Graph Traversals Overview. Just as we had both depth-first (preorder, inorder, and postorder) traversals and a breath-first (level order) traversal for trees, we can generalize these concepts to graphs. Specifically, given a source vertex, we can “visit” vertices in:
- DFS Preorder: order in which DFS is called on each vertex.
- DFS Postorder: order in which we return from DFS calls.
- BFS: order of distance from the source. The lecture calls this “level order” before we banish that term since nobody uses it in the real world for general graphs.
We use the term “depth first”, because we will explore “deeply” first (a la https://xkcd.com/761/]), and use the term “breadth first” because we go wide before we go deep.
If we use BFS on a vertex of a graph that happens to be the root of a tree, we get exactly the same thing as level order traversal.
Topological Sorting. As an example of a graph problem for which DFS is more suitable than BFS, we considered the topological sort problem: Given a directed graph, find an ordering of the vertices that respects the direction of the edges. We can imagine using this to plan, for example a course schedule (albeit at some sort of really terrible university where we only take one class at a time).
The algorithm is simple: Record the DFS postorder from every vertex with in-degree zero (i.e. has no incoming edges). The order you’re after is the reverse of the DFS postorder. Kind of miraculous and amazing IMO (until you think about it enough to realize its trivial, but that takes a while). Note: You can actually perform DFS postorders from any vertex, see B level problems below.
Breadth First Search. Unlike DFS, BFS lends itself more naturally to an iterative solution than a recursive one. When we perform BFS, we visit a source vertex s, then visit every vertex that is one link away from s, then visite very vertex that is two links away from s, and so forth.
To achieve this, we use a simple idea: Create a so-called “fringe” of vertices that we think of as the next vertices to be explored. In the case of BFS, this fringe is a Queue, since we want to visit vertices in the order that we observe them. The pseudocode is as follows:
bfs(s): fringe.enqueue(s) mark(s) while fringe is not empty: dequeue(s) visit(s) for each unmarked neighbor of s: mark(s) enqueue(s)
In class, we discussed how we could use BFS to solve the shortest paths problem: Given a source vertex, find the shortest path from that source to every other vertex. When solving shortest paths, we add additional logic to our BFS traversal, where we also set the edgeTo for every vertex at the same time that it is marked and enqueued.
Problems 1 and 2 from Princeton’s Coursera course. A FIFO queue is a “first-in first-out” queue, i.e. just a regular old queue.
Problems 1 and 2 from Princeton’s Coursera course.
Problem 2a from Princeton’s Fall 2009 final
Suppose we run BFS from a vertex s. The edgeTo array we get back is sometimes known as a ‘Breadth First Paths Tree’. What, if anything, does the BFS tree tell us about the shortest path from v to w, assuming that neither is the source?
Problem 1a and 1b from Princeton’s Spring 2008 final.
Problem 3a from Princeton’s Fall 2010 final.
Problem 2b from Princeton’s Fall 2009 final.
Problem 1c from Princeton’s Spring 2008 final.
Problem 3b from Princeton’s Fall 2010 final.
Problems 4 and 5 from semester’s discussion worksheet.
Adapted from Algorithms textbook 4.2.10: Given a DAG, does there exist a topological order that cannot result from applying a DFS-based algorithm, no matter in what order the vertices adjacent to each vertex are chosen? Prove your answer.
Develop an algorithm that determines whether or not a directed graph contains an Eulerian tour, i.e. a tour that visits every vertex exactly once.
Adapted from Algorithms textbook 4.2.19: Explain why the following algorithm does not necessarily produce a topological order: Run BFS, and label the vertices by increasing distance to their respective source.
Adapted from Algorithms textbook 4.2.27: Show that the number of different V-vertex digraphs with no parallel edges is 2^(V^2). Then compute an upper bound on the percentage of 20-vertex digraphs that could ever be examined by any computer, under the assumptions that every electron in the universe examines a digraph every nanosecond, that the universe has fewer than 1080 electrons, and that the age of the universe will be less than 1020 years.
Just for fun
- Adapted from Algorithms textbook 4.2.40: Run experiments to determine empirically the probability that DepthFirstDirectedPaths finds a path between two randomly chosen vertices and to calculate the average length of the paths found, for various random digraph models.