I have two sets of data. One is nominal form. The other is actual form. The problem is that when I wish to calculate the form error alone. It's a big problem when the two sets of data isn't "on top of each other". That gives errors that also include positional error.
Both curves are read from a series of data. The nominal shape (black) is made up from many different size radius that are tangent to each other. Its the leading edge of an airfoil profile.
I have tried various methods of "Best-Fit" I've found both here and on where ever google took me. But the problem is that they all smooth my "actual" data. So it get modified and is not keeping it's actual form.
Is there any function in scipy or any other python lib that "simply" can fit my two curves together without altering the actual shape?
I wish for the green curve with red dots to lie as much as possible on top of the black.
Might it be possible to calculate the center of gravity of both curves and then move the actual curve in x and y depending on the value difference from the center point? It might not be the ultimate solution, but it would get closer?
Here is a solution assuming that the nominal form can be described as a conic, i.a as solution of the equation ax^2 + by^2 + cxy + dx + ey = 1. Then, a least square fit can be applied to find the coefficients (a, b, c, d, e).
import numpy as np
import matplotlib.pylab as plt
# Generate example data
t = np.linspace(-2, 2.5, 25)
e, theta = 0.5, 0.3 # ratio minor axis/major & orientation angle major axis
c, s = np.cos(theta), np.sin(theta)
x = c*np.cos(t) - s*e*np.sin(t)
y = s*np.cos(t) + c*e*np.sin(t)
# add noise:
xy = 4*np.vstack((x, y))
xy += .08 *np.random.randn(*xy.shape) + np.random.randn(2, 1)
# Least square fit by a generic conic equation
# a*x^2 + b*y^2 + c*x*y + d*x + e*y = 1
x, y = xy
x = x - x.mean()
y = y - y.mean()
M = np.vstack([x**2, y**2, x*y, x, y]).T
b = np.ones_like(x)
# solve M*w = b
w, res, rank, s = np.linalg.lstsq(M, b, rcond=None)
a, b, c, d, e = w
# Get x, y coordinates for the fitted ellipse:
# using polar coordinates
# x = r*cos(theta), y = r*sin(theta)
# for a given theta, the radius is obtained with the 2nd order eq.:
# (a*ct^2 + b*st^2 + c*cs*st)*r^2 + (d*ct + e*st)*r - 1 = 0
# with ct = cos(theta) and st = sin(theta)
theta = np.linspace(-np.pi, np.pi, 97)
ct, st = np.cos(theta), np.sin(theta)
A = a*ct**2 + b*st**2 + c*ct*st
B = d*ct + e*st
D = B**2 + 4*A
radius = (-B + np.sqrt(D))/2/A
# Graph
plt.plot(radius*ct, radius*st, '-k', label='fitted ellipse');
plt.plot(x, y, 'or', label='measured points');
plt.axis('equal'); plt.legend();
plt.xlabel('x'); plt.ylabel('y');
I am trying to solve an issue when two rectangles intersect/overlap each other. when this happens, i want to know if intersection is True or False. I found a solution, however it is written in C or C++. I want to write these code in Python.
Here is the source: http://www.jeffreythompson.org/collision-detection/rect-rect.php
This is literally the first line of python code I've ever written (I do know C++ however)
def rectRect(r1x, r1y, r1w, r1h, r2x, r2y, r2w, r2h):
# are the sides of one rectangle touching the other?
return r1x + r1w >= r2x and \ # r1 right edge past r2 left
r1x <= r2x + r2w and \ # r1 left edge past r2 right
r1y + r1h >= r2y and \ # r1 top edge past r2 bottom
r1y <= r2y + r2h # r1 bottom edge past r2 top
IMHO rectRect is a really bad name for the function, I kept it from the linked code however.
Following is simple class that can perform both rectangle-rectangle intersection as well as point to rectangle intersection. The difference between earlier solution is that following snippet allows even detection of rotated rectangles.
import numpy as np
import matplotlib.pyplot as plt
class Rectangle:
def __init__(self, center: np.ndarray, dims: np.ndarray, angle: float):
self.corners = self.get_rect_points(center, dims, angle)
self.area = dims[0] * dims[1]
#staticmethod
def get_rect_points(center: np.ndarray, dims: np.ndarray, angle: float):
"""
returns four corners of the rectangle.
bottom left is the first conrner, from there it goes
counter clockwise.
"""
center = np.asarray(center)
length, breadth = dims
angle = np.deg2rad(angle)
corners = np.array([[-length/2, -breadth/2],
[length/2, -breadth/2],
[length/2, breadth/2],
[-length/2, breadth/2]])
rot = np.array([[np.cos(angle), np.sin(angle)], [-np.sin(angle), np.cos(angle)]])
corners = rot.dot(corners.T) + center[:, None]
return corners.T
def is_point_in_collision(self, p: np.ndarray):
"""
check if a point is in collision with the rectangle.
"""
def area_triangle(a, b, c):
return abs((b[0] * a[1] - a[0] * b[1]) + (c[0] * b[1] - b[0] * c[1]) + (a[0] * c[1] - c[0] * a[1])) / 2
area = 0
area += area_triangle(self.corners[0], p, self.corners[3])
area += area_triangle(self.corners[3], p, self.corners[2])
area += area_triangle(self.corners[2], p, self.corners[1])
area += area_triangle(self.corners[1], p, self.corners[0])
return area > self.area
def is_intersect(self, rect_2: Rectangle):
"""
check if any of the four corners of the
rectangle is in collision
"""
if not np.all([self.is_point_in_collision(c) for c in rect_2.corners]):
return True
return False
def plot_rect(p1, p2, p3, p4, color='r'):
ax.plot([p1[0], p2[0]], [p1[1], p2[1]], color)
ax.plot([p2[0], p3[0]], [p2[1], p3[1]], color)
ax.plot([p3[0], p4[0]], [p3[1], p4[1]], color)
ax.plot([p4[0], p1[0]], [p4[1], p1[1]], color)
mid_point = 0.5 * (p1 + p3)
plt.scatter(mid_point[0], mid_point[1], marker='*')
plt.xlim([-1, 1])
plt.ylim([-1, 1])
Following are two samples:
Sample 1:
ax = plt.subplot(111)
st = Rectangle((0.067, 0.476),(0.61, 0.41), 90)
gripper = Rectangle((-0.367, 0.476),(0.21,0.16), 45)
plot_rect(*st.corners)
plot_rect(*gripper.corners)
plt.show()
print(f"gripper and rectangle intersect: {st.is_intersect(gripper)}")
Sample 2:
ax = plt.subplot(111)
st = Rectangle((0.067, 0.476),(0.61, 0.41), 90)
gripper = Rectangle((-0.167, 0.476),(0.21,0.16), 45)
plot_rect(*st.corners)
plot_rect(*gripper.corners)
plt.show()
print(f"gripper and rectangle intersect: {st.is_intersect(gripper)}")
I am making a 2D game in pyglet and use both glTranslatef and glScalef:
def background_motion(dt):
if stars.left:
pyglet.gl.glTranslatef(stars.speed, 0, 0)
stars.translation[0] += stars.speed
if stars.right:
pyglet.gl.glTranslatef(-stars.speed, 0, 0)
stars.translation[0] -= stars.speed
if stars.up:
pyglet.gl.glTranslatef(0, -stars.speed, 0)
stars.translation[1] -= stars.speed
if stars.down:
pyglet.gl.glTranslatef(0, stars.speed, 0)
stars.translation[1] += stars.speed
pyglet.clock.schedule_interval(background_motion, 0.05)
#window.event
def on_mouse_scroll(x, y, scroll_x, scroll_y):
if scroll_y > 0:
stars.scale += 0.01
elif scroll_y < 0:
stars.scale -= 0.01
#window.event
def on_draw():
window.clear()
pyglet.gl.glScalef(stars.scale,stars.scale, 1, 1)
stars.image.draw()
for s in game.ships:
s.draw()
pyglet.gl.glPushMatrix()
pyglet.gl.glLoadIdentity()
#HUD Start
overlay.draw(stars.image.x,stars.image.y,game.ships,stars.scale,stars.image.width)
if game.pause:
pause_text.draw()
#HUD End
pyglet.gl.glPopMatrix()
stars.scale = 1
However I also need the cursor coordinates relative to the background. For the movement I simply added the translation onto the x y coordinates which works however only when I don't scale the matrix:
#window.event
def on_mouse_motion(x, y, dx, dy):
if player.course_setting:
player.projected_heading = (x - stars.translation[0],y -stars.translation[1])
How can I get the cursor coordinates accounting for scale?
You'll have to unproject the pointer position. Projection happens as following:
p_eye = M · p
p_clip = P · p_eye
at this point the primitive is clipped, but we can ignore this for the moment. After clipping comes the homogenous divide, which brings the coordinates into NDC space, i.e. the viewport is treated as a cuboid of dimensions [-1,1]×[-1,1]×[0,1]
p_NDC = p_clip / p_clip.w
From there it's mapped into pixel dimensions. I'm going to omit this step here.
Unprojecting is doing these operations in reverse. There's a small trick in there, regarding the homogenous divide, though; this is kind of an "antisymmetric" (not the proper term for this, but it gets across the point) operation, and happens at the end, for each projection and unprojection. Unprojection hence is
p_NDC.w = 1
p_eye' = inv(P)·p_NDC
p' = inv(M)·p_eye'
p = p' / p'.w
All of this has been wrapped into unproject functions for your convenience by GLU (if you insist on using the fixed function matrix stack) or GLM – but not my linmath.h, though.
I want to make a program which plots a Sierpinsky triangle (of any modulo). In order to do it I've used TkInter. The program generates the fractal by moving a point randomly, always keeping it in the sides. After repeating the process many times, the fractal appears.
However, there's a problem. I don't know how to plot points on a canvas in TkInter. The rest of the program is OK, but I had to "cheat" in order to plot the points by drawing small lines instead of points. It works more or less, but it doesn't have as much resolution as it could have.
Is there a function to plot points on a canvas, or another tool to do it (using Python)? Ideas for improving the rest of the program are also welcome.
Thanks. Here's what I have:
from tkinter import *
import random
import math
def plotpoint(x, y):
global canvas
point = canvas.create_line(x-1, y-1, x+1, y+1, fill = "#000000")
x = 0 #Initial coordinates
y = 0
#x and y will always be in the interval [0, 1]
mod = int(input("What is the modulo of the Sierpinsky triangle that you want to generate? "))
points = int(input("How many points do you want the triangle to have? "))
tkengine = Tk() #Window in which the triangle will be generated
window = Frame(tkengine)
window.pack()
canvas = Canvas(window, height = 700, width = 808, bg = "#FFFFFF") #The dimensions of the canvas make the triangle look equilateral
canvas.pack()
for t in range(points):
#Procedure for placing the points
while True:
#First, randomly choose one of the mod(mod+1)/2 triangles of the first step. a and b are two vectors which point to the chosen triangle. a goes one triangle to the right and b one up-right. The algorithm gives the same probability to every triangle, although it's not efficient.
a = random.randint(0,mod-1)
b = random.randint(0,mod-1)
if a + b < mod:
break
#The previous point is dilated towards the origin of coordinates so that the big triangle of step 0 becomes the small one at the bottom-left of step one (divide by modulus). Then the vectors are added in order to move the point to the same place in another triangle.
x = x / mod + a / mod + b / 2 / mod
y = y / mod + b / mod
#Coordinates [0,1] converted to pixels, for plotting in the canvas.
X = math.floor(x * 808)
Y = math.floor((1-y) * 700)
plotpoint(X, Y)
tkengine.mainloop()
If you are wanting to plot pixels, a canvas is probably the wrong choice. You can create a PhotoImage and modify individual pixels. It's a little slow if you plot each individual pixel, but you can get dramatic speedups if you only call the put method once for each row of the image.
Here's a complete example:
from tkinter import *
import random
import math
def plotpoint(x, y):
global the_image
the_image.put(('#000000',), to=(x,y))
x = 0
y = 0
mod = 3
points = 100000
tkengine = Tk() #Window in which the triangle will be generated
window = Frame(tkengine)
window.pack()
the_image = PhotoImage(width=809, height=700)
label = Label(window, image=the_image, borderwidth=2, relief="raised")
label.pack(fill="both", expand=True)
for t in range(points):
while True:
a = random.randint(0,mod-1)
b = random.randint(0,mod-1)
if a + b < mod:
break
x = x / mod + a / mod + b / 2 / mod
y = y / mod + b / mod
X = math.floor(x * 808)
Y = math.floor((1-y) * 700)
plotpoint(X, Y)
tkengine.mainloop()
You can use canvas.create_oval with the same coordinates for the two corners of the bounding box:
from tkinter import *
import random
import math
def plotpoint(x, y):
global canvas
# point = canvas.create_line(x-1, y-1, x+1, y+1, fill = "#000000")
point = canvas.create_oval(x, y, x, y, fill="#000000", outline="#000000")
x = 0 #Initial coordinates
y = 0
#x and y will always be in the interval [0, 1]
mod = int(input("What is the modulo of the Sierpinsky triangle that you want to generate? "))
points = int(input("How many points do you want the triangle to have? "))
tkengine = Tk() #Window in which the triangle will be generated
window = Frame(tkengine)
window.pack()
canvas = Canvas(window, height = 700, width = 808, bg = "#FFFFFF") #The dimensions of the canvas make the triangle look equilateral
canvas.pack()
for t in range(points):
#Procedure for placing the points
while True:
#First, randomly choose one of the mod(mod+1)/2 triangles of the first step. a and b are two vectors which point to the chosen triangle. a goes one triangle to the right and b one up-right. The algorithm gives the same probability to every triangle, although it's not efficient.
a = random.randint(0,mod-1)
b = random.randint(0,mod-1)
if a + b < mod:
break
#The previous point is dilated towards the origin of coordinates so that the big triangle of step 0 becomes the small one at the bottom-left of step one (divide by modulus). Then the vectors are added in order to move the point to the same place in another triangle.
x = x / mod + a / mod + b / 2 / mod
y = y / mod + b / mod
#Coordinates [0,1] converted to pixels, for plotting in the canvas.
X = math.floor(x * 808)
Y = math.floor((1-y) * 700)
plotpoint(X, Y)
tkengine.mainloop()
with a depth of 3 and 100,000 points, this gives:
Finally found a solution: if a 1x1 point is to be placed in pixel (x,y), a command which does it exactly is:
point = canvas.create_line(x, y, x+1, y+1, fill = "colour")
The oval is a good idea for 2x2 points.
Something remarkable about the original program is that it uses a lot of RAM if every point is treated as a separate object.
How can I find the point B(t) along a cubic Bezier curve that is closest to an arbitrary point P in the plane?
I've written some quick-and-dirty code that estimates this for Bézier curves of any degree. (Note: this is pseudo-brute force, not a closed-form solution.)
Demo: http://phrogz.net/svg/closest-point-on-bezier.html
/** Find the ~closest point on a Bézier curve to a point you supply.
* out : A vector to modify to be the point on the curve
* curve : Array of vectors representing control points for a Bézier curve
* pt : The point (vector) you want to find out to be near
* tmps : Array of temporary vectors (reduces memory allocations)
* returns: The parameter t representing the location of `out`
*/
function closestPoint(out, curve, pt, tmps) {
let mindex, scans=25; // More scans -> better chance of being correct
const vec=vmath['w' in curve[0]?'vec4':'z' in curve[0]?'vec3':'vec2'];
for (let min=Infinity, i=scans+1;i--;) {
let d2 = vec.squaredDistance(pt, bézierPoint(out, curve, i/scans, tmps));
if (d2<min) { min=d2; mindex=i }
}
let t0 = Math.max((mindex-1)/scans,0);
let t1 = Math.min((mindex+1)/scans,1);
let d2ForT = t => vec.squaredDistance(pt, bézierPoint(out,curve,t,tmps));
return localMinimum(t0, t1, d2ForT, 1e-4);
}
/** Find a minimum point for a bounded function. May be a local minimum.
* minX : the smallest input value
* maxX : the largest input value
* ƒ : a function that returns a value `y` given an `x`
* ε : how close in `x` the bounds must be before returning
* returns: the `x` value that produces the smallest `y`
*/
function localMinimum(minX, maxX, ƒ, ε) {
if (ε===undefined) ε=1e-10;
let m=minX, n=maxX, k;
while ((n-m)>ε) {
k = (n+m)/2;
if (ƒ(k-ε)<ƒ(k+ε)) n=k;
else m=k;
}
return k;
}
/** Calculate a point along a Bézier segment for a given parameter.
* out : A vector to modify to be the point on the curve
* curve : Array of vectors representing control points for a Bézier curve
* t : Parameter [0,1] for how far along the curve the point should be
* tmps : Array of temporary vectors (reduces memory allocations)
* returns: out (the vector that was modified)
*/
function bézierPoint(out, curve, t, tmps) {
if (curve.length<2) console.error('At least 2 control points are required');
const vec=vmath['w' in curve[0]?'vec4':'z' in curve[0]?'vec3':'vec2'];
if (!tmps) tmps = curve.map( pt=>vec.clone(pt) );
else tmps.forEach( (pt,i)=>{ vec.copy(pt,curve[i]) } );
for (var degree=curve.length-1;degree--;) {
for (var i=0;i<=degree;++i) vec.lerp(tmps[i],tmps[i],tmps[i+1],t);
}
return vec.copy(out,tmps[0]);
}
The code above uses the vmath library to efficiently lerp between vectors (in 2D, 3D, or 4D), but it would be trivial to replace the lerp() call in bézierPoint() with your own code.
Tuning the Algorithm
The closestPoint() function works in two phases:
First, calculate points all along the curve (uniformly-spaced values of the t parameter). Record which value of t has the smallest distance to the point.
Then, use the localMinimum() function to hunt the region around the smallest distance, using a binary search to find the t and point that produces the true smallest distance.
The value of scans in closestPoint() determines how many samples to use in the first pass. Fewer scans is faster, but increases the chances of missing the true minimum point.
The ε limit passed to the localMinimum() function controls how long it continues to hunt for the best value. A value of 1e-2 quantizes the curve into ~100 points, and thus you can see the points returned from closestPoint() popping along the line. Each additional decimal point of precision—1e-3, 1e-4, …—costs about 6-8 additional calls to bézierPoint().
After lots of searching I found a paper that discusses a method for finding the closest point on a Bezier curve to a given point:
Improved Algebraic Algorithm On Point
Projection For Bezier Curves, by
Xiao-Diao Chen, Yin Zhou, Zhenyu Shu,
Hua Su, and Jean-Claude Paul.
Furthermore, I found Wikipedia and MathWorld's descriptions of Sturm sequences useful in understanding the first part of the algoritm, as the paper itself isn't very clear in its own description.
Seeing as the other methods on this page seem to be approximation, this answer will provide a simple numerical solution. It is a python implementation depending on the numpy library to supply Bezier class. In my tests, this approach performed about three times better than my brute-force implementation (using samples and subdivision).
Look at the interactive example here.
Click to enlarge.
I used basic algebra to solve this minimal problem.
Start with the bezier curve equation.
B(t) = (1 - t)^3 * p0 + 3 * (1 - t)^2 * t * p1 + 3 * (1 - t) * t^2 * p2 + t^3 * p3
Since I'm using numpy, my points are represented as numpy vectors (matrices). This means that p0 is a one-dimensional, e.g. (1, 4.2). If you are handling two floating point variables you probably need mutliple equations (for x and y): Bx(t) = (1-t)^3*px_0 + ...
Convert it to a standard form with four coefficients.
You can write the four coefficients by expanding the original equation.
The distance from a point p to the curve B(t) can be calculated using the pythagorean theorem.
Here a and b are the two dimensions of our points x and y. This means that the squared distance D(t) is:
I'm not calculating a square root just now, because it is enough if we compare relative squared distances. All following equation will refer to the squared distance.
This function D(t) describes the distance between the graph and the points. We are interested in the minima in the range of t in [0, 1]. To find them, we have to derive the function twice. The first derivative of the distance function is a 5 order polynomial:
The second derivative is:
A desmos graph let's us examine the different functions.
D(t) has its local minima where d'(t) = 0 and d''(t) >= 0. This means, that we have to find the t for d'(t) = 0'.
black: the bezier curve, green: d(t), purple: d'(t), red:d''(t)
Find the roots of d'(t). I use the numpy library, which takes the coefficients of a polynomial.
dcoeffs = np.stack([da, db, dc, dd, de, df])
roots = np.roots(dcoeffs)
Remove the imaginary roots (keep only the real roots) and remove any roots which are < 0 or > 1. With a cubic bezier, there will probably be about 0-3 roots left.
Next, check the distances of each |B(t) - pt| for each t in roots. Also check the distances for B(0) and B(1) since start and end of the Bezier curve could be the closest points (although they aren't local minima of the distance function).
Return the closest point.
I am attaching the class for the Bezier in python. Check the github link for a usage example.
import numpy as np
# Bezier Class representing a CUBIC bezier defined by four
# control points.
#
# at(t): gets a point on the curve at t
# distance2(pt) returns the closest distance^2 of
# pt and the curve
# closest(pt) returns the point on the curve
# which is closest to pt
# maxes(pt) plots the curve using matplotlib
class Bezier(object):
exp3 = np.array([[3, 3], [2, 2], [1, 1], [0, 0]], dtype=np.float32)
exp3_1 = np.array([[[3, 3], [2, 2], [1, 1], [0, 0]]], dtype=np.float32)
exp4 = np.array([[4], [3], [2], [1], [0]], dtype=np.float32)
boundaries = np.array([0, 1], dtype=np.float32)
# Initialize the curve by assigning the control points.
# Then create the coefficients.
def __init__(self, points):
assert isinstance(points, np.ndarray)
assert points.dtype == np.float32
self.points = points
self.create_coefficients()
# Create the coefficients of the bezier equation, bringing
# the bezier in the form:
# f(t) = a * t^3 + b * t^2 + c * t^1 + d
#
# The coefficients have the same dimensions as the control
# points.
def create_coefficients(self):
points = self.points
a = - points[0] + 3*points[1] - 3*points[2] + points[3]
b = 3*points[0] - 6*points[1] + 3*points[2]
c = -3*points[0] + 3*points[1]
d = points[0]
self.coeffs = np.stack([a, b, c, d]).reshape(-1, 4, 2)
# Return a point on the curve at the parameter t.
def at(self, t):
if type(t) != np.ndarray:
t = np.array(t)
pts = self.coeffs * np.power(t, self.exp3_1)
return np.sum(pts, axis = 1)
# Return the closest DISTANCE (squared) between the point pt
# and the curve.
def distance2(self, pt):
points, distances, index = self.measure_distance(pt)
return distances[index]
# Return the closest POINT between the point pt
# and the curve.
def closest(self, pt):
points, distances, index = self.measure_distance(pt)
return points[index]
# Measure the distance^2 and closest point on the curve of
# the point pt and the curve. This is done in a few steps:
# 1 Define the distance^2 depending on the pt. I am
# using the squared distance because it is sufficient
# for comparing distances and doesn't have the over-
# head of an additional root operation.
# D(t) = (f(t) - pt)^2
# 2 Get the roots of D'(t). These are the extremes of
# D(t) and contain the closest points on the unclipped
# curve. Only keep the minima by checking if
# D''(roots) > 0 and discard imaginary roots.
# 3 Calculate the distances of the pt to the minima as
# well as the start and end of the curve and return
# the index of the shortest distance.
#
# This desmos graph is a helpful visualization.
# https://www.desmos.com/calculator/ktglugn1ya
def measure_distance(self, pt):
coeffs = self.coeffs
# These are the coefficients of the derivatives d/dx and d/(d/dx).
da = 6*np.sum(coeffs[0][0]*coeffs[0][0])
db = 10*np.sum(coeffs[0][0]*coeffs[0][1])
dc = 4*(np.sum(coeffs[0][1]*coeffs[0][1]) + 2*np.sum(coeffs[0][0]*coeffs[0][2]))
dd = 6*(np.sum(coeffs[0][0]*(coeffs[0][3]-pt)) + np.sum(coeffs[0][1]*coeffs[0][2]))
de = 2*(np.sum(coeffs[0][2]*coeffs[0][2])) + 4*np.sum(coeffs[0][1]*(coeffs[0][3]-pt))
df = 2*np.sum(coeffs[0][2]*(coeffs[0][3]-pt))
dda = 5*da
ddb = 4*db
ddc = 3*dc
ddd = 2*dd
dde = de
dcoeffs = np.stack([da, db, dc, dd, de, df])
ddcoeffs = np.stack([dda, ddb, ddc, ddd, dde]).reshape(-1, 1)
# Calculate the real extremes, by getting the roots of the first
# derivativ of the distance function.
extrema = np_real_roots(dcoeffs)
# Remove the roots which are out of bounds of the clipped range [0, 1].
# [future reference] https://stackoverflow.com/questions/47100903/deleting-every-3rd-element-of-a-tensor-in-tensorflow
dd_clip = (np.sum(ddcoeffs * np.power(extrema, self.exp4)) >= 0) & (extrema > 0) & (extrema < 1)
minima = extrema[dd_clip]
# Add the start and end position as possible positions.
potentials = np.concatenate((minima, self.boundaries))
# Calculate the points at the possible parameters t and
# get the index of the closest
points = self.at(potentials.reshape(-1, 1, 1))
distances = np.sum(np.square(points - pt), axis = 1)
index = np.argmin(distances)
return points, distances, index
# Point the curve to a matplotlib figure.
# maxes ... the axes of a matplotlib figure
def plot(self, maxes):
import matplotlib.path as mpath
import matplotlib.patches as mpatches
Path = mpath.Path
pp1 = mpatches.PathPatch(
Path(self.points, [Path.MOVETO, Path.CURVE4, Path.CURVE4, Path.CURVE4]),
fc="none")#, transform=ax.transData)
pp1.set_alpha(1)
pp1.set_color('#00cc00')
pp1.set_fill(False)
pp2 = mpatches.PathPatch(
Path(self.points, [Path.MOVETO, Path.LINETO , Path.LINETO , Path.LINETO]),
fc="none")#, transform=ax.transData)
pp2.set_alpha(0.2)
pp2.set_color('#666666')
pp2.set_fill(False)
maxes.scatter(*zip(*self.points), s=4, c=((0, 0.8, 1, 1), (0, 1, 0.5, 0.8), (0, 1, 0.5, 0.8),
(0, 0.8, 1, 1)))
maxes.add_patch(pp2)
maxes.add_patch(pp1)
# Wrapper around np.roots, but only returning real
# roots and ignoring imaginary results.
def np_real_roots(coefficients, EPSILON=1e-6):
r = np.roots(coefficients)
return r.real[abs(r.imag) < EPSILON]
Depending on your tolerances. Brute force and being accepting of error. This algorithm could be wrong for some rare cases. But, in the majority of them it will find a point very close to the right answer and the results will improve the higher you set the slices. It just tries each point along the curve at regular intervals and returns the best one it found.
public double getClosestPointToCubicBezier(double fx, double fy, int slices, double x0, double y0, double x1, double y1, double x2, double y2, double x3, double y3) {
double tick = 1d / (double) slices;
double x;
double y;
double t;
double best = 0;
double bestDistance = Double.POSITIVE_INFINITY;
double currentDistance;
for (int i = 0; i <= slices; i++) {
t = i * tick;
//B(t) = (1-t)**3 p0 + 3(1 - t)**2 t P1 + 3(1-t)t**2 P2 + t**3 P3
x = (1 - t) * (1 - t) * (1 - t) * x0 + 3 * (1 - t) * (1 - t) * t * x1 + 3 * (1 - t) * t * t * x2 + t * t * t * x3;
y = (1 - t) * (1 - t) * (1 - t) * y0 + 3 * (1 - t) * (1 - t) * t * y1 + 3 * (1 - t) * t * t * y2 + t * t * t * y3;
currentDistance = Point.distanceSq(x,y,fx,fy);
if (currentDistance < bestDistance) {
bestDistance = currentDistance;
best = t;
}
}
return best;
}
You can get a lot better and faster by simply finding the nearest point and recursing around that point.
public double getClosestPointToCubicBezier(double fx, double fy, int slices, int iterations, double x0, double y0, double x1, double y1, double x2, double y2, double x3, double y3) {
return getClosestPointToCubicBezier(iterations, fx, fy, 0, 1d, slices, x0, y0, x1, y1, x2, y2, x3, y3);
}
private double getClosestPointToCubicBezier(int iterations, double fx, double fy, double start, double end, int slices, double x0, double y0, double x1, double y1, double x2, double y2, double x3, double y3) {
if (iterations <= 0) return (start + end) / 2;
double tick = (end - start) / (double) slices;
double x, y, dx, dy;
double best = 0;
double bestDistance = Double.POSITIVE_INFINITY;
double currentDistance;
double t = start;
while (t <= end) {
//B(t) = (1-t)**3 p0 + 3(1 - t)**2 t P1 + 3(1-t)t**2 P2 + t**3 P3
x = (1 - t) * (1 - t) * (1 - t) * x0 + 3 * (1 - t) * (1 - t) * t * x1 + 3 * (1 - t) * t * t * x2 + t * t * t * x3;
y = (1 - t) * (1 - t) * (1 - t) * y0 + 3 * (1 - t) * (1 - t) * t * y1 + 3 * (1 - t) * t * t * y2 + t * t * t * y3;
dx = x - fx;
dy = y - fy;
dx *= dx;
dy *= dy;
currentDistance = dx + dy;
if (currentDistance < bestDistance) {
bestDistance = currentDistance;
best = t;
}
t += tick;
}
return getClosestPointToCubicBezier(iterations - 1, fx, fy, Math.max(best - tick, 0d), Math.min(best + tick, 1d), slices, x0, y0, x1, y1, x2, y2, x3, y3);
}
In both cases you can do the quad just as easily:
x = (1 - t) * (1 - t) * x0 + 2 * (1 - t) * t * x1 + t * t * x2; //quad.
y = (1 - t) * (1 - t) * y0 + 2 * (1 - t) * t * y1 + t * t * y2; //quad.
By switching out the equation there.
While the accepted answer is right, and you really can figure out the roots and compare that stuff. If you really just need to find the nearest point on the curve, this will do it.
In regard to Ben in the comments. You cannot short hand the formula in the many hundreds of control point range, like I did for cubic and quad forms. Because the amount demanded by each new addition of a bezier curve means that you build a Pythagorean pyramids for them, and we're basically dealing with even more and more massive strings of numbers. For quad you go 1, 2, 1, for cubic you go 1, 3, 3, 1. You end up building bigger and bigger pyramids, and end up breaking it down with Casteljau's algorithm, (I wrote this for solid speed):
/**
* Performs deCasteljau's algorithm for a bezier curve defined by the given control points.
*
* A cubic for example requires four points. So it should get at least an array of 8 values
*
* #param controlpoints (x,y) coord list of the Bezier curve.
* #param returnArray Array to store the solved points. (can be null)
* #param t Amount through the curve we are looking at.
* #return returnArray
*/
public static float[] deCasteljau(float[] controlpoints, float[] returnArray, float t) {
int m = controlpoints.length;
int sizeRequired = (m/2) * ((m/2) + 1);
if (returnArray == null) returnArray = new float[sizeRequired];
if (sizeRequired > returnArray.length) returnArray = Arrays.copyOf(controlpoints, sizeRequired); //insure capacity
else System.arraycopy(controlpoints,0,returnArray,0,controlpoints.length);
int index = m; //start after the control points.
int skip = m-2; //skip if first compare is the last control point.
for (int i = 0, s = returnArray.length - 2; i < s; i+=2) {
if (i == skip) {
m = m - 2;
skip += m;
continue;
}
returnArray[index++] = (t * (returnArray[i + 2] - returnArray[i])) + returnArray[i];
returnArray[index++] = (t * (returnArray[i + 3] - returnArray[i + 1])) + returnArray[i + 1];
}
return returnArray;
}
You basically need to use the algorithm directly, not just for the calculation of the x,y which occur on the curve itself, but you also need it to perform actual and proper Bezier subdivision algorithm (there are others but that is what I'd recommend), to calculate not just an approximation as I give by dividing it into line segments, but of the actual curves. Or rather the polygon hull that is certain to contain the curve.
You do this by using the above algorithm to subdivide the curves at the given t. So T=0.5 to cut the curves in half (note 0.2 would cut it 20% 80% through the curve). Then you index the various points at the side of the pyramid and the other side of the pyramid as built from the base. So for example in cubic:
9
7 8
4 5 6
0 1 2 3
You would feed the algorithm 0 1 2 3 as control points, then you would index the two perfectly subdivided curves at 0, 4, 7, 9 and 9, 8, 6, 3. Take special note to see that these curves start and end at the same point. and the final index 9 which is the point on the curve is used as the other new anchor point. Given this you can perfectly subdivide a bezier curve.
Then to find the closest point you'd want to keep subdividing the curve into different parts noting that it is the case that the entire curve of a bezier curve is contained within the hull of the control points. Which is to say if we turn points 0, 1, 2, 3 into a closed path connecting 0,3 that curve must fall completely within that polygon hull. So what we do is define our given point P, then we continue to subdivide curves until such time as we know that the farthest point of one curve is closer than the closest point of another curve. We simply compare this point P to all the control and anchor points of the curves. And discard any curve from our active list whose closest point (whether anchor or control) is further away than the farthest point of another curve. Then we subdivide all the active curves and do this again. Eventually we will have very subdivided curves discarding about half each step (meaning it should be O(n log n)) until our error is basically negligible. At this point we call our active curves the closest point to that point (there could be more than one), and note that the error in that highly subdivided bit of curve is basically equal to a point. Or simply decide the issue by saying whichever of the two anchor point is closest is the closest point to our point P. And we know the error to a very specific degree.
This, though, requires that we actually have a robust solution and do a certainly correct algorithm and correctly find the tiny fraction of curve that will certainly be the closest point to our point. And it should be relatively fast still.
There is also DOM SVG specific implementations of the closest point algorithms from Mike Bostock:
https://bl.ocks.org/mbostock/8027637
https://bl.ocks.org/mbostock/8027835
A solution to this problem would be to get all the possible points on the bezier curve and compare each distance. The number of points can be controlled by the detail variable.
Here is a implementation made in Unity (C#):
public Vector2 FindNearestPointOnBezier(Bezier bezier, Vector2 point)
{
float detail = 100;
List<Vector2> points = new List<Vector2>();
for (float t = 0; t < 1f; t += 1f / detail)
{
// this function can be exchanged for any bezier curve
points.Add(Functions.CalculateBezier(bezier.a, bezier.b, bezier.c, bezier.d, t));
}
Vector2 closest = Vector2.zero;
float minDist = Mathf.Infinity;
foreach (Vector2 p in points)
{
// use sqrMagnitude as it is faster
float dist = (p - point).sqrMagnitude;
if (dist < minDist)
{
minDist = dist;
closest = p;
}
}
return closest;
}
Note that the Bezier class just holds 4 points.
Probably not the best way as it can become very slow depending on the detail.