# Calculate 500 digits of pi

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.

• To check if your digits are correct: eveandersson.com/pi/digits – Nellius Feb 4 '11 at 15:07
• Are we allowed to print more than 500 digits with loss of accuracy after first 500? – Alexandru Feb 4 '11 at 15:27
• @Alexandru, I suppose so but I would prefer to see it truncated. – Thomas O Feb 4 '11 at 17:16
• @Joey no library functions TO CALCULATE PI - I would assume you can use anything from the libraries except the PI constant / function. – Aurel Bílý Feb 5 '11 at 21:27
• Can we use an HTTP library to download a "digits of pi" website? ;-) – dan04 Feb 17 '11 at 5:47

## Golfscript - 29 chars

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


I will post analysis later

• Could you explain how this works? – Thomas O Feb 4 '11 at 13:48
• "I will post analysis later". (waits for 3 years).... – Justin Feb 26 '14 at 3:40
• "I will post analysis later" *waits for more than 6 years* – Erik the Outgolfer Apr 24 '17 at 10:42
• @EriktheOutgolfer I was going to post that. :P – Christopher Apr 25 '17 at 0:53
• "I will post analysis later" (waits for 8 years) – Jono 2906 Oct 4 at 6:02

# 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.

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

# 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


## 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


## bc -l (22 = 5 command line + 17 program)

scale=500
4*a(1)

• The rules say "nor may it call a library function to calculate pi." – Peter Taylor Feb 4 '11 at 20:09
• @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. – Dr. belisarius Feb 4 '11 at 21:52
• I think this is cheating because atan calculates 1/4 pi but it is an interesting solution nonetheless. – Thomas O Feb 5 '11 at 10:55
• @Thomas O: if this is cheating, where's the limit? – J B Mar 17 '11 at 6:53

## Mathematica (17 bytes)

N[ArcCos[-1],500]


## Python3 136

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])


# Mathematica - 50

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


# 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.

• This doesn't deserve a downvote... – Beta Decay Oct 26 '14 at 11:21
• This answer is incorrect, it generates 34247779... which, to my knowledge, is not pi. – orlp Mar 26 '15 at 19:11
• @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. – isaacg Mar 27 '15 at 3:13

# 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


# 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!