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|
% \iffalse meta-comment
%
%% File: l3fp-basics.dtx
%
% Copyright (C) 2011-2014,2016-2019 The LaTeX3 Project
%
% It may be distributed and/or modified under the conditions of the
% LaTeX Project Public License (LPPL), either version 1.3c of this
% license or (at your option) any later version. The latest version
% of this license is in the file
%
% https://www.latex-project.org/lppl.txt
%
% This file is part of the "l3kernel bundle" (The Work in LPPL)
% and all files in that bundle must be distributed together.
%
% -----------------------------------------------------------------------
%
% The development version of the bundle can be found at
%
% https://github.com/latex3/latex3
%
% for those people who are interested.
%
%<*driver>
\documentclass[full,kernel]{l3doc}
\begin{document}
\DocInput{\jobname.dtx}
\end{document}
%</driver>
% \fi
%
% \title{The \textsf{l3fp-basics} package\\
% Floating point arithmetic}
% \author{^^A
% The \LaTeX3 Project\thanks
% {^^A
% E-mail:
% \href{mailto:latex-team@latex-project.org}
% {latex-team@latex-project.org}^^A
% }^^A
% }
% \date{Released 2019-10-24}
%
% \maketitle
%
% \begin{documentation}
%
% \end{documentation}
%
% \begin{implementation}
%
% \section{\pkg{l3fp-basics} Implementation}
%
% \begin{macrocode}
%<*initex|package>
% \end{macrocode}
%
% \begin{macrocode}
%<@@=fp>
% \end{macrocode}
%
% The \pkg{l3fp-basics} module implements addition, subtraction,
% multiplication, and division of two floating points, and the absolute
% value and sign-changing operations on one floating point.
% All operations implemented in this module yield the outcome of
% rounding the infinitely precise result of the operation to the
% nearest floating point.
%
% Some algorithms used below end up being quite similar to some
% described in \enquote{What Every Computer Scientist Should Know About
% Floating Point Arithmetic}, by David Goldberg, which can be found at
% \texttt{http://cr.yp.to/2005-590/goldberg.pdf}.
%
% \begin{macro}[EXP]
% {
% \@@_parse_word_abs:N ,
% \@@_parse_word_logb:N ,
% \@@_parse_word_sign:N ,
% \@@_parse_word_sqrt:N ,
% }
% Unary functions.
% \begin{macrocode}
\cs_new:Npn \@@_parse_word_abs:N
{ \@@_parse_unary_function:NNN \@@_set_sign_o:w 0 }
\cs_new:Npn \@@_parse_word_logb:N
{ \@@_parse_unary_function:NNN \@@_logb_o:w ? }
\cs_new:Npn \@@_parse_word_sign:N
{ \@@_parse_unary_function:NNN \@@_sign_o:w ? }
\cs_new:Npn \@@_parse_word_sqrt:N
{ \@@_parse_unary_function:NNN \@@_sqrt_o:w ? }
% \end{macrocode}
% \end{macro}
%
% \subsection{Addition and subtraction}
%
% We define here two functions, \cs{@@_-_o:ww} and \cs{@@_+_o:ww}, which
% perform the subtraction and addition of their two floating point
% operands, and expand the tokens following the result once.
%
% A more obscure function, \cs{@@_add_big_i_o:wNww}, is used in
% \pkg{l3fp-expo}.
%
% The logic goes as follows:
% \begin{itemize}
% \item \cs{@@_-_o:ww} calls \cs{@@_+_o:ww} to do the work, with the
% sign of the second operand flipped;
% \item \cs{@@_+_o:ww} dispatches depending on the type of floating
% point, calling specialized auxiliaries;
% \item in all cases except summing two normal floating point numbers,
% we return one or the other operands depending on the signs, or
% detect an invalid operation in the case of $\infty - \infty$;
% \item for normal floating point numbers, compare the signs;
% \item to add two floating point numbers of the same sign or of
% opposite signs, shift the significand of the smaller one to match the
% bigger one, perform the addition or subtraction of significands,
% check for a carry, round, and pack using the
% \cs[no-index]{@@_basics_pack_\ldots{}} functions.
% \end{itemize}
% The trickiest part is to round correctly when adding or subtracting
% normal floating point numbers.
%
% \subsubsection{Sign, exponent, and special numbers}
%
% \begin{macro}[EXP]{\@@_-_o:ww}
% The \cs{@@_+_o:ww} auxiliary has a hook: it takes one argument
% between the first \cs{s_@@} and \cs{@@_chk:w}, which is applied to
% the sign of the second operand. Positioning the hook there means
% that \cs{@@_+_o:ww} can still perform the sanity check that it was
% followed by \cs{s_@@}.
% \begin{macrocode}
\cs_new:cpx { @@_-_o:ww } \s_@@
{
\exp_not:c { @@_+_o:ww }
\exp_not:n { \s_@@ \@@_neg_sign:N }
}
% \end{macrocode}
% \end{macro}
%
% \begin{macro}[EXP]{\@@_+_o:ww}
% This function is either called directly with an empty |#1| to
% compute an addition, or it is called by \cs{@@_-_o:ww} with
% \cs{@@_neg_sign:N} as |#1| to compute a subtraction, in which case
% the second operand's sign should be changed. If the
% \meta{types} |#2| and |#4| are the same, dispatch to case |#2| ($0$,
% $1$, $2$, or $3$), where we call specialized functions: thanks to
% \cs{int_value:w}, those receive the tweaked \meta{sign_2}
% (expansion of |#1#5|) as an argument. If the \meta{types} are
% distinct, the result is simply the floating point number with the
% highest \meta{type}. Since case $3$ (used for two \texttt{nan})
% also picks the first operand, we can also use it when \meta{type_1}
% is greater than \meta{type_2}. Also note that we don't need to
% worry about \meta{sign_2} in that case since the second operand is
% discarded.
% \begin{macrocode}
\cs_new:cpn { @@_+_o:ww }
\s_@@ #1 \@@_chk:w #2 #3 ; \s_@@ \@@_chk:w #4 #5
{
\if_case:w
\if_meaning:w #2 #4
#2
\else:
\if_int_compare:w #2 > #4 \exp_stop_f:
3
\else:
4
\fi:
\fi:
\exp_stop_f:
\exp_after:wN \@@_add_zeros_o:Nww \int_value:w
\or: \exp_after:wN \@@_add_normal_o:Nww \int_value:w
\or: \exp_after:wN \@@_add_inf_o:Nww \int_value:w
\or: \@@_case_return_i_o:ww
\else: \exp_after:wN \@@_add_return_ii_o:Nww \int_value:w
\fi:
#1 #5
\s_@@ \@@_chk:w #2 #3 ;
\s_@@ \@@_chk:w #4 #5
}
% \end{macrocode}
% \end{macro}
%
% \begin{macro}[EXP]{\@@_add_return_ii_o:Nww}
% Ignore the first operand, and return the second, but using the sign
% |#1| rather than |#4|. As usual, expand after the floating point.
% \begin{macrocode}
\cs_new:Npn \@@_add_return_ii_o:Nww #1 #2 ; \s_@@ \@@_chk:w #3 #4
{ \@@_exp_after_o:w \s_@@ \@@_chk:w #3 #1 }
% \end{macrocode}
% \end{macro}
%
% \begin{macro}[EXP]{\@@_add_zeros_o:Nww}
% Adding two zeros yields \cs{c_zero_fp}, except if both zeros were
% $-0$.
% \begin{macrocode}
\cs_new:Npn \@@_add_zeros_o:Nww #1 \s_@@ \@@_chk:w 0 #2
{
\if_int_compare:w #2 #1 = 20 \exp_stop_f:
\exp_after:wN \@@_add_return_ii_o:Nww
\else:
\@@_case_return_i_o:ww
\fi:
#1
\s_@@ \@@_chk:w 0 #2
}
% \end{macrocode}
% \end{macro}
%
% \begin{macro}[EXP]{\@@_add_inf_o:Nww}
% If both infinities have the same sign, just return that infinity,
% otherwise, it is an invalid operation. We find out if that invalid
% operation is an addition or a subtraction by testing whether the
% tweaked \meta{sign_2} (|#1|) and the \meta{sign_2} (|#4|) are
% identical.
% \begin{macrocode}
\cs_new:Npn \@@_add_inf_o:Nww
#1 \s_@@ \@@_chk:w 2 #2 #3; \s_@@ \@@_chk:w 2 #4
{
\if_meaning:w #1 #2
\@@_case_return_i_o:ww
\else:
\@@_case_use:nw
{
\exp_last_unbraced:Nf \@@_invalid_operation_o:Nww
{ \token_if_eq_meaning:NNTF #1 #4 + - }
}
\fi:
\s_@@ \@@_chk:w 2 #2 #3;
\s_@@ \@@_chk:w 2 #4
}
% \end{macrocode}
% \end{macro}
%
% \begin{macro}[EXP]{\@@_add_normal_o:Nww}
% \begin{quote}
% \cs{@@_add_normal_o:Nww} \meta{sign_2}
% \cs{s_@@} \cs{@@_chk:w} |1| \meta{sign_1}
% \meta{exp_1} \meta{body_1} |;|
% \cs{s_@@} \cs{@@_chk:w} |1| \meta{initial sign_2}
% \meta{exp_2} \meta{body_2} |;|
% \end{quote}
% We now have two normal numbers to add, and we have to check signs
% and exponents more carefully before performing the addition.
% \begin{macrocode}
\cs_new:Npn \@@_add_normal_o:Nww #1 \s_@@ \@@_chk:w 1 #2
{
\if_meaning:w #1#2
\exp_after:wN \@@_add_npos_o:NnwNnw
\else:
\exp_after:wN \@@_sub_npos_o:NnwNnw
\fi:
#2
}
% \end{macrocode}
% \end{macro}
%
% \subsubsection{Absolute addition}
%
% In this subsection, we perform the addition of two positive normal
% numbers.
%
% \begin{macro}[EXP]{\@@_add_npos_o:NnwNnw}
% \begin{quote}
% \cs{@@_add_npos_o:NnwNnw} \meta{sign_1} \meta{exp_1} \meta{body_1}
% |;| \cs{s_@@} \cs{@@_chk:w} |1| \meta{initial sign_2} \meta{exp_2}
% \meta{body_2} |;|
% \end{quote}
% Since we are doing an addition, the final sign is \meta{sign_1}.
% Start an \cs{@@_int_eval:w}, responsible for computing the exponent:
% the result, and the \meta{final sign} are then given to
% \cs{@@_sanitize:Nw} which checks for overflow. The exponent is
% computed as the largest exponent |#2| or |#5|, incremented if there
% is a carry. To add the significands, we decimate the smaller number by
% the difference between the exponents. This is done by
% \cs{@@_add_big_i:wNww} or \cs{@@_add_big_ii:wNww}. We need to bring
% the final sign with us in the midst of the calculation to round
% properly at the end.
% \begin{macrocode}
\cs_new:Npn \@@_add_npos_o:NnwNnw #1#2#3 ; \s_@@ \@@_chk:w 1 #4 #5
{
\exp_after:wN \@@_sanitize:Nw
\exp_after:wN #1
\int_value:w \@@_int_eval:w
\if_int_compare:w #2 > #5 \exp_stop_f:
#2
\exp_after:wN \@@_add_big_i_o:wNww \int_value:w -
\else:
#5
\exp_after:wN \@@_add_big_ii_o:wNww \int_value:w
\fi:
\@@_int_eval:w #5 - #2 ; #1 #3;
}
% \end{macrocode}
% \end{macro}
%
% \begin{macro}[rEXP]{\@@_add_big_i_o:wNww}
% \begin{macro}[rEXP]{\@@_add_big_ii_o:wNww}
% \begin{quote}
% \cs{@@_add_big_i_o:wNww} \meta{shift} |;| \meta{final sign}
% \meta{body_1} |;| \meta{body_2} |;|
% \end{quote}
% Used in \pkg{l3fp-expo}.
% Shift the significand of the small number, then add with
% \cs{@@_add_significand_o:NnnwnnnnN}.
% \begin{macrocode}
\cs_new:Npn \@@_add_big_i_o:wNww #1; #2 #3; #4;
{
\@@_decimate:nNnnnn {#1}
\@@_add_significand_o:NnnwnnnnN
#4
#3
#2
}
\cs_new:Npn \@@_add_big_ii_o:wNww #1; #2 #3; #4;
{
\@@_decimate:nNnnnn {#1}
\@@_add_significand_o:NnnwnnnnN
#3
#4
#2
}
% \end{macrocode}
% \end{macro}
% \end{macro}
%
% \begin{macro}[rEXP]{\@@_add_significand_o:NnnwnnnnN}
% \begin{macro}[rEXP]
% {\@@_add_significand_pack:NNNNNNN, \@@_add_significand_test_o:N}
% \begin{quote}\raggedright
% \cs{@@_add_significand_o:NnnwnnnnN}
% \meta{rounding digit}
% \Arg{Y'_1} \Arg{Y'_2} \meta{extra-digits} |;|
% \Arg{X_1} \Arg{X_2} \Arg{X_3} \Arg{X_4}
% \meta{final sign}
% \end{quote}
% To round properly, we must know at which digit the rounding
% should occur. This requires to know whether the addition
% produces an overall carry or not. Thus, we do the computation
% now and check for a carry, then go back and do the rounding.
% The rounding may cause a carry in very rare cases such as
% $0.99\cdots 95 \to 1.00\cdots 0$, but this situation always
% give an exact power of $10$, for which it is easy to correct
% the result at the end.
% \begin{macrocode}
\cs_new:Npn \@@_add_significand_o:NnnwnnnnN #1 #2#3 #4; #5#6#7#8
{
\exp_after:wN \@@_add_significand_test_o:N
\int_value:w \@@_int_eval:w 1#5#6 + #2
\exp_after:wN \@@_add_significand_pack:NNNNNNN
\int_value:w \@@_int_eval:w 1#7#8 + #3 ; #1
}
\cs_new:Npn \@@_add_significand_pack:NNNNNNN #1 #2#3#4#5#6#7
{
\if_meaning:w 2 #1
+ 1
\fi:
; #2 #3 #4 #5 #6 #7 ;
}
\cs_new:Npn \@@_add_significand_test_o:N #1
{
\if_meaning:w 2 #1
\exp_after:wN \@@_add_significand_carry_o:wwwNN
\else:
\exp_after:wN \@@_add_significand_no_carry_o:wwwNN
\fi:
}
% \end{macrocode}
% \end{macro}
% \end{macro}
%
% \begin{macro}[rEXP]{\@@_add_significand_no_carry_o:wwwNN}
% \begin{quote}
% \cs{@@_add_significand_no_carry_o:wwwNN}
% \meta{8d} |;| \meta{6d} |;| \meta{2d} |;|
% \meta{rounding digit} \meta{sign}
% \end{quote}
% If there's no carry, grab all the digits again and round. The
% packing function \cs{@@_basics_pack_high:NNNNNw} takes care of the
% case where rounding brings a carry.
% \begin{macrocode}
\cs_new:Npn \@@_add_significand_no_carry_o:wwwNN
#1; #2; #3#4 ; #5#6
{
\exp_after:wN \@@_basics_pack_high:NNNNNw
\int_value:w \@@_int_eval:w 1 #1
\exp_after:wN \@@_basics_pack_low:NNNNNw
\int_value:w \@@_int_eval:w 1 #2 #3#4
+ \@@_round:NNN #6 #4 #5
\exp_after:wN ;
}
% \end{macrocode}
% \end{macro}
%
% \begin{macro}[rEXP]{\@@_add_significand_carry_o:wwwNN}
% \begin{quote}
% \cs{@@_add_significand_carry_o:wwwNN}
% \meta{8d} |;| \meta{6d} |;| \meta{2d} |;|
% \meta{rounding digit} \meta{sign}
% \end{quote}
% The case where there is a carry is very similar. Rounding can even
% raise the first digit from $1$ to $2$, but we don't care.
% \begin{macrocode}
\cs_new:Npn \@@_add_significand_carry_o:wwwNN
#1; #2; #3#4; #5#6
{
+ 1
\exp_after:wN \@@_basics_pack_weird_high:NNNNNNNNw
\int_value:w \@@_int_eval:w 1 1 #1
\exp_after:wN \@@_basics_pack_weird_low:NNNNw
\int_value:w \@@_int_eval:w 1 #2#3 +
\exp_after:wN \@@_round:NNN
\exp_after:wN #6
\exp_after:wN #3
\int_value:w \@@_round_digit:Nw #4 #5 ;
\exp_after:wN ;
}
% \end{macrocode}
% \end{macro}
%
% \subsubsection{Absolute subtraction}
%
% \begin{macro}[EXP]{\@@_sub_npos_o:NnwNnw}
% \begin{macro}[EXP]{\@@_sub_eq_o:Nnwnw, \@@_sub_npos_ii_o:Nnwnw}
% \begin{quote}
% \cs{@@_sub_npos_o:NnwNnw}
% \meta{sign_1} \meta{exp_1} \meta{body_1} |;|
% \cs{s_@@} \cs{@@_chk:w} |1|
% \meta{initial sign_2} \meta{exp_2} \meta{body_2} |;|
% \end{quote}
% Rounding properly in some modes requires to know what the sign of
% the result will be. Thus, we start by comparing the exponents and
% significands. If the numbers coincide, return zero. If the second
% number is larger, swap the numbers and call
% \cs{@@_sub_npos_i_o:Nnwnw} with the opposite of \meta{sign_1}.
% \begin{macrocode}
\cs_new:Npn \@@_sub_npos_o:NnwNnw #1#2#3; \s_@@ \@@_chk:w 1 #4#5#6;
{
\if_case:w \@@_compare_npos:nwnw {#2} #3; {#5} #6; \exp_stop_f:
\exp_after:wN \@@_sub_eq_o:Nnwnw
\or:
\exp_after:wN \@@_sub_npos_i_o:Nnwnw
\else:
\exp_after:wN \@@_sub_npos_ii_o:Nnwnw
\fi:
#1 {#2} #3; {#5} #6;
}
\cs_new:Npn \@@_sub_eq_o:Nnwnw #1#2; #3; { \exp_after:wN \c_zero_fp }
\cs_new:Npn \@@_sub_npos_ii_o:Nnwnw #1 #2; #3;
{
\exp_after:wN \@@_sub_npos_i_o:Nnwnw
\int_value:w \@@_neg_sign:N #1
#3; #2;
}
% \end{macrocode}
% \end{macro}
% \end{macro}
%
% \begin{macro}[EXP]{\@@_sub_npos_i_o:Nnwnw}
% After the computation is done, \cs{@@_sanitize:Nw} checks for
% overflow/underflow. It expects the \meta{final sign} and the
% \meta{exponent} (delimited by |;|). Start an integer expression for
% the exponent, which starts with the exponent of the largest number,
% and may be decreased if the two numbers are very close. If the two
% numbers have the same exponent, call the \texttt{near} auxiliary.
% Otherwise, decimate $y$, then call the \texttt{far} auxiliary to
% evaluate the difference between the two significands. Note that we
% decimate by $1$ less than one could expect.
% \begin{macrocode}
\cs_new:Npn \@@_sub_npos_i_o:Nnwnw #1 #2#3; #4#5;
{
\exp_after:wN \@@_sanitize:Nw
\exp_after:wN #1
\int_value:w \@@_int_eval:w
#2
\if_int_compare:w #2 = #4 \exp_stop_f:
\exp_after:wN \@@_sub_back_near_o:nnnnnnnnN
\else:
\exp_after:wN \@@_decimate:nNnnnn \exp_after:wN
{ \int_value:w \@@_int_eval:w #2 - #4 - 1 \exp_after:wN }
\exp_after:wN \@@_sub_back_far_o:NnnwnnnnN
\fi:
#5
#3
#1
}
% \end{macrocode}
% \end{macro}
%
% \begin{macro}[rEXP]{\@@_sub_back_near_o:nnnnnnnnN}
% \begin{macro}[rEXP]
% {\@@_sub_back_near_pack:NNNNNNw, \@@_sub_back_near_after:wNNNNw}
% \begin{quote}
% \cs{@@_sub_back_near_o:nnnnnnnnN}
% \Arg{Y_1} \Arg{Y_2} \Arg{Y_3} \Arg{Y_4}
% \Arg{X_1} \Arg{X_2} \Arg{X_3} \Arg{X_4}
% \meta{final sign}
% \end{quote}
% In this case, the subtraction is exact, so we discard the
% \meta{final sign} |#9|. The very large shifts of $10^{9}$ and
% $1.1\cdot10^{9}$ are unnecessary here, but allow the auxiliaries to
% be reused later. Each integer expression produces a $10$ digit
% result. If the resulting $16$ digits start with a $0$, then we need
% to shift the group, padding with trailing zeros.
% \begin{macrocode}
\cs_new:Npn \@@_sub_back_near_o:nnnnnnnnN #1#2#3#4 #5#6#7#8 #9
{
\exp_after:wN \@@_sub_back_near_after:wNNNNw
\int_value:w \@@_int_eval:w 10#5#6 - #1#2 - 11
\exp_after:wN \@@_sub_back_near_pack:NNNNNNw
\int_value:w \@@_int_eval:w 11#7#8 - #3#4 \exp_after:wN ;
}
\cs_new:Npn \@@_sub_back_near_pack:NNNNNNw #1#2#3#4#5#6#7 ;
{ + #1#2 ; {#3#4#5#6} {#7} ; }
\cs_new:Npn \@@_sub_back_near_after:wNNNNw 10 #1#2#3#4 #5 ;
{
\if_meaning:w 0 #1
\exp_after:wN \@@_sub_back_shift:wnnnn
\fi:
; {#1#2#3#4} {#5}
}
% \end{macrocode}
% \end{macro}
% \end{macro}
%
% \begin{macro}[rEXP]{\@@_sub_back_shift:wnnnn}
% \begin{macro}[rEXP]
% {
% \@@_sub_back_shift_ii:ww,
% \@@_sub_back_shift_iii:NNNNNNNNw,
% \@@_sub_back_shift_iv:nnnnw
% }
% \begin{quote}
% \cs{@@_sub_back_shift:wnnnn} |;|
% \Arg{Z_1} \Arg{Z_2} \Arg{Z_3} \Arg{Z_4} |;|
% \end{quote}
% This function is called with $\meta{Z_1}\leq 999$. Act with
% \tn{number} to trim leading zeros from \meta{Z_1} \meta{Z_2} (we
% don't do all four blocks at once, since non-zero blocks would then
% overflow \TeX{}'s integers). If the first two blocks are zero, the
% auxiliary receives an empty |#1| and trims |#2#30| from leading
% zeros, yielding a total shift between $7$ and~$16$ to the exponent.
% Otherwise we get the shift from |#1| alone, yielding a result
% between $1$ and~$6$. Once the exponent is taken care of, trim
% leading zeros from |#1#2#3| (when |#1| is empty, the space before
% |#2#3| is ignored), get four blocks of $4$~digits and finally clean
% up. Trailing zeros are added so that digits can be grabbed safely.
% \begin{macrocode}
\cs_new:Npn \@@_sub_back_shift:wnnnn ; #1#2
{
\exp_after:wN \@@_sub_back_shift_ii:ww
\int_value:w #1 #2 0 ;
}
\cs_new:Npn \@@_sub_back_shift_ii:ww #1 0 ; #2#3 ;
{
\if_meaning:w @ #1 @
- 7
- \exp_after:wN \use_i:nnn
\exp_after:wN \@@_sub_back_shift_iii:NNNNNNNNw
\int_value:w #2#3 0 ~ 123456789;
\else:
- \@@_sub_back_shift_iii:NNNNNNNNw #1 123456789;
\fi:
\exp_after:wN \@@_pack_twice_four:wNNNNNNNN
\exp_after:wN \@@_pack_twice_four:wNNNNNNNN
\exp_after:wN \@@_sub_back_shift_iv:nnnnw
\exp_after:wN ;
\int_value:w
#1 ~ #2#3 0 ~ 0000 0000 0000 000 ;
}
\cs_new:Npn \@@_sub_back_shift_iii:NNNNNNNNw #1#2#3#4#5#6#7#8#9; {#8}
\cs_new:Npn \@@_sub_back_shift_iv:nnnnw #1 ; #2 ; { ; #1 ; }
% \end{macrocode}
% \end{macro}
% \end{macro}
%
% \begin{macro}[rEXP]{\@@_sub_back_far_o:NnnwnnnnN}
% \begin{quote}\raggedright
% \cs{@@_sub_back_far_o:NnnwnnnnN}
% \meta{rounding} \Arg{Y'_1} \Arg{Y'_2} \meta{extra-digits} |;|
% \Arg{X_1} \Arg{X_2} \Arg{X_3} \Arg{X_4}
% \meta{final sign}
% \end{quote}
% If the difference is greater than $10^{\meta{expo_x}}$, call the
% \texttt{very_far} auxiliary. If the result is less than
% $10^{\meta{expo_x}}$, call the \texttt{not_far} auxiliary. If it is
% too close a call to know yet, namely if $1 \meta{Y'_1} \meta{Y'_2} =
% \meta{X_1} \meta{X_2} \meta{X_3} \meta{X_4} 0$, then call the
% \texttt{quite_far} auxiliary. We use the odd combination of space
% and semi-colon delimiters to allow the \texttt{not_far} auxiliary to
% grab each piece individually, the \texttt{very_far} auxiliary to use
% \cs{@@_pack_eight:wNNNNNNNN}, and the \texttt{quite_far} to ignore
% the significands easily (using the |;| delimiter).
% \begin{macrocode}
\cs_new:Npn \@@_sub_back_far_o:NnnwnnnnN #1 #2#3 #4; #5#6#7#8
{
\if_case:w
\if_int_compare:w 1 #2 = #5#6 \use_i:nnnn #7 \exp_stop_f:
\if_int_compare:w #3 = \use_none:n #7#8 0 \exp_stop_f:
0
\else:
\if_int_compare:w #3 > \use_none:n #7#8 0 - \fi: 1
\fi:
\else:
\if_int_compare:w 1 #2 > #5#6 \use_i:nnnn #7 - \fi: 1
\fi:
\exp_stop_f:
\exp_after:wN \@@_sub_back_quite_far_o:wwNN
\or: \exp_after:wN \@@_sub_back_very_far_o:wwwwNN
\else: \exp_after:wN \@@_sub_back_not_far_o:wwwwNN
\fi:
#2 ~ #3 ; #5 #6 ~ #7 #8 ; #1
}
% \end{macrocode}
% \end{macro}
%
% \begin{macro}[EXP]{\@@_sub_back_quite_far_o:wwNN}
% \begin{macro}[EXP]{\@@_sub_back_quite_far_ii:NN}
% The easiest case is when $x-y$ is extremely close to a power of
% $10$, namely the first digit of $x$ is $1$, and all others vanish
% when subtracting $y$. Then the \meta{rounding} |#3| and the
% \meta{final sign} |#4| control whether we get $1$ or $0.9999 9999
% 9999 9999$. In the usual round-to-nearest mode, we get $1$
% whenever the \meta{rounding} digit is less than or equal to $5$
% (remember that the \meta{rounding} digit is only equal to $5$ if
% there was no further non-zero digit).
% \begin{macrocode}
\cs_new:Npn \@@_sub_back_quite_far_o:wwNN #1; #2; #3#4
{
\exp_after:wN \@@_sub_back_quite_far_ii:NN
\exp_after:wN #3
\exp_after:wN #4
}
\cs_new:Npn \@@_sub_back_quite_far_ii:NN #1#2
{
\if_case:w \@@_round_neg:NNN #2 0 #1
\exp_after:wN \use_i:nn
\else:
\exp_after:wN \use_ii:nn
\fi:
{ ; {1000} {0000} {0000} {0000} ; }
{ - 1 ; {9999} {9999} {9999} {9999} ; }
}
% \end{macrocode}
% \end{macro}
% \end{macro}
%
% \begin{macro}[rEXP]{\@@_sub_back_not_far_o:wwwwNN}
% In the present case, $x$ and $y$ have different exponents, but
% $y$~is large enough that $x-y$ has a smaller exponent than~$x$.
% Decrement the exponent (with |-1|). Then proceed in a way
% similar to the \texttt{near} auxiliaries seen earlier, but
% multiplying $x$ by~$10$ (|#30| and |#40| below), and with the added
% quirk that the \meta{rounding} digit has to be taken into account.
% Namely, we may have to decrease the result by one unit if
% \cs{@@_round_neg:NNN} returns~$1$. This function expects the
% \meta{final sign}~|#6|, the last digit of |1100000000+#40-#2|, and
% the \meta{rounding} digit. Instead of redoing the computation for
% the second argument, we note that \cs{@@_round_neg:NNN} only cares
% about its parity, which is identical to that of the last digit
% of~|#2|.
% \begin{macrocode}
\cs_new:Npn \@@_sub_back_not_far_o:wwwwNN #1 ~ #2; #3 ~ #4; #5#6
{
- 1
\exp_after:wN \@@_sub_back_near_after:wNNNNw
\int_value:w \@@_int_eval:w 1#30 - #1 - 11
\exp_after:wN \@@_sub_back_near_pack:NNNNNNw
\int_value:w \@@_int_eval:w 11 0000 0000 + #40 - #2
- \exp_after:wN \@@_round_neg:NNN
\exp_after:wN #6
\use_none:nnnnnnn #2 #5
\exp_after:wN ;
}
% \end{macrocode}
% \end{macro}
%
% \begin{macro}[EXP]{\@@_sub_back_very_far_o:wwwwNN}
% \begin{macro}[EXP]{\@@_sub_back_very_far_ii_o:nnNwwNN}
% The case where $x-y$ and $x$ have the same exponent is a bit more
% tricky, mostly because it cannot reuse the same auxiliaries. Shift
% the $y$~significand by adding a leading~$0$. Then the logic is similar
% to the \texttt{not_far} functions above. Rounding is a bit more
% complicated: we have two \meta{rounding} digits |#3| and |#6| (from
% the decimation, and from the new shift) to take into account, and
% getting the parity of the main result requires a computation. The
% first \cs{int_value:w} triggers the second one because the number
% is unfinished; we can thus not use $0$ in place of $2$ there.
% \begin{macrocode}
\cs_new:Npn \@@_sub_back_very_far_o:wwwwNN #1#2#3#4#5#6#7
{
\@@_pack_eight:wNNNNNNNN
\@@_sub_back_very_far_ii_o:nnNwwNN
{ 0 #1#2#3 #4#5#6#7 }
;
}
\cs_new:Npn \@@_sub_back_very_far_ii_o:nnNwwNN #1#2 ; #3 ; #4 ~ #5; #6#7
{
\exp_after:wN \@@_basics_pack_high:NNNNNw
\int_value:w \@@_int_eval:w 1#4 - #1 - 1
\exp_after:wN \@@_basics_pack_low:NNNNNw
\int_value:w \@@_int_eval:w 2#5 - #2
- \exp_after:wN \@@_round_neg:NNN
\exp_after:wN #7
\int_value:w
\if_int_odd:w \@@_int_eval:w #5 - #2 \@@_int_eval_end:
1 \else: 2 \fi:
\int_value:w \@@_round_digit:Nw #3 #6 ;
\exp_after:wN ;
}
% \end{macrocode}
% \end{macro}
% \end{macro}
%
% \subsection{Multiplication}
%
% \subsubsection{Signs, and special numbers}
%
% \begin{macro}[EXP]{\@@_*_o:ww}
% We go through an auxiliary, which is common with \cs{@@_/_o:ww}.
% The first argument is the operation, used for the invalid operation
% exception. The second is inserted in a formula to dispatch cases
% slightly differently between multiplication and division. The third
% is the operation for normal floating points. The fourth is there
% for extra cases needed in \cs{@@_/_o:ww}.
% \begin{macrocode}
\cs_new:cpn { @@_*_o:ww }
{
\@@_mul_cases_o:NnNnww
*
{ - 2 + }
\@@_mul_npos_o:Nww
{ }
}
% \end{macrocode}
% \end{macro}
%
% \begin{macro}[EXP]{\@@_mul_cases_o:nNnnww}
% Split into $10$ cases ($12$ for division).
% If both numbers are normal, go to case $0$
% (same sign) or case $1$ (opposite signs): in both cases, call
% \cs{@@_mul_npos_o:Nww} to do the work. If the first operand is
% \texttt{nan}, go to case $2$, in which the second operand is
% discarded; if the second operand is \texttt{nan}, go to case $3$, in
% which the first operand is discarded (note the weird interaction
% with the final test on signs). Then we separate the case where the
% first number is normal and the second is zero: this goes to cases
% $4$ and $5$ for multiplication, $10$ and $11$ for division.
% Otherwise, we do a computation which
% dispatches the products $0\times 0 = 0\times 1 = 1\times 0 = 0$ to
% case $4$ or $5$ depending on the combined sign, the products
% $0\times\infty$ and $\infty\times0$ to case $6$ or $7$ (invalid
% operation), and the products $1\times\infty = \infty\times1 =
% \infty\times\infty = \infty$ to cases $8$ and $9$. Note that the
% code for these two cases (which return $\pm\infty$) is inserted as
% argument |#4|, because it differs in the case of divisions.
% \begin{macrocode}
\cs_new:Npn \@@_mul_cases_o:NnNnww
#1#2#3#4 \s_@@ \@@_chk:w #5#6#7; \s_@@ \@@_chk:w #8#9
{
\if_case:w \@@_int_eval:w
\if_int_compare:w #5 #8 = 11 ~
1
\else:
\if_meaning:w 3 #8
3
\else:
\if_meaning:w 3 #5
2
\else:
\if_int_compare:w #5 #8 = 10 ~
9 #2 - 2
\else:
(#5 #2 #8) / 2 * 2 + 7
\fi:
\fi:
\fi:
\fi:
\if_meaning:w #6 #9 - 1 \fi:
\@@_int_eval_end:
\@@_case_use:nw { #3 0 }
\or: \@@_case_use:nw { #3 2 }
\or: \@@_case_return_i_o:ww
\or: \@@_case_return_ii_o:ww
\or: \@@_case_return_o:Nww \c_zero_fp
\or: \@@_case_return_o:Nww \c_minus_zero_fp
\or: \@@_case_use:nw { \@@_invalid_operation_o:Nww #1 }
\or: \@@_case_use:nw { \@@_invalid_operation_o:Nww #1 }
\or: \@@_case_return_o:Nww \c_inf_fp
\or: \@@_case_return_o:Nww \c_minus_inf_fp
#4
\fi:
\s_@@ \@@_chk:w #5 #6 #7;
\s_@@ \@@_chk:w #8 #9
}
% \end{macrocode}
% \end{macro}
%
% \subsubsection{Absolute multiplication}
%
% In this subsection, we perform the multiplication
% of two positive normal numbers.
%
% \begin{macro}[EXP]{\@@_mul_npos_o:Nww}
% \begin{quote}
% \cs{@@_mul_npos_o:Nww} \meta{final sign}
% \cs{s_@@} \cs{@@_chk:w} |1| \meta{sign_1} \Arg{exp_1} \meta{body_1} |;|
% \cs{s_@@} \cs{@@_chk:w} |1| \meta{sign_2} \Arg{exp_2} \meta{body_2} |;|
% \end{quote}
% After the computation, \cs{@@_sanitize:Nw} checks for overflow or
% underflow. As we did for addition, \cs{@@_int_eval:w} computes the
% exponent, catching any shift coming from the computation in the
% significand. The \meta{final sign} is needed to do the rounding
% properly in the significand computation. We setup the post-expansion
% here, triggered by \cs{@@_mul_significand_o:nnnnNnnnn}.
%
% This is also used in \pkg{l3fp-convert}.
% \begin{macrocode}
\cs_new:Npn \@@_mul_npos_o:Nww
#1 \s_@@ \@@_chk:w #2 #3 #4 #5 ; \s_@@ \@@_chk:w #6 #7 #8 #9 ;
{
\exp_after:wN \@@_sanitize:Nw
\exp_after:wN #1
\int_value:w \@@_int_eval:w
#4 + #8
\@@_mul_significand_o:nnnnNnnnn #5 #1 #9
}
% \end{macrocode}
% \end{macro}
%
% \begin{macro}[rEXP]{\@@_mul_significand_o:nnnnNnnnn}
% \begin{macro}[EXP]
% {\@@_mul_significand_drop:NNNNNw, \@@_mul_significand_keep:NNNNNw}
% \begin{quote}
% \cs{@@_mul_significand_o:nnnnNnnnn}
% \Arg{X_1} \Arg{X_2} \Arg{X_3} \Arg{X_4} \meta{sign}
% \Arg{Y_1} \Arg{Y_2} \Arg{Y_3} \Arg{Y_4}
% \end{quote}
% Note the three semicolons at the end of the definition. One is for
% the last \cs{@@_mul_significand_drop:NNNNNw}; one is for
% \cs{@@_round_digit:Nw} later on; and one, preceded by
% \cs{exp_after:wN}, which is correctly expanded (within an
% \cs{@@_int_eval:w}), is used by \cs{@@_basics_pack_low:NNNNNw}.
%
% The product of two $16$ digit integers has $31$ or $32$ digits,
% but it is impossible to know which one before computing. The place
% where we round depends on that number of digits, and may depend
% on all digits until the last in some rare cases. The approach is
% thus to compute the $5$ first blocks of $4$ digits (the first one
% is between $100$ and $9999$ inclusive), and a compact version of
% the remaining $3$ blocks. Afterwards, the number of digits is
% known, and we can do the rounding within yet another set of
% \cs{@@_int_eval:w}.
% \begin{macrocode}
\cs_new:Npn \@@_mul_significand_o:nnnnNnnnn #1#2#3#4 #5 #6#7#8#9
{
\exp_after:wN \@@_mul_significand_test_f:NNN
\exp_after:wN #5
\int_value:w \@@_int_eval:w 99990000 + #1*#6 +
\exp_after:wN \@@_mul_significand_keep:NNNNNw
\int_value:w \@@_int_eval:w 99990000 + #1*#7 + #2*#6 +
\exp_after:wN \@@_mul_significand_keep:NNNNNw
\int_value:w \@@_int_eval:w 99990000 + #1*#8 + #2*#7 + #3*#6 +
\exp_after:wN \@@_mul_significand_drop:NNNNNw
\int_value:w \@@_int_eval:w 99990000 + #1*#9 + #2*#8 +
#3*#7 + #4*#6 +
\exp_after:wN \@@_mul_significand_drop:NNNNNw
\int_value:w \@@_int_eval:w 99990000 + #2*#9 + #3*#8 +
#4*#7 +
\exp_after:wN \@@_mul_significand_drop:NNNNNw
\int_value:w \@@_int_eval:w 99990000 + #3*#9 + #4*#8 +
\exp_after:wN \@@_mul_significand_drop:NNNNNw
\int_value:w \@@_int_eval:w 100000000 + #4*#9 ;
; \exp_after:wN ;
}
\cs_new:Npn \@@_mul_significand_drop:NNNNNw #1#2#3#4#5 #6;
{ #1#2#3#4#5 ; + #6 }
\cs_new:Npn \@@_mul_significand_keep:NNNNNw #1#2#3#4#5 #6;
{ #1#2#3#4#5 ; #6 ; }
% \end{macrocode}
% \end{macro}
% \end{macro}
%
% \begin{macro}[rEXP]{\@@_mul_significand_test_f:NNN}
% \begin{quote}
% \cs{@@_mul_significand_test_f:NNN} \meta{sign} |1|
% \meta{digits 1--8} |;| \meta{digits 9--12} |;| \meta{digits 13--16} |;|
% |+| \meta{digits 17--20} |+| \meta{digits 21--24}
% |+| \meta{digits 25--28} |+| \meta{digits 29--32} |;|
% \cs{exp_after:wN} |;|
% \end{quote}
% If the \meta{digit 1} is non-zero, then for rounding we only care
% about the digits $16$ and $17$, and whether further digits are zero
% or not (check for exact ties). On the other hand, if \meta{digit 1}
% is zero, we care about digits $17$ and $18$, and whether further
% digits are zero.
% \begin{macrocode}
\cs_new:Npn \@@_mul_significand_test_f:NNN #1 #2 #3
{
\if_meaning:w 0 #3
\exp_after:wN \@@_mul_significand_small_f:NNwwwN
\else:
\exp_after:wN \@@_mul_significand_large_f:NwwNNNN
\fi:
#1 #3
}
% \end{macrocode}
% \end{macro}
%
% \begin{macro}[EXP]{\@@_mul_significand_large_f:NwwNNNN}
% In this branch, \meta{digit 1} is non-zero. The result is thus
% \meta{digits 1--16}, plus some rounding which depends on the digits
% $16$, $17$, and whether all subsequent digits are zero or not.
% Here, \cs{@@_round_digit:Nw} takes digits $17$ and further (as an
% integer expression), and replaces it by a \meta{rounding digit},
% suitable for \cs{@@_round:NNN}.
% \begin{macrocode}
\cs_new:Npn \@@_mul_significand_large_f:NwwNNNN #1 #2; #3; #4#5#6#7; +
{
\exp_after:wN \@@_basics_pack_high:NNNNNw
\int_value:w \@@_int_eval:w 1#2
\exp_after:wN \@@_basics_pack_low:NNNNNw
\int_value:w \@@_int_eval:w 1#3#4#5#6#7
+ \exp_after:wN \@@_round:NNN
\exp_after:wN #1
\exp_after:wN #7
\int_value:w \@@_round_digit:Nw
}
% \end{macrocode}
% \end{macro}
%
% \begin{macro}[rEXP]{\@@_mul_significand_small_f:NNwwwN}
% In this branch, \meta{digit 1} is zero. Our result is thus
% \meta{digits 2--17}, plus some rounding which depends on the digits
% $17$, $18$, and whether all subsequent digits are zero or not.
% The $8$ digits |1#3| are followed, after expansion of the
% \texttt{small_pack} auxiliary, by the next digit, to form a $9$
% digit number.
% \begin{macrocode}
\cs_new:Npn \@@_mul_significand_small_f:NNwwwN #1 #2#3; #4#5; #6; + #7
{
- 1
\exp_after:wN \@@_basics_pack_high:NNNNNw
\int_value:w \@@_int_eval:w 1#3#4
\exp_after:wN \@@_basics_pack_low:NNNNNw
\int_value:w \@@_int_eval:w 1#5#6#7
+ \exp_after:wN \@@_round:NNN
\exp_after:wN #1
\exp_after:wN #7
\int_value:w \@@_round_digit:Nw
}
% \end{macrocode}
% \end{macro}
%
% \subsection{Division}
%
% \subsubsection{Signs, and special numbers}
%
% Time is now ripe to tackle the hardest of the four elementary
% operations: division.
%
% \begin{macro}[EXP]{\@@_/_o:ww}
% Filtering special floating point is very similar to what we did for
% multiplications, with a few variations. Invalid operation
% exceptions display |/| rather than |*|. In the formula for
% dispatch, we replace |- 2 +| by |-|. The case of normal
% numbers is treated using \cs{@@_div_npos_o:Nww} rather than
% \cs{@@_mul_npos_o:Nww}. There are two additional cases: if the
% first operand is normal and the second is a zero, then the division
% by zero exception is raised: cases $10$ and $11$ of the
% \cs{if_case:w} construction in \cs{@@_mul_cases_o:NnNnww} are
% provided as the fourth argument here.
% \begin{macrocode}
\cs_new:cpn { @@_/_o:ww }
{
\@@_mul_cases_o:NnNnww
/
{ - }
\@@_div_npos_o:Nww
{
\or:
\@@_case_use:nw
{ \@@_division_by_zero_o:NNww \c_inf_fp / }
\or:
\@@_case_use:nw
{ \@@_division_by_zero_o:NNww \c_minus_inf_fp / }
}
}
% \end{macrocode}
% \end{macro}
%
% \begin{macro}[EXP]{\@@_div_npos_o:Nww}
% \begin{quote}
% \cs{@@_div_npos_o:Nww} \meta{final sign}
% \cs{s_@@} \cs{@@_chk:w} |1| \meta{sign_A} \Arg{exp A}
% \Arg{A_1} \Arg{A_2} \Arg{A_3} \Arg{A_4} |;|
% \cs{s_@@} \cs{@@_chk:w} |1| \meta{sign_Z} \Arg{exp Z}
% \Arg{Z_1} \Arg{Z_2} \Arg{Z_3} \Arg{Z_4} |;|
% \end{quote}
% We want to compute $A/Z$. As for multiplication,
% \cs{@@_sanitize:Nw} checks for overflow or underflow; we provide it
% with the \meta{final sign}, and an integer expression in which we
% compute the exponent. We set up the arguments of
% \cs{@@_div_significand_i_o:wnnw}, namely an integer \meta{y} obtained
% by adding $1$ to the first $5$ digits of $Z$ (explanation given soon
% below), then the four \Arg{A_{i}}, then the four \Arg{Z_{i}}, a
% semi-colon, and the \meta{final sign}, used for rounding at the end.
% \begin{macrocode}
\cs_new:Npn \@@_div_npos_o:Nww
#1 \s_@@ \@@_chk:w 1 #2 #3 #4 ; \s_@@ \@@_chk:w 1 #5 #6 #7#8#9;
{
\exp_after:wN \@@_sanitize:Nw
\exp_after:wN #1
\int_value:w \@@_int_eval:w
#3 - #6
\exp_after:wN \@@_div_significand_i_o:wnnw
\int_value:w \@@_int_eval:w #7 \use_i:nnnn #8 + 1 ;
#4
{#7}{#8}#9 ;
#1
}
% \end{macrocode}
% \end{macro}
%
% \subsubsection{Work plan}
%
% In this subsection, we explain how to avoid overflowing \TeX{}'s
% integers when performing the division of two positive normal numbers.
%
% We are given two numbers, $A=0.A_{1}A_{2}A_{3}A_{4}$ and
% $Z=0.Z_{1}Z_{2}Z_{3}Z_{4}$, in blocks of $4$ digits, and we know that
% the first digits of $A_{1}$ and of $Z_{1}$ are non-zero. To compute
% $A/Z$, we proceed as follows.
% \begin{itemize}
% \item Find an integer $Q_{A} \simeq 10^{4} A / Z$.
% \item Replace $A$ by $B = 10^{4} A - Q_{A} Z$.
% \item Find an integer $Q_{B} \simeq 10^{4} B / Z$.
% \item Replace $B$ by $C = 10^{4} B - Q_{B} Z$.
% \item Find an integer $Q_{C} \simeq 10^{4} C / Z$.
% \item Replace $C$ by $D = 10^{4} C - Q_{C} Z$.
% \item Find an integer $Q_{D} \simeq 10^{4} D / Z$.
% \item Consider $E = 10^{4} D - Q_{D} Z$, and ensure
% correct rounding.
% \end{itemize}
% The result is then $Q = 10^{-4} Q_{A} + 10^{-8} Q_{B} + 10^{-12} Q_{C}
% + 10^{-16} Q_{D} + \text{rounding}$. Since the $Q_{i}$ are integers,
% $B$, $C$, $D$, and~$E$ are all exact multiples of $10^{-16}$, in other
% words, computing with $16$ digits after the decimal separator yields
% exact results. The problem is the risk of overflow: in general $B$, $C$,
% $D$, and $E$ may be greater than $1$.
%
% Unfortunately, things are not as easy as they seem. In particular, we
% want all intermediate steps to be positive, since negative results
% would require extra calculations at the end. This requires that
% $Q_{A} \leq 10^{4} A / Z$ \emph{etc.} A reasonable attempt would be
% to define $Q_{A}$ as
% \begin{equation*}
% \cs{int_eval:n} \left\{
% \frac{ A_{1} A_{2} }{ Z_{1} + 1 } - 1 \right\}
% \leq 10^{4} \frac{A}{Z}
% \end{equation*}
% Subtracting $1$ at the end takes care of the fact that \eTeX{}'s
% \cs{@@_int_eval:w} rounds divisions instead of truncating (really,
% $1/2$ would be sufficient, but we work with integers). We add $1$ to
% $Z_{1}$ because $Z_{1} \leq 10^{4}Z < Z_{1}+1$ and we need $Q_{A}$ to
% be an underestimate. However, we are now underestimating $Q_{A}$ too
% much: it can be wrong by up to $100$, for instance when $Z = 0.1$ and
% $A \simeq 1$. Then $B$ could take values up to $10$ (maybe more), and
% a few steps down the line, we would run into arithmetic overflow,
% since \TeX{} can only handle integers less than roughly $2\cdot
% 10^{9}$.
%
% A better formula is to take
% \begin{equation*}
% Q_{A} = \cs{int_eval:n} \left\{
% \frac{ 10 \cdot A_{1} A_{2} }
% { \left\lfloor 10^{-3} \cdot Z_{1} Z_{2} \right\rfloor + 1 }
% - 1 \right\}.
% \end{equation*}
% This is always less than $10^{9} A / (10^{5} Z)$, as we wanted. In
% words, we take the $5$ first digits of $Z$ into account, and the $8$
% first digits of $A$, using $0$ as a $9$-th digit rather than the true
% digit for efficiency reasons. We shall prove that using this formula
% to define all the $Q_{i}$ avoids any overflow. For convenience, let
% us denote
% \begin{equation*}
% y = \left\lfloor 10^{-3} \cdot Z_{1} Z_{2} \right\rfloor + 1,
% \end{equation*}
% so that, taking into account the fact that \eTeX{} rounds ties away
% from zero,
% \begin{align*}
% Q_{A}
% &= \left\lfloor \frac{A_{1}A_{2}0}{y} - \frac{1}{2} \right\rfloor
% \\
% &>\frac{A_{1}A_{2}0}{y} - \frac{3}{2}.
% \end{align*}
% Note that $10^{4}<y\leq 10^{5}$, and $999 \leq Q_{A} \leq 99989$.
% Also note that this formula does not cause an overflow as long as $A <
% (2^{31}-1) / 10^{9} \simeq 2.147\cdots$, since the numerator involves an
% integer slightly smaller than $10^{9} A$.
%
% Let us bound $B$:
% \begin{align*}
% 10^{5} B
% &=
% A_{1}A_{2}0 + 10 \cdot 0.A_{3}A_{4}
% - 10 \cdot Z_{1}.Z_{2}Z_{3}Z_{4} \cdot Q_{A}
% \\
% &<
% A_{1}A_{2}0
% \cdot \left( 1 - 10 \cdot \frac{Z_{1}.Z_{2}Z_{3}Z_{4}}{y} \right)
% + \frac{3}{2} \cdot 10 \cdot Z_{1}.Z_{2}Z_{3}Z_{4} + 10
% \\
% &\leq
% \frac{A_{1}A_{2}0 \cdot (y - 10 \cdot Z_{1}.Z_{2}Z_{3}Z_{4})}{y}
% + \frac{3}{2} y + 10
% \\
% &\leq
% \frac{A_{1}A_{2}0\cdot 1}{y} + \frac{3}{2} y + 10
% \leq
% \frac{10^{9} A}{y} + 1.6\cdot y.
% \end{align*}
% At the last step, we hide $10$ into the second term for later
% convenience. The same reasoning yields
% \begin{align*}
% 10^{5} B &< 10^{9} A/y + 1.6 y, \\
% 10^{5} C &< 10^{9} B/y + 1.6 y, \\
% 10^{5} D &< 10^{9} C/y + 1.6 y, \\
% 10^{5} E &< 10^{9} D/y + 1.6 y. \\
% \end{align*}
% The goal is now to prove that none of $B$, $C$, $D$, and $E$ can go
% beyond $(2^{31}-1) / 10^{9} = 2.147\cdots$.
%
% Combining the various inequalities together with $A<1$, we get
% \begin{align*}
% 10^{5} B &< 10^{9}/y + 1.6 y, \\
% 10^{5} C &< 10^{13}/y^{2} + 1.6 (y + 10^{4}), \\
% 10^{5} D &< 10^{17}/y^{3} + 1.6 (y + 10^{4} + 10^{8}/y), \\
% 10^{5} E &< 10^{21}/y^{4} + 1.6 (y + 10^{4} + 10^{8}/y + 10^{12}/y^{2}). \\
% \end{align*}
% All of those bounds are convex functions of $y$ (since every power of
% $y$ involved is convex, and the coefficients are positive), and thus
% maximal at one of the end-points of the allowed range $10^{4} < y \leq
% 10^{5}$. Thus,
% \begin{align*}
% 10^{5} B &< \mathrm{max} ( 1.16\cdot 10^{5}, 1.7 \cdot 10^{5}), \\
% 10^{5} C &< \mathrm{max} ( 1.32\cdot 10^{5}, 1.77 \cdot 10^{5}), \\
% 10^{5} D &< \mathrm{max} ( 1.48\cdot 10^{5}, 1.777 \cdot 10^{5}), \\
% 10^{5} E &< \mathrm{max} ( 1.64\cdot 10^{5}, 1.7777 \cdot 10^{5}). \\
% \end{align*}
% All of those bounds are less than $2.147\cdot 10^{5}$, and we are thus
% within \TeX{}'s bounds in all cases!
%
% We later need to have a bound on the $Q_{i}$. Their definitions
% imply that $Q_{A} < 10^{9} A/y - 1/2 < 10^{5} A$ and similarly for the
% other $Q_{i}$. Thus, all of them are less than $177770$.
%
% The last step is to ensure correct rounding. We have
% \begin{equation*}
% A/Z = \sum_{i=1}^{4} \left(10^{-4i} Q_{i}\right) + 10^{-16} E/Z
% \end{equation*}
% exactly. Furthermore, we know that the result is in $[0.1,10)$,
% hence will be rounded to a multiple of $10^{-16}$ or of $10^{-15}$, so
% we only need to know the integer part of $E/Z$, and a
% \enquote{rounding} digit encoding the rest. Equivalently, we need to
% find the integer part of $2E/Z$, and determine whether it was an
% exact integer or not (this serves to detect ties). Since
% \begin{equation*}
% \frac{2E}{Z} = 2\frac{10^{5} E}{10^{5} Z}
% \leq 2\frac{10^{5} E}{10^{4}} < 36,
% \end{equation*}
% this integer part is between $0$ and $35$ inclusive. We let \eTeX{}
% round
% \begin{equation*}
% P = \cs{int_eval:n} \left\{
% \frac{2\cdot E_{1}E_{2}}{Z_{1}Z_{2}} \right\},
% \end{equation*}
% which differs from $2E/Z$ by at most
% \begin{equation*}
% \frac{1}{2}
% + 2 \left\lvert \frac{E}{Z} - \frac{E}{10^{-8} Z_{1}Z_{2}}\right\rvert
% + 2 \left\lvert \frac{10^{8} E - E_{1}E_{2}}{Z_{1}Z_{2}}\right\rvert
% < 1,
% \end{equation*}
% ($1/2$ comes from \eTeX{}'s rounding) because each absolute value is
% less than $10^{-7}$. Thus $P$ is either the correct integer part, or
% is off by $1$; furthermore, if $2 E / Z$ is an integer, $P = 2 E / Z$.
% We will check the sign of $2 E - P Z$. If it is negative, then $E / Z
% \in \big((P - 1) / 2, P / 2\big)$. If it is zero, then $E / Z = P /
% 2$. If it is positive, then $E / Z \in \big(P / 2, (P - 1) / 2\big)$.
% In each case, we know how to round to an integer, depending on the
% parity of $P$, and the rounding mode.
%
% \subsubsection{Implementing the significand division}
%
% \begin{macro}[rEXP]{\@@_div_significand_i_o:wnnw}
% \begin{quote}
% \cs{@@_div_significand_i_o:wnnw} \meta{y} |;|
% \Arg{A_1} \Arg{A_2} \Arg{A_3} \Arg{A_4}
% \Arg{Z_1} \Arg{Z_2} \Arg{Z_3} \Arg{Z_4} |;| \meta{sign}
% \end{quote}
% Compute $10^{6} + Q_{A}$ (a $7$~digit number thanks to the shift),
% unbrace \meta{A_1} and \meta{A_2}, and prepare the
% \meta{continuation} arguments for $4$ consecutive calls to
% \cs{@@_div_significand_calc:wwnnnnnnn}. Each of these calls needs
% \meta{y} (|#1|), and it turns out that we need post-expansion there,
% hence the \cs{int_value:w}. Here, |#4| is six brace groups, which
% give the six first |n|-type arguments of the \texttt{calc} function.
% \begin{macrocode}
\cs_new:Npn \@@_div_significand_i_o:wnnw #1 ; #2#3 #4 ;
{
\exp_after:wN \@@_div_significand_test_o:w
\int_value:w \@@_int_eval:w
\exp_after:wN \@@_div_significand_calc:wwnnnnnnn
\int_value:w \@@_int_eval:w 999999 + #2 #3 0 / #1 ;
#2 #3 ;
#4
{ \exp_after:wN \@@_div_significand_ii:wwn \int_value:w #1 }
{ \exp_after:wN \@@_div_significand_ii:wwn \int_value:w #1 }
{ \exp_after:wN \@@_div_significand_ii:wwn \int_value:w #1 }
{ \exp_after:wN \@@_div_significand_iii:wwnnnnn \int_value:w #1 }
}
% \end{macrocode}
% \end{macro}
%
% \begin{macro}[rEXP]{\@@_div_significand_calc:wwnnnnnnn}
% \begin{macro}[rEXP]
% {
% \@@_div_significand_calc_i:wwnnnnnnn,
% \@@_div_significand_calc_ii:wwnnnnnnn,
% }
% \begin{quote}
% \cs{@@_div_significand_calc:wwnnnnnnn} \meta{$10^{6}+{}$Q_{A}} |;|
% \meta{A_1} \meta{A_2} |;| \Arg{A_3} \Arg{A_4}
% \Arg{Z_1} \Arg{Z_2} \Arg{Z_3} \Arg{Z_4}
% \Arg{continuation}
% \end{quote}
% expands to
% \begin{quote}
% \meta{$10^{6}+{}$Q_{A}} \meta{continuation} |;|
% \meta{B_1} \meta{B_2} |;| \Arg{B_3} \Arg{B_4}
% \Arg{Z_1} \Arg{Z_2} \Arg{Z_3} \Arg{Z_4}
% \end{quote}
% where $B = 10^{4} A - Q_{A} \cdot Z$. This function is also used to
% compute $C$, $D$, $E$ (with the input shifted accordingly), and is
% used in \pkg{l3fp-expo}.
%
% We know that $0<Q_{A}<1.8\cdot 10^{5}$, so the product of $Q_{A}$
% with each $Z_{i}$ is within \TeX{}'s bounds. However, it is a
% little bit too large for our purposes: we would not be able to use
% the usual trick of adding a large power of $10$ to ensure that the
% number of digits is fixed.
%
% The bound on $Q_{A}$, implies that $10^{6}+Q_{A}$ starts with the
% digit $1$, followed by $0$ or $1$. We test, and call different
% auxiliaries for the two cases. An earlier implementation did the
% tests within the computation, but since we added a
% \meta{continuation}, this is not possible because the macro has $9$
% parameters.
%
% The result we want is then (the overall power of $10$ is arbitrary):
% \begin{align*}
% &10^{-4} ( \#2 - \#1 \cdot \#5 - 10 \cdot \meta{i} \cdot \#5\#6 )
% + 10^{-8} ( \#3 - \#1 \cdot \#6 - 10 \cdot \meta{i} \cdot \#7 ) \\
% &+ 10^{-12}( \#4 - \#1 \cdot \#7 - 10 \cdot \meta{i} \cdot \#8 )
% + 10^{-16}( - \#1 \cdot \#8 ),
% \end{align*}
% where \meta{i} stands for the $10^{5}$ digit of $Q_{A}$, which is
% $0$ or~$1$, and $\#1$, $\#2$, \emph{etc.\@} are the parameters of
% either auxiliary. The factors of $10$ come from the fact that
% $Q_{A} = 10\cdot 10^{4} \cdot \meta{i} + \#1$. As usual, to combine
% all the terms, we need to choose some shifts which must ensure that
% the number of digits of the second, third, and fourth terms are each
% fixed. Here, the positive contributions are at most $10^{8}$ and
% the negative contributions can go up to $10^{9}$. Indeed, for the
% auxiliary with $\meta{i}=1$, |#1| is at most $80000$, leading to
% contributions of at worse $-8\cdot 10^{8}4$, while the other
% negative term is very small $<10^{6}$ (except in the first
% expression, where we don't care about the number of digits); for the
% auxiliary with $\meta{i}=0$, |#1| can go up to $99999$, but there is
% no other negative term. Hence, a good choice is $2\cdot 10^{9}$,
% which produces totals in the range $[10^{9}, 2.1\cdot 10^{9}]$. We
% are flirting with \TeX{}'s limits once more.
% \begin{macrocode}
\cs_new:Npn \@@_div_significand_calc:wwnnnnnnn 1#1
{
\if_meaning:w 1 #1
\exp_after:wN \@@_div_significand_calc_i:wwnnnnnnn
\else:
\exp_after:wN \@@_div_significand_calc_ii:wwnnnnnnn
\fi:
}
\cs_new:Npn \@@_div_significand_calc_i:wwnnnnnnn
#1; #2;#3#4 #5#6#7#8 #9
{
1 1 #1
#9 \exp_after:wN ;
\int_value:w \@@_int_eval:w \c_@@_Bigg_leading_shift_int
+ #2 - #1 * #5 - #5#60
\exp_after:wN \@@_pack_Bigg:NNNNNNw
\int_value:w \@@_int_eval:w \c_@@_Bigg_middle_shift_int
+ #3 - #1 * #6 - #70
\exp_after:wN \@@_pack_Bigg:NNNNNNw
\int_value:w \@@_int_eval:w \c_@@_Bigg_middle_shift_int
+ #4 - #1 * #7 - #80
\exp_after:wN \@@_pack_Bigg:NNNNNNw
\int_value:w \@@_int_eval:w \c_@@_Bigg_trailing_shift_int
- #1 * #8 ;
{#5}{#6}{#7}{#8}
}
\cs_new:Npn \@@_div_significand_calc_ii:wwnnnnnnn
#1; #2;#3#4 #5#6#7#8 #9
{
1 0 #1
#9 \exp_after:wN ;
\int_value:w \@@_int_eval:w \c_@@_Bigg_leading_shift_int
+ #2 - #1 * #5
\exp_after:wN \@@_pack_Bigg:NNNNNNw
\int_value:w \@@_int_eval:w \c_@@_Bigg_middle_shift_int
+ #3 - #1 * #6
\exp_after:wN \@@_pack_Bigg:NNNNNNw
\int_value:w \@@_int_eval:w \c_@@_Bigg_middle_shift_int
+ #4 - #1 * #7
\exp_after:wN \@@_pack_Bigg:NNNNNNw
\int_value:w \@@_int_eval:w \c_@@_Bigg_trailing_shift_int
- #1 * #8 ;
{#5}{#6}{#7}{#8}
}
% \end{macrocode}
% \end{macro}
% \end{macro}
%
% \begin{macro}[EXP]{\@@_div_significand_ii:wwn}
% \begin{quote}
% \cs{@@_div_significand_ii:wwn} \meta{y} |;|
% \meta{B_1} |;| \Arg{B_2} \Arg{B_3} \Arg{B_4}
% \Arg{Z_1} \Arg{Z_2} \Arg{Z_3} \Arg{Z_4}
% \meta{continuations} \meta{sign}
% \end{quote}
% Compute $Q_{B}$ by evaluating $\meta{B_1}\meta{B_2}0 / y - 1$. The
% result is output to the left, in an \cs{@@_int_eval:w} which we
% start now. Once that is evaluated (and the other $Q_{i}$ also,
% since later expansions are triggered by this one), a packing
% auxiliary takes care of placing the digits of $Q_{B}$ in an
% appropriate way for the final addition to obtain $Q$. This
% auxiliary is also used to compute $Q_{C}$ and $Q_{D}$ with the
% inputs $C$ and $D$ instead of $B$.
% \begin{macrocode}
\cs_new:Npn \@@_div_significand_ii:wwn #1; #2;#3
{
\exp_after:wN \@@_div_significand_pack:NNN
\int_value:w \@@_int_eval:w
\exp_after:wN \@@_div_significand_calc:wwnnnnnnn
\int_value:w \@@_int_eval:w 999999 + #2 #3 0 / #1 ; #2 #3 ;
}
% \end{macrocode}
% \end{macro}
%
% \begin{macro}[rEXP]{\@@_div_significand_iii:wwnnnnn}
% \begin{quote}
% \cs{@@_div_significand_iii:wwnnnnn} \meta{y} |;|
% \meta{E_1} |;| \Arg{E_2} \Arg{E_3} \Arg{E_4}
% \Arg{Z_1} \Arg{Z_2} \Arg{Z_3} \Arg{Z_4} \meta{sign}
% \end{quote}
% We compute $P \simeq 2E/Z$ by rounding $2 E_{1} E_{2}/Z_{1}Z_{2}$.
% Note the first $0$, which multiplies $Q_{D}$ by $10$: we later
% add (roughly) $5\cdot P$, which amounts to adding $P/2 \simeq E/Z$
% to $Q_{D}$, the appropriate correction from a hypothetical $Q_{E}$.
% \begin{macrocode}
\cs_new:Npn \@@_div_significand_iii:wwnnnnn #1; #2;#3#4#5 #6#7
{
0
\exp_after:wN \@@_div_significand_iv:wwnnnnnnn
\int_value:w \@@_int_eval:w ( 2 * #2 #3) / #6 #7 ; % <- P
#2 ; {#3} {#4} {#5}
{#6} {#7}
}
% \end{macrocode}
% \end{macro}
%
% \begin{macro}[rEXP]
% {
% \@@_div_significand_iv:wwnnnnnnn,
% \@@_div_significand_v:NNw,
% \@@_div_significand_vi:Nw
% }
% \begin{quote}
% \cs{@@_div_significand_iv:wwnnnnnnn} \meta{P} |;|
% \meta{E_1} |;| \Arg{E_2} \Arg{E_3} \Arg{E_4}
% \Arg{Z_1} \Arg{Z_2} \Arg{Z_3} \Arg{Z_4} \meta{sign}
% \end{quote}
% This adds to the current expression ($10^{7} + 10\cdot Q_{D}$) a
% contribution of $5 \cdot P + \operatorname{sign}(T)$ with $T = 2 E -
% P Z$. This amounts to adding $P / 2$ to $Q_{D}$, with an extra
% \meta{rounding} digit. This \meta{rounding} digit is $0$ or $5$ if
% $T$ does not contribute, \emph{i.e.,} if $0 = T = 2 E - P Z$, in
% other words if $10^{16} A / Z$ is an integer or half-integer.
% Otherwise it is in the appropriate range, $[1,4]$ or $[6,9]$. This
% is precise enough for rounding purposes (in any mode).
%
% It seems an overkill to compute $T$ exactly as I do here, but I see
% no faster way right now.
%
% Once more, we need to be careful and show that the calculation
% $\#1\cdot\#6\#7$ below does not cause an overflow: naively, $P$ can
% be up to $35$, and $\#6\#7$ up to $10^{8}$, but both cannot happen
% simultaneously. To show that things are fine, we split in two
% (non-disjoint) cases.
% \begin{itemize}
% \item For $P < 10$, the product obeys $P\cdot\#6\#7 < 10^{8} \cdot P
% < 10^{9} $.
% \item For large $P\geq 3$, the rounding error on $P$, which is at
% most $1$, is less than a factor of $2$, hence $P\leq 4E/Z$. Also,
% $\#6\#7 \leq 10^{8} \cdot Z$, hence $P\cdot \#6\#7 \leq 4E\cdot
% 10^{8} < 10^{9}$.
% \end{itemize}
% Both inequalities could be made tighter if needed.
%
% Note however that $P\cdot \#8\#9$ may overflow, since the two
% factors are now independent, and the result may reach $3.5\cdot
% 10^{9}$. Thus we compute the two lower levels separately. The rest
% is standard, except that we use |+| as a separator (ending integer
% expressions explicitly). $T$ is negative if the first character is
% |-|, it is positive if the first character is neither |0| nor |-|.
% It is also positive if the first character is |0| and second
% argument of \cs{@@_div_significand_vi:Nw}, a sum of several terms, is
% also zero. Otherwise, there was an exact agreement: $T = 0$.
% \begin{macrocode}
\cs_new:Npn \@@_div_significand_iv:wwnnnnnnn #1; #2;#3#4#5 #6#7#8#9
{
+ 5 * #1
\exp_after:wN \@@_div_significand_vi:Nw
\int_value:w \@@_int_eval:w -20 + 2*#2#3 - #1*#6#7 +
\exp_after:wN \@@_div_significand_v:NN
\int_value:w \@@_int_eval:w 199980 + 2*#4 - #1*#8 +
\exp_after:wN \@@_div_significand_v:NN
\int_value:w \@@_int_eval:w 200000 + 2*#5 - #1*#9 ;
}
\cs_new:Npn \@@_div_significand_v:NN #1#2 { #1#2 \@@_int_eval_end: + }
\cs_new:Npn \@@_div_significand_vi:Nw #1#2;
{
\if_meaning:w 0 #1
\if_int_compare:w \@@_int_eval:w #2 > 0 + 1 \fi:
\else:
\if_meaning:w - #1 - \else: + \fi: 1
\fi:
;
}
% \end{macrocode}
% \end{macro}
%
% \begin{macro}[EXP]{\@@_div_significand_pack:NNN}
% At this stage, we are in the following situation: \TeX{} is in the
% process of expanding several integer expressions, thus functions at
% the bottom expand before those above.
% \begin{quote}
% \cs{@@_div_significand_test_o:w} $10^{6} + Q_{A}$
% \cs{@@_div_significand_pack:NNN} $10^{6} + Q_{B}$
% \cs{@@_div_significand_pack:NNN} $10^{6} + Q_{C}$
% \cs{@@_div_significand_pack:NNN}
% $10^{7} + 10\cdot Q_{D} + 5 \cdot P + \varepsilon$ |;| \meta{sign}
% \end{quote}
% Here, $\varepsilon = \operatorname{sign}(T)$ is $0$ in case $2E=PZ$,
% $1$ in case $2E>PZ$, which means that $P$ was the correct value, but
% not with an exact quotient, and $-1$ if $2E<PZ$, \emph{i.e.}, $P$
% was an overestimate. The packing function we define now does
% nothing special: it removes the $10^{6}$ and carries two digits (for
% the $10^{5}$'s and the $10^{4}$'s).
% \begin{macrocode}
\cs_new:Npn \@@_div_significand_pack:NNN 1 #1 #2 { + #1 #2 ; }
% \end{macrocode}
% \end{macro}
%
% \begin{macro}[rEXP]{\@@_div_significand_test_o:w}
% \begin{quote}
% \cs{@@_div_significand_test_o:w} |1| |0| \meta{5d} |;|
% ~~\meta{4d} |;| \meta{4d} |;| \meta{5d} |;| \meta{sign}
% \end{quote}
% The reason we know that the first two digits are |1| and |0| is that
% the final result is known to be between $0.1$ (inclusive) and $10$,
% hence $\widetilde{Q_{A}}$ (the tilde denoting the contribution from
% the other $Q_{i}$) is at most $99999$, and $10^{6}+\widetilde{Q_{A}}
% = 10\cdots$.
%
% It is now time to round. This depends on how many digits the final
% result will have.
% \begin{macrocode}
\cs_new:Npn \@@_div_significand_test_o:w 10 #1
{
\if_meaning:w 0 #1
\exp_after:wN \@@_div_significand_small_o:wwwNNNNwN
\else:
\exp_after:wN \@@_div_significand_large_o:wwwNNNNwN
\fi:
#1
}
% \end{macrocode}
% \end{macro}
%
% \begin{macro}[EXP]{\@@_div_significand_small_o:wwwNNNNwN}
% \begin{quote}
% \cs{@@_div_significand_small_o:wwwNNNNwN} |0| \meta{4d} |;|
% ~~\meta{4d} |;| \meta{4d} |;| \meta{5d} |;| \meta{final sign}
% \end{quote}
% Standard use of the functions \cs{@@_basics_pack_low:NNNNNw} and
% \cs{@@_basics_pack_high:NNNNNw}. We finally get to use the
% \meta{final sign} which has been sitting there for a while.
% \begin{macrocode}
\cs_new:Npn \@@_div_significand_small_o:wwwNNNNwN
0 #1; #2; #3; #4#5#6#7#8; #9
{
\exp_after:wN \@@_basics_pack_high:NNNNNw
\int_value:w \@@_int_eval:w 1 #1#2
\exp_after:wN \@@_basics_pack_low:NNNNNw
\int_value:w \@@_int_eval:w 1 #3#4#5#6#7
+ \@@_round:NNN #9 #7 #8
\exp_after:wN ;
}
% \end{macrocode}
% \end{macro}
%
% \begin{macro}[rEXP]{\@@_div_significand_large_o:wwwNNNNwN}
% \begin{quote}
% \cs{@@_div_significand_large_o:wwwNNNNwN} \meta{5d} |;|
% ~~\meta{4d} |;| \meta{4d} |;| \meta{5d} |;| \meta{sign}
% \end{quote}
% We know that the final result cannot reach $10$, hence |1#1#2|,
% together with contributions from the level below, cannot reach
% $2\cdot 10^{9}$. For rounding, we build the \meta{rounding digit}
% from the last two of our $18$ digits.
% \begin{macrocode}
\cs_new:Npn \@@_div_significand_large_o:wwwNNNNwN
#1; #2; #3; #4#5#6#7#8; #9
{
+ 1
\exp_after:wN \@@_basics_pack_weird_high:NNNNNNNNw
\int_value:w \@@_int_eval:w 1 #1 #2
\exp_after:wN \@@_basics_pack_weird_low:NNNNw
\int_value:w \@@_int_eval:w 1 #3 #4 #5 #6 +
\exp_after:wN \@@_round:NNN
\exp_after:wN #9
\exp_after:wN #6
\int_value:w \@@_round_digit:Nw #7 #8 ;
\exp_after:wN ;
}
% \end{macrocode}
% \end{macro}
%
% \subsection{Square root}
%
% \begin{macro}[EXP]{\@@_sqrt_o:w}
% Zeros are unchanged: $\sqrt{-0} = -0$ and $\sqrt{+0} = +0$.
% Negative numbers (other than $-0$) have no real square root.
% Positive infinity, and \texttt{nan}, are unchanged. Finally, for
% normal positive numbers, there is some work to do.
% \begin{macrocode}
\cs_new:Npn \@@_sqrt_o:w #1 \s_@@ \@@_chk:w #2#3#4; @
{
\if_meaning:w 0 #2 \@@_case_return_same_o:w \fi:
\if_meaning:w 2 #3
\@@_case_use:nw { \@@_invalid_operation_o:nw { sqrt } }
\fi:
\if_meaning:w 1 #2 \else: \@@_case_return_same_o:w \fi:
\@@_sqrt_npos_o:w
\s_@@ \@@_chk:w #2 #3 #4;
}
% \end{macrocode}
% \end{macro}
%
% \begin{macro}[EXP]{\@@_sqrt_npos_o:w}
% \begin{macro}[rEXP]
% {\@@_sqrt_npos_auxi_o:wwnnN, \@@_sqrt_npos_auxii_o:wNNNNNNNN}
% Prepare \cs{@@_sanitize:Nw} to receive the final sign~|0| (the
% result is always positive) and the exponent, equal to half of the
% exponent~|#1| of the argument. If the exponent~|#1| is even, find a
% first approximation of the square root of the significand $10^{8}
% a_1 + a_2 = 10^{8} |#2#3| + |#4#5|$ through Newton's method,
% starting at $x = 57234133 \simeq 10^{7.75}$. Otherwise, first shift
% the significand of of the argument by one digit, getting
% $a_1'\in[10^{6}, 10^{7})$ instead of $[10^{7}, 10^{8})$, then use
% Newton's method starting at $17782794 \simeq 10^{7.25}$.
% \begin{macrocode}
\cs_new:Npn \@@_sqrt_npos_o:w \s_@@ \@@_chk:w 1 0 #1#2#3#4#5;
{
\exp_after:wN \@@_sanitize:Nw
\exp_after:wN 0
\int_value:w \@@_int_eval:w
\if_int_odd:w #1 \exp_stop_f:
\exp_after:wN \@@_sqrt_npos_auxi_o:wwnnN
\fi:
#1 / 2
\@@_sqrt_Newton_o:wwn 56234133; 0; {#2#3} {#4#5} 0
}
\cs_new:Npn \@@_sqrt_npos_auxi_o:wwnnN #1 / 2 #2; 0; #3#4#5
{
( #1 + 1 ) / 2
\@@_pack_eight:wNNNNNNNN
\@@_sqrt_npos_auxii_o:wNNNNNNNN
;
0 #3 #4
}
\cs_new:Npn \@@_sqrt_npos_auxii_o:wNNNNNNNN #1; #2#3#4#5#6#7#8#9
{ \@@_sqrt_Newton_o:wwn 17782794; 0; {#1} {#2#3#4#5#6#7#8#9} }
% \end{macrocode}
% \end{macro}
% \end{macro}
%
% \begin{macro}[rEXP]{\@@_sqrt_Newton_o:wwn}
% Newton's method maps $x\mapsto\bigl[(x + [10^{8} a_1 / x])/2\bigr]$
% in each iteration, where $[b/c]$ denotes \eTeX{}'s division. This
% division rounds the real number $b/c$ to the closest integer,
% rounding ties away from zero, hence when $c$~is even,
% $b/c - 1/2 + 1/c \leq [b/c] \leq b/c + 1/2$
% and when $c$~is odd,
% $b/c - 1/2 + 1/(2c) \leq [b/c] \leq b/c + 1/2 - 1/(2c)$.
% For all~$c$, $b/c - 1/2 + 1/(2c) \leq [b/c] \leq b/c + 1/2$.
%
% Let us prove that the method converges when implemented with \eTeX{}
% integer division, for any $10^{6} \leq a_1 < 10^{8}$ and starting
% value $10^{6} \leq x < 10^{8}$. Using the inequalities above and
% the arithmetic--geometric inequality $(x+t)/2 \geq \sqrt{xt}$ for $t
% = 10^{8} a_1 / x$, we find
% \[
% x'
% = \left[\frac{x + [10^{8} a_1 / x]}{2}\right]
% \geq \frac{x + 10^{8} a_1 / x - 1/2 + 1/(2x)}{2}
% \geq \sqrt{10^{8} a_1} - \frac{1}{4} + \frac{1}{4x} \,.
% \]
% After any step of iteration, we thus have $\delta = x - \sqrt{10^{8}
% a_1} \geq -0.25 + 0.25 \cdot 10^{-8}$. The new difference
% $\delta' = x' - \sqrt{10^{8} a_1}$ after one step is bounded above
% as
% \[
% x' - \sqrt{10^{8} a_1}
% \leq \frac{x + 10^{8} a_1 / x + 1/2}{2} + \frac{1}{2}
% - \sqrt{10^{8} a_1}
% \leq \frac{\delta}{2} \frac{\delta}{\sqrt{10^{8} a_1} + \delta}
% + \frac{3}{4} \,.
% \]
% For $\delta > 3/2$, this last expression is
% $\leq\delta/2+3/4<\delta$, hence $\delta$~decreases at each step:
% since all~$x$ are integers, $\delta$~must reach a value
% $-1/4<\delta\leq 3/2$. In this range of values, we get $\delta'
% \leq \frac{3}{4} \frac{3}{2\sqrt{10^{8} a_1}} + \frac{3}{4} \leq
% 0.75 + 1.125 \cdot 10^{-7}$. We deduce that the difference $\delta
% = x - \sqrt{10^{8} a_1}$ eventually reaches a value in the interval
% $[-0.25 + 0.25\cdot 10^{-8}, 0.75 + 11.25 \cdot 10^{-8}]$, whose
% width is $1 + 11 \cdot 10^{-8}$. The corresponding interval for~$x$
% may contain two integers, hence $x$~might oscillate between those
% two values.
%
% However, the fact that $x\mapsto x-1$ and $x-1 \mapsto x$ puts
% stronger constraints, which are not compatible: the first implies
% \[
% x + [10^{8} a_1 / x] \leq 2x - 2
% \]
% hence $10^{8} a_1 / x \leq x - 3/2$, while the second implies
% \[
% x - 1 + [10^{8} a_1 / (x - 1)] \geq 2x - 1
% \]
% hence $10^{8} a_1 / (x - 1) \geq x - 1/2$. Combining the two
% inequalities yields $x^2 - 3x/2 \geq 10^{8} a_1 \geq x - 3x/2 +
% 1/2$, which cannot hold. Therefore, the iteration always converges
% to a single integer~$x$. To stop the iteration when two consecutive
% results are equal, the function \cs{@@_sqrt_Newton_o:wwn} receives
% the newly computed result as~|#1|, the previous result as~|#2|, and
% $a_1$ as~|#3|. Note that \eTeX{} combines the computation of a
% multiplication and a following division, thus avoiding overflow in
% |#3 * 100000000 / #1|. In any case, the result is within $[10^{7},
% 10^{8}]$.
% \begin{macrocode}
\cs_new:Npn \@@_sqrt_Newton_o:wwn #1; #2; #3
{
\if_int_compare:w #1 = #2 \exp_stop_f:
\exp_after:wN \@@_sqrt_auxi_o:NNNNwnnN
\int_value:w \@@_int_eval:w 9999 9999 +
\exp_after:wN \@@_use_none_until_s:w
\fi:
\exp_after:wN \@@_sqrt_Newton_o:wwn
\int_value:w \@@_int_eval:w (#1 + #3 * 1 0000 0000 / #1) / 2 ;
#1; {#3}
}
% \end{macrocode}
% \end{macro}
%
% \begin{macro}[rEXP]{\@@_sqrt_auxi_o:NNNNwnnN}
% This function is followed by $10^{8}+x-1$, which has~$9$ digits
% starting with~$1$, then |;| \Arg{a_1} \Arg{a_2} \meta{a'}. Here, $x
% \simeq \sqrt{10^{8} a_1}$ and we want to estimate the square root of
% $a = 10^{-8} a_1 + 10^{-16} a_2 + 10^{-17} a'$. We set up an
% initial underestimate
% \[
% y = (x - 1) 10^{-8} + 0.2499998875 \cdot 10^{-8} \lesssim \sqrt{a}\,.
% \]
% From the inequalities shown earlier, we know that $y \leq
% \sqrt{10^{-8} a_1} \leq \sqrt{a}$ and that $\sqrt{10^{-8} a_1} \leq
% y + 10^{-8} + 11\cdot 10^{-16}$ hence (using $0.1\leq y\leq
% \sqrt{a}\leq 1$)
% \[
% a - y^2 \leq 10^{-8} a_1 + 10^{-8} - y^2
% \leq (y + 10^{-8} + 11\cdot 10^{-16})^2 - y^2 + 10^{-8}
% < 3.2 \cdot 10^{-8} \,,
% \]
% and $\sqrt{a} - y = (a - y^2)/(\sqrt{a} + y) \leq 16 \cdot 10^{-8}$.
% Next, \cs{@@_sqrt_auxii_o:NnnnnnnnN} is called several times to
% get closer and closer underestimates of~$\sqrt{a}$. By
% construction, the underestimates~$y$ are always increasing, $a - y^2
% < 3.2 \cdot 10^{-8}$ for all. Also, $y<1$.
% \begin{macrocode}
\cs_new:Npn \@@_sqrt_auxi_o:NNNNwnnN 1 #1#2#3#4#5;
{
\@@_sqrt_auxii_o:NnnnnnnnN
\@@_sqrt_auxiii_o:wnnnnnnnn
{#1#2#3#4} {#5} {2499} {9988} {7500}
}
% \end{macrocode}
% \end{macro}
%
% \begin{macro}[rEXP]{\@@_sqrt_auxii_o:NnnnnnnnN}
% This receives a continuation function~|#1|, then five blocks of~$4$
% digits for~$y$, then two $8$-digit blocks and a single digit
% for~$a$. A common estimate of $\sqrt{a} - y = (a - y^2) / (\sqrt{a}
% + y)$ is $(a - y^2)/(2y)$, which leads to alternating overestimates
% and underestimates. We tweak this, to only work with underestimates
% (no need then to worry about signs in the computation). Each step
% finds the largest integer $j\leq 6$ such that $10^{4j}(a-y^2) <
% 2\cdot 10^{8}$, then computes the integer (with \eTeX{}'s rounding
% division)
% \[
% 10^{4j} z =
% \Bigl[\bigl(\lfloor 10^{4j}(a-y^2)\rfloor - 257\bigr)
% \cdot (0.5\cdot 10^{8})
% \Bigm/ \lfloor 10^{8} y + 1\rfloor\Bigr] \,.
% \]
% The choice of~$j$ ensures that $10^{4j} z < 2\cdot 10^{8} \cdot
% 0.5\cdot 10^{8} / 10^{7} = 10^{9}$, thus $10^{9} + 10^{4j} z$ has
% exactly $10$~digits, does not overflow \TeX{}'s integer range, and
% starts with~$1$. Incidentally, since all $a - y^2 \leq 3.2\cdot
% 10^{-8}$, we know that $j\geq 3$.
%
% Let us show that $z$ is an underestimate of $\sqrt{a} - y$. On the
% one hand, $\sqrt{a} - y \leq 16\cdot 10^{-8}$ because this holds for
% the initial~$y$ and values of~$y$ can only increase. On the other
% hand, the choice of~$j$ implies that $\sqrt{a} - y \leq
% 5(\sqrt{a}+y)(\sqrt{a}-y) = 5(a - y^2) < 10^{9-4j}$. For $j=3$, the
% first bound is better, while for larger~$j$, the second bound is
% better. For all $j\in[3,6]$, we find $\sqrt{a}-y < 16\cdot
% 10^{-2j}$. From this, we deduce that
% \[
% 10^{4j} (\sqrt{a}-y)
% = \frac{10^{4j}\bigl(a-y^2-(\sqrt{a}-y)^2\bigr)}{2y}
% \geq \frac{\bigl\lfloor 10^{4j}(a-y^2)\bigr\rfloor-257}
% {2\cdot 10^{-8} \lfloor 10^{8}y+1\rfloor}
% + \frac{1}{2}
% \]
% where we have replaced the bound $10^{4j}(16\cdot 10^{-2j}) = 256$
% by~$257$ and extracted the corresponding term $1/\bigl(2\cdot
% 10^{-8} \lfloor 10^{8}y+1\rfloor\bigr) \geq 1/2$. Given that
% \eTeX{}'s integer division obeys $[b/c] \leq b/c + 1/2$, we deduce
% that $10^{4j} z \leq 10^{4j} (\sqrt{a}-y)$, hence $y+z\leq\sqrt{a}$
% is an underestimate of~$\sqrt{a}$, as claimed. One implementation
% detail: because the computation involves |-#4*#4| |-| |2*#3*#5| |-|
% |2*#2*#6| which may be as low as $-5\cdot 10^{8}$, we need to use
% the \texttt{pack_big} functions, and the \texttt{big} shifts.
% \begin{macrocode}
\cs_new:Npn \@@_sqrt_auxii_o:NnnnnnnnN #1 #2#3#4#5#6 #7#8#9
{
\exp_after:wN #1
\int_value:w \@@_int_eval:w \c_@@_big_leading_shift_int
+ #7 - #2 * #2
\exp_after:wN \@@_pack_big:NNNNNNw
\int_value:w \@@_int_eval:w \c_@@_big_middle_shift_int
- 2 * #2 * #3
\exp_after:wN \@@_pack_big:NNNNNNw
\int_value:w \@@_int_eval:w \c_@@_big_middle_shift_int
+ #8 - #3 * #3 - 2 * #2 * #4
\exp_after:wN \@@_pack_big:NNNNNNw
\int_value:w \@@_int_eval:w \c_@@_big_middle_shift_int
- 2 * #3 * #4 - 2 * #2 * #5
\exp_after:wN \@@_pack_big:NNNNNNw
\int_value:w \@@_int_eval:w \c_@@_big_middle_shift_int
+ #9 000 0000 - #4 * #4 - 2 * #3 * #5 - 2 * #2 * #6
\exp_after:wN \@@_pack_big:NNNNNNw
\int_value:w \@@_int_eval:w \c_@@_big_middle_shift_int
- 2 * #4 * #5 - 2 * #3 * #6
\exp_after:wN \@@_pack_big:NNNNNNw
\int_value:w \@@_int_eval:w \c_@@_big_middle_shift_int
- #5 * #5 - 2 * #4 * #6
\exp_after:wN \@@_pack_big:NNNNNNw
\int_value:w \@@_int_eval:w
\c_@@_big_middle_shift_int
- 2 * #5 * #6
\exp_after:wN \@@_pack_big:NNNNNNw
\int_value:w \@@_int_eval:w
\c_@@_big_trailing_shift_int
- #6 * #6 ;
% (
- 257 ) * 5000 0000 / (#2#3 + 1) + 10 0000 0000 ;
{#2}{#3}{#4}{#5}{#6} {#7}{#8}#9
}
% \end{macrocode}
% \end{macro}
%
% \begin{macro}[rEXP]
% {
% \@@_sqrt_auxiii_o:wnnnnnnnn,
% \@@_sqrt_auxiv_o:NNNNNw,
% \@@_sqrt_auxv_o:NNNNNw,
% \@@_sqrt_auxvi_o:NNNNNw,
% \@@_sqrt_auxvii_o:NNNNNw
% }
% We receive here the difference $a-y^2=d=\sum_i d_i \cdot 10^{-4i}$,
% as \meta{d_2} |;| \Arg{d_3} \ldots{} \Arg{d_{10}}, where each block
% has~$4$ digits, except \meta{d_2}. This function finds the largest
% $j\leq 6$ such that $10^{4j}(a-y^2) < 2\cdot 10^{8}$, then leaves an
% open parenthesis and the integer
% $\bigl\lfloor 10^{4j}(a-y^2)\bigr\rfloor$ in an integer
% expression. The closing parenthesis is provided by the caller
% \cs{@@_sqrt_auxii_o:NnnnnnnnN}, which completes the expression
% \[
% 10^{4j} z =
% \Bigl[\bigl(\lfloor 10^{4j}(a-y^2)\rfloor - 257\bigr)
% \cdot (0.5\cdot 10^{8})
% \Bigm/ \lfloor 10^{8} y + 1\rfloor\Bigr]
% \]
% for an estimate of $10^{4j} (\sqrt{a} - y)$. If $d_2\geq 2$, $j=3$
% and the \texttt{auxiv} auxiliary receives $10^{12} z$. If $d_2\leq
% 1$ but $10^{4} d_2 + d_3 \geq 2$, $j=4$ and the \texttt{auxv}
% auxiliary is called, and receives $10^{16} z$, and so on. In all
% those cases, the \texttt{auxviii} auxiliary is set up to add~$z$
% to~$y$, then go back to the \texttt{auxii} step with continuation
% \texttt{auxiii} (the function we are currently describing). The
% maximum value of $j$ is~$6$, regardless of whether $10^{12} d_2 +
% 10^{8} d_3 + 10^{4} d_4 + d_5 \geq 1$. In this last case, we detect
% when $10^{24} z < 10^{7}$, which essentially means $\sqrt{a} - y
% \lesssim 10^{-17}$: once this threshold is reached, there is enough
% information to find the correctly rounded~$\sqrt{a}$ with only one
% more call to \cs{@@_sqrt_auxii_o:NnnnnnnnN}. Note that the
% iteration cannot be stuck before reaching $j=6$, because for $j<6$,
% one has $2\cdot 10^{8}\leq 10^{4(j+1)}(a-y^2)$, hence
% \[
% 10^{4j} z
% \geq \frac{(20000-257)(0.5\cdot 10^{8})}{\lfloor 10^{8} y + 1\rfloor}
% \geq (20000-257)\cdot 0.5 > 0 \,.
% \]
% \begin{macrocode}
\cs_new:Npn \@@_sqrt_auxiii_o:wnnnnnnnn
#1; #2#3#4#5#6#7#8#9
{
\if_int_compare:w #1 > 1 \exp_stop_f:
\exp_after:wN \@@_sqrt_auxiv_o:NNNNNw
\int_value:w \@@_int_eval:w (#1#2 %)
\else:
\if_int_compare:w #1#2 > 1 \exp_stop_f:
\exp_after:wN \@@_sqrt_auxv_o:NNNNNw
\int_value:w \@@_int_eval:w (#1#2#3 %)
\else:
\if_int_compare:w #1#2#3 > 1 \exp_stop_f:
\exp_after:wN \@@_sqrt_auxvi_o:NNNNNw
\int_value:w \@@_int_eval:w (#1#2#3#4 %)
\else:
\exp_after:wN \@@_sqrt_auxvii_o:NNNNNw
\int_value:w \@@_int_eval:w (#1#2#3#4#5 %)
\fi:
\fi:
\fi:
}
\cs_new:Npn \@@_sqrt_auxiv_o:NNNNNw 1#1#2#3#4#5#6;
{ \@@_sqrt_auxviii_o:nnnnnnn {#1#2#3#4#5#6} {00000000} }
\cs_new:Npn \@@_sqrt_auxv_o:NNNNNw 1#1#2#3#4#5#6;
{ \@@_sqrt_auxviii_o:nnnnnnn {000#1#2#3#4#5} {#60000} }
\cs_new:Npn \@@_sqrt_auxvi_o:NNNNNw 1#1#2#3#4#5#6;
{ \@@_sqrt_auxviii_o:nnnnnnn {0000000#1} {#2#3#4#5#6} }
\cs_new:Npn \@@_sqrt_auxvii_o:NNNNNw 1#1#2#3#4#5#6;
{
\if_int_compare:w #1#2 = 0 \exp_stop_f:
\exp_after:wN \@@_sqrt_auxx_o:Nnnnnnnn
\fi:
\@@_sqrt_auxviii_o:nnnnnnn {00000000} {000#1#2#3#4#5}
}
% \end{macrocode}
% \end{macro}
%
% \begin{macro}[rEXP]
% {\@@_sqrt_auxviii_o:nnnnnnn, \@@_sqrt_auxix_o:wnwnw}
% Simply add the two $8$-digit blocks of~$z$, aligned to the last four
% of the five $4$-digit blocks of~$y$, then call the \texttt{auxii}
% auxiliary to evaluate $y'^{2} = (y+z)^{2}$.
% \begin{macrocode}
\cs_new:Npn \@@_sqrt_auxviii_o:nnnnnnn #1#2 #3#4#5#6#7
{
\exp_after:wN \@@_sqrt_auxix_o:wnwnw
\int_value:w \@@_int_eval:w #3
\exp_after:wN \@@_basics_pack_low:NNNNNw
\int_value:w \@@_int_eval:w #1 + 1#4#5
\exp_after:wN \@@_basics_pack_low:NNNNNw
\int_value:w \@@_int_eval:w #2 + 1#6#7 ;
}
\cs_new:Npn \@@_sqrt_auxix_o:wnwnw #1; #2#3; #4#5;
{
\@@_sqrt_auxii_o:NnnnnnnnN
\@@_sqrt_auxiii_o:wnnnnnnnn {#1}{#2}{#3}{#4}{#5}
}
% \end{macrocode}
% \end{macro}
%
% \begin{macro}[rEXP]
% {\@@_sqrt_auxx_o:Nnnnnnnn, \@@_sqrt_auxxi_o:wwnnN}
% At this stage, $j=6$ and $10^{24} z < 10^{7}$, hence
% \[
% 10^{7} + 1/2 > 10^{24} z + 1/2 \geq
% \bigl(10^{24}(a-y^2) - 258\bigr) \cdot (0.5\cdot 10^{8})
% \Bigm/ (10^{8} y + 1) \,,
% \]
% then $10^{24}(a-y^2) - 258 < 2 (10^{7} + 1/2) (y + 10^{-8})$, and
% \[
% 10^{24}(a-y^2)
% < (10^{7} + 1290.5) (1 + 10^{-8}/y) (2y)
% < (10^{7} + 1290.5) (1 + 10^{-7}) (y + \sqrt{a}) \,,
% \]
% which finally implies $0\leq\sqrt{a}-y < 0.2\cdot 10^{-16}$. In
% particular, $y$~is an underestimate of~$\sqrt{a}$ and $y+0.5\cdot
% 10^{-16}$ is a (strict) overestimate. There is at exactly one
% multiple $m$~of $0.5\cdot 10^{-16}$ in the interval $[y, y+0.5\cdot
% 10^{-16})$. If $m^2>a$, then the square root is inexact and is
% obtained by rounding $m-\epsilon$ to a multiple of $10^{-16}$ (the
% precise shift $0<\epsilon<0.5\cdot 10^{-16}$ is irrelevant for
% rounding). If $m^2=a$ then the square root is exactly~$m$, and
% there is no rounding. If $m^2<a$ then we round $m+\epsilon$. For
% now, discard a few irrelevant arguments |#1|, |#2|, |#3|, and find
% the multiple of $0.5\cdot 10^{-16}$ within $[y, y+0.5\cdot
% 10^{-16})$; rather, only the last $4$~digits |#8| of~$y$ are
% considered, and we do not perform any carry yet. The \texttt{auxxi}
% auxiliary sets up \texttt{auxii} with a continuation function
% \texttt{auxxii} instead of \texttt{auxiii} as before. To prevent
% \texttt{auxii} from giving a negative results $a-m^2$, we compute
% $a+10^{-16}-m^2$ instead, always positive since $m<\sqrt{a}+0.5\cdot
% 10^{-16}$ and $a\leq 1-10^{-16}$.
% \begin{macrocode}
\cs_new:Npn \@@_sqrt_auxx_o:Nnnnnnnn #1#2#3 #4#5#6#7#8
{
\exp_after:wN \@@_sqrt_auxxi_o:wwnnN
\int_value:w \@@_int_eval:w
(#8 + 2499) / 5000 * 5000 ;
{#4} {#5} {#6} {#7} ;
}
\cs_new:Npn \@@_sqrt_auxxi_o:wwnnN #1; #2; #3#4#5
{
\@@_sqrt_auxii_o:NnnnnnnnN
\@@_sqrt_auxxii_o:nnnnnnnnw
#2 {#1}
{#3} { #4 + 1 } #5
}
% \end{macrocode}
% \end{macro}
%
% \begin{macro}[rEXP]
% {\@@_sqrt_auxxii_o:nnnnnnnnw, \@@_sqrt_auxxiii_o:w}
% The difference $0\leq a+10^{-16}-m^2\leq
% 10^{-16}+(\sqrt{a}-m)(\sqrt{a}+m)\leq 2\cdot 10^{-16}$ was just
% computed: its first $8$~digits vanish, as do the next four,~|#1|,
% and most of the following four,~|#2|. The guess~$m$ is an
% overestimate if $a+10^{-16}-m^2 < 10^{-16}$, that is, |#1#2|
% vanishes. Otherwise it is an underestimate, unless
% $a+10^{-16}-m^2=10^{-16}$ exactly. For an underestimate, call the
% \texttt{auxxiv} function with argument~$9998$. For an exact result
% call it with~$9999$, and for an overestimate call it with~$10000$.
% \begin{macrocode}
\cs_new:Npn \@@_sqrt_auxxii_o:nnnnnnnnw 0; #1#2#3#4#5#6#7#8 #9;
{
\if_int_compare:w #1#2 > 0 \exp_stop_f:
\if_int_compare:w #1#2 = 1 \exp_stop_f:
\if_int_compare:w #3#4 = 0 \exp_stop_f:
\if_int_compare:w #5#6 = 0 \exp_stop_f:
\if_int_compare:w #7#8 = 0 \exp_stop_f:
\@@_sqrt_auxxiii_o:w
\fi:
\fi:
\fi:
\fi:
\exp_after:wN \@@_sqrt_auxxiv_o:wnnnnnnnN
\int_value:w 9998
\else:
\exp_after:wN \@@_sqrt_auxxiv_o:wnnnnnnnN
\int_value:w 10000
\fi:
;
}
\cs_new:Npn \@@_sqrt_auxxiii_o:w \fi: \fi: \fi: \fi: #1 \fi: ;
{
\fi: \fi: \fi: \fi: \fi:
\@@_sqrt_auxxiv_o:wnnnnnnnN 9999 ;
}
% \end{macrocode}
% \end{macro}
%
% \begin{macro}[rEXP]{\@@_sqrt_auxxiv_o:wnnnnnnnN}
% This receives $9998$, $9999$ or $10000$ as~|#1| when $m$~is an
% underestimate, exact, or an overestimate, respectively. Then
% comes~$m$ as five blocks of~$4$ digits, but where the last
% block~|#6| may be $0$, $5000$, or~$10000$. In the latter case, we
% need to add a carry, unless $m$~is an overestimate (|#1|~is then
% $10000$). Then comes~$a$ as three arguments. Rounding is done by
% \cs{@@_round:NNN}, whose first argument is the final sign~$0$
% (square roots are positive). We fake its second argument. It
% should be the last digit kept, but this is only used when ties are
% \enquote{rounded to even}, and only when the result is exactly
% half-way between two representable numbers rational square roots of
% numbers with $16$~significant digits have: this situation never
% arises for the square root, as any exact square root of a $16$~digit
% number has at most $8$~significant digits. Finally, the last
% argument is the next digit, possibly shifted by~$1$ when there are
% further nonzero digits. This is achieved by \cs{@@_round_digit:Nw},
% which receives (after removal of the $10000$'s digit) one of $0000$,
% $0001$, $4999$, $5000$, $5001$, or~$9999$, which it converts to $0$,
% $1$, $4$, $5$, $6$, and~$9$, respectively.
% \begin{macrocode}
\cs_new:Npn \@@_sqrt_auxxiv_o:wnnnnnnnN #1; #2#3#4#5#6 #7#8#9
{
\exp_after:wN \@@_basics_pack_high:NNNNNw
\int_value:w \@@_int_eval:w 1 0000 0000 + #2#3
\exp_after:wN \@@_basics_pack_low:NNNNNw
\int_value:w \@@_int_eval:w 1 0000 0000
+ #4#5
\if_int_compare:w #6 > #1 \exp_stop_f: + 1 \fi:
+ \exp_after:wN \@@_round:NNN
\exp_after:wN 0
\exp_after:wN 0
\int_value:w
\exp_after:wN \use_i:nn
\exp_after:wN \@@_round_digit:Nw
\int_value:w \@@_int_eval:w #6 + 19999 - #1 ;
\exp_after:wN ;
}
% \end{macrocode}
% \end{macro}
%
% \subsection{About the sign and exponent}
%
% \begin{macro}[EXP]{\@@_logb_o:w, \@@_logb_aux_o:w}
% The exponent of a normal number is its \meta{exponent} minus one.
% \begin{macrocode}
\cs_new:Npn \@@_logb_o:w ? \s_@@ \@@_chk:w #1#2; @
{
\if_case:w #1 \exp_stop_f:
\@@_case_use:nw
{ \@@_division_by_zero_o:Nnw \c_minus_inf_fp { logb } }
\or: \exp_after:wN \@@_logb_aux_o:w
\or: \@@_case_return_o:Nw \c_inf_fp
\else: \@@_case_return_same_o:w
\fi:
\s_@@ \@@_chk:w #1 #2;
}
\cs_new:Npn \@@_logb_aux_o:w \s_@@ \@@_chk:w #1 #2 #3 #4 ;
{
\exp_after:wN \@@_parse:n \exp_after:wN
{ \int_value:w \int_eval:w #3 - 1 \exp_after:wN }
}
% \end{macrocode}
% \end{macro}
%
% \begin{macro}[EXP]{\@@_sign_o:w}
% \begin{macro}[EXP]{\@@_sign_aux_o:w}
% Find the sign of the floating point: \texttt{nan}, |+0|, |-0|, |+1| or |-1|.
% \begin{macrocode}
\cs_new:Npn \@@_sign_o:w ? \s_@@ \@@_chk:w #1#2; @
{
\if_case:w #1 \exp_stop_f:
\@@_case_return_same_o:w
\or: \exp_after:wN \@@_sign_aux_o:w
\or: \exp_after:wN \@@_sign_aux_o:w
\else: \@@_case_return_same_o:w
\fi:
\s_@@ \@@_chk:w #1 #2;
}
\cs_new:Npn \@@_sign_aux_o:w \s_@@ \@@_chk:w #1 #2 #3 ;
{ \exp_after:wN \@@_set_sign_o:w \exp_after:wN #2 \c_one_fp @ }
% \end{macrocode}
% \end{macro}
% \end{macro}
%
% \begin{macro}[EXP]{\@@_set_sign_o:w}
% This function is used for the unary minus and for \texttt{abs}. It
% leaves the sign of \texttt{nan} invariant, turns negative numbers
% (sign~$2$) to positive numbers (sign~$0$) and positive numbers
% (sign~$0$) to positive or negative numbers depending on~|#1|. It
% also expands after itself in the input stream, just like
% \cs{@@_+_o:ww}.
% \begin{macrocode}
\cs_new:Npn \@@_set_sign_o:w #1 \s_@@ \@@_chk:w #2#3#4; @
{
\exp_after:wN \@@_exp_after_o:w
\exp_after:wN \s_@@
\exp_after:wN \@@_chk:w
\exp_after:wN #2
\int_value:w
\if_case:w #3 \exp_stop_f: #1 \or: 1 \or: 0 \fi: \exp_stop_f:
#4;
}
% \end{macrocode}
% \end{macro}
%
% \subsection{Operations on tuples}
%
% \begin{macro}[EXP]{\@@_tuple_set_sign_o:w}
% \begin{macro}[EXP]{\@@_tuple_set_sign_aux_o:Nnw, \@@_tuple_set_sign_aux_o:w}
% Two cases: |abs(|\meta{tuple}|)| for which |#1| is $0$ (invalid for
% tuples) and |-|\meta{tuple} for which |#1| is $2$. In that case,
% map over all items in the tuple an auxiliary that dispatches to the
% type-appropriate sign-flipping function.
% \begin{macrocode}
\cs_new:Npn \@@_tuple_set_sign_o:w #1
{
\if_meaning:w 2 #1
\exp_after:wN \@@_tuple_set_sign_aux_o:Nnw
\fi:
\@@_invalid_operation_o:nw { abs }
}
\cs_new:Npn \@@_tuple_set_sign_aux_o:Nnw #1#2#3 @
{ \@@_tuple_map_o:nw \@@_tuple_set_sign_aux_o:w #3 }
\cs_new:Npn \@@_tuple_set_sign_aux_o:w #1#2 ;
{
\@@_change_func_type:NNN #1 \@@_set_sign_o:w
\@@_parse_apply_unary_error:NNw
2 #1 #2 ; @
}
% \end{macrocode}
% \end{macro}
% \end{macro}
%
% \begin{macro}[EXP]{\@@_*_tuple_o:ww, \@@_tuple_*_o:ww, \@@_tuple_/_o:ww}
% For \meta{number}|*|\meta{tuple} and \meta{tuple}|*|\meta{number}
% and \meta{tuple}|/|\meta{number}, loop through the \meta{tuple} some
% code that multiplies or divides by the appropriate \meta{number}.
% Importantly we need to dispatch according to the type, and we make
% sure to apply the operator in the correct order.
% \begin{macrocode}
\cs_new:cpn { @@_*_tuple_o:ww } #1 ;
{ \@@_tuple_map_o:nw { \@@_binary_type_o:Nww * #1 ; } }
\cs_new:cpn { @@_tuple_*_o:ww } #1 ; #2 ;
{ \@@_tuple_map_o:nw { \@@_binary_rev_type_o:Nww * #2 ; } #1 ; }
\cs_new:cpn { @@_tuple_/_o:ww } #1 ; #2 ;
{ \@@_tuple_map_o:nw { \@@_binary_rev_type_o:Nww / #2 ; } #1 ; }
% \end{macrocode}
% \end{macro}
%
% \begin{macro}[EXP]{\@@_tuple_+_tuple_o:ww, \@@_tuple_-_tuple_o:ww}
% Check the two tuples have the same number of items and map through
% these a helper that dispatches appropriately depending on the types.
% This means |(1,2)+((1,1),2)| gives |(nan,4)|.
% \begin{macrocode}
\cs_set_protected:Npn \@@_tmp:w #1
{
\cs_new:cpn { @@_tuple_#1_tuple_o:ww }
\s_@@_tuple \@@_tuple_chk:w ##1 ;
\s_@@_tuple \@@_tuple_chk:w ##2 ;
{
\int_compare:nNnTF
{ \@@_array_count:n {##1} } = { \@@_array_count:n {##2} }
{ \@@_tuple_mapthread_o:nww { \@@_binary_type_o:Nww #1 } }
{ \@@_invalid_operation_o:nww #1 }
\s_@@_tuple \@@_tuple_chk:w {##1} ;
\s_@@_tuple \@@_tuple_chk:w {##2} ;
}
}
\@@_tmp:w +
\@@_tmp:w -
% \end{macrocode}
% \end{macro}
%
% \begin{macrocode}
%</initex|package>
% \end{macrocode}
%
% \end{implementation}
%
% \PrintChanges
%
% \PrintIndex
|