1. CS 3343: Analysis of Algorithms
Lecture 18: More Examples on Dynamic Programming

2. Review of Dynamic Programming
We’ve learned how to use DP to solve
a special shortest path problem
the longest subsequence problem
a general sequence alignment
When should I use dynamic programming?
Theory is a little hard to apply
More examples would help

3. Two steps to dynamic programming
Formulate the solution as a recurrence relation of solutions to subproblems.
Specify an order to solve the subproblems so you always have what you need.

4. A special shortest path problemG n
Each edge has a length (cost). We need to get to G from S. Can only move right or down. Aim: find a path with the minimum total length

5. Recursive thinking Suppose we’ve found the shortest path
It must use one of the two edges:
(m, n-1) to (m, n) Case 1
(m-1, n) to (m, n) Case 2
If case 1
find shortest path from (0, 0) to (m, n-1)
SP(0, 0, m, n-1) + dist(m, n-1, m, n) is the overall shortest path
If case 2
find shortest path from (0, 0) to (m-1, n)
We don’t know which case is true
But if we’ve find the two paths, we can compare
Real shortest path is the one with shorter overall length m

6. Recursive formulation
Let F(i, j) = SP(0, 0, i, j). => F(m, n) is length of SP from (0, 0) to (m, n) n
F(m-1, n) + dist(m-1, n, m, n)
F(m, n) = min
F(m, n-1) + dist(m, n-1, m, n)
F(i-1, j) + dist(i-1, j, i, j)
F(i, j) = min
F(i, j-1) + dist(i, j-1, i, j)
Generalize m
Boundary condition: i = 0 or j = 0. Easy to figure out manually.
i = 1 .. m, j = 1 .. n
Number of subproblems = m * n determines structure of DP table
Data dependency determines order to compute
(i, j)

7. Longest Common Subsequence
Given two sequences x[1 . . m] and y[1 . . n], find a longest subsequence common to them both.
“a” not “the” x: A B C D y:
BCBA = LCS(x, y)
functional notation, but not a function

8. Recursive thinking m x n y
Case 1: x[m]=y[n]. There is an optimal LCS that matches x[m] with y[n].
Case 2: x[m] y[n]. At most one of them is in LCS
Case 2.1: x[m] not in LCS
Case 2.2: y[n] not in LCS
Find out LCS (x[1..m-1], y[1..n-1])
Find out LCS (x[1..m-1], y[1..n])
Find out LCS (x[1..m], y[1..n-1])

9. Recursive thinking Case 1: x[m]=y[n] Case 2: x[m]  y[n]
LCS(x, y) = LCS(x[1..m-1], y[1..n-1]) || x[m]
Case 2: x[m]  y[n]
LCS(x, y) = LCS(x[1..m-1], y[1..n]) or
LCS(x[1..m], y[1..n-1]), whichever is longer
Reduce both sequences by 1 char
concatenate
Reduce either sequence by 1 char

10. Recursive formulation
Let c[i, j] be the length of LCS(x[1..i], y[1..j])
=> c[m, n] is the length of LCS(x, y)
c[m–1, n–1] + 1 if x[m] = y[n],
max{c[m–1, n], c[m, n–1]} otherwise.
c[m, n] =
Generalize
c[i–1, j–1] + 1 if x[i] = y[j],
max{c[i–1, j], c[i, j–1]} otherwise.
c[i, j] =
i = 1 .. m
j = 1 .. n
Boundary condition: i = 0 or j = 0. Easy to figure out manually.
Number of subproblems = m * n
Order to compute?
(i, j)

11. Another DP example You work in the fast food business
Your company plans to open up new restaurants in Texas along I-35
Towns along the highway called t1, t2, …, tn
Restaurants at ti has estimated annual profit pi
No two restaurants can be located within 10 miles of each other due to some regulation
Your boss wants to maximize the total profit
You want a big bonus
10 mile

12. Brute-force Each town is either selected or not selected
Test each of the 2n subsets
Eliminate subsets that violate constraints
Compute total profit for each remaining subset
Choose the one with the highest profit
Θ(n 2n)

13. Natural greedy 1 Take first town. Then the next town >= 10 miles
Can you give an example that this algorithm doesn’t return the correct solution?
100k
100k
500k

14. Natural greedy 2
Almost take a town with the highest profit and are not <10 miles of another selected town
Can you give an example that this algorithm doesn’t return the correct solution?
300k
300k
500k

15. A DP algorithm Suppose you’ve already found the optimal solution
It will either include tn or not include tn
Case 1: tn not included in optimal solution
Best solution same as best solution for t1 , …, tn-1
Case 2: tn included in optimal solution
Best solution is pn + best solution for t1 , …, tj , where j < n is the largest index so that dist(tj, tn) ≥ 10

16. Recurrence formulation
Let S(i) be the total profit of the optimal solution when the first i towns are considered (not necessarily selected)
S(n) is the optimal solution to the complete problem
S(n-1)
S(j) + pn j < n & dist (tj, tn) ≥ 10
S(n) = max
S(i-1)
S(j) + pi j < i & dist (tj, ti) ≥ 10
S(i) = max
Generalize
Number of sub-problems: n. Boundary condition: S(0) = 0.
Dependency: i
i-1 j S

17. Example S(i-1) S(j) + pi j < i & dist (tj, ti) ≥ 10 S(i) = max
Distance (mi)
100 5 2 2 6 6 3 6 10 7
dummy 7 3 4 12
Profit (100k) 6 7 9 8 3 3 2 4 12 5
S(i) 6 7 9 9 10 12 12 14 26 26
Optimal: 26
S(i-1)
S(j) + pi j < i & dist (tj, ti) ≥ 10
S(i) = max
Natural greedy 1: = 25
Natural greedy 2: = 24

18. Complexity
Time: (nk), where k is the maximum number of towns that are within 10 miles to the left of any town
In the worst case, (n2)
Can be improved to (n) with some preprocessing tricks
Memory: Θ(n)

19. Knapsack problem Each item has a value and a weight
Objective: maximize value
Constraint: knapsack has a weight limitation
Three versions:
0-1 knapsack problem: take each item or leave it
Fractional knapsack problem: items are divisible
Unbounded knapsack problem: unlimited supplies of each item.
Which one is easiest to solve?
We study the 0-1 problem today.

20. Formal definition (0-1 problem)
Knapsack has weight limit W
Items labeled 1, 2, …, n (arbitrarily)
Items have weights w1, w2, …, wn
Assume all weights are integers
For practical reason, only consider wi < W
Items have values v1, v2, …, vn
Objective: find a subset of items, S, such that iS wi  W and iS vi is maximal among all such (feasible) subsets

21. Naïve algorithms Enumerate all subsets.
Optimal. But exponential time
Greedy 1: take the item with the largest value
Not optimal
Give an example
Greedy 2: take the item with the largest value/weight ratio

22. A DP algorithm Suppose you’ve find the optimal solution S
Case 1: item n is included
Case 2: item n is not included
Total weight limit: W
Total weight limit: W wn wn
Find an optimal solution using items 1, 2, …, n-1 with weight limit W - wn
Find an optimal solution using items 1, 2, …, n-1 with weight limit W

23. Recursive formulation
Let V[i, w] be the optimal total value when items 1, 2, …, i are considered for a knapsack with weight limit w
=> V[n, W] is the optimal solution
V[n, W] = max
V[n-1, W-wn] + vn
V[n-1, W]
Generalize
V[i, w] = max
V[i-1, w-wi] + vi item i is taken
V[i-1, w] item i not taken
V[i-1, w] if wi > w item i not taken
Boundary condition: V[i, 0] = 0, V[0, w] = 0. Number of sub-problems = ?

24. Example n = 6 (# of items) W = 10 (weight limit)
Items (weight, value):

25. w 1 2 3 4 5 6 7 8 9 10 i wi vi 1 2 2 2 4 3 wi 3 3 3
V[i-1, w-wi]
V[i-1, w] 4 5 5 6 6
V[i, w] 5 2 4 6 6 9
V[i-1, w-wi] + vi item i is taken
V[i-1, w] item i not taken
max
V[i, w] =
V[i-1, w] if wi > w item i not taken

26. w 1 2 3 4 5 6 7 8 9 10 i wi vi 1 2 4 3 5 6 9 2 2 2 2 2 2 2 2 2 3 5 2 2 3 5 5 5 5 6 8 3 5 2 3 5 6 8 6 8 9 11 2 3 3 6 9 4 6 7 10 12 13 4 7 10 13 15 9 4 4 6 7 10 13
V[i-1, w-wi] + vi item i is taken
V[i-1, w] item i not taken
max
V[i-1, w] if wi > w item i not taken
V[i, w] =

27. w 1 2 3 4 5 6 7 8 9 10 i wi vi 1 2 4 3 5 6 9 2 2 2 2 2 2 2 2 2 3 5 2 2 3 5 5 5 5 6 8 3 5 2 3 5 6 8 6 8 9 11 2 3 3 6 9 4 7 10 12 13 4 6 7 10 13 9 4 4 6 7 10 13 15
Optimal value: 15
Item: 6, 5, 1
Weight: = 10
Value: = 15

28. Time complexity Θ (nW) Polynomial?
Pseudo-polynomial
Works well if W is small
Consider following items (weight, value):
(10, 5), (15, 6), (20, 5), (18, 6)
Weight limit 35
Optimal solution: item 2, 4 (value = 12). Iterate: 2^4 = 16 subsets
Dynamic programming: fill up a 4 x 35 = 140 table entries
What’s the problem?
Many entries are unused: no such weight combination
Top-down may be better

29. A few more examples

30. Longest increasing subsequence
Given a sequence of numbers
Find a longest subsequence that is non-decreasing
E.g
It has to be a subsequence of the original list
It has to in sorted order
=> It is a subsequence of the sorted list
Original list:
LCS:
Sorted:

31. Events scheduling problem
Time
A list of events to schedule (or shows to see)
ei has start time si and finishing time fi
Indexed such that fi < fj if i < j
Each event has a value vi
Schedule to make the largest value
You can attend only one event at any time
Very similar to the new restaurant location problem
Sort events according to their finish time
Consider: if the last event is included or not

32. Events scheduling problemf9 s8 f8 s7 f7 e8 e3 e4 e5 e7 e9 e1 e2
Time
V(i) is the optimal value that can be achieved when the first i events are considered
V(n) =
V(n-1)
en not selected
max {
V(j) + vn
en selected
j < n and fj < sn

33. Coin change problem
Given some denomination of coins (e.g., 2, 5, 7, 10), decide if it is possible to make change for a value (e.g, 13), or minimize the number of coins
Version 1: Unlimited number of coins for each denomination
Unbounded knapsack problem
Version 2: Use each denomination at most once
0-1 Knapsack problem

34. Use DP algorithm to solve new problems
Directly map a new problem to a known problem
Modify an algorithm for a similar task
Design your own
Think about the problem recursively
Optimal solution to a larger problem can be computed from the optimal solution of one or more subproblems
These sub-problems can be solved in certain manageable order
Works nicely for naturally ordered data such as strings, trees, some special graphs
Trickier for general graphs
The text book has some very good exercises.