pi8.nfo

C语言库函数的源代码,是C语言学习参考的好文档。

+++Date last modified: 05-Jul-1997

This info file is placed into the public domain by its author, Carey
Bloodworth, on Sept 24, 1996.  This info file just describes some of the
thoughts I had while writing the arctangent pi program.


PI programs are  a  carefull  collection  of  compromises.  It's not
possible to write a single program that will  be  suitable  for  all
circumstances.

There are a number  of  things  you  need  to think about before you
actually  start  writing  the  program.   And  many  of   them   are
inter-related.

Questions like:

What  is  the maximum size of the calculation?  It's not possible to
make it so generic that  it  will  be  convenient for a few thousand
digits, but still be capable of generating millions.

Are other people going to have to understand  the  program?   People
who perhaps have little to no pi programming experience?

What type of hardware will the program be run on?  How powerful?

What formula should you use?

What base will you work in?

Do  you want to print out the digits while the calculation is going,
or wait until it's done?

What language will  the  program  be  in?   Some  languages are more
suitable than others, and some formulas are more suitable  for  some
languages than others.

What will the failure point be?


Okay....

It has to be fully in C, since that would make it portable.

We've got a variety  of  bases,  work  sizes, and formulas to choose
from.

For  the base, I'm going to work in some power of 10.  That makes it
easy  to  print.   It does take a bit more memory than some power of
two, but since we are  working  in  C,  there  is  no  advantage  to
actually  working in either one.  So, I might as well go with what's
easy to print.

In an effort to  keep  things  simple,  that limits the selection to
mostly arctangent formulas.

For the arctangent formulas, I'm going to use:

pi=16arctan(1/5)-4arctan(1/239)
pi=32arctan(1/10)-4arctan(1/239)-16arctan(1/515)
pi=8arctan(1/3)+4arctan(1/7)
pi=16arctan(1/5)-4arctan(1/70)+4arctan(1/99)
pi=12arctan(1/4)+4arctan(1/20)+4arctan(1/1985)

That should provide enough  cross  checking  so  the user can always
verify their calculations.  They  aren't  the  most  efficient,  but
arctangents are O(n^2) anyway.


To calculate the limit of the formula, we have to consider  that  we
are working in C, and that our variables need to be able to hold the
maximum divisor times the base size.  Formula wise, it would be:

               BASE*MaxDivisor=MaxUint

For a 16 bit integer and our base of 10, it'd be:

               10*MaxDivisor=65535
               MaxDivisor=6553

To  determine  how  many  digits this would be capable of generating
that are 'guaranteed' correct, you  use the relationship between the
divisor and the number of digits.

               NumDigits=MaxDivisor*log10(term)

The  'term'  is the lowest part of the arctangent relationship.  For
the classic pi=16arctan(1/5)-4arctan(1/239),  we'd  use the 5, since
that is the part that will require the most number of divisions  and
be the part to most likely overflow.

(Note: This formula could be recast to calculate what the MaxDivisor
 would be.  If you do, you'd need to add 4.  That won't be exact, but
 it'll be 'close enough for government work'.)               


So, let's calculate a few limits.

16 bit integers:
For base 10, the max divisor would be 6553
So, the number of digits possible would be:
 3: log10(3) * 6553 = 3126
 4: log10(4) * 6553 = 3945
 5: log10(5) * 6553 = 4580
10: log10(10)* 6553 = 6553

That's way too low.  With assembler, or careful coding, it would  be
possible to go all the way up to  a  max  divisor  size  of  65,535,
allowing about 64k digits  with  the arctan(10) formula.  But that's
still too low.

32 bit integers:

For base 10, the max divisor would be 429496729
So, the number of digits possible would be:
 3: log10(3) * 429496729 = 204,922,018
 4: log10(4) * 429496729 = 258,582,797
 5: log10(5) * 429496729 = 300,205,330
10: log10(10)* 429496729 = 429,496,729

That's more than what I  need!   So,  let's try a larger base.  Say,
100

32 bit integers:
For base 100, the max divisor would be 42949672
So, the number of digits possible would be:
 3: log10(3) * 42949672 = 20,492,201
 4: log10(4) * 42949672 = 25,858,279
 5: log10(5) * 42949672 = 30,020,533
10: log10(10)* 42949672 = 42,949,672

That's still more than I really need.  It would work,  but  it  just
seems a shame to waste the extra space that I'm not going to use.

32 bit integers:
For base 10,000, the max divisor would be 429496
So, the number of digits possible would be:
 3: log10(3) * 429496 = 204,922
 4: log10(4) * 429496 = 258,582
 5: log10(5) * 429496 = 300,205
10: log10(10)* 429496 = 429,496

That's not enough.  It would certainly work for most situations, but
I did sort of set that  arbitrary  limit  of being able to reach one
million.

So, it looks like the best we can do with 32 bit integers is to have
a base of 100.  And of course, a base of 100 fits into an char.


Just for curosity, I'm going to try it with the floating point type.
There  wouldn't  be any point in doing it with a 'float', since that
has only 24 bits of mantissa.  Double's though might be useful.

C's 'double' type, with 53 bits of mantissa.
For base 10,000 the max divisor would be 9e11
So, the number of digits possible would be:
 3: log10(3) * 9e11 = 4e11
 4: log10(4) * 9e11 = 5e11
 5: log10(5) * 9e11 = 6e11
10: log10(10)* 9e11 = 9e11

That's _way_ too many.  So I'm going to go for a base a 1e9.  That's
the largest power of 10 that will fit into a long integer.

C's 'double' type, with 53 bits of mantissa.
For base 1,000,000,000 the max divisor would be 9,007,199
So, the number of digits possible would be:
 3: log10(3) * 9,007,199 = 4,297,526
 4: log10(4) * 9,007,199 = 5,422,874
 5: log10(5) * 9,007,199 = 6,295,761
10: log10(10)* 9,007,199 = 9,007,199

Alright!   That  looks a lot better!  We are getting a lot of digits
from each piece of 'work' that  we  do, and the limit is high enough
to be useful, but low enough to not be rediculous.


Well...!   It looks like we have two possibilities.  32 bit integers
and a base of 100, or  use  the  FPU  and a base of 1e9.  Well, most
computers do have a FPU.  And the larger base  would  make  it  more
efficient.   But  that  would  penalize the few computers that don't
have a FPU.

Fortunately, I don't have to actually decide!  As long as I keep the
code generic, and limit  the  number  of  optimizations, I should be
able to do _both_ forms, selectable with a compile time 'define'!