The Basel Problem
Here are some code snippets in various languages that compute the Basel Problem:
To begin, $\LaTeX$
$$ \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6} = 1.6449340668482264 $$
Python
def pi_squared_over_6(N: int) -> float:
return sum(x**(-2) for x in range(1,N))
Rust
fn pi_squared_over_6(N: u64) -> f64 {
(1..N).map(|x| 1.0 / ((x*x) as f64)).sum()
}
Haskell
piSquaredOver6 :: Integer -> Double
-- no capital N in Haskell :(
piSquaredOver6 n = sum $ map (\x -> 1 / fromIntegral (x * x)) [1..n]
C
double pi_squared_over_6(unsigned int N) {
double sum = 0.0;
for (int i = 1; i < N; i++) {
sum += 1.0 / (i*i);
}
return sum;
}
What’s your favorite solution?
Performance
Let’s see how they compare in performance on an M1 Pro, for $N=10^9$.
Language | Time (ms, $\mu \pm \sigma$) |
---|---|
Rust (parallelized) | $112.6 \pm 3.5$ |
Rust (–release) | $937.9 \pm 0.4$ |
C (-O3) | $995.3 \pm 0.8$ |
Haskell (-O3) | $13454 \pm 205$ |
Python (3.10) | $67720 \pm 0$ |
Fixing Python
The python code took an absurdly long amount of time to run, so lets fix it by taking advantage of numpy, which calls into vectorized C code.
import numpy as np
def pi_squared_over_6(N: int) -> float:
x = np.ones(N)
r = np.arange(1,N)
sq = np.square(r)
div = np.divide(x, sq)
return float(np.sum(div))
A bit better, but when I’m checking btm
, the excessive memory consumption
suggests most of the work done is moving around billions of floats,
not the actual arithmetic. Lets try splitting this into chunks:
def pi_squared_over_6(N: int) -> float:
CHUNKS = 25000
SIZE = N // CHUNKS
s = 0.0
x = np.ones(N // CHUNKS - 1)
for i in range(CHUNKS):
N_tmp = i * SIZE
r = np.arange(N_tmp + 1, N_tmp + SIZE)
sq = np.square(r)
div = np.divide(x, sq)
s += np.sum(div)
# deallocate memory
del sq
del div
del r
return s
Much better! Now it’s running under 2 seconds!