Chudnovsky algorithm

The Chudnovsky algorithm is a fast method for calculating the digits of π, based on Ramanujan's π formulae. Published by the Chudnovsky brothers in 1988,^[1] it was used to calculate π to a billion decimal places.^[2]

It was used in the world record calculations of 2.7 trillion digits of π in December 2009,^[3] 10 trillion digits in October 2011,^[4][5] 22.4 trillion digits in November 2016,^[6] 31.4 trillion digits in September 2018–January 2019,^[7] 50 trillion digits on January 29, 2020,^[8] 62.8 trillion digits on August 14, 2021,^[9] 100 trillion digits on March 21, 2022,^[10] 105 trillion digits on March 14, 2024,^[11] and 202 trillion digits on June 28, 2024.^[12] Recently, the record was broken yet again on April 2nd 2025 with 300 trillion digits of pi.^[13][14] This was done through the usage of the algorithm on y-cruncher.

Algorithm

The algorithm is based on the negated Heegner number ${\displaystyle d=-163}$, the j-function ${\displaystyle j\left({\tfrac {1+i{\sqrt {-163}}}{2}}\right)=-640320^{3}}$, and on the following rapidly convergent generalized hypergeometric series:^[15]$${\displaystyle {\frac {1}{\pi }}=12\sum _{k=0}^{\infty }{\frac {(-1)^{k}(6k)!(545140134k+13591409)}{(3k)!(k!)^{3}(640320)^{3k+3/2}}}}$$A detailed proof of this formula can be found here: ^[16]

This identity is similar to some of Ramanujan's formulas involving π,^[15] and is an example of a Ramanujan–Sato series.

The time complexity of the algorithm is ${\displaystyle O\left(n(\log n)^{3}\right)}$.^[17]

Optimizations

The optimization technique used for the world record computations is called binary splitting.^[18]

Binary splitting

A factor of ${\textstyle 1/{640320^{3/2}}}$ can be taken out of the sum and simplified to$${\displaystyle {\frac {1}{\pi }}={\frac {1}{426880{\sqrt {10005}}}}\sum _{k=0}^{\infty }{\frac {(-1)^{k}(6k)!(545140134k+13591409)}{(3k)!(k!)^{3}(640320)^{3k}}}}$$

Let ${\displaystyle f(n)={\frac {(-1)^{n}(6n)!}{(3n)!(n!)^{3}(640320)^{3n}}}}$, and substitute that into the sum.$${\displaystyle {\frac {1}{\pi }}={\frac {1}{426880{\sqrt {10005}}}}\sum _{k=0}^{\infty }{f(k)\cdot (545140134k+13591409)}}$$

${\displaystyle {\frac {f(n)}{f(n-1)}}}$ can be simplified to ${\displaystyle {\frac {-(6n-1)(2n-1)(6n-5)}{10939058860032000n^{3}}}}$, so$${\displaystyle f(n)=f(n-1)\cdot {\frac {-(6n-1)(2n-1)(6n-5)}{10939058860032000n^{3}}}}$$

${\displaystyle f(0)=1}$ from the original definition of ${\displaystyle f}$, so$${\displaystyle f(n)=\prod _{j=1}^{n}{\frac {-(6j-1)(2j-1)(6j-5)}{10939058860032000j^{3}}}}$$

This definition of ${\displaystyle f}$ is not defined for ${\displaystyle n=0}$, so compute the first term of the sum and use the new definition of ${\displaystyle f}$$${\displaystyle {\frac {1}{\pi }}={\frac {1}{426880{\sqrt {10005}}}}{\Bigg (}13591409+\sum _{k=1}^{\infty }{{\Bigg (}\prod _{j=1}^{k}{\frac {-(6j-1)(2j-1)(6j-5)}{10939058860032000j^{3}}}{\Bigg )}\cdot (545140134k+13591409)}{\Bigg )}}$$

Let ${\displaystyle P(a,b)=\prod _{j=a}^{b-1}{-(6j-1)(2j-1)(6j-5)}}$ and ${\displaystyle Q(a,b)=\prod _{j=a}^{b-1}{10939058860032000j^{3}}}$, so$${\displaystyle {\frac {1}{\pi }}={\frac {1}{426880{\sqrt {10005}}}}{\Bigg (}13591409+\sum _{k=1}^{\infty }{{\frac {P(1,k+1)}{Q(1,k+1)}}\cdot (545140134k+13591409)}{\Bigg )}}$$

Let ${\displaystyle S(a,b)=\sum _{k=a}^{b-1}{{\frac {P(a,k+1)}{Q(a,k+1)}}\cdot (545140134k+13591409)}}$ and ${\displaystyle R(a,b)=Q(a,b)\cdot S(a,b)}$$${\displaystyle \pi ={\frac {426880{\sqrt {10005}}}{13591409+S(1,\infty )}}}$$

${\displaystyle S(1,\infty )}$ can never be computed, so instead compute ${\displaystyle S(1,n)}$ and as ${\displaystyle n}$ approaches ${\displaystyle \infty }$, the ${\displaystyle \pi }$ approximation will get better.$${\displaystyle \pi =\lim _{n\to \infty }{\frac {426880{\sqrt {10005}}}{13591409+S(1,n)}}}$$

From the original definition of ${\displaystyle R}$, ${\displaystyle S(a,b)={\frac {R(a,b)}{Q(a,b)}}}$$${\displaystyle \pi =\lim _{n\to \infty }{\frac {426880{\sqrt {10005}}\cdot Q(1,n)}{13591409Q(1,n)+R(1,n)}}}$$

Recursively computing the functions

Consider a value ${\displaystyle m}$ such that ${\displaystyle a<m<b}$

• ${\displaystyle P(a,b)=P(a,m)\cdot P(m,b)}$

• ${\displaystyle Q(a,b)=Q(a,m)\cdot Q(m,b)}$
• ${\displaystyle S(a,b)=S(a,m)+{\frac {P(a,m)}{Q(a,m)}}S(m,b)}$
• ${\displaystyle R(a,b)=Q(m,b)R(a,m)+P(a,m)R(m,b)}$

Base case for recursion

Consider ${\displaystyle b=a+1}$

• ${\displaystyle P(a,a+1)=-(6a-1)(2a-1)(6a-5)}$
• ${\displaystyle Q(a,a+1)=10939058860032000a^{3}}$
• ${\displaystyle S(a,a+1)={\frac {P(a,a+1)}{Q(a,a+1)}}\cdot (545140134a+13591409)}$
• ${\displaystyle R(a,a+1)=P(a,a+1)\cdot (545140134a+13591409)}$

Python code

# This code is a basic implementation of the Chudnovsky algorithm with binary
# splitting optimizations. It is not meant to be maximally efficient.

from decimal import Decimal, Context, setcontext

setcontext(Context(prec=28))  # increase `prec` for higher precision


def binary_split(a, b):
    if b == a + 1:
        Pab = -(6*a - 1)*(2*a - 1)*(6*a - 5)
        Qab = 10939058860032000 * a**3
        Rab = Pab * (545140134*a + 13591409)

        return Pab, Qab, Rab

    m = (a + b) // 2
    Pam, Qam, Ram = binary_split(a, m)
    Pmb, Qmb, Rmb = binary_split(m, b)

    Pab = Pam * Pmb
    Qab = Qam * Qmb
    Rab = Qmb * Ram + Pam * Rmb

    return Pab, Qab, Rab


def chudnovsky(n):
    _, Q1n, R1n = binary_split(1, n)
    return (426880 * Decimal(10005).sqrt() * Q1n) / (13591409*Q1n + R1n)


# increase argument of `chudnovsky` for higher accuracy
print(chudnovsky(2))  # 3.141592653589793238462643384

Notes

${\displaystyle e^{\pi {\sqrt {163}}}\approx 640320^{3}+743.99999999999925\dots }$

${\displaystyle 640320^{3}/24=10939058860032000}$

${\displaystyle 545140134=163\cdot 127\cdot 19\cdot 11\cdot 7\cdot 3^{2}\cdot 2}$

${\displaystyle 13591409=13\cdot 1045493}$

See also

• Mathematics portal

• Ramanujan–Sato series
• Bailey–Borwein–Plouffe formula
• Borwein's algorithm
• Approximations of π

External links

• How is π calculated to trillions of digits?