Re: Is short-circuit evaluation possible with parfor() ?

On 2026-02-10 00:45, [email protected] wrote:
> Two months ago I got my first sequence approved on oeis.org:
> "Numbers k such that no numbers of the form 1 + (product of k distinct 
> primes of first k+1 primes) are prime."
> https://oeis.org/A391020
>
> The Pari code I provided uses forsubset (although not needed) which 
> cannot be parallelized, and vecprod of primes instead of Pari 
> primorial operator.
>
> Now I have a parallel isok2() version which works superfast:
>
> hermann@x3950-X6:~$ gp -q
> ? default(nbthreads)
> 192
> ? isok2(n) = 
> {s=0;p=prime(n+1)#;export(s,p);parfor(i=2,n+1,ispseudoprime(1+p/prime(i)),r,s+=r);s==0};
> ? isok2(1993)
> 1
> ? ##
>   ***   last result: cpu time 1h, 24min, 36,945 ms, real time 28,186 ms.
> ? isok2(1994)
> 0
> ? ##
>   ***   last result: cpu time 1h, 25min, 5,259 ms, real time 28,404 ms.
> ?
>
> So for numbers of A391020 the complete loop has to be executed in 
> order to conform no prime exists. But for 1994 which does not belong 
> to A391020 computation should be aborted on first prime detection. Is 
> short-circuit evaluation like available for boolean expressions in eg. 
> C++ possible for parfor() somehow? 

See also the docs of parfor:
It  is  allowed  for expr2 to exit the loop using break/next/return.


isok3(n)= {
   my( s=0, p=prime(n+1)# );
   parfor
   ( i=2
   , n+1
   , ispseudoprime(1+p/prime(i))
   , r
   , (s+=r) && break
   );
   !s;
}


? isok(313)
cpu time = 1,382 ms, real time = 1,389 ms.
%25 = 0

? isok2(313)
cpu time = 2,431 ms, real time = 260 ms.
%26 = 0

? isok3(313)
cpu time = 1,077 ms, real time = 117 ms.
%27 = 0

? upto(lim)= [ n |n<-[1..lim], isok3(n) ];

? upto(200)
cpu time = 8,714 ms, real time = 1,086 ms.
%30 = [18, 23, 24, 32, 50, 78, 86, 88, 115, 118, 131, 133, 137, 143, 
145, 146, 149, 152, 153, 157, 159, 162, 165, 190]

BTW, for the OEIS-entry, such an "upto" (or "a_upto") is IMO preferred 
over any code-example that uses print1.

-- Ruud