I have written a function that should calculate all intersectionpoints between a line and all lines that are given to it, and this in 3D. I have these lines parametrized because that seemed to be the easiest way to work with things. The problem is that when I input the variables "t1" and "t2" back into the functions of the lines, there seems to be an inaccuracy that is too big to be acceptable for the thing that I need.
t1 is the parameter for the line of which you would like to know all intersections, so it's written in this form:
x = xo + t1 * dx
y = yo + t1 * dy
z = zo + t1 * dz
Where [xo, yo, zo] represent a point on the line that I call the "origin" and [dx, dy, dz] represents the direction of that line. The other lines are given in the same form and the function I wrote basically solves the following equation:
xo1 + t1 * dx1 = xo2 + t2 * dx2
yo1 + t1 * dy1 = yo2 + t2 * dy2
zo1 + t1 * dz1 = zo2 + t2 * dz2
Where everything is given except for t1 and t2, that's what I'm looking for here. However, I don't think actually finding t1 and t2 is the problem, I do have a solution that gives me some kind of result. As mentioned earlier, the problem is really that when I feed t1 and t2 back into these formulas to get the actual intersectionpoints, that they differ slightly from eachother. I'm talking about differences that are mostly 0.005-0.05 away from eachother in euclidean distance. But in extreme cases it could be up to 0.5 inaccuracy. I am aware that most lines in 3D do not intersect and therefore do not have a solution to these equations, but for the tests that I'm doing right now, I am 100% sure that all of the lines are within the same plane, but some might be parallel to each other. However, these inaccuracies occur for all lines, and I'm really just looking for a solution that gets it accuratly when they do intersect.
Here's the code I have for this:
def lineIntersection(self, lines):
origins = np.zeros((len(lines), 3), dtype=np.float32)
directions = np.zeros((len(lines), 3), dtype=np.float32)
for i in range(0, len(lines)):
origins[i] = lines[i].origin
directions[i] = lines[i].direction
ox = origins[:, 0]
oy = origins[:, 1]
dx = self.origin[0]
dy = self.origin[1]
x1 = directions[:, 0]
y1 = directions[:, 1]
x2 = self.direction[0]
y2 = self.direction[1]
t2 = np.divide((np.subtract(np.add(oy, np.multiply(np.divide(np.subtract(dx, ox), x1), y1)), dy)), np.subtract(y2, np.multiply(np.divide(x2, x1), y1)))
t1 = np.divide((np.add(dx, np.subtract(np.multiply(t2, x2), ox))), x1)
testx1 = np.add(ox, np.multiply(t1, x1))
testx2 = np.add(dx, np.multiply(t2, x2))
testy1 = np.add(oy, np.multiply(t1, y1))
testy2 = np.add(dy, np.multiply(t2, y2))
testz1 = np.add(origins[:, 2], np.multiply(t1, directions[:, 2]))
testz2 = np.add(self.origin[2], np.multiply(t2, self.direction[2]))
arr1 = np.array([testx1, testy1, testz1]).T
arr2 = np.array([testx2, testy2, testz2]).T
diff = np.linalg.norm(np.subtract(arr1, arr2), axis=1)
narr = arr1[diff < 0.05] #Filtering out points that aren't actually intersecting
nt2 = t2[diff < 0.05]
return narr, nt2
This function is located in the "Line" class and has an origin and direction as explained earlier. The input it takes, is an array of objects from the "Line" class.
So to be clear, I'm asking why this doesn't seem to work as precise as I want it to be and how I can fix it. Or, if there are alternatives to calculating intersectionpoints that are really accurate, I would love to hear about it.
Inaccuracy is common case for intersection of lines forming small angle.
I did not checked your algo correctness, but seems you just solve system of three equation with linalg solver.
In case of almost parallel lines intermediate values (determinant) might be small causing significant errors.
Have you tried more robust numeric algorithms like SVD?
But perhaps you really don't need them:
Note that when you are sure that all lines lie in the same plane, you can exploit 2D algorithm - just check what component of dx,dy,dz have the smallest magnitude (check for some distinct lines) and ignore corresponding component - it is similar to projecting of lines onto OXY or OXZ or OYZ plane. 2D code should be much simpler.
For true 3D case there is well-tested vector approach intended to find distance (the shortest line segment) between two skew lines - it is just zero length for intersecting ones. Example here.
Note that det (determinant) magnitude is evaluated to check for parallel (and almost parallel) lines too.
I'm currently working on a python library to extract pen strokes from TrueType fonts
- Here I'm defining a stroke as a midline running between a test point and its reflected point.
I'm using the term reflected point to refer to the closest point on the opposite side of the 'ink' region which under normal circumstances (apart from say at a serif stem) would also have a tangent in the opposite direction to the test point.
I'm working in python using fontTools and a bezier library that I rolled from the processing code described at http://pomax.github.io/bezierinfo/#extremities .
Where I'm stuck stuck at the moment is on how to find the point on a quadratic bezier curve that has a given tangent, my mathematics skills are pretty rudimentary on a good day with a clear head [which it is not rite now] so I was hoping someone with a sharper mind could point out a birds eye overview on how to achieve this.
At the moment the only thing I can think of is to approach it numerically with something similar to the Newton-Raphson root finding algorithm but evaluating the 1st derivative against the target direction values. I am hoping however there is a symbolic solution as this needs to be run on every other curve for each curve in the glyphs contours.
Using the notation given in http://pomax.github.io/bezierinfo/#extremities, a quadratic Bezier curve is given by:
B(t) = P1*(1-t)**2 + 2*P2*(1-t)*t + P3*t**2
Therefore, (by taking the derivative of B with respect to t) the tangent to the curve is given by
B'(t) = -2*P1*(1-t) + 2*P2*(1-2*t) + 2*P3*t
= 2*(P1 - 2*P2 + P3)*t + 2*(-P1 + P2)
Given some tangent direction V, solve
B'(t) = V
for t. If there is a solution, t = ts, then the point on the Bezier curve which has tangent direction V is given by B(ts).
We essentially want to know if two vectors (B'(t) and V) are parallel or anti-parallel. There is a trick to doing that.
Two vectors, X and Y, are perpendicular if their dot product is zero. If X = (a,b) and Y = (c,d) then the dot product of X and Y is given by
a*c + b*d
So, X and Y are parallel if X and Y_perp are perpendicular, where Y_perp is a vector perpendicular to Y.
In 2-dimensions, if in coordinates Y = (a,b) then Y_perp = (-b, a). (There are two perpendicular vectors possible, but this one will do.) Notice that -- using the formula above -- the dot product of Y and Y_perp is
a*(-b) + b*(a) = 0
So indeed, this jibes with the claim that perpendicular vectors have a dot product equal to 0.
Now, to solve our problem: Let
B'(t) = (a*t+b, c*t+d)
V = (e, f)
Then B'(t) is parallel (or anti-parallel) to V if or when
B'(t) is perpendicular to V_perp, which occurs when
dot product((a*t+b, c*t+d), (-f, e)) = 0
-(a*t+b)*f + (c*t+d)*e = 0
We know a, b, c, d, e and f. Solve for t. If t lies between 0 and 1, then B(t) is part of the Bezier curve segment between P1 and P3.
Thanks to #ubutbu for pointing out the solution for me, figured I would post a working implementation in case someone googles upon this question with a need in the future:
def findParallelT(P, V):
"""finds the t value along a quadratic Bezier such that its tangent (1st derivative) is parallel with the direction vector V.
P : a 3x2 matrix containing the control points of B(t).
V : a pair of values representing the direction of interest (magnitude is ignored).
returns 0.0 <= t <= 1.0 or None
Note the result may be in the same direction or flipped 180 degrees from V"""
#refer to answer given by 'unutbu' on the stackoverflow question:
#http://stackoverflow.com/questions/20825173/how-to-find-a-point-if-any-on-quadratic-bezier-with-a-given-tangent-direction
#for explanation.
# also the 'Rearrange It' app at http://www.wolframalpha.com/widgets/view.jsp?id=4be4308d0f9d17d1da68eea39de9b2ce was invaluable.
assert len(P)==3 and len(P[0])==2 and len(P[1])==2 and len(P[2])==2
assert len(V)==2
P1=P[0]
P2=P[1]
P3=P[2]
# B(t) = P1 * (1-t)**2 + 2*P2*(1-t)*t + P3*t**2
# B'(t) = 2 * (1-t) * (P2 - P1) + 2*t*(P3-P2)
# B'(t) = (a*t + b, c*t + d), V = (e, f)
a = -2 * (P2[0] - P1[0]) + 2 * (P3[0]-P2[0])
b = 2 * (P2[0] - P1[0])
c = -2 * (P2[1] - P1[1]) + 2 * (P3[1]-P2[1])
d = 2 * (P2[1] - P1[1])
e = V[0]
f = V[1]
# -(a*t+b)*f + (c*t+d)*e = 0 for parallel... t = (d*e - b*f) / (a*f - c*e)
num = float(d * e - b * f)
den = float(a * f - c * e)
if den == 0:
return None
else:
t = num / den
return t if 0.0 <= t <= 1.0 else None
if __name__ == "__main__":
assert findParallelT([[0.0,0.0], [0.0,25.0], [25.0,25.0]], [25.0,25.0]) == 0.5
assert findParallelT([[0.0,0.0], [0.0,25.0], [25.0,25.0]], [25.0,-25.0]) == None
assert findParallelT([[200.0,200.0], [10.0, 20.0], [300.0, 0.0]], [10.0,0.0]) == None
assert findParallelT([[407.5, 376.5],[321.0,463.0],[321.0,586.0]], [-246.0,0.0] ) == None
assert findParallelT([[617.0, 882.0],[740.0, 882.0], [826.5, 795.5]], [-245.0,0.0]) == 0.0
shadertoy.com has a search function and "bezier" gets you to analytic solutions for calculation (over 2 domains)
The shortest distance of any planar point to a planar quadratic-bezier, by calculating 2 roots of a cubic function
(quadratic Bezier will only have 2 real roots) (euclidean distance increases exponent of polynomial by +1).
3 points on the Bezier that equate to the 3 local extrema of the cubic function, one for each root of the quadratic.
so you just calculate which of the 3 points is closest.
I need an algorithm that can give me positions around a sphere for N points (less than 20, probably) that vaguely spreads them out. There's no need for "perfection", but I just need it so none of them are bunched together.
This question provided good code, but I couldn't find a way to make this uniform, as this seemed 100% randomized.
This blog post recommended had two ways allowing input of number of points on the sphere, but the Saff and Kuijlaars algorithm is exactly in psuedocode I could transcribe, and the code example I found contained "node[k]", which I couldn't see explained and ruined that possibility. The second blog example was the Golden Section Spiral, which gave me strange, bunched up results, with no clear way to define a constant radius.
This algorithm from this question seems like it could possibly work, but I can't piece together what's on that page into psuedocode or anything.
A few other question threads I came across spoke of randomized uniform distribution, which adds a level of complexity I'm not concerned about. I apologize that this is such a silly question, but I wanted to show that I've truly looked hard and still come up short.
So, what I'm looking for is simple pseudocode to evenly distribute N points around a unit sphere, that either returns in spherical or Cartesian coordinates. Even better if it can even distribute with a bit of randomization (think planets around a star, decently spread out, but with room for leeway).
The Fibonacci sphere algorithm is great for this. It is fast and gives results that at a glance will easily fool the human eye. You can see an example done with processing which will show the result over time as points are added. Here's another great interactive example made by #gman. And here's a simple implementation in python.
import math
def fibonacci_sphere(samples=1000):
points = []
phi = math.pi * (3. - math.sqrt(5.)) # golden angle in radians
for i in range(samples):
y = 1 - (i / float(samples - 1)) * 2 # y goes from 1 to -1
radius = math.sqrt(1 - y * y) # radius at y
theta = phi * i # golden angle increment
x = math.cos(theta) * radius
z = math.sin(theta) * radius
points.append((x, y, z))
return points
1000 samples gives you this:
The golden spiral method
You said you couldn’t get the golden spiral method to work and that’s a shame because it’s really, really good. I would like to give you a complete understanding of it so that maybe you can understand how to keep this away from being “bunched up.”
So here’s a fast, non-random way to create a lattice that is approximately correct; as discussed above, no lattice will be perfect, but this may be good enough. It is compared to other methods e.g. at BendWavy.org but it just has a nice and pretty look as well as a guarantee about even spacing in the limit.
Primer: sunflower spirals on the unit disk
To understand this algorithm, I first invite you to look at the 2D sunflower spiral algorithm. This is based on the fact that the most irrational number is the golden ratio (1 + sqrt(5))/2 and if one emits points by the approach “stand at the center, turn a golden ratio of whole turns, then emit another point in that direction,” one naturally constructs a spiral which, as you get to higher and higher numbers of points, nevertheless refuses to have well-defined ‘bars’ that the points line up on.(Note 1.)
The algorithm for even spacing on a disk is,
from numpy import pi, cos, sin, sqrt, arange
import matplotlib.pyplot as pp
num_pts = 100
indices = arange(0, num_pts, dtype=float) + 0.5
r = sqrt(indices/num_pts)
theta = pi * (1 + 5**0.5) * indices
pp.scatter(r*cos(theta), r*sin(theta))
pp.show()
and it produces results that look like (n=100 and n=1000):
Spacing the points radially
The key strange thing is the formula r = sqrt(indices / num_pts); how did I come to that one? (Note 2.)
Well, I am using the square root here because I want these to have even-area spacing around the disk. That is the same as saying that in the limit of large N I want a little region R ∈ (r, r + dr), Θ ∈ (θ, θ + dθ) to contain a number of points proportional to its area, which is r dr dθ. Now if we pretend that we are talking about a random variable here, this has a straightforward interpretation as saying that the joint probability density for (R, Θ) is just c r for some constant c. Normalization on the unit disk would then force c = 1/π.
Now let me introduce a trick. It comes from probability theory where it’s known as sampling the inverse CDF: suppose you wanted to generate a random variable with a probability density f(z) and you have a random variable U ~ Uniform(0, 1), just like comes out of random() in most programming languages. How do you do this?
First, turn your density into a cumulative distribution function or CDF, which we will call F(z). A CDF, remember, increases monotonically from 0 to 1 with derivative f(z).
Then calculate the CDF’s inverse function F-1(z).
You will find that Z = F-1(U) is distributed according to the target density. (Note 3).
Now the golden-ratio spiral trick spaces the points out in a nicely even pattern for θ so let’s integrate that out; for the unit disk we are left with F(r) = r2. So the inverse function is F-1(u) = u1/2, and therefore we would generate random points on the disk in polar coordinates with r = sqrt(random()); theta = 2 * pi * random().
Now instead of randomly sampling this inverse function we’re uniformly sampling it, and the nice thing about uniform sampling is that our results about how points are spread out in the limit of large N will behave as if we had randomly sampled it. This combination is the trick. Instead of random() we use (arange(0, num_pts, dtype=float) + 0.5)/num_pts, so that, say, if we want to sample 10 points they are r = 0.05, 0.15, 0.25, ... 0.95. We uniformly sample r to get equal-area spacing, and we use the sunflower increment to avoid awful “bars” of points in the output.
Now doing the sunflower on a sphere
The changes that we need to make to dot the sphere with points merely involve switching out the polar coordinates for spherical coordinates. The radial coordinate of course doesn't enter into this because we're on a unit sphere. To keep things a little more consistent here, even though I was trained as a physicist I'll use mathematicians' coordinates where 0 ≤ φ ≤ π is latitude coming down from the pole and 0 ≤ θ ≤ 2π is longitude. So the difference from above is that we are basically replacing the variable r with φ.
Our area element, which was r dr dθ, now becomes the not-much-more-complicated sin(φ) dφ dθ. So our joint density for uniform spacing is sin(φ)/4π. Integrating out θ, we find f(φ) = sin(φ)/2, thus F(φ) = (1 − cos(φ))/2. Inverting this we can see that a uniform random variable would look like acos(1 - 2 u), but we sample uniformly instead of randomly, so we instead use φk = acos(1 − 2 (k + 0.5)/N). And the rest of the algorithm is just projecting this onto the x, y, and z coordinates:
from numpy import pi, cos, sin, arccos, arange
import mpl_toolkits.mplot3d
import matplotlib.pyplot as pp
num_pts = 1000
indices = arange(0, num_pts, dtype=float) + 0.5
phi = arccos(1 - 2*indices/num_pts)
theta = pi * (1 + 5**0.5) * indices
x, y, z = cos(theta) * sin(phi), sin(theta) * sin(phi), cos(phi);
pp.figure().add_subplot(111, projection='3d').scatter(x, y, z);
pp.show()
Again for n=100 and n=1000 the results look like:
Further research
I wanted to give a shout out to Martin Roberts’s blog. Note that above I created an offset of my indices by adding 0.5 to each index. This was just visually appealing to me, but it turns out that the choice of offset matters a lot and is not constant over the interval and can mean getting as much as 8% better accuracy in packing if chosen correctly. There should also be a way to get his R2 sequence to cover a sphere and it would be interesting to see if this also produced a nice even covering, perhaps as-is but perhaps needing to be, say, taken from only a half of the unit square cut diagonally or so and stretched around to get a circle.
Notes
Those “bars” are formed by rational approximations to a number, and the best rational approximations to a number come from its continued fraction expression, z + 1/(n_1 + 1/(n_2 + 1/(n_3 + ...))) where z is an integer and n_1, n_2, n_3, ... is either a finite or infinite sequence of positive integers:
def continued_fraction(r):
while r != 0:
n = floor(r)
yield n
r = 1/(r - n)
Since the fraction part 1/(...) is always between zero and one, a large integer in the continued fraction allows for a particularly good rational approximation: “one divided by something between 100 and 101” is better than “one divided by something between 1 and 2.” The most irrational number is therefore the one which is 1 + 1/(1 + 1/(1 + ...)) and has no particularly good rational approximations; one can solve φ = 1 + 1/φ by multiplying through by φ to get the formula for the golden ratio.
For folks who are not so familiar with NumPy -- all of the functions are “vectorized,” so that sqrt(array) is the same as what other languages might write map(sqrt, array). So this is a component-by-component sqrt application. The same also holds for division by a scalar or addition with scalars -- those apply to all components in parallel.
The proof is simple once you know that this is the result. If you ask what's the probability that z < Z < z + dz, this is the same as asking what's the probability that z < F-1(U) < z + dz, apply F to all three expressions noting that it is a monotonically increasing function, hence F(z) < U < F(z + dz), expand the right hand side out to find F(z) + f(z) dz, and since U is uniform this probability is just f(z) dz as promised.
This is known as packing points on a sphere, and there is no (known) general, perfect solution. However, there are plenty of imperfect solutions. The three most popular seem to be:
Create a simulation. Treat each point as an electron constrained to a sphere, then run a simulation for a certain number of steps. The electrons' repulsion will naturally tend the system to a more stable state, where the points are about as far away from each other as they can get.
Hypercube rejection. This fancy-sounding method is actually really simple: you uniformly choose points (much more than n of them) inside of the cube surrounding the sphere, then reject the points outside of the sphere. Treat the remaining points as vectors, and normalize them. These are your "samples" - choose n of them using some method (randomly, greedy, etc).
Spiral approximations. You trace a spiral around a sphere, and evenly-distribute the points around the spiral. Because of the mathematics involved, these are more complicated to understand than the simulation, but much faster (and probably involving less code). The most popular seems to be by Saff, et al.
A lot more information about this problem can be found here
In this example code node[k] is just the kth node. You are generating an array N points and node[k] is the kth (from 0 to N-1). If that is all that is confusing you, hopefully you can use that now.
(in other words, k is an array of size N that is defined before the code fragment starts, and which contains a list of the points).
Alternatively, building on the other answer here (and using Python):
> cat ll.py
from math import asin
nx = 4; ny = 5
for x in range(nx):
lon = 360 * ((x+0.5) / nx)
for y in range(ny):
midpt = (y+0.5) / ny
lat = 180 * asin(2*((y+0.5)/ny-0.5))
print lon,lat
> python2.7 ll.py
45.0 -166.91313924
45.0 -74.0730322921
45.0 0.0
45.0 74.0730322921
45.0 166.91313924
135.0 -166.91313924
135.0 -74.0730322921
135.0 0.0
135.0 74.0730322921
135.0 166.91313924
225.0 -166.91313924
225.0 -74.0730322921
225.0 0.0
225.0 74.0730322921
225.0 166.91313924
315.0 -166.91313924
315.0 -74.0730322921
315.0 0.0
315.0 74.0730322921
315.0 166.91313924
If you plot that, you'll see that the vertical spacing is larger near the poles so that each point is situated in about the same total area of space (near the poles there's less space "horizontally", so it gives more "vertically").
This isn't the same as all points having about the same distance to their neighbours (which is what I think your links are talking about), but it may be sufficient for what you want and improves on simply making a uniform lat/lon grid.
What you are looking for is called a spherical covering. The spherical covering problem is very hard and solutions are unknown except for small numbers of points. One thing that is known for sure is that given n points on a sphere, there always exist two points of distance d = (4-csc^2(\pi n/6(n-2)))^(1/2) or closer.
If you want a probabilistic method for generating points uniformly distributed on a sphere, it's easy: generate points in space uniformly by Gaussian distribution (it's built into Java, not hard to find the code for other languages). So in 3-dimensional space, you need something like
Random r = new Random();
double[] p = { r.nextGaussian(), r.nextGaussian(), r.nextGaussian() };
Then project the point onto the sphere by normalizing its distance from the origin
double norm = Math.sqrt( (p[0])^2 + (p[1])^2 + (p[2])^2 );
double[] sphereRandomPoint = { p[0]/norm, p[1]/norm, p[2]/norm };
The Gaussian distribution in n dimensions is spherically symmetric so the projection onto the sphere is uniform.
Of course, there's no guarantee that the distance between any two points in a collection of uniformly generated points will be bounded below, so you can use rejection to enforce any such conditions that you might have: probably it's best to generate the whole collection and then reject the whole collection if necessary. (Or use "early rejection" to reject the whole collection you've generated so far; just don't keep some points and drop others.) You can use the formula for d given above, minus some slack, to determine the min distance between points below which you will reject a set of points. You'll have to calculate n choose 2 distances, and the probability of rejection will depend on the slack; it's hard to say how, so run a simulation to get a feel for the relevant statistics.
This answer is based on the same 'theory' that is outlined well by this answer
I'm adding this answer as:
-- None of the other options fit the 'uniformity' need 'spot-on' (or not obviously-clearly so). (Noting to get the planet like distribution looking behavior particurally wanted in the original ask, you just reject from the finite list of the k uniformly created points at random (random wrt the index count in the k items back).)
--The closest other impl forced you to decide the 'N' by 'angular axis', vs. just 'one value of N' across both angular axis values ( which at low counts of N is very tricky to know what may, or may not matter (e.g. you want '5' points -- have fun ) )
--Furthermore, it's very hard to 'grok' how to differentiate between the other options without any imagery, so here's what this option looks like (below), and the ready-to-run implementation that goes with it.
with N at 20:
and then N at 80:
here's the ready-to-run python3 code, where the emulation is that same source: " http://web.archive.org/web/20120421191837/http://www.cgafaq.info/wiki/Evenly_distributed_points_on_sphere " found by others. ( The plotting I've included, that fires when run as 'main,' is taken from: http://www.scipy.org/Cookbook/Matplotlib/mplot3D )
from math import cos, sin, pi, sqrt
def GetPointsEquiAngularlyDistancedOnSphere(numberOfPoints=45):
""" each point you get will be of form 'x, y, z'; in cartesian coordinates
eg. the 'l2 distance' from the origion [0., 0., 0.] for each point will be 1.0
------------
converted from: http://web.archive.org/web/20120421191837/http://www.cgafaq.info/wiki/Evenly_distributed_points_on_sphere )
"""
dlong = pi*(3.0-sqrt(5.0)) # ~2.39996323
dz = 2.0/numberOfPoints
long = 0.0
z = 1.0 - dz/2.0
ptsOnSphere =[]
for k in range( 0, numberOfPoints):
r = sqrt(1.0-z*z)
ptNew = (cos(long)*r, sin(long)*r, z)
ptsOnSphere.append( ptNew )
z = z - dz
long = long + dlong
return ptsOnSphere
if __name__ == '__main__':
ptsOnSphere = GetPointsEquiAngularlyDistancedOnSphere( 80)
#toggle True/False to print them
if( True ):
for pt in ptsOnSphere: print( pt)
#toggle True/False to plot them
if(True):
from numpy import *
import pylab as p
import mpl_toolkits.mplot3d.axes3d as p3
fig=p.figure()
ax = p3.Axes3D(fig)
x_s=[];y_s=[]; z_s=[]
for pt in ptsOnSphere:
x_s.append( pt[0]); y_s.append( pt[1]); z_s.append( pt[2])
ax.scatter3D( array( x_s), array( y_s), array( z_s) )
ax.set_xlabel('X'); ax.set_ylabel('Y'); ax.set_zlabel('Z')
p.show()
#end
tested at low counts (N in 2, 5, 7, 13, etc) and seems to work 'nice'
Try:
function sphere ( N:float,k:int):Vector3 {
var inc = Mathf.PI * (3 - Mathf.Sqrt(5));
var off = 2 / N;
var y = k * off - 1 + (off / 2);
var r = Mathf.Sqrt(1 - y*y);
var phi = k * inc;
return Vector3((Mathf.Cos(phi)*r), y, Mathf.Sin(phi)*r);
};
The above function should run in loop with N loop total and k loop current iteration.
It is based on a sunflower seeds pattern, except the sunflower seeds are curved around into a half dome, and again into a sphere.
Here is a picture, except I put the camera half way inside the sphere so it looks 2d instead of 3d because the camera is same distance from all points.
http://3.bp.blogspot.com/-9lbPHLccQHA/USXf88_bvVI/AAAAAAAAADY/j7qhQsSZsA8/s640/sphere.jpg
Healpix solves a closely related problem (pixelating the sphere with equal area pixels):
http://healpix.sourceforge.net/
It's probably overkill, but maybe after looking at it you'll realize some of it's other nice properties are interesting to you. It's way more than just a function that outputs a point cloud.
I landed here trying to find it again; the name "healpix" doesn't exactly evoke spheres...
edit: This does not answer the question the OP meant to ask, leaving it here in case people find it useful somehow.
We use the multiplication rule of probability, combined with infinitessimals. This results in 2 lines of code to achieve your desired result:
longitude: φ = uniform([0,2pi))
azimuth: θ = -arcsin(1 - 2*uniform([0,1]))
(defined in the following coordinate system:)
Your language typically has a uniform random number primitive. For example in python you can use random.random() to return a number in the range [0,1). You can multiply this number by k to get a random number in the range [0,k). Thus in python, uniform([0,2pi)) would mean random.random()*2*math.pi.
Proof
Now we can't assign θ uniformly, otherwise we'd get clumping at the poles. We wish to assign probabilities proportional to the surface area of the spherical wedge (the θ in this diagram is actually φ):
An angular displacement dφ at the equator will result in a displacement of dφ*r. What will that displacement be at an arbitrary azimuth θ? Well, the radius from the z-axis is r*sin(θ), so the arclength of that "latitude" intersecting the wedge is dφ * r*sin(θ). Thus we calculate the cumulative distribution of the area to sample from it, by integrating the area of the slice from the south pole to the north pole.
(where stuff=dφ*r)
We will now attempt to get the inverse of the CDF to sample from it: http://en.wikipedia.org/wiki/Inverse_transform_sampling
First we normalize by dividing our almost-CDF by its maximum value. This has the side-effect of cancelling out the dφ and r.
azimuthalCDF: cumProb = (sin(θ)+1)/2 from -pi/2 to pi/2
inverseCDF: θ = -sin^(-1)(1 - 2*cumProb)
Thus:
let x by a random float in range [0,1]
θ = -arcsin(1-2*x)
with small numbers of points you could run a simulation:
from random import random,randint
r = 10
n = 20
best_closest_d = 0
best_points = []
points = [(r,0,0) for i in range(n)]
for simulation in range(10000):
x = random()*r
y = random()*r
z = r-(x**2+y**2)**0.5
if randint(0,1):
x = -x
if randint(0,1):
y = -y
if randint(0,1):
z = -z
closest_dist = (2*r)**2
closest_index = None
for i in range(n):
for j in range(n):
if i==j:
continue
p1,p2 = points[i],points[j]
x1,y1,z1 = p1
x2,y2,z2 = p2
d = (x1-x2)**2+(y1-y2)**2+(z1-z2)**2
if d < closest_dist:
closest_dist = d
closest_index = i
if simulation % 100 == 0:
print simulation,closest_dist
if closest_dist > best_closest_d:
best_closest_d = closest_dist
best_points = points[:]
points[closest_index]=(x,y,z)
print best_points
>>> best_points
[(9.921692138442777, -9.930808529773849, 4.037839326088124),
(5.141893371460546, 1.7274947332807744, -4.575674650522637),
(-4.917695758662436, -1.090127967097737, -4.9629263893193745),
(3.6164803265540666, 7.004158551438312, -2.1172868271109184),
(-9.550655088997003, -9.580386054762917, 3.5277052594769422),
(-0.062238110294250415, 6.803105171979587, 3.1966101417463655),
(-9.600996012203195, 9.488067284474834, -3.498242301168819),
(-8.601522086624803, 4.519484132245867, -0.2834204048792728),
(-1.1198210500791472, -2.2916581379035694, 7.44937337008726),
(7.981831370440529, 8.539378431788634, 1.6889099589074377),
(0.513546008372332, -2.974333486904779, -6.981657873262494),
(-4.13615438946178, -6.707488383678717, 2.1197605651446807),
(2.2859494919024326, -8.14336582650039, 1.5418694699275672),
(-7.241410895247996, 9.907335206038226, 2.271647103735541),
(-9.433349952523232, -7.999106443463781, -2.3682575660694347),
(3.704772125650199, 1.0526567864085812, 6.148581714099761),
(-3.5710511242327048, 5.512552040316693, -3.4318468250897647),
(-7.483466337225052, -1.506434920354559, 2.36641535124918),
(7.73363824231576, -8.460241422163824, -1.4623228616326003),
(10, 0, 0)]
Take the two largest factors of your N, if N==20 then the two largest factors are {5,4}, or, more generally {a,b}. Calculate
dlat = 180/(a+1)
dlong = 360/(b+1})
Put your first point at {90-dlat/2,(dlong/2)-180}, your second at {90-dlat/2,(3*dlong/2)-180}, your 3rd at {90-dlat/2,(5*dlong/2)-180}, until you've tripped round the world once, by which time you've got to about {75,150} when you go next to {90-3*dlat/2,(dlong/2)-180}.
Obviously I'm working this in degrees on the surface of the spherical earth, with the usual conventions for translating +/- to N/S or E/W. And obviously this gives you a completely non-random distribution, but it is uniform and the points are not bunched together.
To add some degree of randomness, you could generate 2 normally-distributed (with mean 0 and std dev of {dlat/3, dlong/3} as appropriate) and add them to your uniformly distributed points.
OR... to place 20 points, compute the centers of the icosahedronal faces. For 12 points, find the vertices of the icosahedron. For 30 points, the mid point of the edges of the icosahedron. you can do the same thing with the tetrahedron, cube, dodecahedron and octahedrons: one set of points is on the vertices, another on the center of the face and another on the center of the edges. They cannot be mixed, however.
Based on fnord's answer, here is a Unity3D version with added ranges :
Code :
// golden angle in radians
static float Phi = Mathf.PI * ( 3f - Mathf.Sqrt( 5f ) );
static float Pi2 = Mathf.PI * 2;
public static Vector3 Point( float radius , int index , int total , float min = 0f, float max = 1f , float angleStartDeg = 0f, float angleRangeDeg = 360 )
{
// y goes from min (-) to max (+)
var y = ( ( index / ( total - 1f ) ) * ( max - min ) + min ) * 2f - 1f;
// golden angle increment
var theta = Phi * index ;
if( angleStartDeg != 0 || angleRangeDeg != 360 )
{
theta = ( theta % ( Pi2 ) ) ;
theta = theta < 0 ? theta + Pi2 : theta ;
var a1 = angleStartDeg * Mathf.Deg2Rad;
var a2 = angleRangeDeg * Mathf.Deg2Rad;
theta = theta * a2 / Pi2 + a1;
}
// https://stackoverflow.com/a/26127012/2496170
// radius at y
var rY = Mathf.Sqrt( 1 - y * y );
var x = Mathf.Cos( theta ) * rY;
var z = Mathf.Sin( theta ) * rY;
return new Vector3( x, y, z ) * radius;
}
Gist : https://gist.github.com/nukadelic/7449f0872f708065bc1afeb19df666f7/edit
Preview:
# create uniform spiral grid
numOfPoints = varargin[0]
vxyz = zeros((numOfPoints,3),dtype=float)
sq0 = 0.00033333333**2
sq2 = 0.9999998**2
sumsq = 2*sq0 + sq2
vxyz[numOfPoints -1] = array([(sqrt(sq0/sumsq)),
(sqrt(sq0/sumsq)),
(-sqrt(sq2/sumsq))])
vxyz[0] = -vxyz[numOfPoints -1]
phi2 = sqrt(5)*0.5 + 2.5
rootCnt = sqrt(numOfPoints)
prevLongitude = 0
for index in arange(1, (numOfPoints -1), 1, dtype=float):
zInc = (2*index)/(numOfPoints) -1
radius = sqrt(1-zInc**2)
longitude = phi2/(rootCnt*radius)
longitude = longitude + prevLongitude
while (longitude > 2*pi):
longitude = longitude - 2*pi
prevLongitude = longitude
if (longitude > pi):
longitude = longitude - 2*pi
latitude = arccos(zInc) - pi/2
vxyz[index] = array([ (cos(latitude) * cos(longitude)) ,
(cos(latitude) * sin(longitude)),
sin(latitude)])
#robert king It's a really nice solution but has some sloppy bugs in it. I know it helped me a lot though, so never mind the sloppiness. :)
Here is a cleaned up version....
from math import pi, asin, sin, degrees
halfpi, twopi = .5 * pi, 2 * pi
sphere_area = lambda R=1.0: 4 * pi * R ** 2
lat_dist = lambda lat, R=1.0: R*(1-sin(lat))
#A = 2*pi*R^2(1-sin(lat))
def sphere_latarea(lat, R=1.0):
if -halfpi > lat or lat > halfpi:
raise ValueError("lat must be between -halfpi and halfpi")
return 2 * pi * R ** 2 * (1-sin(lat))
sphere_lonarea = lambda lon, R=1.0: \
4 * pi * R ** 2 * lon / twopi
#A = 2*pi*R^2 |sin(lat1)-sin(lat2)| |lon1-lon2|/360
# = (pi/180)R^2 |sin(lat1)-sin(lat2)| |lon1-lon2|
sphere_rectarea = lambda lat0, lat1, lon0, lon1, R=1.0: \
(sphere_latarea(lat0, R)-sphere_latarea(lat1, R)) * (lon1-lon0) / twopi
def test_sphere(n_lats=10, n_lons=19, radius=540.0):
total_area = 0.0
for i_lons in range(n_lons):
lon0 = twopi * float(i_lons) / n_lons
lon1 = twopi * float(i_lons+1) / n_lons
for i_lats in range(n_lats):
lat0 = asin(2 * float(i_lats) / n_lats - 1)
lat1 = asin(2 * float(i_lats+1)/n_lats - 1)
area = sphere_rectarea(lat0, lat1, lon0, lon1, radius)
print("{:} {:}: {:9.4f} to {:9.4f}, {:9.4f} to {:9.4f} => area {:10.4f}"
.format(i_lats, i_lons
, degrees(lat0), degrees(lat1)
, degrees(lon0), degrees(lon1)
, area))
total_area += area
print("total_area = {:10.4f} (difference of {:10.4f})"
.format(total_area, abs(total_area) - sphere_area(radius)))
test_sphere()
This works and it's deadly simple. As many points as you want:
private function moveTweets():void {
var newScale:Number=Scale(meshes.length,50,500,6,2);
trace("new scale:"+newScale);
var l:Number=this.meshes.length;
var tweetMeshInstance:TweetMesh;
var destx:Number;
var desty:Number;
var destz:Number;
for (var i:Number=0;i<this.meshes.length;i++){
tweetMeshInstance=meshes[i];
var phi:Number = Math.acos( -1 + ( 2 * i ) / l );
var theta:Number = Math.sqrt( l * Math.PI ) * phi;
tweetMeshInstance.origX = (sphereRadius+5) * Math.cos( theta ) * Math.sin( phi );
tweetMeshInstance.origY= (sphereRadius+5) * Math.sin( theta ) * Math.sin( phi );
tweetMeshInstance.origZ = (sphereRadius+5) * Math.cos( phi );
destx=sphereRadius * Math.cos( theta ) * Math.sin( phi );
desty=sphereRadius * Math.sin( theta ) * Math.sin( phi );
destz=sphereRadius * Math.cos( phi );
tweetMeshInstance.lookAt(new Vector3D());
TweenMax.to(tweetMeshInstance, 1, {scaleX:newScale,scaleY:newScale,x:destx,y:desty,z:destz,onUpdate:onLookAtTween, onUpdateParams:[tweetMeshInstance]});
}
}
private function onLookAtTween(theMesh:TweetMesh):void {
theMesh.lookAt(new Vector3D());
}
I need to take a list I have created and find the closest two points and print them out. How can I go about comparing each point in the list?
There isn't any need to plot or anything, just compare the points and find the closest two in the list.
import math # 'math' needed for 'sqrt'
# Distance function
def distance(xi,xii,yi,yii):
sq1 = (xi-xii)*(xi-xii)
sq2 = (yi-yii)*(yi-yii)
return math.sqrt(sq1 + sq2)
# Run through input and reorder in [(x, y), (x,y) ...] format
oInput = ["9.5 7.5", "10.2 19.1", "9.7 10.2"] # Original input list (entered by spacing the two points).
mInput = [] # Manipulated list
fList = [] # Final list
for o in oInput:
mInput = o.split()
x,y = float(mInput[0]), float(mInput[1])
fList += [(x, y)] # outputs [(9.5, 7.5), (10.2, 19.1), (9.7, 10.2)]
It is more convenient to rewrite your distance() function to take two (x, y) tuples as parameters:
def distance(p0, p1):
return math.sqrt((p0[0] - p1[0])**2 + (p0[1] - p1[1])**2)
Now you want to iterate over all pairs of points from your list fList. The function iterools.combinations() is handy for this purpose:
min_distance = distance(fList[0], fList[1])
for p0, p1 in itertools.combinations(fList, 2):
min_distance = min(min_distance, distance(p0, p1))
An alternative is to define distance() to accept the pair of points in a single parameter
def distance(points):
p0, p1 = points
return math.sqrt((p0[0] - p1[0])**2 + (p0[1] - p1[1])**2)
and use the key parameter to the built-in min() function:
min_pair = min(itertools.combinations(fList, 2), key=distance)
min_distance = distance(min_pair)
I realize that there are library constraints on this question, but for completeness if you have N points in an Nx2 numpy ndarray (2D system):
from scipy.spatial.distance import pdist
x = numpy.array([[9.5,7.5],[10.2,19.1],[9.7,10.2]])
mindist = numpy.min(pdist(x))
I always try to encourage people to use numpy/scipy if they are dealing with data that is best stored in a numerical array and it's good to know that the tools are out there for future reference.
Note that the math.sqrt function is both slow and, in this case, unnecessary. Try comparing the distance squared to speed it up (sorting distances vs. distance squared will always produce the same ordering):
def distSquared(p0, p1):
return (p0[0] - p1[0])**2 + (p0[1] - p1[1])**2
This might work:
oInput = ["9.5 7.5", "10.2 19.1", "9.7 10.2"]
# parse inputs
inp = [(float(j[0]), float(j[1])) for j in [i.split() for i in oInput]]
# initialize results with a really large value
min_distance = float('infinity')
min_pair = None
# loop over inputs
length = len(inp)
for i in xrange(length):
for j in xrange(i+1, length):
point1 = inp[i]
point2 = inp[j]
if math.hypot(point1[0] - point2[0], point1[1] - point2[0]) < min_distance:
min_pair = [point1, point2]
once the loops are done, min_pair should be the pair with the smallest distance.
Using float() to parse the text leaves room for improvement.
math.hypot is about a third faster than calculating the distance in a handwritten python-function
Your fixed code. No efficient algorithm, just the brute force one.
import math # math needed for sqrt
# distance function
def dist(p1, p2):
return math.sqrt((p2[0] - p1[0]) ** 2 + (p2[1] - p1[1]) ** 2)
# run through input and reorder in [(x, y), (x,y) ...] format
input = ["9.5 7.5", "10.2 19.1", "9.7 10.2"] # original input list (entered by spacing the two points)
points = [map(float, point.split()) for point in input] # final list
# http://en.wikipedia.org/wiki/Closest_pair_of_points
mindist = float("inf")
for p1, p2 in itertools.combinations(points, 2):
if dist(p1, p2) < mindist:
mindist = dist(p1, p2)
closestpair = (p1, p2)
print(closestpair)
First, some notes:
a**2 # squares a
(xi - xii)**2 # squares the expression in parentheses.
mInput doesn't need to be declared in advance.
fList.append((x, y)) is more pythonic than using +=.
Now you have fList. Your distance function can be rewritten to take 2 2-tuple (point) arguments, which I won't bother with here.
Then you can just write:
shortest = float('inf')
for pair in itertools.combinations(fList, 2):
shortest = min(shortest, distance(*pair))
Many of the above questions suggest finding square root using math.sqrt which is slow as well as not a good approach to find square root. In spite of using such approach just recall the basic concepts from school: think of taking the square root of any positive number, x. The square root is then written as a power of one-half: x½. Thus, a fractional exponent indicates that some root is to be taken.
so rather than using math.sqrt((p0[0] - p1[0])**2 + (p0[1] - p1[1])**2)
Use
def distance(a,b):
euclidean_distance = ((b[0]-a[0])**2 + (a[1]-a[1])**2)**0.5
return(euclidean_distance)
Hope it helps