# how to generate pi in c

Ian Smith smithi at nimnet.asn.au
Mon Nov 8 09:01:24 UTC 2010

```In freebsd-questions Digest, Vol 335, Issue 11, Message: 4
On Sat, 06 Nov 2010 01:00:34 -0700 perryh at pluto.rain.com wrote:
> Julian Fagir <gnrp at physik.tu-berlin.de> wrote:
> > > Does anyone has a "generate-pi.c" source code?
> ...
> >   1 #include <stdlib.h>
> >   2 #include <string.h>
> >   3 #include <stdio.h>
> >   4
> >   5 // Change this for a more accurate result.
> >   6 long max = 100000000;
> >   7 double a, b;
> >   8 double pi;
> >   9 long counter;
> >  10 long i;
> >  11
> >  12 int main() {
> >  13     for (i = 0; i< max; i++) {
> >  14         a = drand48();
> >  15         b = drand48();
> >  16         if (a*a + b*b <= 1)
> >  17             counter++;
> >  18     }
> >  19     pi = 4*counter;

Surely that should be 'pi = 4 * counter / max;' otherwise even if the
integer counter were only 1 (of 100000000), pi would already be 4 :)

> >  20
> >  21     printf("%e\n", pi);
> >  22     return(0);
> >  23 }
> ...
> > This approximation is stupid ... Just take 'random' numbers and
> > look whether they are in a circle (that's the a*a + b*b <= 1).
>
> Not stupid, clever.  Very clever.  I rather doubt it resembles what
> the OP had in mind, but it is a brilliant example of what can be
> accomplished when one casts aside any perceived need to adopt a
> conventional approach.

Agreed, quite elegant.  Geometry being a bit rusty, I had to sketch it
to really see it as simply the ratio of the area of the first quadrant
of the unit circle (pi/4) to that of the unit square (1.0), times 4.

> > The detail of this approximation heavily depends on the pseudo-rng
> > you are using, as does its correctness
>
> Perhaps it would be useful in a PRNG test suite?
>
> > (e.g., when your 'rng' always returns 10, pi would be computed to
> > be 10) ...
>
> Bad example.  If abs(drand48()) always exceeded (0.5 * sqrt(2.0)),
> a*a + b*b would always exceed 1.0, thus counter would never be
> incremented and pi would be reported as zero.

That'd be one pretty sad PRNG :) but testing variance of the above
algorithm's result to best known double pi is likely a useful test.

And while a square enclosing a circle, it's hardly squaring the circle:
http://en.wikipedia.org/wiki/Squaring_the_circle .. but an interesting
read nonetheless for unrequited seekers of pi-foo :)

cheers, Ian
```