#500000 sparse full #50000 sparse full
#nb threads times nb iter times nb iter times nb iter times nb iter
-1024 523 27 545 8.61 16
-512 449.426 24 520 9.27 19
-256 440.805 24 480 7.73 15
-128 456.175 22 560 8.64 21
-64 472.862 23 603 7.84 16
-32 830.152 24 920 11.33 18
-16 1234 24 1870 20.47 21
-8 2632.78 23 3589 35.07 26
+1024 523 27 2100.7 8.61 16
+512 449.426 24 1459.35 9.27 19
+256 440.805 24 754.24 7.73 15
+128 456.175 22 718.623 27 8.64 21
+64 472.862 23 715.554 27 7.84 16
+32 830.152 24 1089.61 27 11.33 18
+16 1234 24 1746.53 22 20.47 21
+8 2632.78 23 3112 20 35.07 26
%%\usepackage[french]{babel}
\usepackage{amsmath,amsfonts,amssymb}
\usepackage[ruled,vlined]{algorithm2e}
+%\usepackage[french,boxed,linesnumbered]{algorithm2e}
\usepackage{array,multirow,makecell}
\setcellgapes{1pt}
\makegapedcells
~\\
In this sequential algorithm, one CPU thread executes all the steps. Let us look to the $3^{rd}$ step i.e. the execution of the iterative function, 2 sub-steps are needed. The first sub-step \textit{save}s the solution vector of the previous iteration, the second sub-step \textit{update}s or computes the new values of the roots vector.
-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 $z^{(k+1)}_{i}$, taht is :
+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.
The kernels terminate it computations when all the roots converge. Finally, the solution of the root finding problem is copied back from GPU global memory to CPU memory. We use the communication functions of CUDA for the memory allocation in the GPU \verb=(cudaMalloc())= and for data transfers from the CPU memory to the GPU memory \verb=(cudaMemcpyHostToDevice)=
or from GPU memory to CPU memory \verb=(cudaMemcpyDeviceToHost))=.
%%HIER END MY REVISIONS (SIDER)
-\subsection{Experimental study}
+\section{Experimental study}
-\subsubsection{Definition of the polynomial used}
-We use a polynomial of the following form for which the
-roots are distributed on 2 distinct circles:
+\subsection{Definition of the polynomial used}
+We use two forms of polynomials:
+\paragraph{sparse polynomial}:
+in this following form, the roots are distributed on 2 distinct circles:
\begin{equation}
\forall \alpha_{1} \alpha_{2} \in C,\forall n_{1},n_{2} \in N^{*}; P(z)= (z^{n^{1}}-\alpha_{1})(z^{n^{2}}-\alpha_{2})
\end{equation}
This form makes it possible to associate roots having two
different modules and thus to work on a polynomial constitute
of four non zero terms.
-\\
- An other form of the polynomial to obtain a full polynomial is:
+
+\paragraph{Full polynomial}:
+ the second form used to obtain a full polynomial is:
%%\begin{equation}
%%\forall \alpha_{i} \in C,\forall n_{i}\in N^{*}; P(z)= \sum^{n}_{i=1}(z^{n^{i}}.a_{i})
%%\end{equation}
\begin{equation}
- {\Large \forall a_{i} \in C; p(x)=\sum^{n-1}_{i=1} a_{i}.x^{i}}
+ {\Large \forall a_{i} \in C, i\in N; p(x)=\sum^{n-1}_{i=1} a_{i}.x^{i}}
\end{equation}
-with this formula, we can have until \textit{n} non zero terms.
+with this form, we can have until \textit{n} non zero terms.
-\subsubsection{The study condition}
+\subsection{The study condition}
In order to have representative average values, for each
point of our curves we measured the roots finding of 10
different polynomials.
polynomials. The execution time remains the
element-key which justifies our work of parallelization.
For our tests we used a CPU Intel(R) Xeon(R) CPU
-E5620@2.40GHz and a GPU Tesla C2070 (with 6 Go of ram)
+E5620@2.40GHz and a GPU K40 (with 6 Go of ram)
+
-\subsubsection{Comparative study}
+\subsection{Comparative study}
We initially carried out the convergence of Aberth algorithm with various sizes of polynomial, in second we evaluate the influence of the size of the threads per block....
-\paragraph{Aberth algorithm on CPU and GPU}
+\subsubsection{Aberth algorithm on CPU and GPU}
%\begin{table}[!ht]
% \centering
\end{figure}
-\paragraph{The impact of the thread's number into the convergence of Aberth algorithm}
+\subsubsection{The impact of the thread's number into the convergence of Aberth algorithm}
%\begin{table}[!h]
% \centering
-\paragraph{A comparative study between Aberth and Durand-kerner algorithm}
+\subsubsection{A comparative study between Aberth and Durand-kerner algorithm}
\begin{table}[htbp]
\centering
\begin{tabular} {|R{2cm}|L{2.5cm}|L{2.5cm}|L{1.5cm}|L{1.5cm}|}
