Re: How to efficiently count Proth primes with GP parfor?

[hermann@stamm-wilbrandt.de]

On 2024-03-20 01:53, hermann@stamm-wilbrandt.de wrote:
> ...
> I am doing it not right just now:
>
> ? c=0;parforprime(p=3,4636016641,isProth(p),C,if(C,c+=1));c
>   ***   at top-level: c=0;parforprime(p=3,4636016641,isProth(p),C,if
>   ***                     ^------------------------------------------
>   *** parforprime: mt: please use export(isProth).
>   ***   Break loop: type 'break' to go back to GP prompt
> break>
>
>
> ? export(isProth)
> ? c=0;parforprime(p=3,4636016641,isProth(p),C,if(C,c+=1));c
>   ***   at top-level: c=0;parforprime(p=3,4636016641,isProth(p),C,if
>   ***                     ^------------------------------------------
>   *** parforprime: mt: attempt to change exported variable 'v'.
>   ***   Break loop: type 'break' to go back to GP prompt
> break>
I figured out that variable v has to be local with "my(...)".

? isProth(p)=my(v=valuation(p-1,2));p>>v<2^v;
? export(isProth)
?

I tested the first 3,000 Proth primes:

? c=0;forprime(p=3,323092481,if(isProth(p),c+=1));c
3000
? ##
  ***   last result: cpu time 5,545 ms, real time 5,545 ms.
? c=0;parforprime(p=3,323092481,isProth(p),C,if(C,c+=1));c
3000
? ##
  ***   last result: cpu time 29,142 ms, real time 18,457 ms.
?

1) Why is sequential computation more than 5× faster than parallel? 
Synchronization?


2) Why does gp run with only 336% CPU and not wih more than 1500% or 
3100% ?

top - 02:07:15 up 6 days, 14:38,  2 users,  load average: 0.54, 0.86, 
0.52
Tasks: 457 total,   2 running, 455 sleeping,   0 stopped,   0 zombie
%Cpu(s):  3.9 us,  5.0 sy,  0.0 ni, 91.0 id,  0.0 wa,  0.0 hi,  0.0 si,  
0.0 st
MiB Mem :  31190.9 total,  28409.6 free,    593.4 used,   2187.9 
buff/cache
MiB Swap:   2048.0 total,   1808.3 free,    239.7 used.  30101.4 avail 
Mem

    PID USER      PR  NI    VIRT    RES    SHR S  %CPU  %MEM     TIME+ 
COMMAND
  54630 hermann   20   0  955.0g  25064   6400 R 336.0   0.1  13:12.04 
gp
...


Regards,

Hermann.