Greedy Algorithms - Introduction to Algorithms - Lecture Slides, Slides of Computer Science

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Algorithms

Greedy Algorithms

Review: Dynamic Programming

● Dynamic programming is another strategy for

designing algorithms

● Use when problem breaks down into recurring

small subproblems

Review: Structure of Subproblems

● For the LCS problem:

■ There are few subproblems in total ■ And many recurring instances of each (unlike divide & conquer, where subproblems unique)

● How many distinct problems exist for the LCS

of x[1..m] and y[1..n]?

● A: mn

Memoization

● Memoization is another way to deal with

overlapping subproblems

■ After computing the solution to a subproblem, store in a table ■ Subsequent calls just do a table lookup

● Can modify recursive alg to use memoziation:

■ There are mn subproblems ■ How many times is each subproblem wanted? ■ What will be the running time for this algorithm? The running space?

Review: Dynamic Programming

● Summary of the basic idea:

■ Optimal substructure: optimal solution to problem consists of optimal solutions to subproblems ■ Overlapping subproblems: few subproblems in total, many recurring instances of each ■ Solve bottom-up, building a table of solved subproblems that are used to solve larger ones

● Variations:

■ “Table” could be 3-dimensional, triangular, a tree, etc.

Greedy Algorithms

● A greedy algorithm always makes the choice

that looks best at the moment

■ My everyday examples: ○ Walking to the Corner ○ Playing a bridge hand ■ The hope: a locally optimal choice will lead to a globally optimal solution ■ For some problems, it works

● Dynamic programming can be overkill; greedy

algorithms tend to be easier to code

Activity-Selection

● Formally:

■ Given a set S of n activities s (^) i = start time of activity i f (^) i = finish time of activity i ■ Find max-size subset A of compatible activities

 Assume (wlog) that f 1 ≤ f 2 ≤ … ≤ f (^) n

1

2

3 4 5

6

Activity Selection: Optimal Substructure

● Let k be the minimum activity in A (i.e., the

one with the earliest finish time). Then A - { k }

is an optimal solution to S’ = { i ∈ S : si ≥ f k }

■ In words: once activity #1 is selected, the problem reduces to finding an optimal solution for activity- selection over activities in S compatible with # ■ Proof: if we could find optimal solution B’ to S’ with | B | > | A - { k }|, ○ Then B U { k } is compatible ○ And | B U { k }| > |A|

Greedy Choice Property

● Dynamic programming? Memoize? Yes, but…

● Activity selection problem also exhibits the

greedy choice property:

■ Locally optimal choice ⇒ globally optimal sol’n ■ Them 17.1: if S is an activity selection problem sorted by finish time, then ∃ optimal solution A ⊆ S such that {1} ∈ A ○ Sketch of proof: if ∃ optimal solution B that does not contain {1}, can always replace first activity in B with {1} ( Why? ). Same number of activities, thus optimal.

Activity Selection: A Greedy Algorithm

● So actual algorithm is simple:

■ Sort the activities by finish time ■ Schedule the first activity ■ Then schedule the next activity in sorted list which starts after previous activity finishes ■ Repeat until no more activities

● Intuition is even more simple:

■ Always pick the shortest ride available at the time

Review: The Knapsack Problem

● The famous knapsack problem :

■ A thief breaks into a museum. Fabulous paintings, sculptures, and jewels are everywhere. The thief has a good eye for the value of these objects, and knows that each will fetch hundreds or thousands of dollars on the clandestine art collector’s market. But, the thief has only brought a single knapsack to the scene of the robbery, and can take away only what he can carry. What items should the thief take to maximize the haul?

Review: The Knapsack Problem

● More formally, the 0-1 knapsack problem :

■ The thief must choose among n items, where the i th item worth vi dollars and weighs wi pounds ■ Carrying at most W pounds, maximize value ○ Note: assume v (^) i, w (^) i, and W are all integers ○ “0-1” b/c each item must be taken or left in entirety

● A variation, the fractional knapsack problem :

■ Thief can take fractions of items ■ Think of items in 0-1 problem as gold ingots, in fractional problem as buckets of gold dust

Solving The Knapsack Problem

● The optimal solution to the fractional knapsack

problem can be found with a greedy algorithm

■ How?

● The optimal solution to the 0-1 problem cannot

be found with the same greedy strategy

■ Greedy strategy: take in order of dollars/pound ■ Example: 3 items weighing 10, 20, and 30 pounds, knapsack can hold 50 pounds ○ Suppose item 2 is worth $100. Assign values to the other items so that the greedy strategy will fail

The Knapsack Problem: Greedy Vs. Dynamic

● The fractional problem can be solved greedily

● The 0-1 problem cannot be solved with a

greedy approach

■ As you have seen, however, it can be solved with dynamic programming