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--- courses:cs211:winter2012:journals:jeanpaul:chaptersixsectioniii [2012/03/28 01:48] – [The Problem] mugabej
+++ courses:cs211:winter2012:journals:jeanpaul:chaptersixsectioniii [2012/03/28 02:39] (current) – [Algorithm Analysis] mugabej
@@ Line 8: / Line 8: @@
 >>>>>>>>>>>>>>>>>> (x<sub>1</sub>y<sub>1</sub>),(x<sub>2</sub>y<sub>2</sub>),...,(x<sub>n</sub>y<sub>n</sub>), where x<sub>1</sub> < x<sub>2</sub>,...,< x<sub>n</sub>\\
 >>>>>>>>>>>>>>>>>> Given a line L with equation y = ax + b, we say that an "error" of L with respect to P is the sum of all of its squared distances to the points in P:\\
->>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Error(L,P) = ∑<sub>i =1</sub><sup>n</sup> (y<sub>i</sub> - ax<sub>i</sub>- b) <sup>2</sup>
+>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Error(L,P) = ∑<sub> from i =1 to n</sub> (y<sub>i</sub> - ax<sub>i</sub>- b) <sup>2</sup>\\
+>>>>>>>>>>>>>>>>>> Thus naturally, we are bound to finding the line with minimum error.\\
+>>>>>>>>>>>>>>>>>> The solution turns out to be a line y = ax + b, where:\\
+>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> a = [(n∑<sub>i</sub> x<sub>i</sub>y<sub>i</sub>) - (∑<sub>i</sub>x<sub>i</sub>)(∑<sub>i</sub> y<sub>i</sub>)]/[(n∑<sub>i</sub> x<sub>i</sub><sup>2</sup>) - (∑<sub>i</sub> x<sub>i</sub>)<sup>2</sup>]\\
+>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> And b = (∑<sub>i</sub> y<sub>i</sub>- a∑<sub>i</sub> x<sub>i</sub>)/n\\
+\\
+>>>>>>>>>>>>>>>>>> However, our problem is different from the above mentioned which these (above) formulas solve.\\
+** Formulating our problem**
+\\
+>>>>>>>>>>>>>>>>>> Given a sequence of data points, we need to identify a few points in the sequence at which a discrete change occurs.--> In our specific case, a change from one linear approximation to another.\\
+>>>>>>>>>>>>>>>>>> So, we have a set of points P = {(x<sub>1</sub>,y<sub>1</sub>),(x<sub>2</sub>,y<sub>2</sub>),...,(x<sub>n</sub>,y<sub>n</sub>)} with x<sub>1</sub> < x<sub>2</sub>,..., x<sub>n</sub>.\\
+>>>>>>>>>>>>>>>>>> p<sub>i</sub> denotes the point (x<sub>i</sub>,y<sub>i</sub>)\\
+>>>>>>>>>>>>>>>>>> We must first partition P into some number of segments.\\
+>>>>>>>>>>>>>>>>>> Each segment is a subset of P that represents a contiguous set of x-coordinates of the form {p<sub>i</sub>,p<sub>i+1</sub>,...,p<sub>j-1</sub>,p<sub>j</sub>} for some indices i≤j.\\
+>>>>>>>>>>>>>>>>>> For each segment S in our partition P, we compute the line minimizing the error with respect to the points in S using the formula we found above(where we wrote the value of a and b in the line y = ax + b).\\
+>>>>>>>>>>>>>>>>>> The penalty of a partition is defined to be the sum of:\\
+>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> The number of segments into which we partition P, times a fixed, given multiplier C>0.\\
+>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> 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.