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## CS 3343: Analysis of Algorithms

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**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