Re: parforstep() available, but parsumstep() is missing

[hermann@stamm-wilbrandt.de]

On 2026-01-07 22:14, Bill Allombert wrote:
> On Wed, Jan 07, 2026 at 08:31:31PM +0100, hermann@stamm-wilbrandt.de 
> wrote:
> > ...
> > result is poor, although only 20% isprime() tests are done, parforstep
> > overhead is unacceptable:
> >
> > hermann@7950x:~$ numactl -C 0-15 gp -q
> > ? #
> >    timer = 1 (on)
> > ? parsum(k=1,10^9,isprime(k^2+(k+1)^2))
> > cpu time = 6min, 49,486 ms, real time = 25,606 ms.
> > 68588950
> > ? c=0;parforstep(k=5,10^9,5,isprime(k^2+(k+1)^2),r,c+=r);c
> > cpu time = 5min, 40,308 ms, real time = 2min, 39,913 ms.
> > 22858489
>
> I suggest:
>
> ? parsum(k=1,10^9\5,isprime((5*k)^2+((5*k)+1)^2))
> %7 = 22858489
>   cpu time 3min, 52,056 ms, real time 14,522 ms.
>
> Cheers,
> Bill.
Thank you Bill — as always you show how to do things correctly.

But I committed a fallacy. While I did prevent 40% of isprime() calls, 
those did not take much time (likely because of small divisor testing 
upfront in isprime()). So the runtime does not get reduced by 40%, in 
reality it is only a minimal improvement by 1.374s. But it is nice to 
have learned how to do the rewrite, for other applications a similar 
rewrite might reduce runtime significantly.

? parsum(k=1,10^9,isprime(k^2+(k+1)^2))
cpu time = 6min, 46,742 ms, real time = 25,438 ms.
68588950
? 1 \
+ parsum(k=1,10^9\5,isprime((5*k+0)^2+(5*k+1)^2)) \
+ parsum(k=0,10^9\5-1,isprime((5*k+2)^2+(5*k+3)^2)) \
+ parsum(k=0,10^9\5-1,isprime((5*k+4)^2+(5*k+5)^2))
cpu time = 6min, 24,658 ms, real time = 24,064 ms.
68588950
?

Regards,

Hermann.