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% file pi.tex version 0.993
%
% **** Compute Pi in TeX! ****
%
%
% Author: D. Roegel (roegel@loria.fr)
%
% Version 0.96: 22 July 1996
% First release
%
% Version 0.97: 22 July 1996
% Modified by J. Gelinas (jacquesg@clic.net.ca)
% Added one term to get correct last digits
% Added a second optional argument:
% 100 prints decimals 0..100
% -2 100 prints only decimals 100..101
% Print exact number of decimals
%
% Version 0.98: 23 July 1996
% Modified by D. Roegel, so that the input -3 1 gives 141
% and not 3.141. Also, some unnecessary braces inside \loop...\repeat
% were removed. \ShowResult shortened by introduction of \NextDigit.
%
% Version 0.99: 23 July 1996
% Modified by D. Roegel.
% Improvements following suggestions by J. Gelinas (jacquesg@clic.net.ca).
% Bug corrected, which made some digits false (\fontdimen\firstpos\xb=0pt
% added in \updatefirstpos).
% \N replaced by \count2 in \ShowResult, this enabling calls
% such as \ShowResult{1/\the\N...}{...}.
%
% Version 0.991: 23 July 1996
% Modified by D. Roegel.
% Two \UpdateFirstPos were removed.
%
% Version 0.992: 24 July 1996
% Modified by D. Roegel, following several simplifications suggested
% by J. Gelinas (jacquesg@clic.net.ca).
% \Multiply removed, \Add and \Sub merged.
% \ComputeArcTan shortened.
%
% Version 0.993: 24 July 1996
% Modified by D. Roegel, following several simplifications suggested
% by J. Gelinas (jacquesg@clic.net.ca).
% \ComputeArcTan1/n computes now completely arctan(1/n).
%
%------------------------------------------------------------------------
% This programs uses the formula by John Machin:
%
% Pi=16*arctan(1/5)-4*arctan(1/239)
%
% For arctan(x), we use the development
%
% arctan(x)=\sum_{i=0}^\infty [{x^{2i+1}\over 2i+1} - {x^{2i+3}\over 2i+3}]
%
% One array (\xc) is used to store the partial sum up to {x^{2i+1}\over 2i+1}.
% A second array (\xa) us used to store the value of x^{2i+1}.
% A third array (\xb) is used to store {x^{2i+1}\over 2i+1}.
%
% The result is put in the array \xr.
%
% The implementation of arrays uses a trick shown by Tom Rokicki
% in his game of life program (life.tex).
%
% The number of digits you can compute depends on your implementation
% of TeX. I had no trouble computing 5000 digits.
%
% The last digit can be wrong. And if it is a 0 or a 9, this decimal
% and previous ones can be wrong too. However, the absolute error is
% no greater than one unit of the last digit.
%
% If you want to know more about Pi, check the file sci.math.faq.
% See also http://www.primus.com/staff/paulp/useless/pi.html
% and http://www.tu-chemnitz.de/~arndt/joerg.html.
%
%------------------------------------------------------------------------
%
\newlinechar=`\^^J
\message{^^J***** Computation of Pi with John Machin's formula *****}
\message{^^J** i.e.: pi=16*arctan(1/5)-4*arctan(1/239)}
\message{^^JHow many decimals of pi do you want ? }
\read16to\nbdigits
\newcount\n
\n\nbdigits
\ifnum\n<0
\multiply\n-1
\message{^^JFirst decimal to output ? }
\read16to\firstdigit
\else
\advance\n1
\def\firstdigit{0}
\fi
\newcount\lastdigit
\lastdigit\firstdigit
\advance\lastdigit\n
\advance\lastdigit-1
\newcount\index
\index\lastdigit
\advance\index11
\divide\index4
\def\base{10000}
\def\basesp{10000sp}
\newcount\lastplusone
\lastplusone\index
\advance\lastplusone1
% \index is now the index for the last slot in the arrays
% slot 1 -> integer digits
% slot 2 -> digits 1 to 4
% slot 3 -> digits 5 to 8
% ...
% slot \index -> digits (\index-2) * 4 +1 to (\index-1) * 4
%
\font\xa=cmr10 at 11truept % array for current values of (1/5)^{2n+1}
% and (1/239)^{2n+1}
\fontdimen\lastplusone\xa=0sp % this creates room
\font\xb=cmr10 at 13truept % array for current values of (1/5)^{2n+1}/(2n+1)
% and (1/239)^{2n+1}/(2n+1)
\fontdimen\lastplusone\xb=0sp % this creates room
\font\xc=cmr10 at 15truept % array for current sums of arctan(1/5)
% and arctan(1/239)
\fontdimen\lastplusone\xc=0sp % this creates room
\font\xr=cmr10 at 17truept % array for the result
\fontdimen\lastplusone\xr=0sp % this creates room
% (we have each time allocated one more slot than strictly necessary;
% this avoids a test on \lastplusone in \updatefirstpos)
% \xa, \xb, \xc and \xr are now equal to 0
% Some variables (some of them are not strictly necessary, and might be
% replaced by \count's):
\newcount\dv % will hold dividers
\newcount\firstpos % first non empty slot
\newcount\I % scratch register for loops
\newdimen\carry % for carry (in additions) and borrows (in subtractions)
\newdimen\x % a scratch variable
\newcount\N % counts the terms
\newif\ifcont % flag used to find when an operation on bignums is not done
\newcount\Sdv % value of one digit (used in \ShowResult)
\newcount\dir % toggle for alternating sums
% Initialization of working arrays
\def\InitializeArrays{
{
\I=1
\loop
\fontdimen\I\xa=0sp
\fontdimen\I\xb=0sp
\fontdimen\I\xc=0sp
\advance\I1
\ifnum\I<\lastplusone
\repeat
}
}
% Initialization of the result
\newcount\I
\I=1
\loop
\fontdimen\I\xr=0sp
\advance\I1
\ifnum\I<\lastplusone
\repeat
% divide array #1 by #2 beginning at slot \firstpos and up to \index;
% result is in array #3
% Maximum carry is 9999, so we need to be able to store 99999999.
% \dimen's can hold up to +/- 2,147,483,647 sp and
% \count's up to +/- 2,147,483,647, so it fits.
\def\Divide#1#2into#3{%
\carry0sp
\I\firstpos
{
\loop
\x=\fontdimen\I#1
\multiply\carry\base
\advance\x\carry
\carry\x
\divide\x#2
\fontdimen\I#3=\x
\multiply\x#2
\advance\carry-\x
\advance\I1
\ifnum\I<\lastplusone
\repeat
}
}
% Add or Subtract #2 to #3, depending on #1. array #2 is not modified.
\def\Add#1#2to#3{%
\carry0sp
\I\index
{
\loop
\x=\fontdimen\I#3
\advance\x by #1\fontdimen\I#2
\advance\x by #1\carry
\fontdimen\I#3=\x
\carry\x
\divide\carry\base
\multiply\carry\base
\advance\x-\carry
\divide\carry\base
\ifdim\x<0sp
\advance\x\basesp
\advance\carry1sp
\fi
\fontdimen\I#3=\x
\advance\I-1
\ifnum\I<\firstpos \ifnum\carry=0 \contfalse
\else \conttrue \fi
\else \conttrue
\fi
\ifcont
\repeat
}
}
% Multiply array #1 by #2. Result is in #1.
% This macro is not used in this program, and only remains
% here for didactic and historical reasons.
\def\Multiply#1#2{%
\carry0sp
\I\index
\loop
\x=\fontdimen\I#1
\multiply\x by #2
\advance\x by \carry
\fontdimen\I#1=\x
\carry\x
\divide\carry\base
\multiply\carry\base
\advance\x-\carry
\fontdimen\I#1=\x
\divide\carry\base
\advance\I-1
\ifnum\I<\firstpos \ifnum\carry=0 \contfalse
\else \conttrue \fi
\else \conttrue
\fi
\ifcont
\repeat
}
% update value of \firstpos; in the worst case, \firstpos
% gets increased by 2.
\def\updatefirstpos{
\ifdim\fontdimen\firstpos\xa=0sp
\ifnum\firstpos<\lastplusone
\fontdimen\firstpos\xb=0sp
\advance\firstpos1
\fi
\fi
}
\def\UpdateFirstPos{\updatefirstpos\updatefirstpos}
% Compute arctan(1/#1). Arrays \xa and \xb are used. Result is in \xc.
\def\ComputeArcTan1/#1{%
\firstpos1
\InitializeArrays
% initialize \xa with 1:
\fontdimen1\xa=1sp
\Divide\xa{#1}into\xa
% now, \xa contains 1/#1
\Add1\xa to\xb
% \xb contains 1/#1
\firstpos=2
\dv=#1
\multiply\dv\dv
\N1
\dir-1
\message{^^JI am now computing the following sum: arctan(1/#1)=1/#1}
\Add1\xb to\xc % first term
\loop
\Divide\xa\dv into\xa
\UpdateFirstPos
\ifnum\firstpos<\lastplusone
\advance\N2
\Divide\xa\N into\xb
\ifnum\dir>0
\message{+1/\the\N*1/#1^\the\N}
\else
\message{-1/\the\N*1/#1^\the\N}
\fi
\Add\dir\xb to\xc
\multiply\dir-1
\repeat
}
% Extract the next digit from \count1
\def\NextDigit#1{
\ifnum\count2<\lastdigit
\Sdv\count1
\divide\Sdv#1
\advance\count2 1
\ifnum\count2>\firstdigit
\edef\res{\res\number\Sdv}
\fi
\multiply\Sdv#1
\advance\count1-\Sdv
\fi
}
% Display the result; the digits are scanned one at a time,
% and those between \firstdigit and \lastdigit are extracted
% and saved in \res.
\def\ShowResult#1#2{{
\count0=2
\count2=1
\advance\lastdigit1
\def\res{}
\loop
\count1=\fontdimen\count0#2
\NextDigit{1000}
\NextDigit{100}
\NextDigit{10}
\NextDigit{1}
\advance\count0by 1
\ifnum\count2<\lastdigit
\repeat
\advance\lastdigit-1 % for correct display in next line
\message{^^J#1\res}
}
}
\ComputeArcTan1/5
%\ShowResult{arctan(1/5)=0.}\xc
\message{^^JI multiply it by 4...}
\firstpos2
\Add1\xc to\xc % alternative method is to write \Multiply\xc4
% or to change \ComputeArcTan so that the initial value is 4
\Add1\xc to\xc
\message{done}
% add 4*arctan(1/5) to \xr
\Add1\xc to\xr
%\ShowResult{4*arctan(1/5)=0.}\xr
\ComputeArcTan1/{239}
%\ShowResult{arctan(1/239)=0.}\xc
%\ShowResult{4*arctan(1/239)=0.}\xc
\message{^^JI subtract arctan(1/239) to 4*arctan(1/5)...}
\firstpos2
\Add{-1}\xc to\xr
\message{done}
\message{^^JAnd finally, I multiply it by 4 giving:}
\Add1\xr to\xr
\Add1\xr to\xr
\ifnum\firstdigit=0
\ShowResult{pi[0 .. \number\lastdigit]=3.}\xr
\else
\ShowResult{pi[\firstdigit.. \number\lastdigit]=}\xr
\fi
\bye
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