From 705f945860e432e730f66cdf91d7599b1e3cd3aa Mon Sep 17 00:00:00 2001 From: Kahina Date: Thu, 29 Oct 2015 16:34:39 +0100 Subject: [PATCH] la section 3 --- paper.tex | 18 +++++++++++------- 1 file changed, 11 insertions(+), 7 deletions(-) diff --git a/paper.tex b/paper.tex index bcac7f4..bcf399a 100644 --- a/paper.tex +++ b/paper.tex @@ -349,12 +349,16 @@ 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: -\begin{equation} -\label{R} -R = \exp( \log(DBL\_MAX) / (2*n) ) -\end{equation} - where $DBL\_MAX$ stands for the maximum representable double value. +This solution is applied when the root except the circle unit, represented by the radius $R$ evaluated in C language as: +\begin{verbatim} +R = exp(log(DBL_MAX)/(2*n) ); +\end{verbatim} + +%\begin{equation} + +%R = \exp( \log(DBL\_MAX) / (2*n) ) +%\end{equation} + where \verb=DBL_MAX= stands for the maximum representable \verb=double= value. \section{The implementation of simultaneous methods in a parallel computer} \label{secStateofArt} @@ -539,7 +543,7 @@ Let $K$ be the number of iterations necessary to compute all the roots, so the t \label{eq:T-global} T=\left[n\left(T_{i}(n)+T_{j}\right)+O(n)\right].K \end{equation} -The execution time increases with the increasing of the polynomial degree, which justifies to parallelise these steps in order to reduce the global execution time. In the following, we explain how we did parrallelize these steps on a GPU architecture using the CUDA platform. +The execution time increases with the increasing of the polynomial degree, which justifies to parallelize these steps in order to reduce the global execution time. In the following, we explain how we did parallelize these steps on a GPU architecture using the CUDA platform. \subsubsection{A Parallel implementation with CUDA } On the CPU, both steps 3 and 4 contain the loop \verb=for= and a single thread executes all the instructions in the loop $n$ times. In this subsection, we explain how the GPU architecture can compute this loop and reduce the execution time. -- 2.39.5