Fibonacci series and Dynamic programming

http://functionspace.org/articles/32/Fibonacci-series-and-Dynamic-programming

by aditgupta • 13 years ago

38 28 13 years ago

blt3• 13 years ago

The fastest way to calculate the Fibonacci sequence (with integer math) is a simple loop, just like you'd do it on paper. This version is slower and much more complicated. With so many relevant good examples of recursion and dynamic programming, it boggles my mind to see the Fibonacci sequence used as an example for either.

    def fib(n):
        a, b = 0, 1
        for k in range(n):
            a, b = b, a + b
        return a

cousin_it• 13 years ago

Your method computes fib(n) in O(n) arithmetic operations. There's a way to do it in O(log n) arithmetic operations on integers of the same size. First note that you can obtain the two-element vector (fib(n),fib(n+1)) by applying the matrix A=((0,1),(1,1)) to the vector (fib(n-1),fib(n)). So to get from (fib(0),fib(1)) to (fib(n),fib(n+1)), you need to raise the matrix A to the nth power. That can be done in O(log n) arithmetic operations by using repeated squaring: http://en.wikipedia.org/wiki/Exponentiation_by_squaring.

nicolasp• 13 years ago

The Fibonnacci sequence is (IMO) one of the easiest to understand examples of dynamic programming, so it makes for a good introduction to the concept. Furthermore, the dynamic programming implementation is better if you're going to call it multiple times (depending on the context of course -- if you want the first n number the linear version can be adapted to be even better).

blt3• 13 years ago

The looping version could keep a static array around to eliminate that difference. I think Fibonacci is a bad example because the student will wonder why they are going to the trouble of dynamic programming when it doesn't save any work. A better example would be the recursive computation of N choose K - because it avoids overflow from computing the factorial-based formula directly - or matrix chain multiplication (http://en.wikipedia.org/wiki/Dynamic_programming#Matrix_chai...).

d0mine• 13 years ago

Here's O(log n) solution [1]:

  def fib_sicp(n):
      a, b, p, q = 1, 0, 0, 1
      while n > 0:
          if n % 2 == 0: # even
              oldp = p
              p = p*p + q*q
              q = 2*oldp*q + q*q
              n //= 2
          else:
              olda = a
              a = b*q + a*q + a*p
              b = b*p + olda*q
              n -= 1
      return b

[1]: http://stackoverflow.com/questions/1525521/nth-fibonacci-num...

It is ~20 times faster for n=10000

blt3• 13 years ago

Cool - did not know about that. Wish I could edit my comment.

mjhea0• 13 years ago

Yes - an iterative loop is much faster, because with recursion you have to calculate a number of trivial fibonacci numbers.

https://gist.github.com/mjhea0/5798178

jgeerts• 13 years ago

Reminds me awefully much about the blog post below.

http://agillo.net/getting-groovy-with-fibonacci/

crntaylor• 13 years ago

I can't resist a quick plug for this beautiful O(n) implementation of the fibonacci sequence in Haskell. One line to define the list `fibs` -- an infinite list of fibonacci numbers -- and one line to define the function `fib` which simply takes the n'th element of the list.

    λ> let fibs  = 0 : 1 : zipWith (+) fibs (tail fibs)
    λ> let fib n = fibs!!n
    λ> fib 100
    354224848179261915075

lccarrasco• 13 years ago

I really liked the haskell version when I first learned about it in a programming languages course, I believe I've never been more surprised than after seeing the output of 'take 10 fibs' without knowing how did that single line worked.

taeric3• 13 years ago

Correct me if I'm wrong, but this is also O(n) in space, right? I think you can get a similar scala version in:

    lazy val fib:Stream[BigInt] = 0 #::1 #:: fib.zip(fib.tail).map(x=>x._1+x._2)

This is neat and all, but the imperative version wins by being O(1) in space. Right?

FreeFull• 13 years ago

It is perfectly possible to write a fib function that takes O(1) space functionally too, but of course it'd have a worse time complexity due to the lack of memoisation.

tmoertel5• 13 years ago

No, the Haskell version will operate in constant space provided you don't hold onto the list:

    let fibs = 1 : 1 : zipWith (+) fibs (tail fibs) in fibs !! 100

Since the fibs list is visible only to the expression (fibs !! 100), and since the (!!) operator advances through the list discarding elements until it hits the 100th, those elements are free to be garbage collected. Further, since the elements of the list (after the first two) are generated only when consumed, only a few of the cons cells will ever be live at any point in time.

peter_l_downs2• 13 years ago

Or you could apply some math and come up with a generating function!

http://www.math.ufl.edu/~joelar/generatingfibonacci.pdf

antiterra• 13 years ago

And, if you want to know more about generating functions, you can check out the book "generatingfunctionology" by the recently late Herbert Wilf gratis at http://www.math.upenn.edu/~wilf/DownldGF.html

tmoertel5• 13 years ago

Another great resource on generating functions is Analytic Combinatorics by Philippe Flajolet and Robert Sedgewick:

http://algo.inria.fr/flajolet/Publications/book.pdf

tmoertel5• 13 years ago

Another math-y option is to note that the Fibonacci series satisfies the linear recurrence F[n+2] = F[n+1] + F[n] and can therefore be represented in matrix form. Here's a (trimmed down) Maxima session that gives the details:

    $ maxima
    (%i3) load(eigen)$
    (%i4) A: matrix([1, 1], [1, 0])$
    (%i5) b: columnvector([1, 0])$
    (%i6) fib(n) := (''A^^n . ''b)[2][1];

                              <n>
                      [ 1  1 ]      [ 1 ]
    (%o6) fib(n) := (([      ]    . [   ]) )
                      [ 1  0 ]      [ 0 ]
                                          2
                                            1

    (%i7) makelist(fib(n), n, 10);

    (%o7) [1, 1, 2, 3, 5, 8, 13, 21, 34, 55]

Since the transition matrix A is symmetric, it can be diagonalized using its eigenvector matrix:

    (%i8) similaritytransform(A)$
    (%i9) Adiag: expand(leftmatrix . A . rightmatrix)$

And the n-th power of a diagonal matrix is the same as raising the values on its diagonal to the n-th power:

    (%i10) AdiagN(n) := matrix([''(Adiag[1][1])^n, 0], [0, ''(Adiag[2][2])^n]);

And thus raising A to the n-th power is the same as raising its diagonal counterpart Adiag to that power and then reversing the diagonalization transform (since leftmatrix and rightmatrix are inverses and cancel out for the interior multiplications):

    (%i11) fibdiag(n) := expand(rightmatrix . AdiagN(n) . leftmatrix . b)[2][1]$

Finally, this leads to a closed-form representation for the n-th Fibonacci number:

    (%i12) fibdiag(n);

            sqrt(5)   1 n    1   sqrt(5) n
           (------- + -)    (- - -------)
               2      2      2      2
    (%o12) -------------- - --------------
              sqrt(5)          sqrt(5)

peter_l_downs2• 13 years ago

Very neat, I need to learn more math to truly understand how this works!

tmoertel5• 13 years ago

Perhaps surprisingly, you can convert many recursive algorithms into equivalent iterative versions using a series of simple mechanical steps. For example, if we start with the naive recursive implementation of our beloved fib function,

    def fib(n):
        if n < 2:
            return n
        return fib(n - 1) + fib(n - 2)

we can arrive at the iterative (dynamic-programming) version that follows,

    def fib(n):
        fibn1, fibn2 = (0, 1)
        for _ in xrange(n):
            fibn1, fibn2 = fibn2, fibn1 + fibn2
        return fibn1

through the following steps: https://gist.github.com/tmoertel/5798134

If you're interested in this kind of thing, I'm writing a series of articles on converting recursive algorithms into iterative algorithms. The initial article: http://blog.moertel.com/posts/2013-05-11-recursive-to-iterat...

mechagodzilla• 13 years ago

Here is the Fibonacci sequence calculated by a literal series of mechanical steps: http://www.chrisfenton.com/the-turbo-entabulator/

benjamincburns• 13 years ago

Edit:

Doh - I'm a moron! You're using a recurrence relation, which has the nice property of never computing the same sub-solution twice. I'll leave this here as a reminder to think before I nitpick next time.

Original:

Apologies to nitpick, but your iterative python solution isn't DP. DP hinges on the idea of never computing the answer to the same problem twice. For it to be DP, it would need to memoize subproblem solutions and refer to the solution cache rather than recompute the solution to every subproblem in the tree every time.

This solution fits the bill, although it can probably be written a bit more elegantly:

  fibs = {0: 1, 1: 1}
  def fib(n):
      global fibs
      stack = []
      for a in xrange(n,0,-1):
          if a in fibs:
              stack.reverse()
              for item in stack:
                  fibs[item] = fibs[a] + fibs[a-1]
                  a = item
              break
          else:
              stack.append(a)
      return fibs[n]

dhruvchandna• 13 years ago

In Clojure you can implement this as:

    (def fib-list (map first (iterate (fn [[a b]] [b (+ a b)]) [0N 1N])))

    (nth fib-list 100)
    354224848179261915075N

amckinlay• 13 years ago

Doesn't Clojure have a .memoize() that would be useful here?

pathikrit• 13 years ago

A O(n) Scala one: https://github.com/pathikrit/scalgos/blob/master/src/main/sc...

prakashk• 13 years ago

Perl 6 (from [1]):

    > my @fib := 1, 1, *+* ... *; @fib[99]
    354224848179261915075

    > (1, 1, *+* ... *)[99]
    354224848179261915075

[1] https://justrakudoit.wordpress.com/2010/12/29/perl-6-fibonac...

Fibonacci series and Dynamic programming

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