Tsp using branch and bound program in c

If S is empty that means we visited all nodes, we take distance from that last visited node to node 1 first node. Because after visiting all he has to go back to initial node. Since we are solving this using Dynamic Programming, we know that Dynamic Programming approach contains sub-problems.

Here after reaching i th node finding remaining minimum distance to that i th node is a sub-problem. If we solve recursive equation we will get total n-1 2 n-2 sub-problems, which is O n2 n. Each sub-problem will take O n time finding path to remaining n-1 nodes. Comment below if you found any information incorrect or have doubts regarding Travelling Salesman Problem algorithm. Nicely explained. I was just trying to understand the code to implement this.

What I was not able to understand is why we are adding the return to the same node as well for the minimum comparison. Will the below changed least code not work for all situation? Looping over all subsets of a set is a challenge for Programmers. I tried it for 6 and it fails to find the minimum path. Example cost matrix and found path:. This code is NOT correct.

Also every other site has this same exact code. Some one please share the link to a correct working code for solving TSP using Dynamic Programming approach. Actually this is TSP code,he is making us fool. Watch Tushar Roy video for real Dp implementation. I am really hard to understand your code. The problem is a famous NP-hard problem. There is no polynomial-time known solution for this problem.

Python3 program to implement traveling salesman. Recommended Articles. Article Contributed By :. Easy Normal Medium Hard Expert. Writing code in comment?

Please use ide. Load Comments. We reduce the minimum value in each row from each element in that row. The minimum in each row of cost matrix M is marked by blue [10 2 2 3 4] below.

We then reduce the minimum value in each column from each element in that column. Minimum in each column is marked by blue [1 0 3 0 0]. After reducing the column, we get below the reduced matrix.

This matrix will be further processed by child nodes of the root node to calculate their lower bound. Since we have discovered the root node C0 , the next node to be expanded can be any node from C1 , C2 , C3 , C4. Whichever node has a minimum cost will be expanded further.

So, we have to find out the expanding cost of each node. We try to calculate the lower bound of the path starting at node 1 using the above resulting cost matrix. The lower bound is 0 as the matrix is already in reduced form, i.

Now calculate the lower bound of the path starting at node 2 using the approach discussed earlier. The resultant matrix will be:. Now calculate the lower bound of the path starting at node 3 using the approach discussed earlier. The lower bound of the path starting at node 3 is 0 as it is already in reduced form, i.

Now find a live node with the least estimated cost. Live nodes 1 , 2 , 3 , and 4 have costs 35 , 53 , 25 , and 31 , respectively. The minimum among them is node 3 , having cost So, node 3 will be expanded further, as shown in the state-space tree diagram. After adding its children to the list of live nodes, find a live node with the least cost and expand it.

Continue the search till a leaf is encountered in the space search tree. If a leaf is encountered, then the tour is completed, and we will return to the root node.