1 / 38

380 likes | 557 Views

CS 3343: Analysis of Algorithms. Lecture 19: Introduction to Greedy Algorithms. Outline. Review of DP Greedy algorithms Similar to DP, not an actual algorithm, but a meta algorithm. Two steps to dynamic programming.

Download Presentation
## CS 3343: Analysis of Algorithms

**An Image/Link below is provided (as is) to download presentation**
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.
Content is provided to you AS IS for your information and personal use only.
Download presentation by click this link.
While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

**CS 3343: Analysis of Algorithms**Lecture 19: Introduction to Greedy Algorithms**Outline**• Review of DP • Greedy algorithms • Similar to DP, not an actual algorithm, but a meta algorithm**Two steps to dynamic programming**• Formulate the solution as a recurrence relation of solutions to subproblems. • Specify an order of evaluation for the recurrence so you always have what you need.**Restaurant location problem**• You work in the fast food business • Your company plans to open up new restaurants in Texas along I-35 • Many towns along the highway, call them t1, t2, …, tn • Restaurants at ti has estimated annual profit pi • No two restaurants can be located within 10 miles of each other due to regulation • Your boss wants to maximize the total profit • You want a big bonus 10 mile**A DP algorithm**• Suppose you’ve already found the optimal solution • It will either include tn or not include tn • Case 1: tn not included in optimal solution • Best solution same as best solution for t1 , …, tn-1 • Case 2: tn included in optimal solution • Best solution is pn + best solution for t1 , …, tj , where j < n is the largest index so that dist(tj, tn) ≥ 10**S(n-1)**S(j) + pn j < n & dist (tj, tn) ≥ 10 S(n) = max Generalize S(i-1) S(j) + pi j < i & dist (tj, ti) ≥ 10 S(i) = max Dependency: S j i-1 i Recurrence formulation • Let S(i) be the total profit of the optimal solution when the first i towns are considered(not necessarily selected) • S(n) is the optimal solution to the complete problem Number of sub-problems: n. Boundary condition: S(0) = 0.**Example**Distance (mi) 100 5 2 2 6 6 3 6 10 7 dummy 7 3 4 12 0 Profit (100k) 6 7 9 8 3 2 4 12 5 3 S(i) 6 7 9 9 12 12 14 26 26 10 Optimal: 26 S(i-1) S(j) + pi j < i & dist (tj, ti) ≥ 10 S(i) = max**Complexity**• Time: O(nk), where k is the maximum number of towns that are within 10 miles to the left of any town • In the worst case, O(n2) • Can be reduced to O(n) by pre-processing • Memory: Θ(n)**Knapsack problem**• Each item has a value and a weight • Objective: maximize value • Constraint: knapsack has a weight limitation Three versions: 0-1 knapsack problem: take each item or leave it Fractional knapsack problem: items are divisible Unbounded knapsack problem: unlimited supplies of each item. Which one is easiest to solve? We studied the 0-1 problem.**Formal definition (0-1 problem)**• Knapsack has weight limit W • Items labeled 1, 2, …, n (arbitrarily) • Items have weights w1, w2, …, wn • Assume all weights are integers • For practical reason, only consider wi < W • Items have values v1, v2, …, vn • Objective: find a subset of items, S, such that iS wi W and iS vi is maximal among all such (feasible) subsets**A DP algorithm**• Suppose you’ve find the optimal solution S • Case 1: item n is included • Case 2: item n is not included Total weight limit: W Total weight limit: W wn wn Find an optimal solution using items 1, 2, …, n-1 with weight limit W - wn Find an optimal solution using items 1, 2, …, n-1 with weight limit W**V[n-1, W-wn] + vn**V[n-1, W] V[n, W] = max Generalize V[i-1, w-wi] + vi item i is taken V[i-1, w] item i not taken V[i, w] = max V[i-1, w] if wi > w item i not taken Recursive formulation • Let V[i, w] be the optimal total value when items 1, 2, …, i are considered for a knapsack with weight limit w => V[n, W] is the optimal solution Boundary condition: V[i, 0] = 0, V[0, w] = 0. Number of sub-problems = ?**Example**• n = 6 (# of items) • W = 10 (weight limit) • Items (weight, value): 2 2 4 3 3 3 5 6 2 4 6 9**wi**V[i-1, w-wi] V[i-1, w] i wi vi 0 0 0 0 0 0 0 0 0 0 0 1 2 2 0 2 4 3 0 0 3 3 3 0 4 5 6 5 6 V[i, w] 0 5 2 4 0 6 6 9 V[i-1, w-wi] + vi item i is taken V[i-1, w] item i not taken max V[i, w] = V[i-1, w] if wi > w item i not taken**2**3 5 3 5 6 8 6 8 9 11 4 6 7 10 12 13 9 13 15 V[i-1, w] if wi > w item i not taken 0 0 0 0 0 0 0 0 0 0 0 0 0 2 2 2 2 2 2 2 2 0 0 2 2 3 5 5 5 5 0 0 2 3 5 6 8 0 0 2 3 3 6 9 0 0 4 7 10 0 0 4 4 6 7 10 13 V[i-1, w-wi] + vi item i is taken V[i-1, w] item i not taken max V[i, w] =**3**5 3 5 6 8 6 8 9 11 4 7 10 12 13 9 13 0 0 0 0 0 0 0 0 0 0 0 0 0 2 2 2 2 2 2 2 2 2 0 0 2 2 3 5 5 5 5 0 0 2 3 5 6 8 0 0 2 3 3 6 9 0 0 4 6 7 10 0 0 4 4 6 7 10 13 15 Optimal value: 15 Item: 6, 5, 1 Weight: 6 + 2 + 2 = 10 Value: 9 + 4 + 2 = 15**Time complexity**• Θ (nW) • Polynomial? • Pseudo-polynomial • Works well if W is small • Consider following items (weight, value): (10, 5), (15, 6), (20, 5), (18, 6) • Weight limit 35 • Optimal solution: item 2, 4 (value = 12). Iterate: 2^4 = 16 subsets • Dynamic programming: fill up a 4 x 35 = 140 table entries • What’s the problem? • Many entries are unused: no such weight combination • Top-down may be better**s8**f8 s7 f7 s9 f9 Events scheduling problem • A list of events to schedule • ei has start time si and finishing time fi • Indexed such that fi < fj if i < j • Each event has a value vi • Schedule to make the largest value • You can attend only one event at any time e6 e8 e3 e7 e4 e5 e9 e1 e2 Time**s8**f8 s7 f7 s9 f9 Events scheduling problem • V(i) is the optimal value that can be achieved when the first i events are considered • V(n) = e6 e8 e3 e7 e4 e5 e9 e1 e2 Time V(n-1) en not selected max { V(j) + vn en selected j < n and fj < sn**Restaurant location problem 2**• Now the objective is to maximize the number of new restaurants (subject to the distance constraint) • In other words, we assume that each restaurant makes the same profit, no matter where it is opened 10 mile**A DP Algorithm**• Exactly as before, but pi = 1 for all i S(i-1) S(j) + pi j < i & dist (tj, ti) ≥ 10 S(i) = max S(i-1) S(j) + 1 j < i & dist (tj, ti) ≥ 10 S(i) = max**1**1 1 1 2 2 2 3 4 4 Example Distance (mi) • Natural greedy 1: 1 + 1 + 1 + 1 = 4 • Maybe greedy is ok here? Does it work for all cases? 100 5 2 2 6 6 3 6 10 7 dummy 0 Profit (100k) 1 1 1 1 1 1 1 1 1 1 S(i) Optimal: 4 S(i-1) S(j) + 1 j < i & dist (tj, ti) ≥ 10 S(i) = max**1**1 1 1 2 2 2 3 4 4 Comparison Dist(mi) 100 5 2 2 6 6 3 6 10 7 0 Profit (100k) 1 1 1 1 1 1 1 1 1 1 S(i) Benefit of taking t1 rather than t2? Benefit of waiting to see t2? t1 gives you more choices for the future None! Dist(mi) 100 5 2 2 6 6 3 6 10 7 0 Profit (100k) 6 7 9 8 3 2 4 12 5 3 S(i) 6 7 9 9 12 12 14 26 26 10 Benefit of taking t1 rather than t2? Benefit of waiting to see t2? t1 gives you more choices for the future t2 may have a bigger profit**Moral of the story**• If a better opportunity may come out next, you may want to hold on your decision • Otherwise, grasp the current opportunity immediately because there is no reason to wait …**Greedy algorithm**• For certain problems, DP is an overkill • Greedy algorithm may guarantee to give you the optimal solution • Much more efficient**B**m1 m2 mk A m1 B’ (imaginary) A’ Formal argument • Claim 1: if A = [m1, m2, …, mk] is the optimal solution to the restaurant location problem for a set of towns [t1, …, tn] • m1 < m2 < … < mkare indices of the selected towns • Then B = [m2, m3, …, mk] is the optimal solution to the sub-problem [tj, …, tn], where tj is the first town that are at least 10 miles to the right of tm1 • Proof by contradiction: suppose B is not the optimal solution to the sub-problem, which means there is a better solution B’ to the sub-problem • A’ = mi || B’ gives a better solution than A = mi || B => A is not optimal => contradiction => B is optimal**Implication of Claim 1**• If we know the first town that needs to be chosen, we can reduce the problem to a smaller sub-problem • This is similar to dynamic programming • Optimal substructure**S**S’ Formal argument (cont’d) • Claim 2: for the uniform-profit restaurant location problem, there is an optimal solution that chooses t1 • Proof by contradiction: suppose that no optimal solution can be obtained by choosing t1 • Say the first town chosen by the optimal solution S is ti, i > 1 • Replace ti with t1 will not violate the distance constraint, and the total profit remains the same => S’ is an optimal solution • Contradiction • Therefore claim 2 is valid**Implication of Claim 2**• We can simply choose the first town as part of the optimal solution • This is different from DP • Decisions are made immediately • By Claim 1, we then only need to repeat this strategy to the remaining sub-problem**0**0 Greedy algorithm for restaurant location problem select t1 d = 0; for (i = 2 to n) d = d + dist(ti, ti-1); if (d >= min_dist) select ti d = 0; end end 5 2 2 6 6 3 6 10 7 6 9 15 7 d 5 7 9 15 10 0 0**Complexity**• Time: Θ(n) • Memory: • Θ(n) to store the input • Θ(1) for greedy selection**Events scheduling problem**• Objective: to schedule the maximal number of events • Let vi = 1 for all i and solve by DP, but overkill • Greedy strategy: choose the first-finishing event that is compatible with previous selection (1, 2, 4, 6, 8 for the above example) • Why is this a valid strategy? • Claim 1: optimal substructure • Claim 2: there is an optimal solution that chooses e1 • Proof by contradiction: Suppose that no optimal solution contains e1 • Say the first event chosen is ei => other chosen events start after ei finishes • Replace ei by e1 will result in another optimal solution (e1 finishes earlier than ei) • Contradiction • Simple idea: attend the event that will left you with the most amount of time when finished e6 e8 e3 e7 e4 e5 e9 e1 e2 Time**Knapsack problem**• Each item has a value and a weight • Objective: maximize value • Constraint: knapsack has a weight limitation Three versions: 0-1 knapsack problem: take each item or leave it Fractional knapsack problem: items are divisible Unbounded knapsack problem: unlimited supplies of each item. Which one is easiest to solve? We can solve the fractional knapsack problem using greedy algorithm**Greedy algorithm for fractional knapsack problem**• Compute value/weight ratio for each item • Sort items by their value/weight ratio into decreasing order • Call the remaining item with the highest ratio the most valuable item (MVI) • Iteratively: • If the weight limit can not be reached by adding MVI • Select MVI • Otherwise select MVI partially until weight limit**item**Weight (LB) Value ($) $ / LB 1 2 2 1 2 4 3 0.75 3 3 3 1 4 5 6 1.2 5 2 4 2 6 6 9 1.5 Example • Weight limit: 10**Example**• Weight limit: 10 • Take item 5 • 2 LB, $4 • Take item 6 • 8 LB, $13 • Take 2 LB of item 4 • 10 LB, 15.4**w**w w w Why is greedy algorithm for fractional knapsack problem valid? • Claim: the optimal solution must contain the MVI as much as possible (either up to the weight limit or until MVI is exhausted) • Proof by contradiction: suppose that the optimal solution does not use all available MVI (i.e., there is still w (w < W) pounds of MVI left while we choose other items) • We can replace w pounds of less valuable items by MVI • The total weight is the same, but with value higher than the “optimal” • Contradiction**Elements of greedy algorithm**• Optimal substructure • Locally optimal decision leads to globally optimal solution • For most optimization problems, greedy algorithm will not guarantee an optimal solution • But may give you a good starting point to use other optimization techniques • Starting from next week, we’ll study several problems in graph theory that can actually be solved by greedy algorithm

More Related