# Calculating Fibonacci Numbers, Quickly and Exactly

The well-known Fibonacci series $$F_n$$ can be defined as follows:

$$F_n = \begin{cases} 0 & n = 0 \\ 1 & n = 1 \\ F_{n-2} + F_{n-1} & n \ge 2\\ \end{cases}$$

Let’s use a few facts about matrices to find a quick way to calculate terms in this famous series.

Lemma

Let $$A = \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}$$. Then $$A^n = \begin{bmatrix} F_{n-1} & F_n \\ F_n & F_{n+1} \end{bmatrix}$$.

Proof

The $$n = 1$$ case follows immediately from the definitions of $$A \text{ and } F_n$$.

Suppose the statement is true for n. Then
\begin{align} A^{n+1} & = A^n A \\ & = \begin{bmatrix} F_{n-1} & F_n \\ F_n & F_{n+1} \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix} \\ & = \begin{bmatrix} F_n & F_{n-1} + F_n \\ F_{n+1} & F_n + F_{n+1} \end{bmatrix} \\ & = \begin{bmatrix} F_n & F_{n+1} \\ F_{n+1} & F_{n+2} \end{bmatrix} \end{align}

And our result follows by induction. QED

Now, notice that

\begin{align} \begin{bmatrix} F_{2n-1} & F_{2n} \\ F_{2n} & F_{2n+1} \end{bmatrix} & = A^{2n} \\ & = A^n A^n \\ & = \begin{bmatrix} F_{n-1} & F_n \\ F_n & F_{n+1} \end{bmatrix} \begin{bmatrix} F_{n-1} & F_n \\ F_n & F_{n+1} \end{bmatrix} \\ & = \begin{bmatrix} F_{n-1}^2 + F_n ^ 2 & F_{n-1} F_n + F_n F_{n+1} \\ F_{n-1} F_n + F_n F_{n+1} & F_n^2 + F_{n+1} ^ 2 \end{bmatrix} \end{align}

Using this identity, we can write $$F_{2n}$$ and $$F_{2n+1}$$ in terms of $$F_n$$ and $$F_{n+1}$$.

\begin{align} F_{2n} & = F_{n-1} F_n + F_n F_{n+1} \\ & = (F_{n+1} – F_n) F_n + F_n F_{n+1} \\ & = F_n F_{n+1} – F_n^2 + F_n F_{n+1} \\ & = 2 F_{n+1} F_n – F_n ^ 2 \end{align}

$$F_{2n+1} = F_{n}^2 + F_{n+1}^2$$

We now have a way of calculating $$F_{2n}$$ and $$F_{2n+1}$$ by calculating only a few of the smaller terms in the sequence. This yields some wildly efficient Python code:

 def fib2(N): if N == 0: return (0, 1) half_N, is_N_odd = divmod(N, 2) f_n, f_n_plus_1 = fib2(half_N) f_n_squared = f_n * f_n f_n_plus_1_squared = f_n_plus_1 * f_n_plus_1 f_2n = 2 * f_n * f_n_plus_1 - f_n_squared f_2n_plus_1 = f_n_squared + f_n_plus_1_squared if is_N_odd: return (f_2n_plus_1, f_2n + f_2n_plus_1) return (f_2n, f_2n_plus_1) def fib(N): return fib2(N) 

And it’s fast! It’s $$O(\log N)$$ in fact:

$pypy -m timeit -s 'import expfib' 'expfib.fib(500000)' 10 loops, best of 3: 22.4 msec per loop  Compare to the naÃ¯ve, iterative version, which is $$O(N)$$:  def fib2_linear(N): f_n, f_n_plus_1 = (0, 1) while N > 0: N -= 1 f_n, f_n_plus_1 = f_n_plus_1, f_n + f_n_plus_1 return (f_n, f_n_plus_1) def fib_linear(N): return fib2_linear(N)  $ pypy -m timeit -s 'import myfib' 'myfib.fib_linear(500000)'
10 loops, best of 3: 2.76 sec per loop


This post was inspired by a post by Lee Phillips and as an excuse to play around with MathJax. The best guide I found for getting started is on Stack Exchange.

## 3 thoughts on “Calculating Fibonacci Numbers, Quickly and Exactly”

1. Orion Blastar says:

Great code. I love it!

2. david karapetyan says:

Neat.

3. George says:

It took me a little effort to convince myself that your final algorithm is logarithmic and not linear in time. A naÃ¯ve implementation would be linear, but in a more interesting way than usual: the call tree would have a branch factor of two but would now only be log2(n) deep.

But I see that you take advantage of the two right-hand-side terms always being adjacent to reduce the branch factor to one. Very nice.