Fastest way to calculate Fibonacci number

I was studying linear algebra by Gilbert Strang and i found this formula for computing Fibonacci number fibo

here is my code:

i am taking the floor value of final answer
this program is running till input=10^3 after that it is overflowed and also i am getting slight deviation in value from correct value after n>80 , may be it is because of arithmetic operation
this solution is much faster than dynamic programming one ,so i am sharing in a hope to learn some way of improving this solution from you guys and please let me know if their is a better solution

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That formula is never used in competitive programming as it gives incorrect values for bigger ‘n’ like say, (n>100).

This is the fastest way to calculate the nth fibonacci number : -

Time Complexity:- O(logN) .


Karan Bhaai…? Tumhe pata hai kii ye do formulas ( Fast doubling waale ) kaha se prakat hue…?

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I think using matrix exponentiation to to calculate N'th fibonacci number in O(LogN) is pretty nice and applicable to most cases.


In fact this technique can be used to find T(n) = aT(n-1)+bT(n-2)+cT(n-3)+…+kT(0) for generic recurrence.

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kisi ko pata ho ye kaha se aaya yaa yaad karne kaa koi tareeka pata ho to bata do bhaai log… ratta nhii lagaaya jaaega… :tired_face:

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Lets start with the understanding that Fn = Fn-1 + Fn-2
Consider this beautiful matrix

|1 1|
|1 0|

Lets multiply it with a matrix and see what we get. Here x, y, z are three consecutive fibonacci numbers such that z > y > z.

|z y| |1 1|
|y x| |1 0|

The resulting matrix is
|z+y z+0|
| x+y y+0|

Here z+y is the newest addition to the series! Note that, x+y = z. We have basically created a new matrix but with the same quality of three consecutive fibonaccis.

Considering first three fibs are 0,1,1 - we just need to multiply following matrix as many times as needed with itself to get our result:


Square it and we get 4th fib. Cube it and we get 5th fib.

Now suppose you want me to find out the fib number 36. How much effort will I take?

Matrix square will give me 4th fib.
The above matrix square gives power of 4 -> 6th fib.
The above matrix square will give power of 8 -> 10th fib.
Square of above will give power of 16 -> 18th fib
Square of above will give power of 32 -> 34th fib.
Multiply above with what we got in step number 2 i.e. power of 34 -> 36th fib.

This happens in just 6 steps!
The compexity is visibly O(logn)

Hope it helps!


No…I m still confused plz explain little bit more or any other good video link.

I know the answer but I am busy with something…

Is this the way of talking in a discuss forum?

Not taking any sides, but I am very displeased by how such random comments cause a lot of spam in some threads. Please maintain professionalism.


So sorry @vijju123 ,:pensive:

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it is the simplest way to calculate the fibonacci no
print(a,b,end=" “)
for i in range(50):
print(c,end=” ")

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Hello… I appreciate your effort my friend ; but I think we both are not on the same page… or maybe I was not able to understand something in your explanation…

Actually I understand every single detail about the matrix exponentiation trick… from where it comes ; to its correctness and time complexity of O(log N).

What I don’t know is the method which @anon55659401 mentioned… ( The fast doubling one… ) That from where does those two formulas come from…

I have seen that formula earlier also ; but I forget it each time bcos I don’t know form where it comes ( That’s why they are hard to remember for me ) …

And in my previous comment also ; I replied him ( @anon55659401 ) for that formula only… If you know how do they come ; please share…Thanks…

Ok bro… no problem :+1: ; but whenever you get time ; please re-visit this ; I am eager to know from where they come… Thanks…!

If you know how do they come ; please share…Thanks…

Hi! Here I am using the same matrix exponentiation to derive the fast doubling formula. The explanation is fairly straightforward. Let me know if something needs further explanation!

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Woooo…! Pleased to see how much efforts forum members put in to help others… Was opening the discuss page for past few days just for something like this… It really helped ; Thanks…! :slight_smile:

1 Like It all comes from here @ritesh1340

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long int n;
double d;

This will work

No it does not work fail for large values @anon55659401 look into this

this is very nice :slight_smile:

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