FWIW, on my 2020 era laptop, the following:
#include <stdio.h>
int main(void) {
int a = 1, b = 1, c, n = 99999999;
while (--n) {
c = (a+b) % 9837;
a = b;
b = c;
}
printf("%d\n", a);
return 0;
}
The problem with Fibonacci is there are algorithmic ways to speed things up. The following (using the method at https://en.wikipedia.org/wiki/Fibonacci_sequence to compute Fibonacci numbers recursively in O(log n) arithmetic operations) takes Python 0.1 ms to compute the same value:
import functools
@functools.cache
def F(n):
if n <= 2:
return 1
m = n // 2 # rounds down
if n % 2:
# odd
return (F(m+1)**2 + F(m)**2) % 9837
else:
return ((2*F(m+1) - F(m))*F(m)) % 9837
print(F(99999999))Or recursive hashing?
As mentioned, the topic comes from an end-of-chapter problem set on prefix sums, at https://www.cs.cmu.edu/~guyb/papers/Ble93.pdf .
A prefix sum cam be implemented using a GPU, as demonstrated.
However, using a prefix sum may not be the best way to compute Fibonacci numbers. As demonstrated.
lb-guilherme•4h ago
Someone•4h ago
It wouldn’t surprise me at all if doing those 5 matrix multiplications on a CPU were faster than doing 31 on the GPU.
Also, FTA: “To calculate extremly large fibonacci numbers we need to avoid integer overflow. A simple way to avoid that is to just take everything modulo a given number in the matrix multiplication“
⇒ This also is NOT calculating the Fibonacci numbers, copping out when things get difficult on a GPU.
If you want to compute Fn mod some constant for 1 ≤ n ≤ some large constant fast on a GPU, I think I would compute quite a few of the intermediate matrices using the fast way, then spin up some ‘threads’ to compute the intermediate versions by copying those with the ‘standard’ matrix.
KeplerBoy•3h ago
It only gets interesting if you need a large matmuls or millions of small matmuls in parallel.
meindnoch•1h ago
Unintentionally, the article is a good cautionary tale of how algorithmic knowledge can beat raw hardware power, if you know what you're doing.