Develop Reference

Fading a Line Exponentially

I'd like to fade the values of a line on the Y axis using a gradient that I can control.
This is a simplified version of what I'm trying right now.
y_values = [1,1,1,1,1,1]
for i, y in enumerate(y_values): #i is the index, y is the individual y value
perc = i/(len(y_values)-1) #progress the fade by the percentage of the index divided by the total values
amt = y * perc #y is the target amount to decrease multiplied by index percentage
y -= amt
print(i,y)
This code produces this:
1.0
0.8
0.6
0.4
0.2
0.0
It's creating a linear fade, but how do I increase the fade with an exponential fade like this?
Thank you!

To make exponential fading, you have to provide two coefficients to provide initial factor 1.0 at the start value x1 and desired final factor k at the end of interval x2
y = y * f(x)
f(x) = A * exp(-B * x)
So
f(x1) = 1 = A * exp(B * x1)
f(x2) = k = A * exp(B * x2)
divide the second by the first
k = exp(B * (x2 - x1))
ln(k) = B * (x2 - x1)
so
B = ln(k) / (x2 - x1)
A = exp(B * x1)
Example for x1 = 0, x2 = 60, k = 0.01
B = -4.6/60= -0.076
A = 1
f(x) = exp(-0.076*x)
f(30) = exp(-0.076*20) = 0.1
Python example:
import math
def calcfading(x1, x2, ratio):
B = math.log(ratio) / (x2 - x1)
A = math.exp(B * x1)
return A, B
def coef(x, fade):
return fade[0]*math.exp(x*fade[1])
cosine = [[x, math.cos(x)] for x in range(0, 11)]
print(cosine)
print()
fade = calcfading(0, 10, 0.01)
expcosine = [[s[0], coef(s[0], fade)*s[1]] for s in cosine]
print(expcosine)

How to draw cycloid on curve of other function (other cycloid)?

I would like to draw cycloid that is going on other cycloid but I don't know exactly how to do this. Here is my code.
import numpy as np
import matplotlib.pyplot as plt
import math
from matplotlib import animation
#r = float(input('write r\n'))
#R = float(input('write R\n'))
r = 1
R = 1
x = []
y = []
x2 = []
y2 = []
x3 = []
y3 = []
length=[0]
fig, ax = plt.subplots()
ln, = plt.plot([], [], 'r', animated=True)
f = np.linspace(0, 10*r*math.pi, 1000)
def init():
ax.set_xlim(-r, 12*r*math.pi)
ax.set_ylim(-4*r, 4*r)
return ln,
def update2(frame):
#parametric equations of cycloid
x0 = r * (frame - math.sin(frame))
y0 = r * (1 - math.cos(frame))
x.append(x0)
y.append(y0)
#derivative of cycloid
dx = r * (1 - math.cos(frame))
dy = r * math.sin(frame)
#center of circle
a = dy * dy + dx * dx
b = (-2 * x0 * dy) - (2 * frame * dy * dy) + (2 * y0 * dx) - (2 * frame * dx * dx)
c = (x0 * x0) + (2 * frame * x0 * dy) + (frame * frame * dy * dy) + (y0 * y0) - (2 * frame * y0 * dx) + (frame * frame * dx * dx) -1
t1 = (-b - math.sqrt(b * b - 4 * a * c)) / (2 * a)
#t2 = (-b + math.sqrt(b * b - 4 * a * c)) / (2 * a)
center1x=(x0-dy*(t1-x0))*R
center1y=(y0+dx*(t1-x0))*R
#center2x=(x0-dy*(t2-x0))*R
#center2y=(y0+dx*(t2-x0))*R
#length of cycloid
length.append(math.sqrt(x0*x0 + y0*y0))
dl=sum(length)
param = dl / R
W1x = center1x + R * math.cos(-param)
W1y = center1y + R * math.sin(-param)
#W2x = center2x + R * math.cos(-param)
#W2y = center2y + R * math.sin(-param)
x2.append(W1x)
y2.append(W1y)
#x3.append(W2x)
#y3.append(W2y)
ln.set_data([x, x2], [y, y2])
return ln,
ani = animation.FuncAnimation(fig, update2, frames=f,init_func=init, blit=True, interval = 0.1, repeat = False)
plt.show()
In my function update2 I created parametric equations of first cycloid and then tried to obtain co-ordinates of points of second cycloid that should go on the first one.
My idea is based on that that typical cycloid is moving on straight line, and cycloid that is moving on other curve must moving on tangent of that curve, so center of circle that's creating this cycloid is always placed on normal of curve. From parametric equations of normal I have tried to obtain center of circle that creating cycloid but I think that isn't good way.
My goal is to get something like this:

Here is one way. Calculus gives us the formulas to find the direction angles at any point on the cycloid and the arc lengths along the cycloid. Analytic Geometry tells us how to use that information to find your desired points.
By the way, a plot made by rolling a figure along another figure is called a roulette. My code is fairly simple and could be optimized, but it works now, can be used for other problems, and is broken up to make the math and algorithm easier to understand. To understand my code, use this diagram. The cycloid is the blue curve, the black circles are the rolling circle on the cycloid, point A is an "anchor point" (a point where the rim point touches the cycloid--I wanted to make this code general), and point F is the moving rim point. The two red arcs are the same length, which is what we mean by rolling the circle along the cycloid.
And here is my code. Ask if you need help with the source of the various formulas, but the direction angles and arc lengths use calculus.
"""Numpy-compatible routines for a standard cycloid (one caused by a
circle of radius r above the y-axis rolling along the positive x-axis
starting from the origin).
"""
import numpy as np
def x(t, r):
"""Return the x-coordinate of a point on the cycloid with parameter t."""
return r * (t - np.sin(t))
def y(t, r):
"""Return the y-coordinate of a point on the cycloid with parameter t."""
return r * (1.0 - np.cos(t))
def dir_angle_norm_in(t, r):
"""Return the direction angle of the vector normal to the cycloid at
the point with parameter t that points into the cycloid."""
return -t / 2.0
def dir_angle_norm_out(t, r):
"""Return the direction angle of the vector normal to the cycloid at
the point with parameter t that points out of the cycloid."""
return np.pi - t / 2.0
def arclen(t, r):
"""Return the arc length of the cycloid between the origin and the
point on the cycloid with parameter t."""
return 4.0 * r * (1.0 - np.cos(t / 2.0))
# Roulette problem
def xy_roulette(t, r, T, R):
"""Return the x-y coordinates of a rim point on a circle of radius
R rolling on a cycloid of radius r starting at the anchor point
with parameter T currently at the point with parameter t. (Such a
rolling curve on another curve is called a roulette.)
"""
# Find the coordinates of the contact point P between circle and cycloid
px, py = x(t, r), y(t, r)
# Find the direction angle of PC from the contact point to circle's center
a1 = dir_angle_norm_out(t, r)
# Find the coordinates of the center C of the circle
cx, cy = px + R * np.cos(a1), py + R * np.sin(a1)
# Find cycloid's arc distance AP between anchor and current contact points
d = arclen(t, r) - arclen(T, r) # equals arc PF
# Find the angle φ the circle turned while rolling from the anchor pt
phi = d / R
# Find the direction angle of CF from circle's center to rim point
a2 = dir_angle_norm_in(t, r) - phi # subtract: circle rolls clockwise
# Find the coordinates of the final point F
fx, fy = cx + R * np.cos(a2), cy + R * np.sin(a2)
# Return those coordinates
return fx, fy
import matplotlib.pyplot as plt
r = 1
R = 0.75
T = np.pi / 3
t_array = np.linspace(0, 2*np.pi, 201)
cycloid_x = x(t_array, r)
cycloid_y = y(t_array, r)
roulette_x, roulette_y = xy_roulette(t_array, r, T, R)
fig, ax = plt.subplots()
ax.set_aspect('equal')
ax.axhline(y=0, color='k')
ax.axvline(x=0, color='k')
ax.plot(cycloid_x, cycloid_y)
ax.plot(roulette_x, roulette_y)
plt.show()
And here is the resulting graphic. You can pretty this up as you choose. Note that this only has the circle rolling along one arch of the cycloid. If you clarify what should happen at the cusps, this could be extended.
Or, if you want a smaller circle and a curve that ends at the cusps (here r = 1, T = 0 n = 6 (the number of little arches), and R = 4 * r / np.pi / n),

You can generate coordinates of the center of rolling circle using parallel curve definition. Parametric equations for this center are rather simple (if I did not make mistakes):
for big cycloid:
X = R(t - sin(t))
Y = R(1 - cos(t))
X' = R(1 - cos(t))
Y' = R*sin(t)
parallel curve (center of small circle):
Sqrt(X'^2+Y'^2)=R*Sqrt(1-2*cos(t)+cos^2(t)+sin^2(t)) =
R*Sqrt(2-2*cos(t))=
R*Sqrt(4*sin^2(t/2))=
2*R*sin(t/2)
x(t) = X(t) + r*R*sin(t)/(2R*sin(t/2)) =
R(t - sin(t)) + r*2*sin(t/2)*cos(t/2) / (2*sin(t/2)) =
R(t - sin(t)) + r*cos(t/2)
y(t) = Y(t) - r*R*(1-cos(t))/(2*R*sin(t/2)) =
R(1 - cos(t)) - r*(2*sin^2(t/2)/(2*sin(t/2)) =
R(1 - cos(t)) - r*sin(t/2)
But trajectory of point on the circumference is superposition of center position and rotation around it with angular velocity that depends on length of main cycloid plus rotation of main tangent.
Added from dicussion in comments:
cycloid arc length
L(t) = 4R*(1-cos(t/2))
to use it for small circle rotation, divide by r
tangent rotation derivation
fi(t) = atan(Y'/X') = atan(sin(t)/(1-cos(t)) =
atan(2*sin(t/2)*cos(t/2)/(2(sin^2(t/2))) =
atan(ctg(t/2)) = Pi/2 - t/2
so tangent direction change is proportional to big cycloid parameter
and final result is (perhaps some signs are not correct)
theta(t) = L(t)/r + t/2 + Phase
ox(t) = x(t) + r * cos(theta(t))
oy(t) = y(t) + r * sin(theta(t))

Thanks everyone. Somehow I have managed to accomplish that. Solution is maybe ugly but sufficient for me.
import numpy as np
import matplotlib.pyplot as plt
import math
from matplotlib import animation
r = float(input('write r\n'))
R = float(input('write R\n'))
#r=1
#R=0.1
x = []
y = []
x2 = []
y2 = []
x_1=0
x_2=0
lengthX=[0]
lengthY=[0]
lengthabs=[0]
fig, ax = plt.subplots()
ln, = plt.plot([], [], 'r', animated=True)
f = np.linspace(0, 2*math.pi, 1000)
def init():
ax.set_xlim(-r, 4*r*math.pi)
ax.set_ylim(0, 4*r)
return ln,
def update2(frame):
#cycloid's equations
x0 = r * (frame - math.sin(frame))
y0 = r * (1 - math.cos(frame))
x.append(r * (frame - math.sin(frame)))
y.append(r * (1 - math.cos(frame)))
#arc's length
lengthabs.append(math.sqrt((x0-lengthX[-1])*(x0-lengthX[-1])+(y0-lengthY[-1])*(y0-lengthY[-1])))
lengthX.append(x0)
lengthY.append(y0)
dl=sum(lengthabs)
param = dl / R
#center of circle
center1x = r * (frame - math.sin(frame)) + R * math.cos((frame+2*math.pi) / 2)
center1y = r * (1 - math.cos(frame)) - R * math.sin((frame+2*math.pi) / 2)
if(frame<2*math.pi):
W1x = center1x + R * math.cos(-param)
W1y = center1y + R * math.sin(-param)
else:
W1x = center1x + R * math.cos(param)
W1y = center1y + R * math.sin(param)
x2.append(W1x)
y2.append(W1y)
ln.set_data([x,x2], [y,y2])
return ln,
ani = animation.FuncAnimation(fig, update2, frames=f,init_func=init, blit=True, interval = 0.1, repeat = False)
plt.show()

multithreaded mandelbrot set

Is it possible to change the formula of the mandelbrot set (which is f(z) = z^2 + c by default) to a different one ( f(z) = z^2 + c * e^(-z) is what i need) when using the escape time algorithm and if possible how?
I'm currently using this code by FB36
# Multi-threaded Mandelbrot Fractal (Do not run using IDLE!)
# FB - 201104306
import threading
from PIL import Image
w = 512 # image width
h = 512 # image height
image = Image.new("RGB", (w, h))
wh = w * h
maxIt = 256 # max number of iterations allowed
# drawing region (xa < xb & ya < yb)
xa = -2.0
xb = 1.0
ya = -1.5
yb = 1.5
xd = xb - xa
yd = yb - ya
numThr = 5 # number of threads to run
# lock = threading.Lock()
class ManFrThread(threading.Thread):
def __init__ (self, k):
self.k = k
threading.Thread.__init__(self)
def run(self):
# each thread only calculates its own share of pixels
for i in range(k, wh, numThr):
kx = i % w
ky = int(i / w)
a = xa + xd * kx / (w - 1.0)
b = ya + yd * ky / (h - 1.0)
x = a
y = b
for kc in range(maxIt):
x0 = x * x - y * y + a
y = 2.0 * x * y + b
x = x0
if x * x + y * y > 4:
# various color palettes can be created here
red = (kc % 8) * 32
green = (16 - kc % 16) * 16
blue = (kc % 16) * 16
# lock.acquire()
global image
image.putpixel((kx, ky), (red, green, blue))
# lock.release()
break
if __name__ == "__main__":
tArr = []
for k in range(numThr): # create all threads
tArr.append(ManFrThread(k))
for k in range(numThr): # start all threads
tArr[k].start()
for k in range(numThr): # wait until all threads finished
tArr[k].join()
image.save("MandelbrotFractal.png", "PNG")

From the code I infer that z = x + y * i and c = a + b * i. That corresponds f(z) - z ^2 + c. You want f(z) = z ^2 + c * e^(-z).
Recall that e^(-z) = e^-(x + yi) = e^(-x) * e^i(-y) = e^(-x)(cos(y) - i*sin(y)) = e^(-x)cos(y) - i (e^(-x)sin(y)). Thus you should update your lines to be the following:
x0 = x * x - y * y + a * exp(-x) * cos(y) + b * exp(-x) * sin(y);
y = 2.0 * x * y + a * exp(-x) * sin(y) - b * exp(-x) * cos(y)
x = x0
You might need to adjust maxIt if you don't get the level of feature differentiation you're after (it might take more or fewer iterations to escape now, on average) but this should be the mathematical expression you're after.
As pointed out in the comments, you might need to adjust the criterion itself and not just the maximum iterations in order to get the desired level of differentiation: changing the max doesn't help for ones that never escape.
You can try deriving a good escape condition or just try out some things and see what you get.

Chaotic billiards simulation

I came to ask for some help with maths and programming.
What am I trying to do? I'm trying to implement a simulation of a chaotic billiard system, following the algorithm in this excerpt.
How am I trying it? Using numpy and matplotlib, I implemented the following code
def boundaryFunction(parameter):
return 1 + 0.1 * np.cos(parameter)
def boundaryDerivative(parameter):
return -0.1 * np.sin(parameter)
def trajectoryFunction(parameter):
aux = np.sin(beta - phi) / np.sin(beta - parameter)
return boundaryFunction(phi) * aux
def difference(parameter):
return trajectoryFunction(parameter) - boundaryFunction(parameter)
def integrand(parameter):
rr = boundaryFunction(parameter)
dd = boundaryDerivative (parameter)
return np.sqrt(rr ** 2 + dd ** 2)
##### Main #####
length_vals = np.array([], dtype=np.float64)
alpha_vals = np.array([], dtype=np.float64)
# nof initial phi angles, alpha angles, and nof collisions for each.
n_phi, n_alpha, n_cols, count = 10, 10, 10, 0
# Length of the boundary
total_length, err = integrate.quad(integrand, 0, 2 * np.pi)
for phi in np.linspace(0, 2 * np.pi, n_phi):
for alpha in np.linspace(0, 2 * np.pi, n_alpha):
for n in np.arange(1, n_cols):
nu = np.arctan(boundaryFunction(phi) / boundaryDerivative(phi))
beta = np.pi + phi + alpha - nu
# Determines next impact coordinate.
bnds = (0, 2 * np.pi)
phi_new = optimize.minimize_scalar(difference, bounds=bnds, method='bounded').x
nu_new = np.arctan(boundaryFunction(phi_new) / boundaryDerivative(phi_new))
# Reflection angle with relation to tangent.
alpha_new = phi_new - phi + nu - nu_new - alpha
# Arc length for current phi value.
arc_length, err = integrate.quad(integrand, 0, phi_new)
# Append values to list
length_vals = np.append(length_vals, arc_length / total_length)
alpha_vals = np.append(alpha_vals, alpha)
count += 1
print "{}%" .format(100 * count / (n_phi * n_alpha))
What is the problem? When calculating phi_new, the equation has two solutions (assuming the boundary is convex, which is.) I must enforce that phi_new is the solution which is different from phi, but I don't know how to do that. Are there more issues with the code?
What should the output be? A phase space diagram of S x Alpha, looking like this.
Any help is very appreciated! Thanks in advance.

One way you could try would be (given there really are only two solutions) would be
epsilon = 1e-7 # tune this
delta = 1e-4 # tune this
# ...
bnds = (0, 2 * np.pi)
phi_new = optimize.minimize_scalar(difference, bounds=bnds, method='bounded').x
if abs(phi_new - phi) < epsilon:
bnds_1 = (0, phi - delta)
phi_new_1 = optimize.minimize_scalar(difference, bounds=bnds_1, method='bounded').x
bnds_2 = (phi + delta, 2 * np.pi)
phi_new_2 = optimize.minimize_scalar(difference, bounds=bnds_2, method='bounded').x
if difference(phi_new_1) < difference(phi_new_2):
phi_new = phi_new_1
else:
phi_new = phi_new_2
Alternatively, you could introduce a penalty-term, e.g. delta*exp(eps/(x-phi)^2) with appropriate choices of epsilon and delta.

Finding X and Y axis line intercept points of a Circle - Python

Hey trying to learn how to code and I cant figure this exercise out.
Specifically getting the precise y axis intercept points.
The formula given works for getting the x axis points but I cant figure out how to get the y axis points.
Exercise :
Input : Radius of circle and the y - intercept of the line.
Output : Circle drawn with a horizontal line across the window with the given y intercept. Mark two points of the intersection.
Print the x values of the points of intersection *Formula : x = ± √r^2 - y^2
Code::
from graphics import *
from math import *
def main():
# enter radius and the y intercept of the line
radius = eval(input("Put in radius:: "))
yinter = eval(input("Put in y intersec:: "))
#Draw window + circle + line
win = GraphWin()
win.setCoords(-10.0, -10.0, 10.0, 10.0)
circle = Circle(Point(0.0,0.0), radius)
mcircle = Circle(Point(0.0,0.0), 0.5)
circle.draw(win)
mcircle.draw(win)
line = Line(Point(-10, 0), Point(10, yinter))
line.draw(win)
#Calculate x axis points of intersept
xroot1 = sqrt(radius * radius - yinter * yinter)
xroot2 = -abs(xroot1)
print("Xroot 1 : ", xroot1)
print("Xroot 2 : ", xroot2)
x = 0
yroot1 = sqrt(radius * radius - x * x)
yroot2 = -abs(yroot1)
print("Yroot 1 : ", yroot1)
print("Yroot 2 : ", yroot2)
#mark two points of intersept in red
sc1 = Circle(Point(xroot1, yroot1), 0.3)
sc1.setFill('red')
sc2 = Circle(Point(xroot2, yroot2), 0.3)
sc2.setFill('red')
sc1.draw(win)
sc2.draw(win)
main()
Answer - With Radius of 8 and Y intersect point of 2
Yroot1 = 7.75
Yroot2 = -7.75
Xroot1 = 8.0
Xroot2 = -8.0

I just came up with a subroutine to find intersection points while solving another Zelle-graphics related SO question. There may be ways to simplify the math but I'm going the long way around:
from graphics import *
def intersection(center, radius, p1, p2):
""" find the two points where a secant intersects a circle """
dx, dy = p2.x - p1.x, p2.y - p1.y
a = dx**2 + dy**2
b = 2 * (dx * (p1.x - center.x) + dy * (p1.y - center.y))
c = (p1.x - center.x)**2 + (p1.y - center.y)**2 - radius**2
discriminant = b**2 - 4 * a * c
assert (discriminant > 0), 'Not a secant!'
t1 = (-b + discriminant**0.5) / (2 * a)
t2 = (-b - discriminant**0.5) / (2 * a)
return Point(dx * t1 + p1.x, dy * t1 + p1.y), Point(dx * t2 + p1.x, dy * t2 + p1.y)
def main(win):
center = Point(0.0, 0.0)
# Enter radius
radius = float(input("Put in radius: "))
# Draw circle and center dot
Circle(center, radius).draw(win)
Circle(center, 0.3).draw(win)
# Enter the y intercept of the line
yinter = float(input("Put in y intercept: "))
# Draw line
p1, p2 = Point(-10.0, 0.0), Point(10.0, yinter)
Line(p1, p2).draw(win)
# Mark two points of intercept in red
for i, root in enumerate(intersection(center, radius, p1, p2), start=1):
print("Root {}:".format(i), root)
dot = Circle(root, 0.3)
dot.setFill('red')
dot.draw(win)
win = GraphWin()
win.setCoords(-10.0, -10.0, 10.0, 10.0)
main(win)
win.getMouse()
OUTPUT
NOTE
You can get values input that do not produce a secant, e.g. a radius of 2 and an intercept of 8. Your code doesn't account for this -- the above will simply throw an assertion error if it occurs. But you can upgrade that to an error you can catch and fix.

For the y coordinates you can use a similar formula:
y = ± sqrt(r^2 - x^2)
and the do everything the same with marking the roots.

You should write the code like this:
x = sqrt(r ** 2 - y ** 2)
line = Line(Point(-10, 0), Point(10, yinter))
line.draw(win)
Line(Point(-10,0) is wrong, it should read:
Line(Point(-10,yinter).
Also set Xroot1 and Xroot2 = 0 or -x=0, x=0

def intersection(center, radius, p1, p2):
dx, dy = p2[0] - p1[0], p2[1] - p1[1]
a = dx**2 + dy**2
b = 2 * (dx * (p1[0]- center[0]) + dy * (p1[1] - center[1]))
c = (p1[0] - center[0])**2 + (p1[1] - center[1])**2 - radius**2
K = b**2 - 4 * a * c
return True if K>0 else False
if __name__ == "__main__":
p1 = 10,50
p2 = 100,50
center= 50,150
radius = 600
print(intersection(center, radius, p1, p2))