There are N number of ways to compute square root of a number

we can use the existing library (math) in python to compute the square root

However, there is a constrain in number of decimal digits . Hence, we need an algorithm which can compute a square root and provide n decimal digits .

The following algorithm uses Digit by digit calculation from this wikipedia link

Square Root Computation Algorithms

I have created a python script which will compute square root of 2 with 100 decimal digits .

And i verified the digits from this link

First 100 Digits for SQRT of 2 – Verification

The algorithm is not optimized. However, it gives me the result quickly for first 100 digits

For multiple irrational number , populate the irr list (by default contains 2)

To control number of digits after decimal change decimal variable (by default set to 100)

<br data-mce-bogus="1"> import time import math starttime = time.time() global flag, start, end, decimal flag = True start = 0 end = 1 decimal = 100 def main(): answer = 0 irr = [2] # for each irrational number for num in irr: # Step - 1 : Converting number to string and appending 200 zeros str_num = str(num) for i in range(0, 210): str_num = str_num + "0" # Step - 2: Setting Flag : if lenght of the number is odd we need to take one digit or else two digits if len(str_num) % 2 != 0: flag = True else: flag = False start = 0 end = 0 ap_new_dig = 0 # Step - 3: Logic to take one or 2 digits if flag: start = 0 end = 1 else: start = 0 end = 2 temp_num = str_num[start:end] temp_int_num = int(temp_num) # Step - 4 : Calculate the initial quotiend and remainder quo = int(math.sqrt(temp_int_num)) rem = temp_int_num - int(quo ** 2) # Step - 5 : Here is the main logic to iterate the steps counter = 0 while counter < decimal - 1 : # Step - 6 : Bring down the digits if flag: start = start + 1 end = start + 2 flag = False else: start = start + 2 end = start + 2 new_dig = str_num[start:end] # Step -7 : Add the digits with previous remainder ap_new_dig = str(rem) + new_dig # Step - 8 : Compute a number less than max ap_new_digit double_num = quo * 2 max_val = 0 t = 0 for i in range(1, 10): if int( str(double_num) + str(i) ) * i <= int(ap_new_dig): t = int( str(double_num) + str(i) ) * i max_val = i else: break # Step - 9 : Find new quotient and remainder quo = int ( str(quo) + str(max_val) ) rem = int(ap_new_dig) - t counter = counter + 1 # Final Quotient will be the answer print "Square Root of N is" , quo main() print "Time taken ", time.time() - starttime

Excecution:

>>>

Square Root of N is 1414213562373095048801688724209698078569671875376948073176679737990732478462107038850387534327641572

Time taken 0.0269999504089