Branch And Bound | Set 6 (Traveling Salesman Problem)

Given a set of cities and distance between every pair of cities, the problem is to find the shortest possible tour that visits every city exactly once and returns to the starting point.


For example, consider the graph shown in figure on right side. A TSP tour in the graph is 0-1-3-2-0. The cost of the tour is 10+25+30+15 which is 80.

We have discussed following solutions
1) Naive and Dynamic Programming
2) Approximate solution using MST


Branch and Bound Solution

As seen in the previous articles, in Branch and Bound method, for current node in tree, we compute a bound on best possible solution that we can get if we down this node. If the bound on best possible solution itself is worse than current best (best computed so far), then we ignore the subtree rooted with the node.

Note that the cost through a node includes two costs.
1) Cost of reaching the node from the root (When we reach a node, we have this cost computed)
2) Cost of reaching an answer from current node to a leaf (We compute a bound on this cost to decide whether to ignore subtree with this node or not).

  • In cases of a maximization problem, an upper bound tells us the maximum possible solution if we follow the given node. For example in 0/1 knapsack we used Greedy approach to find an upper bound.
  • In cases of a minimization problem, a lower bound tells us the minimum possible solution if we follow the given node. For example, in Job Assignment Problem, we get a lower bound by assigning least cost job to a worker.

In branch and bound, the challenging part is figuring out a way to compute a bound on best possible solution. Below is an idea used to compute bounds for Traveling salesman problem.

Cost of any tour can be written as below.

Cost of a tour T = (1/2) * ∑ (Sum of cost of two edges
                              adjacent to u and in the
                              tour T) 
                   where u ∈ V
For every vertex u, if we consider two edges through it in T,
and sum their costs.  The overall sum for all vertices would
be twice of cost of tour T (We have considered every edge 

(Sum of two tour edges adjacent to u) >= (sum of minimum weight
                                          two edges adjacent to

Cost of any tour >=  1/2) * ∑ (Sum of cost of two minimum
                              weight edges adjacent to u) 
                   where u ∈ V

For example, consider the above shown graph. Below are minimum cost two edges adjacent to every node.

Node 	Least cost edges   Total cost 	  	 
0 	(0, 1), (0, 2) 	       25
1 	(0, 1), (1, 3)         35
2	(0, 2), (2, 3) 	       45
3 	(0, 3), (1, 3) 	       45

Thus a lower bound on the cost of any tour = 
         1/2(25 + 35 + 45 + 45)
       = 75
Refer this for one more example.

Now we have an idea about computation of lower bound. Let us see how to how to apply it state space search tree. We start enumerating all possible nodes (preferably in lexicographical order)

1. The Root Node: Without loss of generality, we assume we start at vertex “0” for which the lower bound has been calculated above.

Dealing with Level 2: The next level enumerates all possible vertices we can go to (keeping in mind that in any path a vertex has to occur only once) which are, 1, 2, 3… n (Note that the graph is complete). Consider we are calculating for vertex 1, Since we moved from 0 to 1, our tour has now included the edge 0-1. This allows us to make necessary changes in the lower bound of the root.

Lower Bound for vertex 1 = 
   Old lower bound - ((minimum edge cost of 0 + 
                    minimum edge cost of 1) / 2) 
                  + (edge cost 0-1)

How does it work? To include edge 0-1, we add the edge cost of 0-1, and subtract an edge weight such that the lower bound remains as tight as possible which would be the sum of the minimum edges of 0 and 1 divided by 2. Clearly, the edge subtracted can’t be smaller than this.

Dealing with other levels: As we move on to the next level, we again enumerate all possible vertices. For the above case going further after 1, we check out for 2, 3, 4, …n.
Consider lower bound for 2 as we moved from 1 to 1, we include the edge 1-2 to the tour and alter the new lower bound for this node.

Lower bound(2) = 
     Old lower bound - ((second minimum edge cost of 1 + 
                         minimum edge cost of 2)/2)
                     + edge cost 1-2)

Note: The only change in the formula is that this time we have included second minimum edge cost for 1, because the minimum edge cost has already been subtracted in previous level.

// C++ program to solve Traveling Salesman Problem
// using Branch and Bound.
#include <bits/stdc++.h>
using namespace std;
const int N = 4;

// final_path[] stores the final solution ie, the
// path of the salesman.
int final_path[N+1];

// visited[] keeps track of the already visited nodes
// in a particular path
bool visited[N];

// Stores the final minimum weight of shortest tour.
int final_res = INT_MAX;

// Function to copy temporary solution to
// the final solution
void copyToFinal(int curr_path[])
    for (int i=0; i<N; i++)
        final_path[i] = curr_path[i];
    final_path[N] = curr_path[0];

// Function to find the minimum edge cost
// having an end at the vertex i
int firstMin(int adj[N][N], int i)
    int min = INT_MAX;
    for (int k=0; k<N; k++)
        if (adj[i][k]<min && i != k)
            min = adj[i][k];
    return min;

// function to find the second minimum edge cost
// having an end at the vertex i
int secondMin(int adj[N][N], int i)
    int first = INT_MAX, second = INT_MAX;
    for (int j=0; j<N; j++)
        if (i == j)

        if (adj[i][j] <= first)
            second = first;
            first = adj[i][j];
        else if (adj[i][j] <= second &&
                 adj[i][j] != first)
            second = adj[i][j];
    return second;

// function that takes as arguments:
// curr_bound -> lower bound of the root node
// curr_weight-> stores the weight of the path so far
// level-> current level while moving in the search
//         space tree
// curr_path[] -> where the solution is being stored which
//                would later be copied to final_path[]
void TSPRec(int adj[N][N], int curr_bound, int curr_weight,
            int level, int curr_path[])
    // base case is when we have reached level N which
    // means we have covered all the nodes once
    if (level==N)
        // check if there is an edge from last vertex in
        // path back to the first vertex
        if (adj[curr_path[level-1]][curr_path[0]] != 0)
            // curr_res has the total weight of the
            // solution we got
            int curr_res = curr_weight +

            // Update final result and final path if
            // current result is better.
            if (curr_res < final_res)
                final_res = curr_res;

    // for any other level iterate for all vertices to
    // build the search space tree recursively
    for (int i=0; i<N; i++)
        // Consider next vertex if it is not same (diagonal
        // entry in adjacency matrix and not visited
        // already)
        if (adj[curr_path[level-1]][i] != 0 &&
            visited[i] == false)
            int temp = curr_bound;
            curr_weight += adj[curr_path[level-1]][i];

            // different computation of curr_bound for
            // level 2 from the other levels
            if (level==1)
              curr_bound -= ((firstMin(adj, curr_path[level-1]) +
                             firstMin(adj, i))/2);
              curr_bound -= ((secondMin(adj, curr_path[level-1]) +
                             firstMin(adj, i))/2);

            // curr_bound + curr_weight is the actual lower bound
            // for the node that we have arrived on
            // If current lower bound < final_res, we need to explore
            // the node further
            if (curr_bound + curr_weight < final_res)
                curr_path[level] = i;
                visited[i] = true;

                // call TSPRec for the next level
                TSPRec(adj, curr_bound, curr_weight, level+1,

            // Else we have to prune the node by resetting
            // all changes to curr_weight and curr_bound
            curr_weight -= adj[curr_path[level-1]][i];
            curr_bound = temp;

            // Also reset the visited array
            memset(visited, false, sizeof(visited));
            for (int j=0; j<=level-1; j++)
                visited[curr_path[j]] = true;

// This function sets up final_path[] 
void TSP(int adj[N][N])
    int curr_path[N+1];

    // Calculate initial lower bound for the root node
    // using the formula 1/2 * (sum of first min +
    // second min) for all edges.
    // Also initialize the curr_path and visited array
    int curr_bound = 0;
    memset(curr_path, -1, sizeof(curr_path));
    memset(visited, 0, sizeof(curr_path));

    // Compute initial bound
    for (int i=0; i<N; i++)
        curr_bound += (firstMin(adj, i) +
                       secondMin(adj, i));

    // Rounding off the lower bound to an integer
    curr_bound = (curr_bound&1)? curr_bound/2 + 1 :

    // We start at vertex 1 so the first vertex
    // in curr_path[] is 0
    visited[0] = true;
    curr_path[0] = 0;

    // Call to TSPRec for curr_weight equal to
    // 0 and level 1
    TSPRec(adj, curr_bound, 0, 1, curr_path);

// Driver code
int main()
    //Adjacency matrix for the given graph
    int adj[N][N] = { {0, 10, 15, 20},
        {10, 0, 35, 25},
        {15, 35, 0, 30},
        {20, 25, 30, 0}


    printf("Minimum cost : %d\n", final_res);
    printf("Path Taken : ");
    for (int i=0; i<=N; i++)
        printf("%d ", final_path[i]);

    return 0;

Output :

Minimum cost : 80
Path Taken : 0 1 3 2 0 

Time Complexity: The worst case complexity of Branch and Bound remains same as that of the Brute Force clearly because in worst case, we may never get a chance to prune a node. Whereas, in practice it performs very well depending on the different instance of the TSP. The complexity also depends on the choice of the bounding function as they are the ones deciding how many nodes to be pruned.


This article is contributed by Anurag Rai. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to See your article appearing on the GeeksforGeeks main page and help other Geeks.

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