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 problem S G m 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) –SP(0, 0, m, n-1) + dist(m, n-1, m, n) is the overall shortest path 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 n

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) 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 Data dependency determines order to compute (i, j) m n 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

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

8. Recursive thinking 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 x y m n Find out LCS (x[1..m-1], y[1..n-1]) Find out LCS (x[1..m], y[1..n-1]) Find out LCS (x[1..m-1], y[1..n])

9. Recursive thinking Case 1: 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 x y m n Reduce both sequences by 1 char Reduce either sequence by 1 char concatenate

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

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 t 1, t 2, …, t n Restaurants at t i has estimated annual profit p i 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 2 n subsets Eliminate subsets that violate constraints Compute total profit for each remaining subset Choose the one with the highest profit Θ(n 2 n )

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 500k

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

15. A DP algorithm Suppose you’ve already found the optimal solution It will either include t n or not include t n Case 1: t n not included in optimal solution –Best solution same as best solution for t 1, …, t n-1 Case 2: t n included in optimal solution –Best solution is p n + best solution for t 1, …, t j, where j < n is the largest index so that dist(t j, t n ) ≥ 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) + p n j < n & dist (t j, t n ) ≥ 10 S(n) = max S(i-1) S(j) + p i j < i & dist (t j, t i ) ≥ 10 S(i) = max Generalize Number of sub-problems: n. Boundary condition: S(0) = 0. Dependency: i i-1 j S

17. Example Natural greedy 1: 6 + 3 + 4 + 12 = 25 Natural greedy 2: 12 + 9 + 3 = 24 5226663107 6798 3 324125 Distance (mi) Profit (100k) 6799 10 12 1426 S(i) S(i-1) S(j) + p i j < i & dist (t j, t i ) ≥ 10 S(i) = max 100 0 73412 dummy Optimal: 26

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,  (n 2 ) –Can be improved to  (n) with some preprocessing tricks Memory: Θ(n)

19. Knapsack problem 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? Each item has a value and a weight Objective: maximize value Constraint: knapsack has a weight limitation 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 w 1, w 2, …, w n –Assume all weights are integers –For practical reason, only consider w i < W Items have values v 1, v 2, …, v n Objective: find a subset of items, S, such that  i  S w i  W and  i  S v i 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 –Not optimal –Give an example

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 wnwn Total weight limit: W Find an optimal solution using items 1, 2, …, n-1 with weight limit W - w n wnwn 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-w n ] + v n V[n-1, W] Generalize V[i, w] = max V[i-1, w-w i ] + v i item i is taken V[i-1, w] item i not taken V[i-1, w] if w i > 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): 2 4 3 3 5 6 2 4 6 9

25. 0 0 0 0 0 0 00000000000 w012345678910 425 6 4 3 2 1 i 96 65 33 34 22 vivi wiwi max V[i-1, w-w i ] + v i item i is taken V[i-1, w] item i not taken V[i-1, w] if w i > w item i not taken V[i, w] = V[i, w] V[i-1, w]V[i-1, w-w i ] 6 wiwi 5

26. 107400 1310764400 9633200 8653200 555532200 2222222200 00000000000 w0123456789 iwiwi vivi 122 243 333 456 524 669 max V[i-1, w-w i ] + v i item i is taken V[i-1, w] item i not taken V[i-1, w] if w i > w item i not taken V[i, w] = 2 4 3 6 5 6 7 5 9 6 8 10 911 8 3 1213 15

27. 107400 1310764400 9633200 8653200 555532200 2222222200 00000000000 w0123456789 iwiwi vivi 122 243 333 456 524 669 2 4 3 6 5 6 7 5 9 6 8 911 8 3 1213 15 Item: 6, 5, 1 Weight: 6 + 2 + 2 = 10 Value: 9 + 4 + 2 = 15 Optimal 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 1 2 5 3 2 9 4 9 3 5 6 8 Find a longest subsequence that is non- decreasing –E.g. 1 2 5 9 –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: 1 2 5 3 2 9 4 9 3 5 6 8 LCS: Sorted: 1 2 2 3 3 4 5 5 6 8 9 9 1 2 3 4 5 6 8

31. Events scheduling problem A list of events to schedule (or shows to see) –e i has start time s i and finishing time f i –Indexed such that f i < f j if i < j Each event has a value v i 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 Time e1e1 e2e2 e3e3 e4e4 e5e5 e6e6 e7e7 e8e8 e9e9

32. Events scheduling problem Time e1e1 e2e2 e3e3 e4e4 e5e5 e6e6 e7e7 e8e8 e9e9 V(i) is the optimal value that can be achieved when the first i events are considered V(n) = V(n-1) e n not selected e n selected V(j) + v n max { j < n and f j < s n s9s9 f9f9 s8s8 f8f8 s7s7 f7f7

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.