1. Binary Search From solving a problem to verifying an answer

2. Problems using Binary Search Finding a minimum or a maximum value. Change problem from “What is the optimal value?” to “Is optimal less than or equal to X?” Auxiliary function to check if a value to the answer satisfies the problem’s constraints. Answer to “Does value X work as a (not necessarily optimal) answer?” must be distributed as below: (all true followed by all false or vice versa) True True True True True True True True False False False False False False False True True True True True

3. Code f(x) returns true if x is valid answer, false if it is not Find minimum answer Find maximum answer lo will be optimal answer. If f(lo) = false then there is no solution (f is false for all x). while(lo < hi) { mid = lo + (hi-lo)/2; // reduces to (lo+hi)/2 if lo > 0 if( f(x) ) hi = mid; else lo = mid + 1; } while(lo < hi) { mid = lo + (hi-lo+1)/2; // reduces to (lo+hi+1)/2 if lo > 0 if( f(x) ) lo = mid; else hi = mid - 1; }

4. Example 1: Shortest Path Given a directed graph with source A and destination B, find a path that minimizes the maximum edge. Binary Search on the maximum edge weight. Let X be the maximum edge’s value. Delete all edges with weight > X. X works if there is a path from A to B in this graph. Can also be solved using a variant of Dijkstra or a minimum spanning tree.

5. Example 2: Crates You have N books, each with width W_i. You can use at most M crates, and each must have the same width. What’s the smallest width each crate can have if you want to fit in all N books in order? Binary Search on the crate width. Let X be the crate width, and fill books into each crate in the order they are listed. If it requires more than M crates, X is invalid. Otherwise, it is valid.

6. Example 3: More Paths What’s the shortest path between A and B if you are allowed to delete any K edges from the graph, where the path length is the maximum edge weight along the path? Binary Search on the maximum edge weight. Let X be the maximum edge’s value. Set the distance of all edges with weight > X to 1 and all other edge weights to 0. X works if the distance between A and B is X).

7. Problem of the week Farmer John is ranking his N cows. He has decided to rank the cows two at a time, and he may rank cows more than once. He has compiled a list of M decisions A > B, meaning cow A is better than cow B. Find the earliest point in his list where there is a contradiction (such that no sequence of cows satisfies the constraints). 2<=N, M<=10: 10 pts 2<=N, M<=1,000: 15 pts 2<=N, M<=100,000: 25 pts

8. Problem of the week Input format Line 1: Two integers, N and M Line 2…M+1: A B, meaning cow A > cow B Output: Index of earliest contradiction, or -1 if none exist. Sample input 4 5 1 2 2 4 3 4 4 1 1 3 Sample output 4