how to solve travelling salesman problem in java

Traveling Salesman Problem with Genetic Algorithms in Java

• Introduction

Genetic algorithms are a part of a family of algorithms for global optimization called Evolutionary Computation , which is comprised of artificial intelligence metaheuristics with randomization inspired by biology.

In the previous article, Introduction to Genetic Algorithms in Java , we've covered the terminology and theory behind all of the things you'd need to know to successfully implement a genetic algorithm.

• Implementing a Genetic Algorithm

To showcase what we can do with genetic algorithms, let's solve The Traveling Salesman Problem (TSP) in Java.

TSP formulation : A traveling salesman needs to go through n cities to sell his merchandise. There's a road between each two cities, but some roads are longer and more dangerous than others. Given the cities and the cost of traveling between each two cities, what's the cheapest way for the salesman to visit all of the cities and come back to the starting city, without passing through any city twice?

Although this may seem like a simple feat, it's worth noting that this is an NP-hard problem. There's no algorithm to solve it in polynomial time. Genetic algorithms can only approximate the solution.

Because the solution is rather long, I'll be breaking it down function by function to explain it here. If you want to preview and/or try the entire implementation, you can find the IntelliJ project on GitHub .

• Genome Representation

First, we need an individual to represent a candidate solution. Logically, for this we'll use a class to store the random generation, fitness function, the fitness itself, etc.

To make it easier to calculate fitness for individuals and compare them, we'll also make it implement Comparable :

Despite using a class, what our individual essentially is will be only one of its attributes. If we think of TSP, we could enumerate our cities from 0 to n-1 . A solution to the problem would be an array of cities so that the cost of going through them in that order is minimized.

For example, 0-3-1-2-0 . We can store that in an ArrayList because the Collections Framework makes it really convenient, but you can use any array-like structure.

The attributes of our class are as follows:

When it comes to constructors we'll make two - one that makes a random genome, and one that takes an already made genome as an argument:

• Fitness Function

You may have noticed that we called the calculateFitness() method to assign a fitness value to the object attribute during construction. The function works by following the path laid out in the genome through the price matrix, and adding up the cost.

The fitness turns out to be the actual cost of taking a certain path. We'll want to minimize this cost, so we'll be facing a minimization problem:

• The Genetic Algorithm Class

The heart of the algorithm will take place in another class, called TravelingSalesman . This class will perform our evolution, and all of the other functions will be contained within it:

• Generation size is the number of genomes/individuals in each generation/population. This parameter is also often called the population size.
• Genome size is the length of the genome ArrayList , which will be equal to the numberOfCities-1 . The two variables are separated for clarity in the rest of the code. This parameter is also often called the chromosome length.
• Reproduction size is the number of genomes who'll be selected to reproduce to make the next generation. This parameter is also often called the crossover rate.
• Max iteration is the maximum number of generations the program will evolve before terminating, in case there's no convergence before then.
• Mutation rate refers to the frequency of mutations when creating a new generation.
• Travel prices is a matrix of the prices of travel between each two cities - this matrix will have 0s on the diagonal and symmetrical values in its lower and upper triangle.
• Starting city is the index of the starting city.
• Target fitness is the fitness the best genome has to reach according to the objective function (which will in our implementation be the same as the fitness function) for the program to terminate early. Sometimes setting a target fitness can shorten a program if we only need a specific value or better. Here, if we want to keep our costs below a certain number, but don't care how low exactly, we can use it to set that threshold.
• Tournament size is the size of the tournament for tournament selection.
• Selection type will determine the type of selection we're using - we'll implement both roulette and tournament. Here's the enum for SelectionType :

Although the tournament selection method prevails in most cases, there are situations where you'd want to use other methods. Since a lot of genetic algorithms use the same codebase (the individuals and fitness functions change), it's good practice to add more options to the algorithm.

We'll be implementing both roulette and tournament selection:

The crossover for TSP is atypical. Because each genome is a permutation of the list of cities, we can't just crossover two parents conventionally. Look at the following example (the starting city 0 is implicitly the first and last step):

2-4-3|1-6-5

4-6-5|3-1-2

What would happen if we crossed these two at the point denoted with a | ?

2-4-3-3-1-2

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4-6-5-1-6-5

Uh-oh. These don't go through all the cities and they visit some cities twice, violating multiple conditions of the problem.

So if we can't use conventional crossover, what do we use?

The technique we'll be using is called Partially Mapped Crossover or PMX for short. PMX randomly picks one crossover point, but unlike one-point crossover it doesn't just swap elements from two parents, but instead swaps the elements within them. I find that the process is most comprehensible from an illustration, and we can use the example we've previously had trouble with:

As can be seen here, we swap the i th element of one of the parents with the element equivalent in value to the i th element of the other. By doing this, we preserve the properties of permutations. We repeat this process to create the second child as well (with the original values of the parent genomes):

Mutation is pretty straightforward - if we pass a probability check we mutate by swapping two cities in the genome. Otherwise, we just return the original genome:

• Generation Replacement Policies

We're using a generational algorithm, so we make an entirely new population of children:

• Termination

We terminate under the following conditions:

• the number of generations has reached maxIterations
• the best genome's path length is lower than the target path length
• Running time

The best way to evaluate if this algorithm works properly is to generate some random problems for it and evaluate the run-time:

Our average running time is 51972ms, which is about 52 seconds. This is when the input is four cities long, meaning we'd have to wait longer for larger numbers of cities. This may seem like a lot, but implementing a genetic algorithm takes significantly less time than coming up with a perfect solution for a problem.

While this specific problem could be solved using another method, certain problems can't.

For an example, NASA used a genetic algorithm to generate the optimal shape of a spacecraft antenna for the best radiation pattern.

• Genetic Algorithms for Optimizing Genetic Algorithms?

As an interesting aside, genetic algorithms are sometimes used to optimize themselves. You create a genetic algorithm which runs another genetic algorithm, and rates its execution speed and output as its fitness and adjusts its parameters to maximize performance.

A similar technique is used in NeuroEvolution of Augmenting Topologies , or NEAT, where a genetic algorithm is continuously improving a neural network and hinting how to change structure to accommodate new environments.

Genetic algorithms are a powerful and convenient tool. They may not be as fast as solutions crafted specifically for the problem at hand, and we may not have much in the way of mathematical proof of their effectiveness, but they can solve any search problem of any difficulty, and are not too difficult to master and apply. And as a cherry on the top, they're endlessly fascinating to implement when you think of the evolutionary processes they're based on and how you're a mastermind behind a mini-evolution of your own.

If you want to play further with TSP implemented in this article, this is a reminder that you can find it on GitHub . It has some handy functions for printing out generations, travel costs, generating random travel costs for a given number of cities, etc. so you can test out how it works on different sizes of input, or even meddle with the attributes such as mutation rate, tournament size, and similar.

You might also like...

• Introduction to Genetic Algorithms in Java
• Graphs in Java: Representing Graphs in Code
• Bubble Sort in Java
• Simulated Annealing Optimization Algorithm in Java
• Binary Search in Java

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Travelling Salesman Problem (TSP):  

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. 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 know solution for this problem. The following are different solutions for the traveling salesman problem. 

Naive Solution:  

1) Consider city 1 as the starting and ending point.

2) Generate all (n-1)! Permutations of cities. 

3) Calculate the cost of every permutation and keep track of the minimum cost permutation. 

4) Return the permutation with minimum cost. 

Time Complexity: Θ(n!) 

Dynamic Programming:  

Let the given set of vertices be {1, 2, 3, 4,….n}. Let us consider 1 as starting and ending point of output. For every other vertex I (other than 1), we find the minimum cost path with 1 as the starting point, I as the ending point, and all vertices appearing exactly once. Let the cost of this path cost (i), and the cost of the corresponding Cycle would cost (i) + dist(i, 1) where dist(i, 1) is the distance from I to 1. Finally, we return the minimum of all [cost(i) + dist(i, 1)] values. This looks simple so far. 

Now the question is how to get cost(i)? To calculate the cost(i) using Dynamic Programming, we need to have some recursive relation in terms of sub-problems. 

Let us define a term C(S, i) be the cost of the minimum cost path visiting each vertex in set S exactly once, starting at 1 and ending at i . We start with all subsets of size 2 and calculate C(S, i) for all subsets where S is the subset, then we calculate C(S, i) for all subsets S of size 3 and so on. Note that 1 must be present in every subset.

Below is the dynamic programming solution for the problem using top down recursive+memoized approach:-

For maintaining the subsets we can use the bitmasks to represent the remaining nodes in our subset. Since bits are faster to operate and there are only few nodes in graph, bitmasks is better to use.

For example: –  

10100 represents node 2 and node 4 are left in set to be processed

010010 represents node 1 and 4 are left in subset.

NOTE:- ignore the 0th bit since our graph is 1-based

Time Complexity : O(n 2 *2 n ) where O(n* 2 n) are maximum number of unique subproblems/states and O(n) for transition (through for loop as in code) in every states.

Auxiliary Space: O(n*2 n ), where n is number of Nodes/Cities here.

For a set of size n, we consider n-2 subsets each of size n-1 such that all subsets don’t have nth in them. Using the above recurrence relation, we can write a dynamic programming-based solution. There are at most O(n*2 n ) subproblems, and each one takes linear time to solve. The total running time is therefore O(n 2 *2 n ). The time complexity is much less than O(n!) but still exponential. The space required is also exponential. So this approach is also infeasible even for a slightly higher number of vertices. We will soon be discussing approximate algorithms for the traveling salesman problem.

Next Article: Traveling Salesman Problem | Set 2  

References:  

http://www.lsi.upc.edu/~mjserna/docencia/algofib/P07/dynprog.pdf  

http://www.cs.berkeley.edu/~vazirani/algorithms/chap6.pdf  

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Travelling Salesman Problem using Dynamic Programming

• Jun 17, 2023
• 7 Minutes Read
• By Harshita Rajput

Imagine you have to go to a party in another city and return back home as soon as possible. Which algorithm will you apply to reach your home so that you cover the shortest distance? Let us understand Travelling Salesman Problem in this guide.

What is Traveling Salesman Problem?

The Traveling Salesman Problem classic optimization problem in Computer Science. There is a salesman who wants to travel over a number of cities and he has to return back to the original city he started from and he has the choice to start from any of the cities he wants to. During his journey, we need to minimize the total distance traveled by him. 

Here is the main problem statement:

" We will be given a graph that will be represented in the form of an adjacency matrix, say distance, that denotes the distance between 2 cities, say, i and j. The distance[i][j] denotes the distance between 2 cities if the value of dist[i][j] == 0; this implies the given cities are not connected i.e. no direct edge exists between them. We can start from any node and we need to return to the original city through the shortest distance covering all the cities. "

So, basically, we need to complete a cycle in this question, and such a cycle is known as the Hamiltonian cycle. A Hamiltonian cycle is a set of edges such that every node is visited only once and we come back to the original position. So in short, in this problem, we need to return a minimum weight Hamiltonian cycle. We just have to minimize the total cost of the cycle.

Let's first try to get a Hamiltonian cycle with minimum weight from the graph below. Given the following graph and the distance between cities he has traveled, let's find out the shortest path in which he can travel all the cities.

The Hamiltonian cycle with min weight can be:

How does the Algorithm Work?

Let's take an example of a graph say:

• As we have an option to start from any position, we start from A.
• Making a recursion tree and using bit masking. 

As 4 cities are there, the bit mask here would be = 0000 (at the beginning) respectively denoting D, C, B, and A  cities. The bit shows 0 for a particular city if it has not been visited and 1 if already been visited. So if bit mask = 1111 that means all bits are set, which implies all cities have been visited and we can denote this by 1<

We try to find the possible options that exist. This can be done by iterating over the adjacency matrix of A or the other way of doing so can be iterating over other cities and checking if they can be a possible option to travel from A. As we begin, we visit city A, and the bit mask value is now changed to 0001.

Each of these recursive branches denotes one path.

Now, as we move forward from A to B, we add the distance in a variable, say, ans. And we mark city B as visited.

We add the distance to the answer. Now we only visit the cities from B that aren't visited yet so we have 2 options available C and D.

We then move to C and then to D while marking them visited and adding the cost to the and. Now here, we have explored each unvisited city. So now we backtrack, while marking each city as unvisited and go to the original city if a direct edge between them exists. As there is a direct edge from D to A and C to A, the nodes return the cost required to go back to the original source.

This way we explore each available option from the starting node. We also maintain a minimum cost variable that maintains the minimum cost spent to visit each city and return to the original city.

The recursion tree would look like this:

Traveling Salesman Algorithm

Here is the algorithm for Travelling Salesman Problem:

• Define the mask as (1<<n)-1.
• Create a function, say, tsp() having mask and city as arguments. As the mask denotes a set of cities visited so far, we iterate over the mask and get to know which city isn't visited.
• The base case is when we visit all the cities i.e. mask has all of its bits set, we return the distance between the current city and the original city.
• If the city is not visited, we make recursive calls by adding the distance of the original city and the city not visited and try to get the distance remaining from recursive calls
• In the recursive call, we pass a mask with the current city visited and the current city.

Travelling Salesman Problem in Java

Here is the complete Java Program to implement Travelling Salesman Problem:

Time & Space Complexity 

Here we have fixed the source node as we get the same answer whether we start from A or B or any other node because of that cyclic permutation in this case.

The time complexity of the Travelling Salesman Problem comes out to be O(n-1!). The space complexity of this code comes out to be O(n).

Now we can notice that this bitmask can only have 2^n values ranging from 0000 to 1111. The city can take n values. So the total number of distinct states comes out to be (2^n )* n. Also in this approach, some sub-problems were getting overlapped.

To optimize this approach, we use DP. We will create a 2D DP table of the values and we memorize all the results for the given value of mask and city.

Optimized Approach using Dynamic Programming

Here is the simple way to implement the Optimized Approach using Dynamic Programming. Initially we make a 2D array of [2^n][n] and initially put -1 at each position as initially, all states have values = -1.

To avoid overlapping subproblems i.e. avoiding a state which has already been computed, we check dp[bitMask][city]. If this comes out to be -1, implies the city hasn't been visited, else the city has already been visited and we return dp[bitMask][city] = ans.

Here is the Java Code for Travelling Salesman Problem using Dynamic Programming:

The time complexity comes out to be O(n ^2 * (2*n)). As we are using an extra space of (2^n) *n. The space complexity comes out to be O((2^n) *n).

Conclusion 

Travelling Salesman is a classic problem demanding in-depth knowledge of graphs and DP. It is an NP-hard problem. The optimization used in this problem can be implemented in other questions of this type too. This problem is necessary from the interview point of view, so I hope you all understood it well!

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Harshita Rajput

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