Dynamic Programming Assembly-Line Scheduling. Greedy Algorithms

Transcription

1 Chapter 13 Greedy Algorithms Activity Selection Problem 0-1 Knapsack Problem Huffman Code Construction Dynamic Programming Assembly-Line Scheduling C-C Tsai P.1 Greedy Algorithms A greedy algorithm always makes the choice that looks best at the moment. That is, it makes a locally optimal choice in the hope that this choice will lead to a globally optimal solution. Greedy algorithms do not always yield optimal solutions, but for many problems they do. C-C Tsai P.2 1

2 Activity Selection Problem Suppose we have a set S = {a 1, a 2,..., a n } of n proposed activities that with to use a resource. Each activity a i has a start time s i and a finish time f i, where 0 s i < f i <. If selected, activity a i take place during the halfopen time interval [s i, f i ). Activities a i and a j are compatible if the intervals [s i, f i ) and [s j, f j ) do not overlap (i.e., a i and a j are compatible if s i f j or s j f i ). C-C Tsai P.3 Activity Selection problem The activity-selection problem is to select a maximum-size subset of mutually compatible activities. Example: (order in finish time) {a 3, a 9, a 11 } is the subset, but not a maximal. {a 1, a 4, a 8, a 11 } and {a 2, a 4, a 9, a 11 } are largest subsets i s i f i C-C Tsai P.4 2

3 Solutions for Activity Selection Problem We shall solve this problem in several steps. We start by formulating a dynamic programming solution to this program in which we combine optimal solutions to two subproblems to form an optimal solution to the original problem. We shall then observe that we need only consider one choice the greedy choice and that when we make the greedy choice, one of the subproblems is guaranteed to be empty, so that only one nonempty subproblem remains. C-C Tsai P.5 Optimal Substructure of Activity-Selection Problem S ij ={a k S : f i s k < f k s j }, So that S ij is the subset of activities in S that can start after activity a i finishes and finish before activity a j starts. f 0 =0 and s n+1 =. Then S = S 0,n+1, and the ranges for i and j are given by 0 i, j n+1. Maximum-size subset A ij = A ik {a k } A kj An optimal solution to the entire problem is a solution to S = S 0,n+1. C-C Tsai P.6 3

4 A Recursive Solution Let c[i,j] be the number of activities in S ij. c[i,j]=0 if S ij =φ or i j. c[ i, j] = c[ i, k] + c[ k, j] + 1 c[ i, 0 j] = max{ c[ i, k] + c[ k, i< k< j j] + 1} if if S S ij ij = 0 0 C-C Tsai P.7 Converting a dynamic-programming solution to a greedy solution Theorem Consider any nonempty subproblem S ij, and let a m be the activity in S ij with the earliest finish time: f m = min { f k : a k S ij }. then 1. Activity a m is used in some maximum-size subset of mutually compatible activities of S ij. 2. The subproblem S im is empty, so that choosing a m leaves the subproblem S mj as the only one that may be nonempty. C-C Tsai P.8 4

5 A Recursive Greedy Algorithm RECURSIVE-ACTIVITY-SELECTOR (RAS) RAS(s, f, i, j) 1 m i +1 2 while m < j and s m < f i Find the first activity in S ij 3 do m m + 1 4if m < j 5 then return {a m } RAS(s, f, m, j) 6 else return 0 Time complexity is Θ(n) if ordering in finish times. C-C Tsai P.9 Operations of RECURSIVE-ACTIVITY-SELECTOR on the 11 activities given earlier C-C Tsai P.10 5

6 C-C Tsai {a1, a4, a8, a11} is the resulting set. P.11 An Iterative Greedy Algorithm GREEDY-ACTIVITY-SELECTOR(s, f) 1 n length[s] 2 a {a 1 } 3 i 1 4 for m 2 to n 5 do if s m f i 6 then A A {a m } 7 i m 8 return A C-C Tsai P.12 6

7 Elements of Greedy Strategy 1. Determine the optimal substructure of the problem. 2. Develop a recursive solution. 3. Prove that at any stage of the recursion, one of the optimal choices is the greedy choice. Thus, it is always safe to make the greedy choice. 4. Show that all but one of the subproblems induced by having make the greedy choice are empty. 5. Develop a recursive algorithm that implements the greedy strategy. 6. Convert the recursive algorithm to an iterative algorithm. Alternatively, we could have fashioned our optimal substructure with a greedy choice in mind. C-C Tsai P.13 Designing a Greedy Algorithm 1. Cast the optimization problem as one in which we make a choice and are left with one subproblem to solve. 2. Prove that there is always an optimal solution to the original problem that makes the greedy choice, so that the greedy choice is always safe. 3. Demonstrate that, having made the greedy choice, what remains is a subproblem with the property that if we combine an optimal solution to the subproblem with the greedy choice we have made, we arrive at an optimal solution to the original problem. Greedy-choice property: A global optimal solution can be achieved by making a local optimal (greedy) choice. Optimal substructure: An optimal solution to the problem within its optimal solution to subproblem. C-C Tsai P.14 7

8 0-1 Knapsack Problem 0-1 knapsack problem A thief robbing a store finds n items; the i th is worth v i dollars and weights w i pounds. He wants to take as valuable a load as possible, but he can carry at most W pounds at his knapsack. Fractional knapsack problem (v i / w i is the valuable consideration) C-C Tsai P Knapsack Problem The greedy strategy does not work for the 0-1 knapsack. 0-1 knapsack Fractional knapsack C-C Tsai P.16 8

9 Huffman Codes Prefix code: no codeword is also a prefix of some other codeword. a Frequency (in hundred) Fixed length codeword Variable length codeword Fixed length codeword Variable length codeword b c d e f B( T ) = f ( c) dt ( c) C-C Tsai P.17 cost of tree T c C Huffman Codes Huffman codes can be shown that the optimal data compression achievable by a character code can always be achieved with prefix codes. Simple encoding and decoding. An optimal code for a file is always represented by a binary tree. C-C Tsai P.18 9

10 Steps of Constructing Huffman Codes C-C Tsai P.19 Steps of Constructing Huffman Codes Theorem Procedure HUFFMAN produces an optimal prefix code. C-C Tsai P.20 10

11 Constructing a Huffman code HUFFMAN(C ) 1 n C 2 Q C 3 for i 1 to n 1 C 4 do allocate a new node z 5 left[z] x EXTRACT-MIN(Q) 6 right[z] y EXTRACT-MIN(Q) 7 f[z] f[x] + f[y] 8 INSERT(Q,Z) 9 return EXTRACT-MIN(Q) Complexity: O(n log n) C-C Tsai P.21 Dynamic Programming Dynamic programming is typically applied to optimization problems. In such problem there can be many solutions. Each solution has a value, and we wish to find a solution with the optimal value. The development of a dynamic programming algorithm can be broken into a sequence of four steps: 1. Characterize the structure of an optimal solution. 2. Recursively define the value of an optimal solution. 3. Compute the value of an optimal solution in a bottom-up fashion. 4. Construct an optimal solution from computed information. C-C Tsai P.22 11

12 Assembly-Line Scheduling An automobile chassis enters each assembly line, has parts added to it at a number of stations, and a finished auto exits at the end of the line. Each assembly line has n stations, numbered j = 1, 2,...,n. We denote the jth station on line i ( where i is 1 or 2) by S i,j. The jth station on line 1 (S 1,j ) performs the same function as the jth station on line 2 (S 2,j ). The stations were built at different times and with different technologies, however, so that the time required at each station varies, even between stations at the same position on the two different lines. We denote the assembly time required at station S i,j by a i,j. As the coming figure shows, a chassis enters station 1 of one of the assembly lines, and it progresses from each station to the next. There is also an entry time e i for the chassis to enter assembly line i and an exit time x i for the completed auto to exit assembly line i. C-C Tsai P.23 A Manufacturing Problem: Find Fast Way Through A Factory C-C Tsai P.24 12

13 A Manufacturing Problem: Find Fast Way Through A Factory Normally, once a chassis enters an assembly line, it passes through that line only. The time to go from one station to the next within the same assembly line is negligible. Occasionally, a special rush order comes in, and the customer wants the automobile to be manufactured as quickly as possible. For the rush orders, the chassis still passes through the n stations in order, but the factory manager may switch the partially-completed auto from one assembly line to the other after any station. C-C Tsai P.25 A Manufacturing Problem: Find Fast Way Through A Factory The time to transfer a chassis away from assembly line i after having gone through station S ij is t i,j, where i = 1, 2 and j = 1, 2,..., n-1 (since after the nth station, assembly is complete). The problem is to determine which stations to choose from line 1 and which to choose from line 2 in order to minimize the total time through the factory for one auto. C-C Tsai P.26 13

14 A Manufacturing Problem: Find Fast Way Through A Factory C-C Tsai P.27 Step1: The structure of the fastest way through the factory the fast way through station S 1,j is either the fastest way through Station S 1,j-1 and then directly through station S 1,j, or the fastest way through station S 2,j-1, a transfer from line 2 to line 1, and then through station S 1,j. Using symmetric reasoning, the fastest way through station S 2,j is either the fastest way through station S 2,j-1 and then directly through Station S 2,j, or the fastest way through station S 1,j-1, a transfer from line 1 to line 2, and then through Station S 2,j. C-C Tsai P.28 14

15 Step 2: A recursive solution The fast time is denoted by f* = min (f 1 [n]+x 1, f 2 [n]+x 2 ) where n is the station e1 + a1,1 f1[ j] = min( f1[ j 1] + a1, j, f2[ j 1] + t e2 + a2,1 f2[ j] = min( f2[ j 1] + a2, j, f1[ j 1] + t 2, j 1 1, j 1 + a + a 1, j 2, j if j = 1, ) if j 2 if j = 1, ) if j 2 C-C Tsai P.29 Step 3: Computing the fastest times Simple method of recursive algorithm for above equations that running time is exponential in n: Let r i (j) be the number of references made to f i [j] in a recursive algorithm. r 1 (n) = r 2 (n) = 1 r 1 (j) = r 2 (j) = r 1 (j+1) + r 2 (j+1) for j=1, 2,, n-1 Since r i (j)= 2 n-j, thus f 1 [1]=2 n-1 The total number of references to all f i [j] values is Θ(2 n ). We can do much better if we compute the f i [j] values in different order from the recursive way. Observe that for j 2, each value of f i [j] depends only on the values of f 1 [j-1] and f 2 [j-1]. C-C Tsai P.30 15

16 FASTEST-WAY procedure FASTEST-WAY(a, t, e, x, n) 1 f 1 [1] e 1 + a 1,1 ; 2 f 2 [1] e 2 + a 2,1 ; 3 for j 2 to n 4 do if f 1 [j-1] + a 1,j f 2 [j-1] + t 2,j-1 +a 1,j 5 then f 1 [j] f 1 [j-1] + a 1,j ; l 1 [j] 1; 6 else f 1 [j] f 2 [j-1] + t 2,j-1 +a 1,j ; l 1 [j] 2; 7 if f 2 [j-1] + a 2, j f 1 [j-1] + t 1,j-1 +a 2,j 8 then f 2 [j] f 2 [j 1] + a 2,j ; l2[j] 2 9 else f 2 [j] f 1 [j 1] + t 1,j-1 + a 2,j ; l 2 [j] 1; 10 if f 1 [n] + x 1 f 2 [n] + x 2 11 then f * = f 1 [n] + x 1 ; l * = 1 ; 12 else f * = f 2 [n] + x 2 ; l * = 2 ; Θ(n) C-C Tsai P.31 Step 4: Constructing the fastest way through the factory PRINT-STATIONS(l, n) 1 i l* 2 print line i,station n 3 for j n downto 2 4 do i l i [j] 5 print line i,station j 1 l* = 1, use station S 1,6 l 1 [6] = 2, use station S 2,5 l 2 [5] = 2, use station S 2,4 l 2 [4] = 1, use station S 1,3 l 1 [3] = 2, use station S 2,2 l 2 [2] = 1, use station S 1,1 output line 1, station 6 line 2, station 5 line 2, station 4 line 1, station 3 line 2, station 2 line 1, station 1 C-C Tsai P.32 16