Create faster Fibonacci function for n > 100 in MATLAB / octave

Question

I have a function that tells me the nth number in a Fibonacci sequence. The problem is it becomes very slow when trying to find larger numbers in the Fibonacci sequence doe

Answer 1

If you have access to the Symbolic Math Toolbox in MATLAB, you could always just call the Fibonacci function from MuPAD:

>> fib = @(n) evalin(symengine, ['numlib::fibonacci(' num2str(n) ')'])
>> fib(274)
ans =
818706854228831001753880637535093596811413714795418360007

It is pretty fast:

>> timeit(@() fib(274))
ans =
    0.0011

---

Plus you can you go for as large numbers as you want (limited only by how much RAM you have!), it is still blazing fast:

% see if you can beat that!
>> tic
>> x = fib(100000);
>> toc               % Elapsed time is 0.004621 seconds.

% result has more than 20 thousand digits!
>> length(char(x))   % 20899

Here is the full value of fib(100000): http://pastebin.com/f6KPGKBg

Answer 2

If time is important (not programming techniques):

function f = fib(n)
if (n == 1)
   f = 1;
elseif (n == 2)
   f = 2;
else
   fOld = 2;
   fOlder = 1;
   for i = 3 : n
     f = fOld + fOlder;
     fOlder = fOld;
     fOld = f;
   end
end
end

tic;fib(40);toc; ans = 165580141; Elapsed time is 0.000086 seconds.

You could even use uint64. n = 92 is the most you can get from uint64:

tic;fib(92);toc; ans = 12200160415121876738; Elapsed time is 0.001409 seconds.

Because,

fib(93) = 19740274219868223167 > intmax('uint64') = 18446744073709551615

Edit

In order to get fib(n) up to n = 183, It is possible to use two uint64 as one number,

with a special function for summation,

function [] = fib(n)
fL = uint64(0);
fH = uint64(0);
MaxNum = uint64(1e19);
if (n == 1)
   fL = 1;
elseif (n == 2)
   fL = 2;
else   
   fOldH = uint64(0);
   fOlderH = uint64(0);
   fOldL = uint64(2);
   fOlderL = uint64(1);
   for i = 3 : n
      [fL q] = LongSum (fOldL , fOlderL , MaxNum);
      fH = fOldH + fOlderH + q;
      fOlderL = fOldL;
      fOlderH = fOldH;
      fOldL = fL;
      fOldH = fH;
   end
 end
 sprintf('%u',fH,fL)
 end

LongSum is:

function [s q] = LongSum (a, b, MaxNum)
if a + b >= MaxNum
   q = 1;
   if a >= MaxNum
      s = a - MaxNum;
      s = s + b;
   elseif b >= MaxNum
      s = b - MaxNum;
      s = s + a;
   else
      s = MaxNum - a;
      s = b - s;
   end
else
   q = 0;
   s = a + b;
end

Note some complications in LongSum might seem unnecessary, but they are not!

(All the deal with inner if is that I wanted to avoid s = a + b - MaxNum in one command, because it might overflow and store an irrelevant number in s)

Results

tic;fib(159);toc; Elapsed time is 0.009631 seconds.

ans = 1226132595394188293000174702095995

tic;fib(183);toc; Elapsed time is 0.009735 seconds.

fib(183) = 127127879743834334146972278486287885163

However, you have to be careful about sprintf.

I also did it with three uint64, and I could get up to,

tic;fib(274);toc; Elapsed time is 0.032249 seconds.

ans = 1324695516964754142521850507284930515811378128425638237225

(It's pretty much the same code, but I could share it if you are interested).

Note that we have fib(1) = 1 , fib(2) = 2according to question, while it is more common with fib(1) = 1 , fib(2) = 1, first 300 fibs are listed here (thanks to @Rick T).

Answer 3

You can do it in O(log n) time with matrix exponentiation:

X = [0 1
     1 1]

X^n will give you the nth fibonacci number in the lower right-hand corner; X^n can be represented as the product of several matrices X^(2^i), so for example X^11 would be X^1 * X^2 * X^8, i <= log_2(n). And X^8 = (X^4)^2, etc, so at most 2*log(n) matrix multiplications.

Answer 4

Seems like fibonaacci series follows the golden ratio, as talked about in some detail here.

This was used in this MATLAB File-exchange code and I am writing here, just the esssence of it -

sqrt5 = sqrt(5);
alpha = (1 + sqrt5)/2;   %// alpha = 1.618... is the golden ratio
fibs  = round( alpha.^n ./ sqrt5 )

You can feed an integer into n for the nth number in Fibonacci Series or feed an array 1:n to have the whole series.

Please note that this method holds good till n = 69 only.

Answer 5

To reach large numbers you can use symbolic computation. The following works in Matlab R2010b.

syms x y %// declare variables
z = x + y;  %// define formula
xval = '0'; %// initiallize x, y values
yval = '1'; 
for n = 2:300
    zval = subs(z, [x y], {xval yval}); %// update z value
    disp(['Iteration ' num2str(n) ':'])
    disp(zval)
    xval = yval; %// shift values
    yval = zval;
end

Answer 6

The implementation of a fast Fibonacci computation in Python could be as follows. I know this is Python not MATLAB/Octave, however it might be helpful.

Basically, rather than calling the same Fibonacci function over and over again with O(2ⁿ), we are storing Fibonacci sequence on a list/array with O(n):

#!/usr/bin/env python3.5

class Fib:
    def __init__(self,n):
        self.n=n
        self.fibList=[None]*(self.n+1)
        self.populateFibList()
    def populateFibList(self):
        for i in range(len(self.fibList)):
            if i==0:
                self.fibList[i]=0
            if i==1:
                self.fibList[i]=1
            if i>1:
                self.fibList[i]=self.fibList[i-1]+self.fibList[i-2]
    def getFib(self):
        print('Fibonacci sequence up to ', self.n, ' is:')
        for i in range(len(self.fibList)):
            print(i, ' : ', self.fibList[i])
        return self.fibList[self.n]

def isNonnegativeInt(value):
    try:
        if int(value)>=0:#throws an exception if non-convertible to int: returns False
            return True
        else:
            return False
    except:
        return False

n=input('Please enter a non-negative integer: ')

while isNonnegativeInt(n)==False:
    n=input('A non-negative integer is needed: ')

n=int(n) # convert string to int

print('We are using ', n, 'based on what you entered')

print('Fibonacci result is ', Fib(n).getFib())

Output for n=12 would be like:

I tested the runtime for n=100, 300, 1000 and the code is really fast, I don't even have to wait for the output.