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 (s_i) and finish time (f_i). 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.

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 s_i and finish time f_i. Two activities i and j are said to be non-conflicting if s_i ≥ f_j or s_j ≥ f_i. 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

 1 Greedy-Iterative-Activity-Selector(A, s, f): 
 2 
 3     Sort A by finish times stored in f'
 4     
 5     S = {A[1]} 
 6     k = 1
 7     
 8     n = A.length
 9     
10     for i = 2 to n:
11        if s[i] ≥ f[k]: 
12            S = S U {A[i]}
13            k = i
14     
15     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.

• $A$ is an array containing the activities.
• $s$ is an array containing the start times of the activities in $A$.
• $f$ is an array containing the finish times of the activities in $A$.

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 $A$ by using the finish times stored in the array $f$. This operation can be done in $O(n \cdot \log_2 n)$ time, using for example merge sort, heap sort, or quick sort algorithms.

Line 5: Creates a set $S$ to store the selected activities, and initialises it with the first activity $A[1]$. Note that, since the $A$ has already been sorted according to the finish times in $f$, $A[1]$ is the activity with the smallest finish time.

Line 6: Creates a variable $k$ that keeps track of the index of the last selected activity.

Line 10: Starts iterating from the second element of that array $A$ up to its last element.

Line 11: If the start time $s[i]$ of the $ith$ activity ($A[i]$) is greater or equal to the finish time $f[k]$ of the last selected activity ($A[k]$), then $A[i]$ is compatible to the selected activities in the set $S$, and thus it can be added to $S$; this is what is done in line 12.

Line 13: The index of the last selected activity is updated to the just added activity $A[i]$.

Proof of optimality

Let $S=\{1,2,\ldots ,n\}$ be the set of activities ordered by finish time. Thus activity 1 has the earliest finish time.

Suppose A is a subset of S and is an optimal solution, and let activities in A be ordered by finish time. Suppose that the first activity in A is k ≠ 1, that is, this optimal solution does not start with the "greedy choice." We want to show that there is another solution B that begins with the greedy choice, activity 1.

Let $B=(A\setminus \{k\})\cup \{1\}$. Because $f_{1}\leq f_{k}$, the activities in B are disjoint and since B has same number of activities as A, i.e., |A| = |B|, 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, then $A^{\prime }=A\setminus \{1\}$ is an optimal solution to the activity-selection problem $S' = \{i \in S: s_i \geq f_1\}$.

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, 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:^[1]

Consider an optimal solution containing activity $k$. We now have non-overlapping activities on the left and right of $k$. We can recursively find solutions for these two sets because of optimal sub-structure. As we don't know $k$, we can try each of the activities. This approach leads to an $O(n^{3})$ solution. This can be optimized further considering that for each set of activities in $(i,j)$, we can find the optimal solution if we had known the solution for $(i,t)$, where $t$ is the last non-overlapping interval with $j$ in $(i,j)$. This yields an $O(n^{2})$ solution. This can be further optimized considering the fact that we do not need to consider all ranges $(i,j)$ but instead just $(1,j)$. The following algorithm thus yields an $O(n\log n)$ solution:

 1 Weighted-Activity-Selection(S):  // S = list of activities
 2 
 3     sort S by finish time
 4     opt[0] = 0
 5    
 6     for i = 1 to n:
 7         t = binary search to find activity with finish time <= start time for i
 8         opt[i] = MAX(opt[i-1], opt[t] + w(i))
 9         
10     return opt[n]

References

1. Dynamic Programming with introduction to Weighted Activity Selection

External links

• Activity Selection Problem