from green by the same amount as a shift from red may seem greater. Also a shift in saturation from a small amount to zero is the difference between grey and pink, elsewhere the shift would be the difference between two shades of red.
From a programmers point of view, you would need to plot the difference vectors but modified by a proportionality matrix that would adjust the lengths accordingly in various regions of the HSL space - this would be fairly arbitrary and would be based on various colour theory ideas but be tweaked fairly arbitrarily depending on what you wanted to apply this to.
Even better, you could see if anyone has already done such a thing online...
The Wikipedia article on color differences lists a number of color spaces and distance metrics designed to agree with human perception of color distances.
As someone who is color blind I believe it is good to try to add more separation then normal vision. The most common form of color blindness is red/green deficiency. It doesn't mean that you can't see red or green, it means that it is more difficult to see and more difficult to see the differences. So it takes a larger separation before a color blind person can tell the difference.