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\]
closest point on a line
Author
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
it's simpler than that.
Line going through segment[A, B]
point P.
in pseudo code.
set 'segmentClamp' to true if you want the closest point on the segment, not just the line.
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.
Author
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.
be an easier way but all my googling turned up nothing.
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
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
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.
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]
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]
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.
If you are using ints, you'll need to do some clever shifting, so yuo can get a decent precision.
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
. 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
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]
This topic is closed to new replies.
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