More On Dynamic programming Algorithms Shortest path with edge constraint:

1. More On Dynamic programming Algorithms Shortest path with edge constraint: Let G=(V, E) be a directed graph with weighted edges. Let s and v be two vertices in V. Find a shortest path from s to u with exactly k edges. Here kn-1 is part of the input. Solution: Define d(i, v) be the length of the shortest path from s to v with exactly i edges. d(i, v)=min {c(w, v)+d(i-1, w)} wV. Initial values: d(i, s)=0, for i=0, d(i,s)=  for i=1, 2, …, d(0, v)=; d(k, v) will give the length of the shortest path. A backtracking process can give the path.

2. u 5 v 8 8 -2 6 -3 8 7 z 0 -4 2 7 8 8 9 x y • i z, u, x, v, y • 0 0     •  6z 7z   •   14u 4x 2u • 3 4y 2v  9y 23x.

3. Shortest path with delay constraint: Let G=(V, E) be a directed path. Each edge (u, v) has a positive weight edges c(u, v) and a (positive integer) delay delay(u, v). Let s and v be two vertices in V. Find a shortest path from s to u with delay at most k. Solution: Define l(d, v) be the length of the shortest path from s to v with delay = d. l(d, v)= min {c(w, v)+l(d-delay(w, v), w), }. wV & delay(w,v)d Initial values: l(0, s)=0, l(i, s)=  and d(0, v)=; l(k, u) will give the length of the shortest path. A backtracking process can give the path. The previous example is for delay(u, v)=1.

4. u 5 v 8 8 -2 6 -3 8 7 z 0 -4 2 7 8 8 9 x y delay(v,u)=2. For any other edge, it is 1. • d z, u, x, v, y • 0 0     • 1  6z 7z   •   14u 4x 2u • 3 4y   9y 23x.

6. Weighted interval scheduling for a lazy man Input: the same as weighted interval scheduling. Goal: find a set of compatible jobs such that for any two consecutive jobs in the subset either there is some idle time or the total length of the two consecutive jobs is at most k. If k=4, we can choose black, blue, yellow as a subset. But we cannot choose black, blue and green as the subset.

7. Solution: Still the same equation, but use different p(). OPT(j) be the weight of the optimal solution containing at most the first j jobs. OPT(j)=min {OPT(j-1), vj+ OPT(p(j))} Initial Value: OPT(0)=0. P(j) is redefined as: the largest i<j such that job i is compatible with j and either there is some idle time between i and j or fj-si k.

8. K=4 p(b)=0; p( c )=0; p(a)=0; p(e)=0; p(d)=0; p(f)=b; p(g)=c; p(h)=e;

9. k=6 p(b)=0; p( c )=0; p(a)=0; p(e)=b; p(d)=0; p(f)=c; p(g)=c; p(h)=e;

10. Weighted maximum independent set on a path. Given a graph G(V, E), where V={v1, v2, …, vn} and E={(vi, vi+1)|for i=1, 2, …, n-1}. Each vertex vi has a weight wi. Find an independent set with maximum weight. • An independent set in a graph is a subset of vertices in V such that no two vertices in the independent set share an edge. Solution: Define d(i) be the weight of the optimal solution for the first i vertices. d(i)=max {d(i-1), wi+d(i-2)}. d(0)=0; Remarks: The problem is almost identical to the weighted interval scheduling problem if we treat p(i)=i-2.

11. Weighted maximum set on a path. Given a graph G(V, E), where V={v1, v2, …, vn} and E={(vi, vi+1)|for i=1, 2, …, n-1}. Each vertex vi has a weight wi. d(vi, vi+1) is the length of the edge. d(vi, vi+2)=d(vi, vi+1)+d(vi+1, vi+2). Find a subset of vertices with maximum weight such that the distance between any pair of vertices in the subset is at least k. Solution: Define d(i) be the weight of the optimal solution for the first i vertices. d(i)= max {d(i-1), wi+d(p(i))}. d(0)=0; Remarks: p(i) is the biggest j<i such that d(vi,vj)≥k.

12. Another setting of the previous problem. Text book page 307. Consider a highway from west to east. The possible sites for billboards are given by numbers x1, x2, …, xn, each in the interval [0, M]. If you place a billboard at location xi, you receive revenue of ri>0. Regulations imposed by the county’s Highway Department require that no two of the billboards be within less than or equal to 5 miles of each other. How to place billboards at a subset of sites so as to maximize your total revenue, subject to this restriction. . Solution: Define d(i) be the revenue of the optimal solution for the first i sites. d(i)=max { d(i-1), ri+d(p(i))}. d(0)=0; Remarks: p(i) is the biggest j<i such that xi-xj >5. x1 x2 x3 xn

13. Another setting of the previous problem. X1=0, x2=3, x3=6, x4=8, x5=15, x6=25. W1=2, w2=5, w3=8, w4=6, w5=9, w6=12. i 0 1 2 3 4 5 6 p(i): 0 0 0 1 1 4 5 d(i): 0 2, 5 10 10 19 31 d(i)=max { d(i-1), wi+d(p(i))}. d(0)=0; Remarks: p(i) is the biggest j<i such that xi-xj >5. x1 x2 x3 xn

14. Change-making problem: Given an amount n and unlimited quantities of coins of each of the denominations d1, d2, …, dm, find the smallest number of coins that add up to n or indicate that the problem does not have a solution. Solution: Let d(i) be the minimum number of coins used for amount i. Initial values: d(0)=0, d(i)=. equation: d(i) = min d(i-dk)+1. k=1, 2, …, m.

15. Change-making problem: Initial values: d(0)=0, d(i)=. equation: d(i) = min d(i-dk)+1 k=1, 2, …, m & i-dk≥0 d1=2 and d2=5. i=7. i= 0, 1, 2, 3, 4, 5, 6, 7 d(i): 0, , 1, , 2, 1, 3, 2. Backtracking: $5, $2. (how do you get d(7)=2? d(7)=d(2)+5. Print $5 and goto d(2). How do you get d(2)? D(2)=2+d(0). Print out $2 and goto d(0). Whenever reach d(0), we stop. ) for i=3, since d(3)= , there is no solution.

16. LCS for three sequences. Give three sequences X, Y, X. Find the longest common subsequence for three sequences. First try. 1. construct W, the LCS for X and Y 2. Construct U, the LCS for W and Z. 3. U will be the LCS for the tree sequences X, Y and Z. It does not work! Counter Example: X=bbaaa, Y=aaaabb and Z=cbcbc. LCS for X and Y is aaa. LCS for aaa and cbcbc is empty. However, the LCS for X, Y and Z is bb.

17. LCS for three sequences. Give three sequences X, Y, X. Find the longest common subsequence for three sequences. Solution: Define c[i, j, k] be the length of LCS for the first I letters in X, the first j letters in Y and the first k letters in Z. c[i,j,k] is the maximum of the following: c[i-1, j-1, k-1]+1, if Xi=Yj=Zk. c[i-1, j-1, k]; c[i-1, j, k-1] c[i, j-1, k-1] c[i, j, k-1] c[i,j-1, k] c[i-1, j,k]. Initial values: c[i,jk]=0 if i=0 or j=0 or k=0.

18. Exercise 1: Let T be a rooted binary tree, where each internal node in the tree has two children and every node (except the root) in T has a parent. Each leaf in the tree is assigned a letter in ={A, C, G, T}. Figure 1 gives an example. Consider an edge e in T. Assume that every end of e is assigned a letter. The cost of e is 0 if the two letters are identical and the cost is 1 if the two letters are not identical. The problem here is to assign a letter in  to each internal node of T such that the cost of the tree is minimized, where the cost of the tree is the total cost of all edges in the tree. Design a polynomial-time dynamic programming algorithm to solve the problem.

19. A A C Figure 1 A

20. Exercise 2. Give a polynomial time algorithm to find the longest monotonically increasing subsequence of a sequence of n numbers. Can you give a linear space algorithm? (Assume that each integer appears once in the input sequence of n numbers) Example: Consider sequence 1,8, 2,9, 3,10, 4, 5. Both subsequences 1, 2, 3, 4, 5 and 1, 8, 9, 10 are monotonically increasing subsequences. However, 1,2,3, 4, 5 is the longest.

21. Assignment 4. (Due on Friday of Week 14. Drop it in Mail Box 76 ) This time, I can explain the questions, but we will NOT tell you how to solve the problems. Question 1. (20 points) Change-making problem (version 1): Given an amount n and coins of each of the denominations d1, d2, …, dm, where there are k copies of d1 coin and unlimited quantities of other coins, design a dynamic programming algorithm to find the smallest number of coins that add up to n or indicate that the problem does not have a solution. Question 2.(20 points) Change-making problem (version 2): Given an amount n and coins of the denominations d1, d2, …, dm, where there are 2 copies of each di coin, design a dynamic programming algorithm to find the smallest number of coins that add up to n or indicate that the problem does not have a solution. Question 3. (60 points) will be give in week 12.