Python – Calculus VII

Numerical integration visualized

scipy.integrate.quad is a popular method of numerical integration with Python. Let’s see how it chooses the points at which to evaluate the function being integrated. Begin with a simple example, the exponential function.

The blue dots indicate the evaluation points, their y-coordinate being the order of evaluation. So, it begins with x=0, continues with something close to -1, etc. The function, drawn in red, is not to its true vertical scale.

The placement of dots gives away the method of integration: it is the Gauss-Kronrod quadrature with 10 Gauss nodes and 21 Kronrod nodes, abbreviated (G10, K21). The Gauss nodes are included in the Kronrod nodes, and of course the function is not evaluated there again. The evaluation process is slightly out of order in that 0 is a Kronrod node but comes first, followed by 10 Gauss nodes, followed by 10 remaining Kronrod nodes. The process ends there, as the comparison of G10 and K21 results shows the necessary precision was reached.

The square root on [0, 1] is not such a nice function, so (G10, K21) does not reach the required precision at once. The interval is bisected again and again until it does.

Surely the cube root is even worse. Or is it?

The nodes are symmetric about the midpoint of the interval of integration. So for any odd function on an interval symmetric about 0 we get G10 = 0 and K21 = 0, and the process stops at once. To see the effect of the cube root singularity, one has to use a non-symmetric interval such as [-1, 2].

That’s still fewer subdivisions than for the square root: cancellation between the left and right neighborhoods of 0 still helps. Let’s look at smooth functions next.

Rapid oscillations forces subdivisions here. There are also other reasons for subdivision, such as the loss of analyticity:

Although exp(-1/x) is infinitely differentiable on this interval, the fact that it is not analytic at 0 makes it a more difficult integrand that exp(x). Finally, an analytic function with no oscillation which still needs a bunch of subintervals:

This is the standard example used to illustrate the Runge phenomenon. Although we are not interpolating here, numerical integration is also influenced by the function having a small radius of convergence of its Taylor series. After all, G10 can be thought of as degree-9 interpolation of the given function at the Gauss nodes (the zeros of a Legendre polynomial), with the formula returning the integral of the interpolating polynomial.

The code used to plot these things:

import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import quad
global eval_points
f = lambda x: np.exp(x)
def f_int(x):
    eval_points.append(x)
    return f(x)

eval_points = []
a, b = -1, 1
quad(f_int, a, b)
n = len(eval_points)
t = np.linspace(a, b, 1000)
y = f(t)
yy = n*(y - y.min())/(y.max() - y.min())
plt.plot(t, yy, 'r') 
plt.plot(eval_points, np.arange(0, n), '.')
plt.show()

Critical points of a cubic spline

The choice of piecewise polynomials of degree 3 for interpolation is justifiably popular: even-degree splines are algebraically awkward to construct, degree 1 is simply piecewise linear interpolation (not smooth), and degree 5, while feasible, entails juggling too many coefficients. Besides, a cubic polynomial minimizes the amount of wiggling (the integral of second derivative squared) for given values and slopes at the endpoints of an interval. (Recall Connecting dots naturally.)

But the derivative of a cubic spline is a quadratic spline. And one needs the derivative to find the critical points. This results in an awkward example in SciPy documentation, annotated with “(NB: sproot only works for order 3 splines, so we fit an order 4 spline)”.

Although not implemented in SciPy, the task of computing the roots of a quadratic spline is a simple one. Obtaining the roots from the internal representation of a quadratic spline in SciPy (as a linear combination of B-splines) would take some work and reading. But a quadratic polynomial is determined by three values, so sampling it at three points, such as two consecutive knots and their average, is enough.

Quadratic formula with values instead of coefficients

Suppose we know the values of a quadratic polynomial q at -1, 0, 1, and wish to find if it has roots between -1 and 1. Let’s normalize so that q(0)=1, and let x = q(-1), y = q(1). If either x or y is negative, there is definitely a root on the interval. If they are positive, there is still a chance: we need the parabola to be concave up, have a minimum within [-1, 1], and for the minimum to be negative. All of this is easily determined once we note that the coefficients of the polynomial are a = (x+y)/2 – 1, b = (y-x)/2, and c = 1.

The inequality ${(x-y)^2 \ge 8(x+y-2)}$ ensures the suitable sign of the discriminant. It describes a parabola with vertex (1, 1) and focus (2, 2), contained in the first quadrant and tangent to the axes at (4, 0) and (0, 4). Within the orange region there are no real roots.

The line x+y=2, tangent to the parabola at its vertex, separates convex and concave parabolas. While concavity in conjunction with x, y being positive definitely precludes having roots in [-1, 1], slight convexity is not much better: it results in real roots outside of the interval. Here is the complete picture: green means there is a root in [-1, 1], orange means no real roots, red covers the rest.

all_colors — Green = there is a root in the interval [-1, 1]

Back to splines

Since the derivative of a spline is implemented in SciPy (B-splines have a nice formula for derivatives), all we need is a root-finding routine for quadratic splines. Here it is, based on the above observations but using built-in NumPy polynomial solver np.roots to avoid dealing with various special cases for the coefficients.

def quadratic_spline_roots(spl):
    roots = []
    knots = spl.get_knots()
    for a, b in zip(knots[:-1], knots[1:]):
        u, v, w = spl(a), spl((a+b)/2), spl(b)
        t = np.roots([u+w-2*v, w-u, 2*v])
        t = t[np.isreal(t) & (np.abs(t) <= 1)]
        roots.extend(t*(b-a)/2 + (b+a)/2)
    return np.array(roots)

A demonstration, which plots the spline (blue), its critical points (red), and original data points (black) as follows:

import numpy as np
import matplotlib.pyplot as plt
from scipy.interpolate import InterpolatedUnivariateSpline

x = np.arange(7)
y = np.array([3, 1, 1, 2, 2, 4, 3])
f = InterpolatedUnivariateSpline(x, y, k=3)
crit_pts = quadratic_spline_roots(f.derivative())

t = np.linspace(x[0], x[-1], 500)
plt.plot(t, f(t))
plt.plot(x, y, 'kd')
plt.plot(crit_pts, f(crit_pts), 'ro')
plt.show()

The sum of pairwise distances and the square of CDF

Suppose we have ${n}$ real numbers ${x_0,\dots, x_{n-1}}$ and want to find the sum of all distances ${|x_j-x_k|}$ over ${j < k}$ . Why? Maybe because over five years ago, the gradient flow of this quantity was used for "clustering by collision" (part 1, part 2, part 3).

If I have a Python console open, the problem appears to be solved with one line:

>>> 0.5 * np.abs(np.subtract.outer(x, x)).sum()

where the outer difference of x with x creates a matrix of all differences ${x_i-x_j}$ , then absolute values are taken, and then they are all added up. Double-counted, hence the factor of 0.5.

But trying this with, say, one million numbers is not likely to work. If each number takes 8 bytes of memory (64 bits, double precision), then the array x is still pretty small (under 8 MB) but a million-by-million matrix will require over 7 terabytes, and I won’t have that kind of RAM anytime soon.

In principle, one could run a loop adding these values, or store the matrix on a hard drive. Both are going to take forever.

There is a much better way, though. First, sort the numbers in nondecreasing order; this does not require much time or memory (compared to quadratic memory cost of forming a matrix). Then consider the partial sums ${s_k = x_0+\dots+x_k}$ ; the cost of computing them is linear in time and memory. For each fixed ${k}$ , the sum of distances to ${x_j}$ with ${j<k}$ is simply ${kx_k - s_{k-1}}$ , or, equivalently, ${(k+1)x_k - s_k}$ . So, all we have to do is add these up. Still one line of code (after sorting), but a much faster one:

>>> x.sort()
>>> (np.arange(1, n+1)*x - np.cumsum(x)).sum()

For example, x could be a sample from some continuous distribution. Assuming the distribution has a mean (i.e., is not too heavy tailed), the sum of all pairwise distances grows quadratically with n, and its average approaches a finite limit. For the uniform distribution on [0, 1] the computation shows this limit is 1/3. For the standard normal distribution it is 1.128… which is not as recognizable a number.

As ${n\to \infty}$ , the average distance of a sample taken from a distribution converges to the expected value of |X-Y| where X, Y are two independent variables with that distribution. Let’s express this in terms of the probability density function ${p}$ and the cumulative distribution function ${\Phi}$ . By symmetry, we can integrate over ${x> y}$ and double the result:

$\frac12 E|X-Y| = \int_{-\infty}^\infty p(x)\,dx \int_{-\infty}^x (x-y) p(y)\,dy$

Integrate by parts in the second integral: ${p(y) = \Phi'(y)}$ , and the boundary terms are zero.

$\frac12 E|X-Y| = \int_{-\infty}^\infty p(x)\,dx \int_{-\infty}^x \Phi(y)\,dy$

Integrate by parts in the other integral, throwing the derivative onto the indefinite integral and thus eliminating it. There is a boundary term this time.

$\frac12 E|X-Y| = \Phi(\infty) \int_{-\infty}^\infty \Phi(y)\,dy - \int_{-\infty}^\infty \Phi(x)^2\,dx$

Since ${\Phi(\infty) = 1}$ , this simplifies nicely:

$\frac12 E|X-Y| = \int_{-\infty}^\infty \Phi(x) (1-\Phi(x))\,dx$

This is a lot neater than I expected: ${E|X-Y|}$ is simply the integral of ${2\Phi(1-\Phi)}$ . I don’t often see CDF squared, like here. Some examples: for the uniform distribution on [0,1] we get

$E|X-Y| = \int_0^1 2x(1-x)\,dx = \frac13$

and for the standard normal, with ${\Phi(x) = (1+\mathrm{erf}\,(x/\sqrt{2}))/2}$ , it is

$\int_{-\infty}^\infty \frac12 \left(1-\mathrm{erf}\,(x/\sqrt{2}) ^2 \right)\,dx = \frac{2}{\sqrt{\pi}}\approx 1.12838\ldots$

The trick with sorting and cumulative sums can also be used to find, for every point ${x_k}$ , the sum (or average) of distances to all other points. To do this, we don’t sum over ${k}$ but must also add ${|x_j-x_k|}$ for ${j>k}$ . The latter sum is simply ${S - s_k - (n-k-1)x_k}$ where ${S}$ is the total sum. So, all we need is

>>> (2*np.arange(1,n+1)-n)*x - 2*np.cumsum(x) + x.sum()

Unfortunately, the analogous problems for vector-valued sequences are not as easy. If the Manhattan metric is used, we can do the computations for each coordinate separately, and add the results. For the Euclidean metric…

Moderatorship connections between Stack Exchange sites

Consider the graph in which the vertices are Stack Exchange sites, connected if two sites share a moderator. What does this graph look like? As being a moderator correlates with some level of expertise or at least interest in the topic, we can expect the structure of the graph to highlight topical connections between the sites.

Unsurprisingly, there is a dominant component (55 sites) centered at Stack Overflow. (Click the image for the larger version.)

Its radius is 6: every site can be joined to Stack Overflow by a path of length at most 6. The five sites at distance 6 from SO are Lifehacks, Martial Arts, Physical Fitness, Space Exploration, and Travel.

By the triangle inequality, the diameter of the SO component is at most 12. In fact it is exactly 12: it takes 12 steps to go from Space to either Fitness, Lifehacks, or Martial Arts.

The center of the SO component is not unique: Web Applications is another site from which every other one can be reached in 6 steps. And it has an advantage over SO in that only four sites in the component are at distance 6 from Web Apps. Naturally, these are the four sites that realize the diameter 12 of the component.

That said, Stack Overflow is the vertex of highest degree (8), meaning that Stack Overflow moderators also take part in moderating eight other sites.

What about other components? Excluding isolated vertices, we have the following picture of small components.

The largest of these is the “language component”: English Language & Usage, English Language Learners, Japanese Language, Portuguese Language, and less logically, Sustainable Living. Other notable components are Unix (with bioinformatics thrown in) and Mathematics.

Source of the information: on the page listing moderators grouped by users I executed a JavaScript one-liner

JSON.stringify(Array.from(document.querySelectorAll('.mods-summary-list')).filter(e=>e.children.length>1).map(e=>Array.from(e.children).map(x=>x.hostname.split('.')[0])))

which gave a list of connections. The rest was done in Python:

import networkx as nx
from itertools import chain, combinations
connections = # copied JS output
edges = list(chain.from_iterable(combinations(c, 2) for c in connections))
G = nx.Graph()
G.add_edges_from(edges)
components = sorted(list(nx.connected_components(G)), key=len)
main_component = G.subgraph(components[-1])
pos = nx.kamada_kawai_layout(main_component)
nx.draw_networkx(main_component, pos=pos, node_size=100)

I used different layout algorithms for the dominant component and for the rest. The default “spring” layout makes the SO component too crowded, but is okay for small components:

other_components = G.subgraph(chain.from_iterable(components[:-1]))
pos = nx.spring_layout(other_components, k=0.25) 
nx.draw_networkx(other_components, pos=pos, node_size=100)

The rest was done by various functions of the NetworkX library such as

nx.degree(main_component)
nx.center(main_component)
nx.diameter(main_component)
nx.periphery(main_component)
nx.radius(main_component)
nx.shortest_path_length(main_component, source="stackoverflow")
nx.shortest_path_length(main_component, source="webapps")

Experiments with the significance of autocorrelation

Given a sequence of numbers ${x_j}$ of length ${L}$ one may want to look for evidence of its periodic behavior. One way to do this is by computing autocorrelation, the correlation of the sequence with a shift of itself. Here is one reasonable way to do so: for lag values ${\ell=1,\dots, \lfloor L/2 \rfloor}$ compute the correlation coefficient of ${(x_1,\dots, x_{L-\ell}}$ with ${(x_{\ell+1},\dots, x_L)}$ . That the lag does not exceed ${L/2}$ ensures the entire sequence participates in the computation, so we are not making a conclusion about its periodicity after comparing a handful of terms at the beginning and the end. In other words, we are not going to detect periodicity if the period is more than half of the observed time period.

Having obtained the correlation coefficients, pick one with the largest absolute value; call it R. How large does R have to be in order for us to conclude the correlation is not a fluke? The answer depends on the distribution of our data, but an experiment can be used to get some idea of likelihood of large R.

I picked ${x_j}$ independently from the standard normal distribution, and computed ${r}$ as above. After 5 million trials with a sequence of length 100, the distribution of R was as follows:

ac100-5M — Extremal correlation coefficient in a sequence of length 100

Based on this experiment, the probability of obtaining |R| greater than 0.5 is less than 0.0016. So, 0.5 is pretty solid evidence. The probability of ${|R| > 0.6}$ is two orders of magnitude less, etc. Also, |R| is unlikely to be very close to zero unless the data is structured in some strange way. Some kind of correlation ought to be present in the white noise.

Aside: it’s not easy to construct perfectly non-autocorrelated sequences for the above test. For length 5 an example is 1,2,3,2,3. Indeed, (1,2,3,2) is uncorrelated with (2,3,2,3) and (1,2,3) is uncorrelated with (3,2,3). For length 6 and more I can’t construct these without filling them with a bunch of zeros.

Repeating the experiment with sequences of length 1000 shows a tighter distribution of R: now |R| is unlikely to be above 0.2. So, if a universal threshold is to be used here, we need to adjust R based on sequence length.

ac1000-5M — Extremal correlation coefficient in a sequence of length 1000

I did not look hard for statistical studies of this subject, resorting to an experiment. Experimentally obtained p-values are pretty consistent for the criterion ${L^{0.45}|R| > 4}$ . The number of trials was not very large (10000) so there is some fluctuation, but the pattern is clear.

Length, L	P(L^0.45\|R\| > 4)
100	0.002
300	0.0028
500	0.0022
700	0.0028
900	0.0034
1100	0.0036
1300	0.0039
1500	0.003
1700	0.003
1900	0.0042
2100	0.003
2300	0.0036
2500	0.0042
2700	0.0032
2900	0.0043
3100	0.0042
3300	0.0025
3500	0.0031
3700	0.0027
3900	0.0042

Naturally, all this depends on the assumption of independent normal variables.

And this is the approach I took to computing r in Python:

import numpy as np
n = 1000
x = np.random.normal(size=(n,))
acorr = np.correlate(x, x, mode='same')
acorr = acorr[n//2+1:]/(x.var()*np.arange(n-1, n//2, -1))
r = acorr[np.abs(acorr).argmax()]

Recursive randomness of reals: summing a random decreasing sequence

In a comment to Recursive randomness of integers Rahul pointed out a continuous version of the same problem: pick ${x_0}$ uniformly in ${[0,1]}$ , then ${x_1}$ uniformly in ${[0,x_0]}$ , then ${x_2}$ uniformly in ${[0, x_1]}$ , etc. What is the distribution of the sum ${S=\sum_{n=0}^\infty x_n}$ ?

The continuous version turns out to be easier to analyse. To begin with, it’s equivalent to picking uniformly distributed, independent ${y_n\in [0,1]}$ and letting ${x_n = y_0y_1\cdots y_n}$ . Then the sum is

$y_0+y_0y_1 + y_0y_1y_2 + \cdots$

which can be written as

$y_0(1+y_1(1+y_2(1+\cdots )))$

So, ${S}$ is a stationary point of the random process ${X\mapsto (X+1)U}$ where ${U}$ is uniformly distributed in ${[0,1]}$ . Simply put, ${S}$ and ${(S+1)U}$ have the same distribution. This yields the value of ${E[S]}$ in a much simpler way than in the previous post:

${E[S]=E[(S+1)U] = (E[S] + 1) E[U] = (E[S] + 1)/2}$

hence ${E[S]=1}$ .

We also get an equation for the cumulative distribution function ${C(t) = P[S\le t]}$ . Indeed,

$P[S\le t] = P[(S+1)U \le t] = P[S \le t/U-1]$

The latter probability is ${\int_0^1 P[S\le t/u-1]\,du = \int_0^1 C(t/u-1)\,du}$ . Conclusion: ${C(t) = \int_0^1 C(t/u-1)\,du}$ . Differentiate to get an equation for the probability density function ${p(t)}$ , namely ${p(t) = \int_0^1 p(t/u-1)\,du/u}$ . It’s convenient to change the variable of integration to ${v = t/u-1}$ , which leads to

$p(t) = \int_{t-1}^\infty p(v)\,\frac{dv}{v+1}$

Another differentiation turns the integral equation into a delay differential equation,

$p'(t) = - \frac{p(t-1)}{t}$

Looks pretty simple, doesn’t it? Since the density is zero for negative arguments, it is constant on ${[0,1]}$ . This constant, which I’ll denote ${\gamma}$ , is ${\int_{0}^\infty p(v)\,\frac{dv}{v+1}}$ , or simply ${E[1/(S+1)]}$ . I couldn’t get an analytic formula for ${\gamma}$ . My attempt was ${E[1/(S+1)] = \sum_{n=0}^\infty (-1)^n M_n}$ where ${M_n=E[S^n]}$ are the moments of ${S}$ . The moments can be computed recursively using ${E[S^n] = E[(S+1)^n]E[U^n]}$ , which yields

$M_n=\frac{1}{n} \sum_{k=0}^{n-1} \binom{n}{k}M_k$

The first few moments, starting with ${M_0}$ , are 1, 1, 3/2, 17/6, 19/3, 81/5, 8351/80… Unfortunately the series ${\sum_{n=0}^\infty (-1)^n M_n}$ diverges, so this approach seems doomed. Numerically ${\gamma \approx 0.5614}$ which is not far from the Euler-Mascheroni constant, hence the choice of notation.

On the interval (1,2) we have ${p'(t) = -\gamma/t}$ , hence

${p(t) = \gamma(1-\log t)}$ for ${1 \le t \le 2}$ .

The DDE gets harder to integrate after that… on the interval ${[2,3]}$ the solution already involves the dilogarithm (Spence’s function):

$p(t) = \gamma(1+\pi^2/12 - \log t + \log(t-1)\log t + \mathrm{Spence}\,(t))$

following SciPy’s convention for Spence. This is as far as I went… but here is an experimental confirmation of the formulas obtained so far (link to full size image).

match03 — Histogram of 10 million samples and the pdf (red) plotted on [0,3]

To generate a sample from distribution S, I begin with a bunch of zeros and repeat “add 1, multiply by U[0,1]” many times. That’s it.

import numpy as np
import matplotlib.pyplot as plt
trials = 10000000
terms = 10000
x = np.zeros(shape=(trials,))
for _ in range(terms):
    np.multiply(x+1, np.random.uniform(size=(trials,)), out=x)
_ = plt.hist(x, bins=1000, normed=True)
plt.show()

I still want to know the exact value of ${\gamma}$ … after all, it’s also the probability that the sum of our random decreasing sequence is less than 1.

Update

The constant I called “ ${\gamma}$ ” is in fact ${\exp(-\gamma)}$ where ${\gamma}$ is indeed Euler’s constant… This is what I learned from the Inverse Symbolic Calculator after solving the DDE (with initial value 1) numerically, and calculating the integral of the solution. From there, it did not take long to find that

The p.d.f. of S is the Dickman function, normalized to have integral 1. Wikipedia plots it on log scale, but Wolfram Mathworld has a plot identical to the above. Neither one mention the sum of random decreasing sequence.
This exact problem was discussed on Mathematics Stack Exchange in February 2017.
The problem and its solution go back to 1973 paper “A probabilistic approach to differential-difference equations arising in analytic number theory” by J.-M-F. Chamayou, cited in the survey Euler’s constant: Euler’s work and modern developments by J. C. Lagarias.

Oh well. At least I practiced solving delay differential equations in Python. There is no built-in method in SciPy for that, and although there are some modules for DDE out there, I decided to roll my own. The logic is straightforward: solve the ODE on an interval of length 1, then build an interpolating spline out of the numeric solution and use it as the right hand side in the ODE, repeat. I used Romberg’s method for integrating the solution; the integration is done separately on each interval [k, k+1] because of the lack of smoothness at the integers.

import numpy as np
from scipy.integrate import odeint, romb
from scipy.interpolate import interp1d
numpoints = 2**12 + 1
solution = [lambda x: 1]
integrals = [1]
for k in range(1, 15):
    y0 = solution[k-1](k)
    t = np.linspace(k, k+1, numpoints)
    rhs = lambda y, x: -solution[k-1](np.clip(x-1, k-1, k))/x
    y = odeint(rhs, y0, t, atol=1e-15, rtol=1e-13).squeeze()
    solution.append(interp1d(t, y, kind='cubic', assume_sorted=True))
    integrals.append(romb(y, dx=1/(numpoints-1)))
total_integral = sum(integrals)
print("{:.15f}".format(1/total_integral))

As a byproduct, the program found the probabilities of the random sum being in each integer interval:

56.15% in [0,1]
34.46% in [1,2]
8.19% in [2,3]
1.1% in [3,4]
0.1% in [4,5]
less than 0.01% chance of being greater than 5

The Kolakoski-Cantor set

A 0-1 sequence can be interpreted as a point in the interval [0,1]. But this makes the long-term behavior of the sequence practically invisible due to limited resolution of our screens (and eyes). To make it visible, we can also plot the points obtained by shifting the binary sequence to the left (Bernoulli shift, which also goes by many other names). The resulting orbit is often dense in the interval, which doesn’t really help us visualize any patterns. But sometimes we get an interesting complex structure.

kol_set_small — The Kolakoski-Cantor set, KC

The vertical axis here is the time parameter, the number of dyadic shifts. The 0-1 sequence being visualized is the Kolakoski sequence in its binary form, with 0 and 1 instead of 1 and 2. By definition, the n-th run of equal digits in this sequence has length ${x_n+1}$ . In particular, 000 and 111 never occur, which contributes to the blank spots near 0 and 1.

Although the sequence is not periodic, the set is quite stable in time; it does not make a visible difference whether one plots the first 10,000 shifts, or 10,000,000. The apparent symmetry about 1/2 is related to the open problem of whether the Kolakoski sequence is mirror invariant, meaning that together with any finite word (such as 0010) it also contains its complement (that would be 1101).

There are infinitely many forbidden words apart from 000 and 111 (and the words containing those). For example, 01010 cannot occur because it has 3 consecutive runs of length 1, which implies having 000 elsewhere in the sequence. For the same reason, 001100 is forbidden. This goes on forever: 00100100 is forbidden because it implies having 10101, etc.

The number of distinct words of length n in the Kolakoski sequence is bounded by a power of n (see F. M. Dekking, What is the long range order in the Kolakoski sequence?). Hence, the set pictured above is covered by ${O(n^p)}$ intervals of length ${2^{-n}}$ , which implies it (and even its closure) is zero-dimensional in any fractal sense (has Minkowski dimension 0).

The set KC apparently does not have any isolated points; this is also an open problem, of recurrence (whether every word that appears in the sequence has to appear infinitely many times). Assuming this is so, the closure of the orbit is a totally disconnected compact set without isolated points, i.e., a Cantor set. It is not self-similar (not surprising, given it’s zero-dimensional), but its relation to the Bernoulli shift implies a structure resembling self-similarity:

kolhalf — KC is covered by two copies scaled by 1/2

Applying the transformations ${x\mapsto x/2}$ and ${x\mapsto (1+x)/2}$ yields two disjoint smaller copies that cover the original set, but with some spare parts left. The leftover bits exist because not every word in the sequence can be preceded by both 0 and 1.

kolx2 — KC covered by two copies scaled by 2

Applying the transformations ${x\mapsto 2x}$ and ${x\mapsto 2x-1}$ yields two larger copies that cover the original set. There are no extra parts within the interval [0,1] but there is an overlap between the two copies.

The number ${c = \inf KC\approx 0.146778684766479}$ appears several times in the structure of the set: for instance, the central gap is ${((1-c)/2, (1+c)/2)}$ , the second-largest gap on the left has the left endpoint ${(1-c)/4}$ , etc. The Inverse Symbolic Calculator has not found anything about this number. Its binary expansion begins with 0.001 001 011 001 001 101 001 001 101 100… which one can recognize as the smallest binary number that can be written without doing anything three times in a row. (Can’t have 000; also can’t have 001 three times in a row; and 001 010 is not allowed because it contains 01010, three runs of length 1. Hence, the number begins with 001 001 011.) This number is obviously irrational, but other than that…

In conclusion, the Python code used to plot KC.

import numpy as np
import matplotlib.pyplot as plt
n = 1000000
a = np.zeros(n, dtype=int)
j = 0                  
same = False  
for i in range(1, n):
    if same:
        a[i] = a[i-1]    
        same = False
    else:
        a[i] = 1 - a[i-1]
        j += 1            
        same = bool(a[j])
v = np.array([1/2**k for k in range(60, 0, -1)])
b = np.convolve(a, v, mode='valid')
plt.plot(b, np.arange(np.size(b)), '.', ms=2)
plt.show()