1000 digits of pi in Python

Question

I have been thinking about this issue and I can't figure it out. Perhaps you can assist me. The problem is my code isn't working to output 1000 digits of pi in the Python coding language.

Here's my code:

def make_pi():
    q, r, t, k, m, x = 1, 0, 1, 1, 3, 3
    while True:
        if 4 * q + r - t < m * t:
            yield m
            q, r, t, k, m, x = (10*q, 10*(r-m*t), t, k, (10*(3*q+r))//t - 10*m, x)
        else:
            q, r, t, k, m, x = (q*k, (2*q+r)*x, t*x, k+1, (q*(7*k+2)+r*x)//(t*x), x+2)

digits = make_pi()
pi_list = []
my_array = []
for i in range(1000):
    my_array.append(str("hello, I'm an element in an array \n" ))
big_string = "".join(my_array)

print "here is a big string:\n %s" % big_string 

I know this code can be fixed to work, but I'm not sure what to fix... The print statement saying here is a big string and the my_array.append(str("hello, im an element in an array \n)) is just a filler for now. I know how all the code is used to work, but like I said before, I can't get it to shoot out that code.

Answer 1

If you don't want to implement your own algorithm, you can use mpmath.

try:
    # import version included with old SymPy
    from sympy.mpmath import mp
except ImportError:
    # import newer version
    from mpmath import mp

mp.dps = 1000  # set number of digits
print(mp.pi)   # print pi to a thousand places

Reference

Update: Code supports older and newer installations of SymPy (see comment).*

Answer 2

Run this

def make_pi():
    q, r, t, k, m, x = 1, 0, 1, 1, 3, 3
    for j in range(1000):
        if 4 * q + r - t < m * t:
            yield m
            q, r, t, k, m, x = 10*q, 10*(r-m*t), t, k, (10*(3*q+r))//t - 10*m, x
        else:
            q, r, t, k, m, x = q*k, (2*q+r)*x, t*x, k+1, (q*(7*k+2)+r*x)//(t*x), x+2


my_array = []

for i in make_pi():
    my_array.append(str(i))

my_array = my_array[:1] + ['.'] + my_array[1:]
big_string = "".join(my_array)
print "here is a big string:\n %s" % big_string 

And read about yield operator from here: What does the "yield" keyword do?

Here is the answer:

3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679821480865132823066470938446095505822317253594081284811174502841027019385211055596446229489549303819644288109756659334461284756482337

Answer 3

The accepted answer is incorrect, as noted in comments.

The OP's code appears to be based on an implementation of Spigot's algorithm copied from here.

To fix the code per the OP's question (although I renamed the variables and functions to match what they were in the original source), one solution might be:

#!/usr/bin/env python

DIGITS = 1000

def pi_digits(x):
    """Generate x digits of Pi."""
    q,r,t,k,n,l = 1,0,1,1,3,3
    while x >= 0:
        if 4*q+r-t < x*t:
            yield n
            x -= 1
            q,r,t,k,n,l = 10*q, 10*(r-n*t), t, k, (10*(3*q + r))/t-10*n, l
        else:
            q,r,t,k,n,l = q*k, (2*q+r)*l, t*l, k+1, (q*(7*k+2)+r*l)/(t*l), l+2

digits = [str(n) for n in list(pi_digits(DIGITS))]
print("%s.%s\n" % (digits.pop(0), "".join(digits)))

Also, here is a much faster* implementation, also apparently based on Spigot's algorithm:

#!/usr/bin/env python

DIGITS = 1000

def pi_digits(x):
    """Generate x digits of Pi."""
    k,a,b,a1,b1 = 2,4,1,12,4
    while x > 0:
        p,q,k = k * k, 2 * k + 1, k + 1
        a,b,a1,b1 = a1, b1, p*a + q*a1, p*b + q*b1
        d,d1 = a/b, a1/b1
        while d == d1 and x > 0:
            yield int(d)
            x -= 1
            a,a1 = 10*(a % b), 10*(a1 % b1)
            d,d1 = a/b, a1/b1

digits = [str(n) for n in list(pi_digits(DIGITS))]
print("%s.%s\n" % (digits.pop(0), "".join(digits)))

I tested both a few times against this online Pi digit generator.

All credit to this Gist by deeplook.

* Based on testing 10,000 digits, where I got about 7 seconds compared to about 1 second.

Answer 4

For up to 1 million digits of pi use math_pi (note: I am the author of the module)

Install with pip:

pip install math-pi

In Python:

>>> import math_pi
>>> print(math_pi.pi(b=1000))
3.1415926535...

Answer 5

From Fabrice Bellard site: Pi Computation algorithm. Sorry for such a straightforward implementation. 1000 is fast enough (0.1s for me), but 10000 isn't such fast - 71s :-(

import time
from decimal import Decimal, getcontext

def compute(n):
    getcontext().prec = n
    res = Decimal(0)
    for i in range(n):
        a = Decimal(1)/(16**i)
        b = Decimal(4)/(8*i+1)
        c = Decimal(2)/(8*i+4)
        d = Decimal(1)/(8*i+5)
        e = Decimal(1)/(8*i+6)
        r = a*(b-c-d-e)
        res += r
    return res

if __name__ == "__main__":
    t1 = time.time()
    res = compute(1000)
    dt = time.time()-t1
    print(res)
    print(dt)

Answer 6

I was solved with bellow formula 5-6 years ago.

Machin-like formula

Wikipedia: https://en.wikipedia.org/wiki/Machin-like_formula

Sorry for the code quality. Variable names can be meaningless.

#-*- coding: utf-8 -*-

# Author:    Fatih Mert Doğancan
# Date:      02.12.2014

def arccot(x, u):
    sum = ussu = u // x
    n = 3
    sign = -1
    while 1:
        ussu = ussu // (x*x)
        term = ussu // n
        if not term:
            break
        sum += sign * term
        sign = -sign
        n += 2
    return sum

def pi(basamak):
    u = 10**(basamak+10)
    pi = 4 * (4*arccot(5,u) - arccot(239,u))
    return pi // 10**10

if __name__ == "__main__":
    print pi(1000) # 1000 

Answer 7

Here you can check whether your program outputs correct 1000 digits: http://spoj.com/CONSTANT

Of course you can use diff or tc as well but you'd have to copy these 1000 digits from somewhere and there you just submit your program and check whether the score is bigger than 999.

You can try to print even more digits there and thus get more points. Perhaps you'd enjoy it.

Answer 8

I'm not familiar with your algorithm. Is it an implementation of BBP?

In any case, your make_pi is a generator. Try using it in a for loop:

for digit in make_pi():
    print digit

Note that this loop is infinite: make_pi() never throws StopIteration

Answer 9

Here is a different way I found here --> Python pi calculation? to approximate python based on the Chudnovsky brothers formula for generating Pi which I have sightly modified for my program.

def pifunction():
    numberofdigits = int(input("please enter the number of digits of pi that you want to generate"))
    getcontext().prec = numberofdigits

def calc(n):
    t = Decimal(0)
    pi = Decimal(0)
    deno = Decimal(0)
    k = 0
    for k in range(n):
        t = (Decimal(-1)**k)*(math.factorial(Decimal(6)*k))*(13591409+545140134*k)
        deno = math.factorial(3*k)*(math.factorial(k)**Decimal(3))*(640320**(3*k))
        pi += Decimal(t)/Decimal(deno)
    pi = pi * Decimal(12)/Decimal(640320**Decimal(1.5))
    pi = 1/pi
    return str(pi)
print(calc(1))

I hope this helps as you can generate any number of digits of pi that you wish to generate.

Answer 10

wallis formula can get to 3.141592661439964 but a more efficient way is needed to solve this problem.

https://www.youtube.com/watch?v=EZSiQv_G9HM

and now my code

x, y, summing = 2, 3, 4

for count in range (0,100000000):
    summing *= (x/y)
    x += 2
    summing *= (x/y)
    y += 2

print (summing)

Answer 11

Does this do what you want?

i = 0;
pi_str = ""
for x in make_pi():
    pi_str += str(x)
    i += 1
    if i == 1001:
        break

print "pi= %s.%s" % (pi_str[0],pi_str[1:])