Traveling Salesman Problem (TSP) Implementation

Travelling Salesman Problem (TSP): Given a set of cities and distance between every pair of cities, the problem is to find the shortest possible route that visits every city exactly once and returns back to the starting point.
Note the difference between Hamiltonian Cycle and TSP. The Hamiltoninan cycle problem is to find if there exist 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 figure on 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 know solution for this problem.
Examples:

 Output of Given Graph:
 minimum weight Hamiltonian Cycle :
  10 + 25 + 30 + 15 := 80



In this post, implementation of simple solution is discussed.

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

Below is the implementation of above idea

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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 program to test above function
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;
}

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 program to implement traveling salesman
# problem using naive approach.
from sys import maxsize
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
  
    while True:
  
        # store current Path weight(cost)
        current_pathweight = 0
  
        # compute current path weight
        k = s
        for i in range(len(vertex)):
            current_pathweight += graph[k][vertex[i]]
            k = vertex[i]
        current_pathweight += graph[k][s]
  
        # update minimum
        min_path = min(min_path, current_pathweight)
  
        if not next_permutation(vertex):
            break
  
    return min_path
  
# next_permutation implementation
def next_permutation(L):
  
    n = len(L)
  
    i = n - 2
    while i >= 0 and L[i] >= L[i + 1]:
        i -= 1
  
    if i == -1:
        return False
  
    j = i + 1
    while j < n and L[j] > L[i]:
        j += 1
    j -= 1
  
    L[i], L[j] = L[j], L[i]
  
    left = i + 1
    right = n - 1
  
    while left < right:
        L[left], L[right] = L[right], L[left]
        left += 1
        right -= 1
  
    return True
  
# 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))
  
# This code is contributed by
# sanjeev2552

chevron_right



Output:

80


My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : Blinkii, sanjeev2552



Article Tags :
Practice Tags :


5


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.