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
19532821287077577316320149475962563324435429965918733969534051945716\ ... 54385552378378141386675079286837205802043347225419033684684301719893\ 411568996526838242546875 (over two hundred thousand digits, too long to display fully!) real 0m6.614s user 0m5.888s sys 0m0.024s
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
Any thoughts on this command? Does it work on your machine? Can you do the same thing with only 14 characters?
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