Question

Consider procedure schedule in the Scheduling with Deadlines algorithm (Algorithm 4.4). Let \(d\) be the maximum of the deadlines for \(n\) jobs. Modify the procedure so that it adds a job as late as possible to the schedule being built, but no later than its deadline. Do this by initializing \(d+1\) disjoint sets, containing the integers \(0,1, \ldots, d\) Let \(\operatorname{small}\) ( \(\mathrm{S}\) ) be the smallest member of set \(\mathrm{S}\). When a job is scheduled, find the set \(S\) containing the minimum of its deadline and \(n\). If small \((S)=0,\) reject the job. Otherwise, schedule it at time \(\operatorname{small}(S),\) and merge \(S\) with the set containing small \((S)-1 .\) Assuming we use Disjoint Set Data Structure III in Appendix \(C\). show that this version is \(\theta(n \lg m)\), where \(m\) is the minimum of \(d\) and \(n\)

Short answer

The time complexity of the proposed scheduling algorithm is determined by the time complexity of the union operation in disjoint data set represented by size. As it is \(\theta(\lg m)\) and we perform it \(n\) times for \(n\) jobs, the overall time complexity of the algorithm is \(\theta(n \lg m)\), where \(m\) is the smallest of total jobs \(n\) and maximum deadline \(d\).

Step-by-step solution

Identify Important Aspects

The problem mentions that we are using the Disjoint Set Data Structure III as mentioned in Appendix C. Here, \(d\) is the maximum deadline among all jobs, \(n\) is the total number of jobs, and \(m\) is the minimum of \(d\) and \(n\). The job with the minimum time of either its deadline or \(n\) will be scheduled.

Initial Stages of the Algorithm

Initialize \(d+1\) disjoint sets containing integers from \(0\) to \(d\). 'Small' is a function which outputs the smallest member of a set S. Now, schedule a job by finding the set S which contains the minimum of its deadline and \(n\).

Middle Stages of the Algorithm

After finding set S, if the smallest member of the set is \(0\), the job is rejected. If it is not, the job is scheduled at the time given by small(S). Then, set S is merged with the set containing small(S)-1.

Analyze the Time Complexity

Time complexity of a union operation in disjoint set data structure represented by size can be \(\theta(\lg m)\), where \(m\) is the smallest among \(n\) (number of elements) and \(d\) (maximum deadline). As there are \(n\) jobs and for each job the union operation is performed one time, the time complexity of this scheduling algorithm is \(\theta(n \lg m)\).

Key concepts

Scheduling Algorithms

Scheduling algorithms are designed to help organize and allocate limited resources, such as time slots, efficiently. In the context of the exercise, we're dealing with jobs that have specific deadlines. The objective is to complete as many jobs as possible within their constraints, optimizing the use of available time slots. By strategically choosing when to start each job, based on its deadline, you can maximize the number of jobs completed.

The key idea is to place each job as late as possible, while still respecting its deadline. This approach allows more flexibility in the schedule because it makes room for jobs with more stringent deadlines. The process is often compared to arranging fitting pieces into a puzzle, where each piece (or job) must fit into the allocated time without overlapping. In this scenario, efficient scheduling relies heavily on understanding not only the job's deadline but also finding the optimal position for the job within the given timeframe.

Using disjoint sets, discussed further on, enhances this process by helping manage and merge time slots effectively. By ensuring the job is scheduled at the latest possible time without breaching its deadline, more jobs can be accommodated overall.

Disjoint Set Data Structure

The disjoint set data structure, sometimes referred to as union-find, is a crucial component in various computer science algorithms, particularly those that involve partitioning and merging sets. In our scheduling problem, this data structure helps manage scheduling slots efficiently. It is used to track and manage the schedule as jobs are added.

Each job represents a set of time slots, and the disjoint set data structure maintains the connection between these slots. This allows the algorithm to swiftly determine time availability for scheduling each job. When a job is scheduled, it occupies a time slot which is then unified with previous slots using the union operation in the disjoint set data structure.

This method allows for efficient searching and merging operations which are essential in dynamically maintaining the scheduling structure. By reducing the complexity of these operations, the overall process becomes both time and space efficient. This efficiency is partly due to the fact that operations are typically near-constant time thanks to the clever management of set representatives and rank by the disjoint set.

Time Complexity

Time complexity is a measure of the computational effort needed to execute an algorithm, particularly as the size of the input increases. In the context of the scheduling algorithm involving the disjoint set data structure, understanding time complexity helps us predict how the algorithm scales with larger input sizes.

For our exercise, the time complexity is denoted as \( \theta(n \lg m) \), where \( n \) is the number of jobs and \( m \) is the smaller value between the maximum deadline \( d \) and \( n \). This denotes that the operations involved in the scheduling algorithm scale logarithmically with respect to \( m \).

Each job scheduling action involves a union operation in the disjoint set, which typically takes \( \log \) time due to the way these data structures are optimized. Consequently, as the total number of jobs \( n \) increases, the algorithm remains efficient compared to algorithms with higher time complexity sums like quadratic \( n^2 \) or cubic \( n^3 \). This efficiency is crucial in real-world applications where large datasets or numerous time slots must be processed efficiently.

Set Operations

Set operations are fundamental to managing groups of data elements in computer science. In our scheduling context, we use set operations to manage and organize time slots for job scheduling. The main operations in the context of disjoint sets are union, find, and sometimes, path compression.

Union: This operation is fundamental in merging two sets into one. In scheduling, it is used when you need to merge the current time slot with its preceding one once a job is scheduled. This helps in managing availability for future scheduling accurately.

Find: The find operation is used to locate the set to which a particular element belongs. It is crucial for determining the available time slot for a job without conflict.

Path Compression: This is an optimization technique used to flatten the structure of the set, ensuring that future find operations are faster. This operation is important in keeping the data structure efficient, even as more and more union operations are performed.

Together, these operations facilitate efficient management and merging of time slots, ensuring the jobs can be scheduled correctly and optimally. The combination of these operations is what allows the scheduling algorithm to efficiently handle assignment and availability of time slots, reducing unnecessary computations and ensuring quick decision-making.