Add travelling salesman problem.

Author: trekhleb

README.md

@@ -92,6 +92,7 @@ a set of rules that precisely defines a sequence of operations.
   * [Eulerian Path and Eulerian Circuit](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/graph/eulerian-path) - Fleury's algorithm - Visit every edge exactly once
   * [Hamiltonian Cycle](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/graph/hamiltonian-cycle) - Visit every vertex exactly once
   * [Strongly Connected Components](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/graph/strongly-connected-components) - Kosaraju's algorithm
+  * [Travelling Salesman Problem](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/graph/travelling-salesman) - brute force algorithm
 * **Uncategorized**  
   * [Tower of Hanoi](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/uncategorized/hanoi-tower)
   * [N-Queens Problem](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/uncategorized/n-queens)
@@ -105,6 +106,7 @@ algorithm is an abstraction higher than a computer program.
 
 * **Brute Force** - look at all the possibilities and selects the best solution
   * [Maximum Subarray](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/sets/maximum-subarray)
+  * [Travelling Salesman Problem](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/graph/travelling-salesman)
 * **Greedy** - choose the best option at the current time, without any consideration for the future
   * [Unbound Knapsack Problem](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/sets/knapsack-problem)
   * [Dijkstra Algorithm](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/graph/dijkstra) - finding shortest path to all graph vertices

src/algorithms/graph/travelling-salesman/README.md

@@ -0,0 +1,26 @@
+# Travelling Salesman Problem
+
+The travelling salesman problem (TSP) asks the following question: 
+"Given a list of cities and the distances between each pair of 
+cities, what is the shortest possible route that visits each city 
+and returns to the origin city?"
+
+![Travelling Salesman](https://upload.wikimedia.org/wikipedia/commons/1/11/GLPK_solution_of_a_travelling_salesman_problem.svg)
+
+Solution of a travelling salesman problem: the black line shows 
+the shortest possible loop that connects every red dot.
+
+![Travelling Salesman Graph](https://upload.wikimedia.org/wikipedia/commons/3/30/Weighted_K4.svg)
+
+TSP can be modelled as an undirected weighted graph, such that 
+cities are the graph's vertices, paths are the graph's edges, 
+and a path's distance is the edge's weight. It is a minimization 
+problem starting and finishing at a specified vertex after having 
+visited each other vertex exactly once. Often, the model is a 
+complete graph (i.e. each pair of vertices is connected by an 
+edge). If no path exists between two cities, adding an arbitrarily 
+long edge will complete the graph without affecting the optimal tour.
+
+## References
+
+- [Wikipedia](https://en.wikipedia.org/wiki/Travelling_salesman_problem)

src/algorithms/graph/travelling-salesman/__test__/bfTravellingSalesman.test.js

@@ -0,0 +1,51 @@
+import GraphVertex from '../../../../data-structures/graph/GraphVertex';
+import GraphEdge from '../../../../data-structures/graph/GraphEdge';
+import Graph from '../../../../data-structures/graph/Graph';
+import bfTravellingSalesman from '../bfTravellingSalesman';
+
+describe('bfTravellingSalesman', () => {
+  it('should solve problem for simple graph', () => {
+    const vertexA = new GraphVertex('A');
+    const vertexB = new GraphVertex('B');
+    const vertexC = new GraphVertex('C');
+    const vertexD = new GraphVertex('D');
+
+    const edgeAB = new GraphEdge(vertexA, vertexB, 1);
+    const edgeBD = new GraphEdge(vertexB, vertexD, 1);
+    const edgeDC = new GraphEdge(vertexD, vertexC, 1);
+    const edgeCA = new GraphEdge(vertexC, vertexA, 1);
+
+    const edgeBA = new GraphEdge(vertexB, vertexA, 5);
+    const edgeDB = new GraphEdge(vertexD, vertexB, 8);
+    const edgeCD = new GraphEdge(vertexC, vertexD, 7);
+    const edgeAC = new GraphEdge(vertexA, vertexC, 4);
+    const edgeAD = new GraphEdge(vertexA, vertexD, 2);
+    const edgeDA = new GraphEdge(vertexD, vertexA, 3);
+    const edgeBC = new GraphEdge(vertexB, vertexC, 3);
+    const edgeCB = new GraphEdge(vertexC, vertexB, 9);
+
+    const graph = new Graph(true);
+    graph
+      .addEdge(edgeAB)
+      .addEdge(edgeBD)
+      .addEdge(edgeDC)
+      .addEdge(edgeCA)
+      .addEdge(edgeBA)
+      .addEdge(edgeDB)
+      .addEdge(edgeCD)
+      .addEdge(edgeAC)
+      .addEdge(edgeAD)
+      .addEdge(edgeDA)
+      .addEdge(edgeBC)
+      .addEdge(edgeCB);
+
+    const salesmanPath = bfTravellingSalesman(graph);
+
+    expect(salesmanPath.length).toBe(4);
+
+    expect(salesmanPath[0].getKey()).toEqual(vertexA.getKey());
+    expect(salesmanPath[1].getKey()).toEqual(vertexB.getKey());
+    expect(salesmanPath[2].getKey()).toEqual(vertexD.getKey());
+    expect(salesmanPath[3].getKey()).toEqual(vertexC.getKey());
+  });
+});

src/algorithms/graph/travelling-salesman/bfTravellingSalesman.js

@@ -0,0 +1,104 @@
+/**
+ * Get all possible paths
+ * @param {GraphVertex} startVertex
+ * @param {GraphVertex[][]} [paths]
+ * @param {GraphVertex[]} [path]
+ */
+function findAllPaths(startVertex, paths = [], path = []) {
+  // Clone path.
+  const currentPath = [...path];
+
+  // Add startVertex to the path.
+  currentPath.push(startVertex);
+
+  // Generate visited set from path.
+  const visitedSet = currentPath.reduce((accumulator, vertex) => {
+    const updatedAccumulator = { ...accumulator };
+    updatedAccumulator[vertex.getKey()] = vertex;
+
+    return updatedAccumulator;
+  }, {});
+
+  // Get all unvisited neighbors of startVertex.
+  const unvisitedNeighbors = startVertex.getNeighbors().filter((neighbor) => {
+    return !visitedSet[neighbor.getKey()];
+  });
+
+  // If there no unvisited neighbors then treat current path as complete and save it.
+  if (!unvisitedNeighbors.length) {
+    paths.push(currentPath);
+
+    return paths;
+  }
+
+  // Go through all the neighbors.
+  for (let neighborIndex = 0; neighborIndex < unvisitedNeighbors.length; neighborIndex += 1) {
+    const currentUnvisitedNeighbor = unvisitedNeighbors[neighborIndex];
+    findAllPaths(currentUnvisitedNeighbor, paths, currentPath);
+  }
+
+  return paths;
+}
+
+/**
+ * @param {number[][]} adjacencyMatrix
+ * @param {object} verticesIndices
+ * @param {GraphVertex[]} cycle
+ * @return {number}
+ */
+function getCycleWeight(adjacencyMatrix, verticesIndices, cycle) {
+  let weight = 0;
+
+  for (let cycleIndex = 1; cycleIndex < cycle.length; cycleIndex += 1) {
+    const fromVertex = cycle[cycleIndex - 1];
+    const toVertex = cycle[cycleIndex];
+    const fromVertexIndex = verticesIndices[fromVertex.getKey()];
+    const toVertexIndex = verticesIndices[toVertex.getKey()];
+    weight += adjacencyMatrix[fromVertexIndex][toVertexIndex];
+  }
+
+  return weight;
+}
+
+/**
+ * BRUTE FORCE approach to solve Traveling Salesman Problem.
+ *
+ * @param {Graph} graph
+ * @return {GraphVertex[]}
+ */
+export default function bfTravellingSalesman(graph) {
+  // Pick starting point from where we will traverse the graph.
+  const startVertex = graph.getAllVertices()[0];
+
+  // BRUTE FORCE.
+  // Generate all possible paths from startVertex.
+  const allPossiblePaths = findAllPaths(startVertex);
+
+  // Filter out paths that are not cycles.
+  const allPossibleCycles = allPossiblePaths.filter((path) => {
+    /** @var {GraphVertex} */
+    const lastVertex = path[path.length - 1];
+    const lastVertexNeighbors = lastVertex.getNeighbors();
+
+    return lastVertexNeighbors.includes(startVertex);
+  });
+
+  // Go through all possible cycles and pick the one with minimum overall tour weight.
+  const adjacencyMatrix = graph.getAdjacencyMatrix();
+  const verticesIndices = graph.getVerticesIndices();
+  let salesmanPath = [];
+  let salesmanPathWeight = null;
+  for (let cycleIndex = 0; cycleIndex < allPossibleCycles.length; cycleIndex += 1) {
+    const currentCycle = allPossibleCycles[cycleIndex];
+    const currentCycleWeight = getCycleWeight(adjacencyMatrix, verticesIndices, currentCycle);
+
+    // If current cycle weight is smaller then previous ones treat current cycle as most optimal.
+    if (salesmanPathWeight === null || currentCycleWeight < salesmanPathWeight) {
+      salesmanPath = currentCycle;
+      salesmanPathWeight = currentCycleWeight;
+    }
+  }
+
+  // Return the solution.
+  return salesmanPath;
+}