Weighted Interval Scheduling: A Recursive Procedure

In this section we see our first example of dynamic programming applied to the interval scheduling we previously solved. There are two stages to our process: first the recursive procedure then an iterative algorithm. This problem is different from our previous greedy approach because each interval has a certain weight/value associated with it. Our problem is to schedule as many nonoverlapping intervals as possible and maximize the value of the schedule.

The problem has n intervals 1,…n. Each request i has a start time s_i,finish time f_i, and a value v_i. We want to maximize the total value of a set of all mutually compatible intervals.

The Recursive Algorithm

Our previous greedy approach that chose the interval that ends earliest is not the optimal solution for this problem. We will still order the requests i by increasing finish time. P(j) represents the largest index i<j so that intervals i and j are disjoint !!! what do they mean by disjoint??? !!! (The leftmost interval that ends before j begins). If no request is disjoint from j we set p(j) = 0.

Our solution either includes the last interval, n, or it doesn't!

• If n is in our solution then no interval between p(n) and n is included in the solution (because all of those intervals would overlap with n). So if n is in our solution, our solution is n + the optimal solution of the intervals 1 through p(n).
• If n is not in our solution then our solution is the optimal solution of intervals 1 through n-1.

We are looking for the optimal solution of smaller sets of the problem. Opt(j) returns the value of the solution. We are looking for the value of opt(n). So again…

• If j is in our solution then opt(j) = v_j + opt(p(j))
• If j is NOT in our solution then opt(j) = opt(j-1)

6.1 From here we can say that opt(j) = max>(v_j + opt(p(j)), opt(j-1)).

6.2 We determine if an interval is in the optimal solution if and only if v_j + opt(p(j)) >= opt(j-1).

These statements form the recurrence equation that expresses the optimal solution (or its value) in terms of the optimal solutions to smaller subproblems (p. 254)

  Compute-Opt(j)
    If j = 0 
      return 0
    Else
      return max(v<sub>j</sub> + Compute-Opt(p(j)), Compute-Opt(j-1))
    Endif
    

Proof that this algorithm correctly computes opt(j):

By induction. The base case opt(0) = 0 is by definition. No assume for some j>0 Compute-Opt(i) correctly computes opt(i) for all i<j. Our induction hypothesis is Compute-Opt(p(j)) = opt(p(j)) and Compute-Opt(j-1)) = opt(j-1)….so opt(j) = max(v_j + Compute-Opt(p(j)), Compute-Opt(j-1)) = Compute-Opt(j). p. 255

This implementation of the algorithm is exponential in the worst case!!! This is not our goal of polynomial-time solution.

Memoizing the Recursion

We want to eliminate the redundancy calls in Compute-Opt by storing the value in a globally accessible place and call on the previously computed value when we need it. We use an array M[0…n] to store the value of each n as Compute-Opt of that value is called.

Memoized Compute-Opt()….

  M-Compute-Opt(j)
    If j = 0 
      return 0
    Else if M[j] is not empty then
      return M[j]
    Else
      Define M[j] = max(v<sub>j</sub> + Compute-Opt(p(j)), Compute-Opt(j-1))
      return M[j]
    Endif

Analysis:

6.4 the runtime of M-Compute-Opt(n) is O(n) when we assume the intervals are sorted by finish time

A single call of M-Compute-Opt is constant and we call it n number of times (the number of entries + 1).

Finding the Solution…not the value

Our memoized array stores the value of the solution, not the schedule of the intervals. We find it by looking at the values saved in the array after the optimum value has been computed.

  Find-Solution(j)
    If j = 0 
      output nothing
    Else 
      If v<sub>j</sub> + M[p(j)] >= M[j-1] then
        Output j together with the result of Find-Solution(p(j))
      Else
        Output the result of Find-Solution(j-1)
      Endif
    Endif

If we are given the array of optimal values, the Find-Solution algorithm runs in O(n) time.

Principles of Dynamic Programming: Memoization or Iteration over Subproblems

In this section we learn how to solve problems by iterating over subproblems rather then computing solutions recursively. We form an almost equivalent version of the algorithm, but iteratively. This will show us how we will use dynamic programming in later sections

The Algorithm:

We can compute the entries in our array with an iterative algorithm instead of memoized recursion.

  Iterative-Compute-Opt
    M[0] = 0
    For j = 1, 2,...,n
      M[j] = max(v<sub>j</sub> + M[p(j)], M[j-1])
    Endfor

We can prove this algorithm works just like we did in the previous section (p. 255) by induction. The runtime of this algorithm is O(n) because each run is constant and it runs at most n times.

A Basic Outline of Dynamic Programming

We will use this iterative approach to dynamic programming for the rest of Chapter 6 because they are easier to express. To use dynamic programming you need a collection of subproblems derived from the original problem that satisfies these properties:

1. only a polynomial number of subproblems
2. solution to original problem easily computed from solutions to subproblems
3. there is a natural ordering on subproblems from smallest to largest that allows you to determine the solution to a subproblem from solutions to x number of smaller subproblems with an easy to compute recurrence

The most important part is finding a recurrence that links the subproblems.

Segmented Least Squares: Multi-way Choices

In the previous section at each step the subproblem belonged to the solution or it didn't. In this section the recurrence involves “multi-way choices”. This means at each step there are polynomial possibilities to consider for the structure of the optimal solution.

The problem we are trying to solve is finding the line that best fits data plotted on x and y axes. The data we are given is a set of points (x, y) that are in ascending order of x. We assume each point has a unique x value. The error of a line L, with respect to the set of points P is the sum of its squared distances to the points in P. We want to find the line with minimum error, but instead of using just one line, we seek a set of lines through the points that minimized the error using as few lines as possible. This is known as the segmented least squares problem. We have to detect change in the points; identify where the change occurs from one line to another.

We partition the set of points P into subsets that has a segment S. For each subset we compute the line of minimum error. The penalty for each subset is: the number of segments we partition P * a constant C > 0 and the error value of the optimal line through each segment. We want to find a partition of minimum penalty.

Designing the Algorithm

We are seeking to partition n objects. We want a polynomial number of subproblems that yield a solution and we should build up these solutions using recurrence.

The past point p_n belongs to a segment in the optimal partition that begins at some point before p_n at p_i. Let opt(i) equal the optimum solution for points p₁ through p_i. e_i,j represents the minimum error of any line from points p_i to p_j.

6.1 if the last segment of the optimal partition is p_i,…,p_n , then the value of the optimal solution is opt(n) = e_i,n + C + opt(i-1).

To find the best way to produce a segment we find the minimum

6.7 for the subproblem on the points p₁,…,p_j, opt(j) = min(1=<i=<j)(e_i,j + C + opt(i-1).

  Segmented-Least-Squares(n)
    Array M[0...n]
    Set M[0] = 0
    For all pairs i =< j
      compute the least squares error e(i,j) for the segment p(i)...p(j)
    Endfor
    For j = 1,2,...n
      M[j] = min(1=<i=<j)(e(i,j) + C + opt(i-1)
    Endfor
    Return M[n]
    

This algorithms correctness is proved by induction. To find the points that are in a segment and not the value that is stored in the array, we use an algorithm like in the previous section

  Find-Segments(j)
    If j=0 then
      Output nothing
    Else
      Find an i that minimizes e(i,j) + C + M[i-1] 
      Output the segment p(i)...p(j) and the result of Find-Segments(i-1)
    Endif

Analysis:

We perform the value of the least squares errors for n² pairs that we compute at O(n) times, so the runtime of computing e_i,j values is O(n³). Once we have all of the error values, for each value we have to determine the minimum recurrence that takes O(n) for each j, so this is O(n²).

I thought this section was harder to read than the previous two because it was a lot more generalized and assumed you knew more about dynamic programming. I thought this problem was pretty interesting though because I like statistics! Readability: 7/10.

Subset Sums and Knapsacks: Adding a Variable

In this section we solve a problem that has duration and deadline but not have an interval by which they need to be done. The problem we are considering is a machine that processes a set of requests 1 through n where we can only use the machine from 0 to W and each job has a process time w_i. We want to choose the jobs that use the machine as much as possible up to the time W. We can make this problem more general into the knapsack problem where each request has a value and a weight and we want to chose a subset of maximum total value that does not exceed the weight W of the knapsack. We cannot use a greedy approach to solve this problem!

The Algorithm:

One way we could try the algorithm is to consider the first i requests but this does not work. We need more subproblems. To find opt(n) we need the value of opt(n-1) and the best solution of subset n-1 and total allowed weight W-w_n. Opt(i,w) will be the value of the optimal solution of a subset 1 through i with maximum allowed weight w. If n is not in the solution then opt(n,W) = opt(n-1,W). If n is in the solution then opt(n,W) = w_n + opt(n-1, W-w_n). When W is less than w_n then n cannot be in the solution.

  Subset-Sum(n,W)
    Array M[0..n, 0..w]
    initialize M[0,w] = 0 for each w = 0, 1,...W
    for i = 1,2,...,n
      for w = 0,...W
        M[i,w] = opt(i,w) = max(opt(i-1, w),w(i) + opt(i-1,w-w(i))
      Endfor
    Endfor
    Return M[n, W]
    

We prove the algorithms correctness by induction. Instead of a single array, we use a matrix of subproblems to build up the values. We compute each of the values in M in constant time so the runtime is the number of entries in M which is O(nW). It is a polynomial function of n and W, not just n. This is pseudo-polynomial. To get the set of items and not just the value we trace back through the array as we did in previous problems. This can be found in O(n) time.

This section was really interesting to me and seems like something really practical and would be used a lot. I thought like the previous section it cut out a lot of the in depth descriptions but it was a little more easy to read because I understood it better with more experience. Readability 8/10