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Data Structures in JavaScript: Shortest Path Graph Traversal

nielsenjared profile image Jared Nielsen Originally published at jarednielsen.com ใƒป9 min read

At some point in your career (today?!) you will want to learn data structures. It's not just to ace the technical interview and land your dream job. Learning data structures will help you understand how software works and improve your problem-solving skills. In this tutorial, you will learn the breadth-first search (BFS) algorithm with graph data structures in JavaScript.

This article originally published on jarednielsen.com

If you're just joining us, you may want to start with Data Structures in JavaScript: Graph.

If you're new to data structures, you may want to start with Data Structures in JavaScript: Array

Retrieval Practice

Retrieval practice is the surest way to solidify any new learning. Attempt to answer the following questions before proceeding:

  • What is Breadth-First Search?

  • What is the difference between Breadth-First Search and Depth-First Search?

  • What problem(s) does Breadth-First Search solve?

What is Breadth-First Search?

Breadth-First Search is an algorithm that searches a graph for a specific goal by checking all of the edges connected to a vertex before moving on to check the edges of the connected vertices.

What is the Difference Between Breadth-First Search and Depth-First Search?

Breadth-First Search checks all of the vertices adjacent to a given vertex before checking the vertices adjacent to those vertices. Depth-First Search, on the other hand, checks all of the vertices on a path and then backtracks.

What Problem(s) Does Breadth-First Search Solve?

There are a number of specific use cases, such as the Ford-Fulkerson or Cheney's algorithm, for breadth-first search algorithms, but a general application is to find the shortest, or most efficient, path between two vertices.

Let's Get Meta

Programming is problem solving. Both are metacognitive activities. To excel, we want to improve our thinking about thinking. Ask yourself the following questions and keep them back of mind as you proceed:

  • What is pattern recognition?

  • Why is Breadth-First Search used to find the shortest-path?

  • What is a predecessor?

Shortest-Path Graph Traversal in JavaScript

Let's declare our Graph data structure.

class Graph {
    constructor() {
        this.vertices = [];
        this.adjacent = {};
        this.edges = 0;
    }

    addVertex(v) {
        this.vertices.push(v);
        this.adjacent[v] = [];
    }

    addEdge(v, w) {
        this.adjacent[v].push(w);
        this.adjacent[w].push(v);
        this.edges++;
    }

    bfs(goal, root = this.vertices[0]) {
        let adj = this.adjacent;

        const queue = [];
        queue.push(root);

        const discovered = [];
        discovered[root] = true;

        while(queue.length) {
            let v = queue.shift();
            console.log(v);

            if (v === goal) {
                return true;
            }

            for (let i = 0; i < adj[v].length; i++) {
                if (!discovered[adj[v][i]]) {
                    discovered[adj[v][i]] = true;
                    queue.push(adj[v][i]);
                }
            }
        }

        return false;
    }
}
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Next, let's initialize a new Graph and add vertices and edges.

const g = new Graph();

g.addVertex("A");
g.addVertex("B");
g.addVertex("C");
g.addVertex("D");
g.addVertex("E");
g.addVertex("F");
g.addVertex("G");

g.addEdge("A","B");
g.addEdge("A","C");
g.addEdge("A","D");
g.addEdge("B","C");
g.addEdge("B","D");
g.addEdge("C","D");
g.addEdge("C","E");
g.addEdge("D","F");
g.addEdge("F","G");
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Now what?

What do we need our bfs method to return if we want to know the shortest path between the root and the goal?

๐Ÿค”

The vertices and the number of edges that constitute the path.

What data type (or structure) do we want to use to store these values?

It starts with 'a' and rhymes with 'ray'...

To our bfs method, let's add an edges array and initialize our edges array with a key root assigned a value of 0. Why? Because the distance from our root to itself is 0, and we want to be sure we account for all edge cases (no pun intended). We also want to return our edges array if and when we find our goal.

    bfs(goal, root = this.vertices[0]) {
        let adj = this.adjacent;

        const queue = [];
        queue.push(root);

        const discovered = [];
        discovered[root] = true;

        //add edges array and initialize it with root
        const edges = [];
        edges[root] = 0;

        while(queue.length) {
            let v = queue.shift();

            //return edges array if we find our goal
            if (v === goal) {
                return edges;
            }

            for (let i = 0; i < adj[v].length; i++) {
                if (!discovered[adj[v][i]]) {
                    discovered[adj[v][i]] = true;
                    queue.push(adj[v][i]);
                }
            }
        }

        return false;
    }
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Verify that your method works by logging g.bfs("A"));.

Now what?

Let's restate our goal:

Given a graph, a root, and a goal, find the shortest path from the root to the goal.

Let's look at the diagram of our graph...

graph diagram

If we want to find the shortest path between A and E, what do we need to do?

We need to count the edges!

If our bfs is returning our edges array, what does edges need to look like?

[ A: 0, B: 1, C: 1, D: 1, E: 2, F: 2, G: 3]
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Let's write pseudocode!

  • If (while) there are vertices in the queue, check the first vertex.

  • If the first vertex is equal to our goal, return our edge counts.

  • Otherwise, check the vertices adjacent to our vertex.

  • If the adjacent vertices are new to us (not discovered), label them "discovered".

  • Then add the newly discovered vertex to the queue to check later.

  • Then "count" the edges between our starting vertex and our newly discovered vertex.

What do I mean by "count"?

Where have we seen this or something like it before?

What is a very common pattern in iteration?

Incrementation (if that's a word).

count++
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Or:

count += 1
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Let's think this through...

The first vertex we will iterate over is A. We know the value of A in edges is 0. We know the distance between A and it's adjacent vertices, B, C, and D, is 1. We want to add B, C, and D to our edges array and assign them a value of 1.

On the next iteration of our while loop, the value of B in edges will be 1, but the adjacent vertices will be labeled "discovered", so we will skip them and move on to C, where the value stored in edges is also 1. This time, however, E is not "discovered", so we now label it "discovered", push it to the queue and add it to our edges array.

But! It's two edges away from our root? How do we get that value?

From C!

AKA edges[v].

We simply add 1 to the value associated with our current vertex.

            for (let i = 0; i < adj[v].length; i++) {
                if (!discovered[adj[v][i]]) {
                    discovered[adj[v][i]] = true;
                    queue.push(adj[v][i]);
                    edges[adj[v][i]] = edges[v] + 1;
                }
            }
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Passing G to our bfs method, g.bfs("G"), will return the following:

[
  A: 0, B: 1,
  C: 1, D: 1,
  E: 2, F: 2,
  G: 3
]
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If we just want the shortest path to our goal, we can modify our conditional return:

            if (v === goal) {
                return edges[goal];
            }
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Now we need to return the vertices along the path.

Where have we seen this or something like it before?

๐Ÿค”

We can follow the same pattern we used to label vertices "discovered" and to count edges.

    bfs(goal, root = this.vertices[0]) {
        let adj = this.adjacent;

        const queue = [];
        queue.push(root);

        const discovered = [];
        discovered[root] = true;

        const edges = [];
        edges[root] = 0;

        //add vertices array and initialize it with root
        const predecessors = [];
        predecessors[root] = null;

        while(queue.length) {
            let v = queue.shift();

            //refactor our conditional to return both edges & vertices
            if (v === goal) {
                return { 
                    edges,
                    predecessors
                };
            }

            for (let i = 0; i < adj[v].length; i++) {
                if (!discovered[adj[v][i]]) {
                    discovered[adj[v][i]] = true;
                    queue.push(adj[v][i]);
                    edges[adj[v][i]] = edges[v] + 1;
                    //create a key in vertices array with the current vertex
                    //assign it a value of its predecessor
                    predecessors[adj[v][i]] = v;
                }
            }
        }

        return false;
    }
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We declare a predecessors array and create a key with root assigned a value of null. Why null? Because our root doesn't creat a path to itself. We next modify our conditional statement to return both edges and predecessors. Finally, within our while loop, for each adjacent vertex, we add a key to our predecessors array and assign it the value of the predecessor.

Our bfs method, g.bfs("G"), will return the following:

{
  edges: [
    A: 0, B: 1,
    C: 1, D: 1,
    E: 2, F: 2,
    G: 3
  ],
  predecessors: [ A: null, B: 'A', C: 'A', D: 'A', E: 'C', F: 'D', G: 'F' ]
}
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That's not particularly useful, though, is it?

What do we need to do?

We need to write pseudocode!

  • Starting with the goal.

  • Look up its predcessor.

  • Push it to an array.

  • What is the predcessor of the predecessor?

  • Look it up and push it to the array.

  • Repeat until we return to the root.

  • Then reverse the array or pop elements out into a string.

There are several approaches we can take to implement this. We could add a method to our class, or we could build the path outside the class, but my preference is to add a helper function in our bfs method. Let's call it buildPath. What do we want buildPath to return? The path, natch. If we know, as we outlined above, that we need to work with our goal and our root, let's pass those variables to our function along with predecessors. Here's our bfs method:

    bfs(goal, root = this.vertices[0]) {
        let adj = this.adjacent;

        const queue = [];
        queue.push(root);

        const discovered = [];
        discovered[root] = true;

        const edges = [];
        edges[root] = 0;

        const predecessors = [];
        predecessors[root] = null;

        //declare buildPath function
        const buildPath = (goal, root, predecessors) => {
            return path;
        }

        while(queue.length) {
            let v = queue.shift();

            if (v === goal) {
                return { 
                    distance: edges[goal],
                    path: buildPath(goal, root, predecessors)
                };
            }

            for (let i = 0; i < adj[v].length; i++) {
                if (!discovered[adj[v][i]]) {
                    discovered[adj[v][i]] = true;
                    queue.push(adj[v][i]);
                    edges[adj[v][i]] = edges[v] + 1;
                    predecessors[adj[v][i]] = v;
                }
            }
        }

        return false;
    }
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How do we 'look up' the predecessor of our goal?

predecessors[goal];
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What is the predecessor of the predecessor?

predecessors[predecessors[goal]];
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And what's the predecesor of the predecessor of the predecessor?

๐Ÿคจ

While we could go down that rabbit hole, let's find an algorithmic approach:

            let u = predecessors[goal];

            while(u != root) {
                u = predecessors[u];
            }

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Why u?

There is only one like u in the world.

๐Ÿ™„

In graph theory, u is often used for the vertex that precedes a given vertex, v. Here, we use u as a temporary variable to store the value of a given predecessor. With each iteration of our while loop, we reassign the value of u with the predecessor of u. We repeat until u is equal to the value of our root.

That's the crux of it.

What remains?

Our array. Let's name it stack, as a classic implementation of this algorithm uses a Stack data structure for its LIFO semantics. Here's our complete buildPath helper function:

        const buildPath = (goal, root, predecessors) => {
            //declare and initialize a "stack"
            const stack = [];
            //push our goal to the stack
            stack.push(goal);

            let u = predecessors[goal];

            while(u != root) {
                //push each predecssor to the stack
                stack.push(u);
                u = predecessors[u];
            }

            //put the cherry on top
            stack.push(root);

            //LIFO
            let path = stack.reverse().join('-');

            return path;
        }
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Lastly, we need to call our helper function in our conditional statement:

        while(queue.length) {
            let v = queue.shift();

            //refactor to return distance and path
            if (v === goal) {
                return { 
                    distance: edges[goal],
                    path: buildPath(goal, root, predecessors)
                };
            }

            for (let i = 0; i < adj[v].length; i++) {
                if (!discovered[adj[v][i]]) {
                    discovered[adj[v][i]] = true;
                    queue.push(adj[v][i]);
                    edges[adj[v][i]] = edges[v] + 1;
                    predecessors[adj[v][i]] = v;
                }
            }
        }
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Here's our complete bfs method:

    bfs(goal, root = this.vertices[0]) {
        let adj = this.adjacent;

        const queue = [];
        queue.push(root);

        const discovered = [];
        discovered[root] = true;

        const edges = [];
        edges[root] = 0;

        const predecessors = [];
        predecessors[root] = null;

        const buildPath = (goal, root, predecessors) => {
            const stack = [];
            stack.push(goal);

            let u = predecessors[goal];

            while(u != root) {
                stack.push(u);
                u = predecessors[u];
            }

            stack.push(root);

            let path = stack.reverse().join('-');

            return path;
        }


        while(queue.length) {
            let v = queue.shift();

            if (v === goal) {
                return { 
                    distance: edges[goal],
                    path: buildPath(goal, root, predecessors)
                };
            }

            for (let i = 0; i < adj[v].length; i++) {
                if (!discovered[adj[v][i]]) {
                    discovered[adj[v][i]] = true;
                    queue.push(adj[v][i]);
                    edges[adj[v][i]] = edges[v] + 1;
                    predecessors[adj[v][i]] = v;
                }
            }
        }

        return false;
    }
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And now when we call g.bfs("G", "A"), it returns:

{ distance: 3, path: 'A-D-F-G' }
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Can we do better?

Absolutely.

We'll break down classic shortest-path algorithms, such as Djikstra's and the Floyd-Warshall, in the future. Stay tuned!

Reflection

What are your answers to the following questions?

  • What is pattern recognition?

  • Why is Breadth-First Search used to find the shortest-path?

  • What is a predecessor?

What is Pattern Recognition?

According to Wikipedia:

In psychology and cognitive neuroscience, pattern recognition describes cognitive process that matches information from a stimulus with information retrieved from memory

Pattern recognition is an important skill to develop as a programmer. It allows us to quickly and easily recognize opportunities to solve problems with algorithms and automation.

Why is Breadth-First Search Used to Find the Shortest Path?

BFS was always already searching the shortest path. We simply need to record the distances between vertices and the paths taken.

What is a Predecessor?

According to Wikipedia, in graph theory a predecessor is:

A vertex coming before a given vertex in a directed path.

The graph we used in this tutorial was not directed, but we can treat it as such because we are only seaching in one direction.


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