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closest point on a line


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#1 Euronomus   Members   -  Reputation: 122

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Posted 17 April 2007 - 08:23 AM

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\]


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#2 0BZEN   Crossbones+   -  Reputation: 2011

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Posted 17 April 2007 - 08:30 AM

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.

#3 Euronomus   Members   -  Reputation: 122

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Posted 17 April 2007 - 08:37 AM

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.

#4 Álvaro   Crossbones+   -  Reputation: 12509

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Posted 17 April 2007 - 08:39 AM

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



#5 Raghar   Members   -  Reputation: 92

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Posted 17 April 2007 - 09:40 AM

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.

#6 TheAdmiral   Members   -  Reputation: 1118

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Posted 17 April 2007 - 09:59 AM

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]

#7 0BZEN   Crossbones+   -  Reputation: 2011

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Posted 17 April 2007 - 10:14 AM

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.

#8 NerdInHisShoe   Members   -  Reputation: 130

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Posted 17 April 2007 - 12:43 PM

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

#9 Zipster   Crossbones+   -  Reputation: 580

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Posted 17 April 2007 - 01:14 PM

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]

</hilarity>

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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