Greedy Algorithms. CLRS Chapters Introduction to greedy algorithms. Design of data-compression (Huffman) codes
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1 Greedy Algorithms CLRS Chapters Introduction to greedy algorithms Activity-selection problem Design of data-compression (Huffman) codes (Minimum spanning tree problem) (Shortest-path problem) Martin Zachariasen, DIKU May 18,
2 Greedy algorithms: An example Given three types of coins: 1, 2, and 5 units. What is the minimum number of coins we need to pay n units? That is: Find n 1, n 2 and n 5 such that 1 n n n 5 = n and such that n 1 + n 2 + n 5 is minimized. Obvious greedy approach: 1. Pick largest coin c where n c Solve remaining problem with n c units recursively. Problem has optimal substructure. For which of these coin types does the greedy approach work? 1. Coin types: 1, 2 and Coin types: 1, 4 and 5. 2
3 Greedy algorithms General technique/paradigm for solving optimization problems (like dynamic programming). Idea: Make the choice that looks best at the moment. Make a locally optimal choice in hope of getting a globally optimal solution. 1. Cast the optimization problem as one in which we make a choice and are left with a (smaller) subproblem to solve. 2. Prove that there is always an optimal solution that makes the greedy choice. 3. Demonstrate that an optimal solution to the subproblem combined with the greedy choice results in an optimal solution to the original problem. 3
4 Activity-selection problem Given: A set S = {a 1, a 2,..., a n } of n activities. All activities require exclusive use of a common resource. Activity a i has a start time s i and a finishing time f i (for which 0 s i < f i < ). The activity requires the common resource during the time interval [s i, f i [. Find: A maximum-size subset of activities A S that can be scheduled. Since the common resource can only be used by one activity at a time, the chosen activities must be non-overlapping (or mutually compatible). 4
5 Optimal substructure of activity-selection problem (I) (We will first devise a dynamic programming algorithm before proceeding to the greedy algorithm.) Add two fictitious activities: a 0 finishes before any other activity starts: f 0 = 0. a n+1 starts after any other activity finishes: s n+1 =. Assume activities are sorted by finishing time: f 0 f 1 f 2 f n < f n+1 5
6 Optimal substructure of activity-selection problem (II) Consider subproblem: S ij = {a k S : f i s k < f k s j } That is, all activities that start after a i finishes, and finish before a j starts (are between a i and a j ). Observations: S = S 0,n+1. S ij = for i j. a k S ij implies that i < k < j. 6
7 Optimal substructure of activity-selection problem (III) Consider an optimal solution for the subproblem S ij. Assume that activity a k is used by this solution. Generates two subproblems: S ik and S kj. First main observation: The solution to S ik within the optimal solution to S ij must be optimal. (Use cut-and-paste argument.) Second main observation: The solution to S kj within the optimal solution to S ij must be optimal. (Use cut-and-paste argument again.) Thus we have optimal substructure, and can solve the problem recursively. 7
8 Recursive solution to activity-selection problem c[i, j] = maximum-size subset of mutually compatible activities in S ij Optimal value for problem: c[0, n + 1] Recursive computation of c[i, j]: c[i, j] = { 0 if Sij = max ak S ij {c[i, k] + c[k, j] + 1} if S ij Note that the maximum is taken over up to j i 1 possible choices. 8
9 Greedy algorithm for activity-selection problem In the recursive algorithm, we only need to consider the activity a m S ij with the earliest finishing time! Theorem 1. a m is used by some optimal solution to subproblem S ij. 2. Subproblem S im is empty. Proof of 1: In any optimal solution to S ij we may replace the activity with the earliest finishing time with activity a m without creating overlapping activities. Proof of 2: Follows directly from the fact that activity a m has the earliest finishing time among activities in S ij. Thus we may solve the problem in a top-down fashion making greedy choices, solving smaller and smaller subproblems. 9
10 Data-compression codes In many applications we would like to store large files compactly using as few bits as possible. We need a binary character code for each character in the alphabet C: Each character in C is represented by a unique binary string. Fixed-length code: Each character uses the same amount of space (e.g. 7 bits as in the ASCII character system). Variable-length code: Frequent characters use short codes and infrequent characters use longer codes. We only consider prefix codes in which no code is a prefix of some other code. 10
11 Binary-tree representation of prefix codes For any prefix code we may construct a binary tree T that represents the code: Each leaf in T corresponds to a character in C. Edges of the tree are labeled by 0 or 1. The path from the root to a leaf gives the binary code for the leaf. Assume that the frequency f(c) of each character c C is known. For a tree T, let d T (c) denote the depth of the leaf c C. We say that T is an optimal code tree, if the cost B(T) = c C f(c)d T (c) is minimized. (Note that B(T) is the total number of bits needed to perform the encoding.) An optimal code tree is always a full binary tree. 11
12 Huffman prefix code construction Greedy algorithm that constructs an optimal code tree. Builds the tree T in a bottom-up manner by merging subtrees. Main idea: The two characters (or more generally subtrees) that have the lowest frequencies should be siblings as far down in the tree as possible. Merging operation: Join two subtrees by adding a new root z, and connecting the roots x and y of the subtrees to z with edges labeled 0 and 1. The frequency of z becomes the combined frequency of x and y. 12
13 Correctness of Huffman s algorithm Two steps in proof. Let x and y denote any two characters in C having the lowest frequencies. 1. There exists an optimal code tree in which x and y are siblings. Proof: Take any optimal code tree T and let a and b be a pair of siblings in T of maximum depth. Show that exchanging {a, b} with {x, y} in T does not increase the cost of the tree. 2. Let C = C \ {x, y} {z} where z is a new character. The frequencies of characters in C is the same as in C, except that f(z) = f(x) + f(y). Let T be an optimal code tree for C. Replacing leaf node z in T with an internal node having x and y as children results in a optimal code tree T for C. Proof: By contradiction. Assuming that T is not optimal implies that T is not optimal, since B(T) = B(T ) + f(x) + f(y) 13
14 Greedy algorithms: Other applications Fractional/0-1 knapsack problem Minimum spanning tree problem Shortest path problem 14
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