25.1.7. Step-by-Step Example of Longest Path Computation

Let us continue to look at DAGs and in this section we will look at the different problem called identifying the Longest Path in a DAG.

So, recall that the directed acyclic graph is just a directed graph in which there is no directed path from any vertex back to itself, so it is direct and it is acyclic.

A Directed Acyclic Graph (DAG) is a type of graph that has directed edges and does not contain any cycles, meaning there's no path in the graph that starts and ends at the same vertex. This characteristic is important because it guarantees that we can topologically sort the vertices, which allows us to arrange them in a sequence following their dependencies.

Imagine a coursework structure at a university. Each course is a vertex, and each prerequisite (like needing to finish an introductory course before taking an advanced one) is a directed edge. You can't complete a course until you finish its prerequisites – just like in a DAG, you can only proceed forward without going back.

Any directed acyclic graph can be topologically ordered. If you think of the vertices has been 1 to n, you can write out 1 to n as a sequence in such a way that for every pair j, k which is an edge if j, k is an edge in my graph, then j will appear before k in the sequence.

Topological sorting is a process of arranging the vertices of a DAG in a linear order such that for every directed edge (j, k), vertex j comes before vertex k in the ordering. This is crucial for tasks where certain prerequisites must be completed before moving on to the next ones.

Think about planning a wedding. There are several tasks (like booking a venue, sending invitations, and choosing a menu) that depend on each other. For example, you can’t send invitations until you have a date and venue. Topological sorting would help in arranging these tasks in the order they need to be completed.

Supposing we have this DAG and we imagine that the vertices represent courses and the edges are prerequisites... what is the minimum number of semesters that I need to complete this program consisting of these 8 courses, with these prerequisites.

The example involves determining how to complete a set of courses represented by a DAG, where vertices are courses and edges denote prerequisites. The concept here is to figure out the minimum number of semesters needed to finish all courses, taking into account the dependencies. For instance, if a student can take courses with no prerequisites first, they can increase their progress towards graduation.

Imagine you have 8 different classes to complete, and some of these classes require others to be completed first. You might take classes without prerequisites in your first semester, then the next set in your second semester as the dependencies are cleared. This scheduling mimics how you navigate through a project, ensuring that you meet deadlines based on dependent milestones.

Now, this problem corresponds to asking for the longest path in the DAG... this does not help us because after 2 I cannot do it. So, I must wait for everything that has to happen before 8 in order to get the job done.

In this context, the longest path in a DAG relates to the number of semesters needed to complete all courses, where each semester's sequence is governed by prerequisites. Hence, the longest path dictates the maximum number of semesters required to finish all the courses '8', which cannot start until all dependencies are met.

Continuing with the coursework analogy, the longest path could represent a chain of critical projects that must be completed sequentially. If you're working on a multi-phase project, like writing a book, you can’t start editing until the entire manuscript is finished, which defines how long the project lasts.

Therefore, in order to compute the longest path to k, I need to compute the longest path to all its incoming neighbours. Now, if we have arranged the vertices in topological order...

To find the longest path to a specific vertex in the DAG, one must first compute the longest path to all of its incoming neighbors. By processing the vertices in topological order (which ensures that all prerequisite calculations are completed before they are needed), we can efficiently compute the longest path dynamically.

Consider a construction project where some rooms can only be built after others (like roof on top of walls). Each room’s completion time can only be added to when every prior room is finished. In doing so, we get the total project timeline, illustrating how dependencies shape progress.

So, we can actually incrementally compute the value at k as you are going forwards... as we will see in the next example.

Rather than working backward from a vertex to find incoming edges and paths, we can compute the longest path while traversing the graph in the forward direction. This makes calculations efficient as we carry forward the maximum lengths as we cancel out the prerequisites, rather than revisiting previous computations.

Think of it like following a recipe: you don’t backtrack to see how many steps you took after finishing a task (like chopping vegetables) but rather move on to the next step (like sautéing). You build upon the previous steps, continually adding to your preparation without retracing.

So, what this is said is the longest path is 4, now we said that we done in 5 semester...

The process concludes with the realization that the longest path (4) aligns with the calculated number of semesters needed to complete all courses (5). This confirmation reinforces how the longest path aligns with scheduling tasks in a manner that respects their dependencies and maximizes efficiency.

Wrapping up our project analogy, this final accounting would reveal that all tasks needed five major phases where everything else fit within those bounds. It's akin to saying 'it took 5 weeks to complete the project,' emphasizing planning based on dependencies required to achieve that timeline.

So, this pseudo code for longest path as we saw is very similar to what we did for the topological sort...

The pseudo code for finding the longest path in a DAG mirrors that of topological sorting, with the addition of tracking the longest paths associated with each vertex. This allows for maintaining both the order of operations and the cumulative lengths of paths as vertices are processed.

Again with our wedding planning, the algorithm can be understood as a to-do list that not only outlines tasks to be done (like sending invites) but also tracks how long each task takes, ensuring that everything fits together seamlessly for a successful occasion.

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