# Chudnovsky algorithm

The Chudnovsky algorithm is a fast method for calculating the digits of π. It was published by the Chudnovsky brothers in 1989,[1] and was used in the world record calculations of 2.7 trillion digits of π in December 2009,[2] 5 trillion digits in August 2010,[3] 10 trillion digits in October 2011,[4][5] 12.1 trillion digits in December 2013[6] and 22.4 trillion digits of π in November 2016.[7]

The algorithm is based on the negated Heegner number ${\displaystyle d=-163}$, the j-function ${\displaystyle \scriptstyle j\left({\frac {1+{\sqrt {-163}}}{2}}\right)=-640320^{3}}$, and on the following rapidly convergent generalized hypergeometric series:[2]

${\displaystyle {\frac {1}{\pi }}=12\sum _{k=0}^{\infty }{\frac {(-1)^{k}(6k)!(545140134k+13591409)}{(3k)!(k!)^{3}\left(640320\right)^{3k+3/2}}}}$

For a high performance iterative implementation, this can be simplified to

${\displaystyle {\frac {(640320)^{3/2}}{12\pi }}={\frac {426880{\sqrt {10005}}}{\pi }}=\sum _{k=0}^{\infty }{\frac {(6k)!(545140134k+13591409)}{(3k)!(k!)^{3}\left(-262537412640768000\right)^{k}}}}$

There are 3 big integer terms (the multinomial term Mk, the linear term Lk, and the exponential term Xk) that make up the series and π equals the constant C divided by the sum of the series, as below:

${\displaystyle \pi =C\left(\sum _{k=0}^{\infty }{\frac {M_{k}\cdot L_{k}}{X_{k}}}\right)^{-1}}$, where:
${\displaystyle C=426880{\sqrt {10005}},\quad \quad M_{k}={\frac {(6k)!}{(3k)!(k!)^{3}}},\quad \quad L_{k}=545140134k+13591409,\quad \quad X_{k}=(-262537412640768000)^{k}}$

The terms Mk, Lk, and Xk satisfy the following recurrences and can be computed as such:

{\displaystyle {\begin{alignedat}{4}L_{k+1}&=L_{k}+545140134\,\,&&{\textrm {where}}\,\,L_{0}&&=13591409\\[4pt]X_{k+1}&=X_{k}\cdot (-262537412640768000)&&{\textrm {where}}\,\,X_{0}&&=1\\[4pt]M_{k+1}&=M_{k}\cdot \left({\frac {(12k+2)(12k+6)(12k+10)}{(k+1)^{3}}}\right)\,\,&&{\textrm {where}}\,\,M_{0}&&=1\\[4pt]\end{alignedat}}}

The computation of Mk can be further optimized by introducing an additional term Kk as follows:

{\displaystyle {\begin{alignedat}{4}K_{k+1}&=K_{k}+12\,\,&&{\textrm {where}}\,\,K_{0}&&=6\\[4pt]M_{k+1}&=M_{k}\cdot \left({\frac {K_{k}^{3}-16K_{k}}{(k+1)^{3}}}\right)\,\,&&{\textrm {where}}\,\,M_{0}&&=1\\[12pt]\end{alignedat}}}

Note that

${\displaystyle e^{\pi {\sqrt {163}}}\approx 640320^{3}+743.99999999999925\dots }$ and
${\displaystyle 640320^{3}=262537412640768000}$
${\displaystyle 545140134=163\cdot 127\cdot 19\cdot 11\cdot 7\cdot 3^{2}\cdot 2}$
${\displaystyle 13591409=13\cdot 1045493}$

This identity is similar to some of Ramanujan's formulas involving π,[2] and is an example of a Ramanujan–Sato series.

The time complexity of the algorithm is ${\displaystyle O(n\log(n)^{3})}$.[8]

## Example: Python Implementation

π can be computed to any precision using the above algorithm in any environment which supports arbitrary-precision arithmetic. As an example, here is a Python implementation:

from decimal import Decimal as Dec, getcontext as gc

def PI(maxK=70, prec=1008, disp=1007): # parameter defaults chosen to gain 1000+ digits within a few seconds
gc().prec = prec
K, M, L, X, S = 6, 1, 13591409, 1, 13591409
for k in range(1, maxK+1):
M = (K**3 - 16*K) * M // k**3
L += 545140134
X *= -262537412640768000
S += Dec(M * L) / X
K += 12
pi = 426880 * Dec(10005).sqrt() / S
pi = Dec(str(pi)[:disp]) # drop few digits of precision for accuracy
print("PI(maxK=%d iterations, gc().prec=%d, disp=%d digits) =\n%s" % (maxK, prec, disp, pi))
return pi

Pi = PI()
print("\nFor greater precision and more digits (takes a few extra seconds) - Try")
print("Pi = PI(317,4501,4500)")
print("Pi = PI(353,5022,5020)")