Comments on: Fibonacci and filter

By: Loren

Loren — Fri, 22 Feb 2008 11:31:38 +0000

POONKODI-

It just calls it twice. The first call completes (recursion and all) before the second one gets called. This version of recursion is quite wasteful since it doesn’t store and therefore use values that have been previously calculated.

–Loren

By: POONKODI.S

POONKODI.S — Fri, 22 Feb 2008 03:52:16 +0000

hi Loren,
I just wanted to kmow how matlab calls the recursive function when it is repeated twice as in fibonacci series???

By: joe bloggs

joe bloggs — Fri, 08 Feb 2008 14:49:23 +0000

thankyou for the help, it allowed me to cheat!

By: erik

erik — Wed, 15 Nov 2006 20:29:25 +0000

hey loren-
i am interested to hear what you or steve think about the idea of matlab supporting gpgpu (using the stream processing hardware of modern video cards to perform vector/matrix operations up to 50 times faster than on the cpu) from matlab. obvious example related to your article: filtering/fft. wired has a great story on recent exciting results in this field:
http://www.wired.com/news/technology/computers/0,72090-0.html?tw=wn_index_16

(i’m not sending to steve because most of the image processing blog is over my head, so i don’t keep up with it — but it seems like that community would also be very interested.)
-erik

By: Eric Gov

Eric Gov — Mon, 29 May 2006 22:33:52 +0000

Correction to the previous post.

In fact, the 2nd time I run the code I get

Elapsed time is 0.000000 seconds

I believe the 1st time I ended up with 16ms because of the Hard Disk (or memory) read, while the 2nd time the program was running from cache.

Eric

By: Eric Gov

Eric Gov — Mon, 29 May 2006 22:17:39 +0000

What’s wrong with using the closed form expression? Even if you want to calculate the whole series of numbers it will be much faster. For example,

tic;
a=(1+sqrt(5))/2;
nf=102;
b=1-a;
F=(a.^[1:nf]-b.^[1:nf])/sqrt(5);
toc;

gives me report

Elapsed time is 0.016000 seconds

Eric

By: Loren

Loren — Fri, 26 May 2006 13:13:26 +0000

My prealloc results might have been higher because of the way I ran the code. Did you run the preallocation in a script, function, or directly from the command line? That can make a huge difference sometimes, especially command-line vs. file.

By: ravi

ravi — Wed, 24 May 2006 09:42:47 +0000

On my PC, I get the following timings for nf=102 :
prealloc :0.015, filter:0.053ms.
For nf=10000, prealloc:3.12,filter:4.61 ms
That is, the prealloc is faster. I have checked this several times. Can this depend on memory, PC internals and so on?
I found another method described in Roman Maeder’s Mathematica book.
Let m=[1 1;1 0];
The matrix, m^(n-1) gives [f(n) f(n-1);f(n-1) f(n-2)]
where f(n) is fibonacci number for n, f(n-1) for (n-1) and so on.
fib=m^(n-1); fib(1,1) gives the fibonacci number for n. As expected,this computation is fast.
It would be interesting to see how the code can be vectorised with the above method to get the entire series upto n. Any volunteers?

By: Richard

Richard — Tue, 23 May 2006 20:07:34 +0000

If you don’t want all n numbers, the following can give you a single one: Note, this gives f0 = 1, f = 1, f2 = 2, etc…
Also, at around n = 70 (near bitmax) this starts deviating from the results returned by addition based implementations

function fn = fibonnacci(n)
alpha1 = (1 + sqrt(5)) / 2;
alpha2 = (1 – sqrt(5)) / 2;

c1 = (1 – alpha2) / (alpha1 – alpha2);
c2 = (alpha1 – 1) / (alpha1 – alpha2);

fn = round(c1 * alpha1^n + c2 * alpha2^n);

% in fact the alpha2 term is really small, so you don’t
% need it, except for n = 0;
fn = round(c1 * alpha1^n);

By: Loren

Loren — Thu, 18 May 2006 09:13:55 +0000

Michael,

You are correct and I have fixed the text in the blog to reflect that. Thanks for the correction!

–Loren