Chudnovsky algorithm

The Chudnovsky algorithm is a fast method for calculating the digits of $π$ . It was published by the Chudnovsky brothers in 1988,^[1] and was used in the world record calculations of 2.7 trillion digits of $π$ in December 2009,^[2] 10 trillion digits in October 2011,^[3]^[4], 22.4 trillion digits of $π$ in November 2016^[5], and 31.4 trillion digits in September 2018–January 2019.^[6]

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

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

A detailed proof of this formula can be found here:^[7]

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

{\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 M_k, the linear term L_k, and the exponential term X_k) that make up the series and $π$ equals the constant C divided by the sum of the series, as below:

\pi =C\left(\sum _{k=0}^{\infty }{\frac {M_{k}\cdot L_{k}}{X_{k}}}\right)_{k}^{-1}

, where:

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 M_k, L_k, and X_k satisfy the following recurrences and can be computed as such:

{\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 M_k can be further optimized by introducing an additional term K_k as follows:

{\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

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

and

640320^{3}=262537412640768000

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

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 $O(n\log(n)^{3})$ .^[8]

Example: Python Implementation

^{[original research?]}

$π$ 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)")

References

^ Chudnovsky, David; Chudnovsky, Gregory (1988), Approximation and complex multiplication according to ramanujan, Ramanujan revisited: proceedings of the centenary conference
^ ^a ^b ^c Baruah, Nayandeep Deka; Berndt, Bruce C.; Chan, Heng Huat (2009), "Ramanujan's series for 1/ $π$ : a survey", American Mathematical Monthly, 116 (7): 567–587, doi:10.4169/193009709X458555, JSTOR 40391165, MR 2549375
^ Yee, Alexander; Kondo, Shigeru (2011), 10 Trillion Digits of Pi: A Case Study of summing Hypergeometric Series to high precision on Multicore Systems, Technical Report, Computer Science Department, University of Illinois, hdl:2142/28348
^ Aron, Jacob (March 14, 2012), "Constants clash on pi day", New Scientist
^ "22.4 Trillion Digits of Pi". www.numberworld.org.
^ "Google Cloud Topples the Pi Record". www.numberworld.org/.
^ Milla, Lorenz (2018), A detailed proof of the Chudnovsky formula with means of basic complex analysis, arXiv:1809.00533
^ "y-cruncher - Formulas". www.numberworld.org. Retrieved 2018-02-25.

[1] Chudnovsky, David; Chudnovsky, Gregory (1988), Approximation and complex multiplication according to ramanujan, Ramanujan revisited: proceedings of the centenary conference

[baruah-2] Baruah, Nayandeep Deka; Berndt, Bruce C.; Chan, Heng Huat (2009), "Ramanujan's series for 1/ $π$ : a survey", American Mathematical Monthly, 116 (7): 567–587, doi:10.4169/193009709X458555, JSTOR 40391165, MR 2549375

[3] Yee, Alexander; Kondo, Shigeru (2011), 10 Trillion Digits of Pi: A Case Study of summing Hypergeometric Series to high precision on Multicore Systems, Technical Report, Computer Science Department, University of Illinois, hdl:2142/28348

[4] Aron, Jacob (March 14, 2012), "Constants clash on pi day", New Scientist

[5] "22.4 Trillion Digits of Pi". www.numberworld.org.

[6] "Google Cloud Topples the Pi Record". www.numberworld.org/.

[7] Milla, Lorenz (2018), A detailed proof of the Chudnovsky formula with means of basic complex analysis, arXiv:1809.00533

[8] "y-cruncher - Formulas". www.numberworld.org. Retrieved 2018-02-25.

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Chudnovsky algorithm

Example: Python Implementation

See also

References