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This is more of a computational physics problem, and I've asked it on physics stack exchange, but no answers on there. This is, I suppose, a mix of the disciplines on here and there (and maybe even mathematics stack exchange), so finding the right place to post is a task in of itself apparently...
I'm attempting to use Crank-Nicolson scheme to solve the TDSE in 1D. The initial wave is a real Gaussian that has been normalised wrt its probability density. As the solution evolves, a depression grows in the central peak of the real part of the wave, and the imaginary part's central trough is perhaps a bit higher than I expect (image below).
Does this behaviour seem reasonable? I have searched around and not seen questions/figures that are similar. I've tested another person's code from Github and it exhibits the same behaviour, which makes me feel a bit better. But I still think the center peak should just decrease in height and increase in width. The likelihood of me getting a physics-based explanation is relatively low here I'd assume, but a computational-based explanation on errors I may have made is more likely.
I'm happy to give more information, for example my code, or the matrices used in the scheme, etc. Thanks in advance!
Here's a link to GIF of time evolution:
And the part of my code relevant to solving the 1D TDSE:
(pretty much the entire thing except the plotting)
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.animation import FuncAnimation
# Define function for norm.
def normf(dxc, uc, ic):
return sum(dxc * np.square(np.abs(uc[ic, :])))
# Define function for expectation value of position.
def xexpf(dxc, xc, uc, ic):
return sum(dxc * xc * np.square(np.abs(uc[ic, :])))
# Define function for expectation value of squared position.
def xexpsf(dxc, xc, uc, ic):
return sum(dxc * np.square(xc) * np.square(np.abs(uc[ic, :])))
# Define function for standard deviation.
def sdaf(xexpc, xexpsc, ic):
return np.sqrt(xexpsc[ic] - np.square(xexpc[ic]))
# Time t: t0 =< t =< tf. Have N steps at which to evaluate the CN scheme. The
# time interval is dt. decp: variable for plotting to certain number of decimal
# places.
t0 = 0
tf = 20
N = 200
dt = tf / N
t = np.linspace(t0, tf, num = N + 1, endpoint = True)
decp = str(dt)[::-1].find('.')
# Initialise array for filling with norm values at each time step.
norm = np.zeros(len(t))
# Initialise array for expectation value of position.
xexp = np.zeros(len(t))
# Initialise array for expectation value of squared position.
xexps = np.zeros(len(t))
# Initialise array for alternate standard deviation.
sda = np.zeros(len(t))
# Position x: -a =< x =< a. M is an even number. There are M + 1 total discrete
# positions, for the points to be symmetric and centred at x = 0.
a = 100
M = 1200
dx = (2 * a) / M
x = np.linspace(-a, a, num = M + 1, endpoint = True)
# The gaussian function u diffuses over time. sd sets the width of gaussian. u0
# is the initial gaussian at t0.
sd = 1
var = np.power(sd, 2)
mu = 0
u0 = np.sqrt(1 / np.sqrt(np.pi * var)) * np.exp(-np.power(x - mu, 2) / (2 * \
var))
u = np.zeros([len(t), len(x)], dtype = 'complex_')
u[0, :] = u0
# Normalise u.
u[0, :] = u[0, :] / np.sqrt(normf(dx, u, 0))
# Set coefficients of CN scheme.
alpha = dt * -1j / (4 * np.power(dx, 2))
beta = dt * 1j / (4 * np.power(dx, 2))
# Tridiagonal matrices Al and AR. Al to be solved using Thomas algorithm.
Al = np.zeros([len(x), len(x)], dtype = 'complex_')
for i in range (0, M):
Al[i + 1, i] = alpha
Al[i, i] = 1 - (2 * alpha)
Al[i, i + 1] = alpha
# Corner elements for BC's.
Al[M, M], Al[0, 0] = 1 - alpha, 1 - alpha
Ar = np.zeros([len(x), len(x)], dtype = 'complex_')
for i in range (0, M):
Ar[i + 1, i] = beta
Ar[i, i] = 1 - (2 * beta)
Ar[i, i + 1] = beta
# Corner elements for BC's.
Ar[M, M], Ar[0, 0] = 1 - 2*beta, 1 - beta
# Thomas algorithm variables. Following similar naming as in Wiki article.
a = np.diag(Al, -1)
b = np.diag(Al)
c = np.diag(Al, 1)
NT = len(b)
cp = np.zeros(NT - 1, dtype = 'complex_')
for n in range(0, NT - 1):
if n == 0:
cp[n] = c[n] / b[n]
else:
cp[n] = c[n] / (b[n] - (a[n - 1] * cp[n - 1]))
d = np.zeros(NT, dtype = 'complex_')
dp = np.zeros(NT, dtype = 'complex_')
# Iterate over each time step to solve CN method. Maintain boundary
# conditions. Keep track of standard deviation.
for i in range(0, N):
# BC's.
u[i, 0], u[i, M] = 0, 0
# Find RHS.
d = np.dot(Ar, u[i, :])
for n in range(0, NT):
if n == 0:
dp[n] = d[n] / b[n]
else:
dp[n] = (d[n] - (a[n - 1] * dp[n - 1])) / (b[n] - (a[n - 1] * \
cp[n - 1]))
nc = NT - 1
while nc > -1:
if nc == NT - 1:
u[i + 1, nc] = dp[nc]
nc -= 1
else:
u[i + 1, nc] = dp[nc] - (cp[nc] * u[i + 1, nc + 1])
nc -= 1
norm[i] = normf(dx, u, i)
xexp[i] = xexpf(dx, x, u, i)
xexps[i] = xexpsf(dx, x, u, i)
sda[i] = sdaf(xexp, xexps, i)
# Fill in final norm value.
norm[N] = normf(dx, u, N)
# Fill in final position expectation value.
xexp[N] = xexpf(dx, x, u, N)
# Fill in final squared position expectation value.
xexps[N] = xexpsf(dx, x, u, N)
# Fill in final standard deviation value.
sda[N] = sdaf(xexp, xexps, N)
I want to generate two linear chains of 20 monomers each at some distance to each other. The following code generates a single chain. Could someone help me with how to generate the second chain?
The two chains are fixed to a surface i.e the first monomer of the chain is fixed and the rest of the monomers move freely in x-y-z directions but the z component of the monomers should be positive.
Something like this:
import numpy as np
import numba as nb
#import pandas as pd
#nb.jit()
def gen_chain(N):
x = np.zeros(N)
y = np.zeros(N)
z = np.linspace(0, (N)*0.9, num=N)
return np.column_stack((x, y, z)), np.column_stack((x1, y1, z1))
#coordinates = np.loadtxt('2GN_50_T_10.txt', skiprows=199950)
#return coordinates
#nb.jit()
def lj(rij2):
sig_by_r6 = np.power(sigma**2 / rij2, 3)
sig_by_r12 = np.power(sigma**2 / rij2, 6)
lje = 4 * epsilon * (sig_by_r12 - sig_by_r6)
return lje
#nb.jit()
def fene(rij2):
return (-0.5 * K * np.power(R, 2) * np.log(1 - ((np.sqrt(rij2) - r0) / R)**2))
#nb.jit()
def total_energy(coord):
# Non-bonded energy.
e_nb = 0.0
for i in range(N):
for j in range(i - 1):
ri = coord[i]
rj = coord[j]
rij = ri - rj
rij2 = np.dot(rij, rij)
if (rij2 < rcutoff_sq):
e_nb += lj(rij2)
# Bonded FENE potential energy.
e_bond = 0.0
for i in range(1, N):
ri = coord[i]
rj = coord[i - 1] # Can be [i+1] ??
rij = ri - rj
rij2 = np.dot(rij, rij)
e_bond += fene(rij2)
return e_nb + e_bond
#nb.jit()
def move(coord):
trial = np.ndarray.copy(coord)
for i in range(1, N):
while True:
delta = (2 * np.random.rand(3) - 1) * max_delta
trial[i] += delta
#while True:
if trial[i,2] > 0.0:
break
trial[i] -= delta
return trial
#nb.jit()
def accept(delta_e):
beta = 1.0 / T
if delta_e < 0.0:
return True
random_number = np.random.rand(1)
p_acc = np.exp(-beta * delta_e)
if random_number < p_acc:
return True
return False
if __name__ == "__main__":
# FENE potential parameters.
K = 40.0
R = 0.3
r0 = 0.7
# L-J potential parameters
sigma = 0.5716
epsilon = 1.0
# MC parameters
N = 20 # Numbers of monomers
rcutoff = 2.5 * sigma
rcutoff_sq = rcutoff * rcutoff
max_delta = 0.01
n_steps = 100000
T = 10
# MAIN PART OF THE CODE
coord = gen_chain(N)
energy_current = total_energy(coord)
traj = open('2GN_20_T_10.xyz', 'w')
traj_txt = open('2GN_20_T_10.txt', 'w')
for step in range(n_steps):
if step % 1000 == 0:
traj.write(str(N) + '\n\n')
for i in range(N):
traj.write("C %10.5f %10.5f %10.5f\n" % (coord[i][0], coord[i][1], coord[i][2]))
traj_txt.write("%10.5f %10.5f %10.5f\n" % (coord[i][0], coord[i][1], coord[i][2]))
print(step, energy_current)
coord_trial = move(coord)
energy_trial = total_energy(coord_trial)
delta_e = energy_trial - energy_current
if accept(delta_e):
coord = coord_trial
energy_current = energy_trial
traj.close()
I except the chain of particles to collapse into a globule.
There is some problem with the logic of the MC you are implementing.
To perform a MC you need to ATTEMPT a move, evaluate the energy of the new state and then accept/reject according to a random number.
In your code there is not the slightest sign of the attempt to move a particle.
You need to move one (or more of them), evaluate the energy, and then update your coordinates.
By the way, I suppose this is not your entire code. There are many parameters that are not defined like the "k" and the "R0" in your fene potential
The FENE potential models bond interactions. What your code is saying is that all particles within the cutoff are bonded by FENE springs, and that the bonds are not fixed but rather defined by the cutoff. With a r_cutoff = 3.0, larger than equilibrium distance of the LJ well, you are essentially considering that each particle is bonded to potentially many others. You are treating the FENE potential as a non-bonded one.
For the bond interactions you should ignore the cutoff and only evaluate the energy for the actual pairs that are bonded according to your topology, which means that first you need to define a topology. I suggest generating a linear molecule of N atoms in a box big enough to contain the whole stretched molecule, and consider the i-th atom as bonded to the (i-1)-th atom, with i = 2, ..., N. In this way the topology is well defined and persistent. Then consider both interactions separately, non-bonded and bond, and add them at the end.
Something like this, in pseudo-code:
e_nb = 0
for particle i = 1 to N:
for particle j = 1 to i-1:
if (dist(i, j) < rcutoff):
e_nb += lj(i, j)
e_bond = 0
for particle i = 2 to N:
e_bond += fene(i, i-1)
e_tot = e_nb + e_bond
Below you can find a modified version of your code. To make things simpler, in this version there is no box and no boundary conditions, just a chain in free space. The chain is initialized as a linear sequence of particles each distant 80% of R0 from the next, since R0 is the maximum length of the FENE bond. The code considers that particle i is bonded with i+1 and the bond is not broken. This code is just a proof of concept.
#!/usr/bin/python
import numpy as np
def gen_chain(N, R):
x = np.linspace(0, (N-1)*R*0.8, num=N)
y = np.zeros(N)
z = np.zeros(N)
return np.column_stack((x, y, z))
def lj(rij2):
sig_by_r6 = np.power(sigma/rij2, 3)
sig_by_r12 = np.power(sig_by_r6, 2)
lje = 4.0 * epsilon * (sig_by_r12 - sig_by_r6)
return lje
def fene(rij2):
return (-0.5 * K * R0**2 * np.log(1-(rij2/R0**2)))
def total_energy(coord):
# Non-bonded
e_nb = 0
for i in range(N):
for j in range(i-1):
ri = coord[i]
rj = coord[j]
rij = ri - rj
rij2 = np.dot(rij, rij)
if (rij2 < rcutoff):
e_nb += lj(rij2)
# Bonded
e_bond = 0
for i in range(1, N):
ri = coord[i]
rj = coord[i-1]
rij = ri - rj
rij2 = np.dot(rij, rij)
e_bond += fene(rij2)
return e_nb + e_bond
def move(coord):
trial = np.ndarray.copy(coord)
for i in range(N):
delta = (2.0 * np.random.rand(3) - 1) * max_delta
trial[i] += delta
return trial
def accept(delta_e):
beta = 1.0/T
if delta_e <= 0.0:
return True
random_number = np.random.rand(1)
p_acc = np.exp(-beta*delta_e)
if random_number < p_acc:
return True
return False
if __name__ == "__main__":
# FENE parameters
K = 40
R0 = 1.5
# LJ parameters
sigma = 1.0
epsilon = 1.0
# MC parameters
N = 50 # number of particles
rcutoff = 3.5
max_delta = 0.01
n_steps = 10000000
T = 1.5
coord = gen_chain(N, R0)
energy_current = total_energy(coord)
traj = open('traj.xyz', 'w')
for step in range(n_steps):
if step % 1000 == 0:
traj.write(str(N) + '\n\n')
for i in range(N):
traj.write("C %10.5f %10.5f %10.5f\n" % (coord[i][0], coord[i][1], coord[i][2]))
print(step, energy_current)
coord_trial = move(coord)
energy_trial = total_energy(coord_trial)
delta_e = energy_trial - energy_current
if accept(delta_e):
coord = coord_trial
energy_current = energy_trial
traj.close()
The code prints the current configuration at each step, you can just load it up on VMD and see how it behaves. The bonds will not show correctly at first on VMD, you must use a bead representation for the particles and define the bonds manually or with a script within VMD. In any case, you don't need to see the bonds to notice that the chain does not collapse.
Please bear in mind that if you want to simulate a chain at a certain density, you need to be careful to generate the correct topology. I recommend the EMC package to efficiently generate polymers at the desired thermodynamic conditions. It is by no means a trivial problem, especially for larger chains.
By the way, your code had an error in the FENE energy evaluation. rij2 is already squared, you squared it again.
Below you can see how the total energy as a function of the number of steps behaves for T = 1.0, N = 20, rcutoff = 3.5, and also the last current configuration after 10 thousand steps.
And below for N = 50, T = 1.5, max_delta = 0.01, K = 40, R = 1.5, rcutoff = 3.5, and 10 million steps. This is the last current configuration.
The full "trajectory", which isn't really a trajectory since this is MC, you can find here (it's under 6 MB).
I am trying to implement the S1 measure (Spectral Measure of Sharpness - Section III-A) from this paper. Here we have to calculate slope (alpha) of the magnitude spectrum for an image in order to measure sharpness. I am able to write the other part of the algorithm, but unable to calculate the slope. Here is my code. Function 'alpha' is where I calculate the magnitude_spectrum and I think using this we can calculate the slope but am not sure how to do that -
def aplha(image_block):
img_float32 = np.float32(image_block)
dft = cv2.dft(img_float32, flags = cv2.DFT_COMPLEX_OUTPUT)
dft_shift = np.fft.fftshift(dft)
magnitude_spectrum = 20*np.log(cv2.magnitude(dft_shift[:,:,0],dft_shift[:,:,1]))
return output (??)
Rest of the code:
def S1_calc(alpha):
tou1 = -3
tou2 = 2
output = 1 - (1 / (1 + np.exp(tou1 * (alpha - tou2))))
return output
def lx(image_block):
b = 0.7656
k = 0.0364
y = 2.2
return np.power((b + k * image_block), y)
def contrast(lx_val):
T1 = 5
T2 = 2
max_val = np.max(lx_val)
min_val = np.min(lx_val)
mean_val = np.mean(lx_val)
return (((max_val - min_val) < T1) or (mean_val < T2))
def image_gray(image_RGB):
output = (0.2989 * image_RGB[:,:,0] +
0.5870 * image_RGB[:,:,1] +
0.1140 * image_RGB[:,:,2])
return output
def S1(gray_image, m = 32, d = 24):
### SPECTRAL MEASURE OF SHARPNESS ###
# m = each block size
# d = overlapping pixels of neighbouring blocks
h,w = gray_image.shape
output = gray_image.copy()
row = 0
while (row < h):
col = 0
while (col < w):
top = row
bottom = min(row + m, h)
left = col
right = min(col + m, w)
image_block = gray_image[top : bottom, left : right]
lx_val = lx(image_block)
contrast_bool = contrast(lx_val)
if contrast_bool==True:
output[top : bottom, left : right] = 0
else:
alpha_val = aplha(image_block)
output[top : bottom, left : right] = S1_calc(alpha_val)
col = col + m - d
row = row + m - d
return output
Am using jupyter notebook, python 3.6
You could check this MATLAB code. See also another MATLAB code.
According to the latter one, we need to know freq and power value, and then we could fit these two var with a linear function, the slope of the line is what we need. We could get the slope with np.polyfit.
Now, our question is how to get the freq of a image, you could do this:
from skimage.data import camera
import numpy as np
image = camera()
height, width = image.shape
u, v = np.meshgrid(np.arange(height // 2), np.arange(width // 2))
freq = np.round(np.sqrt(u**2 + v**2)).astype(np.int64)
Now freq should be the same shape as fft transform of the input image. You need to sum all value of the magnitude_spectrum where they have the same freq, like this:
freq_uniq = np.unique(freq.flatten())
y = []
for value in f_uniq:
y.append(magnitude_spectrum[f == value].sum())
y = np.array(y)
Finally, you could just fit freq_uniq and y and get the slope. You might need to scale them with np.log first.
I am trying to calculate the euclidean distance and direction from a source coordinate within a numpy array.
Graphic Example
Here is what I was able to come up with, however it is relatively slow for large arrays. Euclidean Distance and Direction based on source coordinates rely heavily on the index of each cell. that is why I am looping each row and column. I have looked into scipy cdist, pdist, and np linalg.
import numpy as np
from math import atan, degrees, sqrt
from timeit import default_timer
def euclidean_from_source(input_array, y_index, x_index):
# copy arrays
distance = np.empty_like(input_array, dtype=float)
direction = np.empty_like(input_array, dtype=int)
# loop each row
for i, row in enumerate(X):
# loop each cell
for c, cell in enumerate(row):
# get b
b = x_index - i
# get a
a = y_index - c
hypotenuse = sqrt(a * a + b * b) * 10
distance[i][c] = hypotenuse
direction[i][c] = get_angle(a, b)
return [distance, direction]
def calibrate_angle(a, b, angle):
if b > 0 and a > 0:
angle+=90
elif b < 0 and a < 0:
angle+=270
elif b > 0 > a:
angle+=270
elif a > 0 > b:
angle+=90
return angle
def get_angle(a, b):
# get angle
if b == 0 and a == 0:
angle = 0
elif b == 0 and a >= 0:
angle = 90
elif b == 0 and a < 0:
angle = 270
elif a == 0 and b >= 0:
angle = 180
elif a == 0 and b < 0:
angle = 360
else:
theta = atan(b / a)
angle = degrees(theta)
return calibrate_angle(a, b, angle)
if __name__ == "__main__":
dimension_1 = 5
dimension_2 = 5
X = np.random.rand(dimension_1, dimension_2)
y_index = int(dimension_1/2)
x_index = int(dimension_2/2)
start = default_timer()
distance, direction = euclidean_from_source(X, y_index, x_index)
print('Total Seconds {{'.format(default_timer() - start))
print(distance)
print(direction)
UPDATE
I was able to use the broadcasting function to do exactly what I needed, and at a fraction of the speed. however I am still figuring out how to calibrate the angle to 0, 360 throughout the matrix (modulus will not work in this scenario).
import numpy as np
from math import atan, degrees, sqrt
from timeit import default_timer
def euclidean_from_source_update(input_array, y_index, x_index):
size = input_array.shape
center = (y_index, x_index)
x = np.arange(size[0])
y = np.arange(size[1])
# use broadcasting to get euclidean distance from source point
distance = np.multiply(np.sqrt((x - center[0]) ** 2 + (y[:, None] - center[1]) ** 2), 10)
# use broadcasting to get euclidean direction from source point
direction = np.rad2deg(np.arctan2((x - center[0]) , (y[:, None] - center[1])))
return [distance, direction]
def euclidean_from_source(input_array, y_index, x_index):
# copy arrays
distance = np.empty_like(input_array, dtype=float)
direction = np.empty_like(input_array, dtype=int)
# loop each row
for i, row in enumerate(X):
# loop each cell
for c, cell in enumerate(row):
# get b
b = x_index - i
# get a
a = y_index - c
hypotenuse = sqrt(a * a + b * b) * 10
distance[i][c] = hypotenuse
direction[i][c] = get_angle(a, b)
return [distance, direction]
def calibrate_angle(a, b, angle):
if b > 0 and a > 0:
angle+=90
elif b < 0 and a < 0:
angle+=270
elif b > 0 > a:
angle+=270
elif a > 0 > b:
angle+=90
return angle
def get_angle(a, b):
# get angle
if b == 0 and a == 0:
angle = 0
elif b == 0 and a >= 0:
angle = 90
elif b == 0 and a < 0:
angle = 270
elif a == 0 and b >= 0:
angle = 180
elif a == 0 and b < 0:
angle = 360
else:
theta = atan(b / a)
angle = degrees(theta)
return calibrate_angle(a, b, angle)
if __name__ == "__main__":
dimension_1 = 5
dimension_2 = 5
X = np.random.rand(dimension_1, dimension_2)
y_index = int(dimension_1/2)
x_index = int(dimension_2/2)
start = default_timer()
distance, direction = euclidean_from_source(X, y_index, x_index)
print('Total Seconds {}'.format(default_timer() - start))
start = default_timer()
distance2, direction2 = euclidean_from_source_update(X, y_index, x_index)
print('Total Seconds {}'.format(default_timer() - start))
print(distance)
print(distance2)
print(direction)
print(direction2)
Update 2
Thanks everyone for the responses, after testing methods, these two methods were the fastest and produced the results I needed. I am still open to any optimizations you guys can think of.
def get_euclidean_direction(input_array, y_index, x_index):
rdist = np.arange(input_array.shape[0]).reshape(-1, 1) - x_index
cdist = np.arange(input_array.shape[1]).reshape(1, -1) - y_index
direction = np.mod(np.degrees(np.arctan2(rdist, cdist)), 270)
direction[y_index:, :x_index]+= -90
direction[y_index:, x_index:]+= 270
direction[y_index][x_index] = 0
return direction
def get_euclidean_distance(input_array, y_index, x_index):
size = input_array.shape
center = (y_index, x_index)
x = np.arange(size[0])
y = np.arange(size[1])
return np.multiply(np.sqrt((x - center[0]) ** 2 + (y[:, None] - center[1]) ** 2), 10)
This operation is extremely easy to vectorize. For one thing, a and b don't need to be computed in 2D at all, since they only depend on one direction in the array. The distance can be computed with np.hypot. Broadcasting will convert the shape into the correct 2D form.
Your angle function is almost exactly equivalent to applying np.degrees to np.arctan2.
It's unclear why you label your rows with xand columns with y instead of the standard way of doing it, but as long as you're consistent, it should be fine.
So here's the vectorized version:
def euclidean_from_source(input_array, c, r):
rdist = np.arange(input_array.shape[0]).reshape(-1, 1) - r
# Broadcasting doesn't require this second reshape
cdist = np.arange(input_array.shape[1]).reshape(1, -1) - c
distance = np.hypot(rdist, cdist) * 10
direction = np.degrees(np.arctan2(rdist, cdist))
return distance, direction
I will leave it as an exercise for the reader to determine if any additional processing is necessary to fine-tune the angle, and if so, to implement it in a vectorized manner.
Might be easier to just pass the corrdinate you want to measure as an array or tuple. Also, while it might take a bit more memory, I think using np.indices might be a bit faster for the calculation (as it allows np.einsum to do its magic).
def euclidean_from_source(input_array, coord):
grid = np.indices(input_array.shape)
grid -= np.asarray(coord)[:, None, None]
distance = np.einsum('ijk, ijk -> jk', grid, grid) ** .5
direction = np.degrees(np.arctan2(grid[0], grid[1]))
return distance, direction
This method is also a bit more extensible to n-d (although obviously the angle calculations would be a bit trickier
I am new to programming, so I hope my stupid questions do not bug you.
I am now trying to calculate the poisson sphere distribution(a 3D version of the poisson disk) using python and then plug in the result to POV-RAY so that I can generate some random distributed packing rocks.
I am following these two links:
[https://github.com/CodingTrain/Rainbow-Code/blob/master/CodingChallenges/CC_33_poisson_disc/sketch.js#L13]
[https://www.cs.ubc.ca/~rbridson/docs/bridson-siggraph07-poissondisk.pdf]
tl;dr
0.Create an n-dimensional grid array and cell size = r/sqrt(n) where r is the minimum distance between each sphere. All arrays are set to be default -1 which stands for 'without point'
1.Create an initial sample. (it should be placed randomly but I choose to put it in the middle). Put it in the grid array. Also, intialize an active array. Put the initial sample in the active array.
2.While the active list is not empty, pick a random index. Generate points near it and make sure the points are not overlapping with nearby points(only test with the nearby arrays). If no sample can be created near the 'random index', kick the 'random index' out. Loop the process.
And here is my code:
import math
from random import uniform
import numpy
import random
radius = 1 #you can change the size of each sphere
mindis = 2 * radius
maxx = 10 #you can change the size of the container
maxy = 10
maxz = 10
k = 30
cellsize = mindis / math.sqrt(3)
nrofx = math.floor(maxx / cellsize)
nrofy = math.floor(maxy / cellsize)
nrofz = math.floor(maxz / cellsize)
grid = []
active = []
default = numpy.array((-1, -1, -1))
for fillindex in range(nrofx * nrofy * nrofz):
grid.append(default)
x = uniform(0, maxx)
y = uniform(0, maxy)
z = uniform(0, maxz)
firstpos = numpy.array((x, y, z))
firsti = maxx // 2
firstj = maxy // 2
firstk = maxz // 2
grid[firsti + nrofx * (firstj + nrofy * firstk)] = firstpos
active.append(firstpos)
while (len(active) > 0) :
randindex = math.floor(uniform(0,len(active)))
pos = active[randindex]
found = False
for attempt in range(k):
offsetx = uniform(mindis, 2 * mindis)
offsety = uniform(mindis, 2 * mindis)
offsetz = uniform(mindis, 2 * mindis)
samplex = offsetx * random.choice([1,-1])
sampley = offsety * random.choice([1,-1])
samplez = offsetz * random.choice([1,-1])
sample = numpy.array((samplex, sampley, samplez))
sample = numpy.add(sample, pos)
xcoor = math.floor(sample.item(0) / cellsize)
ycoor = math.floor(sample.item(1) / cellsize)
zcoor = math.floor(sample.item(2) / cellsize)
attemptindex = xcoor + nrofx * (ycoor + nrofy * zcoor)
if attemptindex >= 0 and attemptindex < nrofx * nrofy * nrofz and numpy.all([sample, default]) == True and xcoor > 0 and ycoor > 0 and zcoor > 0 :
test = True
for testx in range(-1,2):
for testy in range(-1, 2):
for testz in range(-1, 2):
testindex = (xcoor + testx) + nrofx * ((ycoor + testy) + nrofy * (zcoor + testz))
if testindex >=0 and testindex < nrofx * nrofy * nrofz :
neighbour = grid[testindex]
if numpy.all([neighbour, sample]) == False:
if numpy.all([neighbour, default]) == False:
distance = numpy.linalg.norm(sample - neighbour)
if distance > mindis:
test = False
if test == True and len(active)<len(grid):
found = True
grid[attemptindex] = sample
active.append(sample)
if found == False:
del active[randindex]
for printout in range(len(grid)):
print("<" + str(active[printout][0]) + "," + str(active[printout][1]) + "," + str(active[printout][2]) + ">")
print(len(grid))
My code seems to run forever.
Therefore I tried to add a print(len(active)) in the last of the while loop.
Surprisingly, I think I discovered the bug as the length of the active list just keep increasing! (It is supposed to be the same length as the grid) I think the problem is caused by the active.append(), but I can't figure out where is the problem as the code is literally the 90% the same as the one made by Mr.Shiffman.
I don't want to free ride this but I have already checked again and again while correcting again and again for this code :(. Still, I don't know where the bug is. (why do the active[] keep appending!?)
Thank you for the precious time.