Re: 3^k performance

* Ruud H.G. van Tol [2022-11-26 08:47]:
> 
> A022330_1(n)=1+n+sum(k=1,n,logint(3^k,2))
> 
> A022330_2(n)=my(t=1);1+n+sum(k=1,n,logint(t*=3,2))
> 
> ? A022330_1(10^5)
> cpu time = 8,322 ms, real time = 8,352 ms.
> %15 = 7924941755
> 
> ? A022330_2(10^5)
> cpu time = 215 ms, real time = 217 ms.
> %16 = 7924941755
> 
> So about a 40x difference.
> 
> Is that worthwhile to look into?

This is expected:

(09:02) gp > for(k=1,10^5,3^k)
time = 8,409 ms.
(09:02) gp > my(t=1);for(k=1,10^5,t*=3)
time = 175 ms.

Assume x is a small integer (here x = 3).

Computing x^k as x * t where t = x^(k-1) is already known involves a
single multiplication by x, which is done in linear tine wrt the size of t,
i.e. about s = k * log2(x).

Computing x^k from scratch involves essentially log2(k) consecutive
squarings of an increasingly large integer [ and also a few
multiplications by x whose cost is negligible ]. The final squaring
only, namely (x^r)^2 where r = k >> 1, has a cost which is superlinear
in the size of x^r, i.e about s / 2. Concretely :

(09:18) gp > t = 3^(10^5 >> 1);
(09:18) gp > for(i=1,10^4,t^2)
time = 1,805 ms.
(09:19) gp > T = t^2;
(09:19) gp > for(i=1,10^4,T^2)
time = 4,726 ms.

So doubling the size of t multiplies the squaring time by about 2.62 in
this range, which would correspond to a complexity M(s) ~ s^1.39 
(where 1.39 ~ log2(2.62)) for the cost of a multiplication of size s.

Now compare s and (s / 2)^1.39 for s = log2(3^10^5), the actual size
you're interested in:

(09:20) gp > s = log(3^10^5)/log(2);
(09:20) gp > (s/2)^1.39 / s
%3 = 40.698247890865989130716669156266722278

So a 40 x difference is quite expected !

> How to best approach that?

Never compute from scratch when previous results can be reused :-)

Cheers,

    K.B.
--
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/
`