![David Luebke 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?](https://image.slidesharecdn.com/dynamicprogramming-170208045617/85/Dynamic-programming-in-Algorithm-Analysis-8-320.jpg)
![David Luebke 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?
−−
=+−−
=
otherwise]),1[],1,[max(
],[][if1]1,1[
],[
jicjic
jyixjic
jic](https://image.slidesharecdn.com/dynamicprogramming-170208045617/85/Dynamic-programming-in-Algorithm-Analysis-10-320.jpg)
Dynamic programming in Algorithm Analysis
- 1. David Luebke 1 CS 332: Algorithms Dynamic Programming
- 2. David Luebke 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 2m 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. David Luebke 3 Review: Analysis Of Dynamic Tables ● Let ci = cost of ith insert ● ci = 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. David Luebke 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 nnnnc n j j n i i 3)12(2 lg 01 <−+=+≤ ∑∑ ==
- 5. David Luebke 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. David Luebke 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. David Luebke 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. David Luebke 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. David Luebke 9 LCS Algorithm ● Brute-force algorithm: 2m subsequences of x to check against n elements of y: O(n 2m ) ● 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. David Luebke 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? −− =+−− = otherwise]),1[],1,[max( ],[][if1]1,1[ ],[ jicjic jyixjic jic
- 11. David Luebke 11 The End