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The 1 millionth fibonacci number

Terminal - The 1 millionth fibonacci number
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
2009-09-10 09:00:44
User: Escher
Functions: echo time
-135
The 1 millionth fibonacci number

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

Alternatives

There is 1 alternative - vote for the best!

Terminal - Alternatives
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
2009-09-10 08:58:47
User: Escher
Functions: echo time
-136

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

Know a better way?

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What others think

My C version is faster. Stop posting duplicates.

Comment by hank 276 weeks and 3 days ago

Stop it Escher. You're just spamming now. Quit posting this and more variants. We get it already!

Comment by linuxrawkstar 276 weeks and 3 days ago

Your C version is slower,in fact ~47secs vs my ~36secson a Core 2 Duo. I have put a message for people like you in capitals in the description section. No go away little buzzing insect.

Comment by Escher 276 weeks and 3 days ago

Don't be too hard on this MrMerry/Escher degenerate... Desperate attempts to get some attention on the internet is his only way of communicating with the world outside his mama's basement...

Comment by sklm 276 weeks and 3 days ago

Please only post comments if you have required skillz to compete against me. I know it's depressing for you to witness true greatness and realise how insignificant you are, but that's life sklm, you always were and always will be an irrelevant ant.

Comment by Escher 276 weeks and 3 days ago

"Please only post comments if you have required skillz to compete against me. I know it's depressing for you to witness true greatness and realise how insignificant you are, but that's life sklm, you always were and always will be an irrelevant ant."

LMAO!!!

Comment by sklm 276 weeks and 3 days ago

Well now. Someone has a voting script...

Comment by SuperFly 276 weeks and 3 days ago

can this be made faster by using multiple CPUs ?

Comment by alperyilmaz 276 weeks and 2 days ago

Not really, though it can be made faster by using the recurrence relations from the wikipedia article more efficiently. Check for an update soon :)

For the record, mathematica uses an algorithm which is about 4 times faster.

Comment by Escher 276 weeks and 2 days ago

I don't think this was a voting script; I'd like to think this is a reward for arrogance.

Escher: a truly great person is humble. What bothers you is that the reverse is not necessarily true. And as long as that bothers you, you will be what you are, and will not grow further.

Comment by sitaram 276 weeks and 1 day ago

Don't be so arrogant, dude !

It's just a shell command... I don't care for these numbers in any shell script... useless for me, sorry.

Comment by alvinx 275 weeks and 6 days ago

Your point of view

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