For any two positive integer number m and n, **GCD** ( greatest common divisor) is the largest integer number which divides them evenly.

So for the following two numbers 8 and 12, 4 is the largest number which divides them evenly hence 4 is the GCD of 8 & 12. Since 1 is the divisor of every number, so there is always at least one common divisor for a pair of numbers.

In order to calculate GCD for a pair number m & n programmatically. Typically one would go through all the factors of m and n respectively and store them in a list then extract the common factors in both the list and find out the largest factor. This will work although this isn’t the most efficient way of calculating GCD, for two really large numbers.

## Euclid algorithm

Euclid, a Greek mathematician in 300 B.C. discovered an extremely efficient way of calculating GCD for a given pair of numbers. Euclid observed that for a pair of numbers m & n assuming `m>n`

and n is not a divisor of m.

Number m can be written as `m = qn + r`

, where q in the **quotient** and r is the **reminder**. Recursive substitution of r with q and q with m until the remainder is 0 will ultimately deliver the GCD for the pair since `gcd(n,0) = n`

.

Mathematically i.e.

`gcd(m, n) == gcd(n, m % n)`

We can verify this algorithm by taking the same two numbers 12 & 8, having a common divisor d = 4

```
# m = qn + r
12 = q * 8 + r
# q = 1 & n = 8 & r =4
12 = 8 + 4
#Substituiting m with n and q with r
#q =2 & n = 4 & r =0
8 = 4*2 + 0
#Substituiting m with n and q with r
GCD = 4
```

Euclid algorithm remarkably increases the efficiency of the program calculating GCD as the reminder keeps on decreasing resulting in saving the precious computer cycles.

## Program to find GCD In Python

```
def gcd(m,n):
if m< n:
(m,n) = (n,m)
if(m%n) == 0:
return n
else:
return (gcd(n, m % n)) # recursion taking place
# calling function with parameters and printing it out
print(gcd(8,12))
```

This is a simple recursive function, which will return the greatest common divisor when the condition is met, we just need to pass the numbers as parameters in the function call.

The same function can also be made on the top of a while loop with the controlling statement being (m%n != 0) as follows:

```
def gcd(m,n):
if m< n:
(m,n) = (n,m)
while (m % n != 0):
(m, n) = (n, m % n)
return n
# calling function with parameters and printing it out
print(gcd(8,12))
```

Both recursive functions return the GCD for a given pair of numbers efficiently even if the numbers are huge.