Commands using time (30)

  • 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


    -1
    time cat /proc/cpuinfo |grep proc|wc -l|xargs seq|parallel -N 0 echo "2^2^20" '|' bc
    kostis · 2018-12-06 05:36:55 0
  • 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


    -2
    time cat /proc/cpuinfo |grep proc|wc -l|xargs seq|parallel -N 0 echo "scale=4000\; a\(1\)\*4" '|' bc -l
    kostis · 2018-12-06 05:15:24 2

  • -9
    Convert UNIX time to human readable date
    bytor4232 · 2009-03-30 05:03:31 4
  • 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


    -135
    time echo 'n=1000000;m=(n+1)/2;a=0;b=1;i=0;while(m){e[i++]=m%2;m/=2};while(i--){c=a*a;a=c+2*a*b;b=c+b*b;if(e[i]){t=a;a+=b;b=t}};if(n%2)a*a+b*b;if(!n%2)a*(a+2*b)' | bc
    Escher · 2009-09-10 09:00:44 11
  • 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


    -136
    time echo 'n=70332;m=(n+1)/2;a=0;b=1;i=0;while(m){e[i++]=m%2;m/=2};while(i--){c=a*a;a=c+2*a*b;b=c+b*b;if(e[i]){t=a;a+=b;b=t}};if(n%2)a*a+b*b;if(!n%2)a*(a+2*b)' | bc
    Escher · 2009-09-10 08:58:47 5
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