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Cyclic colormaps comparison

I started this post looking for a diffusion map on Python, that I didn't find. Then I continued following an example on manifold learning by Jake Vanderplas on a different dataset. It worked nicely,

/images/manifold_learning_toroidal_helix.svg

but the colormap used is Spectral, that is divergent. This made me think about using a cyclic colormap, and ended up in this StackOverflow question. And I decided to compare some cyclic colormaps.

I picked up colormaps from different sources

  • cmocean:

    • phase

  • Paraview:

    • hue_L60

    • erdc_iceFire

    • nic_Edge

  • Colorcet:

    • colorwheel

    • cyclic_mrybm_35_75_c68

    • cyclic_mygbm_30_95_c78

and, of course, hsv. You can download the colormaps in text format from here.

Comparison

For all the examples below the following imports are done:

from __future__ import division, print_function
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.colors import LinearSegmentedColormap
from colorspacious import cspace_convert

Color map test image

Peter Kovesi proposed a way to compare cyclic colormaps on a paper on 2015. It consists of a spiral ramp with an undulation. In polar coordinates it reads

\begin{equation*} c = (A \rho^p \sin(n \theta) + \theta)\, \mathrm{mod}\, 2\pi \end{equation*}

with \(A\) the amplitude of the oscilation, \(\rho\) the normalized radius in [0, 1], \(p\) a positive number, and \(n\) the number of cycles.

And the following function creates the grid in Python

def circle_sine_ramp(r_max=1, r_min=0.3, amp=np.pi/5, cycles=50,
                     power=2, nr=50, ntheta=1025):
r, t = np.mgrid[r_min:r_max:nr*1j, 0:2*np.pi:ntheta*1j]
r_norm = (r - r_min)/(r_max - r_min)
vals = amp * r_norm**power * np.sin(cycles*t) + t
vals = np.mod(vals, 2*np.pi)
return t, r, vals

The following is the result

/images/sine_helix_cyclic_cmap.png

Colorblindness test

t, r, vals = circle_sine_ramp(cycles=0)
cmaps = ["hsv",
         "cmocean_phase",
         "hue_L60",
         "erdc_iceFire",
         "nic_Edge",
         "colorwheel",
         "cyclic_mrybm",
         "cyclic_mygbm"]
severity = [0, 50, 50, 50]
for colormap in cmaps:
    data = np.loadtxt(colormap + ".txt")
    fig = plt.figure()
    for cont in range(4):
        cvd_space = {"name": "sRGB1+CVD",
             "cvd_type": cvd_type[cont],
             "severity": severity[cont]}
        data2 = cspace_convert(data, cvd_space, "sRGB1")
        data2 = np.clip(data2, 0, 1)
        cmap = LinearSegmentedColormap.from_list('my_colormap', data2)
        ax = plt.subplot(2, 2, 1 + cont, projection="polar")
        ax.pcolormesh(t, r, vals, cmap=cmap)
        ax.set_xticks([])
        ax.set_yticks([])
    plt.suptitle(colormap)
    plt.tight_layout()
    plt.savefig(colormap + ".png", dpi=300)
/images/hsv_eval.png

hsv colormap comparison for different color vision deficiencies.

/images/cmocean_phase_eval.png

Phase colormap comparison for different color vision deficiencies.

/images/hue_L60_eval.png

hue_L60 colormap comparison for different color vision deficiencies.

/images/erdc_iceFire_eval.png

erdc_iceFire colormap comparison for different color vision deficiencies.

/images/nic_Edge_eval.png

nic_Edge colormap comparison for different color vision deficiencies.

/images/colorwheel_eval.png

Colorwheel colormap comparison for different color vision deficiencies.

/images/cyclic_mrybm_eval.png

Cyclic_mrybm colormap comparison for different color vision deficiencies.

/images/cyclic_mygbm_eval.png

Cyclic_mygbm colormap comparison for different color vision deficiencies.

Randomly generated cyclic colormaps

What if we generate some random colormaps that are cyclic? How would they look like?

Following are the snippet and resulting colormaps.

from __future__ import division, print_function
import numpy as np
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
from matplotlib.colors import LinearSegmentedColormap


# Next line to silence pyflakes. This import is needed.
Axes3D

plt.close("all")
fig = plt.figure()
fig2 = plt.figure()
nx = 4
ny = 4
azimuths = np.arange(0, 361, 1)
zeniths = np.arange(30, 70, 1)
values = azimuths * np.ones((30, 361))
for cont in range(nx * ny):
    np.random.seed(seed=cont)
    rad = np.random.uniform(0.1, 0.5)
    center = np.random.uniform(rad, 1 - rad, size=(3, 1))
    mat = np.random.rand(3, 3)
    rot_mat, _ = np.linalg.qr(mat)
    t = np.linspace(0, 2*np.pi, 256)
    x = rad*np.cos(t)
    y = rad*np.sin(t)
    z = 0.0*np.cos(t)
    X = np.vstack((x, y, z))
    X = rot_mat.dot(X) + center

    ax = fig.add_subplot(ny, nx, 1 + cont, projection='polar')
    cmap = LinearSegmentedColormap.from_list('my_colormap', X.T)
    ax.pcolormesh(azimuths*np.pi/180.0, zeniths, values, cmap=cmap)
    ax.set_xticks([])
    ax.set_yticks([])

    ax2 = fig2.add_subplot(ny, nx, 1 + cont, projection='3d')
    ax2.plot(X[0, :], X[1, :], X[2, :])
    ax2.set_xlim(0, 1)
    ax2.set_ylim(0, 1)
    ax2.set_zlim(0, 1)
    ax2.view_init(30, -60)
    ax2.set_xticks([0, 0.5, 1.0])
    ax2.set_yticks([0, 0.5, 1.0])
    ax2.set_zticks([0, 0.5, 1.0])
    ax2.set_xticklabels([])
    ax2.set_yticklabels([])
    ax2.set_zticklabels([])


fig.savefig("random_cmaps.png", dpi=300, transparent=True)
fig2.savefig("random_cmaps_traj.svg", transparent=True)
/images/random_cmaps.png

16 randomly generated colormaps.

/images/random_cmaps_traj.svg

Trajectories in RGB space for the randomly generated colormaps.

A good idea would be to take these colormaps and optimize some perceptual parameters such as lightness to get some usable ones.

Conclusions

I probably would use phase, colorwheel, or mrybm in the future.

/images/toroidal_helix_phase.svg

Initial image using phase.

/images/toroidal_helix_colorwheel.svg

Initial image using colorwheel.

/images/toroidal_helix_mrybm.svg

Initial image using mrybm.

References

  1. Peter Kovesi. Good Colour Maps: How to Design Them. arXiv:1509.03700 [cs.GR] 2015

Probability that a random tetrahedron over a sphere contains its center

I got interested in this problem watching the YouTube channel 3Blue1Brown, by Grant Sanderson, where he explains a way to tackle the problem that is just … elegant!

I can't emphasize enough how much I like this channel. For example, his approach to linear algebra in Essence of linear algebra is really good. I mention it, just in case you don't know it.

The problem

Let's talk business now. The problem was originally part of the 53rd Putnam competition on 1992 and was stated as

Four points are chosen at random on the surface of a sphere. What is the probability that the center of the sphere lies inside the tetrahedron whose vertices are at the four points? (It is understood that each point is in- dependently chosen relative to a uniform distribution on the sphere.)

As shown in the mentioned video, the probability is \(1/8\). Let's come with an algorithm to obtain this result —approximately, at least.

The proposed approach

The approach that we are going to use is pretty straightforward. We are going to obtain a sample of (independent) random sets, with four points each, and check how many of them satisfy the condition of being inside the tetrahedron with the points as vertices.

For this approach to work, we need two things:

  1. A way to generate random numbers uniformly distributed. This is already in numpy.random.uniform, so we don't need to do much about it.

  2. A way to check if a point is inside a tetrahedron.

Checking that a point is inside a tetrahedron

To find if a point is inside a tetrahedron, we could compute the barycentric coordinates for that point and check that all of them are have the same sign. Equivalently, as proposed here, we can check that the determinants of the matrices

\begin{equation*} M_0 = \begin{bmatrix} x_0 &y_0 &z_0 &1\\ x_1 &y_1 &z_1 &1\\ x_2 &y_2 &z_2 &1\\ x_3 &y_3 &z_3 &1 \end{bmatrix}\, , \end{equation*}
\begin{equation*} M_1 = \begin{bmatrix} x &y &z &1\\ x_1 &y_1 &z_1 &1\\ x_2 &y_2 &z_2 &1\\ x_3 &y_3 &z_3 &1 \end{bmatrix}\, , \end{equation*}
\begin{equation*} M_2 = \begin{bmatrix} x_0 &y_0 &z_0 &1\\ x &y & &1\\ x_2 &y_2 &z_2 &1\\ x_3 &y_3 &z_3 &1 \end{bmatrix}\, , \end{equation*}
\begin{equation*} M_3 = \begin{bmatrix} x_0 &y_0 &z_0 &1\\ x_1 &y_1 &z_1 &1\\ x &y &z &1\\ x_3 &y_3 &z_3 &1 \end{bmatrix}\, , \end{equation*}
\begin{equation*} M_4 = \begin{bmatrix} x_0 &y_0 &z_0 &1\\ x_1 &y_1 &z_1 &1\\ x_2 &y_2 &z_2 &1\\ x &y &z &1 \end{bmatrix}\, , \end{equation*}

have the same sign. In this case, \((x, y, z)\) is the point of interest and \((x_i, y_i, z_i)\) are the coordinates of each vertex.

The algorithm

Below is a Python implementation of the approach discussed before

from __future__ import division, print_function
from numpy import (random, pi, cos, sin, sign, hstack,
                   column_stack, logspace)
from numpy.linalg import det
import matplotlib.pyplot as plt


def in_tet(x, y, z, pt):
    """
    Determine if the point pt is inside the
    tetrahedron with vertices coordinates x, y, z
    """
    mat0 = column_stack((x, y, z, [1, 1, 1, 1]))
    det0 = det(mat0)
    for cont in range(4):
        mat = mat0.copy()
        mat[cont] = hstack((pt, 1))
        if sign(det(mat)*det0) < 0:
            inside = False
            break
        else:
            inside = True
    return inside


#%% Computation
prob = []
random.seed(seed=2)
N_min = 1
N_max = 5
N_vals = logspace(N_min, N_max, 100, dtype=int)
for N in N_vals:
    inside_cont = 0
    for cont_pts in range(N):
        phi = random.uniform(low=0.0, high=2*pi, size=4)
        theta = random.uniform(low=0.0, high=pi, size=4)
        x = sin(theta)*cos(phi)
        y = sin(theta)*sin(phi)
        z = cos(theta)
        if in_tet(x, y, z, [0, 0, 0]):
            inside_cont += 1

    prob.append(inside_cont/N)


#%% Plotting
plt.figure(figsize=(4, 3))
plt.hlines(0.125, 10**N_min, 10**N_max, color="#3f3f3f")
plt.semilogx(N_vals, prob, "o", alpha=0.5)
plt.xlabel("Number of trials")
plt.ylabel("Computed probability")
plt.tight_layout()
plt.show()

As expected, when the number of samples is sufficiently large, the estimated probability is close to the theoretical value: 0.125. This can be seen in the following figure.

Computed probability for different sample sizes