Code Golf: Leibniz formula for Pi

Answer 1

Most of the current answers assume that they'll get 5 digits accuracy within some number of iterations and this number is hardcoded into the program. My understanding of the question was that the program itself is supposed to figure out when it's got an answer accurate to 5 digits and stop there. On that assumption here's my C# solution. I haven't bothered to minimise the number of characters since there's no way it can compete with some of the answers already out there, so I thought I'd make it readable instead. :)

    private static double GetPi()
    {
        double acc = 1, sign = -1, lastCheck = 0;

        for (double div = 3; ; div += 2, sign *= -1)
        {
            acc += sign / div;

            double currPi = acc * 4;
            double currCheck = Math.Round(currPi, 5);

            if (currCheck == lastCheck)
                return currPi;

            lastCheck = currCheck;
        }
    }

Answer 2

Ruby, 33 characters

(0..1e6).inject{|a,b|2/(0.5-b)-a}

Answer 3

Here's a solution in MUMPS.

pi(N)
 N X,I
 S X=1 F I=3:4:N-2 S X=X-(1/I)+(1/(I+2))
 Q 4*X

Parameter N indicates how many repeated fractions to use. That is, if you pass in 5 it will evaluate 4 * (1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11)

Some empirical testing showed that N=272241 is the lowest value that gives a correct value of 3.14159 when truncated to 5 decimal points. You have to go to N=852365 to get a value that rounds to 3.14159.

Answer 4

Javascript:

a=0,b=-1,d=-4,c=1e6;while(c--)a+=(d=-d)/(b+=2)

In javascript. 51 characters. Obviously not going to win but eh. :P

Edit -- updated to be 46 characters now, thanks to Strager. :)

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UPDATE (March 30 2010)

A faster (precise only to 5 decimal places) 43 character version by David Murdoch

for(a=0,b=1,d=4,c=~4e5;++c;d=-d)a-=d/(b-=2)

Answer 5

52 chars in Python:

print 4*sum(((-1.)**i/(2*i+1)for i in xrange(5**8)))

(51 dropping the 'x' from xrange.)

36 chars in Octave (or Matlab):

l=0:5^8;disp((-1).^l*(4./(2.*l+1))')

(execute "format long;" to show all the significant digits.) Omitting 'disp' we reach 30 chars:

octave:5> l=0:5^8;(-1).^l*(4./(2.*l+1))'
ans = 3.14159009359631

Answer 6

Ruby, 41 chars (using irb):

s=0;(3..3e6).step(4){|i|s+=8.0/i/(i-2)};s

Or this slightly longer, non-irb version:

s=0;(3..3e6).step(4){|i|s+=8.0/i/(i-2)};p s

This is a modified Leibniz:

1. Combine pairs of terms. This gives you 2/3 + 2/35 + 2/99 + ...
2. Pi becomes 8 * (1/(1 * 3) + 1/(5 * 7) + 1/(9 * 11) + ...)