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## closest point on a line

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### #1Euronomus  Members

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\]


### #20BZEN  Members

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.

### #3Euronomus  Members

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.

### #4Álvaro  Members

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))

### #5Raghar  Members

Posted 17 April 2007 - 09:40 AM

Quote:
 Original post by EuronomusI iterate through all the points in the line checking the distancefor 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 washoping 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.

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].

[Edited by - TheAdmiral on April 18, 2007 6:59:59 AM]

### #70BZEN  Members

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.

### #8NerdInHisShoe  Members

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]

### #9Zipster  Members

Posted 17 April 2007 - 01:14 PM

Quote:
 Original post by oliiiVector 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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