Time Complexity of Euclidean Algorithm

In this article, we will discuss the time complexity of the Euclidean Algorithm which is O(log(min(a, b)) and it is achieved.

Euclid’s Algorithm: It is an efficient method for finding the GCD(Greatest Common Divisor) of two integers. The time complexity of this algorithm is O(log(min(a, b)). Recursively it can be expressed as:

gcd(a, b) = gcd(b, a%b),
where, a and b are two integers.

Proof: Suppose, a and b are two integers such that a >b then according to Euclid’s Algorithm:

gcd(a, b) = gcd(b, a%b)

Use the above formula repetitively until reach a step where b is 0. At this step, the result will be the GCD of the two integers, which will be equal to a. So, after observing carefully, it can be said that the time complexity of this algorithm would be proportional to the number of steps required to reduce b to 0.

Let’s assume, the number of steps required to reduce b to 0 using this algorithm is N.

gcd(a, b) ------> N steps

Now, if the Euclidean Algorithm for two numbers a and b reduces in N steps then, a should be at least f_{(N + 2)} and b should be at least f_{(N + 1)}.

gcd(a, b) ——> N steps
Then, a >= f_{(N + 2)} and b >= f_{(N + 1)}
where, f_N is the N^th term in the Fibonacci series(0, 1, 1, 2, 3, …) and N >= 0.

To prove the above statement by using the Principle of Mathematical Induction(PMI):

Base Case:
- Let’s assume a = 2 and b = 1. Then, gcd(2, 1) will reduce to gcd(1, 0) in 1 step, i.e., N = 1.
- This means 2 should be at least f₃ and 1 should be at least f₂ and f₃ = 2 and f₂ = 1.
- This implies, a is at least f_{(N + 2)} and b is at least f_{(N + 1)}.
- It can be concluded that the statement holds true for the Base Case.
Inductive Step: Assume that the statement holds true for the (N – 1)^th Step. So, Below are the steps to prove it for the N^th Step:

gcd(b, a%b) ——> (N – 1) steps
Then,
b >= f_{(N – 1 + 2)} i.e., b >= f_{(N + 1)}
a%b >= f_{(N – 1 + 1)} i.e., a%b >= f_N

It can also be written as:

a = floor(a/b)*b + a%b

floor(a/b)*b means highest multiple which is closest to b.

ex – floor(5/2)*2 = 4. If we then add 5%2=1, we will get a(=5) back.

Now, (a/b) would always be greater than 1 ( as a >= b). So, from the above result, it is concluded that:

a >= b + (a%b)
This implies, a >= f_{(N + 1)} + f_N

It is known that each number is the sum of the two preceding terms in a Fibonacci series. This implies f_{(N + 2)} = f_{(N + 1)} + f_N.

a >= f_{(N + 2)} and b >= f_{(N + 1)}

Since the above statement holds true for the inductive step as well. This proves that the statement is correct.
Before proceeding further, Look at the Binet Formula:

Binet Formula:

f_N = {((1 + √5)/2)^N – ((1 – √5)/2)^N}/√5
or
f_N ≈ ∅^N

where, ∅ is known as the golden ratio(∅≈1.618), and f_N is the N^th Fibonacci Number.
Now, it is already stated that the time complexity will be proportional to N i.e., the number of steps required to reduce b to 0.
So, to prove the time complexity, it is known that:

f_N ≈ ∅^N
N ≈ log_∅(f_N)

Now, from the above statement, it is proved that using the Principle of Mathematical Induction, it can be said that if the Euclidean algorithm for two numbers a and b reduces in N steps then, a should be at least f_{(N + 2)}and b should be at least f_{(N + 1)}.

From the above two results, it can be concluded that: