Fast Fibonacci Algorithms (2015) | Hacker News Reader

Fast Fibonacci Algorithms (2015)

nayuki.io

102 pointsold_sound28 comments

It's interesting to note that if you add a modulus to the calculation (i.e. if you calculate F(n) mod M), you can calculate ridiculously large values using the matrix/doubling method. You can, for instance, calculate the last 10 digits of the Googolth Fibonacci number instantly (since you only need O(log n) multiplications, it's just a few hundred for 10^100 ). Doing it the "naive" dynamic programming way, it would take you far longer than the age of the universe.

It's a neat illustration of asymptotic running times.

Correct! Calculating Fibonacci modulo a number is used in quite a few Project Euler problems.

The fast exponentiation algorithm actually generalizes:

http://kukuruku.co/hub/algorithms/automatic-algorithms-optim...

That is amazing. (Go look! A Python decorator that edits the bytecode...)

There is a log time way to do this, use Binet's formula (https://en.m.wikipedia.org/wiki/Fibonacci_number#Closed-form...), but do the arithmetic in the field Q[√5].

Edit: Someone's implementation: http://ideone.com/NWQe38

Edit: constant time -> log time

That's log(n) time due to the exponentiation.

Oops, yes log(n), I'll correct the comment, thanks.

It's log(n) multiplications, each of which requires something like O(n log n log log n) time [1]; so the actual complexity is O(n (log n)^2 log log n) ish.

[1] Schönhage-Strassen multiplication, one of many bignum multiplication algorithms with sub-quadratic complexity. Other algorithms in common use have worse asymptotic performance.

Don't know why you were downvoted, you're absolutely correct. At first glance your n and my n are different, with mine (n1) being the input to the function and yours representing the number of digits in a multiplicand (n2). n1 and n2 are linearly related, though, since the exponentiation process increases the number of digits in a linear fashion.

It's also worth noting that since the Fibonacci sequence grows exponentially, the number of digits is the result will be linear in n. Therefore, the memory requirement is actually O(n), and writing the result to the screen will be O(n).

> At first glance your n and my n are different

Right; as you sketch out, there's some constants that disappear into the O() notation.

for really big numbers there are the generalisations of karatsuba and toom-cook for higher numbers of splits, and beyond that there are the FFT multiplication methods which i believe are the fastest general case methods for very large numbers.

also the laddering scheme used for the 'exponentiation' can be in different forms, the left-to-right or right-to-left form or even a Montgomery ladder...

i seemed to recall there is a 'fastest known' method for generating fibonacci or lucas numbers... but google is not helping me.

pretty sure i had seen it in this book: https://www.amazon.co.uk/Prime-Numbers-Computational-Carl-Po... i'll have to check when i have access to it next. :)

The BigInteger implementation in java uses better multiplication algorithms in Java 1.8 then in the version of 1.6 tested [1], switching to toom-cook[2] for even larger numbers. So that in practice Fibonacci in java becomes an benchmark of allocation rate.

[1]: http://grepcode.com/file/repository.grepcode.com/java/root/j...

[2]:https://en.wikipedia.org/wiki/Toom%E2%80%93Cook_multiplicati...

For anyone interested in Fibonacci numbers: Fibonacci Quarterly has been published for more than fifty years.

http://www.fq.math.ca/

Isn't there a constant time implementation using the Golden Mean? What is the advantage of these algorithms over the constant time one? (i.e. Binet's formula)

There is no constant time implementation. Binet's formula uses exponentiation, which gives us log(n) time.

O(1)? Having all the answers pre-cached would be the only way to do that.

Unless there is some kind of quantum computer algorithm that fits this problem well.

The nth Fibonacci number has O(n) digits, so even reading a cached pre-computed result from memory is O(n).

The "compute the eigenvalues of Matrix([[1,1],[1,0]]) raised to the power N" approach to computing the Nth Fibonacci number is only constant time if you assume exponentiation is constant time, which it is not on any computer or realistic model of computation.

I still think this is a really cool technique though, even if it only works in "math land." http://bit.ly/fib_using_eigenvals

Constant time fibonacci is derived as an example for linear basis change in the superb book "Vector Calculus, Linear Algebra, and Differential Forms: A Unified Approach".

It's all contained in the sample pages http://matrixeditions.com/VC5.Chap2.219-221.pdf

However, even though the formula is O(1), the runtime of the algorithm might be larger.

Simply because a standard 64bit integer might be too small. And the BigNum implementation needs more time to multiply all the bits.

I really like the style of that book! I want to get it, but I'm hesitating at the $87 pricetag. Looks very nice, though....

No, but I go into it a bit in https://lee-phillips.org/lispmath/

So, seems like matrix exponentiation is faster than computing eigenvalues / eigenvectors in practice? Take a look: http://mathoverflow.net/questions/62904/complexity-of-eigenv...

Product of Lucas Numbers Algorithm - "A fast algorithm for computing large Fibonacci numbers" -https://pdfs.semanticscholar.org/9652/9e1fedb0a372b825215c7b...

Also wrote a gem that implements this algorithm here https://github.com/chaitanyav/fibonacci

That's nice. Here's a (recursive) Python implementation:

http://ideone.com/z1sUp4

fast doubling for fibonacci is mentioned in sicp. Cheers for sicp

Previous discussion: https://news.ycombinator.com/item?id=9315346

the continued fraction of (1+√5)/2 is [1;1,1,1,1,1,1,...] it goes on forever.

there are more advanced formulas in this article of Curtis McMullen http://www.math.harvard.edu/~ctm/papers/home/text/papers/cf/...

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