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Traveling Salesman Problem using Branch And Bound

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.  

Euler1

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 Travelling salesman problem. Cost of any tour can be written as below.  

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

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.   

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.  

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.   

Output :    

The rounding is being done in this line of code:

In the Branch and Bound TSP algorithm, we compute a lower bound on the total cost of the optimal solution by adding up the minimum edge costs for each vertex, and then dividing by two. However, this lower bound may not be an integer. To get an integer lower bound, we can use rounding.

In the above code, the curr_bound variable holds the current lower bound on the total cost of the optimal solution. When we visit a new vertex at level level, we compute a new lower bound new_bound by taking the sum of the minimum edge costs for the new vertex and its two closest neighbors. We then update the curr_bound variable by rounding new_bound to the nearest integer.

If level is 1, we round down to the nearest integer. This is because we have only visited one vertex so far, and we want to be conservative in our estimate of the total cost of the optimal solution. If level is greater than 1, we use a more aggressive rounding strategy that takes into account the fact that we have already visited some vertices and can therefore make a more accurate estimate of the total cost of the optimal solution.

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. References:   http://lcm.csa.iisc.ernet.in/dsa/node187.html

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travelling salesman problem using branch and bound algorithm

Travelling Salesman Problem using Branch and Bound

Given a set of cities and the distance 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.

For example, consider the following graph . A TSP tour in the graph is A —> B —> C —> D —> B —> A . The cost of the tour is 10 + 25 + 40 + 25 + 10 = 100 .

TSP Tour

This post discusses the Travelling Salesman Problem using Branch and Bound.

  The term Branch and Bound refer to all state-space search methods in which all the children of an E–node are generated before any other live node can become the E–node. E–node is the node, which is being expended. State–space tree can be expended in any method, i.e., BFS or DFS . Both start with the root node and generate other nodes. A node that has been generated and whose children are not yet been expanded is called a live-node. A node is called a dead node, which has been generated, but it cannot be expanded further. In this method, we expand the most promising node, which means the node which promises that expanding or choosing it will give us the optimal solution. So, we prepare the tree starting from the root and then expand it.

We have a cost matrix defined by:

For example, consider the following cost matrix M ,

Travelling Salesman Problem – Cost Matrix

Following is the state-space tree for the above TSP problem, which shows the optimal solution marked in green:

State–Space Tree: TSP Problem

As we can see from the above diagram, every node has a cost associated with it. When we go from the city i to city j , the cost of a node j will be the sum of the parent node i , the cost of an edge (i, j) , and the lower bound of the path starting at node j .

As the root node is the first node to be expanded, it doesn’t have any parent. So, the cost will be only the lower bound of the path starting at the root.

Now, how do we calculate the lower bound of the path starting at any node?

In general, to get the lower bound of the path starting from the node, we reduce each row and column so that there must be at least one zero in each row and Column. We need to reduce the minimum value from each element in each row and column.

Let’s start from the root node.

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.

Travelling Salesman Problem – Step 1

After reducing the row, we get the below reduced matrix.

Travelling Salesman Problem – Step 2

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.

Travelling Salesman Problem – Step 3

The total expected cost at the root node is the sum of all reductions.

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.

The parent node C0 has below reduced matrix:

Travelling Salesman Problem – Step 4

Let’s consider an edge from 0 —> 1 .

1. As we add an edge (0, 1) to our search space, set outgoing edges for city 0 to INFINITY and all incoming edges to city 1 to INFINITY . We also set (1, 0) to INFINITY .

So in a reduced matrix of the parent node, change all the elements in row 0 and column 1 and at index (1, 0) to INFINITY (marked in red).

Travelling Salesman Problem – Step 5

The resulting cost matrix is:

Travelling Salesman Problem – Step 6

2. 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.e., all rows and all columns have zero value.

Therefore, for node 1, the cost will be:

Let’s consider an edge from 0 —> 2

1. Change all the elements in row 0 and column 2 and at index (2, 0) to INFINITY (marked in red).

Travelling Salesman Problem – Step 7

2. Now calculate the lower bound of the path starting at node 2 using the approach discussed earlier. The resultant matrix will be:

Travelling Salesman Problem – Step 9

Therefore, for node 2, the cost will be

Let’s consider an edge from 0 —> 3 .

1. Change all the elements in row 0 and column 3 and at index (3, 0) to INFINITY (marked in red).

Travelling Salesman Problem – Step 10

2. 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.e., all rows and all columns have zero value.

Therefore, for node 3, the cost will be

Similarly, we calculate the cost of 0 —> 4 . Its cost will be 31 .

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 25 . 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.

  Following is the C++ implementation of the above idea:

Download    Run Code

Output: 1 —> 3 3 —> 4 4 —> 2 2 —> 5 5 —> 1   Total cost is 34

  The proposed method, which uses Branch and Bound, is better because it prepares the matrices in different steps. At each step, the cost matrix is calculated. From the initial point, we know that what can be the minimum cost of the tour. The cost of the initial stages is not an exact cost, but it gives some idea because it is the approximated cost. At each step, it gives us a strong reason for which node we should travel the next and which one not. It gives this fact in terms of the cost of expanding a particular node.

  References:

1. https://www.seas.gwu.edu/~bell/csci212/Branch_and_Bound.pdf 2. http://research.ijcaonline.org/volume65/number5/pxc3885866.pdf

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Travelling Salesman Problem using Branch and Bound approach

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In this article we will briefly discuss about the travelling salesman problem and the branch and bound method to solve the same.

What is the problem statement ?

Travelling Salesman Problem is based on a real life scenario, where a salesman from a company has to start from his own city and visit all the assigned cities exactly once and return to his home till the end of the day. The exact problem statement goes like this, " 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 to the starting point. "

There are two important things to be cleared about in this problem statement,

  • Visit every city exactly once
  • Cover the shortest path

What is branch and bound ?

Before going into the details of the branch and bound method lets focus on some important terms for the same,

State Space Search Method - Remember the word sample space from the probability theory ? That was a set of all the possible sample outputs. In a similar way, state space means, a set of states that a problem can be in. The set of states forms a graph where two states are connected if there is an operation that can be performed to transform the first state into the second. In a lighter note, this is just a set of some objects, which has some properties/characteristics, like in this problem, we have a node, it has a cost associated to it. The entire state space can be represented as a tree known as state space tree, which has the root and the leaves as per the normal tree, which are interms the elements of the statespace having the given graph node and a cost associated to it.

E-node - Expanded node or E node is the node which is been expanded. As we know a tree can be expanded using both BFS(Bredth First Search) and DFS(Depth First Search),all the expanded nodes are known as E-nodes .

Live-node - A node which has been generated and all of whose children are not yet been expanded is called a live-node.

Dead-node - If a node can't be expanded further, it's known as a dead-node.

The word, Branch and Bound refers to all the state space search methods in which we generate the childern of all the expanded nodes, before making any live node as an expanded one. In this method, we find the most promising node and expand it. The term promising node means, choosing a node that can expand and give us an optimal solution. We start from the root and expand the tree untill unless we approach an optilmal (minimum cost in case of this problem) solution.

How to get the cost for each node in the state space tree ?

To get further in branch and bound, we need to find the cost at the nodes at first. The cost is found by using cost matrix reduction , in accordance with two accompanying steps row reduction & column reduction .

In general to get the optimal(lower bound in this problem) cost starting from the node, we reduce each row and column in such a way that there must be atleast one 0 in each row and column. For doing this, we just need to reduce the minimum value from each row and column.

For example, suppose we have a cost matrix as shown below,

capture-given-graph

Let's start from node N0,;lets consider N0 as our first live node.

First of all we will perform the row operations and to do the same, we need to subtract the minimum value in each row from each element in that row. The minimums and the row reduced matrix is shown below,

Capture-row-reduction

And after row reduction,

Capture-row-reduced

Now we need to perform the column operation in the same manner and we will get a final matrix as below,

Capture-final-reduced

The total cost at any node is calculated by making a sum of all reductions, hence the cost for the node N0 can be depicted as,

cost = 10 + 2 + 2 + 3 + 4 + 1 + 3 = 25

Now as we have founded the cost for the root node, it's the time for the expansion, and to do so, we have four more nodes namely, N1 , N2 , N3 , N4 .

Let's add the edge from N0 to N1 in our search space. To do so, we need to set outgoing routes for city N0 to INF as well as all incoming routes to city N1 to INF.Also we will set the (N0,N1) position of the cost matrix as INF. By doing so, we are just focusing on the cost of the node N1, that's an E-node. In an easier note, we have just forgotten that the graph has a N0 node, but we are focusing on something that the graph has been started from the N1 node. Following the transformation we have something as below,

Capture-reduced-N1

Now it's the time to repeat the lower bound finding(row-reduction and the column-reduction) part that we have performed earlier as well, following which we will get that the matrix is already in reduced form, i.e. all rows and all columns have zero value.

Now it's the time to calculate the cost.

cost = cost of node N0 + cost of the (N0,N1) position(before setting INF) + lower bound of the path starting at N1 = 25 + 10 + 0 = 35

Hence the cost for the E-Node N1 is 35. In a similar way we can find that, For N2 the cost is 53, For N3 the cost is 25, And for N4 the cost is 31.

Now we have a part of state space tree ready, whcih can be shown as below,

Capture-state-space-tree

How to get towards the optimal solution ?

To get the optimal solution, we have to choose the next live node from the list of the extended nodes (E-nodes) and repeat the same procedure of extending the tree further by calculating the lower bound. Here in the partially built space search tree above we have the Node N3 as the node having the minimum cost and we can consider it as the live node and extend the tree further. This step is repeated until unless we find a dead node and can not extend the tree further. The cost of the dead node (leaf node) will be the answer. The entire space search tree can be drawn as follow,

Capture-branch-tree

And the final cost is 28 , that's the minimum cost for a salesman to start from N0 and return to N0 by covering all the cities.

The Main Code

Now after knowing the entire process, this thing is easier to code. We have tried something new this time by attaching some more datastructures and objects to print the path as well. Have a look at the following code in order to understand it,

Code Explanation :

We would briefly go through the three main functions of the algorithm and try to understand how the algorithm is working, which is pretty similar to what we have learbed above.

reduce_row()

As the name suggests, this function is used to reduce the desired row. In this function we are using an awesome builtin function namely fill_n() available in the C++ STL. Have a look at the syntax,

This function makes it easy to fill an entire row with the desired value(INF here).

Now it's time to run two loops, one for the minimum value finding for each row and one for making the reduction. Have a look at the below snippet of the code.

You may guess the run time complexity of the above function is O(n^2) because the loop has two iterations embeded one inside the other. The run time complexity of the fill_n() function is Linear, i.e O(n) .

reduce_column()

As the name suggests, this function is used to reduce the desired column, in a similar fashion to what we have done above.

This function makes it easy to fill an entire column with the desired value(INF here).

cost_calculation()

This function need to get the reduced matrix as a parameter, perform the row reduction, the column reduction successively and then calculate the cost.

This can been seen below.

The time complexity of the program is O(n^2) as explained above for the row and column reduction functions.

What is the final time complexity ?

Now this thing is tricky and need a deeper understanding of what we are doing. We are actually creating all the possible extenstions of E-nodes in terms of tree nodes. Which is nothing but a permutation . Suppose we have N cities, then we need to generate all the permutations of the (N-1) cities, excluding the root city. Hence the time complexity for generating the permutation is O((n-1)!) , which is equal to O(2^(n-1)) .

Hence the final time complexity of the algorithm can be O(n^2 * 2^n) .

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Traveling salesman problem

This web page is a duplicate of https://optimization.mccormick.northwestern.edu/index.php/Traveling_salesman_problems

Author: Jessica Yu (ChE 345 Spring 2014)

Steward: Dajun Yue, Fengqi You

The traveling salesman problem (TSP) is a widely studied combinatorial optimization problem, which, given a set of cities and a cost to travel from one city to another, seeks to identify the tour that will allow a salesman to visit each city only once, starting and ending in the same city, at the minimum cost. 1

  • 2.1 Graph Theory
  • 2.2 Classifications of the TSP
  • 2.3 Variations of the TSP
  • 3.1 aTSP ILP Formulation
  • 3.2 sTSP ILP Formulation
  • 4.1 Exact algorithms
  • 4.2.1 Tour construction procedures
  • 4.2.2 Tour improvement procedures
  • 5 Applications
  • 7 References

travelling salesman problem using branch and bound algorithm

The origins of the traveling salesman problem are obscure; it is mentioned in an 1832 manual for traveling salesman, which included example tours of 45 German cities but gave no mathematical consideration. 2 W. R. Hamilton and Thomas Kirkman devised mathematical formulations of the problem in the 1800s. 2

It is believed that the general form was first studied by Karl Menger in Vienna and Harvard in the 1930s. 2,3

Hassler Whitney, who was working on his Ph.D. research at Harvard when Menger was a visiting lecturer, is believed to have posed the problem of finding the shortest route between the 48 states of the United States during either his 1931-1932 or 1934 seminar talks. 2 There is also uncertainty surrounding the individual who coined the name “traveling salesman problem” for Whitney’s problem. 2

The problem became increasingly popular in the 1950s and 1960s. Notably, George Dantzig, Delber R. Fulkerson, and Selmer M. Johnson at the RAND Corporation in Santa Monica, California solved the 48 state problem by formulating it as a linear programming problem. 2 The methods described in the paper set the foundation for future work in combinatorial optimization, especially highlighting the importance of cutting planes. 2,4

In the early 1970s, the concept of P vs. NP problems created buzz in the theoretical computer science community. In 1972, Richard Karp demonstrated that the Hamiltonian cycle problem was NP-complete, implying that the traveling salesman problem was NP-hard. 4

Increasingly sophisticated codes led to rapid increases in the sizes of the traveling salesman problems solved. Dantzig, Fulkerson, and Johnson had solved a 48 city instance of the problem in 1954. 5 Martin Grötechel more than doubled this 23 years later, solving a 120 city instance in 1977. 5 Enoch Crowder and Manfred W. Padberg again more than doubled this in just 3 years, with a 318 city solution. 5

In 1987, rapid improvements were made, culminating in a 2,392 city solution by Padberg and Giovanni Rinaldi. In the following two decades, David L. Appelgate, Robert E. Bixby, Vasek Chvátal, & William J. Cook led the cutting edge, solving a 7,397 city instance in 1994 up to the current largest solved problem of 24,978 cities in 2004. 5

Description

Graph theory.

{\displaystyle G=(V,E)}

In the context of the traveling salesman problem, the verticies correspond to cities and the edges correspond to the path between those cities. When modeled as a complete graph, paths that do not exist between cities can be modeled as edges of very large cost without loss of generality. 6 Minimizing the sum of the costs for Hamiltonian cycle is equivalent to identifying the shortest path in which each city is visiting only once.

Classifications of the TSP

The TRP can be divided into two classes depending on the nature of the cost matrix. 3,6

{\displaystyle C}

  • Applies when the distance between cities is the same in both directions

{\displaystyle \exists ~i,j:c_{ij}\neq c_{ji}}

  • Applies when there are differences in distances (e.g. one-way streets)

An ATSP can be formulated as an STSP by doubling the number of nodes. 6

Variations of the TSP

{\displaystyle u}

Formulation

{\displaystyle n}

The objective function is then given by

{\displaystyle {\text{min}}\sum _{i}\sum _{j}c_{ij}y_{ij}}

To ensure that the result is a valid tour, several contraints must be added. 1,3

{\displaystyle \sum _{j}y_{ij}=1,~~\forall i=0,1,...,n-1}

There are several other formulations for the subtour elimnation contraint, including circuit packing contraints, MTZ constraints, and network flow constraints.

aTSP ILP Formulation

The integer linear programming formulation for an aTSP is given by

{\displaystyle {\begin{aligned}{\text{min}}&~~\sum _{i}\sum _{j}c_{ij}y_{ij}\\{\text{s.t}}&~~\sum _{j}y_{ij}=1,~~i=0,1,...,n-1\\&~~\sum _{i}y_{ij}=1,~~j=0,1,...,n-1\\&~~\sum _{i}\sum _{j}y_{ij}\leq |S|-1~~S\subset V,2\leq |S|\leq n-2\\&~~y_{ij}\in \{0,1\},~\forall i,j\in E\\\end{aligned}}}

sTSP ILP Formulation

The symmetric case is a special case of the asymmetric case and the above formulation is valid. 3, 6 The integer linear programming formulation for an sTSP is given by

{\displaystyle {\begin{aligned}{\text{min}}&~~\sum _{i}\sum _{j}c_{ij}y_{ij}\\{\text{s.t}}&~~\sum _{i<k}y_{ik}+\sum _{j>k}y_{kj}=2,~~k\in V\\&~~\sum _{i}\sum _{j}y_{ij}\leq |S|-1~~S\subset V,3\leq |S|\leq n-3\\&~~y_{ij}\in \{0,1\}~\forall i,j\in E\\\end{aligned}}}

Exact algorithms

{\displaystyle O(n!)}

Branch-and-bound algorithms are commonly used to find solutions for TSPs. 7 The ILP is first relaxed and solved as an LP using the Simplex method, then feasibility is regained by enumeration of the integer variables. 7

Other exact solution methods include the cutting plane method and branch-and-cut. 8

Heuristic algorithms

Given that the TSP is an NP-hard problem, heuristic algorithms are commonly used to give a approximate solutions that are good, though not necessarily optimal. The algorithms do not guarantee an optimal solution, but gives near-optimal solutions in reasonable computational time. 3 The Held-Karp lower bound can be calculated and used to judge the performance of a heuristic algorithm. 3

There are two general heuristic classifications 7 :

  • Tour construction procedures where a solution is gradually built by adding a new vertex at each step
  • Tour improvement procedures where a feasbile solution is improved upon by performing various exchanges

The best methods tend to be composite algorithms that combine these features. 7

Tour construction procedures

{\displaystyle k}

Tour improvement procedures

{\displaystyle t}

Applications

The importance of the traveling salesman problem is two fold. First its ubiquity as a platform for the study of general methods than can then be applied to a variety of other discrete optimization problems. 5 Second is its diverse range of applications, in fields including mathematics, computer science, genetics, and engineering. 5,6

travelling salesman problem using branch and bound algorithm

Suppose a Northwestern student, who lives in Foster-Walker , has to accomplish the following tasks:

  • Drop off a homework set at Tech
  • Work out a SPAC
  • Complete a group project at Annenberg

Distances between buildings can be found using Google Maps. Note that there is particularly strong western wind and walking east takes 1.5 times as long.

It is the middle of winter and the student wants to spend the least possible time walking. Determine the path the student should take in order to minimize walking time, starting and ending at Foster-Walker.

Start with the cost matrix (with altered distances taken into account):

Method 1: Complete Enumeration

All possible paths are considered and the path of least cost is the optimal solution. Note that this method is only feasible given the small size of the problem.

From inspection, we see that Path 4 is the shortest. So, the student should walk 2.28 miles in the following order: Foster-Walker → Annenberg → SPAC → Tech → Foster-Walker

Method 2: Nearest neighbor

Starting from Foster-Walker, the next building is simply the closest building that has not yet been visited. With only four nodes, this can be done by inspection:

  • Smallest distance is from Foster-Walker is to Annenberg
  • Smallest distance from Annenberg is to Tech
  • Smallest distance from Tech is to Annenberg ( creates a subtour, therefore skip )
  • Next smallest distance from Tech is to Foster-Walker ( creates a subtour, therefore skip )
  • Next smallest distance from Tech is to SPAC
  • Smallest distance from SPAC is to Annenberg ( creates a subtour, therefore skip )
  • Next smallest distance from SPAC is to Tech ( creates a subtour, therefore skip )
  • Next smallest distance from SPAC is to Foster-Walker

So, the student would walk 2.54 miles in the following order: Foster-Walker → Annenberg → Tech → SPAC → Foster-Walker

Method 3: Greedy

With this method, the shortest paths that do not create a subtour are selected until a complete tour is created.

  • Smallest distance is Annenberg → Tech
  • Next smallest is SPAC → Annenberg
  • Next smallest is Tech → Annenberg ( creates a subtour, therefore skip )
  • Next smallest is Anneberg → Foster-Walker ( creates a subtour, therefore skip )
  • Next smallest is SPAC → Tech ( creates a subtour, therefore skip )
  • Next smallest is Tech → Foster-Walker
  • Next smallest is Annenberg → SPAC ( creates a subtour, therefore skip )
  • Next smallest is Foster-Walker → Annenberg ( creates a subtour, therefore skip )
  • Next smallest is Tech → SPAC ( creates a subtour, therefore skip )
  • Next smallest is Foster-Walker → Tech ( creates a subtour, therefore skip )
  • Next smallest is SPAC → Foster-Walker ( creates a subtour, therefore skip )
  • Next smallest is Foster-Walker → SPAC

So, the student would walk 2.40 miles in the following order: Foster-Walker → SPAC → Annenberg → Tech → Foster-Walker

travelling salesman problem using branch and bound algorithm

As we can see in the figure to the right, the heuristic methods did not give the optimal solution. That is not to say that heuristics can never give the optimal solution, just that it is not guaranteed.

Both the optimal and the nearest neighbor algorithms suggest that Annenberg is the optimal first building to visit. However, the optimal solution then goes to SPAC, while both heuristic methods suggest Tech. This is in part due to the large cost of SPAC → Foster-Walker. The heuristic algorithms cannot take this future cost into account, and therefore fall into that local optimum.

We note that the nearest neighbor and greedy algorithms give solutions that are 11.4% and 5.3%, respectively, above the optimal solution. In the scale of this problem, this corresponds to fractions of a mile. We also note that neither heuristic gave the worst case result, Foster-Walker → SPAC → Tech → Annenberg → Foster-Walker.

Only tour building heuristics were used. Combined with a tour improvement algorithm (such as 2-opt or simulated annealing), we imagine that we may be able to locate solutions that are closer to the optimum.

The exact algorithm used was complete enumeration, but we note that this is impractical even for 7 nodes (6! or 720 different possibilities). Commonly, the problem would be formulated and solved as an ILP to obtain exact solutions.

  • Vanderbei, R. J. (2001). Linear programming: Foundations and extensions (2nd ed.). Boston: Kluwer Academic.
  • Schrijver, A. (n.d.). On the history of combinatorial optimization (till 1960).
  • Matai, R., Singh, S., & Lal, M. (2010). Traveling salesman problem: An overview of applications, formulations, and solution approaches. In D. Davendra (Ed.), Traveling Salesman Problem, Theory and Applications . InTech.
  • Junger, M., Liebling, T., Naddef, D., Nemhauser, G., Pulleyblank, W., Reinelt, G., Rinaldi, G., & Wolsey, L. (Eds.). (2009). 50 years of integer programming, 1958-2008: The early years and state-of-the-art surveys . Heidelberg: Springer.
  • Cook, W. (2007). History of the TSP. The Traveling Salesman Problem . Retrieved from http://www.math.uwaterloo.ca/tsp/history/index.htm
  • Punnen, A. P. (2002). The traveling salesman problem: Applications, formulations and variations. In G. Gutin & A. P. Punnen (Eds.), The Traveling Salesman Problem and its Variations . Netherlands: Kluwer Academic Publishers.
  • Laporte, G. (1992). The traveling salesman problem: An overview of exact and approximate algorithms. European Journal of Operational Research, 59 (2), 231–247.
  • Goyal, S. (n.d.). A suvey on travlling salesman problem.

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Travelling Salesman Problem: Python, C++ Algorithm

Alyssa Walker

What is the Travelling Salesman Problem (TSP)?

Travelling Salesman Problem (TSP) is a classic combinatorics problem of theoretical computer science. The problem asks to find the shortest path in a graph with the condition of visiting all the nodes only one time and returning to the origin city.

The problem statement gives a list of cities along with the distances between each city.

Objective: To start from the origin city, visit other cities only once, and return to the original city again. Our target is to find the shortest possible path to complete the round-trip route.

Example of TSP

Here a graph is given where 1, 2, 3, and 4 represent the cities, and the weight associated with every edge represents the distance between those cities.

Example of TSP

The goal is to find the shortest possible path for the tour that starts from the origin city, traverses the graph while only visiting the other cities or nodes once, and returns to the origin city.

For the above graph, the optimal route is to follow the minimum cost path: 1-2-4-3-1. And this shortest route would cost 10+25+30+15 =80

Different Solutions to Travelling Salesman Problem

Different Solutions to Travelling Salesman Problem

Travelling Salesman Problem (TSP) is classified as a NP-hard problem due to having no polynomial time algorithm. The complexity increases exponentially by increasing the number of cities.

There are multiple ways to solve the traveling salesman problem (tsp). Some popular solutions are:

The brute force approach is the naive method for solving traveling salesman problems. In this approach, we first calculate all possible paths and then compare them. The number of paths in a graph consisting of n cities is n! It is computationally very expensive to solve the traveling salesman problem in this brute force approach.

The branch-and-bound method: The problem is broken down into sub-problems in this approach. The solution of those individual sub-problems would provide an optimal solution.

This tutorial will demonstrate a dynamic programming approach, the recursive version of this branch-and-bound method, to solve the traveling salesman problem.

Dynamic programming is such a method for seeking optimal solutions by analyzing all possible routes. It is one of the exact solution methods that solve traveling salesman problems through relatively higher cost than the greedy methods that provide a near-optimal solution.

The computational complexity of this approach is O(N^2 * 2^N) which is discussed later in this article.

The nearest neighbor method is a heuristic-based greedy approach where we choose the nearest neighbor node. This approach is computationally less expensive than the dynamic approach. But it does not provide the guarantee of an optimal solution. This method is used for near-optimal solutions.

Algorithm for Traveling Salesman Problem

We will use the dynamic programming approach to solve the Travelling Salesman Problem (TSP).

Before starting the algorithm, let’s get acquainted with some terminologies:

  • A graph G=(V, E), which is a set of vertices and edges.
  • V is the set of vertices.
  • E is the set of edges.
  • Vertices are connected through edges.
  • Dist(i,j) denotes the non-negative distance between two vertices, i and j.

Let’s assume S is the subset of cities and belongs to {1, 2, 3, …, n} where 1, 2, 3…n are the cities and i, j are two cities in that subset. Now cost(i, S, j) is defined in such a way as the length of the shortest path visiting node in S, which is exactly once having the starting and ending point as i and j respectively.

For example, cost (1, {2, 3, 4}, 1) denotes the length of the shortest path where:

  • Starting city is 1
  • Cities 2, 3, and 4 are visited only once
  • The ending point is 1

The dynamic programming algorithm would be:

  • Set cost(i, , i) = 0, which means we start and end at i, and the cost is 0.
  • When |S| > 1, we define cost(i, S, 1) = ∝ where i !=1 . Because initially, we do not know the exact cost to reach city i to city 1 through other cities.
  • Now, we need to start at 1 and complete the tour. We need to select the next city in such a way-

cost(i, S, j)=min cost (i, S−{i}, j)+dist(i,j) where i∈S and i≠j

For the given figure, the adjacency matrix would be the following:

Algorithm for Traveling Salesman Problem

Let’s see how our algorithm works:

Step 1) We are considering our journey starting at city 1, visit other cities once and return to city 1.

Step 2) S is the subset of cities. According to our algorithm, for all |S| > 1, we will set the distance cost(i, S, 1) = ∝. Here cost(i, S, j) means we are starting at city i, visiting the cities of S once, and now we are at city j. We set this path cost as infinity because we do not know the distance yet. So the values will be the following:

Cost (2, {3, 4}, 1) = ∝ ; the notation denotes we are starting at city 2, going through cities 3, 4, and reaching 1. And the path cost is infinity. Similarly-

cost(3, {2, 4}, 1) = ∝

cost(4, {2, 3}, 1) = ∝

Step 3) Now, for all subsets of S, we need to find the following:

cost(i, S, j)=min cost (i, S−{i}, j)+dist(i,j), where j∈S and i≠j

That means the minimum cost path for starting at i, going through the subset of cities once, and returning to city j. Considering that the journey starts at city 1, the optimal path cost would be= cost(1, {other cities}, 1).

Let’s find out how we could achieve that:

Now S = {1, 2, 3, 4}. There are four elements. Hence the number of subsets will be 2^4 or 16. Those subsets are-

1) |S| = Null:

2) |S| = 1:

{{1}, {2}, {3}, {4}}

3) |S| = 2:

{{1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}}

4) |S| = 3:

{{1, 2, 3}, {1, 2, 4}, {2, 3, 4}, {1, 3, 4}}

5) |S| = 4:

{{1, 2, 3, 4}}

As we are starting at 1, we could discard the subsets containing city 1.

The algorithm calculation:

1) |S| = Φ:

cost (2, Φ, 1) = dist(2, 1) = 10

cost (3, Φ, 1) = dist(3, 1) = 15

cost (4, Φ, 1) = dist(4, 1) = 20

cost (2, {3}, 1) = dist(2, 3) + cost (3, Φ, 1) = 35+15 = 50

cost (2, {4}, 1) = dist(2, 4) + cost (4, Φ, 1) = 25+20 = 45

cost (3, {2}, 1) = dist(3, 2) + cost (2, Φ, 1) = 35+10 = 45

cost (3, {4}, 1) = dist(3, 4) + cost (4, Φ, 1) = 30+20 = 50

cost (4, {2}, 1) = dist(4, 2) + cost (2, Φ, 1) = 25+10 = 35

cost (4, {3}, 1) = dist(4, 3) + cost (3, Φ, 1) = 30+15 = 45

cost (2, {3, 4}, 1) = min [ dist[2,3]+Cost(3,{4},1) = 35+50 = 85,

dist[2,4]+Cost(4,{3},1) = 25+45 = 70 ] = 70

cost (3, {2, 4}, 1) = min [ dist[3,2]+Cost(2,{4},1) = 35+45 = 80,

dist[3,4]+Cost(4,{2},1) = 30+35 = 65 ] = 65

cost (4, {2, 3}, 1) = min [ dist[4,2]+Cost(2,{3},1) = 25+50 = 75

dist[4,3]+Cost(3,{2},1) = 30+45 = 75 ] = 75

cost (1, {2, 3, 4}, 1) = min [ dist[1,2]+Cost(2,{3,4},1) = 10+70 = 80

dist[1,3]+Cost(3,{2,4},1) = 15+65 = 80

dist[1,4]+Cost(4,{2,3},1) = 20+75 = 95 ] = 80

So the optimal solution would be 1-2-4-3-1

Algorithm for Traveling Salesman Problem

Pseudo-code

Implementation in c/c++.

Here’s the implementation in C++ :

Implementation in Python

Academic solutions to tsp.

Computer scientists have spent years searching for an improved polynomial time algorithm for the Travelling Salesman Problem. Until now, the problem is still NP-hard.

Though some of the following solutions were published in recent years that have reduced the complexity to a certain degree:

  • The classical symmetric TSP is solved by the Zero Suffix Method.
  • The Biogeography‐based Optimization Algorithm is based on the migration strategy to solve the optimization problems that can be planned as TSP.
  • Multi-Objective Evolutionary Algorithm is designed for solving multiple TSP based on NSGA-II.
  • The Multi-Agent System solves the TSP of N cities with fixed resources.

Application of Traveling Salesman Problem

Travelling Salesman Problem (TSP) is applied in the real world in both its purest and modified forms. Some of those are:

  • Planning, logistics, and manufacturing microchips : Chip insertion problems naturally arise in the microchip industry. Those problems can be planned as traveling salesman problems.
  • DNA sequencing : Slight modification of the traveling salesman problem can be used in DNA sequencing. Here, the cities represent the DNA fragments, and the distance represents the similarity measure between two DNA fragments.
  • Astronomy : The Travelling Salesman Problem is applied by astronomers to minimize the time spent observing various sources.
  • Optimal control problem : Travelling Salesman Problem formulation can be applied in optimal control problems. There might be several other constraints added.

Complexity Analysis of TSP

So the total time complexity for an optimal solution would be the Number of nodes * Number of subproblems * time to solve each sub-problem. The time complexity can be defined as O(N 2 * 2^N).

  • Space Complexity: The dynamic programming approach uses memory to store C(S, i), where S is a subset of the vertices set. There is a total of 2 N subsets for each node. So, the space complexity is O(2^N).

Next, you’ll learn about Sieve of Eratosthenes Algorithm

  • Linear Search: Python, C++ Example
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  • Heap Sort Algorithm (With Code in Python and C++)
  • Kadence’s Algorithm: Largest Sum Contiguous Subarray
  • Radix Sort Algorithm in Data Structure
  • Doubly Linked List: C++, Python (Code Example)
  • Singly Linked List in Data Structures
  • Adjacency List and Matrix Representation of Graph

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Travelling Salesman Problem in C and C++

Here you will learn about Travelling Salesman Problem (TSP) with example and also get a program that implements Travelling Salesman Problem in C and C++.

Let say there are some villages (1, 2, 3, 4, 5). To work with worst case let assume each villages connected with every other villages. And there is a Salesman living in village 1 and he has to sell his things in all villages by travelling and he has to come back to own village 1.

He has to travel each village exactly once, because it is waste of time and energy that revisiting same village. This is same as visiting each node exactly once, which is Hamiltonian Circuit . But our problem is bigger than Hamiltonian cycle because this is not only just finding Hamiltonian path, but also we have to find shortest path.

Finally the problem is we have to visit each vertex exactly once with minimum edge cost in a graph.

Brute Force Approach takes O (n n ) time, because we have to check (n-1)! paths (i.e all permutations) and have to find minimum among them.

The correct approach for this problem is solving using Dynamic Programming.

Dynamic Programming can be applied only if main problem can be divided into sub-problems. Let’s check that.

Travelling Salesman Problem (TSP) Using Dynamic Programming

Example problem.

Travelling Salesman Problem (TSP)

Above we can see a complete directed graph and cost matrix which includes distance between each village. We can observe that cost matrix is symmetric that means distance between village 2 to 3 is same as distance between village 3 to 2.

Here problem is travelling salesman wants to find out his tour with minimum cost.

Say it is T (1,{2,3,4}), means, initially he is at village 1 and then he can go to any of {2,3,4}. From there to reach non-visited vertices (villages) becomes a new problem. Here we can observe that main problem spitted into sub-problem, this is property of dynamic programming.

Note: While calculating below right side values calculated in bottom-up manner. Red color values taken from below calculations.

T ( 1, {2,3,4} ) = minimum of

= { (1,2) + T (2,  {3,4} )     4+ 6 =10

= { (1,3)  + T (3, {2,4} )     1+ 3 =4

= { (1,4) + T (4, {2,3} )     3+ 3 =6

Here minimum of above 3 paths is answer but we know only values of (1,2) , (1,3) , (1,4) remaining thing which is T ( 2, {3,4} ) …are new problems now. First we have to solve those and substitute here.

T (2, {3,4} )   = minimum of

=  { (2,3) + T (3, {4} )     2+ 5 =7

= { (2,4) + T {4, {3} )     1+ 5 =6

T (3, {2,4} )   = minimum of

=  { (3,2) + T (2, {4} )     2+ 1 =3

= { (3,4) + T {4, {2} )     5+ 1 =6

T (4, {2,3} )   = minimum of

=  { (4,2) + T (2, {3} )     1+ 2 =3

= { (4,3) + T {3, {2} )     5+ 2 =7

T ( 3, {4} ) =  (3,4) + T (4, {} )     5+0=5

T ( 4, {3} ) =  (4,3) + T (3, {} )     5+0=5

T ( 2, {4} ) =  (2,4) + T (4, {} )     1+0=1

T ( 4, {2} ) =  (4,2) + T (2, {} )     1+0 = 1

T ( 2, {3} ) =  (2,3) + T (3, {} )     2+0 = 2

T ( 3, {2} ) =  (3,2) + T (2, {} )     2+0=2

Here T ( 4, {} ) is reaching base condition in recursion, which returns 0 (zero ) distance.

This is where we can find final answer,

= { (1,2) + T (2,  {3,4} )     4+ 6 =10 in this path we have to add +1 because this path ends with 3. From there we have to reach 1 so 3->1 distance 1 will be added total distance is 10+1=11

= { (1,3)  + T (3, {2,4} )     1+ 3 =4 in this path we have to add +3 because this path ends with 3. From there we have to reach 1 so 4->1 distance 3 will be added total distance is 4+3=7

= { (1,4) + T (4, {2,3} )     3+ 3 =6 in this path we have to add +1 because this path ends with 3. From there we have to reach 1 so 3->1 distance 1 will be added total distance is 6+1=7

Minimum distance is 7 which includes path 1->3->2->4->1.

After solving example problem we can easily write recursive equation.

Recursive Equation

T (i , s) = min ( ( i , j) + T ( j , S – { j }) ) ;  S!= Ø   ; j € S ;

S is set that contains non visited vertices

=  ( i, 1 ) ;  S=Ø, This is base condition for this recursive equation.

T (i, S) means We are travelling from a vertex “i” and have to visit set of non-visited vertices  “S” and have to go back to vertex 1 (let we started from vertex 1).

( i, j ) means cost of path from node i  to node j

If we observe the first recursive equation from a node we are finding cost to all other nodes (i,j) and from that node to remaining using recursion ( T (j , {S-j}))

But it is not guarantee that every vertex is connected to other vertex then we take that cost as infinity. After that we are taking minimum among all so the path which is not connected get infinity in calculation and won’t be consider.

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.

Time Complexity

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).

Therefore total time complexity is O (n2 n ) * O (n) = O (n 2 2 n )

Space complexity is also number of sub-problems which is O (n2 n )

Program for Travelling Salesman Problem in C

Enter the number of villages: 4

Enter the Cost Matrix

Enter Elements of Row: 1 0 4 1 3

Enter Elements of Row: 2 4 0 2 1

Enter Elements of Row: 3 1 2 0 5

Enter Elements of Row: 4 3 1 5 0 The cost list is: 0 4 1 3 4 0 2 1 1 2 0 5 3 1 5 0

The Path is: 1—>3—>2—>4—>1

Minimum cost is 7

Program for Travelling Salesman Problem in C++

Comment below if you found any information incorrect or have doubts regarding Travelling Salesman Problem algorithm.

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travelling salesman problem using branch and bound 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 ?

int least(int c) { int i,nc=999; int min=999,kmin;

for(i=0;i < n;i++) { if((ary[c][i]!=0)&&(completed[i]==0)) if(ary[c][i] < min) /* REPLACED */ { min=ary[c][i]; /* REPLACED */ kmin=ary[c][i]; nc=i; } } if(min!=999) cost+=kmin; return nc; }

travelling salesman problem using branch and bound algorithm

Sir can u please explain this function

travelling salesman problem using branch and bound algorithm

hellow mam your code is not work properly (for selecting minimum path) because i insert a cost matrix input 0 7 3 4 0 2 9 1 0 this cost matrix currect answer is==>8 and also travel a vertex in 1–>3–>2–>1 But your code is only work with a order wise selection example it will travel only with 1–>2–>3–>1. etc…………….

travelling salesman problem using branch and bound algorithm

Are you for real? Are you that dumb?

this is an undirected graph. 1->3->2->1 is exactly the same as 1->2->3->1.

go ahead, bob your head.

travelling salesman problem using branch and bound algorithm

Hi Good explanation (: But… is it posible to do TSP problem in C without the recursion?

travelling salesman problem using branch and bound algorithm

Your Dynamic TSP-Code might not work correctly for more than 4 cities. Looping over all subsets of a set is a challenge for Programmers.

travelling salesman problem using branch and bound algorithm

Quote: Your Dynamic TSP-Code might not work correctly for more than 4 cities.

It doesn’t. I tried it for 6 and it fails to find the minimum path. Example cost matrix and found path:

The cost list is: 0 5 9 12 4 8 5 0 4 7 9 7 9 4 0 5 5 11 12 7 5 0 10 14 4 9 5 10 0 12 8 7 11 14 12 0

The Path is: 1—>5—>3—>2—>6—>4—>1 (cost 46)

But the path 1->2->3->4->5->6->1 has cost 44

travelling salesman problem using branch and bound algorithm

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.

travelling salesman problem using branch and bound algorithm

Actually this is TSP code,he is making us fool.Watch Tushar Roy video for real Dp implementation.

travelling salesman problem using branch and bound algorithm

Thank you so much for this resource. Why do these sites go on copying I don’t understand.

travelling salesman problem using branch and bound algorithm

Yes, you are correct…..

travelling salesman problem using branch and bound algorithm

min=ary[i][0]+ary[c][i]; has to be min=ary[i][c]+ary[c][i]; this migth solve the issue with the above code

travelling salesman problem using branch and bound algorithm

Thannnnnnnnnnnnnks

travelling salesman problem using branch and bound algorithm

it didnt solve the code…

travelling salesman problem using branch and bound algorithm

Is the code written using dynamic approach?

NO,it is greedy ,this not for TSP,it for MST

travelling salesman problem using branch and bound algorithm

Isn’t this the branch and bound method?

travelling salesman problem using branch and bound algorithm

what if I do not want him to go back to starting node ?

travelling salesman problem using branch and bound algorithm

I’m pretty sure that this is just another implementation of the nearest neighbor algorithm….

travelling salesman problem using branch and bound algorithm

it will be better if you could add more explanation about these above functions such as takeInput(), least(), minCost(). I am really hard to understand your code.

travelling salesman problem using branch and bound algorithm

Nice..can i ask you something..how we want to assign a value of the array with specific value..is that possible for an array consists 2 value..its more like we put the coordinate in one array..

travelling salesman problem using branch and bound algorithm

Thank you friend. I was trying to implement one here and yours came to save my work. Now I’m sorry in the heuristic way. hugs Att. Anderson Itacoatiara – Amazonas – Brazil

travelling salesman problem using branch and bound algorithm

I ran this for 10 cities. It ran fine, but total cost for my matrix of random costs was 138, which is higher than the 125 cost with another program which gave a result of 1 10 9 8 7 6 5 4 3 2 1, which is clearly not a valid calculation. Sigh…

Well, the thought was there, just not carried to correct completion.

travelling salesman problem using branch and bound algorithm

The explanation is solid but the code is wrong. In each recursion step only the closest next hop in regards to the starting city is calculated, but you really have to check ALL sub-problems. The recursion doesn’t do anything special here and could as well have been a for-loop. Just check the following matrix where the start point 1 has a large cost to the furthest city 4:

“The cost list is: 0 1 1 99 1 0 1 1 1 1 0 1 99 1 1 0

The Path is: 1—>2—>3—>4—>1

Minimum cost is 102”

When obviously this could have been just 4 cost with 1->2->4->3->1

travelling salesman problem using branch and bound algorithm

Dude checkout your code it does not work for all case; int adj_matx[4][4] = {{0,10,15,20},{10,0,35,25},{15,35,0,30},{20,25,30,0}}; //ans: 80 int adj_matx[4][4] = {{0,4,1,3},{4,0,2,1},{1,2,0,5},{3,1,5,0}}; //ans: 7 int adj_matx[5][5] = {{0,100,300,100,75},{100,0,50,75,125},{300,50,0,100,125},{100,75,100,0,50},{75,125,125,50,0}}; //ans: 375 int adj_matx[4][4] = {{0,2,1,3},{2,0,4,100},{1,4,0,2},{3,100,2,0}}; //ans: 11 int adj_matx[4][4] = {{0,2,1,4},{2,0,4,3},{1,4,0,2},{4,3,2,0}}; //ans: 8 int adj_matx[4][4] = {{0,5,6,3},{5,0,3,6},{6,3,0,7},{3,6,7,0}}; //ans: 18 int adj_matx[5][5] = {{0,6,9,100,10},{6,0,11,100,100},{9,11,0,100,14},{100,100,100,0,8},{10,100,14,8,0}}; //ans:57

for the last case if starting node is 1 then path is 1-5-4-3-2-1 and cost is 135

travelling salesman problem using branch and bound algorithm

I second that. Saurabh is correct.

———————-T ( 1,{ 2 3 4 5 })——————— Pairwise cost { 6 9 100 10 } Subproblem cost { 129 128 39 125 } Sum cost { 135 137 139 135 } Sub Paths Printing Matrix 5 4 3 2 2 4 5 3 2 3 5 4 2 3 4 5 Choosing subpath 0 Path Vector { 5 4 3 2 1 }

travelling salesman problem using branch and bound algorithm

Time complexity of given code is n!

travelling salesman problem using branch and bound algorithm

Your Program is good but it is not working for more than 4 cities.

travelling salesman problem using branch and bound algorithm

I have never commented on any website. But i was compelled to do so this time. I have been reading your blog for a long time and i find explanations and code far easier than other websites. It’s amazing and very helpful.

U r finding this code for TSP simple bczz it is completely wrong.This is code of MST,using greedy.

travelling salesman problem using branch and bound algorithm

0 9 8 8 12 0 13 6 10 9 0 5 20 15 10 0

for this matrix the solution should be 35 (1-2-4-3-1)but by using this code it give 40(1-3-4-2-1). and also this approach is not dynamic it is greedy.

travelling salesman problem using branch and bound algorithm

Can any one write code to display all possible paths and their respective sum of that path

travelling salesman problem using branch and bound algorithm

Why we declare a value 999 ?

travelling salesman problem using branch and bound algorithm

Replace: min=ary[i][0]+ary[c][i]; to: min=ary[i][c]+ary[c][i];

now works :))

travelling salesman problem using branch and bound algorithm

No it will not

travelling salesman problem using branch and bound algorithm

hello can you pls give program travelling sales man using branch and bound

travelling salesman problem using branch and bound algorithm

The Algorithm has this result : The cost list is: 0 10 15 20 10 0 35 25 15 35 0 30 20 25 30 0

The Path is: 1—>;2—>;3—>;4—>;1

Minimum cost is 95 But the correct minimum cost is 80 and the correct path is 1–>2–>4–>3–>1

travelling salesman problem using branch and bound algorithm

not showing optimal path.

travelling salesman problem using branch and bound algorithm

Function least should have a prototype error occurs here so pls check it out

travelling salesman problem using branch and bound algorithm

The code is totally wrong and all the explanation is being plagarized.

travelling salesman problem using branch and bound algorithm

It is not working correctly for testcase 4 0 5 15 15 5 0 3 7 15 3 0 10 15 7 10 0 Output is : 1—>2—>4—>3—>1 cost 37 Output should be: 1—>2—>3—>4—>1 cost 33

travelling salesman problem using branch and bound algorithm

Crazy bro.. Code is not working bro….

travelling salesman problem using branch and bound algorithm

can any one know program of TSP then pls share

travelling salesman problem using branch and bound algorithm

Hi, I’m sorry but your code is wrong. Please make the required changes or at least remove the code.

travelling salesman problem using branch and bound algorithm

I mean how can you be so wrong?You wrote the explanation so nicely(probably you copied that too) and then this shit code which is present all over the internet just for wasting our time.Dont create such rubbish posts if you dont know how to implement the algorithm.

travelling salesman problem using branch and bound algorithm

The code is really helpful but I request you to add comments. A commented code helps a way better.

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Solving the traveling salesman problem with a distributed branch-and-bound algorithm on a 1024 processor network

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International Conference on Optimization and Applications

OPTIMA 2021: Optimization and Applications pp 136–148 Cite as

Problem-Specific Branch-and-Bound Algorithms for the Precedence Constrained Generalized Traveling Salesman Problem

  • Michael Khachay   ORCID: orcid.org/0000-0003-3555-0080 13 , 15 ,
  • Stanislav Ukolov   ORCID: orcid.org/0000-0002-9946-6446 14 &
  • Alexander Petunin   ORCID: orcid.org/0000-0003-2540-1305 13 , 14  
  • Conference paper
  • First Online: 05 November 2021

740 Accesses

2 Citations

Part of the book series: Lecture Notes in Computer Science ((LNTCS,volume 13078))

The Generalized Traveling Salesman Problem (GTSP) is a well-known combinatorial optimization problem having numerous valuable practical applications in operations research. In the Precedence Constrained GTSP (PCGTSP), any feasible tour is restricted to visit all the clusters according to some given partial order. Unlike the common setting of the GTSP, the PCGTSP appears still weakly studied in terms of algorithmic design and implementation. To the best of our knowledge, all the known algorithmic results for this problem can be exhausted by Salmans’s general branching framework, a few MILP models, and the PCGLNS meta-heuristic proposed by the authors recently. In this paper, we present the first problem-specific branch-and-bound algorithm designed with an extension of Salman’s approach and exploiting PCGLNS as a powerful primal heuristic. Using the public PCGTSPLIB testbench, we evaluate the performance of the proposed algorithm against the classic Held-Karp dynamic programming scheme with branch-and-bound node fathoming strategy and Gurobi state-of-the-art solver armed by our recently proposed MILP model and PCGLNS-based warm start.

  • Generalized Traveling Salesman Problem
  • Precedence constraints
  • Branch-and-bound algorithm

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The only evident exception is made for the last arc \((v_m,v_1)\) closing the tour T .

In our dynamic programming, this bound is tight.

Balas, E., Simonetti, N.: Linear time dynamic-programming algorithms for new classes of restricted TSPs: a computational study. INFORMS J. Comput. 13 (1), 56–75 (2001). https://doi.org/10.1287/ijoc.13.1.56.9748

Article   MathSciNet   MATH   Google Scholar  

Castelino, K., D’Souza, R., Wright, P.K.: Toolpath optimization for minimizing airtime during machining. J. Manuf. Syst. 22 (3), 173–180 (2003). https://doi.org/10.1016/S0278-6125(03)90018-5 . http://www.sciencedirect.com/science/article/pii/S0278612503900185

Chentsov, A.G., Khachai, M.Y., Khachai, D.M.: An exact algorithm with linear complexity for a problem of visiting megalopolises. Proc. Steklov Inst. Math. 295 (1), 38–46 (2016). https://doi.org/10.1134/S0081543816090054

Chentsov, A., Khachay, M., Khachay, D.: Linear time algorithm for precedence constrained asymmetric generalized traveling salesman problem. IFAC-PapersOnLine 49 (12), 651–655 (2016). 8th IFAC Conference on Manufacturing Modelling, Management and Control MIM 2016. https://doi.org/10.1016/j.ifacol.2016.07.767 . http://www.sciencedirect.com/science/article/pii/S2405896316310485

Chentsov, A.G., Chentsov, P.A., Petunin, A.A., Sesekin, A.N.: Model of megalopolises in the tool path optimisation for CNC plate cutting machines. Int. J. Prod. Res. 56 (14), 4819–4830 (2018). https://doi.org/10.1080/00207543.2017.1421784

Article   Google Scholar  

Dewil, R., Küçükoǧlu, I., Luteyn, C., Cattrysse, D.: A critical review of multi-hole drilling path optimization. Arch. Comput. Methods Eng. 26 (2), 449–459 (2019). https://doi.org/10.1007/s11831-018-9251-x

Feremans, C., Grigoriev, A., Sitters, R.: The geometric generalized minimum spanning tree problem with grid clustering. 4OR 4 (4), 319–329 (2006). https://doi.org/10.1007/s10288-006-0012-6

Fischetti, M., González, J.J.S., Toth, P.: A branch-and-cut algorithm for the symmetric generalized traveling salesman problem. Oper. Res. 45 (3), 378–394 (1997). https://doi.org/10.1287/opre.45.3.378

Gutin, G., Karapetyan, D.: A memetic algorithm for the generalized traveling salesman problem. Nat. Comput. 9 (1), 47–60 (2010). https://doi.org/10.1007/s11047-009-9111-6

Gutin, G., Punnen, A.P.: The Traveling Salesman Problem and Its Variations. Springer, Boston (2007). https://doi.org/10.1007/b101971

Book   MATH   Google Scholar  

Held, M., Karp, R.M.: A dynamic programming approach to sequencing problems. J. Soc. Ind. Appl. Math. 10 (1), 196–210 (1962). http://www.jstor.org/stable/2098806

Helsgaun, K.: Solving the equality generalized traveling salesman problem using the Lin-Kernighan-Helsgaun algorithm. Math. Program. Comput. 7 , 269–287 (2015). https://doi.org/10.1007/s12532-015-0080-8

Karapetyan, D., Gutin, G.: Efficient local search algorithms for known and new neighborhoods for the generalized traveling salesman problem. Eur. J. Oper. Res. 219 (2), 234–251 (2012). https://doi.org/10.1016/j.ejor.2012.01.011 . https://www.sciencedirect.com/science/article/pii/S0377221712000288

Khachai, M.Y., Neznakhina, E.D.: Approximation schemes for the generalized traveling salesman problem. Proc. Steklov Inst. Math. 299 (1), 97–105 (2017). https://doi.org/10.1134/S0081543817090127

Khachay, M., Kudriavtsev, A., Petunin, A.: PCGLNS: a heuristic solver for the precedence constrained generalized traveling salesman problem. In: Olenev, N., Evtushenko, Y., Khachay, M., Malkova, V. (eds.) OPTIMA 2020. LNCS, vol. 12422, pp. 196–208. Springer, Cham (2020). https://doi.org/10.1007/978-3-030-62867-3_15

Chapter   Google Scholar  

Khachay, M., Neznakhina, K.: Towards tractability of the Euclidean generalized traveling salesman problem in grid clusters defined by a grid of bounded height. In: Eremeev, A., Khachay, M., Kochetov, Y., Pardalos, P. (eds.) OPTA 2018. CCIS, vol. 871, pp. 68–77. Springer, Cham (2018). https://doi.org/10.1007/978-3-319-93800-4_6

Khachay, M., Neznakhina, K.: Complexity and approximability of the Euclidean generalized traveling salesman problem in grid clusters. Ann. Math. Artif. Intell. 88 (1), 53–69 (2019). https://doi.org/10.1007/s10472-019-09626-w

Kudriavtsev, A., Khachay, M.: PCGLNS: adaptive heuristic solver for the Precedence Constrained GTSP (2020). https://github.com/AndreiKud/PCGLNS/

Laporte, G., Semet, F.: Computational evaluation of a transformation procedure for the symmetric generalized traveling salesman problem. INFOR: Inf. Syst. Oper. Res. 37 (2), 114–120 (1999). https://doi.org/10.1080/03155986.1999.11732374

Makarovskikh, T., Panyukov, A., Savitskiy, E.: Mathematical models and routing algorithms for economical cutting tool paths. Int. J. Prod. Res. 56 (3), 1171–1188 (2018). https://doi.org/10.1080/00207543.2017.1401746

Morin, T.L., Marsten, R.E.: Branch-and-bound strategies for dynamic programming. Oper. Res. 24 (4), 611–627 (1976). http://www.jstor.org/stable/169764

Noon, C.E., Bean, J.C.: An efficient transformation of the generalized traveling salesman problem. INFOR: Inf. Syst. Oper. Res. 31 (1), 39–44 (1993). https://doi.org/10.1080/03155986.1993.11732212

Article   MATH   Google Scholar  

Papadimitriou, C.: Euclidean TSP is NP-complete. Theor. Comput. Sci. 4 , 237–244 (1977)

Salman, R., Carlson, J.S., Ekstedt, F., Spensieri, D., Torstensson, J., Söderberg, R.: An industrially validated CMM inspection process with sequence constraints. Procedia CIRP 44 , 138–143 (2016). 6th CIRP Conference on Assembly Technologies and Systems (CATS). https://doi.org/10.1016/j.procir.2016.02.136 . http://www.sciencedirect.com/science/article/pii/S2212827116004182

Salman, R., Ekstedt, F., Damaschke, P.: Branch-and-bound for the precedence constrained generalized traveling salesman problem. Oper. Res. Lett. 48 (2), 163–166 (2020). https://doi.org/10.1016/j.orl.2020.01.009

Smith, S.L., Imeson, F.: GLNS: an effective large neighborhood search heuristic for the generalized traveling salesman problem. Comput. Oper. Res. 87 , 1–19 (2017). https://doi.org/10.1016/j.cor.2017.05.010

Srivastava, S., Kumar, S., Garg, R., Sen, P.: Generalized traveling salesman problem through n sets of nodes. CORS J. 7 (2), 97–101 (1969)

MathSciNet   MATH   Google Scholar  

Steiner, G.: On the complexity of dynamic programming for sequencing problems with precedence constraints. Ann. Oper. Res. 256 , 103–123 (1990). https://doi.org/10.1007/BF02248587

Ukolov, S., Khachay, M.: Branch-and-bound algorithm for the Precedence Constrained GTSP (2021). https://github.com/ukoloff/PCGTSP-BnB

Yuan, Y., Cattaruzza, D., Ogier, M., Semet, F.: A branch-and-cut algorithm for the generalized traveling salesman problem with time windows. Eur. J. Oper. Res. 286 (3), 849–866 (2020). https://doi.org/10.1016/j.ejor.2020.04.024 . https://www.sciencedirect.com/science/article/pii/S0377221720303581

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Acknowledgments

The work was performed as a part of research carried out in the Ural Mathematical Center with the financial support of the Ministry of Science and Higher Education of the Russian Federation (Agreement number 075-02-2021-1383).

All the computations were performed on supercomputer ‘Uran’ at Krasovsky Institute of Mathematics and Mechanics.

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Michael Khachay & Alexander Petunin

Ural Federal University, Ekaterinburg, Russia

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Omsk Technical University, Omsk, Russia

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Khachay, M., Ukolov, S., Petunin, A. (2021). Problem-Specific Branch-and-Bound Algorithms for the Precedence Constrained Generalized Traveling Salesman Problem. In: Olenev, N.N., Evtushenko, Y.G., Jaćimović, M., Khachay, M., Malkova, V. (eds) Optimization and Applications. OPTIMA 2021. Lecture Notes in Computer Science(), vol 13078. Springer, Cham. https://doi.org/10.1007/978-3-030-91059-4_10

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  1. Traveling Salesman Problem using Branch And Bound

    Traveling Salesman Problem using Branch And Bound. 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.

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    In order to solve the problem using branch n bound, we use a level order. First, we will observe in which order, the nodes are generated. While creating the node, we will calculate the cost of the node simultaneously. If we find the cost of any node greater than the upper bound, we will remove that node.

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    If salesman starting city is A, then a TSP tour in the graph is-A → B → D → C → A Cost of the tour = 10 + 25 + 30 + 15 = 80 units In this article, we will discuss how to solve travelling salesman problem using branch and bound approach with example. PRACTICE PROBLEM BASED ON TRAVELLING SALESMAN PROBLEM USING BRANCH AND BOUND APPROACH ...

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    To get further in branch and bound, we need to find the cost at the nodes at first. The cost is found by using cost matrix reduction, in accordance with two accompanying steps row reduction & column reduction. In general to get the optimal (lower bound in this problem) cost starting from the node, we reduce each row and column in such a way ...

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    The problem involves determining the sequence in which the cities should be visited by a salesperson so that the resulting trip covers the shortest possible distance and each city is visited exactly once. Solution of a traveling salesman problem: the black line shows the shortest possible loop that connects every red dot. Source: Wikipedia.

  7. PDF Branch-and-bound algorithm for the traveling salesman problem

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  8. Travelling Salesman Problem

    Travelling Salesman Problem (TSP) is an interesting problem. Problem is defined as "given n cities and distance between each pair of cities, find out the path which visits each city exactly once and come back to starting city, with the constraint of minimizing the travelling distance.". TSP has many practical applications.

  9. PDF Solving Traveling Salesman Problems Using Branch and Bound Methods

    we will examine the search for solving TSP problem using branch and bound methods. Keywords: Traveling Salesman Problem, The Greedy heuristic, Branch and Bound. 1. Introduction. The traveling salesman problem (TSP) is to find the shortest hamiltonian cycle in a graph. This problem is -hard NP and thus interesting. There are a number of ...

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  11. Traveling salesman problem using Branch and Bound

    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 re...

  12. Travelling Salesman Problem Using Branch and Bound Algorithm ...

    In this video, we will discuss about Travelling Salesman Problem and and how to solve Travelling Salesman Problem using Branch and Bound Algorithm.Topics cov...

  13. Travelling Salesman Problem: Python, C++ Algorithm

    The branch-and-bound method: The problem is broken down into sub-problems in this approach. The solution of those individual sub-problems would provide an optimal solution. This tutorial will demonstrate a dynamic programming approach, the recursive version of this branch-and-bound method, to solve the traveling salesman problem.

  14. Results for the close-enough traveling salesman problem with a branch

    The Close-Enough Traveling Salesman Problem is a generalization of the Traveling Salesman Problem that requires a salesman to just get close enough to each customer instead of visiting the exact location of each customer. In this paper, we propose improvements to an existing branch-and-bound (B &B) algorithm for this problem that finds and proves optimality of solutions by examining partial ...

  15. Fast Branch and Bound Algorithm for the Travelling Salesman Problem

    Branch and Bound (Branch and Bound, BnB, branch & bound) is an approach advised for designing exact algorithms solving \ (\mathcal {NP}\) -hard combinatorial optimization and discrete problems. Branch and Bound was introduced by Land and Doig in 1960 [ 10 ]. Until the late 1970s, it was the state-of-the-art method for almost all big and complex ...

  16. An Algorithm for the Traveling Salesman Problem

    A "branch and bound" algorithm is presented for solving the traveling salesman problem. The set of all tours feasible solutions is broken up into increasingly small subsets by a procedure called branching. For each subset a lower bound on the length of the tours therein is calculated. Eventually, a subset is found that contains a single ...

  17. Travelling Salesman Problem in C and C++

    Here you will learn about Travelling Salesman Problem (TSP) with example and also get a program that implements Travelling Salesman Problem in C and C++. ... hello can you pls give program travelling sales man using branch and bound. Reply. javid. November 30, 2018 at 1:09 pm ... all over the internet just for wasting our time.Dont create such ...

  18. Branch-and-Bound Algorithm for Symmetric Travelling Salesman Problem

    Abstract. In this paper a branch-and-bound algorithm for the Symmetric Travelling Salesman Problem (STSP) is presented. The algorithm is based on the 1-tree Lagrangian relaxation. A new branching strategy is suggested in which the algorithm branches on the 1-tree edge belonging to the vertex with maximum degree in the 1-tree and having the ...

  19. PDF An algorithm for the traveling salesman problem

    and a reduced matrix C[X] such that for all t<grX. z(t) = w[X] + z(t-a) Theorem 3; Step 1 leaves the node condition satisfied for X=T. Step 2a leaves the node condition satisfied for X. -18-. If X satisfies the node condition, then after. Step 2, X(l) satisfies it.

  20. The Branch-and-Bound Algorithm for the Traveling Salesman Problem is

    The thesis that the class of direct algorithms is broad and includes many classical combinatorial algorithms, including the branch-and-bound algorithm for the traveling salesman problem proposed by J.D.C. Little, K.G. Murty, D.W. Sweeney, and C. Karel in 1963, was the main motivation for these articles.

  21. Solving the traveling salesman problem with a distributed branch-and

    This paper is the first to present a parallelization of a highly efficient best-first branch-and-bound algorithm to solve large symmetric traveling salesman problems on a massively parallel computer containing 1024 processors. The underlying sequential branch-and-bound algorithm is based on 1-tree relaxation. The parallelization of the branch-and-bound algorithm is fully distributed. Every ...

  22. Problem-Specific Branch-and-Bound Algorithms for the Precedence

    The Generalized Traveling Salesman Problem (GTSP) is a well-known combinatorial optimization problem introduced in the seminal paper [] by S. Srivastava et al. and attracted the attention of many researchers (see the survey in []).In the GTSP, for a given weighted digraph \(G=(V,E,c)\) and partition \(V_1\cup \ldots \cup V_m\) of the nodeset V into non-empty mutually disjoint clusters, it is ...