Upcoming

• Week 09: finish dynamic programming
• Week 10: network flow & applications
• Week 11: midterm 2
• Weeks 12, 13: NP completeness

Overview

1. Finishing Weighted Interval Scheduling
2. The Knapsack Problem

Weighted Interval Scheduling

Input:

1. A set $R$ of $n$ intervals $r_1 = [s_1, t_1], r_2 = [s_2, t_2], \ldots, r_n = [s_n, t_n]$
2. For each interval $r \in R$, a weight $w(r) > 0$
   • e.g., weight = profit from serving request $r$

Output. A collection of intervals from $R$ that is

1. feasible no two intervals overlap
2. maximum weight choice maximizes sum of $w(r)$ for chosen $r$

Note: equivalent to (unweighted) interval scheduling when all weights are the same

Example w/ Predecessors

A Recursive Solution

  MaxWeightSchedule(w, p, n):
    if n = 0 then return 0
    opt-n <- w[n] + MaxWeightSchedule(w, p, p[n])
    opt-no-n <- MaxWeightSchedule(w, p, n-1)
    return Max(opt-n, opt-no-n)

Recursion to Iteration

Idea. Store array max:

• max[i] is maximum weight of schedule consisting of intervals $r_1, r_2, \ldots, r_i$

Question. How to initialize/update max values?

Iterative Solution

  IMaxWeightSchedule(w, p)
    max <- new array of size n+1 
    max[0] <- 0
    for i = 1 up to n do
      max[i] <- Max(w[i] + max[p[i]], max[i-1])
    endfor
    return max[n]

Overall Running Time?

Steps: (assume $n$ intervals)

1. Sort intervals by end time

2. Compute array p

3. Run IMaxWeightSchedule(w, p)

Total?

Exercise

Update IMaxWeightSchedule to return the actual schedule of maximum weight, not just the weight itself.

The Knapsack Problem

Knapsack Motivation

In weighted interval scheduling each request had:

• start time
• end time
• value

Goal: to service set of non-overlapping requests to maximize total value

Knapsack Problem

Input:

1. A set $R$ of $n$ requests, each having
   • duration (weight) $b_r$
   • value $v_r$
2. Total time (weight) budget $B$

Output: A set $S$ of requests to service with

1. sum of durations of requests in $S$ is at most $B$
2. sum of values of requests is maximized

Example

Recurrence Relation?

Previous technique. Express relationship between optimal solutions that

1. service the final request $r_n$
2. do not service final request $r_n$

Question. How can we express this relationship for the knapsack problem?

A Recurrence Relation

Subtlety. Must keep track of remaining budget

• if $r_n$ is not serviced, remaining budget is $B$
• if $r_n$ is serviced, remaining budget is $B - b_n$

Computing Optimal Values

Assume. All durations $b_i$ are integers at most $B$.

• we will revisit this assumption later

Compute. To compute $\opt(n, B)$:

• Generate a two dimensional array max where max[j, C] stores the value $\opt(j, C)$

Questions.

1. How to initialize max?
2. How to update max?

Example

Pseudocode

  FindMax(R, n, B):
    max <- new 2d array of dimensions n+1, B+1
    set max[0, C] <- 0 for C = 0 to B
    for j from 1 to n
      (b, v) <- i-th request in R
      for C from 0 to B
        if b <= C then 
          max[j, C] <- Max(v + max[j-1, C-b], max[j-1,C])
        else
          max[j, C] <- max[j-1, C]
    return max[n, B]

Correctness

Claim. For all $j$ and $C$, $\max[j, C] = \opt(j, C)$

Proof. Induction on $j$.

Base case $j = 0$. Optimal subset of size $0$ has value $0$.

Running Time?

  FindMax(R, n, B):
    max <- new 2d array of dimensions n+1, B+1
    set max[0, C] <- 0 for C = 0 to B
    for j from 1 to n
      (b, v) <- i-th request in R
      for C from 0 to B
        if b <= C then 
          max[j, C] <- Max(v + max[j-1, C-b], max[j-1,C])
        else
          max[j, C] <- max[j-1, C]
    return max[n, B]

Conclusion

For the knapsack problem with $n$ requests and budget $B$, we can find compute $\opt(n, B)$ in $O(B n)$ time.

• assuming the duration of each request is an integer

For Next Time

Lecture 25 Ticket. Modify FindMax procedure to return the subset $S$ whose value is $\opt(n, B)$.