Traveling Salesman Problem (TSP) Implementation

Last Updated : 31 Jan, 2023

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 Problem
int 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 Code
int 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 Problem
static 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 Code
public 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 permutations
V = 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 Problem
const 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 indices
const 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 inclusive
const 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 array
const 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 Code
const 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

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.

Nishant_Singh

Improve

Introduction to Graph Coloring

Introduction and Approximate Solution for Vertex Cover Problem

BFS and DFS on Graph

Cycle in a Graph

Shortest Paths in Graph

Minimum Spanning Tree in Graph

Topological Sorting in Graph

Connectivity of Graph

Maximum flow in a Graph

Some must do problems on Graph

Traveling Salesman Problem (TSP) Implementation

C++

Java

Python3

C#

Javascript

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