![Dynamic Programming Example: Longest Common Subsequence Longest common subsequence ( LCS ) problem: Given two sequences x[1..m] and y[1..n], find the longest subsequence which occurs in both Ex: x = {A B C B D A B }, y = {B D C A B A} {B C} and {A A} are both subsequences of both What is the LCS? Brute-force algorithm: For every subsequence of x, check if it’s a subsequence of y How many subsequences of x are there? What will be the running time of the brute-force alg?](https://image.slidesharecdn.com/23-1224487945718578-8/85/lecture-23-8-320.jpg)
![Finding LCS Length Define c[ i,j ] to be the length of the LCS of x[1.. i ] and y[1.. j ] What is the length of LCS of x and y? Theorem: What is this really saying?](https://image.slidesharecdn.com/23-1224487945718578-8/85/lecture-23-10-320.jpg)
lecture 23
- 1. CS 332: Algorithms Dynamic Programming
- 2. Review: Amortized Analysis To illustrate amortized analysis we examined dynamic tables 1. Init table size m = 1 2. Insert elements until number n > m 3. Generate new table of size 2 m 4. Reinsert old elements into new table 5. (back to step 2) What is the worst-case cost of an insert? What is the amortized cost of an insert?
- 3. Review: Analysis Of Dynamic Tables Let c i = cost of i th insert c i = i if i-1 is exact power of 2, 1 otherwise Example: Operation Table Size Cost Insert(1) 1 1 1 Insert(2) 2 1 + 1 2 Insert(3) 4 1 + 2 Insert(4) 4 1 Insert(5) 8 1 + 4 Insert(6) 8 1 Insert(7) 8 1 Insert(8) 8 1 Insert(9) 16 1 + 8 1 2 3 4 5 6 7 8 9
- 4. Review: Aggregate Analysis n Insert() operations cost Average cost of operation = (total cost)/(# operations) < 3 Asymptotically, then, a dynamic table costs the same as a fixed-size table Both O(1) per Insert operation
- 5. Review: Accounting Analysis Charge each operation $3 amortized cost Use $1 to perform immediate Insert() Store $2 When table doubles $1 reinserts old item, $1 reinserts another old item We’ve paid these costs up front with the last n /2 Insert()s Upshot: O(1) amortized cost per operation
- 6. Review: Accounting Analysis Suppose must support insert & delete, table should contract as well as expand Table overflows double it (as before) Table < 1/4 full halve it Charge $3 for Insert (as before) Charge $2 for Delete Store extra $1 in emptied slot Use later to pay to copy remaining items to new table when shrinking table What if we halve size when table < 1/8 full?
- 7. Dynamic Programming Another strategy for designing algorithms is dynamic programming A metatechnique, not an algorithm (like divide & conquer) The word “programming” is historical and predates computer programming Use when problem breaks down into recurring small subproblems
- 8. Dynamic Programming Example: Longest Common Subsequence Longest common subsequence ( LCS ) problem: Given two sequences x[1..m] and y[1..n], find the longest subsequence which occurs in both Ex: x = {A B C B D A B }, y = {B D C A B A} {B C} and {A A} are both subsequences of both What is the LCS? Brute-force algorithm: For every subsequence of x, check if it’s a subsequence of y How many subsequences of x are there? What will be the running time of the brute-force alg?
- 9. LCS Algorithm Brute-force algorithm: 2 m subsequences of x to check against n elements of y: O( n 2 m ) We can do better: for now, let’s only worry about the problem of finding the length of LCS When finished we will see how to backtrack from this solution back to the actual LCS Notice LCS problem has optimal substructure Subproblems: LCS of pairs of prefixes of x and y
- 10. Finding LCS Length Define c[ i,j ] to be the length of the LCS of x[1.. i ] and y[1.. j ] What is the length of LCS of x and y? Theorem: What is this really saying?
- 11. The End