Travelling Salesman Problem (TSP) : Given a set of cities and distances between every pair of cities, the problem is to find the shortest possible route that visits every city exactly once and returns to the starting point.
Note the difference between Hamiltonian Cycle and TSP. The Hamiltonian cycle problem is to find if there exists a tour that visits every city exactly once. Here we know that Hamiltonian Tour exists (because the graph is complete) and in fact, many such tours exist, the problem is to find a minimum weight Hamiltonian Cycle.
For example, consider the graph shown in the figure on the right side. A TSP tour in the graph is 1-2-4-3-1. The cost of the tour is 10+25+30+15 which is 80.
The problem is a famous NP-hard problem. There is no polynomial-time known solution for this problem.
Examples:
Output of Given Graph: minimum weight Hamiltonian Cycle : 10 + 25 + 30 + 15 := 80
In this post, the implementation of a simple solution is discussed.
- Consider city 1 as the starting and ending point. Since the route is cyclic, we can consider any point as a starting point.
- Generate all (n-1)! permutations of cities.
- Calculate the cost of every permutation and keep track of the minimum cost permutation.
- Return the permutation with minimum cost.
Below is the implementation of the above idea
C++
// CPP program to implement traveling salesman// problem using naive approach.#include <bits/stdc++.h>using namespace std;#define V 4// implementation of traveling Salesman Problemint travllingSalesmanProblem(int graph[][V], int s){ // store all vertex apart from source vertex vector<int> vertex; for (int i = 0; i < V; i++) if (i != s) vertex.push_back(i); // store minimum weight Hamiltonian Cycle. int min_path = INT_MAX; do { // store current Path weight(cost) int current_pathweight = 0; // compute current path weight int k = s; for (int i = 0; i < vertex.size(); i++) { current_pathweight += graph[k][vertex[i]]; k = vertex[i]; } current_pathweight += graph[k][s]; // update minimum min_path = min(min_path, current_pathweight); } while ( next_permutation(vertex.begin(), vertex.end())); return min_path;}// Driver Codeint main(){ // matrix representation of graph int graph[][V] = { { 0, 10, 15, 20 }, { 10, 0, 35, 25 }, { 15, 35, 0, 30 }, { 20, 25, 30, 0 } }; int s = 0; cout << travllingSalesmanProblem(graph, s) << endl; return 0;} |
Java
// Java program to implement // traveling salesman problem // using naive approach.import java.util.*;class GFG{ static int V = 4;// implementation of traveling // Salesman Problemstatic int travllingSalesmanProblem(int graph[][], int s){ // store all vertex apart // from source vertex ArrayList<Integer> vertex = new ArrayList<Integer>(); for (int i = 0; i < V; i++) if (i != s) vertex.add(i); // store minimum weight // Hamiltonian Cycle. int min_path = Integer.MAX_VALUE; do { // store current Path weight(cost) int current_pathweight = 0; // compute current path weight int k = s; for (int i = 0; i < vertex.size(); i++) { current_pathweight += graph[k][vertex.get(i)]; k = vertex.get(i); } current_pathweight += graph[k][s]; // update minimum min_path = Math.min(min_path, current_pathweight); } while (findNextPermutation(vertex)); return min_path;}// Function to swap the data // present in the left and right indices public static ArrayList<Integer> swap( ArrayList<Integer> data, int left, int right) { // Swap the data int temp = data.get(left); data.set(left, data.get(right)); data.set(right, temp); // Return the updated array return data; } // Function to reverse the sub-array // starting from left to the right // both inclusive public static ArrayList<Integer> reverse( ArrayList<Integer> data, int left, int right) { // Reverse the sub-array while (left < right) { int temp = data.get(left); data.set(left++, data.get(right)); data.set(right--, temp); } // Return the updated array return data; } // Function to find the next permutation // of the given integer array public static boolean findNextPermutation( ArrayList<Integer> data) { // If the given dataset is empty // or contains only one element // next_permutation is not possible if (data.size() <= 1) return false; int last = data.size() - 2; // find the longest non-increasing // suffix and find the pivot while (last >= 0) { if (data.get(last) < data.get(last + 1)) { break; } last--; } // If there is no increasing pair // there is no higher order permutation if (last < 0) return false; int nextGreater = data.size() - 1; // Find the rightmost successor // to the pivot for (int i = data.size() - 1; i > last; i--) { if (data.get(i) > data.get(last)) { nextGreater = i; break; } } // Swap the successor and // the pivot data = swap(data, nextGreater, last); // Reverse the suffix data = reverse(data, last + 1, data.size() - 1); // Return true as the // next_permutation is done return true; } // Driver Codepublic static void main(String args[]){ // matrix representation of graph int graph[][] = {{0, 10, 15, 20}, {10, 0, 35, 25}, {15, 35, 0, 30}, {20, 25, 30, 0}}; int s = 0; System.out.println( travllingSalesmanProblem(graph, s));}}// This code is contributed by adityapande88 |
Python3
# Python3 program to implement traveling salesman # problem using naive approach. from sys import maxsize from itertools import permutationsV = 4# implementation of traveling Salesman Problem def travellingSalesmanProblem(graph, s): # store all vertex apart from source vertex vertex = [] for i in range(V): if i != s: vertex.append(i) # store minimum weight Hamiltonian Cycle min_path = maxsize next_permutation=permutations(vertex) for i in next_permutation: # store current Path weight(cost) current_pathweight = 0 # compute current path weight k = s for j in i: current_pathweight += graph[k][j] k = j current_pathweight += graph[k][s] # update minimum min_path = min(min_path, current_pathweight) return min_path # Driver Code if __name__ == "__main__": # matrix representation of graph graph = [[0, 10, 15, 20], [10, 0, 35, 25], [15, 35, 0, 30], [20, 25, 30, 0]] s = 0 print(travellingSalesmanProblem(graph, s)) |
C#
// C# program to implement// traveling salesman problem// using naive approach.using System;using System.Collections.Generic;class GFG { static int V = 4; // implementation of traveling Salesman Problem static int travllingSalesmanProblem(int[, ] graph, int s) { List<int> vertex = new List<int>(); for (int i = 0; i < V; i++) if (i != s) vertex.Add(i); // store minimum weight // Hamiltonian Cycle. int min_path = Int32.MaxValue; do { // store current Path weight(cost) int current_pathweight = 0; int k = s; // compute current path weight for (int i = 0; i < vertex.Count; i++) { current_pathweight += graph[k, vertex[i]]; k = vertex[i]; } current_pathweight += graph[k, s]; // update minimum min_path = Math.Min(min_path, current_pathweight); } while (findNextPermutation(vertex)); return min_path; } // Function to swap the data resent in the left and // right indices public static List<int> swap(List<int> data, int left, int right) { // Swap the data int temp = data[left]; data[left] = data[right]; data[right] = temp; // Return the updated array return data; } // Function to reverse the sub-array starting from left // to the right both inclusive public static List<int> reverse(List<int> data, int left, int right) { // Reverse the sub-array while (left < right) { int temp = data[left]; data[left++] = data[right]; data[right--] = temp; } // Return the updated array return data; } // Function to find the next permutation of the given // integer array public static bool findNextPermutation(List<int> data) { // If the given dataset is empty // or contains only one element // next_permutation is not possible if (data.Count <= 1) return false; int last = data.Count - 2; // find the longest non-increasing // suffix and find the pivot while (last >= 0) { if (data[last] < data[last + 1]) break; last--; } // If there is no increasing pair // there is no higher order permutation if (last < 0) return false; int nextGreater = data.Count - 1; // Find the rightmost successor // to the pivot for (int i = data.Count - 1; i > last; i--) { if (data[i] > data[last]) { nextGreater = i; break; } } // Swap the successor and // the pivot data = swap(data, nextGreater, last); // Reverse the suffix data = reverse(data, last + 1, data.Count - 1); // Return true as the // next_permutation is done return true; } // Driver Code public static void Main(string[] args) { // matrix representation of graph int[, ] graph = new int[4, 4] { { 0, 10, 15, 20 }, { 10, 0, 35, 25 }, { 15, 35, 0, 30 }, { 20, 25, 30, 45 } }; int s = 0; Console.WriteLine( travllingSalesmanProblem(graph, s)); }}// This code is contributed by Tapesh(tapeshdua420) |
Javascript
// JavaScript program to implement// traveling salesman problem// using naive approach.const V = 4;// implementation of traveling// Salesman Problemconst travllingSalesmanProblem = (graph, s) => { // store all vertex apart // from source vertex let vertex = []; for (let i = 0; i < V; i++) { if (i !== s) { vertex.push(i); } } // store minimum weight // Hamiltonian Cycle. let min_path = Number.MAX_VALUE; do { // store current Path weight(cost) let current_pathweight = 0; // compute current path weight let k = s; for (let i = 0; i < vertex.length; i++) { current_pathweight += graph[k][vertex[i]]; k = vertex[i]; } current_pathweight += graph[k][s]; // update minimum min_path = Math.min(min_path, current_pathweight); } while (findNextPermutation(vertex)); return min_path;}// Function to swap the data// present in the left and right indicesconst swap = (data, left, right) => { // Swap the data let temp = data[left]; data[left] = data[right]; data[right] = temp; // Return the updated array return data;}// Function to reverse the sub-array// starting from left to the right// both inclusiveconst reverse = (data, left, right) => { // Reverse the sub-array while (left < right) { let temp = data[left]; data[left++] = data[right]; data[right--] = temp; } // Return the updated array return data;}// Function to find the next permutation// of the given integer arrayconst findNextPermutation = (data) => { // If the given dataset is empty // or contains only one element // next_permutation is not possible if (data.length <= 1) { return false; } let last = data.length - 2; // find the longest non-increasing // suffix and find the pivot while (last >= 0) { if (data[last] < data[last + 1]) { break; } last--; } // If there is no increasing pair // there is no higher order permutation if (last < 0) { return false; } let nextGreater = data.length - 1; // Find the rightmost successor // to the pivot for (let i = data.length - 1; i > last; i--) { if (data[i] > data[last]) { nextGreater = i; break; } } // Swap the successor and // the pivot data = swap(data, nextGreater, last); // Reverse the suffix data = reverse(data, last + 1, data.length - 1); // Return true as the // next_permutation is done return true;}// Driver Codeconst graph = [[0, 10, 15, 20], [10, 0, 35, 25], [15, 35, 0, 30], [20, 25, 30, 0]];let s = 0;console.log(travllingSalesmanProblem(graph, s));// This code is contributed by ishankhandelwals. |
Output
80
Time complexity: O(n!) where n is the number of vertices in the graph. This is because the algorithm uses the next_permutation function which generates all the possible permutations of the vertex set.
Auxiliary Space: O(n) as we are using a vector to store all the vertices.
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