1. Lecture 8: Dynamic Programming Shang-Hua Teng

2. First Example: n choose k Many combinatorial problems require the calculation of the binomial coefficient. This is equivalent to given n objects how many different ways can you choose k of them. Mathematically we have or,

3. int choose(n,k) if (n = = k) return 1; if (n < k) or (k < 0) return 0; return choose(n-1, k-1) + choose(n - k, k); Consider Divide and Conquer How long does this take? T(n,k) = T(n-1, k-1) + T(n-k, k)

4. Remember Pascal’s Triangle function choose (n,k) 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 By adding two numbers from previous row we can calculate the triangle quickly.

5. Consider triangle shape 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 012345012345 0 1 2 3 4 5 Looks like a two dimensional array. Create the triangle in this array and output A(i,j) for

6. Do it algorithmically Create an 2-d array A of size (n+1) by (k +1 ). Set A[i, 0]  1 for i = 0,1,2,.., n Set A[i, i ]  1 for i = 0,1,2,...., n for i = 1 to n for j = 1 to k A[ i, j ] = A[ i-1, j ] + A[ i –1, j-1 ]; Output A[n,k] Runtime is O(nk)

7. Matrix Chain Multiplication - Review Recall how to multiply matrices. Given two matrices The product is Time is A B C X = d e e f d f

8. Matrix Chaining contd. What about multiplying multiple matrices? We know matrix multiplication is associative. Can we use this to help minimize calculations ? Sure, consider If we multiply we do 4000 operations But instead if we trywe do 1575 operations

9. How do we parenthesize Brute force - try all ways to parenthesize Find out which has smallest number of operations by doing multiplication, then pick the best one. But how many ways can we parenthesize these? Equivalent to the number of ways to make a binary tree with n nodes. This is see Catalan numbers. See homework Problem 12-4, ex15.2-3 in the book.

10. Be greedy We just learned that a greedy algorithm can sometimes work, let’s try. Why not try starting with the product with the most operations. Consider If we do our greedy method we would compute This is 2000 operations. Try instead, which takes 1000 operations.

11. Another Greedy way Select the product with the fewest operations. Consider Here the new greedy method would compute This is 109989+9900+108900=228789 operations Try Here we only need 9999+89991+89100=189090

12. Optimality of Subproblems Consider general question again, we want to parenthesize Now suppose somehow we new that the last multiplication we need to do was at the i th position. Then the problem is the same as knowing the best way to parenthesize and

13. Overlapping Subproblems We know how to break problem into smaller pieces, why doesn’t divide and conquer work? We need to optimize each piece why doesn’t greedy work. Again suppose we know where the final multiply is and we want to generalize the cost with a recurrence, consider But notice these subproblems overlap.

14. Dynamic Programming Since subproblems overlap we can not use divide and conquer method. Instead work from the “bottom up.” We know how to parenthesize one matrix, two matrices, we can build from there…

15. Matrix Chain Order

16. What have we noticed Optimal solution depends on optimal subproblems Subproblems overlap. So a “top down” divide and conquer doesn’t seem to work. Somehow we need to build up from bottom by remembering solution from subproblem.

17. Another Example Biologists need to measure how similar strands of DNA are to determine how closely related an organism is to another. They do this by considering DNA as strings of letters A,C,G,T and then comparing similarities in the strings. Formally they look at common subsequences in the strings. Example X = AGTCAACGTT, Y=GTTCGACTGTG Both S = AGTG and S’=GTCACGT are subsequences How to do find these efficiently?

18. Brute Force if |X| = m, |Y| = n, then there are 2 m subsequences of x; we must compare each with Y (n comparisons) So the running time of the brute-force algorithm is O(n 2 m ) Notice that the LCS problem has optimal substructure: solutions of subproblems are parts of the final solution. Subproblems: “find LCS of pairs of prefixes of X and Y”

19. Some observations

20. Setup First we’ll find the length of LCS, along the way we will leave “clues” on finding subsequence. Define X i, Y j to be the prefixes of X and Y of length i and j respectively. Define c[i,j] to be the length of LCS of X i and Y j Then the length of LCS of X and Y will be c[m,n]

21. The recurrence The subproblems overlap, to find LCS we need to find LCS of c[i, j-1] and of c[i-1, j]

22. LCS-Length(X, Y) m = length(X), n = length(Y) for i = 1 to m do c[i, 0] = 0 for j = 0 to n do c[0, j] = 0 for i = 1 to m do for j = 1 to n do if ( x i = = y j ) then c[i, j] = c[i - 1, j - 1] + 1 else if c[i - 1, j]>=c[i, j - 1] then c[i, j] = c[i - 1, j] else c[i, j] = c[i, j - 1] return c and b