Trapezoid rule and Romberg integration

This post will look at two numerical integration methods, the trapezoid rule and Romberg’s algorithm, and memoization. This post is a continuation of ideas from the recent posts on Lobatto integration and memoization.

Although the trapezoid rule is not typically very accurate, it can be in special instances, and Romberg combined it with extrapolation to create a very accurate method.

Trapezoid rule

The trapezoid is the simplest numerical integration method. The only thing that could be simpler is Riemann sums. By replacing rectangles of Riemann sums with trapezoids, you can make the approximation error an order of magnitude smaller.

The trapezoid rule is crude, and hardly recommended for practical use, with two exceptions. It can be remarkably efficient for periodic functions and for analytic functions that decay double exponentially. The trapezoid rule works so well in these cases that it’s common to transform a general function so that it has one of these forms so the trapezoid rule can be applied.

To be clear, the trapezoid rule for a given step size h may not be very accurate. But for periodic and double exponential functions the error decreases exponentially as h decreases.

Here’s an implementation of the trapezoid rule that follows the derivation directly.

    def trapezoid1(f, a, b, n):
        integral = 0
        h = (b-a)/n
        for i in range(n):
            integral += 0.5*h*(f(a + i*h) + f(a + (i+1)*h))
        return integral

This code approximates the integral of f(x) over [a, b] by adding up the areas of n trapezoids. Although we want to keep things simple, a slight change would make this code twice as efficient.

    def trapezoid2(f, a, b, n):
        integral = 0.5*( f(a) + f(b) )
        h = (b-a)/n
        for i in range(1, n):
            integral += f(a + i*h)
        return h*integral

Now we’re not evaluating f twice at every interior point.

Estimating error

Suppose you’ve used the trapezoid rule once, then you decide to use it again with half as large a step size in order to compare the results. If the results are the same within your tolerance, then presumably you have your result. Someone could create a function where this comparison would be misleading, where the two results agree but both are way off. But this is unlikely to happen in practice. As Einstein said, God is subtle but he is not malicious.

If you cut your step size h in half, you double your number of integration points. So if you evaluated your integrand at n points the first time, you’ll evaluate it at 2n points the second time. But half of these points are duplicates. It would be more efficient to save the function evaluations from the first integration and reuse them in the second integration, only evaluating your function at the n new integration points.

It would be most efficient to write your code to directly save previous results, but using memoization would be easier and still more efficient than redundantly evaluating your integrand. We’ll illustrate this with Python code.

Now let’s integrate exp(cos(x)) over [0, π] with 4 and then 8 steps.

    from numpy import exp, cos, pi

    print( trapezoid2(f, 0, pi, 4) )
    print( trapezoid2(f, 0, pi, 8) )

This prints


So this suggests we’ve already found our integral to six decimal places. Why so fast? Because we’re integrating a periodic function. If we repeat our experiment with exp(x) we see that we don’t even get one decimal place agreement. The code

    print( trapezoid2(exp, 0, pi, 4 ) )
    print( trapezoid2(exp, 0, pi, 8 ) )



Eliminating redundancy

The function trapezoid2 eliminated some redundancy, but we still have redundant function evaluations when we call this function twice as we do above. When we call trapezoid2 with n = 4, we do 5 function evaluations. When we call it again with n = 8 we do 9 function evaluations, 4 of which we’ve done before.

As we did in the Lobatto quadrature example, we will have our integrand function sleep for 10 seconds to make the function calls obvious, and we will add memoization to have Python to cache function evaluations for us.

    from time import sleep, time
    from functools import lru_cache

    def f(x):
        return exp(cos(x))

    t0 = time()
    trapezoid2(f, 0, pi, 4)
    t1 = time()
    print(t1 - t0)
    trapezoid2(f, 0, pi, 8)
    t2 = time()
    print(t2 - t1)

This shows that the first integration takes 50 seconds and the second requires 40 seconds. The first integration requires 5 function evaluations and the second requires 9, but the latter is faster because it only requires 4 new function evaluations.

Romberg integration

In the examples above, we doubled the number of integration intervals and compared results in order to estimate our numerical integration error. A natural next step would be to double the number of intervals again. Maybe by comparing three integrations we can see a pattern and project what the error would be if we did more integrations.

Werner Romberg took this a step further. Rather than doing a few integrations and eye-balling the results, he formalized the inference using Richardson extrapolation to project where the integrations are going. Specifically, his method applies the trapezoid rule at 2m points for increasing values of m. The method stops when either the maximum value of m has been reached or the difference between successive integral estimates is within tolerance. When Romberg’s method is appropriate, it converges very quickly and there is no need for m to be large.

To illustrate Romberg’s method, let’s go back to the example of integrating exp(x) over [0, π]. If we were to use the trapezoid rule repeatedly, we would get these results.

     Steps    Results
         1  37.920111
         2  26.516336
         4  23.267285
         8  22.424495
        16  22.211780
        32  22.158473

This doesn’t look promising. We don’t appear to have even the first decimal correct. But Romberg’s method applies Richardson extrapolation to the data above to produce a very accurate result.

    from scipy.integrate import romberg

    r = romberg(exp, 0, pi, divmax = 5) 
    print("exact:   ", exp(pi) - 1)
    print("romberg: ", r)

This produces

    exact:    22.1406926327792
    romberg:  22.1406926327867

showing that although none of the trapezoid rule estimates are good to more than 3 significant figures, the extrapolated estimate is good 12 figures, almost to 13 figures.

If you pass the argument show=True to romberg you can see the inner workings of the integration, including a report that the integrand was evaluated 33 times, i.e. 1 + 2m times where m is given by divmax.

It seems mysterious how Richardson extrapolation could take the integral estimates above, good to three figures, and produce an estimate good to twelve figures. But if we plot the error in each estimate on a log scale it becomes more apparent what’s going on.

Plot of error in Romberg integration

The errors follow nearly a straight line, and so the extrapolated error is “nearly” negative infinity. That is, since the log errors nearly follow a straight line going down, polynomial extrapolation produces a value whose log error is very large and negative.

More on numerical integration

Python and the Tell-Tale Heart

I was browsing through SciPy documentation this evening and ran across a function in scipy.misc called electrocardiogram. What?!

It’s an actual electrocardiogram, sampled at 360 Hz. Presumably it’s included as convenient example data. Here’s a plot of the first five seconds.

ECG plot

I wrote a little code using it to turn the ECG into an audio file.

from numpy import int16, iinfo
from import write
from scipy.misc import electrocardiogram

def to_integer(signal):
    # Take samples in [-1, 1] then scale to 16-bit integers
    m = iinfo(int16).max
    M = max(abs(signal))
    return int16(signal*m/M)

ecg = electrocardiogram()
write("heartbeat.wav", 360, to_integer(ecg))

I had to turn the volume way up to hear it, and that made me think of Edgar Allan Poe’s story The Tell-Tale Heart.

I may be doing something wrong. According to the documentation for the write function, I shouldn’t need to convert the singal to integers. I should just be able to leave the signal as floating point and normalize it to [-1, 1] by dividing by the largest absolute value in the signal. But when I do that, the output file will not play.

Related posts

Why HIPAA matters even if you’re not a “covered entity”


medical data

The HIPAA privacy rule only applies to “covered entities.” This generally means insurance plans, healthcare clearinghouses, and medical providers. If your company is using heath information but isn’t a covered entity per the HIPAA statute, there are a couple reasons you might still need to pay attention to HIPAA [1].

The first is that state laws may be broader than federal laws. For example, the Texas Medical Records Privacy Act extends the definition of covered entity to any business “assembling, collecting, analyzing, using, evaluating, storing, or transmitting protected health information.” So even if the US government does not consider your business to be a covered entity, the State of Texas might.

The second is that more recent privacy laws look to HIPAA. For example, it’s not clear yet what exactly California’s new privacy legislation CCPA will mean in practice, even though the law went into effect at the beginning of the year. Because HIPAA is well established and guidance documentation, companies needing to comply with CCPA are looking to HIPAA for precedent.

The connection between CCPA and HIPAA may be formalized into more than an analogy. There is a proposed amendment to CCPA that would introduce HIPAA-like expert determination for CCPA.

If you would like to discuss HIPAA deidentification or data privacy more generally, let’s talk.

More on HIPAA

[1] I advise lawyers on statistical matters, but I am not a lawyer. Nothing here should be considered legal advice. Ask your legal counsel if you need to comply with HIPAA, or with state laws analogous to HIPAA.


Scaling and memoization

The previous post explained that Lobatto’s integration method is more efficient than Gaussian quadrature when the end points of the interval need to be included as integration points. It mentioned that this is an advantage when you need to integrate over a sequence of contiguous intervals, say [1, 2] then [2, 3], because the function being integrated only needs to be evaluated at the common end points once.

This occurs in application, for example, when numerically solving differential equations. An integral might need to be evaluated at a sequence of contiguous intervals, one for each time step.

This post will illustrate the time savings from the combination of Lobatto integration and memoization. i.e. caching function evaluations. Some languages have a built-in feature for memoization. In other languages you may need to write your own memoization code.


In order to integrate functions over intervals other than [-1, 1] we need a change of variables to rescale the domain. We’ll incorporate that in the code below.

Here is our code to integrate a function f over an interval [a, b].

    def lobatto(f, a, b):
        # Change integration interval to [-1, 1]
        c = (b-a)/2
        d = (b+a)/2
        # Multiply by c becase c is the
        # Jacobian of the change of variables
        return c*integrate(lambda x: f(c*x + d),
            lobatto_points, lobatto_weights)

Reducing function evaluations

Next, we create a function to integrate which takes 10 second to evaluate. This is an artificial example, but the time required for numerical integration often is dominated by function evaluations. Here we choose an example that makes this obvious.

    from time import sleep, time

    def my_slow_function(x):
        return x**3 + x**2

The following code integrates my_slow_function over three contiguous intervals.

    t0 = time()
    lobatto(my_slow_function, 1, 2)
    lobatto(my_slow_function, 2, 3)
    lobatto(my_slow_function, 3, 4)
    t1 = time()
    print("Elapsed time: ", t1-t0)

This code takes 150 seconds because each integration requires five function evaluations at 10 seconds each.

However, by adding one line of code we can reduce the run time to 130 seconds. We add the decorator functools.lru_cache() to ask Python to cache evaluations of our integrand.

    def my_slow_function(x):
        return x**3 + x**2

Now the three integrations above take 130 seconds because my_slow_function is only evaluated at 2 and 3 one time each.

You could write your own code to cache function evaluations, and that might be worthwhile if efficiency is a priority, but it’s easy to let the language do it for you.

More on memoization

Lobatto integration

A basic idea in numerical integration is that if a method integrates polynomials exactly, it should do well on polynomial-like functions [1]. The higher the degree of polynomial it integrates exactly, the more accurate we expect it will be on functions that behave like polynomials.

The best known example of this is Gaussian quadrature. However, this post shows why for some applications you might want to use Lobatto quadrature instead. Lobatto quadrature is similar in spirit to Gaussian quadrature, but allocates points to solve a slightly different optimization problem.

Gaussian quadrature

If you need to integrate a function over an interval, it matters a great deal where you choose to evaluate the function. The optimal choice, in term of what polynomials you can integrate exactly, is to use Gaussian quadrature. By evaluating the integrand at n optimally chosen points you can integrate polynomials of degree 2n – 1 or less exactly.

So if Gaussian integration is optimal, why would you want to do anything else? The Gaussian integration points are all interior to the interval of integration. In some applications, you have to evaluate your integrand at the end points, and so you want to optimize subject to a constraint: how can you best allocate your integration points subject to the constraint that two of your integration points are the two ends of the interval? The solution is Lobatto quadrature, also called Gauss-Lobatto quadrature.

Lobatto quadrature

By evaluating the integrand at n points, two of which are the end points, Lobatto’s method exactly integrates polynomials of degree 2n-3.

Suppose for whatever reason you already know the value of the integrand evaluated at the end points. You’ve got m more function evaluations to spend. If you use those m points for Gaussian quadrature, you can exactly integrate polynomials of degree

2m – 1

or less. But if you use Lobatto quadrature, your m interior evaluations plus your two known values at the end points give you a total of m+2 function evaluations, and so can integrate polynomials of degree

2(m + 2) – 3 = 2m + 1

or less exactly, two degrees higher than if we had used Gaussian quadrature.

Next, suppose you only know the value of the function you’re integrating at one end point. Say you’ve already integrate f(x) over the interval [1, 2] using Lobatto quadrature, and now you want to integrate over [2, 3]. You already know the value f(2) from your previous integration.

Suppose you have m new function evaluations you can afford, and you don’t know f(3). If you use Lobatto quadrature, f(3) has to come out of your function evaluation budget, so you can afford m – 1 interior integration points. You then know the value of f(x) at m+1 points: f(2) came for free, you evaluated f(x) at m – 1 interior points and at x = 3. This lets you exactly integrate polynomials of degree

2(m + 1) – 3 = 2m – 1

or less, the same as Gaussian quadrature. But if you then need to integrate over [3, 4], knowing f(3) gives you a head start on the next integration, and so on.

Weights and integration points

You can look up the weights and integration points for Gaussian quadrature and Lobatto quadrature in, for example, Abramowitz and Stegun.

There is a nice symmetry between the two integration methods: Gaussian quadrature uses integration points based on the zeros of Legendre polynomials, and weights that depend on the derivatives of these polynomials. Lobatto quadrature is the other way around: integration points are given by the zeros of derivatives of Legendre polynomials, and the weights involve the Legendre polynomials themselves.

Python example

Here we’ll implement the Gauss and Lobatto rules of order five. Most of the code is data on integration points and weights.

    gauss_points = [
    gauss_weights = [
    lobatto_points = [
    lobatto_weights = [

The integration points and weights are symmetrical, so you could make the code more compact at the expense of making it a little more complicated. Putting the + in front of positive integration points is a little unconventional, but it emphasizes the symmetry by making the positive and negative weights align vertically.

Here’s our integration code:

    def integrate(f, xs, ws):
        return sum(f(xs[i])*ws[i] for i in range(5))

where we pass it the function to integrate and either Gauss data or Lobatto data.

The following verifies that with 5 integration points, Gauss should be able to exactly integrate a 9th order polynomial, and Lobatto should be able to integrate a 7th order polynomial.

    print( integrate(lambda x: x**9 + 1,
               gauss_points, gauss_weights) )

    print( integrate(lambda x: x**7 + 1,
               lobatto_points, lobatto_weights) )

Both print 2.0 as expected. The integral of an odd function over [-1, 1] is zero, and the integral of 1 over the same interval is 2.

Now let’s use both to integrate cosine over [-1, 1].

    print( integrate(cos, gauss_points, gauss_weights) )
    print( integrate(cos, lobatto_points, lobatto_weights) )

The exact integral is 2 sin(1). Here are the results.

    Exact:   1.682941969
    Gauss:   1.682941970
    Lobatto: 1.682942320

So Gauss is correct to 8, almost 9, decimal places, and Lobatto is correct to 5 decimal places.

Next, let’s hard code a 3rd order Gauss rule for comparison.

    def gauss3(f):
        k = 0.6**0.5
        s = f(-k)*5 + f(0)*8 + f(k)*5
        return s/9

We can verify that it integrates fifth order polynomials exactly:

    print(gauss3(lambda x: x**5 + 1))

and we can use it to integrate cosine:


This returns 1.68300, a little less accurate then the Lobatto rule above, illustrating that typically Lobatto will be more accurate than Gauss with the same number of function evaluations interior to the interval of integration.

More on numerical integration

[1] Polynomials can’t have horizontal asymptotes, for example, and so we should not be surprised that a method that integrates high order polynomials exactly could still do poorly on, say, a normal probability density.

Runge-Kutta methods and Butcher tableau

If you know one numerical method for solving ordinary differential equations, it’s probably Euler’s method. If you know two methods, the second is probably 4th order Runge-Kutta. It’s standard in classes on differential equations or numerical analysis to present Euler’s method as conceptually simple but inefficient introduction, then to present Runge-Kutta as a complicated but efficient alternative.

Runge-Kutta methods are a huge family of numerical methods with a wide variety of trade-offs: efficiency, accuracy, stability, etc. Euler’s method is a member of the Runge-Kutta family as are countless other variations. You could devote a career to studying Runge-Kutta methods, and some people have.

Beneath the complexity and variety, all Runge-Kutta methods have a common form that can be summarized by a matrix and two vectors. For explicit Runge-Kutta methods (ERK) the matrix is triangular, and for implicit Runge-Kutta methods (IRK) the matrix is full.

This summary of an RK method is known as a Butcher tableau, named after J. C. Butcher who classified RK methods.

“The” Runge-Kutta method

For example, let’s start with what students often take to be “the” Runge-Kutta method. This method approximates solutions to a differential equation of the form

y' = f(t, y)


y_{n+1} = y_n + \frac{h}{6}\left( k_{n1} + 2k_{n2} + 2k_{n3} + k_{n4}\right)


k_{n1} &=& f(t_n, y_n) \\ k_{n2} &=& f(t_n + 0.5h, y_n + 0.5hk_{n1}) \\ k_{n3} &=& f(t_n + 0.5h, y_n + 0.5hk_{n2}) \\ k_{n4} &=& f(t_n + h, y_n + hk_{n3}) \\

The Butcher tableau for this ERK method is

\begin{array} {c|cccc} 0\\ 1/2 & 1/2\\ 1/2 &0 &1/2 \\ 1& 0& 0& 1\\ \hline & 1/6 & 1/3 & 1/3 &1/6 \end{array}

The numbers along the left side are the coefficients of h in the first argument of f.

The numbers along the bottom are the coefficients of the ks in the expression for the value of y at the next step.

The numbers in the middle of the array are the coefficients of the ks in second argument  of f. Because this is an explicit method, each k only depends on the previous ks, and so the table of coefficients has a triangular form.

Runge-Kutta 3/8 rule

The method above is the most common 4th order ERK rule, there is another known as the 3/8 rule. It is a little less efficient and a little more accurate. A step of this rule is given by

y_{n+1} = y_n + \frac{h}{8}\left( k_{n1} + 3k_{n2} + 3k_{n3} + k_{n4}\right)


\begin{align*} k_{n1} &= f(t_n, y_n) \\ k_{n2} &= f(t_n + \frac{h}{3}, y_n + \frac{h}{3}k_{n1}) \\ k_{n3} &= f(t_n +\frac{2h}{3}, y_n -\frac{h}{3}k_{n1} + k_{n2}) \\ k_{n4} &= f(t_n + h, y_n + h k_{n1} - h k_{n2} + hk_{n3}) \end{align*}

This method is summarized in the following Butcher tableau.

\begin{array} {c|cccc} 0\\ 1/3 & 1/3\\ 2/3 & -1/3 &1 \\ 1& 1& -1 & 1\\ \hline & 1/8 & 3/8 & 3/8 &1/8 \end{array}

This example makes it a little easier to see what’s going on since none of the coefficients in the triangular array are zero. Full detail is given in the section below.

General Explicit Runge-Kutta

The most general form of an ERK rule with s steps is

y_{n+1} = y_n + h \sum_{i-1}^s b_i k_{ni}


k_{ni} = f\left(x_n + c_i h, y_n + h \sum_{j=1}^{i-1} a_{ij} k_{nj}\right)

and the Butcher tableau is

\begin{array} {c|ccccc} 0\\ c_2 & a_{21}\\ c_3 & a_{31} & a_{32} \\ \vdots & \vdots & & \ddots \\ c_s& a_{s1}& a_{s2} & \cdots & a_{s,s-1}\\ \hline & b_1 & b_2 & \cdots & b_{s-1} & b_s \end{array}

General implicit Runge-Kutta

With explicit (ERK) methods, each k depends only on its predecessors. With implicit (IRK) methods each k potentially depends on each of the others. The matrix in the tableau is full, not triangular, and one must solve for the ks.


k_{ni} = f\left(x_n + c_i h, y_n + h \sum_{j=1}^s a_{ij} k_{nj}\right)

with the sum going all the way up to s, and the Butcher tableau is

\begin{array} {c|ccccc} c_1 & a_{11} & a_{12} & \cdots & a_{1s} \\ c_2 & a_{21} & a_{22} & \cdots & a_{2s} \\ \vdots & \vdots & & \ddots & \vdots \\ c_s& a_{s1}& a_{s2} & \cdots & a_{s,s}\\ \hline & b_1 & b_2 & \cdots & b_{s} \end{array}

Implicit methods are more complicated to implement, and require more computation for a given step size. However, they are more stable for stiff differential equations and may allow larger steps. Implicit methods are less efficient when they’re not needed, and more efficient when they are needed.

Back to Euler’s method

I said at the top of the post that Euler’s method was a special case of Runge-Kutta. The Butcher tableau for the explicit (forward) Euler method is simply

 \begin{array} {c|c} 0 & 0\\ \hline & 1\end{array}

and the tableau for the implicit (backward) Euler method is just

\begin{array} {c|c} 1 & 1\\ \hline & 1\end{array}

In this post I say more about these two methods and compare their stability.

More on differential equations

Plotting complex functions

I wrote a blog post of sorts, spread over several tweets, about plotting functions of a complex variable.

Here are a couple of the images from the tweet stream. First Weierstrass’s elliptic function.

Complex plot of Weierstrass elliptic function

And a phase plot for z5.

Phase plot for fifth power function

Related posts

The orbit that put men on the moon

Richard Arenstorf (1929–2014) discovered a stable periodic orbit between the Earth and the Moon which was used as the basis for the Apollo missions.

His orbit is a special case of the three body problem where two bodies are orbiting in a plane, i.e. the Earth and the Moon, along with a third body of negligible mass relative to the other bodies, i.e. the satellite.

The system of differential equations for the Arenstorf orbit are

\begin{align*} x'' &= x + 2y' - \mu' \frac{x+\mu}{D_1} - \mu \frac{x - \mu'}{D_2} \\ y'' &= y - 2x' - \mu' \,\frac{y}{\,D_1} \,\,\,\,\,\,- \mu \frac{y}{\,D_2} \\ \end{align*}


\begin{align*} D_1 &= ((x + \mu)^2 \,+ y^2)^{3/2} \\ D_2 &= ((x - \mu')^2 + y^2)^{3/2} \end{align*}

Here the Earth is at the origin and the Moon is initially at (0, 1). The mass of the Moon is μ = 0.012277471 and the mass of the Earth is μ’ = 1-μ.

The initial conditions are

\begin{align*} x(0) &= \phantom{-}0.994 \\ x'(0) &= \phantom{-}0 \\ y(0) &= \phantom{-}0 \\ y'(0) &= -2.001585106 \end{align*}

Here’s a plot of the orbit.

Arenstorf orbit

I found the equations above in [1] which sites the original paper [2]. Note that the paper was written in 1963, seven years before the Apollo missions. Also, before leaving NASA Arenstorf mapped out a rescue orbit. This orbit was later used on Apollo 13.

Richard Arenstorf

I was fortunate to do my postdoc at Vanderbilt before Arenstorf retired and was able to sit in on an introductory course he taught on orbital mechanics. His presentation was leisurely and remarkably clear.

His course was old-school “hard analysis,” much more concrete than the abstract “soft analysis” I had studied in graduate school. He struck me as a 19th century mathematician transported to the 20th century.

He scoffed at merely measurable functions. “Have you ever seen a function that wasn’t analytic?” This would have been heresy at my alma mater.

When I asked him about “Arenstorf’s theorem” from a recently published book I was reading, he said that he didn’t recognize it. I forget now how it was stated, maybe involving Banach spaces and/or manifolds. Arenstorf was much more concrete. He wanted to help put a man on the Moon, not see how abstractly he could state his results.

More orbital mechanics posts

[1] Hairer, Nørsett, and Wanner. Solving Ordinary Differential Equations I: Nonstiff Problems. Springer-Verlag 1987.

[2] Richard F. Arenstorf. Periodic Solutions of the Restricted Three Body Problem Representing Analytic Continuations of Keplerian Elliptic Motion. American Journal of Mathematics, Vol. 85, No. 1 (Jan., 1963), pp. 27–35.

Behold! The Brusselator!

Having watched a few episodes of Phineas and Ferb, when I see “Brusselator” I imagine Dr. Doofenschmertz saying “Behold! The Brusselator!”

But the Brusselator is considerably older than Phineas and Ferb. It goes back to Belgian scientists René Lefever and Grégoire Nicolis in 1971 [1] who combined “Brussels” and “oscillator” to name the system after their institution, Université Libre de Bruxelles. It’s a dynamical system that has its origins in modeling chemical reactions.

\begin{align*} x' &= A + x^2 y - (B+1)x \\ y' &= Bx - x^2 y \end{align*}

The phase plot below shows that as you start from multiple initial conditions, you always end up on the same limit cycle.

Brusselator phase plot

Here’s the Python code that produced the plot.

    from scipy import linspace
    from scipy.integrate import solve_ivp
    import matplotlib.pyplot as plt

    A, B = 1, 3

    def brusselator(t, z):
        x, y = z
        return [A + x*x*y - (B+1)*x, B*x - x*x*y]

    a, b = 0, 10
    t = linspace(a, b, 1000)

    for x0 in range(0, 6):
        for y0 in [0, 3]:
            sol = solve_ivp(brusselator, [a, b], [x0, y0], t_eval=t)
            plt.plot(sol.y[0], sol.y[1], ":", color="tab:blue")

More dynamical systems posts

[1] R. Lefever and G. Nicholis. Chemical instabilities and sustained oscillations. Journal of Theoretical Biology. Volume 30, Issue 2, February 1971, Pages 267-284

TestU01 small crush test suite

In recent posts I’ve written about using RNG test suites on the output of the μRNG entropy extractor. This is probably the last post in the series. I’ve looked at NIST STS, PractRand, and DIEHARDER before. In this post I’ll be looking at TestU01.

TestU01 includes three batteries of tests: Small Crush, Crush, and Big Crush. The entropy extractor failed the smallest of the three, so I didn’t go on to the larger suites. Small Crush isn’t small; it used over 22 billion 32-bit samples as input, about 0.84 GB of data. Crush uses two orders of magnitude more data, and Big Crush uses another order of magnitude more data than Crush.

SmallCrush consists of 10 tests:

  • smarsa_BirthdaySpacings
  • sknuth_Collision
  • sknuth_Gap
  • sknuth_SimpPoker
  • sknuth_CouponCollector
  • sknuth_MaxOft
  • svaria_WeightDistrib
  • smarsa_MatrixRank
  • sstring_HammingIndep
  • swalk_RandomWalk1

The test names begin with s, followed by a prefix indicating the origin of the test. For example, knuth refers to Donald Knuth’s tests in volume 2 of TAOCP and marsa refers to George Marsaglia. The remainder of the name is more descriptive, such as SimpPoker for Knuth’s simple poker test.

The output of the entropy extractor failed four of the tests, failure being defined as producing a p-value less than 10-300. The other tests passed without issue, meaning they returned p-values in the range [0.001, 0.999].

Recall from earlier posts that μRNG entropy extractor takes three possibly biased bit streams and produces an unbiased bit stream, provided each of the input streams has min-entropy of at least 1/3. I produced biased streams by taking the bitwise OR of two consecutive values, producing a stream with probability 0.75 of being a 1 and probability 0.25 of being a 0. The result passed all STS and DIEHARDER tests, but failed some PractRand and Test01 SmallCrush tests. This is consistent with the generally held opinion that STS and DIEHARDER are relatively weak tests and PractRand and TestU01 are more rigorous tests.

I applied the entropy extractor to PCG without creating a biased stream, and the result passed PractRand and TestIU01 SmallCrush. Presumably it would have passed STS and DIEHARDER as well. This confirms that the extractor does no harm to a high-quality stream of pseudorandom bits. It largely removes the bias from biased streams, enough to pass the easier two test suites but not enough to pass the two more demanding test suites.

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