Traveling Salesman Problem (TSP) Implementation

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.

  1. Consider city 1 as the starting and ending point. Since the route is cyclic, we can consider any point as a starting point.
  2. Generate all (n-1)! permutations of cities.
  3. Calculate the cost of every permutation and keep track of the minimum cost permutation.
  4. Return the permutation with minimum cost.

Below is the implementation of the above idea 
 

C++

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// 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;
}

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Java

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// 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

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Python3

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# 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))

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Output

80


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