[Algorithm] Breadth First JavaScript Search Algorithm for Graphs
Breadth first search is a graph search algorithm that starts at one node and visits neighboring nodes as widely as possible before going further down any other path. This algorithm requires the use of a queue to keep track of which nodes to visit, so it might be worth your time to brush up on that data structure before watching this lesson.
const {createQueue} = require('./queue');
function createNode(key) {
let children = [];
return {
key,
children,
addChild(child) {
children.push(child)
}
}
}
function createGraph(directed = false) {
const nodes = [];
const edges = [];
return {
nodes,
edges,
directed,
addNode(key) {
nodes.push(createNode(key))
},
getNode (key) {
return nodes.find(n => n.key === key)
},
addEdge (node1Key, node2Key) {
const node1 = this.getNode(node1Key);
const node2 = this.getNode(node2Key);
node1.addChild(node2);
if (!directed) {
node2.addChild(node1);
}
edges.push(`${node1Key}${node2Key}`)
},
print() {
return nodes.map(({children, key}) => {
let result = `${key}`;
if (children.length) {
result += ` => ${children.map(n => n.key).join(' ')}`
}
return result;
}).join('\n')
},
/**
* Breadth First Search
*/
bfs (startNodeKey = "", visitFn = () => {}) {
/**
* Keytake away:
* 1. Using Queue to get next visit node
* 2. Enqueue the node's children for next run
* 3. Hashed visited map for keep tracking visited node
*/
const startNode = this.getNode(startNodeKey);
// create a hashed map to check whether one node has been visited
const visited = this.nodes.reduce((acc, curr) => {
acc[curr.key] = false;
return acc;
}, {});
// Create a queue to put all the nodes to be visited
const queue = createQueue();
queue.enqueue(startNode);
// start process
while (!queue.isEmpty()) {
const current = queue.dequeue();
// check wheather the node exists in hashed map
if (!visited[current.key]) {
visitFn(current);
visited[current.key] = true;
// process the node's children
current.children.map(n => {
if (!visited[n.key]) {
queue.enqueue(n);
}
});
}
}
}
}
}
const graph = createGraph(true)
graph.addNode('Kyle')
graph.addNode('Anna')
graph.addNode('Krios')
graph.addNode('Tali')
graph.addEdge('Kyle', 'Anna')
graph.addEdge('Anna', 'Kyle')
graph.addEdge('Kyle', 'Krios')
graph.addEdge('Kyle', 'Tali')
graph.addEdge('Anna', 'Krios')
graph.addEdge('Anna', 'Tali')
graph.addEdge('Krios', 'Anna')
graph.addEdge('Tali', 'Kyle')
console.log(graph.print())
const nodes = ['a', 'b', 'c', 'd', 'e', 'f']
const edges = [
['a', 'b'],
['a', 'e'],
['a', 'f'],
['b', 'd'],
['b', 'e'],
['c', 'b'],
['d', 'c'],
['d', 'e']
]
const graph2 = createGraph(true)
nodes.forEach(node => {
graph2.addNode(node)
})
edges.forEach(nodes => {
graph2.addEdge(...nodes)
})
graph2.bfs('a', node => {
console.log(node.key) //a,b,e,f,d,c
})
A more general function:
bfs (startNodeKey, predFn = () => {}, cb = () => {}) {
const startNode = this.getNode(startNodeKey);
const visited = createVistedMap(this.nodes);
const queue = createQueue();
startNode.children.forEach((n) => {
queue.enqueue(n);
});
while (!queue.isEmpty()) {
const current = queue.dequeue();
if (!visited[current.key]) {
if (predFn(current)) return cb(current);
else {
visited[current.key] = true;
}
}
}
cb(null)
},
let graph3 = createGraph(true)
const tyler = {key: 'tyler', dog: false};
const henry = {key: 'henry', dog: false};
const john = {key: 'john', dog: false};
const aimee = {key: 'aimee', dog: true};
const peggy = {key: 'peggy', dog: false};
const keli = {key: 'keli', dog: false};
const claire = {key: 'claire', dog: false}; graph3.addNode('tyler', tyler);
graph3.addNode('henry', henry);
graph3.addNode('john', john);
graph3.addNode('claire', claire);
graph3.addNode('aimee', aimee);
graph3.addNode('peggy', peggy)
graph3.addNode('keli', keli); graph3.addEdge('tyler', 'henry')
graph3.addEdge('tyler', 'john')
graph3.addEdge('tyler', 'aimee')
graph3.addEdge('henry', 'keli')
graph3.addEdge('henry', 'peggy')
graph3.addEdge('john', 'john')
graph3.addEdge('keli', 'claire') graph3.bfs2('tyler', (node) => {
return node.dog;
}, (node) => {
if (node) console.log(`${node.key} has a dog`)
else console.log('Tyler friends has no dog')
})
Time Complexity: O(V+E) where V is number of vertices in the graph and E is number of edges in the graph.
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