You could have that little benchmark run on all cores in parallel, as a multi-core benchmark or stress test First find the number of cores, then have parallel iterate over that in, well, parallel Show Sample Output
Broken in two parts, first get the number of cores with cat /proc/cpuinfo |grep proc|wc -l and create a integer sequence with that number (xargs seq), then have GNU parallel loop that many times over the given command. Cheers! Show Sample Output
EDIT: Trolling crap removed ;)
takes approx 6 secs on a Core 2 Duo @ 2GHz, and 15 secs on atom based netbooks!
uses monoid (a,b).(x,y)=(ax+bx+ay,ax+by) with identity (0,1), and recursion relations:
F(2n-1)=Fn*Fn+F(n-1)*F(n-1)
F(2n)=(Fn+2*F(n-1))*Fn
then apply fast exponentiation to (1,0)^n = (Fn,F(n-1))
.
Note that: (1,0)^-1=(1,-1) so (a,b).(1,0) = (a+b,a) and (a,b)/(1,0)=(a,b).(1,0)^-1=(b,a-b)
So we can also use a NAF representation to do the exponentiation,http://en.wikipedia.org/wiki/Non-adjacent_form , it's also very fast (about the same, depends on n):
time echo 'n=1000000;m=(n+1)/2;a=0;b=1;i=0;while(m>0){z=0;if(m%2)z=2-(m%4);m=(m-z)/2;e[i++]=z};while(i--){c=a*a;a=c+2*a*b;b=c+b*b;if(e[i]>0){t=a;a+=b;b=t};if(e[i]<0){t=a;a=b;b=t-b}};if(n%2)a*a+b*b;if(!n%2)a*(a+2*b)' | bc
Show Sample Output
Calculates nth Fibonacci number for all n>=0, (much faster than matrix power algorithm from http://everything2.com/title/Compute+Fibonacci+numbers+FAST%2521 )
n=70332 is the biggest value at http://bigprimes.net/archive/fibonacci/ (corresponds to n=70331 there), this calculates it in less than a second, even on a netbook.
UPDATE: Now even faster! Uses recurrence relation for F(2n), see http://en.wikipedia.org/wiki/Fibonacci_number#Matrix_form
n is now adjusted to match Fn at wikipedia, so bigprimes.net table is offset by 1.
UPDATE2: Probably fastest possible now ;), uses a simple monoid operation:
uses monoid (a,b).(x,y)=(ax+bx+ay,ax+by) with identity (0,1), and recursion relations:
F(2n-1)=Fn*Fn+F(n-1)*F(n-1)
F(2n)=Fn*(2*F(n-1)+Fn)
then apply fast exponentiation to (1,0)^n = (Fn,F(n-1))
.
Note that: (1,0)^-1=(1,-1) so (a,b).(1,0) = (a+b,a) and (a,b)/(1,0)=(a,b).(1,0)^-1=(b,a-b)
So we can also use a NAF representation to do the exponentiation,http://en.wikipedia.org/wiki/Non-adjacent_form , it's also very fast (about the same, depends on n):
time echo 'n=70332;m=(n+1)/2;a=0;b=1;i=0;while(m>0){z=0;if(m%2)z=2-(m%4);m=(m-z)/2;e[i++]=z};while(i--){c=a*a;a=c+2*a*b;b=c+b*b;if(e[i]>0){t=a;a+=b;b=t};if(e[i]<0){t=a;a=b;b=t-b}};if(n%2)a*a+b*b;if(!n%2)a*(a+2*b)' | bc
Show Sample Output
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