Euronomus 122 Report post Posted April 17, 2007 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\] 0 Share this post Link to post Share on other sites
oliii 2196 Report post Posted April 17, 2007 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. 1 Share this post Link to post Share on other sites
Euronomus 122 Report post Posted April 17, 2007 Thank you, the whole time I was messing with this I kept thinking there had tobe an easier way but all my googling turned up nothing. 0 Share this post Link to post Share on other sites
alvaro 21272 Report post Posted April 17, 2007 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 minimizedist(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 0 Share this post Link to post Share on other sites
Raghar 96 Report post Posted April 17, 2007 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. 0 Share this post Link to post Share on other sites
TheAdmiral 1122 Report post Posted April 17, 2007 Simpler still? For a normalised ray about the origin, the answer ismin_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] 0 Share this post Link to post Share on other sites
oliii 2196 Report post Posted April 17, 2007 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. 0 Share this post Link to post Share on other sites
NerdInHisShoe 130 Report post Posted April 17, 2007 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 productSo 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] = 012 - 4k + 6 -36k = 0-40k = -18k = 0.45[-1 + 0.9, 3 + 2.7][-0.1,5.7] is your point 0 Share this post Link to post Share on other sites
Zipster 2377 Report post Posted April 17, 2007 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] 0 Share this post Link to post Share on other sites