What is the fastest way to get the value of π?

The Monte Carlo method, as mentioned, applies some great concepts but it is, clearly, not the fastest, not by a long shot, not by any reasonable measure. Also, it all depends on what kind of accuracy you are looking for. The fastest π I know of is the one with the digits hard coded. Looking at Pi and Pi[PDF], there are a lot of formulae.

Here is a method that converges quickly — about 14 digits per iteration. PiFast, the current fastest application, uses this formula with the FFT. I'll just write the formula, since the code is straightforward. This formula was almost found by Ramanujan and discovered by Chudnovsky. It is actually how he calculated several billion digits of the number — so it isn't a method to disregard. The formula will overflow quickly and, since we are dividing factorials, it would be advantageous then to delay such calculations to remove terms.

where,

Below is the Brent–Salamin algorithm. Wikipedia mentions that when a and b are "close enough" then (a + b)² / 4t will be an approximation of π. I'm not sure what "close enough" means, but from my tests, one iteration got 2 digits, two got 7, and three had 15, of course this is with doubles, so it might have an error based on its representation and the true calculation could be more accurate.

let pi_2 iters =
    let rec loop_ a b t p i =
        if i = 0 then a,b,t,p
        else
            let a_n = (a +. b) /. 2.0 
            and b_n = sqrt (a*.b)
            and p_n = 2.0 *. p in
            let t_n = t -. (p *. (a -. a_n) *. (a -. a_n)) in
            loop_ a_n b_n t_n p_n (i - 1)
    in 
    let a,b,t,p = loop_ (1.0) (1.0 /. (sqrt 2.0)) (1.0/.4.0) (1.0) iters in
    (a +. b) *. (a +. b) /. (4.0 *. t)

Lastly, how about some pi golf (800 digits)? 160 characters!

int a=10000,b,c=2800,d,e,f[2801],g;main(){for(;b-c;)f[b++]=a/5;for(;d=0,g=c*2;c-=14,printf("%.4d",e+d/a),e=d%a)for(b=c;d+=f[b]*a,f[b]=d%--g,d/=g--,--b;d*=b);}

I really like this program, because it approximates π by looking at its own area.

IOCCC 1988 : westley.c

#define _ -F<00||--F-OO--;
int F=00,OO=00;main(){F_OO();printf("%1.3f\n",4.*-F/OO/OO);}F_OO()
{
            _-_-_-_
       _-_-_-_-_-_-_-_-_
    _-_-_-_-_-_-_-_-_-_-_-_
  _-_-_-_-_-_-_-_-_-_-_-_-_-_
 _-_-_-_-_-_-_-_-_-_-_-_-_-_-_
 _-_-_-_-_-_-_-_-_-_-_-_-_-_-_
_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_
_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_
_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_
_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_
 _-_-_-_-_-_-_-_-_-_-_-_-_-_-_
 _-_-_-_-_-_-_-_-_-_-_-_-_-_-_
  _-_-_-_-_-_-_-_-_-_-_-_-_-_
    _-_-_-_-_-_-_-_-_-_-_-_
        _-_-_-_-_-_-_-_
            _-_-_-_
}

Here's a general description of a technique for calculating pi that I learnt in high school.

I only share this because I think it is simple enough that anyone can remember it, indefinitely, plus it teaches you the concept of "Monte-Carlo" methods -- which are statistical methods of arriving at answers that don't immediately appear to be deducible through random processes.

Draw a square, and inscribe a quadrant (one quarter of a semi-circle) inside that square (a quadrant with radius equal to the side of the square, so it fills as much of the square as possible)

Now throw a dart at the square, and record where it lands -- that is, choose a random point anywhere inside the square. Of course, it landed inside the square, but is it inside the semi-circle? Record this fact.

Repeat this process many times -- and you will find there is a ratio of the number of points inside the semi-circle versus the total number thrown, call this ratio x.

Since the area of the square is r times r, you can deduce that the area of the semi circle is x times r times r (that is, x times r squared). Hence x times 4 will give you pi.

This is not a quick method to use. But it's a nice example of a Monte Carlo method. And if you look around, you may find that many problems otherwise outside your computational skills can be solved by such methods.

In the interests of completeness, a C++ template version, which, for an optimised build, will compute an approximation of PI at compile time, and will inline to a single value.

#include <iostream>

template<int I>
struct sign
{
    enum {value = (I % 2) == 0 ? 1 : -1};
};

template<int I, int J>
struct pi_calc
{
    inline static double value ()
    {
        return (pi_calc<I-1, J>::value () + pi_calc<I-1, J+1>::value ()) / 2.0;
    }
};

template<int J>
struct pi_calc<0, J>
{
    inline static double value ()
    {
        return (sign<J>::value * 4.0) / (2.0 * J + 1.0) + pi_calc<0, J-1>::value ();
    }
};


template<>
struct pi_calc<0, 0>
{
    inline static double value ()
    {
        return 4.0;
    }
};

template<int I>
struct pi
{
    inline static double value ()
    {
        return pi_calc<I, I>::value ();
    }
};

int main ()
{
    std::cout.precision (12);

    const double pi_value = pi<10>::value ();

    std::cout << "pi ~ " << pi_value << std::endl;

    return 0;
}

Note for I > 10, optimised builds can be slow, likewise for non-optimised runs. For 12 iterations I believe there are around 80k calls to value() (in the absence of memoisation).

There's actually a whole book dedicated (amongst other things) to fast methods for the computation of \pi: 'Pi and the AGM', by Jonathan and Peter Borwein (available on Amazon).

I studied the AGM and related algorithms quite a bit: it's quite interesting (though sometimes non-trivial).

Note that to implement most modern algorithms to compute \pi, you will need a multiprecision arithmetic library (GMP is quite a good choice, though it's been a while since I last used it).

The time-complexity of the best algorithms is in O(M(n)log(n)), where M(n) is the time-complexity for the multiplication of two n-bit integers (M(n)=O(n log(n) log(log(n))) using FFT-based algorithms, which are usually needed when computing digits of \pi, and such an algorithm is implemented in GMP).

Note that even though the mathematics behind the algorithms might not be trivial, the algorithms themselves are usually a few lines of pseudo-code, and their implementation is usually very straightforward (if you chose not to write your own multiprecision arithmetic :-) ).

The BBP formula allows you to compute the nth digit - in base 2 (or 16) - without having to even bother with the previous n-1 digits first :)

Instead of defining pi as a constant, I always use acos(-1).

This is a "classic" method, very easy to implement. This implementation in python (not the fastest language) does it:

from math import pi
from time import time


precision = 10**6 # higher value -> higher precision
                  # lower  value -> higher speed

t = time()

calc = 0
for k in xrange(0, precision):
    calc += ((-1)**k) / (2*k+1.)
calc *= 4. # this is just a little optimization

t = time()-t

print "Calculated: %.40f" % calc
print "Constant pi: %.40f" % pi
print "Difference: %.40f" % abs(calc-pi)
print "Time elapsed: %s" % repr(t)

You can find more information here.

Anyway, the fastest way to get a precise as-much-as-you-want value of pi in python is:

from gmpy import pi
print pi(3000) # the rule is the same as 
               # the precision on the previous code

Here is the piece of source for the gmpy pi method, I don't think the code is as useful as the comment in this case:

static char doc_pi[]="\
pi(n): returns pi with n bits of precision in an mpf object\n\
";

/* This function was originally from netlib, package bmp, by
 * Richard P. Brent. Paulo Cesar Pereira de Andrade converted
 * it to C and used it in his LISP interpreter.
 *
 * Original comments:
 * 
 *   sets mp pi = 3.14159... to the available precision.
 *   uses the gauss-legendre algorithm.
 *   this method requires time o(ln(t)m(t)), so it is slower
 *   than mppi if m(t) = o(t**2), but would be faster for
 *   large t if a faster multiplication algorithm were used
 *   (see comments in mpmul).
 *   for a description of the method, see - multiple-precision
 *   zero-finding and the complexity of elementary function
 *   evaluation (by r. p. brent), in analytic computational
 *   complexity (edited by j. f. traub), academic press, 1976, 151-176.
 *   rounding options not implemented, no guard digits used.
*/
static PyObject *
Pygmpy_pi(PyObject *self, PyObject *args)
{
    PympfObject *pi;
    int precision;
    mpf_t r_i2, r_i3, r_i4;
    mpf_t ix;

    ONE_ARG("pi", "i", &precision);
    if(!(pi = Pympf_new(precision))) {
        return NULL;
    }

    mpf_set_si(pi->f, 1);

    mpf_init(ix);
    mpf_set_ui(ix, 1);

    mpf_init2(r_i2, precision);

    mpf_init2(r_i3, precision);
    mpf_set_d(r_i3, 0.25);

    mpf_init2(r_i4, precision);
    mpf_set_d(r_i4, 0.5);
    mpf_sqrt(r_i4, r_i4);

    for (;;) {
        mpf_set(r_i2, pi->f);
        mpf_add(pi->f, pi->f, r_i4);
        mpf_div_ui(pi->f, pi->f, 2);
        mpf_mul(r_i4, r_i2, r_i4);
        mpf_sub(r_i2, pi->f, r_i2);
        mpf_mul(r_i2, r_i2, r_i2);
        mpf_mul(r_i2, r_i2, ix);
        mpf_sub(r_i3, r_i3, r_i2);
        mpf_sqrt(r_i4, r_i4);
        mpf_mul_ui(ix, ix, 2);
        /* Check for convergence */
        if (!(mpf_cmp_si(r_i2, 0) && 
              mpf_get_prec(r_i2) >= (unsigned)precision)) {
            mpf_mul(pi->f, pi->f, r_i4);
            mpf_div(pi->f, pi->f, r_i3);
            break;
        }
    }

    mpf_clear(ix);
    mpf_clear(r_i2);
    mpf_clear(r_i3);
    mpf_clear(r_i4);

    return (PyObject*)pi;
}

EDIT: I had some problems with cut and paste and indentation, you can find the source here.

If by fastest you mean fastest to type in the code, here's the golfscript solution:

;''6666,-2%{2+.2/@*\/10.3??2*+}*`1000<~\;

Pi is exactly 3! [Prof. Frink (Simpsons)]

Joke, but here's one in C# (.NET-Framework required).

using System;
using System.Text;

class Program {
    static void Main(string[] args) {
        int Digits = 100;

        BigNumber x = new BigNumber(Digits);
        BigNumber y = new BigNumber(Digits);
        x.ArcTan(16, 5);
        y.ArcTan(4, 239);
        x.Subtract(y);
        string pi = x.ToString();
        Console.WriteLine(pi);
    }
}

public class BigNumber {
    private UInt32[] number;
    private int size;
    private int maxDigits;

    public BigNumber(int maxDigits) {
        this.maxDigits = maxDigits;
        this.size = (int)Math.Ceiling((float)maxDigits * 0.104) + 2;
        number = new UInt32[size];
    }
    public BigNumber(int maxDigits, UInt32 intPart)
        : this(maxDigits) {
        number[0] = intPart;
        for (int i = 1; i < size; i++) {
            number[i] = 0;
        }
    }
    private void VerifySameSize(BigNumber value) {
        if (Object.ReferenceEquals(this, value))
            throw new Exception("BigNumbers cannot operate on themselves");
        if (value.size != this.size)
            throw new Exception("BigNumbers must have the same size");
    }

    public void Add(BigNumber value) {
        VerifySameSize(value);

        int index = size - 1;
        while (index >= 0 && value.number[index] == 0)
            index--;

        UInt32 carry = 0;
        while (index >= 0) {
            UInt64 result = (UInt64)number[index] +
                            value.number[index] + carry;
            number[index] = (UInt32)result;
            if (result >= 0x100000000U)
                carry = 1;
            else
                carry = 0;
            index--;
        }
    }
    public void Subtract(BigNumber value) {
        VerifySameSize(value);

        int index = size - 1;
        while (index >= 0 && value.number[index] == 0)
            index--;

        UInt32 borrow = 0;
        while (index >= 0) {
            UInt64 result = 0x100000000U + (UInt64)number[index] -
                            value.number[index] - borrow;
            number[index] = (UInt32)result;
            if (result >= 0x100000000U)
                borrow = 0;
            else
                borrow = 1;
            index--;
        }
    }
    public void Multiply(UInt32 value) {
        int index = size - 1;
        while (index >= 0 && number[index] == 0)
            index--;

        UInt32 carry = 0;
        while (index >= 0) {
            UInt64 result = (UInt64)number[index] * value + carry;
            number[index] = (UInt32)result;
            carry = (UInt32)(result >> 32);
            index--;
        }
    }
    public void Divide(UInt32 value) {
        int index = 0;
        while (index < size && number[index] == 0)
            index++;

        UInt32 carry = 0;
        while (index < size) {
            UInt64 result = number[index] + ((UInt64)carry << 32);
            number[index] = (UInt32)(result / (UInt64)value);
            carry = (UInt32)(result % (UInt64)value);
            index++;
        }
    }
    public void Assign(BigNumber value) {
        VerifySameSize(value);
        for (int i = 0; i < size; i++) {
            number[i] = value.number[i];
        }
    }

    public override string ToString() {
        BigNumber temp = new BigNumber(maxDigits);
        temp.Assign(this);

        StringBuilder sb = new StringBuilder();
        sb.Append(temp.number[0]);
        sb.Append(System.Globalization.CultureInfo.CurrentCulture.NumberFormat.CurrencyDecimalSeparator);

        int digitCount = 0;
        while (digitCount < maxDigits) {
            temp.number[0] = 0;
            temp.Multiply(100000);
            sb.AppendFormat("{0:D5}", temp.number[0]);
            digitCount += 5;
        }

        return sb.ToString();
    }
    public bool IsZero() {
        foreach (UInt32 item in number) {
            if (item != 0)
                return false;
        }
        return true;
    }

    public void ArcTan(UInt32 multiplicand, UInt32 reciprocal) {
        BigNumber X = new BigNumber(maxDigits, multiplicand);
        X.Divide(reciprocal);
        reciprocal *= reciprocal;

        this.Assign(X);

        BigNumber term = new BigNumber(maxDigits);
        UInt32 divisor = 1;
        bool subtractTerm = true;
        while (true) {
            X.Divide(reciprocal);
            term.Assign(X);
            divisor += 2;
            term.Divide(divisor);
            if (term.IsZero())
                break;

            if (subtractTerm)
                this.Subtract(term);
            else
                this.Add(term);
            subtractTerm = !subtractTerm;
        }
    }
}

Use the Machin-like formula

176 * arctan (1/57) + 28 * arctan (1/239) - 48 * arctan (1/682) + 96 * arctan(1/12943) 

[; \left( 176 \arctan \frac{1}{57} + 28 \arctan \frac{1}{239} - 48 \arctan \frac{1}{682} + 96 \arctan \frac{1}{12943}\right) ;], for you TeX the World people.

Implemented in Scheme, for instance:

(+ (- (+ (* 176 (atan (/ 1 57))) (* 28 (atan (/ 1 239)))) (* 48 (atan (/ 1 682)))) (* 96 (atan (/ 1 12943))))

With doubles:

4.0 * (4.0 * Math.Atan(0.2) - Math.Atan(1.0 / 239.0))

This will be accurate up to 14 decimal places, enough to fill a double (the inaccuracy is probably because the rest of the decimals in the arc tangents are truncated).

Also Seth, it's 3.141592653589793238463, not 64.

Calculate PI at compile-time with D.

( Copied from DSource.org )

/** Calculate pi at compile time
 *
 * Compile with dmd -c pi.d
 */
module calcpi;

import meta.math;
import meta.conv;

/** real evaluateSeries!(real x, real metafunction!(real y, int n) term)
 *
 * Evaluate a power series at compile time.
 *
 * Given a metafunction of the form
 *  real term!(real y, int n),
 * which gives the nth term of a convergent series at the point y
 * (where the first term is n==1), and a real number x,
 * this metafunction calculates the infinite sum at the point x
 * by adding terms until the sum doesn't change any more.
 */
template evaluateSeries(real x, alias term, int n=1, real sumsofar=0.0)
{
  static if (n>1 && sumsofar == sumsofar + term!(x, n+1)) {
     const real evaluateSeries = sumsofar;
  } else {
     const real evaluateSeries = evaluateSeries!(x, term, n+1, sumsofar + term!(x, n));
  }
}

/*** Calculate atan(x) at compile time.
 *
 * Uses the Maclaurin formula
 *  atan(z) = z - z^3/3 + Z^5/5 - Z^7/7 + ...
 */
template atan(real z)
{
    const real atan = evaluateSeries!(z, atanTerm);
}

template atanTerm(real x, int n)
{
    const real atanTerm =  (n & 1 ? 1 : -1) * pow!(x, 2*n-1)/(2*n-1);
}

/// Machin's formula for pi
/// pi/4 = 4 atan(1/5) - atan(1/239).
pragma(msg, "PI = " ~ fcvt!(4.0 * (4*atan!(1/5.0) - atan!(1/239.0))) );

This version (in Delphi) is nothing special, but it is at least faster than the version Nick Hodge posted on his blog :). On my machine, it takes about 16 seconds to do a billion iterations, giving a value of 3.1415926525879 (the accurate part is in bold).

program calcpi;

{$APPTYPE CONSOLE}

uses
  SysUtils;

var
  start, finish: TDateTime;

function CalculatePi(iterations: integer): double;
var
  numerator, denominator, i: integer;
  sum: double;
begin
  {
  PI may be approximated with this formula:
  4 * (1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 .......)
  //}
  numerator := 1;
  denominator := 1;
  sum := 0;
  for i := 1 to iterations do begin
    sum := sum + (numerator/denominator);
    denominator := denominator + 2;
    numerator := -numerator;
  end;
  Result := 4 * sum;
end;

begin
  try
    start := Now;
    WriteLn(FloatToStr(CalculatePi(StrToInt(ParamStr(1)))));
    finish := Now;
    WriteLn('Seconds:' + FormatDateTime('hh:mm:ss.zz',finish-start));
  except
    on E:Exception do
      Writeln(E.Classname, ': ', E.Message);
  end;
end.

Back in the old days, with small word sizes and slow or non-existent floating-point operations, we used to do stuff like this:

/* Return approximation of n * PI; n is integer */
#define pi_times(n) (((n) * 22) / 7)

For applications that don't require a lot of precision (video games, for example), this is very fast and is accurate enough.

If you want to compute an approximation of the value of π (for some reason), you should try a binary extraction algorithm. Bellard's improvement of BBP gives does PI in O(N^2).

If you want to obtain an approximation of the value of π to do calculations, then:

PI = 3.141592654

Granted, that's only an approximation, and not entirely accurate. It's off by a little more than 0.00000000004102. (four ten-trillionths, about ⁴/_{10,000,000,000}).

If you want to do math with π, then get yourself a pencil and paper or a computer algebra package, and use π's exact value, π.

If you really want a formula, this one is fun:

π = -i ln(-1)

The Chudnovsky algorithm is pretty fast if you don't mind performing a square root and a couple inverses. It converges to double precision in just 2 iterations.

/*
    Chudnovsky algorithm for computing PI
*/

#include <iostream>
#include <cmath>
using namespace std;

double calc_PI(int K=2) {

    static const int A = 545140134;
    static const int B = 13591409;
    static const int D = 640320;

    const double ID3 = 1./ (double(D)*double(D)*double(D));

    double sum = 0.;
    double b   = sqrt(ID3);
    long long int p = 1;
    long long int a = B;

    sum += double(p) * double(a)* b;

    // 2 iterations enough for double convergence
    for (int k=1; k<K; ++k) {
        // A*k + B
        a += A;
        // update denominator
        b *= ID3;
        // p = (-1)^k 6k! / 3k! k!^3
        p *= (6*k)*(6*k-1)*(6*k-2)*(6*k-3)*(6*k-4)*(6*k-5);
        p /= (3*k)*(3*k-1)*(3*k-2) * k*k*k;
        p = -p;

        sum += double(p) * double(a)* b;
    }

    return 1./(12*sum);
}

int main() {

    cout.precision(16);
    cout.setf(ios::fixed);

    for (int k=1; k<=5; ++k) cout << "k = " << k << "   PI = " << calc_PI(k) << endl;

    return 0;
}

Results:

k = 1   PI = 3.1415926535897341
k = 2   PI = 3.1415926535897931
k = 3   PI = 3.1415926535897931
k = 4   PI = 3.1415926535897931
k = 5   PI = 3.1415926535897931

Basically the C version of paperclip optimizer's answer, and much more simpilified:

#include <stdio.h>
#include <math.h>

double calc_PI(int K) {
    static const int A = 545140134;
    static const int B = 13591409;
    static const int D = 640320;
    const double ID3 = 1.0 / ((double) D * (double) D * (double) D);
    double sum = 0.0;
    double b = sqrt(ID3);
    long long int p = 1;
    long long int a = B;
    sum += (double) p * (double) a * b;
    for (int k = 1; k < K; ++k) {
        a += A;
        b *= ID3;
        p *= (6 * k) * (6 * k - 1) * (6 * k - 2) * (6 * k - 3) * (6 * k - 4) * (6 * k - 5);
        p /= (3 * k) * (3 * k - 1) * (3 * k - 2) * k * k * k;
        p = -p;
        sum += (double) p * (double) a * b;
    }
    return 1.0 / (12 * sum);
}

int main() {
    for (int k = 1; k <= 5; ++k) {
        printf("k = %i, PI = %.16f\n", k, calc_PI(k));
    }
}

But for more simplification, this algorithm takes Chudnovsky's formula, which I can fully simplify if you don't really understand the code.

Summary: We will get a number from 1 to 5 and add it in to a function we will use to get PI. Then 3 numbers are given to you: 545140134 (A), 13591409 (B), 640320 (D). Then we will use D as a double multiplying itself 3 times into another double (ID3). We will then take the square root of ID3 into another double (b) and assign 2 numbers: 1 (p), the value of B (a). Take note that C is case-insensitive. Then a double (sum) will be created by multiplying the value's of p, a and b, all in doubles. Then a loop up until the number given for the function will start and add up A's value to a, b's value gets multiplied by ID3, p's value will be multiplied by multiple values that I hope you can understand and also gets divided by multiple values as well. The sum will add up by p, a and b once again and the loop will repeat until the value of the loop's number is greater or equal to 5. Later, the sum is multiplied by 12 and returned by the function giving us the result of PI.

Okay, that was long, but I guess you will understand it...

I think the value of pi is the ratio between the circumference and radius of the circle.

It can be simply achieved by a regular math calculation

Better Approach

To get the output of standard constants like pi or the standard concepts, we should first go with the builtins methods available in the language that you are using. It will return a value in the fastest and best way. I am using python to run the fastest way to get the value of pi.

• pi variable of the math library. The math library stores the variable pi as a constant.

math_pi.py

import math
print math.pi

Run the script with time utility of linux /usr/bin/time -v python math_pi.py

Output:

Command being timed: "python math_pi.py"
User time (seconds): 0.01
System time (seconds): 0.01
Percent of CPU this job got: 91%
Elapsed (wall clock) time (h:mm:ss or m:ss): 0:00.03

• Use arc cos method of math

acos_pi.py

import math
print math.acos(-1)

Run the script with time utility of linux /usr/bin/time -v python acos_pi.py

Output:

Command being timed: "python acos_pi.py"
User time (seconds): 0.02
System time (seconds): 0.01
Percent of CPU this job got: 94%
Elapsed (wall clock) time (h:mm:ss or m:ss): 0:00.03

• use BBP formula

bbp_pi.py

from decimal import Decimal, getcontext
getcontext().prec=100
print sum(1/Decimal(16)**k * 
          (Decimal(4)/(8*k+1) - 
           Decimal(2)/(8*k+4) - 
           Decimal(1)/(8*k+5) -
           Decimal(1)/(8*k+6)) for k in range(100))

Run the script with time utility of linux /usr/bin/time -v python bbp_pi.py

Output:

Command being timed: "python c.py"
User time (seconds): 0.05
System time (seconds): 0.01
Percent of CPU this job got: 98%
Elapsed (wall clock) time (h:mm:ss or m:ss): 0:00.06

So the best way is to use builtin methods provided by the language because they are the fastest and best to get the output. In python use math.pi

I tried several ways; Manchin, AGM and Chudnovsky Bros. Chudnovsky with Binary Split was the fastest:
My github : https://github.com/Overboard-code/Pi-Pourri

My Binary Split Chudnovsky is about twice the speed of the builtin gmpy2.const_pi()

I would greatly appreciate performance tips as well. I am not sure my code is very well written. 100% accurate but maybe could be faster?

I tried gmpy2.const_pi() to 100 million digits and it took 300 seconds vs. 150 seconds for the Chudnovsky on the same machine. pi.txt and pi2.txt were the same.

I got to a billion digits on my old i7 16GB laptop in about an hour.

Here is a snippet of the fastest of the 11 methods I tried:

class PiChudnovsky:
    """Version of Chudnovsky Bros using Binary Splitting 
        So far this is the winner for fastest time to a million digits on my older intel i7
    """
    A = mpz(13591409)
    B = mpz(545140134)
    C = mpz(640320)
    D = mpz(426880)
    E = mpz(10005)
    C3_24  = pow(C, mpz(3)) // mpz(24)
    #DIGITS_PER_TERM = math.log(53360 ** 3) / math.log(10)  #=> 14.181647462725476
    DIGITS_PER_TERM = 14.181647462725476
    MMILL = mpz(1000000)

    def __init__(self,ndigits):
        """ Initialization
        :param int ndigits: digits of PI computation
        """
        self.ndigits = ndigits
        self.n      = mpz(self.ndigits // self.DIGITS_PER_TERM + 1)
        self.prec   = mpz((self.ndigits + 1) * LOG2_10)
        self.one_sq = pow(mpz(10),mpz(2 * ndigits))
        self.sqrt_c = isqrt(self.E * self.one_sq)
        self.iters  = mpz(0)
        self.start_time = 0

    def compute(self):
        """ Computation """
        try:
            self.start_time = time.time()
            logging.debug("Starting {} formula to {:,} decimal places"
                .format(name,ndigits) )
            __, q, t = self.__bs(mpz(0), self.n)  # p is just for recursion
            pi = (q * self.D * self.sqrt_c) // t
            logging.debug('{} calulation Done! {:,} iterations and {:.2f} seconds.'
                .format( name, int(self.iters),time.time() - self.start_time))
            get_context().precision= int((self.ndigits+10) * LOG2_10)
            pi_s = pi.digits() # digits() gmpy2 creates a string 
            pi_o = pi_s[:1] + "." + pi_s[1:]
            return pi_o,int(self.iters),time.time() - self.start_time
        except Exception as e:
            print (e.message, e.args)
            raise

    def __bs(self, a, b):
        """ PQT computation by BSA(= Binary Splitting Algorithm)
        :param int a: positive integer
        :param int b: positive integer
        :return list [int p_ab, int q_ab, int t_ab]
        """
        try:
            self.iters += mpz(1)
            if self.iters % self.MMILL  == mpz(0):
                logging.debug('Chudnovsky ... {:,} iterations and {:.2f} seconds.'
                    .format( int(self.iters),time.time() - self.start_time))
            if a + mpz(1) == b:
                if a == mpz(0):
                    p_ab = q_ab = mpz(1)
                else:
                    p_ab = mpz((mpz(6) * a - mpz(5)) * (mpz(2) * a - mpz(1)) * (mpz(6) * a - mpz(1)))
                    q_ab = pow(a,mpz(3)) * self.C3_24
                t_ab = p_ab * (self.A + self.B * a)
                if a & 1:
                    t_ab *= mpz(-1)
            else:
                m = (a + b) // mpz(2)
                p_am, q_am, t_am = self.__bs(a, m)
                p_mb, q_mb, t_mb = self.__bs(m, b)
                p_ab = p_am * p_mb
                q_ab = q_am * q_mb
                t_ab = q_mb * t_am + p_am * t_mb
            return [p_ab, q_ab, t_ab]
        except Exception as e:
            print (e.message, e.args)
            raise

Here is the output of 100,000,000 digits less than 3 minutes:

python pi-pourri.py -a 10 -d 100,000,000 
[INFO] 2022-09-19 14:00:09,970 <module>: MainProcess Computing π to 100,000,000 digits.
[DEBUG] 2022-09-19 14:00:19,644 compute: MainProcess Starting   Chudnovsky brothers  1988 
    π = (Q(0, N) / 12T(0, N) + 12AQ(0, N))**(C**(3/2))
 formula to 100,000,000 decimal places
[DEBUG] 2022-09-19 14:00:24,787 __bs: MainProcess Chudnovsky ... 1,000,000 iterations and 5.14 seconds.
[DEBUG] 2022-09-19 14:00:31,436 __bs: MainProcess Chudnovsky ... 2,000,000 iterations and 11.79 seconds.
[DEBUG] 2022-09-19 14:00:37,021 __bs: MainProcess Chudnovsky ... 3,000,000 iterations and 17.38 seconds.
[DEBUG] 2022-09-19 14:00:46,985 __bs: MainProcess Chudnovsky ... 4,000,000 iterations and 27.34 seconds.
[DEBUG] 2022-09-19 14:00:52,687 __bs: MainProcess Chudnovsky ... 5,000,000 iterations and 33.04 seconds.
[DEBUG] 2022-09-19 14:00:59,797 __bs: MainProcess Chudnovsky ... 6,000,000 iterations and 40.15 seconds.
[DEBUG] 2022-09-19 14:01:05,536 __bs: MainProcess Chudnovsky ... 7,000,000 iterations and 45.89 seconds.
[DEBUG] 2022-09-19 14:01:22,491 __bs: MainProcess Chudnovsky ... 8,000,000 iterations and 62.85 seconds.
[DEBUG] 2022-09-19 14:01:29,731 __bs: MainProcess Chudnovsky ... 9,000,000 iterations and 70.09 seconds.
[DEBUG] 2022-09-19 14:01:35,720 __bs: MainProcess Chudnovsky ... 10,000,000 iterations and 76.08 seconds.
[DEBUG] 2022-09-19 14:01:45,824 __bs: MainProcess Chudnovsky ... 11,000,000 iterations and 86.18 seconds.
[DEBUG] 2022-09-19 14:01:52,042 __bs: MainProcess Chudnovsky ... 12,000,000 iterations and 92.40 seconds.
[DEBUG] 2022-09-19 14:01:59,230 __bs: MainProcess Chudnovsky ... 13,000,000 iterations and 99.59 seconds.
[DEBUG] 2022-09-19 14:02:05,309 __bs: MainProcess Chudnovsky ... 14,000,000 iterations and 105.67 seconds.
[DEBUG] 2022-09-19 14:02:48,926 compute: MainProcess    Chudnovsky brothers  1988 
    π = (Q(0, N) / 12T(0, N) + 12AQ(0, N))**(C**(3/2))
 calulation Done! 14,102,733 iterations and 149.28 seconds.
[INFO] 2022-09-19 14:03:16,038 <module>: MainProcess Last 5 digits of π were 51592 as expected at offset 99,999,995
[INFO] 2022-09-19 14:03:16,587 <module>: MainProcess Calculated π to 100,000,000 digits using a formula of:
 10     Chudnovsky brothers  1988 
    π = (Q(0, N) / 12T(0, N) + 12AQ(0, N))**(C**(3/2))
 
[DEBUG] 2022-09-19 14:03:16,587 <module>: MainProcess Wrote 100,000,000 digits of π to file pi.txt in 0:00:00.548809
[INFO] 2022-09-19 14:03:16,587 <module>: MainProcess Calculation took 14,102,733 iterations and 0:02:56.375900.