## Video transcripts

Hi! And welcome to this lesson about graph traversal.

In many cases, we’re interested in **accessibility**. Accessibility is finding how to reach a target vertex from an initial vertex, following a path in the graph. To obtain paths from the initial vertex to all the accessible ones, I’m going to show you how to build a tree, obtained from the initial graph, that will convey just enough information to guide us to any accessible vertex from the initial one.

### Neighbors and spanning trees

Recall that a graph is a couple composed of a set of vertices and a set of edges . Now we need to define a few more things.

As we’ve seen in the previous lesson, a tree is a connected graph with no cycles.

In this lesson, we’re going to see how to find a path between two vertices of a connected graph, so that we’re able to navigate between any two points.

By definition, a spanning tree will include exactly one path between those two points.

We will compare two approaches: the **depth-first search** and the **breadth-first search**.

### Depth-first search

The main idea of a depth-first search is to keep exploring the graph, jumping from a given vertex to one of its unexplored neighbors. Sometimes, all neighbors of the vertex we’re in have already been explored, in which case we turn back, following the trail in the opposite direction until we find an unexplored vertex to jump to.

DFS is a simple yet efficient strategy to ensure that all accessible vertices are explored.

### Breadth-first search

The other approach I will describe here is the breadth-first search. When using a breadth-first search, or BFS, we explore the graph by gradually increasing the number of hops from the initial vertex.

A BFS is a cautious strategy in which we try to keep the distances to the initial vertex as small as possible during exploration.

### Comparison between BFS and DFS

In the previous example, the spanning trees obtained by a BFS and a DFS are very different.

Here’s another example using the graph we considered earlier in this lesson.

Because a BFS is performed by gradually increasing the hops from , a spanning tree obtained from a BFS will always provide the shortest path from to any other vertex in the graph, provided that the graph is unweighted. This can be seen clearly in this example.

### Concluding words

In the next lesson, we will see how the difference between a BFS and a DFS directly relates to the type of memory structure that can be used to implement these two algorithms.

## Properties of the BFS

In order to understand this section properly, you should first study the proposed implementation of the BFS here.

When writing an algorithm, it is important to answer three questions:

- Will it terminate?
- Is it correct?
- How complex is it?

### Termination

The algorithm only visits once each vertex that can be attained. Since the number of vertices is finite, the algorithm will eventually terminate.

### Correctness

We want to show that the BFS algorithm returns a shortest path from a vertex to all other vertices that can be attained from through graph edges.

We are going to show that:

- All vertices are visited in increasing distance to .
- Obtained paths are of minimum length.
- All attainable vertices are visited.

#### All vertices are visited in increasing distance to *u*

Let’s assume by the contradiction the first moment when a vertex is added to the FIFO, and there already exists a vertex inside it such that .

Let’s note the vertex, neighbor of , that adds to the FIFO. There can be two cases:

- Either is closer to than and we reach a contradiction. Indeed, when was added, there was a vertex in the FIFO that was already farther. Therefore, this is not the first moment when this situation happens.
- Or the distance from to is the same as the one from to . We therefore conclude that is closer from than . Let be a predecessor of on a shortest path from to . has not been visited, otherwise would have added it to the FIFO. By recurrence, we see that a direct neighbor of hasn’t been added to the FIFO, thus . cannot thus be closer from than .

#### Obtained paths are of minimum length

The proof is performed by induction. We will show that, at any moment, all paths in the FIFO are of minimal length.

This is trivially true for starting vertex , as it is at distance 0 from itself.

As shown previously, all vertices are visited in increasing distance to . This ensures that adding a vertex to the FIFO is always performed by a neighbor that is a predecessor along a shortest path initiated from .

#### All attainable vertices are visited

By contradiction, let be an attainable vertex, but not yet visited, at minimal distance from . Let us consider a shortest path from to . Let be the predecessor of along that path.

Thus, has not been visited, otherwise it would have been added to the FIFO. Therefore, is attainable and not yet visited, and is closer from than . We reach a contradiction.

### Complexity

The algorithm only visits once each vertex, and checks the list of neighbors each time. Therefore, its complexity is , where is the number of edges in the graph.

## Properties of the DFS

### Termination

The algorithm only visits once each vertex that can be attained. Since the number of vertices is finite, the algorithm will eventually terminate.

### Correctness

We want to show that the algorithm is such that it will visit all the vertices attainable from .

To verify this property, we first remark that all the visited vertices during a DFS are indeed attainable. This can easily be shown by induction ( is trivially attainable from itself, then neighboring vertices are also attainable).

Then, we have to verify that all attainable vertices are visited by the DFS. Let’s proceed by contradiction. A vertex which is attainable is, by definition, such that there exists a path from to in the graph. On such a path, we consider the vertex just before . This vertex cannot be visited, otherwise would have been visited. By induction, we conclude that has not been visited, and reach a contradiction.

### Complexity

The DFS search only checks twice each edge of the graph, with a number of elemental operations which is proportional to that quantity. Therefore, complexity is , where is the number of edges in the graph.

## To go further

- Beam search: a variation of the BFS to accelerate it while sacrificing the guarantee it will find the shortest path.
- IDDFS: a mixed approach between DFS and BFS that takes the best of both worlds.