Babylonian method for square root

Algorithm: 
This method can be derived from (but predates) Newton–Raphson method. 
 

1 Start with an arbitrary positive start value x (the closer to the 
 root, the better).
2 Initialize y = 1.
3. Do following until desired approximation is achieved.
 a) Get the next approximation for root using average of x and y
 b) Set y = n/x

Implementation: 
 

Output : 

Square root of 50 is 7.071068

Time Complexity: O(n^1/2)

Auxiliary Space: O(1)
Example:

n = 4 /*n itself is used for initial approximation*/
Initialize x = 4, y = 1
Next Approximation x = (x + y)/2 (= 2.500000), 
y = n/x (=1.600000)
Next Approximation x = 2.050000,
y = 1.951220
Next Approximation x = 2.000610,
y = 1.999390
Next Approximation x = 2.000000, 
y = 2.000000
Terminate as (x - y) > e now.

If we are sure that n is a perfect square, then we can use following method. The method can go in infinite loop for non-perfect-square numbers. For example, for 3 the below while loop will never terminate. 
 

Output : 
 

 root of 49 is 7

Time Complexity: O(n^1/2)

Auxiliary Space: O(1)
References; 
https://en.wikipedia.org/wiki/Square_root 
https://en.wikipedia.org/wiki/Babylonian_method#Babylonian_method
Asked by Snehal
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