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closest point on a line

Started by Euronomus Apr 17, 2007 at 2:23 PM 7 replies 43.8k views
Original Post
Euronomus
Euronomus
I'm trying to find how, given a line and a point, to find the closest point on the line to that point. I searched around and could'nt find a standard way, so I've written a function using Bresenhams line alogorithm. I iterate through all the points in the line checking the distance for each point and when the current distance is greater than the previous I return the previous point. It works for most cases but not all and at this point I think I've just been staring at it for to long, so I was hoping someone could point out my problem or point me to a more consice method. this works line = 299, 200, 300, 300 point = 345, 400 this does'nt line = 300, 200, 300, 300 point = 345, 400

import math

def find_closest(line, point):
    x0, y0, x1, y1 = line
    x2, y2 = point
    steep = abs(y1-y0) > abs(x1-x0)
    if steep:
        x0,y0 = y0,x0
        x1,y1 = y1,x1
    if x0 > x1:
        x0,x1 = y1,x0
        y0,y1 = y1,y0
    deltax = x1-y0
    deltay = abs(y1-y0)
    err = 0
    if y0 < y1:
        ystep = 1
    else:
        ystep = -1
    y = x0
    old = None
    for x in range(x0, x1):
        if steep:
            d = math.sqrt(math.pow((y-y2),2)+math.pow((x-x2), 2))
            if old is None:
                old = (d, (y, x))
            elif d > old[0]:
                return old[1]
            old = (d, (y, x))
        else:
            d = math.sqrt(math.pow((x-x2),2)+math.pow((y-y2), 2))
            if old is None:
                 old = (d, (x, y))
            elif d > old[0]:
                return old[1]
            old =  (d, (x, y))
        err += deltay
        if err*2 > deltax:
            y += ystep
            err -= deltax
    return old[1]
if __name__ == '__main__':
    import Tkinter
    class Test(Tkinter.Frame):
        def __init__(self):
            Tkinter.Frame.__init__(self)
            self.grid()
            self.canv = Tkinter.Canvas(self, height=450, width=450, bg='white')
            self.line = Tkinter.Entry(self)
            self.point = Tkinter.Entry(self)
            self.gobutton = Tkinter.Button(self, text='go', command=self.run)
            self.canv.grid(row=0, column=0, columnspan=3)
            self.line.grid(row=1, column=0)
            self.point.grid(row=1, column=1)
            self.gobutton.grid(row=1, column=2)
            self.line.insert('end', '300, 200, 300, 300')
            self.point.insert('end', '345, 400')
        def run(self):
            line = [int(a) for a in self.line.get().split(',')]
            point = [int(a) for a in self.point.get().split(',')]
            x0, y0, x1, y1 = line
            x2, y2 = point
            x3, y3 = find_closest(line, point)
            self.canv.delete('all')
            self.canv.create_line(x0, y0, x1, y1, fill='red')
            self.canv.create_line(x2, y2, x3, y3, fill='blue')
    t = Test()
    t.mainloop()
[source\]
oliii
oliii
it's simpler than that.

Line going through segment[A, B]
point P.

in pseudo code.

Vector GetClosetPoint(Vector A, Vector B, Vector P, bool segmentClamp){    Vector AP = P - A:    Vector AB = B - A;    float ab2 = AB.x*AB.x + AB.y*AB.y;    float ap_ab = AP.x*AB.x + AP.y*AB.y;    float t = ap_ab / ab2;    if (segmentClamp)    {         if (t < 0.0f) t = 0.0f;         else if (t > 1.0f) t = 1.0f;    }    Vector Closest = A + AB * t;    return Closest;}


set 'segmentClamp' to true if you want the closest point on the segment, not just the line.
Everything is better with Metal.
Euronomus
Euronomus
Thank you, the whole time I was messing with this I kept thinking there had to
be an easier way but all my googling turned up nothing.
alvaro
alvaro
For this you need to know about the dot product. Write the line in "ray" form: A point on the line can be expressed as P+t*v, where P is a point in the line, t is a real number and v is a vector along the direction of the line.

If your point is X, you want to minimize
dist(P+t*v , X) = (P+t*v-X).(P+t*v-X) = (P-X + t*v).(P-X + t*v) = (P-X).(P-X) + 2*t*(P-X).(v) + t^2*(v).(v)

If you plot that as a function of t, you get a parabola that achieves its minimum at t=-((P-X).(v))/((v).v()). This can be rewritten as ((X-P).(v))/((v).(v))

Your answer is P+((X-P).(v))/((v).(v))*v

Raghar
Raghar
Quote:
Original post by Euronomus
I iterate through all the points in the line checking the distance
for each point and when the current distance is greater than the previous
I return the previous point. It works for most cases but not all and
at this point I think I've just been staring at it for to long, so I was
hoping someone could point out my problem or point me to a more consice method.


http://www.geometryalgorithms.com/

Look at this site, they have also other quite important algorithms.
TheAdmiral
TheAdmiral
Simpler still? For a normalised ray about the origin, the answer is

min_dist = |ray × point|

That's the modulus of a vector cross-product if it's not clear. Turning an arbitrary line into such a ray is trivial, but I can't think of a way to restrict the line to a segment without losing out to oliii's method.

Edit: I get it now. This is just the distance to the closest point, whereas we need the point itself [pig].

Admiral

[Edited by - TheAdmiral on April 18, 2007 6:59:59 AM]
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oliii
oliii
Note that you will be working with floating point values. Especially for 't'. It's near the range [0, 1] usually.

If you are using ints, you'll need to do some clever shifting, so yuo can get a decent precision.
Everything is better with Metal.
NerdInHisShoe
NerdInHisShoe
Wouldn't the cloest point on the line be when the vector from the point on the line to your point in space is perpenciular to the line?

. is dot product

So if your line was [-1,3] + k[2,6] and point [5,4]

[2,6] . [5-(-1+2k),4-(3+6k)] = 0
[2,6] . [6 - 2k, 1 - 6k] = 0
12 - 4k + 6 -36k = 0
-40k = -18
k = 0.45

[-1 + 0.9, 3 + 2.7]
[-0.1,5.7] is your point
Johnny was a chemist's son by Johnny is no more, for what Johnny thought was H2O was HO4
Zipster
Zipster
Quote:
Original post by oliii
Vector GetClosetPoint(Vector A, Vector B, Vector P, bool segmentClamp)

But what if the point isn't in the closet, or doesn't want to come out [wink]

Here's my solution:
Vector GetClosestPoint(Vector A, Vector B, Vector P, bool segmentClamp){   float min_x = 0.0f, max_x = 0.0f;    for(float y = -3.40282e038; y <= 3.40282e038; y += 1.19209e-007)      for(float x = -3.40282e038; x <= 3.40282e038; x += 1.19209e-007)        if(point_on_line(A, B, Vector(x,y), segmentClamp))           if(length(Vector(x,y) - P) < length(Vector(min_x,min_y) - P))           {              min_x = x;              min_y = y;           }    return Vector(min_x,min_y);}

[totally]



In all seriousness though, the projection method is going to be your best bet. A direct algebraic method based on the dot product works too, but a lot better on paper than in code. As a matter of fact, the projection method might be better on paper too [smile]

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