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 so take x=n/2) .
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/2; //n/2 can be closest one.
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.
For a perfect square number this e would be equal to 0.
Friday, September 2, 2011
Method to find Square Root
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Another solution for Perfect Square is :
ReplyDeleteSquare numbers are sum of consecutive odd numbers like
1+3+5+7=16 so add numbers till you get the required number --- O(sqrt(n)).
1 + 3 + 5 + 7 +9 =25 Count of odd numbers = num = 5. Complexity O(sqrt (n)) .
(or)
Si=sum of i terms of AP i =1...n/2 as n/2>sqrt(n)
Si can be obtained in O(1) i.e. sum of AP
Do a binary search on those n/2 S items.
complexity=log(n/2)+O(sqrt(n)).