• FEATURED
• FEATURED
• FEATURED
• FEATURED
• FEATURED

View more

View more

View more

### Image of the Day Submit

IOTD | Top Screenshots

### The latest, straight to your Inbox.

Subscribe to GameDev.net Direct to receive the latest updates and exclusive content.

# closest point on a line

Old topic!

Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.

8 replies to this topic

### #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  GDNet+

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  GDNet+

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]

Old topic!

Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.