Question
Consider the following variation on the Activity Selection Problem.You have a resource that may be used for activities 24 hours a day,ever day.Activities repeat that may be used for activities repeat on a daily basis. As in the original problem, each activity has a start time and as end time.If an activity is selected, it will exclusively use the resource during the duration between the start and end time(i.e., no other activity may be scheduled during this time). Note that jobs can begin before midnight and end midnight; this situation differentiates it from the original Activity Selection Problem.
Given a list of n activities, your goal is to accept as many activities as possible for a daily scheduled of activities(i.e., every day the schedule of activities will be identical). Provide an algorithm for activity selection with running time polynomial in n. You may assume for simplicity that no two jobs have same start or end times.
An Example: Consider the following four activities, where each activity is specified by(start-time, end-time) pair(using 24-time).
(18:00,6:00),(21:00,4:00),(3:00,14:00),(13:00,19:00).
The optimal solution would be to pick the two jobs(21:00,4:00) and (13:00,19:00), which can be scheduled without overlapping.
Explanation / Answer
An activity-selection is the problem of scheduling a resource among several competing activity. Statement: Given a set S of n activities with and start time, Si and fi, finish time of an ith activity. Find the maximum size set of mutually compatible activities. Compatible Activities Activities i and j are compatible if the half-open internal [si, fi) and [sj, fj) do not overlap, that is, i and j are compatible if si = fj and sj = fi Greedy Algorithm for Selection Problem I. Sort the input activities by increasing finishing time. f1 = f2 = . . . = fn II Call GREEDY-ACTIVITY-SELECTOR (Sif) n = length [s] A={i} j = 1 FOR i = 2 to n do if si = fj then A= AU{i} j = i Return A Operation of the algorithm Let 11 activities are given S = {p, q, r, s, t, u, v, w, x, y, z} start and finished times for proposed activities are (1, 4), (3, 5), (0, 6), 5, 7), (3, 8), 5, 9), (6, 10), (8, 11), (8, 12), (2, 13) and (12, 14). A = {p} Initialization at line 2 A = {p, s} line 6 - 1st iteration of FOR - loop A = {p, s, w} line 6 -2nd iteration of FOR - loop A = {p, s, w, z} line 6 - 3rd iteration of FOR-loop Out of the FOR-loop and Return A = {p, s, w, z} Analysis Part I requires O(nlgn) time (use merge of heap sort). Part II requires Theta(n) time assuming that activities were already sorted in part I by their finish time. CORRECTNESS Note that Greedy algorithm do not always produce optimal solutions but GREEDY-ACTIVITY-SELECTOR does. Theorem: Algorithm GREED-ACTIVITY-SELECTOR produces solution of maximum size for the activity-selection problem. Proof Idea: Show the activity problem satisfied I. Greedy choice property. II. Optimal substructure property Proof: I. Let S = {1, 2, . . . , n} be the set of activities. Since activities are in order by finish time. It implies that activity 1 has the earliest finish time. Suppose, A is a subset of S is an optimal solution and let activities in A are ordered by finish time. Suppose, the first activity in A is k. If k = 1, then A begins with greedy choice and we are done (or to be very precise, there is nothing to proof here). If k not=1, we want to show that there is another solution B that begins with greedy choice, activity 1. Let B = A - {k} U {1}. Because f1 == fi}. why? If we could find a solution B` to S` with more activities then A`, adding 1 to B` would yield a solution B to S with more activities than A, there by contradicting the optimality. Dynamic-Programming Algorithm for the Activity-Selection Problem Problem: Given a set of activities to among lecture halls. Schedule all the activities using minimal lecture halls. In order to determine which activity should use which lecture hall, the algorithm uses the GREEDY-ACTIVITY-SELECTOR to calculate the activities in the first lecture hall. If there are some activities yet to be scheduled, a new lecture hall is selected and GREEDY-ACTIVITY-SELECTOR is called again. This continues until all activities have been scheduled. LECTURE-HALL-ASSIGNMENT (s,f) n = length [s) FOR i = 1 to n DO HALL [i] = NIL k = 1 WHILE (Not empty (s)) Do HALL [k] = GREEDY-ACTIVITY-SELECTOR (s, t, n) k = k + 1 RETURN HALL Following changes can be made in the GREEDY-ACTIVITY-SELECTOR (s, f) (CLR). j = first (s) A = i FOR i = j + 1 to n DO IF s(i) not= "-" THEN IF GREED-ACTIVITY-SELECTOR (s,f,n) j = first (s) A = i = j + 1 to n IF s(i] not = "-" THEN IF s[i] >= f[j]| THEN A = A U {i} s[i] = "-" j = i return A