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:

/*Returns the square root of n. Note that the function */
float
squareRoot(
float
n)
{
/*We are using n itself as initial approximation
This can definitely be improved */
float
x = n;
float
y = 1;
float
e = 0.000001;
/* e decides the accuracy level*/
while
(x  y > e)
{
x = (x + y)/2;
y = n/x;
}
return
x;
}
/* Driver program to test above function*/
int
main()
{
int
n = 50;
printf
(
"Square root of %d is %f"
, n, squareRoot(n));
getchar
();
}
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 nonperfectsquare numbers. For example, for 3 the below while loop will never terminate.

/*Returns the square root of n. Note that the function
will not work for numbers which are not perfect squares*/
unsigned
int
squareRoot(
int
n)
{
int
x = n;
int
y = 1;
while
(x > y)
{
x = (x + y)/2;
y = n/x;
}
return
x;
}
/* Driver program to test above function*/
int
main()
{
int
n = 49;
printf
(
" root of %d is %d"
, n, squareRoot(n));
getchar
();
}
Disclaimer: This does not belong to TechCodeBit, its an article taken from the below
source and credits.
source and credits:http://www.geeksforgeeks.org/squarerootofaperfectsquare/
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