travelling salesman problem brute force java

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Travelling Salesman Problem (TSP) : Given a set of cities and distances between every pair of cities, the problem is to find the shortest possible route that visits every city exactly once and returns to the starting point.  Note the difference between Hamiltonian Cycle and TSP. The Hamiltonian cycle problem is to find if there exists a tour that visits every city exactly once. Here we know that Hamiltonian Tour exists (because the graph is complete) and in fact, many such tours exist, the problem is to find a minimum weight Hamiltonian Cycle.  For example, consider the graph shown in the figure on the right side. A TSP tour in the graph is 1-2-4-3-1. The cost of the tour is 10+25+30+15 which is 80. The problem is a famous NP-hard problem. There is no polynomial-time known solution for this problem.   

Examples: 

In this post, the implementation of a simple solution is discussed.

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

Below is the implementation of the above idea 

Time complexity:  O(n!) where n is the number of vertices in the graph. This is because the algorithm uses the next_permutation function which generates all the possible permutations of the vertex set.  Auxiliary Space: O(n) as we are using a vector to store all the vertices.

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Travelling Salesman Problem (Basics + Brute force approach)

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In this article we will start our discussion by understanding the problem statement of The Travelling Salesman Problem perfectly and then go through the naive bruteforce approach for solving the problem using a mathematical concept known as " permutation "

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

Visualizing the problem

Steps To Solve the Problem

There are few classical and easy steps that we must follow to solve the TSP problem,

• Finding Adjacent matrix of the graph, which will act as an input.
• performing the shortest_path algorithm, by coding out a function.
• Understanding C++ STL on using next_permutation.

Step-1 - Finding Adjacent Matrix Of the Graph

You will need a two dimensional array for getting the Adjacent Matrix of the given graph. Here are the steps;

• Get the total number of nodes and total number of edges in two variables namely num_nodes and num_edges.
• Create a multidimensional array edges_list having the dimension equal to num_nodes * num_nodes
• Run a loop num_nodes time and take two inputs namely first_node and second_node * everytime as two nodes having an edge between them and place the edges_list[first_node][second_node] position equal to 1.
• Finally after the loop executes we have an adjacent matrix available i.e edges_list.

Time Complexity - O(V^2) , space complexity - O(V^2) , where V is the number of nodes

Step - 2 - Performing The Shortest Path Algorithm

The most important step in designing the core algorithm is this one, let's have a look at the pseudocode of the algorithm below.

• Considering a starting source city, from where the salesman will strat. We can consider any city as the starting point and by default we have considered 0 as the starting point here.
• Generating the permutation of the rest cities. Suppose we have total N nodes and we have considered one node as the source, then we need to generate the rest (N-1)! (Factorial of N-1) permutations.
• We need to calculate the edge sum (path sum) for every permutation and take a track of the minumum path sum with each permutation.
• Return the minimum edge cost.

Step - 3 - Understanding next_permutation in C++ STL

It's a good practise to understand the functions from Standard Template Library on what they take as arguement, their working mechanism and their output. In this algorithm we have used a function named next_permutation() , which takes two Bidirectional Iterators namely, (here vector::iterator) nodes.begin() and nodes.end() .

This functions returns a Boolean Type (i.e. either true or false ).

Working Mechanism :

This function rearranges the objects in [nodes.begin(),nodes.end()] , where the [] represents both iterator inclusive, in a lexicographical order . If there exits a greater lexicographical arrangement than the current arrangement then the function returns true else it returns false.

Lexicographical order is also known as dictionary order in mathematics.

The Main Code

The time complexity of the algorithm is dependent upon the number of nodes. If the number of nodes is n then the time complexity will be proportional to n! (factorial of n) i.e. O(n!) .

The most amount of space in this graph algorithm is taken by the adjacent matrix which is a n * n two dimensional matrix, where n is the number of nodes. Hence the space complexity is O(n^2) .

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Brute force solution to the Travelling Salesman Problem.

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The Travelling Salesman Problem (TSP) asks the question:

Suppose you are a salesman who needs to travel between multiple cities. Given their location, what is the shortest possible route you could take to visit each city once and return to the origin?

This problem has been historically difficult to solve. Although there are many approximation algorithms, the only exact solution (that we know of) runs in time O(n!) . This is the brute-force method of checking every permutation of cities and returning route with the shortest distance.

This program visualises the brute-force method on a slight variation of the TSP. The variation being, you do not need to start and end in the same city.

The circles represent cities and the lines connecting them represent the route. The brighter lines are the best route that the algorithm has discovered so far and the dimmer lines are the routes that the algorithm is currently checking. The dimmmer lines will disappear once the algorithm has found the best route.

To create a city, left click anywhere in the window. Click and drag any of the cities to move it. Right click on any of the cities to delete it. The program will begin calculating the best route as soon as you complete any of these actions.

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Algorithms for the Travelling Salesman Problem

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• The Travelling Salesman Problem (TSP) is a classic algorithmic problem in the field of computer science and operations research, focusing on optimization. It seeks the shortest possible route that visits every point in a set of locations just once.
• The TSP problem is highly applicable in the logistics sector , particularly in route planning and optimization for delivery services. TSP solving algorithms help to reduce travel costs and time.
• Real-world applications often require adaptations because they involve additional constraints like time windows, vehicle capacity, and customer preferences.
• Advances in technology and algorithms have led to more practical solutions for real-world routing problems. These include heuristic and metaheuristic approaches that provide good solutions quickly.
• Tools like Routific use sophisticated algorithms and artificial intelligence to solve TSP and other complex routing problems, transforming theoretical solutions into practical business applications.

The Traveling Salesman Problem (TSP) is the challenge of finding the shortest path or shortest possible route for a salesperson to take, given a starting point, a number of cities (nodes), and optionally an ending point. It is a well-known algorithmic problem in the fields of computer science and operations research, with important real-world applications for logistics and delivery businesses.

There are obviously a lot of different routes to choose from, but finding the best one — the one that will require the least distance or cost — is what mathematicians and computer scientists have spent decades trying to solve.

It’s much more than just an academic problem in graph theory. Finding more efficient routes using route optimization algorithms increases profitability for delivery businesses, and reduces greenhouse gas emissions because it means less distance traveled.

In theoretical computer science, the TSP has commanded so much attention because it’s so easy to describe yet so difficult to solve. The TSP is known to be a combinatorial optimization problem that’s an NP-hard problem, which means that the number of possible solution sequences grows exponential with the number of cities. Computer scientists have not found any algorithm for solving travelling salesman problems in polynomial time, and therefore rely on approximation algorithms to try numerous permutations and select the shortest route with the minimum cost.

The main problem can be solved by calculating every permutation using a brute force approach, and selecting the optimal solution. However, as the number of destinations increases, the corresponding number of roundtrips grows exponentially, soon surpassing the capabilities of even the fastest computers. With 10 destinations, there can be more than 300,000 roundtrip permutations. With 15 destinations, the number of possible routes could exceed 87 billion.

For larger real-world travelling salesman problems, when manual methods such as Google Maps Route Planner or Excel route planner no longer suffice, businesses rely on approximate solutions that are sufficiently optimized by using fast tsp algorithms that rely on heuristics. Finding the exact optimal solution using dynamic programming is usually not practical for large problems.

Three popular Travelling Salesman Problem Algorithms

Here are some of the most popular solutions to the Travelling Salesman Problem:

1. The brute-force approach

TThe brute-force approach, also known as the naive approach, calculates and compares all possible permutations of routes or paths to determine the shortest unique solution. To solve the TSP using the brute-force approach, you must calculate the total number of routes and then draw and list all the possible routes. Calculate the distance of each route and then choose the shortest one — this is the optimal solution. 

This is only feasible for small problems, rarely useful beyond theoretical computer science tutorials.

2. The branch and bound method

The branch and bound algorithm starts by creating an initial route, typically from the starting point to the first node in a set of cities. Then, it systematically explores different permutations to extend the route beyond the first pair of cities, one node at a time. Each time a new node is added, the algorithm calculates the current path's length and compares it to the optimal route found so far. If the current path is already longer than the optimal route, it "bounds" or prunes that branch of the exploration, as it would not lead to a more optimal solution.

This pruning is the key to making the algorithm efficient. By discarding unpromising paths, the search space is narrowed down, and the algorithm can focus on exploring only the most promising paths. The process continues until all possible routes are explored, and the shortest one is identified as the optimal solution to the traveling salesman problem. Branch and bound is an effective greedy approach for tackling NP-hard optimization problems like the traveling salesman problem.

3. The nearest neighbor method

To implement the nearest neighbor algorithm, we begin at a randomly selected starting point. From there, we find the closest unvisited node and add it to the sequencing. Then, we move to the next node and repeat the process of finding the nearest unvisited node until all nodes are included in the tour. Finally, we return to the starting city to complete the cycle.

While the nearest neighbor approach is relatively easy to understand and quick to execute, it rarely finds the optimal solution for the traveling salesperson problem. It can be significantly longer than the optimal route, especially for large and complex instances. Nonetheless, the nearest neighbor algorithm serves as a good starting point for tackling the traveling salesman problem and can be useful when a quick and reasonably good solution is needed.

This greedy algorithm can be used effectively as a way to generate an initial feasible solution quickly, to then feed into a more sophisticated local search algorithm, which then tweaks the solution until a given stopping condition. ‍

How route optimization algorithms work to solve the Travelling Salesman Problem.

Academic tsp solutions.

Academics have spent years trying to find the best solution to the Travelling Salesman Problem The following solutions were published in recent years:

• Machine learning speeds up vehicle routing : MIT researchers apply Machine Learning methods to solve large np-complete problems by solving sub-problems.
• Zero Suffix Method : Developed by Indian researchers, this method solves the classical symmetric TSP. 
• Biogeography‐based Optimization Algorithm : This method is designed based on the animals’ migration strategy to solve the problem of optimization. 
• Meta-Heuristic Multi Restart Iterated Local Search (MRSILS): The proponents of this research asserted that the meta-heuristic MRSILS is more efficient than the Genetic Algorithms when clusters are used. 
• Multi-Objective Evolutionary Algorithm : This method is designed for solving multiple TSP based on NSGA-II. 
• Multi-Agent System : This system is designed to solve the TSP of N cities with fixed resource. 

Real-world TSP applications

Despite the complexity of solving the Travelling Salesman Problem, approximate solutions — often produced using artificial intelligence and machine learning — are useful in all verticals.

For example, TSP solutions can help the logistics sector improve efficiency in the last mile. Last mile delivery is the final link in a supply chain, when goods move from a transportation hub, like a depot or a warehouse, to the end customer. Last mile delivery is also the leading cost driver in the supply chain. It costs an average of $10.1, but the customer only pays an average of $8.08 because companies absorb some of the cost to stay competitive. So bringing that cost down has a direct effect on business profitability.

Minimizing costs in last mile delivery is essentially a Vehicle Routing Problem (VRP). VRP, a generalized version of the travelling salesman problem, is one of the most widely studied problems in mathematical optimization. Instead of one best path, it deals with finding the most efficient set of routes or paths. The problem may involve multiple depots, hundreds of delivery locations, and several vehicles. As with the travelling salesman problem, determining the best solution to VRP is NP-complete.

Real-life TSP and VRP solvers

While academic solutions to TSP and VRP aim to provide the optimal solution to these NP-hard problems, many of them aren’t practical when solving real world problems, especially when it comes to solving last mile logistical challenges.

That’s because academic solvers strive for perfection and thus take a long time to compute the optimal solutions – hours, days, and sometimes years. If a delivery business needs to plan daily routes, they need a route solution within a matter of minutes. Their business depends on delivery route planning software so they can get their drivers and their goods out the door as soon as possible. Another popular alternative is to use Google maps route planner .

Real-life TSP and VRP solvers use route optimization algorithms that find near-optimal solutions in a fraction of the time, giving delivery businesses the ability to plan routes quickly and efficiently.

If you want to know more about real-life TSP and VRP solvers, check out the resources below 👇

Route Optimization API - TSP Solver

Route Optimization API - VRP Solver

Frequently Asked Questions

What is a hamiltonian cycle, and why is it important in solving the travelling salesman problem.

A Hamiltonian cycle is a complete loop that visits every vertex in a graph exactly once before returning to the starting vertex. It's crucial for the TSP because the problem essentially seeks to find the shortest Hamiltonian cycle that minimizes travel distance or time.

What role does linear programming play in solving the Travelling Salesman Problem?

Linear programming (LP) is a mathematical method used to optimize a linear objective function, subject to linear equality and inequality constraints. In the context of TSP, LP can help in formulating and solving relaxation of the problem to find bounds or approximate solutions, often by ignoring the integer constraints (integer programming being a subset of LP) to simplify the problem.

What is recursion, and how does it apply to the Travelling Salesman Problem?

Recursion involves a function calling itself to solve smaller sub-problems of the same type as the larger problem. In TSP, recursion is used in methods like the "Divide and Conquer" strategy, breaking down the problem into smaller, manageable subsets, which can be particularly useful in dynamic programming solutions. It reduces redundancy and computation time, making the problem more manageable.

Why is understanding time complexity important when studying the Travelling Salesman Problem?

Time complexity refers to the computational complexity that describes the amount of computer time it takes to solve a problem. For TSP, understanding time complexity is crucial because it helps predict the performance of different algorithms, especially given that TSP is NP-hard and solutions can become impractically slow as the number of cities increases.

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Solving the Vehicle Routing Problem (2024)

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How Route Optimization Impacts Our Earth

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Travelling Salesman Problem (Dynamic Approach)

Travelling salesman dynamic programming algorithm, implementation.

Travelling salesman problem is the most notorious computational problem. We can use brute-force approach to evaluate every possible tour and select the best one. For n number of vertices in a graph, there are (n−1)! number of possibilities. Thus, maintaining a higher complexity.

However, instead of using brute-force, using the dynamic programming approach will obtain the solution in lesser time, though there is no polynomial time algorithm.

Let us consider a graph G = (V,E) , where V is a set of cities and E is a set of weighted edges. An edge e(u, v) represents that vertices u and v are connected. Distance between vertex u and v is d(u, v) , which should be non-negative.

Suppose we have started at city 1 and after visiting some cities now we are in city j . Hence, this is a partial tour. We certainly need to know j , since this will determine which cities are most convenient to visit next. We also need to know all the cities visited so far, so that we don't repeat any of them. Hence, this is an appropriate sub-problem.

For a subset of cities S $\epsilon$ {1,2,3,...,n} that includes 1 , and j $\epsilon$ S, let C(S, j) be the length of the shortest path visiting each node in S exactly once, starting at 1 and ending at j .

When |S|> 1 , we define 𝑪 C(S,1)= $\propto$ since the path cannot start and end at 1.

Now, let express C(S, j) in terms of smaller sub-problems. We need to start at 1 and end at j . We should select the next city in such a way that

$$C\left ( S,j \right )\, =\, min\, C\left ( S\, -\, \left\{j \right\},i \right )\, +\, d\left ( i,j \right )\: where\: i\: \epsilon \: S\: and\: i\neq j$$

There are at the most 2 n .n sub-problems and each one takes linear time to solve. Therefore, the total running time is O(2 n .n 2 ) .

In the following example, we will illustrate the steps to solve the travelling salesman problem.

From the above graph, the following table is prepared.

$$Cost\left ( 2,\Phi ,1 \right )\, =\, d\left ( 2,1 \right )\,=\,5$$

$$Cost\left ( 3,\Phi ,1 \right )\, =\, d\left ( 3,1 \right )\, =\, 6$$

$$Cost\left ( 4,\Phi ,1 \right )\, =\, d\left ( 4,1 \right )\, =\, 8$$

$$Cost(i,s)=min\left\{Cos\left ( j,s-(j) \right )\, +\,d\left [ i,j \right ] \right\}$$

$$Cost(2,\left\{3 \right\},1)=d[2,3]\, +\, Cost\left ( 3,\Phi ,1 \right )\, =\, 9\, +\, 6\, =\, 15$$

$$Cost(2,\left\{4 \right\},1)=d[2,4]\, +\, Cost\left ( 4,\Phi ,1 \right )\, =\, 10\, +\, 8\, =\, 18$$

$$Cost(3,\left\{2 \right\},1)=d[3,2]\, +\, Cost\left ( 2,\Phi ,1 \right )\, =\, 13\, +\, 5\, =\, 18$$

$$Cost(3,\left\{4 \right\},1)=d[3,4]\, +\, Cost\left ( 4,\Phi ,1 \right )\, =\, 12\, +\, 8\, =\, 20$$

$$Cost(4,\left\{3 \right\},1)=d[4,3]\, +\, Cost\left ( 3,\Phi ,1 \right )\, =\, 9\, +\, 6\, =\, 15$$

$$Cost(4,\left\{2 \right\},1)=d[4,2]\, +\, Cost\left ( 2,\Phi ,1 \right )\, =\, 8\, +\, 5\, =\, 13$$

$$Cost(2,\left\{3,4 \right\},1)=min\left\{\begin{matrix} d\left [ 2,3 \right ]\,+ \,Cost\left ( 3,\left\{ 4\right\},1 \right )\, =\, 9\, +\, 20\, =\, 29 \\ d\left [ 2,4 \right ]\,+ \,Cost\left ( 4,\left\{ 3\right\},1 \right )\, =\, 10\, +\, 15\, =\, 25 \\ \end{matrix}\right.\, =\,25$$

$$Cost(3,\left\{2,4 \right\},1)=min\left\{\begin{matrix} d\left [ 3,2 \right ]\,+ \,Cost\left ( 2,\left\{ 4\right\},1 \right )\, =\, 13\, +\, 18\, =\, 31 \\ d\left [ 3,4 \right ]\,+ \,Cost\left ( 4,\left\{ 2\right\},1 \right )\, =\, 12\, +\, 13\, =\, 25 \\ \end{matrix}\right.\, =\,25$$

$$Cost(4,\left\{2,3 \right\},1)=min\left\{\begin{matrix} d\left [ 4,2 \right ]\,+ \,Cost\left ( 2,\left\{ 3\right\},1 \right )\, =\, 8\, +\, 15\, =\, 23 \\ d\left [ 4,3 \right ]\,+ \,Cost\left ( 3,\left\{ 2\right\},1 \right )\, =\, 9\, +\, 18\, =\, 27 \\ \end{matrix}\right.\, =\,23$$

$$Cost(1,\left\{2,3,4 \right\},1)=min\left\{\begin{matrix} d\left [ 1,2 \right ]\,+ \,Cost\left ( 2,\left\{ 3,4\right\},1 \right )\, =\, 10\, +\, 25\, =\, 35 \\ d\left [ 1,3 \right ]\,+ \,Cost\left ( 3,\left\{ 2,4\right\},1 \right )\, =\, 15\, +\, 25\, =\, 40 \\ d\left [ 1,4 \right ]\,+ \,Cost\left ( 4,\left\{ 2,3\right\},1 \right )\, =\, 20\, +\, 23\, =\, 43 \\ \end{matrix}\right.\, =\, 35$$

The minimum cost path is 35.

Start from cost {1, {2, 3, 4}, 1}, we get the minimum value for d [1, 2]. When s = 3, select the path from 1 to 2 (cost is 10) then go backwards. When s = 2, we get the minimum value for d [4, 2]. Select the path from 2 to 4 (cost is 10) then go backwards.

When s = 1, we get the minimum value for d [4, 2] but 2 and 4 is already selected. Therefore, we select d [4, 3] (two possible values are 15 for d [2, 3] and d [4, 3], but our last node of the path is 4). Select path 4 to 3 (cost is 9), then go to s = ϕ step. We get the minimum value for d [3, 1] (cost is 6).

Following are the implementations of the above approach in various programming languages −