1. Weighted Interval Scheduling

2. Optimal Substructure
Is the solution equal to the “sum” of smaller solutions? or
Does the solution to the whole problem define solutions to the sub-problems?

3. Shortest Path: Optimal Substructure?w v
Shortest path from node u to node v goes through node w
Is this the shortest path from w to v?
Is this the shortest path from u to w?
Yes – improving one improves the whole

4. Longest Path: Optimal Substructure?w v
Longest path from node u to node v goes through node w
Is this the longest path from u to w?

5. The (u,v) path does define the longest path from u to w
True
False 10

6. The (u,v) path does define the longest path from u to w
True
False 10

7. Longest path from u to v u w v u v v
Longest path from u to v is not composed of the longest path from u to w plus the longest path from w to v. u w
Longest path from u to w?

8. Interval Scheduling Problem
Input: A list of intervals, each with a:
Start time si
Finish time fi
Output: A subset of non-overlapping intervals
Goal: The largest possible subset

9. Weighted Interval Scheduling Problem
Input: A list of intervals, each with a:
Start time si
Finish time fi
Weight wi
Output: A subset of non-overlapping intervals
Goal: The largest possible weight sum

10. Weighted “Classroom” Scheduling Problem
Input: A list of classes, each with a:
Start time si
Finish time fi
Enrollment wi
Output: A set of (non-overlapping) classes to be scheduled in the same room
Goal: The largest possible enrollment

11. (Unweighted) Interval Scheduling
Solution: Sort by finish time, keep taking next available

12. Weighted Interval Scheduling6 1 3
100 8 2 16
100 1
Optimal weight: = 203

13. Notice: If we have already sorted by f, we know that p(i) < i
Notation
Number the objects by finish time
Consider object i
p(i) = Largest j such that fj < si
(Largest j such that j does not overlap i)
Notice: If we have already sorted by f, we know that p(i) < i
(fp(i) < si < fi )

14. Weighted Interval Scheduling
Note: we are numbering from 1, not 0
(It will be less confusing later on.)
7:6
2:1
5:3
9:100
1:8
4:2
8:16
3:100
6:1
p(6) = 4
What is p(9)?
p(8) = 5
p(7) =
Undefined

15. What is p(9)? 1 2 3 4 5 6 7 8
5:3
2:1
4:2
1:8
3:100
6:1
8:16
9:100
7:6 10

16. What is p(9)? 1 2 3 4 5 6 7 8
5:3
2:1
4:2
1:8
3:100
6:1
8:16
9:100
7:6 10

17. Dynamic Programming
We will solve the (weighted) problem by dynamic programming
Allows us to try a limited number of potential solutions at once
More than one solution (as opposed to a greedy approach)
But still a limited number (as opposed to brute force)

18. DP Hints We follow the “steps” that apply to all DP problems
Order objects
Concentrate on optimal score (not solution)
Use the optimal substructure property to create a recursion
Use overlapping subproblems property to avoid a recursive encoding

19. Step 1: Order Objects
A DP solution usually requires that we place some order on the objects
In this case, we will still want to order by finish time.
Assume this already has been done:
so if i < j  fi  fj

20. Step 2: Concentrate on optimal score
We will start by looking for the optimal score only
In this case: what is the highest weight total we can achieve? 
Define: opt(i) = Optimal score using only elements 1 to i
Finding the optimal score will be easier than trying to find the optimal solution
Finding the optimal solution will actually be an easy second step

21. Step 3: Exploit Optimal Substructure
Is the solution equal to the “sum” of smaller solutions? or
Does the solution to the whole problem define solutions to the sub-problems?

22. Optimal Substructure opt(9) = 100+100+20+7 = 227 optimal = 100+100?6 1 3 20
100
100 4 3 7
optimal = ?
optimal = 7+20?
Yes – since opt(n) = ( )+(7+20)

23. Optimal Substruture opt(9) = 100+3+4+5 = 112 opt(8) = 100+3+4 = 107?6 1 3 5 8 2 4
100 1
opt(8) = = 107?
Yes – so opt(9) = opt(8) + 5

24. Optimal Substruture opt(9) = 100+100+7 = 207 opt(8) = 100+100+7 = 207?6 1 3 2
100
100 4 3 7
opt(8) = = 207?
Yes – so opt(9) = opt(8)

25. Using Optimal Substructure
We want to find the value of opt(n)
Interval n is used in the optimal solution
Two possibilities:
Interval n is not used in the optimal solution
This is a tautology.
(Or it isn’t.)

26. Using Optimal Substructure
We can make use of optimal substructure to solve the problem
Compute best solution possible when using interval n
Compute best solution possible when not using interval n
use whichever is better

27. Case 1: interval n is used
If we know interval n is used, what else do we need to do?
p(n) is the last interval we can still use
(if j > p(n), then fj > sn)
Claim: opt(n) = opt(p(n)) + wn

28. Case 1: interval n is used
Claim: opt(n) = opt(p(n)) + wn
Must have opt(n)  opt(p(n)) + wn
Take solution for opt(p(n))
Add interval n (must be possible)
Resulting total: opt(p(n)) + wn
Must have opt(n)  opt(p(n)) + wn
Remove interval n from optimal
New weight: opt(n) – wn
Picked from {1, 2, …, p(n)}
…Hence must be  opt(p(n))
Result: opt(n) – wn  opt(p(n))
Results comes from the optimal sub-structure property

29. Case 1: Illustrated opt(9) = 100+3+100 = 203, p(9) = 68 2 16
100 1
opt(9) = = 203, p(9) = 6
opt(6) = = 103
opt(9) = opt(6) + w9

30. If interval n is in the optimal solution: then opt(n) = opt(p(n)) + wn
Case 1: Recap
If interval n is in the optimal solution: then opt(n) = opt(p(n)) + wn

31. If we know the interval item is not used, what else do we need to do?
Case 2: Item n is not used
If we know the interval item is not used, what else do we need to do?
Claim: opt(n) = opt(n-1)
If item n is not used, then opt(n) and opt(n-1) must correspond to the same solution score

32. The optimal solution clearly doesn’t use interval 9
Case 2: Illustrated
7:6
2:1
5:3
9:2
1:8
4:2
8:1
3:100
6:100
opt(9) = = 203
The optimal solution clearly doesn’t use interval 9
opt(8) = = 203
opt(9) = opt(8)
So we get the same solution if we remove 9, considering only intervals 1 to 8

33. Summing Up If n is in the optimal solution: opt(n) = opt(p(n))+wn
If n is not in the optimal solution: opt(n) = opt(n-1)
Either we use item n, or we don’t.
opt(p(n))+ wn
opt(n) = max
opt(n-1)
opt(0) = 0

34. Recursive Formula int WIS(array W, array P, int n) { if n==0 return 0
int c1 = WIS(W, P, n-1)
int c2 = WIS(W, P, P[n])+W[n]
return max(c1, c2)
Bad news: runtime is terrible
Good news: Overlapping sub-problems

35. Step 5: Bottom-Up Calculation
opt(p(n))+ wn
opt(n) = max
opt(0) = 0
opt(n-1)
You can definitely calculate opt(i) if you know opt(i) i < n
Calculate opt(1), then opt(2), then …

36. Execution opt(7) = max{opt(0) + w7, opt(6)} = 1038 8
100
100
103
103
103
119
203
opt(7) = max{opt(0) + w7, opt(6)} = 103
opt(8) = max{opt(5) + w8, opt(7)} = 119
opt(8) = max{opt(6) + w9, opt(8)} = 203
opt(6) = max{opt(4) + w6, opt(5)} = 103
opt(2) = max{opt(0) + w2, opt(1)} = 8
opt(1) = max{opt(0) + w1, opt(0)} = 8
opt(3) = max{opt(0) + w3, opt(2)} = 100
opt(4) = max{opt(2) + w4, opt(3)} = 100
opt(5) = max{opt(3) + w5, opt(4)} = 100

37. Trace-back 7:6 (p = 0) 5:3 (p = 3) 2:1 (p = 0) 9:100 (p = 6)8 8
100
100
100
100
103
103
103
103
103
103
103
119
203
203
203

38. Maintains Solutions
7:6 (p = 0)
5:3 (p = 3)
2:1 (p = 0)
9:100 (p = 6)
1:8 (p = 0)
4:2 (p = 2)
8:16 (p = 5)
3:100 (p = 0)
6:1 (p = 4) 8 8 8
100
100
100
103
103
103
119
119
203
As we continue, we track up to n different solutions
-- more than a greedy algorithm, less than brute force

39. Runtime We need to fill in values into n boxes8 8 8
100
100
100
103
103
103
119
119
203
We need to fill in values into n boxes
Each box requires O(1) time to fill (if we know p())
Total runtime: O(n)

40. Bottom-Up Code What is the runtime of the WIS algorithm?
int WIS(array W, array P, int n)
array opt, opt[0] = 0
for i  1 to n
c1  opt[P[i]] + W[i]
c2  opt[i-1]
opt[i]  max{c1,c2}
return opt[n]
array WIS_Sol(array opt, array P, int n)
array solution
i  n
while (i > 0)
if opt[i] == opt[i-1]
i = i-1
else
solution.push(i)
i  p[i]
return solution
What is the runtime of the WIS algorithm?

41. Runtime of WIS?
O(1)
O(log n)
O(n)
O(n log n)
O(n2) 9

42. Runtime of WIS?
O(1)
O(log n)
O(n)
O(n log n)
O(n2) 9