The Nth Fibonacci Number in O(log N) | Hacker News Reader

The Nth Fibonacci Number in O(log N)

kukuruku.co

51 pointsskazka1645 comments

It's easy to compute things quickly if you pretend that you can do arbitary-precision arithmetic in constant time. Want a quadrillion digits of Pi? Easy, just use the Brent–Salamin AGM algorithm and you'll get them in about 500 arithmetic operations. NP-complete problems? No problem at all, rephrase them as subset-sum problems and use the polynomial-integer-operations dynamic programming algorithm. I don't think you can solve PSPACE in a polynomial number of integer operations, but I can't see how to prove it immediately.

In "real" terms, computing the Nth fibonacci number takes O(M(N)) = O(N log N 2^(O(log* N))) operations.

Minor pedantry: the best proved bound for M(N) is O(N log N 8^(log* N)). You can easily pretend that log* N is constant, though...

Oops, forgot the O() inside the exponential. Fixed, thanks.

Out of curiosity, what line of work are you in given that you can seemingly pull this stuff from memory? All I read was "lorem ipsum..." until you got to Big O.

This is actually just standard theory. They teach this in every undergrad algorithms class. A lot of it is just getting good intuition for these sorts of problems.

For example, the nth Fibonacci number has n log_2 \phi bits, which means that to simply list the digits in the nth number, it takes O(n) time.

So from there it's pretty easy to see that actually this algorithm can't "really" operate in O(log n), or at least, something must be up that causes us to produce that analysis.

Another bit of intuition that might help you see where this bound comes from: what's actually true is that it takes O(log n) _matrix multiplications_ to get the result. So if you see that and you see that listing the digits of the nth number of the sequence is O(n), it starts to point at the fact that there's a hidden cost to the multiplications.

It's Colin Percival: author of tarsnap, former BSD security officer and considered something of an authority on crytography. I presume his maths skills are therefore suitably well polished to be able to pull this stuff from memory.

I want to dispel the rumor that this is magic. This is standard material in undergrad algorithms classes. See, for example, Jeff Erickson's algorithms notes here[1]. It's literally the first page of the chapter on dynamic programming.

[1] http://web.engr.illinois.edu/~jeffe/teaching/algorithms/note...

I appreciate that it's not magic and agree that it should be demystified.

Having said that, doing it once in the past in an undergrad class and remembering it 10 years later when a relevant article on some website is posted are two different things.

Yeah, but "kids these days" only want to know Java or Angular.js and bawl at the first sight of math.

Psh, who wants to learn Java?

I think "know" over "learn" used by the grandparent is important there; probably they were made to learn it for coursework, they want to be allowed to continue only having to know that much and not learn further.

More relevant than the above, my doctoral thesis was all about algorithms, and as an undergraduate student I published research in large integer arithmetic.

He's in cryptography, he also started university at age 13.

https://news.ycombinator.com/item?id=35076

Another similar "algorithm": sleep sort.

Basically, take a list of integers, and for each element n create a thread that sleeps n seconds, after which it appends n to the result list.

Assuming the original list is dense (there are no gaps between integers), we can sort it in O(n)!

Even if the computation could be done by magic, simply writing the result to memory would take O(n) operations, since the nth Fibonacci number has O(n) bits.

An issue here is that the the nth Fibonacci number has O(N) bits, so once you move into arbitrary arithmetic calculations your +'s and *'s will become O(N) and O(N logN) respectively.

Factoring into powers of 2 seems to me like an unnecessary complication. It's possible to calculate an arbitrary power in O(log N) time without memoization.

  def __get_matrix_power(self, M, p):
    if p == 1:
      return M
    if p % 2 == 1: # odd power
      return self.__multiply_matrices(M, self.__get_matrix_power(M, p - 1))
    else: # even power
      K = self.__get_matrix_power(M, int(p/2))
      return self.__multiply_matrices(K, K)

Also working on pairs of consecutive fibonacci numbers (f_n, f_(n+1)) instead of the matrix [[f_(n+1) f_n] [f_n f_(n-1)]] makes this much simpler.

  def fib(n):
    def fib2(n):
      # returns (f_n, f_(n+1))
      if n == 0:
        return 0, 1
      if n % 2:
        a, b = fib2(n-1)
        return b, a+b
      a, b = fib2(n//2)
      return b*a + a*(b-a), a*a + b*b
    return fib2(n)[0]

As mentioned, the actual number has O(N) bits and thus can't actually be computed in O(log N) time. So if you just want an approximation, you may as well use Binet's formula. [1]

However, one thing this algorithm can do that Binet's formula can't is compute the N-th Fibonacci number modulo some constant M in O(log N) time. So you could very efficiently compute, say, the last 12 digits of F_123456789. A lot of programming contests (TopCoder, Google Code Jam, etc.) will ask for the answer modulo some constant to take advantage of this trick.

[1] http://en.wikipedia.org/wiki/Fibonacci_number#Closed-form_ex...

A really neat observation here is that Binet's formula and the matrix exponentiation trick are arguably the same thing.

How so? You can think of the operations of Binet's formula as happening in the field Q(sqrt{5}). Since sqrt{5} is a square root, every element of this field has a unique representation of the form a + b sqrt{5} with a and b rational, and so the field can be said to have dimension 2 over the rationals. When you squint at how the field operations work in this representation, you'll find that you end up doing something that looks very similar to the 2x2 matrix multiplications trick; with a bit of rearrangement, it's the same thing - except that the matrix multiplication trick is usually derived via dynamic programming, and Binet's formula is usually derived via power series.

The fact that things often fit together so nicely is perhaps the most beautiful aspect of mathematics.

I remember there is an equation that could get the Nth fibonacci number directly. (Maybe in the exercises of Concrete Mathematics by Knuth).

It requires arbitrary precision arithmetic. It is known as Binet's formula.

It's also in SICP, but depend on what you consider the basic step is, it's also O(logN). As someone else mentions, it requires arbitrary precision calculation, and exponential calculation is O(logN).

If you diagonalize the matrix in the problem, the formula pops out pretty quickly.

Nice catch, is that from Knuth's book?

Diagonalization is a fairly standard mathematical trick for doing matrix powers easily. If X is some matrix that can be diagonalised as

  X = P^(-1) D P

then

  X^n = (P^(-1) D P)^n = P^(-1) D^n P

https://en.wikipedia.org/wiki/Fibonacci_number#Closed-form_e...

If ever there was a way to impress an interviewer it is to bust out this baby when asked to make a fibonacci function.

Personally, I think the matrix powers method is more impressive. You're using the matrix as a state engine to get to whatever state you intend to reach. That has more applications than just the closed form, which a lot of people know about, and you most likely just happened to memorize.

Don't get me wrong, it'd still be impressive "Woah, he memorized this!" but seeing someone do the matrix trick would be even more impressive, to me.

fair enough.

The closed form is just the diagonalized version of the matrix power, though.

I honestly didn't know that until it was mentioned in another part of this thread. My point was that someone can memorize the formula. Chances are, they'll see the matrix form and actually know how to use that in other situations.

Now, if you derived the formula on the board from the matrix, that'd be incredibly impressive. The idea here is how the person's brain thinks. When I see them write an equation on the board, I think "Oh, they memorized it" when I see them step through the process of creating a matrix and doing matrix multiplication, I think "Wow, they /really/ know their stuff". If they just threw up the matrix and couldn't explain it, I wouldn't be as impressed. Same as if they wrote the equation and couldn't explain it.

At the end of the day, if they could explain either approach, I'd be impressed, I think the matrix form just lends itself to more likely be in a way that the person can explain it

It would be impressive if you could derive or prove it, not if you'd just memorized it.

not even a little bit impressed?

The closed form solution is not that difficult to derive at all if you put the problem as a reoccurrence equation with boundary conditions.

or just calculate the eigenvalues & eigenvectors of Q.

I wrote about this last December:

http://blog.richardkiss.com/?p=398

The matrix math is easier, and the Python code is about a dozen lines.

This brings back some memories. I remember talking about the different ways of calculating the Fibonacci numbers in an old comp.lang.java usenet thread about recursion, memoization and dynamic programming from 2007,

http://coding.derkeiler.com/Archive/Java/comp.lang.java.prog...

The thread discusses the exponential, linear and logarithmic algorithms.

In Python:

    def fib(n):
        import math
        r5 = math.sqrt(5)
        x = pow((1 + r5) / 2, n)
        x -= pow((1 - r5) / 2, n)
        return x / r5

  $ python fib.py 4
  3.0

  $ python fib.py 5
  5.0

  $ python fib.py 50
  12586269025.0

  $ python fib.py 5000
  OverflowError: (34, 'Result too large')

So we add arbitrary precision. Uglier, but supports any n:

    def fib(n):
        from math import sqrt
        from decimal import Decimal
        r5 = Decimal(sqrt(5))
        x = pow((1 + r5) / 2, n)
        x -= pow((1 - r5) / 2, n)
        return x / r5

  $ python fib.py 4
  3.000000000000000242931577139

  $ python fib.py 5
  5.000000000000000607328942851

  $ python fib.py 50000
  1.077773489309106593392984005E+10449

  $ python fib.py 50309230
  8.147201098193506535089177522E+10514006

Or does it?

  $ python fib.py 50309230390
  decimal.Overflow: above Emax

This was fun. Computing fib(50309230390) is left as an exercise for the reader.

Wolfram alpha verifies this is correct for Fib 50,000: http://www.wolframalpha.com/input/?i=Fib+50000

This shows nothing but one's ability to code-up a math formula. The blog post goes into much detail about how to solve the problem using matrices and memoization.

Edit: Looks like sillysaurus3 edited his comment.

I wasn't trying to one-up anyone. I thought it was interesting, so I figured others might find it interesting too. In particular, it was surprising to find that there's a closed-form solution for the Fibonacci sequence. Sorry.

Don't be sorry! I found your comment very interesting and was also surprised to find that there exists a closed-form solution.

The whole book is pretty fascinating, "Mathematics for Computer Science" by Lehman and Leighton: http://www.cs.princeton.edu/courses/archive/spring10/cos433/...

The formula came from section 17.3.2, which goes into detail about how to derive it.

Thanks. I like the way they introduce what a proof is in chapter 1, I feel like they have a good, subtle sense of humor. I don't have a mathematical background so it might be a tough read for me, but it seems like a great resource.

Was the purpose of this post to help people prepare for interviews? Does code always have to show off one's ability? This code inspired me to learn about Binet's formula and for that reason I think it was valuable.

returning int(round(x)) could act as a bandaid for the slight error in the results.

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