# 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, fN is the Nth 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 Nth 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 >= fN

• 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) + fN

• 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) + fN.

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:

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

• where, is known as the golden ratio(∅≈1.618), and fN is the Nth 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:

fN ≈ ∅N
N ≈ log(fN)

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:

=> fN+1 ≈ min(a, b)
=> N+1 ≈ logmin(a, b)

=> O(N) = O(N+1) =  log(min(a, b))

Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!