Dynamic Programming

Dynamic Programming Day 3: Examples

Longest Common Subsequence • Given two sequences of characters A, B • Find the longest sequence C such that C is a subsequence of A as well as of B • Example: • ABCDGH • ZBAEDH • Length of LCS = 3

Longest Common Subsequence Iterative DP 0 0 1 1 1 1

Longest Common SubsequenceIterative DP

Longest Common Subsequence Length (UVA 10405) – Iterative DP int a[1010][1010]; int main() { ... memset(a, 0, sizeof(a)); for(int i = 0; i < str1.size(); i++) { for(int j = 0; j < str2.size(); j++) { if (str1[i] == str2[j]) a[i+1][j+1] >?= a[i][j] + 1; a[i+1][j+1] >?= a[i+1][j]; a[i+1][j+1] >?= a[i][j+1]; } } cout << a[str1.size()][str2.size()] << endl; }

Longest Common Subsequence Length (UVA 10405) - Memoization int lcs(int a, int b) { if (cache[a][b] >= 0) return cache[a][b]; if (a == str1.size() || b == str2.size()) return 0; int ret = 0; if (str1[a] == str2[b]) ret = 1 + lcs(a+1, b+1); return cache[a][b] = max(ret, lcs(a, b + 1), lcs(a + 1, b)); }

Longest Common SubsequencePath Extraction (UVA 531) • It is wonderful that we know how to compute the length of the longest common subsequence, BUT: • How do we find out what the subsequence really is? • In a more general case, we often need to find out not only the “quality” of the optimal solution, but also what it is • How to do this? Let’s look at our DP table again.

Longest Common SubsequencePath Extraction (UVA 531)

Longest Common SubsequencePath Extraction (UVA 531) • We can trace our way back through the DP table: vector<string> path; int i = str1.size() - 1, j = str2.size() - 1; while (i >= 0 && j >= 0) { if (a[i+1][j+1] == a[i][j+1]) i--; else if (a[i+1][j+1] == a[i+1][j]) j--; else { path.push_back(str1[i]); i--; j--; } } reverse(path.begin(), path.end());

Critical Mass(UVA 580) • 3 types of boxes: lead and uranium • Stacking boxes on top of each other • Stack is dangerous if 3 or more uranium boxes on top of each other • How many dangerous stacks of height H are there? • For example, 3 dangerous stacks if H = 4: • LUUU • UUUL • UUUU

Step 1: Identify the sub-problem • Modify the definition of a dangerous stack • Stack is dangerous if: • It contains 3 adjacent uranium boxes OR • It contains K adjacent uranium boxes at the bottom (K <= 3)

Step 2: Form a Recursive Solution int doit(int height, int bottom) { if (bottom == 0) return 1 << height; if (height == 0) return 0; return doit(height - 1, 3) + doit(height - 1, bottom - 1); }

Step 3: Recursion -> DP int doit(int height, int bottom) { if (bottom == 0) return 1 << height; if (height == 0) return 0; if (cache[height][bottom] >= 0) return cache[height][bottom]; return cache[height][bottom] = doit(height - 1, 3) + doit(height - 1, bottom - 1); }

Headmaster’s Headache(UVA 10817)

Headmaster’s Headache(UVA 10817) • S subjects, 1 <= S <= 8 • M teachers you have, each teaches some of the subjects, 1 <= M <= 20 • N applicants from whom you can hire, each teaches some of the subjects 1 <= N <= 100 • Each teacher and each applicant has a salary they require • What is the minimum budget the headmaster needs so that each subject is taught by at least 2 teachers?

Headmaster’s Headache(UVA 10817)

Questions to Think About • What are different possible representation of the sub-problem? What are the advantages and disadvantages of each? • Would you prefer to solve this problem using memoization or iterative DP? What are the advantages and disadvantages of each?

Mailbox Manufacturer’s Problem (UVA 882) • You have 1 <= k <= 10 mailboxes • At most 100 crackers fit into a mailbox • If a mailbox can withstand x fire-crackers, it can also withstand x - 1 fire-crackers • After explosion, a mailbox is either totally destroyed or unharmed • How many fire-crackers do you need to guarantee you will find out the smallest number of crackers that blow up a mailbox?

Dynamic Programming