Re: Give up with MT, need help!!!

[Karim Belabas <Karim.Belabas@math.u-bordeaux.fr>]

* Bill Allombert [2023-06-14 21:30]:
[...]
> But this is not the right way to compute this. You should use
> the recurrence formula for the factorial
> k! = k*(k-1)! and not recompute k!^3, (3*k)!, (6*k)! for all k from scratch.
> (3*k)!= (3*k)*(3*k-1)*(3*k-2)*(3*(k-1))!
> etc.
> Also you should factor out 640320^(3/2) to avoid recomputing it.

... and use binary splitting to have the right complexity
(a multithreaded version remains possible, but harder to get right).

Cheers,

    K.B.
-- 
Pr. Karim Belabas, U. Bordeaux, Vice-président en charge du Numérique
Institut de Mathématiques de Bordeaux UMR 5251 - (+33) 05 40 00 29 77
http://www.math.u-bordeaux.fr/~kbelabas/