X-Git-Url: https://bilbo.iut-bm.univ-fcomte.fr/and/gitweb/ThesisAli.git/blobdiff_plain/9edfdcf87db0146ae33e8ba7bd87fea713043167..9b7796a5f0b1031f41f762d6d46eb69475bb764c:/CHAPITRE_02.tex

diff --git a/CHAPITRE_02.tex b/CHAPITRE_02.tex
index fdd5ded..ea8bba5 100644
--- a/CHAPITRE_02.tex
+++ b/CHAPITRE_02.tex
@@ -164,7 +164,7 @@ GAF is developed by Xu et al. \cite{GAF}, it uses geographic location informatio
 \label{gaf1}
 \end{figure}
 
-For two adjacent squares grids, (for example, A and B in figure~\ref{gaf1}) all sensor nodes inside A can communicate with sensor nodes inside B and vice versa. Therefore, all the sensor nodes are equivalent from the point of view the routing. The size of the fixed grid is based on the radio communication range $R_c$. It is supposed that the fixed grid is square with $r$ units on a side as shown in figure~\ref{gaf1}. The distance between the farthest sensor nodes in two adjacent squares, such as B and C in figure~\ref{gaf1}, should not be greater than the radio communication range $R_c$. For instance, the sensor node \textbf{2} of grid B can communicate with the sensor node \textbf{5} of square grid C. Thus, 
+For two adjacent squares grids, (for example, A and B in Figure~\ref{gaf1}) all sensor nodes inside A can communicate with sensor nodes inside B and vice versa. Therefore, all the sensor nodes are equivalent from the point of view the routing. The size of the fixed grid is based on the radio communication range $R_c$. It is supposed that the fixed grid is square with $r$ units on a side as shown in Figure~\ref{gaf1}. The distance between the farthest sensor nodes in two adjacent squares, such as B and C in Figure~\ref{gaf1}, should not be greater than the radio communication range $R_c$. For instance, the sensor node \textbf{2} of grid B can communicate with the sensor node \textbf{5} of square grid C. Thus, 
 
 
 \begin{eqnarray}
@@ -202,7 +202,7 @@ one sensor node (based on the remaining energy of sensor nodes inside the fixed
 DESK is a novel distributed heuristic to ensure that the energy consumption among the sensors is balanced  and the lifetime  maximized  while  the coverage  requirement is satisfied~\cite{DESK}. This heuristic  works in  rounds, it requires  only  one-hop neighbor information, and each  sensor decides its status (Active or  Sleep) based on the perimeter coverage model from~\cite{ref133}. 
 
 %DESK is based on the result from \cite{ref133}. 
-In DESK \cite{ref133}, the whole area is K-covered if and only if the perimeters of all sensors are K-covered. The coverage level of a sensor $s_i$ is determined by calculating the angle corresponding to the arc that each of its neighbors covers its perimeter. Figure~\ref{figp}~(a) illuminates such arcs whilst Figure~\ref{figp}~(b) shows the angles corresponding with those arcs in the range [0,2$ \pi $]. According to figure~\ref{figp}~(a) and (b), the coverage level of sensor $s_i$ can be calculated as follows.
+In DESK \cite{ref133}, the whole area is K-covered if and only if the perimeters of all sensors are K-covered. The coverage level of a sensor $s_i$ is determined by calculating the angle corresponding to the arc that each of its neighbors covers its perimeter. Figure~\ref{figp}~(a) illuminates such arcs whilst Figure~\ref{figp}~(b) shows the angles corresponding with those arcs in the range [0,2$ \pi $]. According to Figure~\ref{figp}~(a) and (b), the coverage level of sensor $s_i$ can be calculated as follows.
 %via traversing the range from 0 to  2$ \pi $.
 For each sensor $s_j$ such that $d(s_i,s_j)$ $<$ $2R_s$, we calculate the angle of $s_i$'s arc, denoted by [$\alpha_{j,L}$, $\alpha_{j,R}$], which is perimeter covered by $s_j$, where $\alpha= arccos(d(s_i, s_j)/2R_s)$ and $d(s_i,s_j)$ is the Euclidean distance between $s_i$ and $s_j$. After that, we locate the points $\alpha_{j,L}$ and $\alpha_{j,R}$ of each neighboring sensor $s_j$ of $s_i$ on the line segment $[0, 2\pi]$. These points are sorted in ascending order into a list L. We traverse the line segment from 0 to $2\pi$ by visiting each element in the sorted list L from the left to the right and determine the perimeter coverage of $s_i$. Whenever an element $\alpha_{j,L}$ is traversed, the level of perimeter coverage should be increased by one. Whenever an element $\alpha_{j,R}$ is traversed, the level of perimeter coverage should be decreased by one.