PiCalc - Calculate digits of pi on your iPod

Contributor Joined: 26 Jul 2007

As the title hints, this podzilla2 module uses a cross-compiled pi-agm binary to calculates digits of pi. It uses the terminal module to allow for input.

Due to a bug in the terminal module, backspacing will not work (I would link to a trac ticket if trac wasn't down - if you have the time to fix it, a patch would be greatly welcomed). If you make an input mistake, just exit and re-enter the window and re-input things carefully. Once the calculations are done, pi-agm will exit and the calculated values will be saved in a text file located where-ever your current working directory is (e.g. mine saved to "/pi8k.txt").

Here's the readme:

ReadMe from Keripo.txt wrote:

Last updated: Feb 18, 2008
~Keripo

Calculates digits of pi via pi-agm. Uses the terminal module for input.
Saves output to working directory (most likely root "/pi##.txt").
Was able to calculate 8192 digits in 37 seconds (used "8k", "2", "-1").

pi-agm is written by Carey Bloodworth.
http://web.archive.org/web/20030803090725/http://www.bloodworth.org/
Source code downloaded from myownlittleworld.com
http://myownlittleworld.com/miscellaneous/computers/files/pi-programs/source/pi-agm.tgz

To compile pi-agm, extract the original source code, replace the
"${LD} ${LDFLAGS} ${OUT_FILE} .o ${MATH_LIBS}" line of your intended
target in src/Makefile with "${CC} ${LDFLAGS} -o ${OUT_FILE} .o ${MATH_LIBS}"
and compile with "make fft-hartley CC=arm-elf-gcc LDFLAGS=-elf2flt"
(you can replace fft-hartley with whichever type you want). This will give
an unoptimized pi-agm binary that you can run on iPodLinux. I may port this
as a proper podzilla2 module later when I have time.

(Note that when I say "when I have time", it doesn't necessarily mean anytime soon ; P

Download:
http://ipodlinux.org/Image:Picalc-1.0.tar.gz

Knock yourselves out Wink

~Keripo

Joined: 17 Sep 2006 Location: /mnt/zowki

Wouldnt it just be easier to make the ipod calculate 22 divide by 7 and tell it when to stop?

Moderator Joined: 30 Oct 2005

zowki wrote:

Wouldnt it just be easier to make the ipod calculate 22 divide by 7 and tell it when to stop?

22/7 is not pi. It's an approximation of pi.

Joined: 17 Sep 2006 Location: /mnt/zowki

joeyjwc wrote:

zowki wrote:

Wouldnt it just be easier to make the ipod calculate 22 divide by 7 and tell it when to stop?

22/7 is not pi. It's an approximation of pi.

Wow, I never knew that! I always thought it was! So what equation actually does calculate pi?

Edit: According to wikipedia:

Quote:

it cannot be expressed as a fraction m/n, where m and n are integers. Consequently its decimal representation never ends or repeats. Beyond being irrational, it is a transcendental number, which means that no finite sequence of algebraic operations on integers (powers, roots, sums, etc.) could ever produce it.

Its impossible to have an equation for it... So how can it be calculated to be more exact?

Contributor Joined: 11 Apr 2006 Location: Ontario, Canada

Look at the pi-agm source code and you'll find various different algorithms ; ) Hartley's FFT is the fastest in pi-agm but theres also instructions on how to create your own algorithm/pi-agm build in the source code (my build is the recommended fft-hartley target though you can also build the other targets - haven't tried any other though and all of them are optimized for Intels I think so performance may vary on the iPod)

Joined: 17 Sep 2006 Location: /mnt/zowki

Looked at agm.c and this is what I found:

Code:

/*
** The AGM itself
**
** A[0]=1 B[0]=1/sqrt(2) Sum=1
**
** n=1..inf
** A[n] = (A[n-1] + B[n-1])/2
** B[n] = Sqrt(A[n-1]*B[n-1])
** C[n] = (A[n-1]-B[n-1])/2 or:
** C[n]^2 = A[n]^2 - B[n]^2 or:
** C[n]^2 = 4A[n+1]*C[n+1]
** Sum = Sum - C[n]^2*(2^(n+1))
** PI[n] = 4A[n+1]^2 / Sum
**
** However, it's not implemented that way. We can save a full
** sized multiplication (with two numbers) by rearranging it,
** and keeping the squares of the A and B. Also, we can do the
** first pass a little differently since A and A^2 will both be 1.
**
** A[0]=1 A[0]^2=1
** B[0]=1/sqrt(2) B[0]^2=0.5
**
** First pass:
**
** A[1] = (A[0] + B[0])/2
** B[1]^2 = A[0]*B[0] ; Since A[0]==1, B[1]^2=B[0]
** C[1]^2 = ((A[0]^2+B[0]^2)/2-B[1]^2)/2
**
** Remainging passes:
**
** C[n] = (A[n-1]-B[n-1])/2; C[n] is actually temp usage of C[n]^2
** A[n] = A[n-1] - C[n]
** C[n]^2 = C[n]*C[n]
** B[n]^2 = (A[n-1]^2+B[n-1]^2-4C[n]^2)/2
**
** Then the rest of the formula is done the same:
**
** A[n]^2 = C[n]^2 + B[n]^2
** B[n] = sqrt(B[n]^2)
** Sum = Sum - C[n]^2*(2^(n+1))
**
** It is an unusual arrangment, but they are all derived directly
** from the standard AGM formulas as given by Salamin.
**
*/

That is really confusing! I'm only in grade 8 maths!

Contributor Joined: 11 Apr 2006 Location: Ontario, Canada

Read more books then. Or, you know, google more Wink

I myself haven't even looked at any of the algorithms and doubt I'd be able to fully understand them myself. And theres really no need to other than out of curiousity. If you're really all that curious, try asking your math teacher (though s/he probably won't have an answer). No need to post your every question here ; P

Joined: 02 Nov 2005 Location: /dev/null

Wasn't there a chain fraction to calculate it?

Joined: 17 Sep 2006 Location: /mnt/zowki

megabyte wrote:

Wasn't there a chain fraction to calculate it?

According to wikipedia Pi can not be represented in a fraction as it would still be continuous.

Like this:

Or the Gregory-Leibniz series:

But it seems as if wikipedia is contradicting itself because I looked under Computing Pi and I found it actually can be written as a formula. Here are two famous ones:

Moderator Joined: 30 Oct 2005

This makes much more sense if you have studied series. @zowki: Wikipedia is not contradicting itself. The "formula" you found is just an infinite series where you sum each term. That's what that big sigma means.

Joined: 02 Nov 2005 Location: /dev/null

(for those less mathematical: a sigma = a sum of iterations. Under it you state the iteration variable, in this case K, and what value does it start with, in this case 0. Over the sigma is the maximum of iterations, in this case infinity. After the sigma you will find what to do with K in each iteration, thus what to sum up)