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--- courses:cs211:winter2012:journals:jeanpaul:chaptersixsectioniii [2012/03/28 02:24] – [The Problem] mugabej
+++ courses:cs211:winter2012:journals:jeanpaul:chaptersixsectioniii [2012/03/28 02:39] (current) – [Algorithm Analysis] mugabej
@@ Line 27: / Line 27: @@
 >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> For each segment, the error value of the optimal line through that segment.\\
 >>>>>>>>>>>>>>>>>> Our goal is to find the a partition of minimum penalty.\\
+=====  Designing the Algorithm =====
+>>>>>>>>>>>>>>>>>> INPUT: n, p<sub>1</sub>,...,p<sub>N</sub>, C\\
+>>>>>>>>>>>>>>>>>> Segmented-Least-Squares()\\
+>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> M[0] = 0\\
+>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> e[0][0] = 0\\
+>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> for j = 1 to n\\
+>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> for i = 1 to j\\
+>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> e[i][j] = least square error for the segment p<sub>i</sub>,...,p<sub>j</sub>\\
+>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> for j = 1 to n\\
+>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> M[j] = min <sub>1 ≤ i ≤ j</sub> (e[i][j] + c + M[i-1])\\
+>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> return M[n]\\
+\\
+Just like the Weighted Interval Scheduling Problem, we need to find an algorithm that gives us the solution itself.\\
+** Algorithm to give for the solution itself**
+\\
+>>>>>>>>>>>>>>>>>> FindSegments(j):\\
+>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> if j = 0:\\
+>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> output nothing\\
+>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> else:\\
+>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Find an i that minimizes e<sub>i,j</sub> + C + M[i-1]\\
+>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Output the segment {p<sub>i</sub>, …, p<sub>j</sub>} and the result of \\
+>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> FindSegments(i-1)
+>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> End if
+=====  Algorithm Analysis =====
+\\
+The running time of the algorithm is O(n<sup>2</sup>) since filling in the array M[j] takes us O(n) time for each j.\\
+I give this section 7/10.