Differences

This shows you the differences between two versions of the page.

--- courses:cs211:winter2018:journals:boyese:chapter6 [2018/03/27 21:16] – [Section 6.1: Weighted Interval Scheduling: A Recursive Procedure] boyese
+++ courses:cs211:winter2018:journals:boyese:chapter6 [2018/03/27 23:00] (current) – [Section 6.3: Segmented Least Squares: Multi-way Choices] boyese
@@ Line 1: / Line 1: @@
 ====Section 6.1: Weighted Interval Scheduling: A Recursive Procedure====
+Weighted Interval Scheduling
+  * Job j starts at s<sub>j</sub>, finishes at f<sub>j</sub>, and has weight or value v<sub>j</sub>
+  * Two jobs are compatible if they don't overlap
+  * Goal: find maximum weight subset of mutually compatible jobs
+  * Notation: Label jobs by finishing time: f<sub>1</sub>≤ f<sub>2</sub> ≤ ... ≤ f<sub>n</sub>
+  * Definition: p(j) = largest index i	< j such that job i is compatible with j
+The recursive algorithm that computes the optimal schedule, now with weights for importance of each request/job:
 <code>
 Compute-Opt(j)
@@ Line 10: / Line 18: @@
 </code>
+The correctness of the algorithm follows directly by induction on j:
+Compute-0pt(j) correctly computes OPT(j) for each j = 1, 2, ..., n.
+But, unfortunately, if we really implemented the algorithm Compute-Opt as just written, it would take exponential time to run in the worst case. To come up with a better algorithm we must implement the concept of memoization, a technique that involves storing values that have already been computed. Below is the new algorithm that runs in O(n) time at worst.
 <code>
@@ Line 23: / Line 34: @@
 </code>
+So far we have simply computed the value of an optimal solution; but presumably we want a full optimal set of intervals as well. It would be easy to extend M-Compute-0pt so as to keep track of an optimal solution in addition to its value. This extension of the algorithm is shown below.
 <code>
@@ Line 37: / Line 49: @@
 </code>
+Since Find-Solution calls itself recursively only on strictly smaller values, it makes a total of O(n) recursive calls; and since it spends constant time per call, we have Find-Solution returns an optimal solution in O(n) time.
 ====Section 6.2: Principles of Dynamic Programming: Memoization or Iteration over Subproblems====
+Once we have the array M, the problem is solved: M[n] contains the value of the optimal solution on the full instance, and Find-Solution can be used to trace back through M efficiently and return an optimal solution itself. We can directly compute the entries in M by an iterative algorithm, rather than using memoized recursion. This algorithm is below:
+<code>
+Iterative-Compute-Opt
+    M[O] = 0
+    For j = l, 2 .....n
+        M[j] = max(v<sub>j</sub> + M[p(j)], M[j - 1])
+    Endfor
+</code>
+The running time of Iterative-Compute-0pt is clearly O(n), since it explicitly runs for n iterations and spends constant time in each.
+To set about developing an algorithm based on dynamic programming, one needs a collection of subproblems derived from the original problem that satisfies a few basic properties:
+  * There are only a polynomial number of subproblems
+  * The solution to the original problem can be easily computed from the solutions to the subproblems
+  * There is a natural ordering on subproblems from "smallest" to "largest" together with an easy-to-compute recurrence that allows one to determine the solution to a subproblem from the solutions to some number of smaller subproblems
 ====Section 6.3: Segmented Least Squares: Multi-way Choices====
+In the problem we consider here, the recurrence will involve what might be called "multi-way choices": at each step, we have a polynomial number of possibilities to consider for the structure of the optimal solution. This problem involves finding a line of best fit for a collection of points on a graph. This process is used frequently in statistics and called regression, but in this case the points do not fall along a straight line. Any single line through the points on a polynomial graph would have a terrible error; but if we use two lines, we could achieve quite a small error. So we could try formulating a new problem as follows: Rather than seek a single line of best fit, we are allowed to pass an arbitrary set of lines through the points, and we seek a set of lines that minimizes the error. We need a problem formulation that requires us to fit the points well, using as few lines as possible. We now formulate a problem--the Segmented Least Squares Problem--that captures these issues quite cleanly.
+Suppose we let OPT(i) denote the optimum solution for the points p<sub>1</sub>, ..., p<sub>i</sub> and we let e<sub>i, j</sub> denote the minimum error of any line with respect to p<sub>i</sub>, p<sub>i + 1</sub>, ..., p<sub>j</sub>. Then our observation above says that if  the last segment of the optimal partition is p<sub>i</sub>, ..., p<sub>n</sub>, then the value of the optimal solution is OPT(n) = e<sub>i, n</sub> + C + OPT(i - 1). The algorithm is as follows:
+<code>
+Segmented-Least-Squares(n)
+    Array M[0... n]
+    Set M[0] = 0
+    For all pairs i < j
+        Compute the least squares error e<sub>i, j</sub> for the segment p<sub>i</sub>, ..., p<sub>j</sub>
+    Endfor
+    For j = 1, 2, ..., n
+        Use the recurrence (6.7) to compute M[j]
+    Endfor
+    Return M[n]
+</code>
+The algorithm to compute an optimum partition is as follows:
+<code>
+Find-Segments(j)
+    If j = 0 then
+        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>1</sub>]} and the result of
+            Find-Segments (i - 1)
+    Endif
+</code>
+**Analyzing The Run Time**
+There are O(n<sup>2</sup>) pairs (i, j) for which this computation is needed; and for each pair (i, j); we can use the formula given at the beginning of this section to compute e<sub>i, j</sub> in O(n) time. Thus the total running time to compute all e<sub>i, j</sub> values is O(n<sup>3</sup>). The algorithm has n iterations, for values j = 1, ..., n. For each value of j, we have to determine the minimum in the recurrence (6.7) to fill in the array entry M[j]; this takes time O(n) for each j, for a total of O(n<sup>2</sup>). Thus the running time is O(n<sup>2</sup>) once all the e<sub>i, j</sub> values have been determined.