Fix zlib usage with TNG

[alexxy/gromacs.git] / docs / manual / implement.tex

%
% This file is part of the GROMACS molecular simulation package.
%
% Copyright (c) 2013,2014, by the GROMACS development team, led by
% Mark Abraham, David van der Spoel, Berk Hess, and Erik Lindahl,
% and including many others, as listed in the AUTHORS file in the
% top-level source directory and at http://www.gromacs.org.
%
% GROMACS is free software; you can redistribute it and/or
% modify it under the terms of the GNU Lesser General Public License
% as published by the Free Software Foundation; either version 2.1
% of the License, or (at your option) any later version.
%
% GROMACS is distributed in the hope that it will be useful,
% but WITHOUT ANY WARRANTY; without even the implied warranty of
% MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the GNU
% Lesser General Public License for more details.
%
% You should have received a copy of the GNU Lesser General Public
% License along with GROMACS; if not, see
% http://www.gnu.org/licenses, or write to the Free Software Foundation,
% Inc., 51 Franklin Street, Fifth Floor, Boston, MA  02110-1301  USA.
%
% If you want to redistribute modifications to GROMACS, please
% consider that scientific software is very special. Version
% control is crucial - bugs must be traceable. We will be happy to
% consider code for inclusion in the official distribution, but
% derived work must not be called official GROMACS. Details are found
% in the README & COPYING files - if they are missing, get the
% official version at http://www.gromacs.org.
%
% To help us fund GROMACS development, we humbly ask that you cite
% the research papers on the package. Check out http://www.gromacs.org.

\chapter{Some implementation details}
In this chapter we will present some implementation details. This is
far from complete, but we deemed it necessary to clarify some things
that would otherwise be hard to understand.

\section{Single Sum Virial in {\gromacs}}
\label{sec:virial}
The \normindex{virial} $\Xi$ can be written in full tensor form as:
\beq
\Xi~=~-\half~\sum_{i < j}^N~\rvij\otimes\Fvij
\eeq
where $\otimes$ denotes the {\em direct product} of two vectors.\footnote
{$({\bf u}\otimes{\bf v})^{\ab}~=~{\bf u}_{\al}{\bf v}_{\be}$} When this is 
computed in the inner loop of an MD program 9 multiplications and 9
additions are needed.\footnote{The calculation of 
Lennard-Jones and Coulomb forces is about 50 floating point operations.}

Here it is shown how it is possible to extract the virial calculation
from the inner loop~\cite{Bekker93b}.

\subsection{Virial}
In a system with \index{periodic boundary conditions}, the
periodicity must be taken into account for the virial:
\beq
\Xi~=~-\half~\sum_{i < j}^{N}~\rnij\otimes\Fvij
\eeq
where $\rnij$ denotes the distance vector of the
{\em nearest image} of atom $i$ from atom $j$. In this definition we add
a {\em shift vector} $\delta_i$ to the position vector $\rvi$ 
of atom $i$. The difference vector $\rnij$ is thus equal to:
\beq
\rnij~=~\rvi+\delta_i-\rvj
\eeq
or in shorthand:
\beq
\rnij~=~\rni-\rvj
\eeq
In a triclinic system, there are 27 possible images of $i$; when a truncated 
octahedron is used, there are 15 possible images.

\subsection{Virial from non-bonded forces}
Here the derivation for the single sum virial in the {\em non-bonded force} 
routine is given. $i \neq j$ in all formulae below.
\newcommand{\di}{\delta_{i}}
\newcommand{\qrt}{\frac{1}{4}}
\bea
\Xi     
&~=~&-\half~\sum_{i < j}^{N}~\rnij\otimes\Fvij                          \\
&~=~&-\qrt\sum_{i=1}^N~\sum_{j=1}^N ~(\rvi+\di-\rvj)\otimes\Fvij        \\
&~=~&-\qrt\sum_{i=1}^N~\sum_{j=1}^N ~(\rvi+\di)\otimes\Fvij-\rvj\otimes\Fvij    \\
&~=~&-\qrt\left(\sum_{i=1}^N~\sum_{j=1}^N ~(\rvi+\di)\otimes\Fvij~-~\sum_{i=1}^N~\sum_{j=1}^N ~\rvj\otimes\Fvij\right)  \\
&~=~&-\qrt\left(\sum_{i=1}^N~(\rvi+\di)\otimes\sum_{j=1}^N~\Fvij~-~\sum_{j=1}^N ~\rvj\otimes\sum_{i=1}^N~\Fvij\right)   \\
&~=~&-\qrt\left(\sum_{i=1}^N~(\rvi+\di)\otimes\Fvi~+~\sum_{j=1}^N ~\rvj\otimes\Fvj\right)       \\
&~=~&-\qrt\left(2~\sum_{i=1}^N~\rvi\otimes\Fvi+\sum_{i=1}^N~\di\otimes\Fvi\right)
\eea
In these formulae we introduced:
\bea
\Fvi&~=~&\sum_{j=1}^N~\Fvij                                     \\
\Fvj&~=~&\sum_{i=1}^N~\Fvji
\eea
which is the total force on $i$ with respect to $j$. Because we use Newton's Third Law:
\beq
\Fvij~=~-\Fvji
\eeq
we must, in the implementation, double the term containing the shift $\delta_i$.

\subsection{The intra-molecular shift (mol-shift)}
For the bonded forces and SHAKE it is possible to make a {\em mol-shift}
list, in which the periodicity is stored. We simple have an array {\tt mshift}
in which for each atom an index in the {\tt shiftvec} array is stored.

The algorithm to generate such a list can be derived from graph theory,
considering each particle in a molecule as a bead in a graph, the bonds 
as edges.
\begin{enumerate}
\item[1]        Represent the bonds and atoms as bidirectional graph
\item[2]        Make all atoms white
\item[3]        Make one of the white atoms black (atom $i$) and put it in the
                central box
\item[4]        Make all of the neighbors of $i$ that are currently 
                white, gray 
\item[5]        Pick one of the gray atoms (atom $j$), give it the
                correct periodicity with respect to any of 
                its black neighbors
                and make it black
\item[6]        Make all of the neighbors of $j$ that are currently 
                white, gray
\item[7]        If any gray atom remains, go to [5]
\item[8]        If any white atom remains, go to [3]
\end{enumerate}
Using this algorithm we can 
\begin{itemize}
\item   optimize the bonded force calculation as well as SHAKE 
\item   calculate the virial from the bonded forces
        in the single sum method again
\end{itemize}

Find a representation of the bonds as a bidirectional graph.

\subsection{Virial from Covalent Bonds}
Since the covalent bond force gives a contribution to the virial, we have:
\bea
b       &~=~&   \|\rnij\|                                       \\
V_b     &~=~&   \half k_b(b-b_0)^2                              \\
\Fvi    &~=~&   -\nabla V_b                                     \\
        &~=~&   k_b(b-b_0)\frac{\rnij}{b}                       \\
\Fvj    &~=~&   -\Fvi
\eea
The virial contribution from the bonds then is:
\bea
\Xi_b   &~=~&   -\half(\rni\otimes\Fvi~+~\rvj\otimes\Fvj)       \\
        &~=~&   -\half\rnij\otimes\Fvi
\eea

\subsection{Virial from SHAKE}
An important contribution to the virial comes from shake. Satisfying 
the constraints a force {\bf G} that is exerted on the particles ``shaken.'' If this
force does not come out of the algorithm (as in standard SHAKE) it can be
calculated afterward (when using {\em leap-frog}) by:
\bea
\Delta\rvi&~=~&\rvi(t+\Dt)-
[\rvi(t)+{\bf v}_i(t-\frac{\Dt}{2})\Dt+\frac{\Fvi}{m_i}\Dt^2]   \\
{\bf G}_i&~=~&\frac{m_i\Delta\rvi}{\Dt^2}
\eea
This does not help us in the general case. Only when no periodicity
is needed (like in rigid water) this can be used, otherwise
we must add the virial calculation in the inner loop of SHAKE.

When it {\em is} applicable the virial can be calculated in the single sum way:
\beq
\Xi~=~-\half\sum_i^{N_c}~\rvi\otimes\Fvi
\eeq
where $N_c$ is the number of constrained atoms.

%Another method is the Non-Iterative shake as proposed (and implemented)
%by Yoneya. In this algorithm the Lagrangian multipliers are solved in a 
%matrix equation, and given these multipliers it is easy to get the periodicity
%in the virial afterwards. 

%More...


\section{Optimizations}
Here we describe some of the algorithmic optimizations used 
in {\gromacs}, apart from parallelism. 
One of these, the implementation of the 
1.0/sqrt(x) function is treated separately in \secref{sqrt}.
The most important other optimizations are described below.

\subsection{Inner Loops for Water}
\label{sec:waterloops}
{\gromacs} uses special inner loops to calculate non-bonded
interactions for water molecules with other atoms, and yet
another set of loops for interactions between pairs of
water molecules. There highly optimized loops for two types of water models.
For three site models similar to
SPC~\cite{Berendsen81}, {\ie}:
\begin{enumerate}
\item   There are three atoms in the molecule.
\item   The whole molecule is a single charge group.
\item   The first atom has Lennard-Jones (\secref{lj}) and 
        Coulomb (\secref{coul}) interactions.
\item   Atoms two and three have only Coulomb interactions, 
        and equal charges.
\end{enumerate}
These loops also works for the SPC/E~\cite{Berendsen87} and 
TIP3P~\cite{Jorgensen83} water models.
And for four site water models similar to TIP4P~\cite{Jorgensen83}:
\begin{enumerate}
\item   There are four atoms in the molecule.
\item   The whole molecule is a single charge group.
\item   The first atom has only Lennard-Jones (\secref{lj}) interactions.
\item   Atoms two and three have only Coulomb (\secref{coul}) interactions, 
        and equal charges.
\item   Atom four has only Coulomb interactions.
\end{enumerate}

The benefit of these implementations is that there are more floating-point
operations in a single loop, which implies that some compilers
can schedule the code better. However, it turns out that even
some of the most advanced compilers have problems with scheduling, 
implying that manual tweaking is necessary to get optimum 
\normindex{performance}.
This may include common-sub-expression elimination, or moving
code around. 

\subsection{Fortran Code}
Unfortunately, on a few platforms \normindex{Fortran} compilers are
still better than C-compilers. For some machines ({\eg} SGI
Power Challenge) the difference may be up to a factor of 3, in the
case of vector computers this may be even larger. Therefore, some of
the routines that take up a lot of computer time have been translated
into Fortran and even assembly code for Intel and AMD x86 processors.
In most cases, the Fortran or assembly loops should be selected 
automatically by the {\tt configure} script when appropriate, but you can
also tweak this by setting options to the {\tt configure} script.

\section{Computation of the 1.0/sqrt function}
\label{sec:sqrt}
\subsection{Introduction}
The {\gromacs} project started with the development of a $1/\sqrt{x}$
processor that calculates:
\begin{equation}
Y(x) = \frac{1}{ \sqrt{x} }
\end{equation}
As the project continued, the {\intel} processor was used to implement
{\gromacs}, which now turned into almost a full software project.  The
$1/\sqrt{x}$ processor was implemented using a Newton-Raphson
iteration scheme for one step. For this it needed look-up tables to
provide the initial approximation. The $1/\sqrt{x}$ function makes it
possible to use two almost independent tables for the exponent seed
and the fraction seed with the IEEE floating-point representation.

\subsection{General}
According to~\cite{Bekker87} the $1/\sqrt{x}$ function can be evaluated using
the Newton-Raphson iteration scheme. The inverse function is:
\begin{equation}
X(y) = \frac{1}{y^{2}}
\end{equation}
So instead of calculating:
\begin{equation}
Y(a) = q
\end{equation}
the equation:
\begin{equation}
X(q) - a = 0
\label{eqn:inversef}
\end{equation}
can now be solved using Newton-Raphson. An iteration is performed by
calculating:
\begin{equation}
y_{n+1} = y_{n} - \frac{f(y_{n})}{f'(y_{n})}
\label{eqn:nr}
\end{equation}
The absolute error $\varepsilon$, in this approximation is defined by:
\begin{equation}
\varepsilon \equiv y_{n} - q
\end{equation}
Using Taylor series expansion to estimate the error results in:
\begin{equation}
\varepsilon _{n+1} = - \frac{\varepsilon _{n}^{2}}{2}
                       \frac{ f''(y_{n})}{f'(y_{n})}
\label{eqn:taylor}
\end{equation}
according to~\cite{Bekker87} equation (3.2). This is an estimation of the
absolute error.

\subsection{Applied to floating-point numbers}
Floating-point numbers in IEEE 32 bit single-precision format have a nearly
constant relative error of $\Delta x / x = 2^{-24}$. As seen earlier in the
Taylor series expansion equation (\eqnref{taylor}), the error in every
iteration step is absolute and in general dependent of $y$. If the error is
expressed as a relative error $\varepsilon_{r}$ the following holds:
\begin{equation}
\varepsilon _{{r}_{n+1}} \equiv \frac{\varepsilon_{n+1}}{y}
\end{equation}
and so:
\begin{equation}
\varepsilon _{{r}_{n+1}} =
- ( \frac{\varepsilon _{n}}{y} )^{2} y \frac{ f''}{2f'}
\end{equation}
For the function $f(y) = y^{-2}$ the term $y f''/2f'$ is constant (equal
to $-3/2$) so the relative error $\varepsilon _{r_{n}}$ is independent of $y$.
\begin{equation}
\varepsilon _{{r}_{n+1}} =
\frac{3}{2} (\varepsilon_{r_{n}})^{2}
\label{eqn:epsr}
\end{equation}

The conclusion of this is that the function $1/\sqrt{x}$ can be
calculated with a specified accuracy.

\begin{figure}
\begin{center}
\newcommand{\twltt}{\tt}
\setlength{\unitlength}{0.0080in}
\begin{picture}(489,176)(40,390)
\thicklines
\put(180,505){$\underbrace{\hspace{2.68in}}$}
\put( 60,505){$\underbrace{\hspace{0.88in}}$}
\put( 40,510){\framebox(480,30){}}
\put( 45,505){\vector( 0,-1){ 15}}
\put( 40,540){\dashbox{4}(0,0){}}
\multiput(250,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(220,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(190,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(280,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(310,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(340,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(370,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(400,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(430,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(460,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(490,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(205,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(235,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(265,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(295,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(325,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(355,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(385,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(415,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(445,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(475,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(505,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\put( 40,510){\framebox(480,30){}}
\put( 40,540){\dashbox{4}(0,0){}}
\multiput(250,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(220,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(190,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(160,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(130,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(100,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(280,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(310,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(340,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(370,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(400,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(430,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(460,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(490,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput( 70,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\put( 55,540){\line( 0,-1){ 30}}
\multiput( 85,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(115,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(145,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(205,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(235,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(265,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(295,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(325,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(355,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(385,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(415,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(445,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(475,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(505,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\put(175,540){\line( 0,-1){ 30}}
\put(175,540){\line( 0,-1){ 30}}
\multiput(145,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(115,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput( 85,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\put( 55,540){\line( 0,-1){ 30}}
\multiput( 70,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(100,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(130,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\multiput(160,540)(0.00000,-8.57143){4}{\line( 0,-1){  4.286}}
\put(345,470){\makebox(0,0)[lb]{\raisebox{0pt}[0pt][0pt]{\twltt $F$}}}
\put(110,470){\makebox(0,0)[lb]{\raisebox{0pt}[0pt][0pt]{\twltt $E$}}}
\put( 40,470){\makebox(0,0)[lb]{\raisebox{0pt}[0pt][0pt]{\twltt $S$}}}
\put(505,545){\makebox(0,0)[lb]{\raisebox{0pt}[0pt][0pt]{\twltt $0$}}}
\put(160,545){\makebox(0,0)[lb]{\raisebox{0pt}[0pt][0pt]{\twltt $23$}}}
\put( 40,545){\makebox(0,0)[lb]{\raisebox{0pt}[0pt][0pt]{\twltt $31$}}}
\put(140,400){\makebox(0,0)[lb]{\raisebox{0pt}[0pt][0pt]{\twltt $Value=(-1)^{S}(2^{E-127})(1.F)$}}}
\put(505,545){\makebox(0,0)[lb]{\raisebox{0pt}[0pt][0pt]{\twltt $0$}}}
\put(160,545){\makebox(0,0)[lb]{\raisebox{0pt}[0pt][0pt]{\twltt $23$}}}
\put( 40,545){\makebox(0,0)[lb]{\raisebox{0pt}[0pt][0pt]{\twltt $31$}}}
\put(140,400){\makebox(0,0)[lb]{\raisebox{0pt}[0pt][0pt]{\twltt $Value=(-1)^{S}(2^{E-127})(1.F)$}}}
\end{picture}
\end{center}
\caption{IEEE single-precision floating-point format}
\label{fig:ieee}
\end{figure}

\subsection{Specification of the look-up table}
To calculate the function $1/\sqrt{x}$ using the previously mentioned
iteration scheme, it is clear that the first estimation of the solution must
be accurate enough to get precise results. The requirements for the
calculation are
\begin{itemize}
\item Maximum possible accuracy with the used IEEE format
\item Use only one iteration step for maximum speed
\end{itemize}

The first requirement states that the result of $1/\sqrt{x}$ may have a
relative error $\varepsilon_{r}$ equal to the $\varepsilon_{r}$ of a IEEE 32
bit single-precision floating-point number. From this, the $1/\sqrt{x}$
of the initial approximation can be derived, rewriting the definition of
the relative error for succeeding steps (\eqnref{epsr}):
\begin{equation}
\frac{\varepsilon_{n}}{y} =
\sqrt{\varepsilon_ {r_{n+1}} \frac{2f'}{yf''}}
\end{equation}
So for the look-up table the needed accuracy is:
\begin{equation}
\frac{\Delta Y}{Y} = \sqrt{\frac{2}{3} 2^{-24}}
\label{eqn:accy}
\end{equation}
which defines the width of the table that must be $\geq 13$ bit.

At this point the relative error, $\varepsilon_{r_{n}}$, of the look-up table
is known. From this the maximum relative error in the argument can be 
calculated as follows. The absolute error $\Delta x$ is defined as:
\begin{equation}
\Delta x \equiv \frac{\Delta Y}{Y'}
\end{equation}
and thus:
\begin{equation}
\frac{\Delta x}{Y} = \frac{\Delta Y}{Y} (Y')^{-1}
\end{equation}
and thus:
\begin{equation}
\Delta x = constant \frac{Y}{Y'}
\end{equation}
For the $1/\sqrt{x}$ function, $Y / Y' \sim x$ holds, so
$\Delta x / x = constant$. This is a property of the used floating-point
representation as earlier mentioned. The needed accuracy of the argument of the
look-up table follows from:
\begin{equation}
\frac{\Delta x}{x} = -2 \frac{\Delta Y}{Y}
\end{equation}
So, using the floating-point accuracy (\eqnref{accy}):
\begin{equation}
\frac{\Delta x}{x} = -2 \sqrt{\frac{2}{3} 2^{-24}}
\end{equation}
This defines the length of the look-up table which should be $\geq 12$ bit.

\subsection{Separate exponent and fraction computation}
The used IEEE 32 bit single-precision floating-point format specifies
that a number is represented by a exponent and a fraction. The previous 
section specifies for every possible floating-point number the look-up table 
length and width. Only the size of the fraction of a floating-point number 
defines the accuracy. The conclusion from this can be that the size of the 
look-up table is length of look-up table, earlier specified, times the size of 
the exponent ($2^{12}2^{8}, 1Mb$). The $1/\sqrt{x}$  function has the 
property that the exponent is independent of the fraction. This becomes clear 
if the floating-point representation is used. Define:
\begin{equation}
x \equiv (-1)^{S}(2^{E-127})(1.F)
\label{eqn:fpdef}
\end{equation}
See \figref{ieee}, where $0 \leq S \leq 1$, $0 \leq E \leq 255$,
$1 \leq 1.F < 2$ and $S$, $E$, $F$ integer (normalization conditions). 
The sign bit ($S$) can be omitted because $1/\sqrt{x}$ is only defined 
for $x > 0$. The $1/\sqrt{x}$ function applied to $x$ results in:
\begin{equation}
y(x) = \frac{1}{\sqrt{x}}
\end{equation}
or:
\begin{equation}
y(x) = \frac{1}{\sqrt{(2^{E-127})(1.F)}}
\end{equation}
This can be rewritten as:
\begin{equation}
y(x) = (2^{E-127})^{-1/2}(1.F)^{-1/2}
\label{eqn:yx}
\end{equation}
Define:
\begin{equation}
(2^{E'-127}) \equiv (2^{E-127})^{-1/2}
\end{equation}
\begin{equation}
1.F'\equiv (1.F)^{-1/2}
\end{equation}
Then $\frac{1}{\sqrt{2}} < 1.F' \leq 1$ holds, so the condition
$1 \leq 1.F' < 2$, which is essential for normalized real representation, is
not valid anymore. By introducing an extra term, this can be corrected.
Rewrite the $1/\sqrt{x}$ function applied to floating-point numbers (\eqnref{yx}) as:
\begin{equation}
y(x) = (2^{\frac{127-E}{2}-1}) (2(1.F)^{-1/2})
\end{equation}
and:
\begin{equation}
(2^{E'-127}) \equiv (2^{\frac{127-E}{2}-1})
\label{eqn:exp}
\end{equation}
\begin{equation}
1.F'\equiv 2(1.F)^{-1/2}
\label{eqn:frac}
\end{equation}
Then $\sqrt{2} < 1.F \leq 2$ holds. This is not the exact valid range as
defined for normalized floating-point numbers in \eqnref{fpdef}. 
The value $2$  causes the problem. By mapping this value on the nearest
representation $< 2$, this can be solved. The small error that is introduced
by this approximation is within the allowable range. 

The integer representation of the exponent is the next problem. Calculating
$(2^{\frac{127-E}{2}-1})$ introduces a fractional result if $(127-E) = odd$.
This is again easily accounted for by splitting up the calculation into an
odd and an even part. For $(127-E) = even$ $E'$ in equation (\eqnref{exp})
can be exactly calculated in integer arithmetic as a function of $E$.
\begin{equation}
E' = \frac{127-E}{2} + 126
\end{equation}

For $(127-E) = odd$ equation (\eqnref{yx}) can be rewritten as:
\begin{equation}
y(x) = (2^{\frac{127-E-1}{2}})(\frac{1.F}{2})^{-1/2}
\end{equation}
Thus:
\begin{equation}
E' = \frac{126-E}{2} + 127
\end{equation}
which also can be calculated exactly in integer arithmetic.
{\bf Note} that the fraction is automatically corrected for its range earlier
mentioned, so the exponent does not need an extra correction.

The conclusions from this are:
\begin{itemize}
\item The fraction and exponent look-up table are independent. The fraction
look-up table exists of two tables (odd and even exponent) so the odd/even
information of the exponent (lsb bit) has to be used to select the right
table.
\item The exponent table is an 256 x 8 bit table, initialized for $odd$
and $even$.% as specified before
\end{itemize}

\subsection{Implementation}
The look-up tables can be generated by a small C program, which uses
floating-point numbers and operations with IEEE 32 bit single-precision format.
Note that because of the $odd$/$even$ information that is needed, the
fraction table is twice the size earlier specified (13 bit i.s.o. 12 bit).
%\figref{expgen} in the appendix shows how to fill the exponent table,
%\figref{fractgen} shows how to fill the fraction table.

The function according to~\eqnref{nr} has to be implemented. 
Applied to the $1/\sqrt{x}$ function, equation~\eqnref{inversef} leads to:
\begin{equation}
f = a - \frac{1}{y^{2}}
\end{equation}
and so:
\begin{equation}
f' = \frac{2}{y^{3}}
\end{equation}
so:
\begin{equation}
y_{n+1} = y_{n} - \frac{ a - \frac{1}{y^{2}_{n}} }{ \frac{2}{y^{3}_{n}} }
\end{equation}
or:
\begin{equation}
y_{n+1} = \frac{y_{n}}{2} (3 - a y^{2}_{n})
\end{equation}
Where $y_{0}$ can be found in the look-up tables, and $y_{1}$ gives the result
to the maximum accuracy. 
%In \figref{as_implementation}) the assembler implementation is given. 
It is clear that only one iteration extra (in double 
precision) is needed for a double-precision result.

% LocalWords:  Virial virial triclinic intra mol mshift shiftvec sqrt SPC lj yf
% LocalWords:  coul Fortran SGI AMD Raphson IEEE taylor epsr accy ieee yx fpdef
% LocalWords:  lsb nr inversef src formulae GROMACS