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Write a program to calculate the first 500 digits of pi, meeting the rules below:

  • It must be less than 500 characters in length.
  • It cannot include "pi", "math.pi" or similar pi constants, nor may it call a library function to calculate pi.
  • It may not use the digits "3", "1" and "4" consecutively.
  • It must execute in a reasonable time (under 1 minute) on a modern computer.

The shortest program wins.

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  • \$\begingroup\$ To check if your digits are correct: eveandersson.com/pi/digits \$\endgroup\$ – Nellius Feb 4 '11 at 15:07
  • \$\begingroup\$ Are we allowed to print more than 500 digits with loss of accuracy after first 500? \$\endgroup\$ – Alexandru Feb 4 '11 at 15:27
  • \$\begingroup\$ @Alexandru, I suppose so but I would prefer to see it truncated. \$\endgroup\$ – Thomas O Feb 4 '11 at 17:16
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    \$\begingroup\$ Can we use an HTTP library to download a "digits of pi" website? ;-) \$\endgroup\$ – dan04 Feb 17 '11 at 5:47
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    \$\begingroup\$ Came here hoping to get something nice and concise for generating arbitrary length approximations of pi in python... unfortunately @Soulman's python solution is apparently tuned for 500 digits; replacing 500 with 1000 gives an incorrect answer. I wonder if there is a good way of phrasing an alternative challenge that would produce a nice short function that is generally useful for generating an arbitrary number of digits? \$\endgroup\$ – Don Hatch Apr 28 '16 at 8:21

11 Answers 11

Active Oldest Votes
11
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Golfscript - 29 chars

6666,-2%{2+.2/@*\/9)499?2*+}*

I will post analysis later

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  • 6
    \$\begingroup\$ Could you explain how this works? \$\endgroup\$ – Thomas O Feb 4 '11 at 13:48
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    \$\begingroup\$ "I will post analysis later". (waits for 3 years).... \$\endgroup\$ – Justin Feb 26 '14 at 3:40
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    \$\begingroup\$ "I will post analysis later" *waits for more than 6 years* \$\endgroup\$ – Erik the Outgolfer Apr 24 '17 at 10:42
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    \$\begingroup\$ "I will post analysis later" (waits for 8 years) \$\endgroup\$ – Lyxal Oct 4 '19 at 6:02
  • 2
    \$\begingroup\$ "I will post analysis later" (waits for 9 years) \$\endgroup\$ – Srivaths Feb 18 at 15:11
8
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Mathematica (34 chars): (without "cheating" with trig)

N[2Integrate[[1-x^2]^.5,-1,1],500]

So, to explain the magic here:
Integrate[function, lower, upper] gives you the area under the curve "function" from "lower" to "upper". In this case, that function is [1-x^2]^.5, which is a formula that describes the top half of a circle with radius 1. Because the circle has a radius of 1, it does not exist for values of x lower than -1 or higher than 1. Therefore, we are finding the area of half of a circle. When we multiply by 2, then we get the area inside of a circle of radius 1, which is equal to pi.

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  • \$\begingroup\$ Perhaps you should insert, in your answer, an explanation of why this works (for them non-math folks). \$\endgroup\$ – Justin Feb 26 '14 at 3:31
  • \$\begingroup\$ wonderful idea. I will see to it presently. I'll give a basic explanation of the math involved. \$\endgroup\$ – Stack Tracer Feb 26 '14 at 3:31
  • \$\begingroup\$ Maybe you could shorten it: change sqrt[1-x^2] to (1-x^2)^.5) \$\endgroup\$ – Justin Feb 26 '14 at 3:33
  • \$\begingroup\$ and I can remove the * after the 2. Mathematica is wonderful. \$\endgroup\$ – Stack Tracer Feb 26 '14 at 3:36
4
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Python (83 chars)

P=0
B=10**500
i=1666
while i:d=2*i+1;P=(P*i%B+(P*i/B+3*i)%d*B)/d;i-=1
print'3.%d'%P
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3
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PARI/GP, 14

\p500
acos(-1)

You can avoid trig by replacing the second line with

gamma(.5)^2

or

(6*zeta(2))^.5

or

psi(3/4)-psi(1/4)

or

4*intnum(x=0,1,(1-x^2)^.5)

or

sumalt(k=2,(-1)^k/(2*k-3))*4
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2
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Python3 136

Uses Madhava's formula.

from decimal import *
D=Decimal
getcontext().prec=600
p=D(3).sqrt()*sum(D(2-k%2*4)/3**k/(2*k+1)for k in range(1100))
print(str(p)[:502])

Python3 164

Uses this formula.

from decimal import *
D=Decimal
getcontext().prec=600
p=sum(D(1)/16**k*(D(4)/(8*k+1)-D(2)/(8*k+4)-D(1)/(8*k+5)-D(1)/(8*k+6))for k in range(411))
print(str(p)[:502])
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2
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bc -l (22 = 5 command line + 17 program) 

scale=500
4*a(1)
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  • 5
    \$\begingroup\$ The rules say "nor may it call a library function to calculate pi." \$\endgroup\$ – Peter Taylor Feb 4 '11 at 20:09
  • \$\begingroup\$ @Peter The problem I guess, is that "library function" is not always a well defined term, and it only get worse when you say "to calculate Pi", as you may use it to calculate intermediate results, for example Sqrt() in Alexandru's answer. \$\endgroup\$ – Dr. belisarius Feb 4 '11 at 21:52
  • \$\begingroup\$ I think this is cheating because atan calculates 1/4 pi but it is an interesting solution nonetheless. \$\endgroup\$ – Thomas O Feb 5 '11 at 10:55
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    \$\begingroup\$ @Thomas O: if this is cheating, where's the limit? \$\endgroup\$ – J B Mar 17 '11 at 6:53
2
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Mathematica (17 bytes)

N[ArcCos[-1],500]

Proof of validity.

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1
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Mathematica - 50

½ = 1/2; 2/Times @@ FixedPointList[(½ + ½ #)^½~N~500 &, ½^½]
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1
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Pyth, 21

u+/*GHhyHy^T500r^3T1Z

Uses this algorithm: pi = 2 + 1/3*(2 + 2/5*(2 + 3/7*(2 + 4/9*(2 + ...)))) found in the comments of the Golfscript answer.

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  • \$\begingroup\$ This doesn't deserve a downvote... \$\endgroup\$ – Beta Decay Oct 26 '14 at 11:21
  • \$\begingroup\$ This answer is incorrect, it generates 34247779... which, to my knowledge, is not pi. \$\endgroup\$ – orlp Mar 26 '15 at 19:11
  • \$\begingroup\$ @orlp The r operation was recently changed in a way which broke this answer. Change the 1 to a 0, and it will work in current Pyth. \$\endgroup\$ – isaacg Mar 27 '15 at 3:13
0
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Axiom, 80 bytes

digits(503);v:=1./sqrt(3);6*reduce(+,[(-1)^k*v^(2*k+1)/(2*k+1)for k in 0..2000])

for reference https://tuts4you.com/download.php?view.452; it would be an approssimation to 6*arctg(1/sqrt(3))=%pi and it would use serie expansion for arctg

  3.1415926535 8979323846 2643383279 5028841971 6939937510 5820974944 592307816
  4 0628620899 8628034825 3421170679 8214808651 3282306647 0938446095 505822317
  2 5359408128 4811174502 8410270193 8521105559 6446229489 5493038196 442881097
  5 6659334461 2847564823 3786783165 2712019091 4564856692 3460348610 454326648
  2 1339360726 0249141273 7245870066 0631558817 4881520920 9628292540 917153643
  6 7892590360 0113305305 4882046652 1384146951 9415116094 3305727036 575959195
  3 0921861173 8193261179 3105118548 0744623799 6274956735 1885752724 891227938
  1 8301194913 01
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0
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JavaScript, 68 bytes

i=1n;x=3n*(10n**520n);p=x;while(x>0){x=x*i/((i+1n)*4n);i+=2n;p+=x/i}

Try it online!

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