Termination of the Euclidean Algorithm if $~a < 2^n~$ with $~b>a~$

Question

Show for all $~n ∈ \mathbb Z^+~$ that if $~a < 2^n~$, then the Euclidean algorithm applied to $~a,~ b~$ with $~b > a~$ will terminate in at most $~2n − 1~$ steps.

So I spoke to my professor and he said to consider that if we begin with a remainder of $~r_1~$ and then consider the remainder from the second iteration $~r_2~$ to compare the value of $~r_2~$ to the initial size of $~a~$.

I'm not really looking for a complete proof, I just kind of need guidance on how to begin looking at this problem.

I'm still new to mathematical proofs so I just need to get a feeling for how to structure this proof or maybe a general guiding clue.

Answer 1

Let us define:

$$a_0=a<b=b_0$$

$$a_{n+1}=(b_n\bmod a_n)$$

$$b_{n+1}=a_n$$

There is actually an improved bound you can get by observing the worst case scenario where $a_{n+1}=b_n-a_n$ on every step. By subsituting $b_n=a_{n-1}$, one reaches the formula:

$$a_{n-1}=a_n+a_{n+1}$$

which is the Fibonacci sequence in reverse! That is to say, the the Euclidean algorithm runs slowest in the case of $a=F_n<F_{n+1}=b$ and takes $n$ steps. Since we know that $F_n\ge\varphi^n/\sqrt5$, it follows that the Euclidean algorithm takes $\lceil\log_\varphi(\sqrt5\min\{a,b\})\rceil$ steps in the worst case, where $\varphi=\frac{1+\sqrt5}2$ is the golden ratio.

In your case specifically, we know that it will take less than $\left\lceil n\log_\varphi(2)+\frac12\log_\varphi(5)\right\rceil\approx\lceil1.44n+1.67\rceil$ steps.