chore: Added BellmanFord (#679)

* Added BellmanFord

* Add References for BellmanFord

* Style code using standard.js

* Add tests and modify code

* Fixed BellmanFord test file

* Add BellmanFord and tests
This commit is contained in:
Mayank Mamgain
2021-09-09 16:46:52 +05:30
committed by GitHub
parent 8d72872738
commit d49cf9fead
2 changed files with 90 additions and 0 deletions

56
Graphs/BellmanFord.js Normal file
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/*
The BellmanFord algorithm is an algorithm that computes shortest paths
from a single source vertex to all of the other vertices in a weighted digraph.
It also detects negative weight cycle.
Complexity:
Worst-case performance O(VE)
Best-case performance O(E)
Worst-case space complexity O(V)
Reference:
https://en.wikipedia.org/wiki/BellmanFord_algorithm
https://cp-algorithms.com/graph/bellman_ford.html
*/
/**
*
* @param graph Graph in the format (u, v, w) where
* the edge is from vertex u to v. And weight
* of the edge is w.
* @param V Number of vertices in graph
* @param E Number of edges in graph
* @param src Starting node
* @param dest Destination node
* @returns Shortest distance from source to destination
*/
function BellmanFord (graph, V, E, src, dest) {
// Initialize distance of all vertices as infinite.
const dis = Array(V).fill(Infinity)
// initialize distance of source as 0
dis[src] = 0
// Relax all edges |V| - 1 times. A simple
// shortest path from src to any other
// vertex can have at-most |V| - 1 edges
for (let i = 0; i < V - 1; i++) {
for (let j = 0; j < E; j++) {
if ((dis[graph[j][0]] + graph[j][2]) < dis[graph[j][1]]) { dis[graph[j][1]] = dis[graph[j][0]] + graph[j][2] }
}
}
// check for negative-weight cycles.
for (let i = 0; i < E; i++) {
const x = graph[i][0]
const y = graph[i][1]
const weight = graph[i][2]
if ((dis[x] !== Infinity) && (dis[x] + weight < dis[y])) {
return null
}
}
for (let i = 0; i < V; i++) {
if (i === dest) return dis[i]
}
}
export { BellmanFord }

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import { BellmanFord } from '../BellmanFord.js'
test('Test Case 1', () => {
const V = 5
const E = 8
const destination = 3
const graph = [[0, 1, -1], [0, 2, 4],
[1, 2, 3], [1, 3, 2],
[1, 4, 2], [3, 2, 5],
[3, 1, 1], [4, 3, -3]]
const dist = BellmanFord(graph, V, E, 0, destination)
expect(dist).toBe(-2)
})
test('Test Case 2', () => {
const V = 6
const E = 9
const destination = 4
const graph = [[0, 1, 3], [0, 3, 6],
[0, 5, -1], [1, 2, -3],
[1, 4, -2], [5, 2, 5],
[2, 3, 1], [4, 3, 5], [5, 4, 2]]
const dist = BellmanFord(graph, V, E, 0, destination)
expect(dist).toBe(1)
})
test('Test Case 3', () => {
const V = 4
const E = 5
const destination = 1
const graph = [[0, 3, -1], [0, 2, 4],
[3, 2, 2], [3, 1, 5],
[2, 1, -1]]
const dist = BellmanFord(graph, V, E, 0, destination)
expect(dist).toBe(0)
})