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Nt1330 Unit 9 Final Paper

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Initially, the cell $c_4$ enclosing node $n_2$ (As shown in Figure~\ref{f:data structure access}(a)) is accessed to obtain the information of extending edges $e_1$, $e_2$, and $e_6$ that connect with $n_2$. Having obtained the edge information, the distances from nodes $n_2$ to $n_1$, $n_5$, and $n_6$ are computed. Then, obtained the object information, we can know that there have objects on the $e_1$ and $e_2$. For $e_1$, the distance from node $n_2$ to object $t_1$ are computed, the edge is $e_1$, the end node of the $e_1$ is $n_1$ , and node $n_1$ is belong to the cell $c_8$ (As shown in Figure~\ref{f:The cell index structure}(b)). So as $e_2$. For $e_6$, there have no object on this edge, so we kept the $d(n_2,n_6)$ as $d(n_2,o)$. Then, the three nodes with their information are enqueued. The visited cell …show more content…

Although there have same value of $d(q,o)$ between queue entry \{8,14,$e_{10}$,$n_{10}$,$c_6$\} and \{8,8,$e_6$,$n_6$,$c_1$\}, we kept the \{8,14,$e_{10}$,$n_{10}$,$c_6$\} priority in \{8,14,$e_{10}$,$n_{10}$,$c_6$\}. Because the $d(q,o)$ is not equal to $d(q,n_i)$ means there has objects on the edge of entry \{8,14,$e_{10}$,$n_{10}$,$c_6$\}. Again, accessing the cell $c_8$, the object $h_2$ is accessed, we put the object $h_2$, the distance between $h_2$ and $q$ ($d(h_2,q)$), and cell $c_8$ into the table $T_{vo}$ and table $T_{vc}$ shown as Figure~\ref{f:data structure access}(d) and Figure~\ref{f:data structure access}(e). Then, the cell is accessed to get the information of edge $e_3$ and $e_7$ (which connect with $n_1$) and to compute the distance information (as shown in Figure~\ref{f:data structure access}(c)). There has two cells ($c_4$ and $c_8$) are accessed, we using the {\bf Pruning rules} which is described in section 4.1.2. After the {pruning rules}, the distance between $t_2$ and $h_2$ is smaller or equal than $d$ ($d(t_2,h_2) \le d$), so we kept this information in the table

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