correction des equations

diff --git a/paper.tex b/paper.tex
index 0ff4207f68d8e410d2fe7ebb99f16ae008e07966..9fee850fbfa1c35aff39f3254619f1b4267ae742 100644 (file)
--- a/paper.tex
+++ b/paper.tex
@@ -115,14 +115,14 @@ The root finding problem consists in finding the values of all the $n$ values of
 \end{equation}
 
 The problem of finding a root is equivalent to that of solving a fixed-point problem. To see this, consider the fixed-point problem of finding the $n$-dimensional
 \end{equation}
 
 The problem of finding a root is equivalent to that of solving a fixed-point problem. To see this, consider the fixed-point problem of finding the $n$-dimensional

-vector $x$ such that
+vector $x$ such that :

 \begin{center}
 $x=g(x)$
 \end{center}
 where $g : C^{n}\longrightarrow C^{n}$. Usually, we can easily
 rewrite this fixed-point problem as a root-finding problem by
 setting $f(x) = x-g(x)$ and likewise we can recast the
 \begin{center}
 $x=g(x)$
 \end{center}
 where $g : C^{n}\longrightarrow C^{n}$. Usually, we can easily
 rewrite this fixed-point problem as a root-finding problem by
 setting $f(x) = x-g(x)$ and likewise we can recast the

-root-finding problem into a fixed-point problem by setting
+root-finding problem into a fixed-point problem by setting :

 \begin{center}
 $g(x)= f(x)-x$.
 \end{center}
 \begin{center}
 $g(x)= f(x)-x$.
 \end{center}


@@ -147,10 +147,10 @@ approximation of all the roots, starting with the Durand-Kerner (DK)
 method:
 %%\begin{center}
 \begin{equation}
 method:
 %%\begin{center}
 \begin{equation}

- Z_i^{k+1}=Z_{i}^k-\frac{P(Z_i^k)}{\prod_{i\neq j}(Z_i^k-Z_j^k)}
+ z_i^{k+1}=z_{i}^k-\frac{P(z_i^k)}{\prod_{i\neq j}(z_i^k-z_j^k)}

 \end{equation}
 %%\end{center}
 \end{equation}
 %%\end{center}

-where $Z_i^k$ is the $i^{th}$ root of the polynomial $P$ at the
+where $z_i^k$ is the $i^{th}$ root of the polynomial $P$ at the

 iteration $k$.
 
 
 iteration $k$.
 
 


@@ -164,11 +164,11 @@ in the following form by Ehrlich~\cite{Ehrlich67} and
 Aberth~\cite{Aberth73} uses a different iteration formula given as fellows :
 %%\begin{center}
 \begin{equation}
 Aberth~\cite{Aberth73} uses a different iteration formula given as fellows :
 %%\begin{center}
 \begin{equation}

- Z_i^{k+1}=Z_i^k-\frac{1}{{\frac {P'(Z_i^k)} {P(Z_i^k)}}-{\sum_{i\neq j}(Z_i^k-Z_j^k)}}.
+ z_i^{k+1}=z_i^k-\frac{1}{{\frac {P'(z_i^k)} {P(z_i^k)}}-{\sum_{i\neq j}\frac{1}{(z_i^k-z_j^k)}}}.

 \end{equation}
 %%\end{center}
 \end{equation}
 %%\end{center}

-where $P'(Z)$ is the polynomial derivative of $P$ evaluated in the
-point $Z$.
+where $P'(z)$ is the polynomial derivative of $P$ evaluated in the
+point $z$.

 
 Aberth, Ehrlich and Farmer-Loizou~\cite{Loizon83} have proved that
 the Ehrlich-Aberth method (EA) has a cubic order of convergence for simple roots whereas the Durand-Kerner has a quadratic order of convergence.
 
 Aberth, Ehrlich and Farmer-Loizou~\cite{Loizon83} have proved that
 the Ehrlich-Aberth method (EA) has a cubic order of convergence for simple roots whereas the Durand-Kerner has a quadratic order of convergence.


@@ -201,8 +201,7 @@ where it takes up to 3300 seconds to obtain the same results, the authors show a
 Very few works had been since this last work until the appearing of
 the Compute Unified Device Architecture (CUDA)~\cite{CUDA10}, a
 parallel computing platform and a programming model invented by
 Very few works had been since this last work until the appearing of
 the Compute Unified Device Architecture (CUDA)~\cite{CUDA10}, a
 parallel computing platform and a programming model invented by

-NVIDIA. The computing power of GPUs (Graphics Processing Unit) has exceeded that of
-of CPUs. However, CUDA adopts a totally new computing architecture to use the
+NVIDIA. The computing power of GPUs (Graphics Processing Unit) has exceeded that of CPUs. However, CUDA adopts a totally new computing architecture to use the

 hardware resources provided by GPU in order to offer a stronger
 computing ability to the massive data computing.
 
 hardware resources provided by GPU in order to offer a stronger
 computing ability to the massive data computing.
 


@@ -214,9 +213,9 @@ the sequential implementation on a single CPU for high degree
 polynomials of about 48000. To our knowledge, it is the first time such high degree polynomials are numerically solved.
 
 
 polynomials of about 48000. To our knowledge, it is the first time such high degree polynomials are numerically solved.
 
 

-In this paper, we focus on the implementation of the Aberth method for
-high degree polynomials on GPU. The paper is organised as fellows. Initially, we recall the Aberth method in Section.\ref{sec1}. Improvements for the Aberth method are proposed in Section.\ref{sec2}. Related work to the implementation of simultaneous methods using a parallel approach is presented in Section.\ref{secStateofArt}.
-In Section.4 we propose a parallel implementation of the Aberth method on GPU and discuss it. Section 5 presents and investigates our implementation and experimental study results. Finally, Section 6 concludes this paper and gives some hints for future research directions in this topic.  
+In this paper, we focus on the implementation of the Ehrlich-Aberth method for
+high degree polynomials on GPU. The paper is organized as fellows. Initially, we recall the Ehrlich-Aberth method in Section \ref{sec1}. Improvements for the Ehrlich-Aberth method are proposed in Section \ref{sec2}. Related work to the implementation of simultaneous methods using a parallel approach is presented in Section \ref{secStateofArt}.
+In Section \ref{sec5} we propose a parallel implementation of the Ehrlich-Aberth method on GPU and discuss it. Section \ref{sec6} presents and investigates our implementation and experimental study results. Finally, Section\ref{sec7} 6 concludes this paper and gives some hints for future research directions in this topic.  

 
 \section{The Sequential Aberth method}
 \label{sec1}
 
 \section{The Sequential Aberth method}
 \label{sec1}


@@ -240,10 +239,10 @@ Differentiating the rational function $R_{i}(z)$ and applying the
 Newton method, we have:
 
 \begin{equation}
 Newton method, we have:
 
 \begin{equation}

-\frac{R_{i}(z)}{R_{i}^{'}(z)}= \frac{p(z)}{p^{'}(z)-p(z)\frac{w_{i}(z)}{w_{i}^{'}(z)}}= \frac{p(z)}{p^{'}(z)-p(z) \sum _{j=1,j \neq i}^{n}\frac{1}{z-x_{i}}}, i=1,2,...,n
+\frac{R_{i}(z)}{R_{i}^{'}(z)}= \frac{p(z)}{p^{'}(z)-p(z)\frac{w_{i}(z)}{w_{i}^{'}(z)}}= \frac{p(z)}{p^{'}(z)-p(z) \sum _{j=1,j \neq i}^{n}\frac{1}{z-x_{j}}}, i=1,2,...,n

 \end{equation}
 \end{equation}

-
-Substituting $x_{j}$ for z we obtain the Aberth iteration method.
+where R_{i}^{'}(z)is the rational function derivative of F evaluated in the point z 
+Substituting $x_{j}$ for $z_{j}$ we obtain the Aberth iteration method.

 
 In the fellowing we present the main stages of the running of the Aberth method.
 
 
 In the fellowing we present the main stages of the running of the Aberth method.
 


@@ -264,7 +263,7 @@ The initial guess is very important since the number of steps needed by the iter
 a given approximation strongly depends on it.
 In~\cite{Aberth73} the Aberth iteration is started by selecting $n$
 equi-spaced points on a circle of center 0 and radius r, where r is
 a given approximation strongly depends on it.
 In~\cite{Aberth73} the Aberth iteration is started by selecting $n$
 equi-spaced points on a circle of center 0 and radius r, where r is

-an upper bound to the moduli of the zeros. Later, Bini et al.~\cite{Bini96}
+an upper bound to the moduli of the zeros. Later, Bini and al.~\cite{Bini96}

 performed this choice by selecting complex numbers along different
 circles and relies on the result of~\cite{Ostrowski41}.
 
 performed this choice by selecting complex numbers along different
 circles and relies on the result of~\cite{Ostrowski41}.
 


@@ -281,14 +280,14 @@ u_{i}=2.|a_{i}|^{\frac{1}{i}};
 v_{i}=\frac{|\frac{a_{n}}{a_{i}}|^{\frac{1}{n-i}}}{2}.
 \end{equation}
 
 v_{i}=\frac{|\frac{a_{n}}{a_{i}}|^{\frac{1}{n-i}}}{2}.
 \end{equation}
 

-\subsection{Iterative Function $H_{i}$}
+\subsection{Iterative Function $H_{i}(z^{k})$}

 The operator used by the Aberth method is corresponding to the
 following equation which will enable the convergence towards
 polynomial solutions, provided all the roots are distinct.
 
 \begin{equation}
 The operator used by the Aberth method is corresponding to the
 following equation which will enable the convergence towards
 polynomial solutions, provided all the roots are distinct.
 
 \begin{equation}

-H_{i}(z)=z_{i}-\frac{1}{\frac{p^{'}(z_{i})}{p(z_{i})}-\sum_{j\neq
-i}{\frac{1}{z_{i}-z_{j}}}}
+H_{i}(z^{k+1})=z_{i}^{k}-\frac{1}{\frac{p^{'}(z_{i}^{k})}{p(z_{i}^{k})}-\sum_{j=1,j\neq
+i}^{j=n}{\frac{1}{z_{i}^{k}-z_{j}^{k}}}}

 \end{equation}
 
 \subsection{Convergence Condition}
 \end{equation}
 
 \subsection{Convergence Condition}


@@ -299,7 +298,7 @@ converges sufficiently when :
 \begin{equation}
 \label{eq:Aberth-Conv-Cond}
 \forall i \in
 \begin{equation}
 \label{eq:Aberth-Conv-Cond}
 \forall i \in

-[1,n];\frac{z_{i}^{(k)}-z_{i}^{(k-1)}}{z_{i}^{(k)}}<\xi
+[1,n];\frac{z_{i}^{k}-z_{i}^{k-1}}{z_{i}^{k}}<\xi

 \end{equation}
 
 
 \end{equation}
 
 


@@ -332,22 +331,22 @@ propose to use the logarithm and the exponential of a complex in order to comput
 Using the logarithm (eq.~\ref{deflncomplex}) and the exponential (eq.~\ref{defexpcomplex}) operators, we can replace any multiplications and divisions with additions and subtractions. Consequently, computations
 manipulate lower absolute values and the roots for large polynomial's degrees can be looked for successfully~\cite{Karimall98}.
 
 Using the logarithm (eq.~\ref{deflncomplex}) and the exponential (eq.~\ref{defexpcomplex}) operators, we can replace any multiplications and divisions with additions and subtractions. Consequently, computations
 manipulate lower absolute values and the roots for large polynomial's degrees can be looked for successfully~\cite{Karimall98}.
 

-Applying this solution for the Aberth method we obtain the
+Applying this solution for the Ehrlich-Aberth method we obtain the

 iteration function with logarithm:
 %%$$ \exp \bigl(  \ln(p(z)_{k})-ln(\ln(p(z)_{k}^{'}))- \ln(1- \exp(\ln(p(z)_{k})-ln(\ln(p(z)_{k}^{'})+\ln\sum_{i\neq j}^{n}\frac{1}{z_{k}-z_{j}})$$
 \begin{equation}
 \label{Log_H2}
 iteration function with logarithm:
 %%$$ \exp \bigl(  \ln(p(z)_{k})-ln(\ln(p(z)_{k}^{'}))- \ln(1- \exp(\ln(p(z)_{k})-ln(\ln(p(z)_{k}^{'})+\ln\sum_{i\neq j}^{n}\frac{1}{z_{k}-z_{j}})$$
 \begin{equation}
 \label{Log_H2}

-H_{i}(z)=z_{i}^{k}-\exp \left(\ln \left(
-p(z_{k})\right)-\ln\left(p(z_{k}^{'})\right)- \ln
-\left(1-Q(z_{k})\right)\right),
+H_{i}(z^{k+1})=z_{i}^{k}-\exp \left(\ln \left(
+p(z_{i}^{k})\right)-\ln\left(p'(z^{k}_{i})\right)- \ln
+\left(1-Q(z^{k}_{i})\right)\right),

 \end{equation}
 
 where:
 
 \begin{equation}
 \label{Log_H1}
 \end{equation}
 
 where:
 
 \begin{equation}
 \label{Log_H1}

-Q(z_{k})=\exp\left( \ln (p(z_{k}))-\ln(p(z_{k}^{'}))+\ln \left(
-\sum_{k\neq j}^{n}\frac{1}{z_{k}-z_{j}}\right)\right).
+Q(z^{k}_{i})=\exp\left( \ln (p(z^{k}_{i}))-\ln(p'(z^{k}^{i}))+\ln \left(
+\sum_{k\neq j}^{n}\frac{1}{z^{k}_{i}-z^{k}_{j}}\right)\right).

 \end{equation}
 
 This solution is applied when the root except the circle unit, represented by the radius $R$ evaluated as:
 \end{equation}
 
 This solution is applied when the root except the circle unit, represented by the radius $R$ evaluated as:


@@ -466,7 +465,8 @@ provides two read-only memory spaces, the constant space and the
 texture space, which reside in external DRAM, and are accessed via
 read-only caches.
 
 texture space, which reside in external DRAM, and are accessed via
 read-only caches.
 

-\subsection{ The implementation of Aberth method on GPU}
+\section{ The implementation of Aberth method on GPU}
+\label{sec5}

 %%\subsection{A CUDA implementation of the Aberth's method }
 %%\subsection{A GPU implementation of the Aberth's method }
 
 %%\subsection{A CUDA implementation of the Aberth's method }
 %%\subsection{A GPU implementation of the Aberth's method }
 


@@ -510,15 +510,15 @@ In this sequential algorithm, one CPU thread  executes all the steps. Let us loo
 There exists two ways to execute the iterative function that we call a Jacobi one and a Gauss-Seidel one. With the Jacobi iteration, at iteration $k+1$ we need all the previous values $z^{(k)}_{i}$ to compute the new values $z^{(k+1)}_{i}$, that is :
 
 \begin{equation}
 There exists two ways to execute the iterative function that we call a Jacobi one and a Gauss-Seidel one. With the Jacobi iteration, at iteration $k+1$ we need all the previous values $z^{(k)}_{i}$ to compute the new values $z^{(k+1)}_{i}$, that is :
 
 \begin{equation}

-H(i,z^{k+1})=\frac{p(z^{(k)}_{i})}{p'(z^{(k)}_{i})-p(z^{(k)}_{i})\sum^{n}_{j=1 j\neq i}\frac{1}{z^{(k)}_{i}-z^{(k)}_{j}}}, i=1,...,n.
+H_{i}(z^{k+1})=\frac{p(z^{k}_{i})}{p'(z^{k}_{i})-p(z^{k}_{i})\sum^{n}_{j=1 j\neq i}\frac{1}{z^{k}_{i}-z^{k}_{j}}}, i=1,...,n.

 \end{equation}
 
 With the Gauss-seidel iteration, we have:
 \begin{equation}
 \label{eq:Aberth-H-GS}
 \end{equation}
 
 With the Gauss-seidel iteration, we have:
 \begin{equation}
 \label{eq:Aberth-H-GS}

-H(i,z^{k+1})=\frac{p(z^{(k)}_{i})}{p'(z^{(k)}_{i})-p(z^{(k)}_{i})(\sum^{i-1}_{j=1}\frac{1}{z^{(k)}_{i}-z^{(k+1)}_{j}}+\sum^{n}_{j=i+1}\frac{1}{z^{(k)}_{i}-z^{(k)}_{j}})}, i=1,...,n.
+H_{i}(z^{k+1})=\frac{p(z^{k}_{i})}{p'(z^{k}_{i})-p(z^{k}_{i})(\sum^{i-1}_{j=1}\frac{1}{z^{k}_{i}-z^{k+1}_{j}}+\sum^{n}_{j=i+1}\frac{1}{z^{k}_{i}-z^{k}_{j}})}, i=1,...,n.

 \end{equation}
 \end{equation}

-
+%%Here a finiched my revision %%

 Using Equation.~\ref{eq:Aberth-H-GS} for the update sub-step of $H(i,z^{k+1})$, we expect the Gauss-Seidel iteration to converge more quickly because, just as its ancestor (for solving linear systems of equations), it uses the most fresh computed roots $z^{k+1}_{i}$.
 
 The $4^{th}$ step of the algorithm checks the convergence condition using Equation.~\ref{eq:Aberth-Conv-Cond}.
 Using Equation.~\ref{eq:Aberth-H-GS} for the update sub-step of $H(i,z^{k+1})$, we expect the Gauss-Seidel iteration to converge more quickly because, just as its ancestor (for solving linear systems of equations), it uses the most fresh computed roots $z^{k+1}_{i}$.
 
 The $4^{th}$ step of the algorithm checks the convergence condition using Equation.~\ref{eq:Aberth-Conv-Cond}.


@@ -609,7 +609,7 @@ The kernels terminate it computations when all the roots converge. Finally, the
 or from GPU memory to CPU memory \verb=(cudaMemcpyDeviceToHost))=. 
 %%HIER END MY REVISIONS (SIDER)
 \section{Experimental study}
 or from GPU memory to CPU memory \verb=(cudaMemcpyDeviceToHost))=. 
 %%HIER END MY REVISIONS (SIDER)
 \section{Experimental study}

-
+\label{sec6}

 \subsection{Definition of the used polynomials }
 We study two categories of polynomials : the sparse polynomials and the full polynomials.
 \paragraph{A sparse polynomial}: is a polynomial for which only some coefficients are not null. We use in the following polonymial for which the roots are distributed on 2 distinct circles :
 \subsection{Definition of the used polynomials }
 We study two categories of polynomials : the sparse polynomials and the full polynomials.
 \paragraph{A sparse polynomial}: is a polynomial for which only some coefficients are not null. We use in the following polonymial for which the roots are distributed on 2 distinct circles :


@@ -731,7 +731,7 @@ This figure show the execution time of the both algorithm EA and DK with sparse
 
 
 \section{Conclusion and perspective}
 
 
 \section{Conclusion and perspective}

-
+\label{sec7}

 \bibliography{mybibfile}
 
 \end{document}
 \bibliography{mybibfile}
 
 \end{document}