
Activity selection problem

The activity selection problem is a combinatorial optimization problem concerning the selection of non-conflicting activities to perform within a given time frame, given a set of activities each marked by a start time (si) and finish time (fi). The problem is to select the maximum number of activities that can be performed by a single person or machine, assuming that a person can only work on a single activity at a time. The activity selection problem is also known as the Interval scheduling maximization problem (ISMP), which is a special type of the more general Interval Scheduling problem.
A classic application of this problem is in scheduling a room for multiple competing events, each having its own time requirements (start and end time), and many more arise within the framework of operations research.
Formal definition.
Assume there exist "n" activities with each of them being represented by a start time "si" and finish time "fi". Two activities "i" and "j" are said to be non-conflicting if "si" ≥ "fj" or "sj" ≥ "fi". The activity selection problem consists in finding the maximal solution set (S) of non-conflicting activities, or more precisely there must exist no solution set S' such that |S'| > |S| in the case that multiple maximal solutions have equal sizes.
Optimal solution.
The activity selection problem is notable in that using a greedy algorithm to find a solution will always result in an optimal solution. A pseudocode sketch of the iterative version of the algorithm and a proof of the optimality of its result are included below.
Algorithm.
Greedy-Iterative-Activity-Selector(A, s, f): 
 Sort A by finish times stored in 
 S = {A[1]} 
 k = 1
 n = A.length
 for i = 2 to n:
 if s[i] ≥ f[k]: 
 k = i
 return S
Explanation.
Line 1: This algorithm is called "Greedy-Iterative-Activity-Selector", because it is first of all a greedy algorithm, and then it is iterative. There's also a recursive version of this greedy algorithm.
Note that these arrays are indexed starting from 1 up to the length of the corresponding array.
Line 3: Sorts in "increasing order of finish times" the array of activities formula_1 by using the finish times stored in the array formula_4. This operation can be done in formula_8 time, using for example merge sort, heap sort, or quick sort algorithms.
Line 4: Creates a set formula_9 to store the "selected activities", and initialises it with the activity formula_10 that has the earliest finish time.
Line 5: Creates a variable formula_11 that keeps track of the index of the last selected activity.
Line 9: Starts iterating from the second element of that array formula_1 up to its last element.
Lines 10,11: If the "start time" formula_13 of the formula_14 activity (formula_15) is greater or equal to the "finish time" formula_16 of the "last selected activity" (formula_17), then formula_15 is compatible to the selected activities in the set formula_9, and thus it can be added to formula_9.
Line 12: The index of the last selected activity is updated to the just added activity formula_15.
Proof of optimality.
Let formula_22 be the set of activities ordered by finish time. Assume that formula_23 is an optimal solution, also ordered by finish time; and that the index of the first activity in "A" is formula_24, i.e., this optimal solution "does not" start with the greedy choice. We will show that formula_25, which begins with the greedy choice (activity 1), is another optimal solution. Since formula_26, and the activities in A are disjoint by definition, the activities in B are also disjoint. Since "B" has the same number of activities as "A", that is, formula_27, "B" is also optimal.
Once the greedy choice is made, the problem reduces to finding an optimal solution for the subproblem. If "A" is an optimal solution to the original problem "S" containing the greedy choice, then formula_28 is an optimal solution to the activity-selection problem formula_29.
Why? If this were not the case, pick a solution "B"′ to "S"′ with more activities than "A"′ containing the greedy choice for "S"′. Then, adding 1 to "B"′ would yield a feasible solution "B" to "S" with more activities than "A", contradicting the optimality.
Weighted activity selection problem.
The generalized version of the activity selection problem involves selecting an optimal set of non-overlapping activities such that the total weight is maximized. Unlike the unweighted version, there is no greedy solution to the weighted activity selection problem. However, a dynamic programming solution can readily be formed using the following approach:
Consider an optimal solution containing activity . We now have non-overlapping activities on the left and right of . We can recursively find solutions for these two sets because of optimal sub-structure. As we don't know , we can try each of the activities. This approach leads to an formula_30 solution. This can be optimized further considering that for each set of activities in formula_31, we can find the optimal solution if we had known the solution for formula_32, where is the last non-overlapping interval with in formula_31. This yields an formula_34 solution. This can be further optimized considering the fact that we do not need to consider all ranges formula_31 but instead just formula_36. The following algorithm thus yields an formula_37 solution:
Weighted-Activity-Selection(S): // S = list of activities
 sort S by finish time
 opt[0] = 0 // opt[j] represents optimal solution (sum of weights of selected activities) for S[1,2..,j]
 for i = 1 to n:
 t = binary search to find activity with finish time <= start time for i
 // if there are more than one such activities, choose the one with last finish time
 opt[i] = MAX(opt[i-1], opt[t] + w(i))
 return opt[n]