How does Lamé's Theorem give us an order-of-growth estimate for Euclid's Algorithm?

Question

I'm reading SICP, and I don't understand how Lamé's Theorem gives us an estimate for the order-of-growth of Euclid's algorithm (the relevant passage is below). It would make sense to me if the assertion was that the smaller of the two inputs grows as the logarithm (to the base phi) of k. Intuitively, it makes sense to me that the number of steps k would increase with the smaller of the two inputs n, but it seems like all we've proven is that the smaller of the two inputs n grows with the number of steps k. Since all we've proven is that n must be at least Fib(k), I don't see how we can make any claims that k must grow with n, only that n must grow with k.

Answer 1

In my edition of SICP (first edition from 1987), the final sentence in that paragraph contained indeed O(log n), not Theta(log n), which fits to your observation - the provided proof literally just gives an upper bound for the order of growth of the algorithm.

However, in the second edition (online version here) it seems authors had replaced O by Theta. I guess they did so because log n is indeed a lower bound for the order of growth of the Euclidean algorithm.

Look first what we are measuring here: the Euclidean algorithm takes two parameters, n and m with n<=m. Let us define the number of steps as E(n,m). To make this dependend just on n, we set

 f(n) := max_{m>=n}( E(n,m) )

This maximum exists because of Lamés theorem, which is saying, when n < Fib(k+1), then E(n,m) must be <=k for all m, which means f(n)<=k. When we speak about O(log(n)) or Theta(log(n)) of the "number of steps of the Euclidean algorithm", we actually mean the order of growth of this function f.

On the other hand, when Fib(k) <= n < F(k+1), it is simple to see that f(n) must be also greater or equal to k: just set m:= n +Fib(k-1). Starting the Euclidean algorithm with (n,m) leads to (Fib(k-1), Fib(k)) in the first step, and then to the Fibonacci sequence in reverse order, hence requiring k steps.

So this gives you the lower bound for f(n), and Lamés theorem the identical upper bound.