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--- courses:cs211:winter2012:journals:ian:chapter6 [2012/03/28 04:11] – 6.2 sturdyi
+++ courses:cs211:winter2012:journals:ian:chapter6 [2012/03/28 04:27] (current) – sturdyi
@@ Line 12: / Line 12: @@
    * A note, at least, on the polynomial number of subproblems (and why the iterative setup is not always equivalent to the
 recursive): in Abstract Algebra we recently got an extra-credit problem implementing a dynamic algorithm finding the number of three-free permutations of the first n natural numbers (rather convenient timing, really). The algorithm determined, given a set of numbers yet to be placed, whether each one of them could be placed next; for each one that could, the algorithm was recursively applied to determine the number of valid solutions that placed that number next. There are really 2^n subproblems, but most of them prove inaccessible (I was able to calculate up to n=62 despite a rather memory inefficient memoization structure). An iterative algorithm there would be exponential time; the recursive algorithm was not great, but still quite workable (I do not know of a tighter bound than O(2^n) space and O(n^2*2^n) time, but it is experimentally much lower).
-   * No complaints).
+   * No complaints.
+====== 6.3 Segmented Least Squares =======
+   * Dynamic programming does not require binary choice. Segmented least squares attempts to determine the natural number of kinks in a set of points, minimizing the sum of deviations from the solution and a penalty for each break. One can thus make a pass through recursively determining the best way to fit a line to subsets of the point. OLS takes O(n^3) and the rest O(n^2).
+   * No questions.
+   * No comments.
+   * No complaints.
+====== 6.4 Subset Sums and Knapsacks ======
+   * The knapsack problem is the selection of a subset (without replacement) of values that has sum as close as possible to a target. Here one cannot recurse merely over items, since unlike in weighted interval schedule the subproblems for a given choice are not independent of the choices already made (they affect the target). Therefore, one needs to define subproblems in two variables, the set of goods to be considered and the weight still to fill (but one can still make the choice sequentially, so that there are n sets defined by their greatest element). Total running time is O(nW), the result of filling in nW matrix entries in O(1) time each. The knapsack problem is a generalization, maximizing the sum of value subject to a weight constraint. However, the same recurrence holds, with merely a different definition of optimality, and it can also be solved in O(nW).
+   * No questions.
+   * No insights.
+   * Again with the introduction of the false starts first? I can see their reasons, but still do not much like it.