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courses:cs211:winter2018:journals:patelk:chapter6 [2018/03/26 16:16] – [6.2 Principles of Dynamic Programming: Memoization or Iteration over Subproblems] patelkcourses:cs211:winter2018:journals:patelk:chapter6 [2018/03/26 16:42] (current) – [6.3 Segmented Least Squares: Multi-way Choices] patelk
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   * but also can be difficult to think about recurrences in the absence of the "smaller" subproblems that they build on   * but also can be difficult to think about recurrences in the absence of the "smaller" subproblems that they build on
  
 +==== Personal Thoughts ====
  
 +The section was short and easy to process, as the information was clear and not overly complicated. I think this section is a good set up for the section that follows, as it sets up a nice foundation. The next section will probably do a better job of applying this conceptual idea to a problem or problems.
  
 +Readability: 8.0
 +Interesting: 6.5
 +
 +
 +----
 +
 +
 +===== 6.3 Segmented Least Squares: Multi-way Choices =====
 +
 +  * Here, the recurrence will involve "multi-way choices," meaning it is not necessarily binary. 
 +  * Each step has a polynomial number of possibilities to consider for the structure of the optimal solution
 +
 +__The Problem__
 +  * Example: line of best fit
 +    * Suppose our data consists of a set P of n points in the plane denoted (x1,y1), (x2,y2),...,(xn,yn) and x1<x2<...<xn
 +    * Given a line L defined by the equation y = ax+b, we say that the error of L with respect to P is the sum of its squared "distances" to the points in P.
 +  * Goal: find a line with minimum error
 +    * But what if using more than one line is better than using just one line of best fit
 +  * Modified Goal: fit the points well, using as few lines as possible
 +    * Change Detection: given a sequence of data points, we want to identify a few points in the sequence at which discrete change occurs
 +
 +  * //Formulating The Problem//
 +    * partition P into some number of segments
 +    * Each segment is a subset of P that represents a contiguous set of x-coordinates
 +    * For each segment S in our partition of P, we compute the line minimizing the error with respect to the points in S
 +      * The penalty of a partition is defined to be a sum of the following terms:
 +          - number of segments * a fixed, given multiplies (C>0)
 +          - 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
 +
 +
 +----
  
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