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--- courses:cs211:winter2018:journals:patelk:chapter6 [2018/03/26 16:29] – [6.2 Principles of Dynamic Programming: Memoization or Iteration over Subproblems] patelk
+++ courses:cs211:winter2018:journals:patelk:chapter6 [2018/03/26 16:42] (current) – [6.3 Segmented Least Squares: Multi-way Choices] patelk
@@ Line 129: / Line 129: @@
           - for each segment, the error value of the optimal line through that segment
     * as we increase the number of segments, we reduce the penalty in terms of 2, but we increase the term in terms of 1
+__Designing the Algorithm__
+  * Polynomial number of subproblems, solutions yield a solution to the original problem, build up solutions using a recurrence
+  * For segmented least squares, the last point pn belongs to a single segment in the optimal partition, and the segment begins at some earlier point pi.
+  * If we know the identity of the last segment, pi,...,pn, then we could remove those points from consideration and recursively sold the problem on the remaining points p1,...,pi-1
+{{:courses:cs211:winter2018:journals:patelk:lineofbestfit.png?nolink&400|}}
+  * If the last segment of the optimal partition is pi,...,pn, then the value of the optimal solution is OPT(n) = error of i through n + C + OPT(i-1)
+  * For the subproblem on the points p1,...,pj,
+{{:courses:cs211:winter2018:journals:patelk:lineofbestfitstatement.png?nolink&400|}}
+{{:courses:cs211:winter2018:journals:patelk:segmented-least-squares-alg.png?nolink&400|}}
+  * we can trace back through array M to compute an optimum partition:
+{{:courses:cs211:winter2018:journals:patelk:find-segments-alg.png?nolink&400|}}
+**Analyzing the Algorithm**
+  * Runtime of Segmented-Least-Squares: O(n³)
+    * compute the values of all the least-squares errors e of i through j (eij)
+    * O(n²) pairs (i,j)
+      * For each pair, we can compute the error e of i through j in O(n) time
+  * Algorithm has n iterations j=1,...,n.
+    * For each value, we have to determing the minimum in the recurrence to fill in the array entry M[j]
+      * Takes O(n) for each j -> total of O(n²)
+  * Runtime is O(n²), once all error values have been determined.
+==== Personal Thoughts ====
+The section did a really good job laying out the algorithms in a way that was easy to follow. I think the least squares problem is also inherently easier to understand because we have worked with similar problems in math classes before. Overall, I think the algorithms and the data structures needed to solve this problem are pretty intuitive.
+Readability: 8.0
+Interesting: 6.5
+----