Practice: Computational reproducibility

Computational reproducibility

Warm-upHuman

Which of these reports state their evidence honestly? (Choose all that apply.)

Choose all that apply

"With seed 7, NumPy 1.26.4 and

10^6

samples, our Monte Carlo estimate is

0.7466 \pm 0.0011

(95% confidence interval).""We checked the conjecture for all

n \le 10^9

; it is therefore true.""Interval arithmetic with outward rounding shows

1.41421356 \lt \sqrt 2 \lt 1.41421357

.""Our formula agrees with the computed value to 15 digits, which proves it.""Summed in double precision; the result is exact because floating point is deterministic."

Computational reproducibility

Warm-upComputational

In IEEE double precision ( $p = 53$ ), let x = 1e16. What does (x + 1) - x evaluate to?

Computational reproducibility

CoreHuman

In double precision, 1 - math.cos(1e-8) evaluates to 0.0, although $1 - \cos(10^{-8}) \approx 5 \times 10^{-17}$ . Which explanation is correct?

Choose one

math.cos(1e-8) is correctly rounded to 1.0, because doubles next to

1

are about

10^{-16}

apart; the subtraction is then exact and reveals that earlier rounding.Subtracting two nearly equal numbers is itself an inaccurate floating-point operation.The library's cosine function has a bug.No floating-point formula can compute this quantity accurately.

Computational reproducibility

CoreHuman

Choose one

Because of cancellation: the exact sum can be

0

while the partial sums, and hence the rounding errors, are large.Because absolute values make the proof easier, but the stronger bound is also true.Because the bound must also hold for complex numbers.Because the rounding errors

\delta_k

might all have the same sign.

Computational reproducibility

CoreHuman

Using interval multiplication, compute $[-1, 2] \cdot [-1, 2]$ . What is the lower endpoint of the result?

Computational reproducibility

CoreHuman

A program you do not trust outputs: $\gcd(1071, 462) = 21$ , with certificate $x = -3$ , $y = 7$ . What can you conclude?

Choose one

The claim is proved:

1071\cdot(-3) + 462\cdot 7 = 21

, and

21

divides both

1071 = 51 \cdot 21

and

462 = 22 \cdot 21

.Nothing, unless we rerun Euclid's algorithm ourselves.The certificate is invalid, because Bézout coefficients must be positive.The certificate only shows that

21

is a common divisor.

Computational reproducibility

CoreFormal

In the Lean module, fsum x δ n is the floating-point recursive sum of x 0, …, x n with rounding errors δ. Which statement correctly formalises Theorem 2?

Choose one

∀ u, 0 ≤ u → (∀ k, |δ k| ≤ u) → |fsum x δ n - ∑ i ∈ range (n+1), x i| ≤ ((1+u)^n - 1) * ∑ i ∈ range (n+1), |x i|

|fsum x δ n - ∑ i ∈ range (n+1), x i| ≤ ((1+u)^n - 1) * ∑ i ∈ range (n+1), |x i| with no hypothesis on δ

∀ u, 0 ≤ u → (∀ k, |δ k| ≤ u) → |fsum x δ n - ∑ i ∈ range (n+1), x i| ≤ u * |∑ i ∈ range (n+1), x i|

∃ δ, fsum x δ n = ∑ i ∈ range (n+1), x i

Computational reproducibility

Warm-upHuman

You want a colleague to be able to regenerate every figure of your paper in five years' time. Which is the best plan?

Choose one

A versioned repository with a lock file of exact dependency versions, recorded seeds, tests, one command that rebuilds everything, and a release archived with a DOI.A requirements.txt saying numpy>=1.20 and the notebook as it was last run.High-resolution images of all the figures in the supplementary material.A detailed description of the algorithms in the paper; the code is not needed.

Computational reproducibility

StretchComputational

Implement Kahan's compensated summation kahan_sum(xs). Keep a running sum s and a compensation c for the low-order bits lost so far: for each x, set y = x - c, t = s + y, c = (t - s) - y, s = t. The tests compare your result with the exact rational sum.

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