The Fastest Way to Compute the Nth Fibonacci Number: The Doubling Method
Introduction
The Fibonacci Sequence is defined as follows:
\[ \begin{align} \mathrm{F}_{0} = 0\\ \mathrm{F}_{1} = 1\\ \mathrm{F}_{n} = \mathrm{F}_{n  2} + \mathrm{F}_{n  1}. \end{align} \]
That is, each Fibonacci number \(\mathrm{F}_{n}\) is the sum of the previous two Fibonacci numbers, except the very first two numbers which are defined to be 0 and 1. ^{1}
From the definition above, it appears that computing \(\mathrm{F}_{n}\) requires one to always compute \(\mathrm{F}_{n  2}\) and \(\mathrm{F}_{n  1}\). This is false: enter the “doubling method”. ^{2} ^{3}
The Genesis of the Doubling Method
The doubling method uses a couple of mathematical formulas derived from matrix multiplication as it applies to calculating Fibonacci numbers; it can be seen as an improvement over the matrix multiplication method, although it does not use matrix multplication itself. The matrix multiplication method uses the following formula:
\[ \begin{equation} \begin{bmatrix} 1 & 1\\ 1 & 0 \end{bmatrix}^n = \begin{bmatrix} \mathrm{F}_{n + 1} & \mathrm{F}_{n}\\ \mathrm{F}_{n} & \mathrm{F}_{n  1} \end{bmatrix}. \end{equation} \]
This result is quite interesting in its own right; to find \(\mathrm{F}_{n}\) you only need to raise the matrix
\[ \begin{bmatrix} 1 & 1\\ 1 & 0 \end{bmatrix} \]
to the \(n\)th power. To be more precise, this method is matrix exponentiation. The only downside is that much of the answer is wasted — we don’t care about \(\mathrm{F}_{n  1}\), not to mention how \(\mathrm{F}_{n}\) is redundantly computed twice.
Thinking in Terms of \(\mathrm{F}_{n}\)
What if we could find \(\mathrm{F}_{n}\) not by multiplying or adding some numbers, but by multiplying and adding other Fibonacci terms? Of course, we’re not talking about adding \(\mathrm{F}_{n  2}\) and \(\mathrm{F}_{n  1}\) because that would be too slow. Let’s have a look at the matrix identity again (reversed for easier reading):
\[ \begin{equation} \begin{bmatrix} \mathrm{F}_{n + 1} & \mathrm{F}_{n}\\ \mathrm{F}_{n} & \mathrm{F}_{n  1} \end{bmatrix} = \begin{bmatrix} 1 & 1\\ 1 & 0 \end{bmatrix}^n. \end{equation} \]
If we substitute in \(2n\) for \(n\), we get
\[ \begin{align} \begin{bmatrix} \mathrm{F}_{2n + 1} & \mathrm{F}_{2n}\\ \mathrm{F}_{2n} & \mathrm{F}_{2n  1} \end{bmatrix} & = \begin{bmatrix} 1 & 1\\ 1 & 0 \end{bmatrix}^{2n} \\ & = \bigg(\begin{bmatrix} 1 & 1\\ 1 & 0 \end{bmatrix}^{n}\bigg)^2 \end{align} \]
and we can substitute in our matrix identity from above to rewrite this as
\[ \begin{align} & = \bigg(\begin{bmatrix} \mathrm{F}_{n + 1} & \mathrm{F}_{n}\\ \mathrm{F}_{n} & \mathrm{F}_{n  1} \end{bmatrix}\bigg)^2 \end{align} \]
and carry out the squaring to get
\[ \begin{align} & = \begin{bmatrix} {{\mathrm{F}_{n + 1}}^2 + {\mathrm{F}_{n}}^2} & {{\mathrm{F}_{n + 1}\mathrm{F}_{n}} + {\mathrm{F}_{n}\mathrm{F}_{n  1}}}\\ {{\mathrm{F}_{n}\mathrm{F}_{n + 1}} + {\mathrm{F}_{n  1}\mathrm{F}_{n}}} & {{\mathrm{F}_{n}}^2 + {\mathrm{F}_{n  1}}^2} \end{bmatrix}. \end{align} \]
The top right and bottom left terms are identical; we can also rewrite them to be a bit simpler.
\[ \begin{align} {{\mathrm{F}_{n + 1}\mathrm{F}_{n}} + {\mathrm{F}_{n}\mathrm{F}_{n  1}}} & = \mathrm{F}_{n}(\mathrm{F}_{n + 1} + \mathrm{F}_{n  1}) \\ & = \mathrm{F}_{n}[\mathrm{F}_{n + 1} + (\mathrm{F}_{n + 1}  \mathrm{F}_{n})] \\ & = \mathrm{F}_{n}(2\mathrm{F}_{n + 1}  \mathrm{F}_{n}). \end{align} \]
This simplication achieves an important task — it obviates \(\mathrm{F}_{n  1}\) by cleverly defining it as \(\mathrm{F}_{n + 1}  \mathrm{F}_{n}\). Putting everything together, whe have
\[ \begin{align} \begin{bmatrix} \mathrm{F}_{2n + 1} & \mathrm{F}_{2n}\\ \mathrm{F}_{2n} & \mathrm{F}_{2n  1} \end{bmatrix} & = \begin{bmatrix} {{\mathrm{F}_{n + 1}}^2 + {\mathrm{F}_{n}}^2} & {\mathrm{F}_{n}(2\mathrm{F}_{n + 1}  \mathrm{F}_{n})}\\ {\mathrm{F}_{n}(2\mathrm{F}_{n + 1}  \mathrm{F}_{n})} & {{\mathrm{F}_{n}}^2 + {\mathrm{F}_{n  1}}^2} \end{bmatrix} \end{align} \]
where the first row (or column) gives us two very useful identities
\[ \begin{align} \mathrm{F}_{2n} & = {\mathrm{F}_{n}(2\mathrm{F}_{n + 1}  \mathrm{F}_{n})} \\ \mathrm{F}_{2n + 1} & = {{\mathrm{F}_{n}}^2 + {\mathrm{F}_{n + 1}}^2}. \end{align} \]
As these identities form the heart of the doubling method, let’s call them the doubling identities.
And now we just need one more piece to formulate our doubling method; we need to borrow an idea from number theory. Given any positive integer \(n\), it is the same as either \(2m\) (even) or \(2m + 1\) (odd), where \(m = \lfloor\frac{n}{2}\rfloor\); for our purposes, let us call this property the “halving property”.
Whereas the doubling identities allow us to “double” our way into bigger numbers, the halving property allows us to halve our way down to smaller and smaller numbers. The marriage of these two concepts gives rise to the doubling method.
The Doubling Method
To compute the \(n\)th Fibonacci term we break \(n\) itself down into its halves (\(2m\)) recursively, until we go down to \(n = 0\). At this point we multiply our way back up using the doubling identities. Because halving and doubling by themselves always calculate \(\mathrm{F}_{2m}\), we have to manually return \(\mathrm{F}_{2m + 1}\) if our current sequence index number \(n\) is odd.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 

Line 12 is where we do the halving. We use the rightshift operator to do this. Lines 13 and 14 are our doubling identities (I use the leftshift operator here because it feels more natural to me). The ifcondition on line 15 returns \(\mathrm{F}_{2m + 1}\) if \(n\) was odd, and \(\mathrm{F}_{2m}\) otherwise.
For comparison, here is an iterative version. On the one hand it avoids Python’s recursion limit, but the downside is a small loss of elegance (we have to loop twice — first to build up the halving/doubling points, and again for the main loop).
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 

Conclusion
I hope you enjoyed reading about this method of calculationg Fibonacci numbers as much as I enjoyed learning the math behind it. This algorithm can be sped up if it uses a faster multiplication algorithm as a
and b
get very large (e.g., Karatsuba multiplication). ^{4} Time complexity is \(\Theta(\log{n})\); it reminds me of the binary search algorithm, in how the problem space is halved repeatedly. Neat!
We can choose to define the first two terms as 1 and 1 instead, but this distinction is needlessly arbitrary.↩
There is actually a known formula for our purposes, where \[ \mathrm{F}_{n} = \frac{\varphi^n  (\varphi)^{n}}{2\varphi  1}\] and \(\varphi = \frac{1 + \sqrt{5}}{2} \approx 1.6180339887\cdots\) (the golden ratio). Unfortunately this requires arbitraryprecision floating point calculations.↩
For more discussion, see https://www.nayuki.io/page/fastfibonaccialgorithms.↩
Python already uses Karatsuba multiplication natively for large integers.↩