Python - Find PI to the Nth Digit

🐍 Enter a number and have the program generate π (pi) up to that many decimal places. Keep a limit to how far the program will go.

7 Different Solutions

Solution 1

from decimal import *



def pipric(n):

    getcontext().prec = n+1

    return Decimal(22)/Decimal(7)

if __name__ == '__main__':

    correct_input = False

    while not correct_input:

        print('Precision must be between 1 and 51')

        precision = int(input('Number of decimal places: '))

        if 51 >= precision > 0:

            correct_input = True

    print(pipric(precision))

Solution 2

def pipric(n):

    getcontext().prec = n+1

    return '%.*f' % (n, 22/7)



if __name__ == '__main__':

    correct_input = False

    while not correct_input:

        print('Precision must be between 1 and 51')

        precision = int(input('Number of decimal places: '))

        if 51 >= precision > 0:

            correct_input = True

    print(pipric(precision))

Solution 3: More precise

#!/usr/bin/env python3
# https://github.com/MrBlaise/learnpython/blob/master/Numbers/pi.py
# Find PI to the Nth Digit
# Have the user enter a number 'n'
# and print out PI to the 'n'th digit

def calcPi(limit):  # Generator function
    """
    Prints out the digits of PI
    until it reaches the given limit
    """

    q, r, t, k, n, l = 1, 0, 1, 1, 3, 3

    decimal = limit
    counter = 0

    while counter != decimal + 1:
            if 4 * q + r - t < n * t:
                    # yield digit
                    yield n
                    # insert period after first digit
                    if counter == 0:
                            yield '.'
                    # end
                    if decimal == counter:
                            print('')
                            break
                    counter += 1
                    nr = 10 * (r - n * t)
                    n = ((10 * (3 * q + r)) // t) - 10 * n
                    q *= 10
                    r = nr
            else:
                    nr = (2 * q + r) * l
                    nn = (q * (7 * k) + 2 + (r * l)) // (t * l)
                    q *= k
                    t *= l
                    l += 2
                    k += 1
                    n = nn
                    r = nr


def main():  # Wrapper function

    # Calls CalcPi with the given limit
    pi_digits = calcPi(int(input(
        "Enter the number of decimals to calculate to: ")))

    i = 0

    # Prints the output of calcPi generator function
    # Inserts a newline after every 40th number
    for d in pi_digits:
            print(d, end='')
            i += 1
            if i == 40:
                print("")
                i = 0

if __name__ == '__main__':
    main()

Solution 4

#https://github.com/rlingineni/PythonPractice/blob/master/piCalc/pi.py
import math 

def CalculatePi(roundVal):

  somepi = round(math.pi,roundVal);
  pi = str(somepi)
  someList = list(pi)
  return somepi;
roundTo = input('Enter the number of digits you want after the decimal for Pi: ')
try:
 roundint = int(roundTo);
 print(CalculatePi(roundint));
except:
 print("You did not enter an integer");

Solution 5

# generate pi to nth digit
# Chudnovsky algorihtm to find pi to n-th digit
# from https://en.wikipedia.org/wiki/Chudnovsky_algorithm
# https://github.com/microice333/Python-projects/blob/master/n_digit_pi.py

import decimal
def compute_pi(n):
    decimal.getcontext().prec = n + 1
    C = 426880 * decimal.Decimal(10005).sqrt()
    K = 6.
    M = 1.
    X = 1
    L = 13591409
    S = L
    for i in range(1, n):
        M = M * (K ** 3 - 16 * K) / ((i + 1) ** 3)
        L += 545140134
        X *= -262537412640768000
        S += decimal.Decimal(M * L) / X
    pi = C / S
    return pi


while True:
    n = int(input("Please type number between 0-1000: "))
    if n >= 0 and n <= 1000:
        break
print(compute_pi(n))

Solution 6

#coding:utf-8
"""
Pi = SUM k=0 to infinity 16^-k [ 4/(8k+1) - 2/(8k+4) - 1/(8k+5) - 1/(8k+6) ]
ref: https://www.math.hmc.edu/funfacts/ffiles/20010.5.shtml
https://github.com/Flowerowl/Projects/blob/master/solutions/numbers/find_pi_to_the_nth_digit.py
"""
from __future__ import division
import math
from decimal import Decimal as D
from decimal import getcontext

getcontext().prec = 400
MAX = 10000
pi = D(0)

for k in range(MAX):
    pi += D(math.pow(16, -k)) * (D(4/(8*k+1)) - D(2/(8*k+4)) - D(1/(8*k+5)) - D(1/(8*k+6)))

print('PI >>>>>>>>>>' , pi)

Solution 7

""""Find PI to the Nth Digit - 
https://bitbucket.org/desertwebdesigns/learn_python/src/master/Numbers/pi.py?fileviewer=file-view-default


Enter a number and have the program generate PI up to that many decimal places.
Keep a limit to how far the program will go."""

import math

precision = int(input("How many spaces? "))

while precision > 50:
   print("Number to large")
   precision = int(raw_input("How many spaces? "))
else:
   print('%.*f' % (precision, math.pi))

References

• https://docs.python.org/2/library/decimal.html

• https://www.w3resource.com/projects/python/python-projects-1.php