An earlier post described how to compute Fibonacci numbers in a single arithmetic expression.

Faré Rideau, the author of a page of Fibonacci computations in Lisp, suggested in a private email a simple and efficient variant, that I believe is novel.

For $X$ large enough, $\mathrm{Fib}_n = (X^{n+1}\ \mathrm{mod}\ (X^2-X-1))\ \mathrm{mod}\ X$.

That means you can compute Fibonacci numbers efficiently with a simple program:

for n in range(1, 21):
    X = 1<<(n+2)
    print(pow(X, n+1, X*X-X-1) % X)

This blog post describes how this method works, gives a few ways to think about it, easily infers the fast Fibonacci doubling formulas, provides a nice alternative to Binet’s formula relating the golden ratio and Fibonacci numbers, and expands the method to generalized Fibonacci recurrences, including a near one-line solution to the problem of counting how many ways to reach the end-square of a 100-square game using a single six-sided dice.

Who would disagree that the run-time of mergesort is $O(n\mathrm{log},n)$ and it’s asymptotically optimal? Not many programmers I reckon, except perhaps to question whether it’s talking about a model of computation that’s not sufficiently close to a real computer, for example a quantum computer or one that performs arbitrary operations in parallel (possibly involving sticks of spaghetti).

However, if you try to understand how to formalize what it means for a sort to run in $O(n\mathrm{log},n)$ and for it to be optimal, it’s surprisingly difficult to find a suitable computational model, that is, an abstraction of a computer which elides all but the important details of the computer: the operations it can perform, and how the memory works.

In this post, I’ll look at some of the most common computational models used in both practice and theory, and find out that they’re all flawed in one way or another, and in fact in all of them either mergesort doesn’t run in $O(n\mathrm{log},n)$ or there’s asymptotically faster sorts.