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determine count of Proth primes until environment n
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\\ detemine count of Proth primes until environment n | |
\\ environment l (loops) allows to tune work distribution | |
\\ | |
\\ based on Karim's isProth2(): | |
\\ https://pari.math.u-bordeaux.fr/archives/pari-users-2403/msg00064.html | |
\\ and Bill's parallel implementation: | |
\\ https://pari.math.u-bordeaux.fr/archives/pari-users-2403/msg00067.html | |
\\ | |
isProth2(p) = !(p >> (valuation(p-1,2)<<1)); | |
export(isProth2); | |
nbt=default(nbthreads); | |
doit(n,l)={ | |
my(c=0,B=n\nbt\if(l,l,1)); | |
parfor(ii=0, | |
(n+B-1)\B, | |
my(cc=0,i=ii*B+1); | |
forprime(p=i, | |
min(i+B-1,n), | |
if(isProth2(p),cc++));cc, | |
C, | |
c+=C | |
); | |
c | |
}; | |
n=eval(getenv("n")); | |
gettime(); | |
t0=getwalltime(); | |
r=doit(n,eval(getenv("l"))); | |
wt=getwalltime()-t0; | |
tt=gettime(); | |
print(nbt); | |
printf("%s × %.3f = %s",strtime(wt),tt/wt/1.0,strtime(tt)); | |
print(r); | |
t0=getwalltime(); | |
c=0;forprime(p=3,n,if(isProth2(p),c++)); | |
wts=getwalltime()-t0; | |
printf("%s (%.3f×)",strtime(wts),wts/wt/1.0); | |
print(c); |
For 2.5× n even closer to factor 32×, but the other factor 13.046× seems to be maxed out:
hermann@7950x:~$ n=10^11 l=32 gp -q < proth.gp
32
1min, 10,058 ms × 31.784 = 37min, 6,695 ms
39170
15min, 13,952 ms (13.046×)
39170
hermann@7950x:~$
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Bill's original work partitioning corresponds to l=1:
better factor 30.404× of overall time divided by walltime for l=2:
l=4 slightly better 31.190×:
even better 31.422× for l=8:
Factor of parallel walltime by sequential time gets better, from 12.315× up to 13.076×.
Amdahl's law is limiting speedup.