I have been searching for weeks for a way to compute shortest paths in JavaScript. I have been playing with the book Data Structures and Algorithms by Groner (aptly named) at https://github.com/loiane/javascript-datastructures-algorithms/tree/master/chapter09.
The trouble I keep on finding is that the code is so customized that it is almost impossible to rewrite to produce the desired results. I would like to be able to get the shortest path from any given vertex to any other, rather than, as Groner codes it, just the list of everything from A. I would like to be able to get, for example, the path from F to B, or from C to A.
The full code is here: http://jsfiddle.net/8cn7e2x8/
Can anyone help?
var graph = new Graph();
var myVertices = ['A','B','C','D','E','F'];
for (var i=0; i<myVertices.length; i++) {
graph.addVertex(myVertices[i]);
}
graph.addEdge('A', 'B');
graph.addEdge('B', 'C');
graph.addEdge('B', 'E');
graph.addEdge('C', 'D');
graph.addEdge('C', 'E');
graph.addEdge('C', 'G');
graph.addEdge('D', 'E');
graph.addEdge('E', 'F');
graph.dfs();
console.log('********* sortest path - BFS ***********');
var shortestPathA = graph.BFS(myVertices[0]);
//from A to all other vertices
var fromVertex = myVertices[0];
for (i = 1; i < myVertices.length; i++) {
var toVertex = myVertices[i],
path = new Stack();
for (var v = toVertex; v !== fromVertex; v = shortestPathA.predecessors[v]) {
path.push(v);
}
path.push(fromVertex);
var s = path.pop();
while (!path.isEmpty()) {
s += ' - ' + path.pop();
}
console.log(s);
}
Let us begin by remarking that breadth-first search (BFS) computes the shortest paths from a given source vertex if the graph is unweighted. In other words, we consider the length of a path to be the number of edges in the path.
Here is an easy way to construct an unweighted graph:
function Graph() {
var neighbors = this.neighbors = {}; // Key = vertex, value = array of neighbors.
this.addEdge = function (u, v) {
if (neighbors[u] === undefined) { // Add the edge u -> v.
neighbors[u] = [];
}
neighbors[u].push(v);
if (neighbors[v] === undefined) { // Also add the edge v -> u so as
neighbors[v] = []; // to implement an undirected graph.
} // For a directed graph, delete
neighbors[v].push(u); // these four lines.
};
return this;
}
Note that we have implemented an undirected graph. As mentioned in the inline comments, you can modify the code to construct a directed graph by deleting four lines from the addEdge
function.
This implementation of BFS would work equally well on a directed graph:
function bfs(graph, source) {
var queue = [ { vertex: source, count: 0 } ],
visited = { source: true },
tail = 0;
while (tail < queue.length) {
var u = queue[tail].vertex,
count = queue[tail++].count; // Pop a vertex off the queue.
print('distance from ' + source + ' to ' + u + ': ' + count);
graph.neighbors[u].forEach(function (v) {
if (!visited[v]) {
visited[v] = true;
queue.push({ vertex: v, count: count + 1 });
}
});
}
}
To find the shortest path between two given vertices and display the vertices along the path, we have to remember the predecessor of each vertex as we explore the graph:
function shortestPath(graph, source, target) {
if (source == target) { // Delete these four lines if
print(source); // you want to look for a cycle
return; // when the source is equal to
} // the target.
var queue = [ source ],
visited = { source: true },
predecessor = {},
tail = 0;
while (tail < queue.length) {
var u = queue[tail++], // Pop a vertex off the queue.
neighbors = graph.neighbors[u];
for (var i = 0; i < neighbors.length; ++i) {
var v = neighbors[i];
if (visited[v]) {
continue;
}
visited[v] = true;
if (v === target) { // Check if the path is complete.
var path = [ v ]; // If so, backtrack through the path.
while (u !== source) {
path.push(u);
u = predecessor[u];
}
path.push(u);
path.reverse();
print(path.join(' → '));
return;
}
predecessor[v] = u;
queue.push(v);
}
}
print('there is no path from ' + source + ' to ' + target);
}
The following snippet demonstrates these operations on the graph that you gave in your question. First we find the shortest paths to all vertices reachable from A
. Then we find the shortest path from B
to G
and from G
to A
.
function Graph() {
var neighbors = this.neighbors = {}; // Key = vertex, value = array of neighbors.
this.addEdge = function (u, v) {
if (neighbors[u] === undefined) { // Add the edge u -> v.
neighbors[u] = [];
}
neighbors[u].push(v);
if (neighbors[v] === undefined) { // Also add the edge v -> u in order
neighbors[v] = []; // to implement an undirected graph.
} // For a directed graph, delete
neighbors[v].push(u); // these four lines.
};
return this;
}
function bfs(graph, source) {
var queue = [ { vertex: source, count: 0 } ],
visited = { source: true },
tail = 0;
while (tail < queue.length) {
var u = queue[tail].vertex,
count = queue[tail++].count; // Pop a vertex off the queue.
print('distance from ' + source + ' to ' + u + ': ' + count);
graph.neighbors[u].forEach(function (v) {
if (!visited[v]) {
visited[v] = true;
queue.push({ vertex: v, count: count + 1 });
}
});
}
}
function shortestPath(graph, source, target) {
if (source == target) { // Delete these four lines if
print(source); // you want to look for a cycle
return; // when the source is equal to
} // the target.
var queue = [ source ],
visited = { source: true },
predecessor = {},
tail = 0;
while (tail < queue.length) {
var u = queue[tail++], // Pop a vertex off the queue.
neighbors = graph.neighbors[u];
for (var i = 0; i < neighbors.length; ++i) {
var v = neighbors[i];
if (visited[v]) {
continue;
}
visited[v] = true;
if (v === target) { // Check if the path is complete.
var path = [ v ]; // If so, backtrack through the path.
while (u !== source) {
path.push(u);
u = predecessor[u];
}
path.push(u);
path.reverse();
print(path.join(' → '));
return;
}
predecessor[v] = u;
queue.push(v);
}
}
print('there is no path from ' + source + ' to ' + target);
}
function print(s) { // A quick and dirty way to display output.
s = s || '';
document.getElementById('display').innerHTML += s + '<br>';
}
window.onload = function () {
var graph = new Graph();
graph.addEdge('A', 'B');
graph.addEdge('B', 'C');
graph.addEdge('B', 'E');
graph.addEdge('C', 'D');
graph.addEdge('C', 'E');
graph.addEdge('C', 'G');
graph.addEdge('D', 'E');
graph.addEdge('E', 'F');
bfs(graph, 'A');
print();
shortestPath(graph, 'B', 'G');
print();
shortestPath(graph, 'G', 'A');
};
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