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Travelling salesman problem explained

• Published on April 25, 2024
• by Oguzhan Uyar
• Last updated: 1 min ago

Revolutionize your understanding of the Travelling Salesman Problem (TSP); a mind-boggling conundrum that has compelled academia and industry alike. In the upcoming lines, we decode the key concepts, algorithms, and anticipated solutions for 2024 to this age-old dilemma.

Now, picture the TSP as a globetrotting traveling salesman who’s whirlwind journey. He must stop at every city once, only the origin city once, and find the quickest shortest possible route back home. If daunting to visualize, consider this: the possible number of routes in the problem concerning just 20 cities exceeds the number of atoms in the observable universe.

Fathom the sheer magnitude?

So, what is the Travelling Salesman Problem, and why has it remained unsolved for years? Let’s snap together the puzzle of this notorious problem that spans mathematics, computer science, and beyond. Welcome to an insightful voyage into the astonishing world of the TSP.

Understanding the Travelling Salesman Problem (TSP): Key Concepts and Algorithms

Defining the travelling salesman problem: a comprehensive overview.

The Travelling Salesman Problem, often abbreviated as TSP, has a strong footing in the realm of computer science. To the untrained eye, it presents itself as a puzzle: a salesperson or traveling salesman must traverse through a number of specific cities before an ending point and return to their ending point as of origin, managing to do so in the shortest possible distance. But this problem is not simply a conundrum for those fond of riddles; it holds immense significance in the broad field of computer science and optimization.

The sheer computational complexity of TSP is what sets it apart, and incidentally, why it is considered a challenging problem to solve. Its complexity derives from the factorial nature of the problem: whenever a new city is added, the total number of possibilities increases exponentially. Thus, as the problem scope enlarges, it swiftly becomes computationally prohibitive to simply calculate all possible solutions to identify an optimal shortest route through. Consequently, developing effective and efficient algorithms to solve the TSP has become a priority in the world of computational complexity.

Polynomial Time Solution: The TSP can be solved by a deterministic Turing machine in polynomial time, which means that the number of steps to solve the problem can be at most 1.5 times the optimal global solution.

One such algorithm is the dynamic programming approach that can solve TSP problems in polynomial time. The approach uses a recursive formula to compute the shortest possible route that visits all other nodes in the cities exactly once and ends at all nodes in the starting point or city. Moreover, linear programming and approximation algorithms have also been used to find near-optimal solutions.

Unraveling TSP Algorithms: From Brute Force to Heuristics

A multitude of algorithms have been developed to contend with the TSP. The brute force method, for example, entails considering the shortest distance for all possible permutations of cities, calculating the total distance for six cities along each route, and selecting the shortest distance for one. While brute force promises an optimal solution, it suffers from exponential computational complexity, rendering it unfeasible for large datasets.

TSP Complexity: A TSP with just 10 cities has 362,880 possible routes , making brute force infeasible for larger instances.

On the other hand, we have heuristic algorithms that generate good, albeit non-optimal, solutions in reasonable timeframes. The greedy algorithm, for instance, initiates from a starting city and looks for the shortest distance from other nodes to the next node minimizes the distance, and guarantees speed but is not necessarily an optimal solution.

Algorithmic Potential: Local Solutions 4/3 Times Optimal Global: The theoretical conjecture suggests an algorithm that can provide a local solution within 4/3 times the optimal global solution.

Record Local TSP Solution: The record for a local solution to the Traveling Salesman Problem (TSP) is 1.4 times the optimal global solution, achieved in September 2012 by Andr´as Seb˝o and Jens Vygen.

Exploring further, we encounter more refined heuristic solutions such as the genetic algorithm and simulated annealing algorithm. These algorithms deploy probabilistic rules, with the former incorporating principles of natural evolution and the latter being inspired by the cooling process in metallurgy. They differ significantly from each other in terms of how they search for solutions, but when compared to brute force, they often offer a promising trade-off between quality and computational effort.

Certainly, the TSP is far more than a problem to puzzle over during a Sunday afternoon tea. It’s a complex computational task with real-world implications, and unraveling it requires more than brute force; it demands the application of sophisticated algorithms designed to balance efficiency and quality of results. Nevertheless, with a better understanding of the problem statement and its dynamics, you can unlock the potential to not just solve problems faster, but to do so more intelligently.

Practical Solutions to the Travelling Salesman Problem

Implementing tsp solutions: a step-by-step guide, a comprehensive process for implementing tsp solutions.

Developing complete, practical solutions for the Travelling Salesman Problem (TSP) requires a good grasp of specific algorithms. Visual aids, for example, such as finding all the edges leading out of a given city, can help to simplify the process.

TSP's Computer Science Quest for Shortest Routes: The TSP involves finding the shortest route to visit a set of cities exactly once before returning to the starting city, aiming to minimize travel distance.

One popular approach to TSP problem solving is the dynamic programming approach, which involves breaking down the problem into smaller sub-problems and recursively solving them. Other approaches include approximation algorithms, which generate near-optimal solutions to small problems in polynomial time, and bound algorithms, which aim to find an upper bound on the optimal solution given graph top.

TSP Variants: ASTP and STSP The TSP can be divided into two types: the asymmetric traveling salesman problem (ASTP) and the symmetric traveling salesman problem (STSP)

Practical code implementation

Before diving into code manipulations, it’s worth noting that practical implementation varies based on the specifics of the problem and the chosen algorithm. Let’s further deconstruct these notions by examining the steps for code implementation for a pre-determined shortest path first. It’s crucial to efficiently concede each point of the shortest path, call other nodes, connect each node, and conclude each route. Hereby, I shall delve into the practicalities of coding from an output standpoint, concentrating on readability, scalability, and execution efficiency.

TSP Instance Size: The size of the TSP instances used in the studies is 100 nodes.

Cluster Quantity in TSP Instances: The number of clusters in the TSP instances used in the studies is 10 .

The Quantity of Instances per Cluster in Studies: The number of instances in each cluster used in the studies is 100.

Optimizing TSP Solutions: Tips and Tricks

Tsp solutions optimization techniques.

An optimized solution for the TSP is often accomplished using advanced algorithms and techniques. Optimization techniques allow professionals to solve more intricate problems, rendering them invaluable tools when handling large datasets and attempting to achieve the best possible solution. Several approaches can be taken, such as heuristic and metaheuristic approaches, that can provide near-optimal solutions with less use of resources.

Expert Advice for Maximum Efficiency Minimum Cost

Even the most proficient problem solvers can gain from learning expert tips and tricks. These nuggets of wisdom often provide the breakthrough needed to turn a good solution into a great one. The TSP is no exception. Peering into the realm of experts can provide novel and creative ways to save time, increase computation speed, manage datasets, and achieve the most efficient shortest route around.

By comprehending and implementing these solutions to the Travelling Salesman Problem, one can unravel the complex web of nodes and paths. This equips any problem-solver with invaluable insights that make navigating through real-world TSP applications much more manageable.

Real-World Applications of the Travelling Salesman Problem

Tsp in logistics and supply chain management.

The Travelling Salesman Problem (TSP) is not confined to theoretical mathematics or theoretical computer science either, it shines in real-world applications too. One of its most potent applications resides in the field of logistics and supply chain management.

With globalization, the importance of more efficient routes for logistics and supply chain operations has risen dramatically. Optimizing routes and decreasing costs hold the key to success in this arena. Here’s where TSP comes into play.

In the vast complexity of supply chain networks, routing can be a colossal task. TSP algorithms bring clarity to the chaos, eliminating redundant paths and pointing to the optimal route covering the most distance, shortest path, and least distance between all the necessary points—leading to a drastic dip in transportation costs and delivery times.

Real-World Examples

Consider the case of UPS – the multinational package delivery company. They’ve reportedly saved hundreds of millions of dollars yearly by implementing route optimization algorithms originating from the Travelling Salesman Problem. The solution, named ORION, helped UPS reduce the distance driven by their drivers roughly by 100 million miles annually.

TSP in GIS and Urban Planning: Optimal Route

TSP’s contributions to cities aren’t confined to logistics; it is invaluable in Geographic Information Systems (GIS) and urban planning as well. A city’s efficiency revolves around transportation. The better the transport networks and systems, the higher the city’s productivity. And as city authorities continuously strive to upgrade transportation systems, they find an ally in TSP.

Advanced GIS systems apply TSP to design more efficient routes and routing systems for public transportation. The aim of the route is to reach maximum locations with the least travel time and least distance, ensuring essential amenities are accessible to everyone in the city.

The city of Singapore, known for its efficient public transportation system, owes part of its success to similar routing algorithms. The Land Transport Authority uses TSP-related solutions to plan bus routes, reducing travel durations and enhancing accessibility across the city.

Delving Deeper: The Complexity and History of the Travelling Salesman Problem

Understanding the complexity of tsp.

TSP is fascinating not just for its inherent practicality but also for the complexity it brings along. TSP belongs to a class of problems known as NP-hard, a category that houses some of the most complex problems in computer science. But what makes TSP so gnarled?

Unravel the Intricacies: Why is TSP complex?

TSP is a combinatorial optimization problem, which simply means that it’s all about figuring out the most often optimal solution among a vast number of possible solutions or combinations of approximate solutions. The real challenge comes from an innocent-sounding feature of TSP: As the number of cities (we call them nodes) increases, the complexity heightens exponentially, not linearly. The number of possible routes takes a dramatic upward turn as you add more nodes, clearly exemplifying why TSP is no walk-in-the-park problem.

NP-hard and TSP: A Connection Explored

In computer science, we use the terms P and NP for classifying problems. P stands for problems where a solution can be found in ‘polynomial time’. NP stands for ‘nondeterministic polynomial time’, which includes problems where a solution can be verified in polynomial time. The concept of ‘NP-hardness’ applies to TSP. Any problem that can be ‘reduced’ to an NP problem in polynomial time is described as NP-hard. In simpler terms, it’s harder than the hardest problems of NP. TSP holds a famed position in this class, making it a noteworthy example of an NP-hard problem.

A Look Back: The History of the Travelling Salesman Problem

TSP is not a new kid on the block. Its roots can be traced back to the 1800s, and the journey since then has been nothing short of a compelling tale of mathematical and computational advancements.

The Origins and Evolution of TSP

Believe it or not, the Travelling Salesman Problem was first formulated as a mathematical problem in the 1800s. This was way before the advent of modern computers. Mathematicians and logisticians were intrigued by the problem and its implications. However, it wasn’t until the mid-20th century that the problem started to gain more widespread attention. Over the years, the TSP has transformed from a theoretical model to a practical problem that helps solve real-world issues.

A Classic Shortest Route Problem Since 1930s: The TSP is a classic optimization problem within the field of operations research, first studied during the 1930s.

Milestones in TSP Research

Looking at all the cities and milestones in TSP research, the story is truly impressive. From some of the initial heuristic algorithms to solve smaller instances of TSP, to geometric methods and then approximation algorithms, TSP research has seen a lot. More recently, we have also seen practical solutions to TSP using quantum computing — a testament to how far we’ve come. Each of these milestones signifies an innovative shift in how we understand and solve TSP.

Wrapping Up The Journey With Algorithms and Solutions

The Travelling Salesman Problem (TSP) remains a complex enigma of business logistics, yet, the advent of sophisticated algorithms and innovative solutions are paving avenues to  optimize routing, reduce travel costs further,  and enhance customer interactions.

Navigating the TSP intricacies is no longer a daunting challenge but a worthwhile investment in refining operational efficiency and delivering unparalleled customer experiences. With advanced toolsets and intelligent systems, businesses are closer to making more informed, strategic decisions.

Now, it’s your turn. Reflect on your current logistics complexities. How can your business benefit from implementing these key concepts and algorithms? Consider where you could incorporate these practices to streamline and revolutionize your daily operations.

Remember, in the world of dynamic business operations, mastering the TSP is not just an option, but a strategic imperative. As you move forward into 2024, embrace the art of solving the TSP. Unravel the potential, decipher the complexities, and unlock new horizons for your business.

Ready for the challenge?

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The Traveling Salesman Problem

The Traveling Salesman Problem states that you are a salesperson and you must visit a number of cities or towns.

Rules : Visit every city only once, then return back to the city you started in.

Goal : Find the shortest possible route.

Except for the Held-Karp algorithm (which is quite advanced and time consuming, (\(O(2^n n^2)\)), and will not be described here), there is no other way to find the shortest route than to check all possible routes.

This means that the time complexity for solving this problem is \(O(n!)\), which means 720 routes needs to be checked for 6 cities, 40,320 routes must be checked for 8 cities, and if you have 10 cities to visit, more than 3.6 million routes must be checked!

Note: "!", or "factorial", is a mathematical operation used in combinatorics to find out how many possible ways something can be done. If there are 4 cities, each city is connected to every other city, and we must visit every city exactly once, there are \(4!= 4 \cdot 3 \cdot 2 \cdot 1 = 24\) different routes we can take to visit those cities.

The Traveling Salesman Problem (TSP) is a problem that is interesting to study because it is very practical, but so time consuming to solve, that it becomes nearly impossible to find the shortest route, even in a graph with just 20-30 vertices.

If we had an effective algorithm for solving The Traveling Salesman Problem, the consequences would be very big in many sectors, like for example chip design, vehicle routing, telecommunications, and urban planning.

Checking All Routes to Solve The Traveling Salesman Problem

To find the optimal solution to The Traveling Salesman Problem, we will check all possible routes, and every time we find a shorter route, we will store it, so that in the end we will have the shortest route.

Good: Finds the overall shortest route.

Bad: Requires an awful lot of calculation, especially for a large amount of cities, which means it is very time consuming.

How it works:

• Check the length of every possible route, one route at a time.
• Is the current route shorter than the shortest route found so far? If so, store the new shortest route.
• After checking all routes, the stored route is the shortest one.

Such a way of finding the solution to a problem is called brute force .

Brute force is not really an algorithm, it just means finding the solution by checking all possibilities, usually because of a lack of a better way to do it.

Speed: {{ inpVal }}

Finding the shortest route in The Traveling Salesman Problem by checking all routes (brute force).

Progress: {{progress}}%

n = {{vertices}} cities

{{vertices}}!={{posRoutes}} possible routes

Show every route: {{showCompares}}

The reason why the brute force approach of finding the shortest route (as shown above) is so time consuming is that we are checking all routes, and the number of possible routes increases really fast when the number of cities increases.

Finding the optimal solution to the Traveling Salesman Problem by checking all possible routes (brute force):

Using A Greedy Algorithm to Solve The Traveling Salesman Problem

Since checking every possible route to solve the Traveling Salesman Problem (like we did above) is so incredibly time consuming, we can instead find a short route by just going to the nearest unvisited city in each step, which is much faster.

Good: Finds a solution to the Traveling Salesman Problem much faster than by checking all routes.

Bad: Does not find the overall shortest route, it just finds a route that is much shorter than an average random route.

• Visit every city.
• The next city to visit is always the nearest of the unvisited cities from the city you are currently in.
• After visiting all cities, go back to the city you started in.

This way of finding an approximation to the shortest route in the Traveling Salesman Problem, by just going to the nearest unvisited city in each step, is called a greedy algorithm .

Finding an approximation to the shortest route in The Traveling Salesman Problem by always going to the nearest unvisited neighbor (greedy algorithm).

As you can see by running this simulation a few times, the routes that are found are not completely unreasonable. Except for a few times when the lines cross perhaps, especially towards the end of the algorithm, the resulting route is a lot shorter than we would get by choosing the next city at random.

Finding a near-optimal solution to the Traveling Salesman Problem using the nearest-neighbor algorithm (greedy):

Other Algorithms That Find Near-Optimal Solutions to The Traveling Salesman Problem

In addition to using a greedy algorithm to solve the Traveling Salesman Problem, there are also other algorithms that can find approximations to the shortest route.

These algorithms are popular because they are much more effective than to actually check all possible solutions, but as with the greedy algorithm above, they do not find the overall shortest route.

Algorithms used to find a near-optimal solution to the Traveling Salesman Problem include:

• 2-opt Heuristic: An algorithm that improves the solution step-by-step, in each step removing two edges and reconnecting the two paths in a different way to reduce the total path length.
• Genetic Algorithm: This is a type of algorithm inspired by the process of natural selection and use techniques such as selection, mutation, and crossover to evolve solutions to problems, including the TSP.
• Simulated Annealing: This method is inspired by the process of annealing in metallurgy. It involves heating and then slowly cooling a material to decrease defects. In the context of TSP, it's used to find a near-optimal solution by exploring the solution space in a way that allows for occasional moves to worse solutions, which helps to avoid getting stuck in local minima.
• Ant Colony Optimization: This algorithm is inspired by the behavior of ants in finding paths from the colony to food sources. It's a more complex probabilistic technique for solving computational problems which can be mapped to finding good paths through graphs.

Time Complexity for Solving The Traveling Salesman Problem

To get a near-optimal solution fast, we can use a greedy algorithm that just goes to the nearest unvisited city in each step, like in the second simulation on this page.

Solving The Traveling Salesman Problem in a greedy way like that, means that at each step, the distances from the current city to all other unvisited cities are compared, and that gives us a time complexity of \(O(n^2) \).

But finding the shortest route of them all requires a lot more operations, and the time complexity for that is \(O(n!)\), like mentioned earlier, which means that for 4 cities, there are 4! possible routes, which is the same as \(4 \cdot 3 \cdot 2 \cdot 1 = 24\). And for just 12 cities for example, there are \(12! = 12 \cdot 11 \cdot 10 \cdot \; ... \; \cdot 2 \cdot 1 = 479,001,600\) possible routes!

See the time complexity for the greedy algorithm \(O(n^2)\), versus the time complexity for finding the shortest route by comparing all routes \(O(n!)\), in the image below.

But there are two things we can do to reduce the number of routes we need to check.

In the Traveling Salesman Problem, the route starts and ends in the same place, which makes a cycle. This means that the length of the shortest route will be the same no matter which city we start in. That is why we have chosen a fixed city to start in for the simulation above, and that reduces the number of possible routes from \(n!\) to \((n-1)!\).

Also, because these routes go in cycles, a route has the same distance if we go in one direction or the other, so we actually just need to check the distance of half of the routes, because the other half will just be the same routes in the opposite direction, so the number of routes we need to check is actually \( \frac{(n-1)!}{2}\).

But even if we can reduce the number of routes we need to check to \( \frac{(n-1)!}{2}\), the time complexity is still \( O(n!)\), because for very big \(n\), reducing \(n\) by one and dividing by 2 does not make a significant change in how the time complexity grows when \(n\) is increased.

To better understand how time complexity works, go to this page .

Actual Traveling Salesman Problems Are More Complex

The edge weight in a graph in this context of The Traveling Salesman Problem tells us how hard it is to go from one point to another, and it is the total edge weight of a route we want to minimize.

So far on this page, the edge weight has been the distance in a straight line between two points. And that makes it much easier to explain the Traveling Salesman Problem, and to display it.

But in the real world there are many other things that affects the edge weight:

• Obstacles: When moving from one place to another, we normally try to avoid obstacles like trees, rivers, houses for example. This means it is longer and takes more time to go from A to B, and the edge weight value needs to be increased to factor that in, because it is not a straight line anymore.
• Transportation Networks: We usually follow a road or use public transport systems when traveling, and that also affects how hard it is to go (or send a package) from one place to another.
• Traffic Conditions: Travel congestion also affects the travel time, so that should also be reflected in the edge weight value.
• Legal and Political Boundaries: Crossing border for example, might make one route harder to choose than another, which means the shortest straight line route might be slower, or more costly.
• Economic Factors: Using fuel, using the time of employees, maintaining vehicles, all these things cost money and should also be factored into the edge weights.

As you can see, just using the straight line distances as the edge weights, might be too simple compared to the real problem. And solving the Traveling Salesman Problem for such a simplified problem model would probably give us a solution that is not optimal in a practical sense.

It is not easy to visualize a Traveling Salesman Problem when the edge length is not just the straight line distance between two points anymore, but the computer handles that very well.

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Traveling Salesperson Problem

This section presents an example that shows how to solve the Traveling Salesperson Problem (TSP) for the locations shown on the map below.

The following sections present programs in Python, C++, Java, and C# that solve the TSP using OR-Tools

Create the data

The code below creates the data for the problem.

The distance matrix is an array whose i , j entry is the distance from location i to location j in miles, where the array indices correspond to the locations in the following order:

The data also includes:

• The number of vehicles in the problem, which is 1 because this is a TSP. (For a vehicle routing problem (VRP), the number of vehicles can be greater than 1.)
• The depot : the start and end location for the route. In this case, the depot is 0, which corresponds to New York.

Other ways to create the distance matrix

In this example, the distance matrix is explicitly defined in the program. It's also possible to use a function to calculate distances between locations: for example, the Euclidean formula for the distance between points in the plane. However, it's still more efficient to pre-compute all the distances between locations and store them in a matrix, rather than compute them at run time. See Example: drilling a circuit board for an example that creates the distance matrix this way.

Another alternative is to use the Google Maps Distance Matrix API to dynamically create a distance (or travel time) matrix for a routing problem.

Create the routing model

The following code in the main section of the programs creates the index manager ( manager ) and the routing model ( routing ). The method manager.IndexToNode converts the solver's internal indices (which you can safely ignore) to the numbers for locations. Location numbers correspond to the indices for the distance matrix.

The inputs to RoutingIndexManager are:

• The number of rows of the distance matrix, which is the number of locations (including the depot).
• The number of vehicles in the problem.
• The node corresponding to the depot.

Create the distance callback

To use the routing solver, you need to create a distance (or transit) callback : a function that takes any pair of locations and returns the distance between them. The easiest way to do this is using the distance matrix.

The following function creates the callback and registers it with the solver as transit_callback_index .

The callback accepts two indices, from_index and to_index , and returns the corresponding entry of the distance matrix.

Set the cost of travel

The arc cost evaluator tells the solver how to calculate the cost of travel between any two locations — in other words, the cost of the edge (or arc) joining them in the graph for the problem. The following code sets the arc cost evaluator.

In this example, the arc cost evaluator is the transit_callback_index , which is the solver's internal reference to the distance callback. This means that the cost of travel between any two locations is just the distance between them. However, in general the costs can involve other factors as well.

You can also define multiple arc cost evaluators that depend on which vehicle is traveling between locations, using the method routing.SetArcCostEvaluatorOfVehicle() . For example, if the vehicles have different speeds, you could define the cost of travel between locations to be the distance divided by the vehicle's speed — in other words, the travel time.

Set search parameters

The following code sets the default search parameters and a heuristic method for finding the first solution:

The code sets the first solution strategy to PATH_CHEAPEST_ARC , which creates an initial route for the solver by repeatedly adding edges with the least weight that don't lead to a previously visited node (other than the depot). For other options, see First solution strategy .

Add the solution printer

The function that displays the solution returned by the solver is shown below. The function extracts the route from the solution and prints it to the console.

The function displays the optimal route and its distance, which is given by ObjectiveValue() .

Solve and print the solution

Finally, you can call the solver and print the solution:

This returns the solution and displays the optimal route.

Run the programs

When you run the programs, they display the following output.

In this example, there's only one route because it's a TSP. But in more general vehicle routing problems, the solution contains multiple routes.

Save routes to a list or array

As an alternative to printing the solution directly, you can save the route (or routes, for a VRP) to a list or array. This has the advantage of making the routes available in case you want to do something with them later. For example, you could run the program several times with different parameters and save the routes in the returned solutions to a file for comparison.

The following functions save the routes in the solution to any VRP (possibly with multiple vehicles) as a list (Python) or an array (C++).

You can use these functions to get the routes in any of the VRP examples in the Routing section.

The following code displays the routes.

For the current example, this code returns the following route:

As an exercise, modify the code above to format the output the same way as the solution printer for the program.

Complete programs

The complete TSP programs are shown below.

Example: drilling a circuit board

The next example involves drilling holes in a circuit board with an automated drill. The problem is to find the shortest route for the drill to take on the board in order to drill all of the required holes. The example is taken from TSPLIB, a library of TSP problems.

Here's scatter chart of the locations for the holes:

The following sections present programs that find a good solution to the circuit board problem, using the solver's default search parameters. After that, we'll show how to find a better solution by changing the search strategy .

The data for the problem consist of 280 points in the plane, shown in the scatter chart above. The program creates the data in an array of ordered pairs corresponding to the points in the plane, as shown below.

Compute the distance matrix

The function below computes the Euclidean distance between any two points in the data and stores it in an array. Because the routing solver works over the integers, the function rounds the computed distances to integers. Rounding doesn't affect the solution in this example, but might in other cases. See Scaling the distance matrix for a way to avoid possible rounding issues.

Add the distance callback

The code that creates the distance callback is almost the same as in the previous example. However, in this case the program calls the function that computes the distance matrix before adding the callback.

Solution printer

The following function prints the solution to the console. To keep the output more compact, the function displays just the indices of the locations in the route.

Main function

The main function is essentially the same as the one in the previous example , but also includes a call to the function that creates the distance matrix.

Running the program

The complete programs are shown in the next section . When you run the program, it displays the following route:

Here's a graph of the corresponding route:

The OR-Tools library finds the above tour very quickly: in less than a second on a typical computer. The total length of the above tour is 2790.

Here are the complete programs for the circuit board example.

Changing the search strategy

The routing solver does not always return the optimal solution to a TSP, because routing problems are computationally intractable. For instance, the solution returned in the previous example is not the optimal route.

To find a better solution, you can use a more advanced search strategy, called guided local search , which enables the solver to escape a local minimum — a solution that is shorter than all nearby routes, but which is not the global minimum. After moving away from the local minimum, the solver continues the search.

The examples below show how to set a guided local search for the circuit board example.

For other local search strategies, see Local search options .

The examples above also enable logging for the search. While logging isn't required, it can be useful for debugging.

When you run the program after making the changes shown above, you get the following solution, which is shorter than the solution shown in the previous section .

For more search options, see Routing Options .

The best algorithms can now routinely solve TSP instances with tens of thousands of nodes. (The record at the time of writing is the pla85900 instance in TSPLIB, a VLSI application with 85,900 nodes. For certain instances with millions of nodes, solutions have been found guaranteed to be within 1% of an optimal tour.)

Scaling the distance matrix

Since the routing solver works over the integers, if your distance matrix has non-integer entries, you have to round the distances to integers. If some distances are small, rounding can affect the solution.

To avoid any issue with rounding, you can scale the distance matrix: multiply all entries of the matrix by a large number — say 100. This multiplies the length of any route by a factor of 100, but it doesn't change the solution. The advantage is that now when you round the matrix entries, the rounding amount (which is at most 0.5), is very small compared to the distances, so it won't affect the solution significantly.

If you scale the distance matrix, you also need to change the solution printer to divide the scaled route lengths by the scaling factor, so that it displays the unscaled distances of the routes.

Except as otherwise noted, the content of this page is licensed under the Creative Commons Attribution 4.0 License , and code samples are licensed under the Apache 2.0 License . For details, see the Google Developers Site Policies . Java is a registered trademark of Oracle and/or its affiliates.

Last updated 2023-01-16 UTC.

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

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.

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

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

• 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

Formulation

The objective function is then given by

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

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

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

Exact algorithms

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

Tour improvement procedures

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

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

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.

Navigation menu

12.9 Traveling Salesperson Problem

Learning objectives.

After completing this section, you should be able to:

• Distinguish between brute force algorithms and greedy algorithms.
• List all distinct Hamilton cycles of a complete graph.
• Apply brute force method to solve traveling salesperson applications.
• Apply nearest neighbor method to solve traveling salesperson applications.

We looked at Hamilton cycles and paths in the previous sections Hamilton Cycles and Hamilton Paths . In this section, we will analyze Hamilton cycles in complete weighted graphs to find the shortest route to visit a number of locations and return to the starting point. Besides the many routing applications in which we need the shortest distance, there are also applications in which we search for the route that is least expensive or takes the least time. Here are a few less common applications that you can read about on a website set up by the mathematics department at the University of Waterloo in Ontario, Canada:

• Design of fiber optic networks
• Minimizing fuel expenses for repositioning satellites
• Development of semi-conductors for microchips
• A technique for mapping mammalian chromosomes in genome sequencing

Before we look at approaches to solving applications like these, let's discuss the two types of algorithms we will use.

Brute Force and Greedy Algorithms

An algorithm is a sequence of steps that can be used to solve a particular problem. We have solved many problems in this chapter, and the procedures that we used were different types of algorithms. In this section, we will use two common types of algorithms, a brute force algorithm and a greedy algorithm . A brute force algorithm begins by listing every possible solution and applying each one until the best solution is found. A greedy algorithm approaches a problem in stages, making the apparent best choice at each stage, then linking the choices together into an overall solution which may or may not be the best solution.

To understand the difference between these two algorithms, consider the tree diagram in Figure 12.187 . Suppose we want to find the path from left to right with the largest total sum. For example, branch A in the tree diagram has a sum of 10 + 2 + 11 + 13 = 36 10 + 2 + 11 + 13 = 36 .

To be certain that you pick the branch with greatest sum, you could list each sum from each of the different branches:

A : 10 + 2 + 11 + 13 = 36 10 + 2 + 11 + 13 = 36

B : 10 + 2 + 11 + 8 = 31 10 + 2 + 11 + 8 = 31

C : 10 + 2 + 15 + 1 = 28 10 + 2 + 15 + 1 = 28

D : 10 + 2 + 15 + 6 = 33 10 + 2 + 15 + 6 = 33

E : 10 + 7 + 3 + 20 = 40 10 + 7 + 3 + 20 = 40

F : 10 + 7 + 3 + 14 = 34 10 + 7 + 3 + 14 = 34

G : 10 + 7 + 4 + 11 = 32 10 + 7 + 4 + 11 = 32

H : 10 + 7 + 4 + 5 = 26 10 + 7 + 4 + 5 = 26

Then we know with certainty that branch E has the greatest sum.

Now suppose that you wanted to find the branch with the highest value, but you only were shown the tree diagram in phases, one step at a time.

After phase 1, you would have chosen the branch with 10 and 7. So far, you are following the same branch. Let’s look at the next phase.

After phase 2, based on the information you have, you will choose the branch with 10, 7 and 4. Now, you are following a different branch than before, but it is the best choice based on the information you have. Let’s look at the last phase.

After phase 3, you will choose branch G which has a sum of 32.

The process of adding the values on each branch and selecting the highest sum is an example of a brute force algorithm because all options were explored in detail. The process of choosing the branch in phases, based on the best choice at each phase is a greedy algorithm. Although a brute force algorithm gives us the ideal solution, it can take a very long time to implement. Imagine a tree diagram with thousands or even millions of branches. It might not be possible to check all the sums. A greedy algorithm, on the other hand, can be completed in a relatively short time, and generally leads to good solutions, but not necessarily the ideal solution.

Example 12.42

Distinguishing between brute force and greedy algorithms.

A cashier rings up a sale for $4.63 cents in U.S. currency. The customer pays with a $5 bill. The cashier would like to give the customer $0.37 in change using the fewest coins possible. The coins that can be used are quarters ($0.25), dimes ($0.10), nickels ($0.05), and pennies ($0.01). The cashier starts by selecting the coin of highest value less than or equal to $0.37, which is a quarter. This leaves $ 0.37 − $ 0.25 = $ 0.12 $ 0.37 − $ 0.25 = $ 0.12 . The cashier selects the coin of highest value less than or equal to $0.12, which is a dime. This leaves $ 0.12 − $ 0.10 = $ 0.02 $ 0.12 − $ 0.10 = $ 0.02 . The cashier selects the coin of highest value less than or equal to $0.02, which is a penny. This leaves $ 0.02 − $ 0.01 = $ 0.01 $ 0.02 − $ 0.01 = $ 0.01 . The cashier selects the coin of highest value less than or equal to $0.01, which is a penny. This leaves no remainder. The cashier used one quarter, one dime, and two pennies, which is four coins. Use this information to answer the following questions.

• Is the cashier’s approach an example of a greedy algorithm or a brute force algorithm? Explain how you know.
• The cashier’s solution is the best solution. In other words, four is the fewest number of coins possible. Is this consistent with the results of an algorithm of this kind? Explain your reasoning.
• The approach the cashier used is an example of a greedy algorithm, because the problem was approached in phases and the best choice was made at each phase. Also, it is not a brute force algorithm, because the cashier did not attempt to list out all possible combinations of coins to reach this conclusion.
• Yes, it is consistent. A greedy algorithm does not always yield the best result, but sometimes it does.

Your Turn 12.42

The traveling salesperson problem.

Now let’s focus our attention on the graph theory application known as the traveling salesperson problem (TSP) in which we must find the shortest route to visit a number of locations and return to the starting point.

Recall from Hamilton Cycles , the officer in the U.S. Air Force who is stationed at Vandenberg Air Force base and must drive to visit three other California Air Force bases before returning to Vandenberg. The officer needed to visit each base once. We looked at the weighted graph in Figure 12.192 representing the four U.S. Air Force bases: Vandenberg, Edwards, Los Angeles, and Beal and the distances between them.

Any route that visits each base and returns to the start would be a Hamilton cycle on the graph. If the officer wants to travel the shortest distance, this will correspond to a Hamilton cycle of lowest weight. We saw in Table 12.11 that there are six distinct Hamilton cycles (directed cycles) in a complete graph with four vertices, but some lie on the same cycle (undirected cycle) in the graph.

Since the distance between bases is the same in either direction, it does not matter if the officer travels clockwise or counterclockwise. So, there are really only three possible distances as shown in Figure 12.193 .

The possible distances are:

So, a Hamilton cycle of least weight is V → B → E → L → V (or the reverse direction). The officer should travel from Vandenberg to Beal to Edwards, to Los Angeles, and back to Vandenberg.

Finding Weights of All Hamilton Cycles in Complete Graphs

Notice that we listed all of the Hamilton cycles and found their weights when we solved the TSP about the officer from Vandenberg. This is a skill you will need to practice. To make sure you don't miss any, you can calculate the number of possible Hamilton cycles in a complete graph. It is also helpful to know that half of the directed cycles in a complete graph are the same cycle in reverse direction, so, you only have to calculate half the number of possible weights, and the rest are duplicates.

In a complete graph with n n vertices,

• The number of distinct Hamilton cycles is ( n − 1 ) ! ( n − 1 ) ! .
• There are at most ( n − 1 ) ! 2 ( n − 1 ) ! 2 different weights of Hamilton cycles.

TIP! When listing all the distinct Hamilton cycles in a complete graph, you can start them all at any vertex you choose. Remember, the cycle a → b → c → a is the same cycle as b → c → a → b so there is no need to list both.

Example 12.43

Calculating possible weights of hamilton cycles.

Suppose you have a complete weighted graph with vertices N, M, O , and P .

• Use the formula ( n − 1 ) ! ( n − 1 ) ! to calculate the number of distinct Hamilton cycles in the graph.
• Use the formula ( n − 1 ) ! 2 ( n − 1 ) ! 2 to calculate the greatest number of different weights possible for the Hamilton cycles.
• Are all of the distinct Hamilton cycles listed here? How do you know? Cycle 1: N → M → O → P → N Cycle 2: N → M → P → O → N Cycle 3: N → O → M → P → N Cycle 4: N → O → P → M → N Cycle 5: N → P → M → O → N Cycle 6: N → P → O → M → N
• Which pairs of cycles must have the same weights? How do you know?
• There are 4 vertices; so, n = 4 n = 4 . This means there are ( n − 1 ) ! = ( 4 − 1 ) ! = 3 ⋅ 2 ⋅ 1 = 6 ( n − 1 ) ! = ( 4 − 1 ) ! = 3 ⋅ 2 ⋅ 1 = 6 distinct Hamilton cycles beginning at any given vertex.
• Since n = 4 n = 4 , there are ( n − 1 ) ! 2 = ( 4 − 1 ) ! 2 = 6 2 = 3 ( n − 1 ) ! 2 = ( 4 − 1 ) ! 2 = 6 2 = 3 possible weights.
• Yes, they are all distinct cycles and there are 6 of them.
• Cycles 1 and 6 have the same weight, Cycles 2 and 4 have the same weight, and Cycles 3 and 5 have the same weight, because these pairs follow the same route through the graph but in reverse.

TIP! When listing the possible cycles, ignore the vertex where the cycle begins and ends and focus on the ways to arrange the letters that represent the vertices in the middle. Using a systematic approach is best; for example, if you must arrange the letters M, O, and P, first list all those arrangements beginning with M, then beginning with O, and then beginning with P, as we did in Example 12.42.

Your Turn 12.43

The brute force method.

The method we have been using to find a Hamilton cycle of least weight in a complete graph is a brute force algorithm, so it is called the brute force method . The steps in the brute force method are:

Step 1: Calculate the number of distinct Hamilton cycles and the number of possible weights.

Step 2: List all possible Hamilton cycles.

Step 3: Find the weight of each cycle.

Step 4: Identify the Hamilton cycle of lowest weight.

Example 12.44

Applying the brute force method.

On the next assignment, the air force officer must leave from Travis Air Force base, visit Beal, Edwards, and Vandenberg Air Force bases each exactly once and return to Travis Air Force base. There is no need to visit Los Angeles Air Force base. Use Figure 12.194 to find the shortest route.

Step 1: Since there are 4 vertices, there will be ( 4 − 1 ) ! = 3 ! = 6 ( 4 − 1 ) ! = 3 ! = 6 cycles, but half of them will be the reverse of the others; so, there will be ( 4 − 1 ) ! 2 = 6 2 = 3 ( 4 − 1 ) ! 2 = 6 2 = 3 possible distances.

Step 2: List all the Hamilton cycles in the subgraph of the graph in Figure 12.195 .

To find the 6 cycles, focus on the three vertices in the middle, B, E, and V . The arrangements of these vertices are BEV, BVE, EBV, EVB, VBE , and VEB . These would correspond to the 6 cycles:

1: T → B → E → V → T

2: T → B → V → E → T

3: T → E → B → V → T

4: T → E → V → B → T

5: T → V → B → E → T

6: T → V → E → B → T

Step 3: Find the weight of each path. You can reduce your work by observing the cycles that are reverses of each other.

1: 84 + 410 + 207 + 396 = 1097 84 + 410 + 207 + 396 = 1097

2: 84 + 396 + 207 + 370 = 1071 84 + 396 + 207 + 370 = 1071

3: 370 + 410 + 396 + 396 = 1572 370 + 410 + 396 + 396 = 1572

4: Reverse of cycle 2, 1071

5: Reverse of cycle 3, 1572

6: Reverse of cycle 1, 1097

Step 4: Identify a Hamilton cycle of least weight.

The second path, T → B → V → E → T , and its reverse, T → E → V → B → T , have the least weight. The solution is that the officer should travel from Travis Air Force base to Beal Air Force Base, to Vandenberg Air Force base, to Edwards Air Force base, and return to Travis Air Force base, or the same route in reverse.

Your Turn 12.44

Now suppose that the officer needed a cycle that visited all 5 of the Air Force bases in Figure 12.194 . There would be ( 5 − 1 ) ! = 4 ! = 4 × 3 × 2 × 1 = 24 ( 5 − 1 ) ! = 4 ! = 4 × 3 × 2 × 1 = 24 different arrangements of vertices and ( 5 − 1 ) ! 2 = 4 ! 2 = 24 2 = 12 ( 5 − 1 ) ! 2 = 4 ! 2 = 24 2 = 12 distances to compare using the brute force method. If you consider 10 Air Force bases, there would be ( 10 − 1 ) ! = 9 ! = 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 362 , 880 ( 10 − 1 ) ! = 9 ! = 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 362 , 880 different arrangements and ( 10 − 1 ) ! 2 = 9 ! 2 = 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 2 = 181 , 440 ( 10 − 1 ) ! 2 = 9 ! 2 = 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 2 = 181 , 440 distances to consider. There must be another way!

The Nearest Neighbor Method

When the brute force method is impractical for solving a traveling salesperson problem, an alternative is a greedy algorithm known as the nearest neighbor method , which always visit the closest or least costly place first. This method finds a Hamilton cycle of relatively low weight in a complete graph in which, at each phase, the next vertex is chosen by comparing the edges between the current vertex and the remaining vertices to find the lowest weight. Since the nearest neighbor method is a greedy algorithm, it usually doesn’t give the best solution, but it usually gives a solution that is "good enough." Most importantly, the number of steps will be the number of vertices. That’s right! A problem with 10 vertices requires 10 steps, not 362,880. Let’s look at an example to see how it works.

Suppose that a candidate for governor wants to hold rallies around the state. They plan to leave their home in city A , visit cities B, C, D, E , and F each once, and return home. The airfare between cities is indicated in the graph in Figure 12.196 .

Let’s help the candidate keep costs of travel down by applying the nearest neighbor method to find a Hamilton cycle that has a reasonably low weight. Begin by marking starting vertex as V 1 V 1 for "visited 1st." Then to compare the weights of the edges between A and vertices adjacent to A : $250, $210, $300, $200, and $100 as shown in Figure 12.197 . The lowest of these is $100, which is the edge between A and F .

Mark F as V 2 V 2 for "visited 2nd" then compare the weights of the edges between F and the remaining vertices adjacent to F : $170, $330, $150 and $350 as shown in Figure 12.198 . The lowest of these is $150, which is the edge between F and D .

Mark D as V 3 V 3 for "visited 3rd." Next, compare the weights of the edges between D and the remaining vertices adjacent to D : $120, $310, and $270 as shown in Figure 12.199 . The lowest of these is $120, which is the edge between D and B .

So, mark B as V 4 V 4 for "visited 4th." Finally, compare the weights of the edges between B and the remaining vertices adjacent to B : $160 and $220 as shown in Figure 12.200 . The lower amount is $160, which is the edge between B and E .

Now you can mark E as V 5 V 5 and mark the only remaining vertex, which is C , as V 6 V 6 . This is shown in Figure 12.201 . Make a note of the weight of the edge from E to C , which is $180, and from C back to A , which is $210.

The Hamilton cycle we found is A → F → D → B → E → C → A . The weight of the circuit is $ 100 + $ 150 + $ 120 + $ 160 + $ 180 + $ 210 = $ 920 $ 100 + $ 150 + $ 120 + $ 160 + $ 180 + $ 210 = $ 920 . This may or may not be the route with the lowest cost, but there is a good chance it is very close since the weights are most of the lowest weights on the graph and we found it in six steps instead of finding 120 different Hamilton cycles and calculating 60 weights. Let’s summarize the procedure that we used.

Step 1: Select the starting vertex and label V 1 V 1 for "visited 1st." Identify the edge of lowest weight between V 1 V 1 and the remaining vertices.

Step 2: Label the vertex at the end of the edge of lowest weight that you found in previous step as V n V n where the subscript n indicates the order the vertex is visited. Identify the edge of lowest weight between V n V n and the vertices that remain to be visited.

Step 3: If vertices remain that have not been visited, repeat Step 2. Otherwise, a Hamilton cycle of low weight is V 1 → V 2 → ⋯ → V n → V 1 V 1 → V 2 → ⋯ → V n → V 1 .

Example 12.45

Using the nearest neighbor method.

Suppose that the candidate for governor wants to hold rallies around the state but time before the election is very limited. They would like to leave their home in city A , visit cities B , C , D , E , and F each once, and return home. The airfare between cities is not as important as the time of travel, which is indicated in Figure 12.202 . Use the nearest neighbor method to find a route with relatively low travel time. What is the total travel time of the route that you found?

Step 1: Label vertex A as V 1 V 1 . The edge of lowest weight between A and the remaining vertices is 85 min between A and D .

Step 2: Label vertex D as V 2 V 2 . The edge of lowest weight between D and the vertices that remain to be visited, B, C, E , and F , is 70 min between D and F .

Repeat Step 2: Label vertex F as V 3 V 3 . The edge of lowest weight between F and the vertices that remain to be visited, B, C, and E , is 75 min between F and C .

Repeat Step 2: Label vertex C as V 4 V 4 . The edge of lowest weight between C and the vertices that remain to be visited, B and E , is 100 min between C and B .

Repeat Step 2: Label vertex B as V 5 V 5 . The only vertex that remains to be visited is E . The weight of the edge between B and E is 95 min.

Step 3: A Hamilton cycle of low weight is A → D → F → C → B → E → A . So, a route of relatively low travel time is A to D to F to C to B to E and back to A . The total travel time of this route is: 85 min + 70 min + 75 min + 100 min + 95 min + 90 min = 515 min or 8 hrs 35 min 85 min + 70 min + 75 min + 100 min + 95 min + 90 min = 515 min or 8 hrs 35 min

Your Turn 12.45

Check your understanding, section 12.9 exercises.

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The travelling salesman problem [and how to solve it]

The Travelling Salesman Problem (TSP) refers to the challenge of finding the shortest route between any multiple stops on a map.

Brendan Tobin , Caroline Hassett

The travelling salesman problem (TSP) refers to the efforts of a door-to-door salesman trying to find the shortest and/or quickest way to serve all of the stops on his list of visits in a given time period (usually a working day).

Although it was once the problem of a salesperson, today there are far more workers that are faced with it. In recent years, the explosion of eCommerce and online shopping has seen more and more deliveries made directly to consumers' homes than ever before. For the delivery drivers, it can mean delivering to over 100 individual locations on any given day.

So, getting the travelling salesman problem solved is mission-critical!

In this article, we’ll look at the problem in a bit more detail, as well as discuss the importance of it for various industries including couriers and last mile carriers. If you are struggling with routing salespeople or delivery teams then we have the solution for you . We’ll also take a look at some of the solutions to the problem and discuss which route optimization software is the most suited to solving your own TSP problem.

In this article you will learn:

What is the travelling salesman problem?

• Why the travelling salesman problem is important
• Possible solutions to the travelling salesman problem
• How to practically address the TSP
• How SmartRoutes helps to solve the TSP

The Travelling Salesman Problem (TSP) refers to the challenge of finding the shortest route between any multiple stops on a map. It takes its name from the historical challenge faced by door-to-door salespeople who needed to visit as many potential customers in the shortest time possible to increase their sales numbers.

The TSP has become a renowned algorithmic problem in fields of study such as physics, computer sciences, and latterly in logistics and operations research.

Why is the travelling salesman problem important?

When it comes to business, time is money and no matter what businesses you're in, if it involves travelling between multiple locations by vehicle, you need to ensure that it’s doing so in the most efficient way possible.

Consider this for a minute:

According to Route Consultant , a FedEx Ground Driver in the United States does an average of 100 miles or less per day, and completes between 150-230 stops a day, which includes pickups.

Believe it or not, there are trillions (yes, trillions) of unique ways that each of these stops could be served. Of course, some of these are most obviously not the most efficient routes, but actually determining which one is can be a challenge that is still beyond practical human capabilities.

What began as a field sales problem many moons ago is now a very real issue in the supply chain logistics industry. For any business that has more than a couple of vehicles, ensuring that they are all delivering goods in the most efficient manner can be the difference between having a viable business model, or not.

By attempting to solve the TSP, we can find the most efficient routes for delivery drivers to take when out on the road. This results in less distance being travelled, less fuel being used and fewer hours driven. This means a double bonus for the balance sheet in any delivery-based business operation.

But the travelling salesman problem (TSP) isn’t just important for businesses' financial viability. With the ever-increasing importance of reducing our carbon footprint, reducing the distances travelled directly contributes to the amount of fossil fuels being burned. We all have a part to play, and the ability to effect such important change through problem solving is a bonus for the greater society too. In fact, a recent effort to reduce the time taken for a politician to tour the UK shows how difficult the problem is.

The benefits of solving this problem (TSP) also trickles down to the customers who rely on timely and efficient delivery of goods. Whether they need the deliveries for the operations of their own businesses, or whether they rely on on-time delivery of medication for personal illness, it has real-world implications for everyone that relies on it. Believe it or not, customer experience (or as it’s correctly termed in the last mile, ‘delivery experience’) is another big reason why the TSP is so important.

SmartRoutes Route Planning Software

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Possible solutions to the Travelling Salesman Problem

There are literally entire books dedicated to trying to solve the TSP , so it is important to note that what we list below are ‘possible’ solutions rather than ‘actual solutions’. The TSP is what is referred to as a ‘NP-hard’ problem, which means that there literally is no known solution to it.

However, some of the more notable or practical answers to the problem are as follows:

1. The ‘Brute-Force’ check

The trouble with the travelling salesman problem (TSP), is that technically, it is solvable. You just need to work out every possible sequence that a number of stops can be served. Right?

Well, sure, but the number of routes possible grows so rapidly that even the most powerful of modern algorithmic computers fail at the task. The trouble with creating an optimized route for 10 stops isn’t the problem per se, it’s creating the route fast.

Well, this is what the brute-force method attempts, or as it’s also known, the ‘naive approach’. It calculates all the possible permutations, finds the shortest one possible and then selects the route that is actually the shortest.

2. The branch and bound approach

The branch and bound approach pays attention to finding the lowest possible cost for the remaining pathways that the initial stages of approaches like the brute-force provide.

Sometimes referred to as Parallel TSP , it assumes that we don’t want to try every single ‘possible’ route. Even using common sense, would probably narrow it down somewhat. For example, we’re not going to go to the furthest drop-off point first and then return to the one nearest to the depot. By using nodes, and attaching ‘costs’ to each node, the algorithm estimates how likely a given choice is to lead to a solution to the problem.

3. The Nearest Neighbour Method

With the nearest neighbour approach, the algorithm begins with the start location of the route, and then adds the stop nearest to it. A list of locations visited is already served, and if a stop has been served it is skipped and the next stop is added. Once the final stop is reached, you return to the start location again.

The benefit of the nearest neighbour method is that it is relatively quick to do, and also more practical than the other 2 approaches outlined. The downside, however, is that it isn't always as accurate at providing the optimized route as the other two approaches.

How to practically address the travelling salesman problem

Well, if you’ve managed to read this far, you know that the answer is unsolvable in a literal sense but in a practical sense, it is!

While algorithms might not give you the very quickest or cheapest route every single time, they can do so much more effectively than the human brain can. As we’ve seen with the nearest neighbour method, it is possible to use computational power to automatically eliminate the bulk of routes that would most obviously not be efficient. Furthermore, the same computers can also then select which route is the quickest (or in mathematical terms has the lowest cost) from the remaining options.

As technology advances, and machine learning and automation become more powerful than ever, route planning and optimization can be done in a matter of seconds. There are essentially two processes that are complete: the mathematical equation and the automated selection of the best route. This has been a game changer for the road transport industry, particularly with larger volumes of packed goods being distributed in smaller quantities in recent years.

How SmartRoutes Solves the Travelling Salesman Problem

SmartRoutes is what is referred to as delivery management software. As a total delivery software solution, it helps businesses to manage everything from order management and fulfilment, to the planning of routes for delivery drivers and the capture of proof-of-delivery for end customers.

At the core of the solution, however, is route optimization . This means that we have developed the solution with on clear overall aim:

To create the quickest and most efficient routes for delivering goods by road as quickly as is technically possible.

Now as we’ve discussed, there isn't a cut-and-dry solution to the travelling salesman problem. But here at SmartRoutes, we took what is regarded as the best approach, added the best technology and then set about accounting for every other factor that plays a role in the delivery of goods by delivery vehicles.

Some of the most obvious ones include the size of fleets, the types of roads, real-time driver analytics, and whether time or distance between stops is a better indicator of efficiency based on field-testing.

However, we have also accounted for some less obvious factors in our own algorithm. For example, our algorithm is weighted to prefer routes that have less right turns for users in the US and the UK.

Why? Well, because crossing traffic is actually a time-suck when it comes to delivering large volumes of goods. By turning left, drivers don’t have to wait for oncoming traffic to pass before continuing their route. Of course, this is also much safer for drivers which is another reason why we adopted this within our routing algorithm.

If you're looking to create a great delivery experience for your customers then look no further than SmartRoutes. You can try it out with our 7-day free trial or book a demo today. We can get you up and running in minutes and help you to make the most of it from the outset.

Frequently asked questions

1. what is the traveling salesman problem (tsp), and why is it important.

The traveling salesman problem (TSP) is a classic optimization problem in which a salesperson must find the shortest possible route that visits a set of given cities exactly once and returns to the starting city. It's important because it has applications in various fields like logistics, transportation, and manufacturing, where efficient routing and cost-saving are crucial.

2. Are there practical applications of TSP-solving algorithms?

Yes, TSP-solving algorithms have numerous practical applications. For instance, they are used in optimizing delivery routes for couriers and delivery services, planning circuit design on computer chips, and even in DNA sequencing.

3. What are some common algorithms to solve the traveling salesman problem?

• Brute-Force Method: Exhaustively explores all possible routes.
• Nearest Neighbour: Select the nearest unvisited city at each step.
• The Branch and Bound Approach: This pays attention to finding the lowest possible cost for the remaining pathways.

4. What are the limitations and challenges associated with solving the traveling salesman problem?

Solving the TSP can be challenging due to its computational complexity. As the number of cities increases, the problem becomes exponentially harder to solve. Additionally, approximating an optimal solution for large datasets can be time-consuming. Researchers are continually working on algorithms to address these challenges and find efficient solutions for real-world TSP instances.

If you enjoyed this blog, you might also be interested in:

• Customer stories
• What's New!
• Integrations
• Route planning
• Route optimization
• Delivery fleet tracking
• Customer experience
• Proof of delivery
• Delivery driver app
• Order management
• Terms & Privacy

Algorithms for the Travelling Salesman Problem

Printables.com

• 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.

Related articles

Liked this article? See below for more recommended reading!

Solving the Vehicle Routing Problem (2024)

How To Optimize Multi-Stop Routes With Google Maps (2024)

How Route Optimization Impacts Our Earth

What is a Travelling Salesman Problem (TSP)? Explained!

• Author: Diksha Bhandari
• Read Time: 9 min
• Published: September 14, 2023
• Last Update: November 8th, 2023

Table of Contents

• Asset Tracking
• Dispatch Management
• Driver Behaviour
• Electric Vehicle
• Employee Management
• Field Sales And Service
• Fleet Management
• Fuel Management
• GPS Software
• GPS Trackers and Hardware
• Last Mile Delivery
• Lead Management
• Leave And Attendance
• Roster Management
• Route Optimisation
• Route Planning
• Sensor Integration
• Task and Workflow
• Tech and Beyond
• TrackoBit Industries
• TrackoField
• TrackoField Industries
• TrackoMile Industries
• Vehicle Tracking
• Video telematics
• What's New

Want to know what a travelling salesman problem (TSP) is? Need solutions to real-life TSP challenges. Learn here. 

Do you also look for the shortest route on Google Maps before embarking on a trip?

I am sure, you know multiple routes to reach the office, the mall, or your desired location, but checking on the internet before leaving home has become a ritual. It becomes all the more important to refer to maps when you have to pick up friends or colleagues along the way.

‘ADD STOPS’

Yes, you are right! 💯

That’s what solving the TSP challenge using software means!

What is a Travelling Salesman Problem (TSP)?

The traveling salesman problem is the popular combinatorial optimisation challenge in mathematics and computer science. The prime objective of the problem is to determine the shortest possible route a salesperson must take to cover a set of locations in one go and then return to the starting point.

Addressing travelling salesman challenges and their optimisation are more relevant in this time and age, especially in the supply chain, logistics and delivery space.

TSP may result in delayed deliveries and slimming profits as it’s not easy for delivery agents to choose the most viable and cost-effective route in real-time.

What are Traveling Salesman Problem Challenges to Solve?

When a salesperson is in the field hopping from one client site to another, finding out the best and the shortest route is an added pressure on the agent. In today’s day and age, distance isn’t the only factor that defines efficiency. There are several factors, such as time, fuel consumption, capacity, etc. that together define efficiency.

However, addressing the travelling salesman challenges involves mitigating a few unavoidable challenges along the way that field agents face themselves.

1. Time Constraints

Sales agents often have a tight schedule with multiple deliveries to make with a short TAT. Similarly, in TSP, optimising routes to minimise travel time is a fundamental challenge.

2. Last-minute Changes

Eleventh-hour changes are not a new concept for salespeople. They encounter urgent visits and last-minute cancellations a lot. Similarly, TSP solutions must be adaptable to handle dynamic scenarios and route modifications.

3. Resource Efficiency

Just as salespersons aim at reducing fuel costs and ensuring on-time deliveries, TSP solutions such as TrackoMile must strive for resource optimisation by reducing travel distances and delivery TAT.

4. Objective Diversification

While solving the travelling salesman problem (TSP) , optimising multiple objectives such as cost, time, and environmental factors adds complexity as solutions need to balance conflicting goals.

5. Combinatorial Complexity

TSP is a combinatorial optimisation problem, which means it involves complicated mathematical calculations with numerous variables. Sometimes, complex scenarios get further intricate as multiple variables are involved.

6. Adaptability and Scalability

Similarly, how sales agents adjust to the routes on the fly, the route algorithm must be flexible and responsive to real-time changes such as spiking volumes, vehicle breakdown or traffic slow down. A TSP solution must have a good appetite to handle large datasets and complex networks.

Also Read 4 Key Solutions for Fuel Management System 2023

Top 5 Solutions to The Travelling Salesman Problem

The traveling salesman problem solutions offer various trade-offs between computational intricacies and the quality of the resolution, allowing practitioners to choose the best-suited approach based on their needs and problems.

Here are the Top 5 solutions to the Traveling Salesman Problem (TSP) :

1. Brute Force Algorithm

The Brute Force algorithm is a straight approach to solving the Traveling Salesman Problem (TSP). It systematically explores all possible routes to identify the shortest one among them all. While it guarantees an optimal solution, its downside lies in its major time complexity, making it practical only for small TSP challenges.

2. Nearest Neighbour Algorithm

The Nearest Neighbour method is the simplest heuristic for the TSP. It starts from the first location and repeatedly selects the closest unvisited location to form a tour. Although it is quick to implement this method, it may always yield the optimal solution for it prioritises proximity over other factors.

3. Genetic Algorithm

This technique or method draws inspiration from nature itself. They evolve TSP solutions through selection, crossovers and mutation. They pick the best routes and mix them up. This creates new routes that might be even better. Then, they keep the best ones and repeat the mixing and picking process. Survival of the fittest in the true sense.

4. Ant Colony Optimisation (ACO)

Ants have a tendency to leave pheromones on the shorter routes they find, calling fellow ants on the same route. They keep leaving more pheromones on the shorter routes they find. Over time, the collective behaviour of the ants causes them to converge on the shortest route. Inspired by the nature of ants, ACO finds the shortest route by analysing the trails of data left by artificial ants based on the strength of these data trails.

5. Dynamic Programming

Dynamic Programming is like solving a puzzle, step-by-step, by breaking it into smaller pieces. In TSP challenges, it finds the best route to visit all locations. It begins with figuring out the shortest route between two locations; then it builds on that to find ways to more locations. It’s a smart TSP solution for small scenarios but may require significant memory resources for larger and more complex problems.

What Are Real-world Travelling Salesman Problem Applications?

The Traveling Salesman Problem (TSP) has a wide array of applications across various domains due to its relevance in optimising routes and sequences. Here are several crucial real-word TSP applications and implementations in the real world.

1. TSP implementation in Logistics and Delivery Services

The logistics and supply chain sectors have the widest TSP applications.

• Courier, Express & Parcel : Companies like FedEx, UPS, and DHL rely on TSP algorithms to optimise delivery routes for their fleet of delivery trucks. By finding the most efficient sequence of stops, they minimise fuel consumption , reduce delivery TAT, and save on operational overheads too.
• On-demand Delivery : Food delivery companies, instant grocery delivery apps and at-home appointment platforms like Swiggy, BlinkIt and UrbanCompany, respectively, leverage TSP solutions to ensure timely delivery. Enhancing the customer experience and increasing the number of deliveries each rider can make.

2. TSP Applications in Transportation and Urban Planning Waste collection routes, Traffic light synchronisation, optic cable installation, etc. are some areas where TSP Solutions works like a knight in shining armour. Other real-world TSP applications include

• Public Transport : City planners and public transport agencies use TSP principles to design bus, tram and train routes that reduce travel for passengers.
• Emergency Service Dispatch : Ambulance services, Police PCR vans employ TSP algorithms to dispatch vehicles quickly and efficiently in response to emergency calls. Finding the shortest route to reach the incident location can save lives.
• Urban Mobility Solution : In the era of ride-sharing and on-demand mobility apps like Uber, Ola, Lyft, etc., real-world TSP applications become prominent. TSP solutions optimise the route to destinations, ensuring quick and cost-effective transportation.

Other significant real-life applications of the Travelling Salesman Problem are

• TSP in Healthcare and Medical Research – for DNA sequencing and understanding genetic patterns and diseases.
• TSP in Manufacturing and Production – In circuit board manufacturing and job scheduling of technicians.
• TSP in Robotics and Autonomous Vehicles -Self-driving cars and drones use TSP-like algorithms for efficient navigation.

Solving the Travelling Salesman Problem – Last Mile Delivery Route Optimisation

Route optimisation is the key to efficient last-mile delivery . In order to attain flawless route optimisation, the software must solve the traveling salesman problem every step of the way.

Why it’s essential to solve TSP for Last Mile Delivery?

In simple and minimal words, solving TSP problems helps in many ways:

• Saves Time : It makes deliveries faster, so your customers get orders sooner.
• Customer Satisfaction : Fast deliveries give you an edge over the competition and enhance customer experience too.
• Saves Money : It reduces fuel wastage and vehicle wear, making deliveries cheaper.
• Environment Friendly : It lowers pollution by using fewer vehicles and shorter routes.
• Happy Staff: Drivers and dispatchers have less stress and can finish their work faster.

How do we solve the travelling salesman problem for last-mile delivery?

Solving TSP challenges for Last-mile delivery is like solving a big jigsaw puzzle. There are a hundred thousand addresses to visit daily. The software must find the shortest and most optimised route to them and come back to the starting point at the end.

• Our route optimisation software , TrackoMile, leverages capacity management , routing algorithms and robust rule engines to present the most optimal combination of delivery addresses. Thereby giving the most optimally planned routes or trips.
• All delivery managers have to do is upload the CSV file of the addresses or integrate TrackoMile to their CRM to fetch the delivery addresses. Now trip allocation, route optimisation, dispatch and everything else happen in a few clicks.
• ETA when the delivery is en route, POD when the order is delivered successfully, and trip analysis, are added features to simplify overall operations.

The Vehicle Routing Problem is very similar to TSP, with wide applications in logistics, delivery services and transportation. While TSP focuses on finding the shortest route for a single traveller visiting various locations, VRP deals with multiple vehicles serving multiple customers, considering added constraints like vehicle capacity, TATs and more.

How Can AI Help in Solving Traveling Salesman Problem (TSP)?

AI or Artificial Intelligence are becoming the driving force for business growth across various industrial sectors. AI particularly aids in solving the Traveling Salesman Problem(TSP) in the logistics and delivery sector by employing advanced algorithms and techniques. What are a few tricks up AI’s sleeves that help in automating TSP resolution? Let’s find out!

1. Advanced Algorithms

AI algorithms such as Genetic Algorithms, ACO, simulated annealing and a few others mentioned above, tackle complex Travelling Salesman Problem scenarios.

2. Machine Learning

Gathering information from historical data and optimising routes based on real-time insights is what AI is best for. Machine learning models are trained to adapt to changing conditions, like traffic, weather and delivery constraints, to provide a more accurate plan of action.

3. Parallel Computing

AIi enables the use of a parallel computing process, which means solving multiple segments of TSP simultaneously. This accelerates the problem-solving process for large-scale challenges.

4. Heuristic Improvement

TSP Heuristics powered by AI can groom initial solutions, gradually improving their results over time. These heuristics can be applied iteratively by AI to reach better results.

5. Hybrid Approaches

Applying hybrid algorithms is not a new technique to refine techniques and produce more accurate results. AI on top of it singles out data-oriented combinations that work the best in varied use cases.

Wrapping Up!

The travelling salesman problem’s importance lies in its real-world applications. Whether optimising delivery routes, planning manufacturing processes or organising circuit board drilling, finding the most efficient way to cover multiple locations is crucial to minimise costs and save time.

The TSP problems have evolved over the years, and so have TSP algorithms, heuristics and solutions. With the advent of advanced technologies such as GPS and machine learning, TSP continues to adapt and find new applications in emerging fields, cementing its status as a fundamental problem in optimization theory and a valuable tool for various industries. Mobility automation software like Trackobit, TrackoMile and TrackoField resort to TSP heuristics to solve challenges along the way.

Read Blog – Best Delivery Route Planner Apps for 2023

Traveling Salesman Problem FAQs

What is tsp.

TSP is an abbreviation for Traveling Salesman Problem. It’s the routing problem that deals with finding the shortest route to travel to a combination of locations in the most optimal manner.

Is Travelling Salesman Problem Solvable?

Yes, the Traveling Salesman Problem (TSP) is solvable, but the time to find the solution can grow proportionately with an increase in the number of locations. With the evolution of travelling salesman problem applications, various TSP algorithms and heuristics, their hybrids have also emerged.

Wh at is the objective of TSP?

The objective of the Traveling Salesman Problem (TSP) is to find the shortest possible route that covers all given locations or waypoints and comes back to the starting point with the least resource utilisation.

Diksha Bhandari

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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: ?(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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Erica Klarreich

Computer Scientists Break the 'Traveling Salesperson' Record

When Nathan Klein started graduate school two years ago, his advisers proposed a modest plan: to work together on one of the most famous, long-standing problems in theoretical computer science.

Even if they didn’t manage to solve it, they figured, Klein would learn a lot in the process. He went along with the idea. “I didn’t know to be intimidated,” he said. “I was just a first-year grad student—I don’t know what’s going on.”

Now, in a paper posted online in July, Klein and his advisers at the University of Washington, Anna Karlin and Shayan Oveis Gharan, have finally achieved a goal computer scientists have pursued for nearly half a century: a better way to find approximate solutions to the traveling salesperson problem.

This optimization problem, which seeks the shortest (or least expensive) round trip through a collection of cities, has applications ranging from DNA sequencing to ride-sharing logistics. Over the decades, it has inspired many of the most fundamental advances in computer science, helping to illuminate the power of techniques such as linear programming. But researchers have yet to fully explore its possibilities—and not for want of trying.

The traveling salesperson problem “isn’t a problem, it’s an addiction,” as Christos Papadimitriou, a leading expert in computational complexity, is fond of saying.

Most computer scientists believe that there is no algorithm that can efficiently find the best solutions for all possible combinations of cities. But in 1976, Nicos Christofides came up with an algorithm that efficiently finds approximate solutions—round trips that are at most 50 percent longer than the best round trip. At the time, computer scientists expected that someone would soon improve on Christofides’ simple algorithm and come closer to the true solution. But the anticipated progress did not arrive.

“A lot of people spent countless hours trying to improve this result,” said Amin Saberi of Stanford University.

Now Karlin, Klein and Oveis Gharan have proved that an algorithm devised a decade ago beats Christofides’ 50 percent factor, though they were only able to subtract 0.2 billionth of a trillionth of a trillionth of a percent. Yet this minuscule improvement breaks through both a theoretical logjam and a psychological one. Researchers hope that it will open the floodgates to further improvements.

“This is a result I have wanted all my career,” said David Williamson of Cornell University, who has been studying the traveling salesperson problem since the 1980s.

The traveling salesperson problem is one of a handful of foundational problems that theoretical computer scientists turn to again and again to test the limits of efficient computation. The new result “is the first step towards showing that the frontiers of efficient computation are in fact better than what we thought,” Williamson said.

While there is probably no efficient method that always finds the shortest trip, it is possible to find something almost as good: the shortest tree connecting all the cities, meaning a network of connections (or “edges”) with no closed loops. Christofides’ algorithm uses this tree as the backbone for a round-trip tour, adding extra edges to convert it into a round trip.

Any round-trip route must have an even number of edges into each city, since every arrival is followed by a departure. It turns out that the reverse is also true—if every city in a network has an even number of connections then the edges of the network must trace a round trip.

The shortest tree connecting all the cities lacks this evenness property, since any city at the end of a branch has just one connection to another city. So to turn the shortest tree into a round trip, Christofides (who died last year) found the best way to connect pairs of cities that have odd numbers of edges. Then he proved that the resulting round trip will never be more than 50 percent longer than the best possible round trip.

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In doing so, he devised perhaps the most famous approximation algorithm in theoretical computer science—one that usually forms the first example in textbooks and courses.

“Everybody knows the simple algorithm,” said Alantha Newman of Grenoble Alpes University and the National Center for Scientific Research in France. And when you know it, she said, “you know the state of the art”—at least, you did until this past July.

Computer scientists have long suspected that there should be an approximation algorithm that outperforms Christofides’ algorithm. After all, his simple and intuitive algorithm isn’t always such an effective way to design a traveling salesperson route, since the shortest tree connecting the cities may not be the best backbone you could choose. For instance, if this tree has many branches, each city at the end of a branch will need to be matched with another city, potentially forming lots of expensive new connections.

In 2010, Oveis Gharan, Saberi and Mohit Singh of the Georgia Institute of Technology started wondering if it might be possible to improve on Christofides’ algorithm by choosing not the shortest tree connecting all the cities, but a random tree from a carefully chosen collection. They took inspiration from an alternate version of the traveling salesperson problem in which you are allowed to travel along a combination of paths—maybe you get to Denver via 3/4 of the route from Chicago to Denver plus 1/4 of the route from Los Angeles to Denver.

Unlike the regular traveling salesperson problem, this fractional problem can be solved efficiently. And while fractional routes don’t make physical sense, computer scientists have long believed that the best fractional route should be a rough guide to the contours of good ordinary routes.

So to create their algorithm, Oveis Gharan, Saberi and Singh defined a random process that picks a tree connecting all the cities, so that the probability that a given edge is in the tree equals that edge’s fraction in the best fractional route. There are many such random processes, so the researchers chose one that tends to produce trees with many evenly connected cities. After this random process spits out a specific tree, their algorithm plugs it into Christofides’ scheme for matching cities with odd numbers of edges, to convert it into a round trip.

This method seemed promising, not just to the three researchers but to other computer scientists. “The intuition is simple,” said Ola Svensson of the Swiss Federal Institute of Technology Lausanne. But “to prove it turns out to be a different beast.”

The following year, though, Oveis Gharan, Saberi and Singh managed to prove that their algorithm beats Christofides’ algorithm for “graphical” traveling salesperson problems—ones where the distances between cities are represented by a network (not necessarily including all connections) in which every edge has the same length. But the researchers couldn’t figure out how to extend their result to the general traveling salesperson problem, in which some edges may be vastly longer than others.

“If we have to add a super expensive edge to the matching then we’re screwed, basically,” Karlin said.

Nevertheless, Oveis Gharan emerged from that collaboration with an unshakable belief that their algorithm should beat Christofides’ algorithm for the general traveling salesperson problem. “I never had a doubt,” he said.

Oveis Gharan kept turning the problem over in his mind over the years that followed. He suspected that a mathematical discipline called the geometry of polynomials, little known in the theoretical computer science world, might have the tools he needed. So when Karlin came to him two years ago suggesting that they co-advise a brilliant new graduate student named Nathan Klein who had double-majored in math and cello, he said, “OK, let’s give it a try—I have this interesting problem.”

Karlin thought that, if nothing else, it would be a fun opportunity to learn more about the geometry of polynomials. “I really didn’t think we would be able to solve this problem,” she said.

She and Oveis Gharan had no hesitation about throwing Klein into the deep end of computer science research. Oveis Gharan had himself cut his teeth on the traveling salesperson problem as a graduate student back in 2010. And the two advisers agreed about the merits of assigning hard problems to graduate students, especially during their first couple of years, when they are not under pressure to get results.

The three dived into an intense collaboration. “It’s all I was thinking about for two years,” Klein said.

They spent the first year solving a simplified version of the problem, to get a sense of the challenges they were facing. But even after they accomplished that, the general case still felt like a “moonshot,” Klein said.

Still, they had gotten a feel for their tools—in particular, the geometry of polynomials. A polynomial is a combination of terms made out of numbers and variables raised to powers, such as 3 x 2 y + 8 xz 7 . To study the traveling salesperson problem, the researchers distilled a map of cities down to a polynomial that had one variable for each edge between cities, and one term for each tree that could connect all the cities. Numerical factors then weighted these terms to reflect each edge’s value in the fractional solution to the traveling salesperson problem.

This polynomial, they found, has a coveted property called “real stability,” which means that the complex numbers that make the polynomial evaluate to zero never lie in the upper half of the complex plane. The nice thing about real stability is that it stays in force even when you make many kinds of changes to your polynomial. So, for example, if the researchers wanted to focus on particular cities, they could use a single variable to represent all the different edges leading into a city, and they could set the variables for edges they didn’t care about equal to 1. As they manipulated these simplified polynomials, the results of their manipulations still had real stability, opening the door to a wide assortment of techniques.

This approach enabled the researchers to get a handle on questions like how often the algorithm would be forced to connect two distant cities. In a nearly 80-page analysis, they managed to show that the algorithm beats out Christofides’ algorithm for the general traveling salesperson problem (the paper has yet to be peer-reviewed, but experts are confident that it’s correct). Once the paper was completed, Oveis Gharan dashed off an email to Saberi, his old doctoral adviser. “I guess I can finally graduate,” he joked.

While the improvement the researchers established is vanishingly small, computer scientists hope this breakthrough will inspire rapid further progress. That’s what happened back in 2011 when Oveis Gharan, Saberi and Singh figured out the graphical case. Within a year, other researchers had come up with radically different algorithms that greatly improved the approximation factor for the graphical case, which has now been lowered to 40 percent instead of Christofides’ 50 percent.

“When they announced their result [about the graphical case], … that made us think that it’s possible. It made us work for it,” said Svensson, one of the researchers who made further progress in that case. He’s been trying for many years to beat Christofides’ algorithm for the general traveling salesperson problem. “I will try again now I know it’s possible,” he said.

Over the decades, the traveling salesperson problem has launched many new methods into prominence. Oveis Gharan hopes that it will now play that role for the geometry of polynomials, for which he has become an eager evangelist. In the decade or so since he started learning about this approach, it has helped him prove a wide range of theorems. The tool has “shaped my whole career,” he said.

The new traveling salesperson result highlights the power of this approach, Newman said. “Definitely it’s an inspiration to look at it more closely.”

Klein will now have to find a new problem to obsess over. “It’s a bit sad to lose the problem, because it just built up so many structures in my head, and now they’re all kind of gone,” he said. But he couldn’t have asked for a more satisfying introduction to computer science research. “I felt like we pushed back a little bit on something that was unknown.”

Original story reprinted with permission from Quanta Magazine , an editorially independent publication of the Simons Foundation whose mission is to enhance public understanding of science by covering research developments and trends in mathematics and the physical and life sciences.

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Is it made out of tin foil , paper 2024/626, exponential quantum speedup for the traveling salesman problem.

The traveling salesman problem is the problem of finding out the shortest route in a network of cities, that a salesman needs to travel to cover all the cities, without visiting the same city more than once. This problem is known to be $NP$-hard with a brute-force complexity of $O(N^N)$ or $O(N^{2N})$ for $N$ number of cities. This problem is equivalent to finding out the shortest Hamiltonian cycle in a given graph, if at least one Hamiltonian cycle exists in it. Quantum algorithms for this problem typically provide with a quadratic speedup only, using Grover's search, thereby having a complexity of $O(N^{N/2})$ or $O(N^N)$. We present a bounded-error quantum polynomial-time (BQP) algorithm for solving the problem, providing with an exponential speedup. The overall complexity of our algorithm is $O(N^3\log(N)\kappa/\epsilon + 1/\epsilon^3)$, where the errors $\epsilon$ are $O(1/{\rm poly}(N))$, and $\kappa$ is the not-too-large condition number of the matrix encoding all Hamiltonian cycles.

Note: 6 pages, 2 figures, comments welcome.

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12.10: Traveling Salesperson Problem

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Learning Objectives

After completing this section, you should be able to:

• Distinguish between brute force algorithms and greedy algorithms.
• List all distinct Hamilton cycles of a complete graph.
• Apply brute force method to solve traveling salesperson applications.
• Apply nearest neighbor method to solve traveling salesperson applications.

We looked at Hamilton cycles and paths in the previous sections Hamilton Cycles and Hamilton Paths. In this section, we will analyze Hamilton cycles in complete weighted graphs to find the shortest route to visit a number of locations and return to the starting point. Besides the many routing applications in which we need the shortest distance, there are also applications in which we search for the route that is least expensive or takes the least time. Here are a few less common applications that you can read about on a website set up by the mathematics department at the University of Waterloo in Ontario, Canada:

• Design of fiber optic networks
• Minimizing fuel expenses for repositioning satellites
• Development of semi-conductors for microchips
• A technique for mapping mammalian chromosomes in genome sequencing

Before we look at approaches to solving applications like these, let's discuss the two types of algorithms we will use.

Brute Force and Greedy Algorithms

An algorithm is a sequence of steps that can be used to solve a particular problem. We have solved many problems in this chapter, and the procedures that we used were different types of algorithms. In this section, we will use two common types of algorithms, a brute force algorithm and a greedy algorithm . A brute force algorithm begins by listing every possible solution and applying each one until the best solution is found. A greedy algorithm approaches a problem in stages, making the apparent best choice at each stage, then linking the choices together into an overall solution which may or may not be the best solution.

To understand the difference between these two algorithms, consider the tree diagram in Figure 12.214. Suppose we want to find the path from left to right with the largest total sum. For example, branch A in the tree diagram has a sum of 10 + 2 + 11 + 13 = 36 10 + 2 + 11 + 13 = 36 .

To be certain that you pick the branch with greatest sum, you could list each sum from each of the different branches:

A : 10 + 2 + 11 + 13 = 36 10 + 2 + 11 + 13 = 36

B : 10 + 2 + 11 + 8 = 31 10 + 2 + 11 + 8 = 31

C : 10 + 2 + 15 + 1 = 28 10 + 2 + 15 + 1 = 28

D : 10 + 2 + 15 + 6 = 33 10 + 2 + 15 + 6 = 33

E : 10 + 7 + 3 + 20 = 40 10 + 7 + 3 + 20 = 40

F : 10 + 7 + 3 + 14 = 34 10 + 7 + 3 + 14 = 34

G : 10 + 7 + 4 + 11 = 32 10 + 7 + 4 + 11 = 32

H : 10 + 7 + 4 + 5 = 26 10 + 7 + 4 + 5 = 26

Then we know with certainty that branch E has the greatest sum.

Now suppose that you wanted to find the branch with the highest value, but you only were shown the tree diagram in phases, one step at a time.

After phase 1, you would have chosen the branch with 10 and 7. So far, you are following the same branch. Let’s look at the next phase.

After phase 2, based on the information you have, you will choose the branch with 10, 7 and 4. Now, you are following a different branch than before, but it is the best choice based on the information you have. Let’s look at the last phase.

After phase 3, you will choose branch G which has a sum of 32.

The process of adding the values on each branch and selecting the highest sum is an example of a brute force algorithm because all options were explored in detail. The process of choosing the branch in phases, based on the best choice at each phase is a greedy algorithm. Although a brute force algorithm gives us the ideal solution, it can take a very long time to implement. Imagine a tree diagram with thousands or even millions of branches. It might not be possible to check all the sums. A greedy algorithm, on the other hand, can be completed in a relatively short time, and generally leads to good solutions, but not necessarily the ideal solution.

Example 12.42

Distinguishing between brute force and greedy algorithms.

A cashier rings up a sale for $4.63 cents in U.S. currency. The customer pays with a $5 bill. The cashier would like to give the customer $0.37 in change using the fewest coins possible. The coins that can be used are quarters ($0.25), dimes ($0.10), nickels ($0.05), and pennies ($0.01). The cashier starts by selecting the coin of highest value less than or equal to $0.37, which is a quarter. This leaves $ 0.37 − $ 0.25 = $ 0.12 $ 0.37 − $ 0.25 = $ 0.12 . The cashier selects the coin of highest value less than or equal to $0.12, which is a dime. This leaves $ 0.12 − $ 0.10 = $ 0.02 $ 0.12 − $ 0.10 = $ 0.02 . The cashier selects the coin of highest value less than or equal to $0.02, which is a penny. This leaves $ 0.02 − $ 0.01 = $ 0.01 $ 0.02 − $ 0.01 = $ 0.01 . The cashier selects the coin of highest value less than or equal to $0.01, which is a penny. This leaves no remainder. The cashier used one quarter, one dime, and two pennies, which is four coins. Use this information to answer the following questions.

• Is the cashier’s approach an example of a greedy algorithm or a brute force algorithm? Explain how you know.
• The cashier’s solution is the best solution. In other words, four is the fewest number of coins possible. Is this consistent with the results of an algorithm of this kind? Explain your reasoning.
• The approach the cashier used is an example of a greedy algorithm, because the problem was approached in phases and the best choice was made at each phase. Also, it is not a brute force algorithm, because the cashier did not attempt to list out all possible combinations of coins to reach this conclusion.
• Yes, it is consistent. A greedy algorithm does not always yield the best result, but sometimes it does.

Your Turn 12.42

The traveling salesperson problem.

Now let’s focus our attention on the graph theory application known as the traveling salesperson problem (TSP) in which we must find the shortest route to visit a number of locations and return to the starting point.

Recall from Hamilton Cycles, the officer in the U.S. Air Force who is stationed at Vandenberg Air Force base and must drive to visit three other California Air Force bases before returning to Vandenberg. The officer needed to visit each base once. We looked at the weighted graph in Figure 12.219 representing the four U.S. Air Force bases: Vandenberg, Edwards, Los Angeles, and Beal and the distances between them.

Any route that visits each base and returns to the start would be a Hamilton cycle on the graph. If the officer wants to travel the shortest distance, this will correspond to a Hamilton cycle of lowest weight. We saw in Table 12.11 that there are six distinct Hamilton cycles (directed cycles) in a complete graph with four vertices, but some lie on the same cycle (undirected cycle) in the graph.

a → b → c → d → a

a → b → d → c → a

a → c → b → d → a

a → d → c → b → a

a → c → d → b → a

a → d → b → c → a

Since the distance between bases is the same in either direction, it does not matter if the officer travels clockwise or counterclockwise. So, there are really only three possible distances as shown in Figure 12.220.

The possible distances are:

396 + 410 + 106 + 159 = 1071 207 + 410 + 439 + 159 = 1215 396 + 439 + 106 + 207 = 1148 396 + 410 + 106 + 159 = 1071 207 + 410 + 439 + 159 = 1215 396 + 439 + 106 + 207 = 1148

So, a Hamilton cycle of least weight is V → B → E → L → V (or the reverse direction). The officer should travel from Vandenberg to Beal to Edwards, to Los Angeles, and back to Vandenberg.

Finding Weights of All Hamilton Cycles in Complete Graphs

Notice that we listed all of the Hamilton cycles and found their weights when we solved the TSP about the officer from Vandenberg. This is a skill you will need to practice. To make sure you don't miss any, you can calculate the number of possible Hamilton cycles in a complete graph. It is also helpful to know that half of the directed cycles in a complete graph are the same cycle in reverse direction, so, you only have to calculate half the number of possible weights, and the rest are duplicates.

In a complete graph with n n vertices,

• The number of distinct Hamilton cycles is ( n − 1 ) ! ( n − 1 ) ! .
• There are at most ( n − 1 ) ! 2 ( n − 1 ) ! 2 different weights of Hamilton cycles.

TIP! When listing all the distinct Hamilton cycles in a complete graph, you can start them all at any vertex you choose. Remember, the cycle a → b → c → a is the same cycle as b → c → a → b so there is no need to list both.

Example 12.43

Calculating possible weights of hamilton cycles.

Suppose you have a complete weighted graph with vertices N, M, O , and P .

• Use the formula ( n − 1 ) ! ( n − 1 ) ! to calculate the number of distinct Hamilton cycles in the graph.
• Use the formula ( n − 1 ) ! 2 ( n − 1 ) ! 2 to calculate the greatest number of different weights possible for the Hamilton cycles.
• Are all of the distinct Hamilton cycles listed here? How do you know? Cycle 1: N → M → O → P → N Cycle 2: N → M → P → O → N Cycle 3: N → O → M → P → N Cycle 4: N → O → P → M → N Cycle 5: N → P → M → O → N Cycle 6: N → P → O → M → N
• Which pairs of cycles must have the same weights? How do you know?
• There are 4 vertices; so, n = 4 n = 4 . This means there are ( n − 1 ) ! = ( 4 − 1 ) ! = 3 ⋅ 2 ⋅ 1 = 6 ( n − 1 ) ! = ( 4 − 1 ) ! = 3 ⋅ 2 ⋅ 1 = 6 distinct Hamilton cycles beginning at any given vertex.
• Since n = 4 n = 4 , there are ( n − 1 ) ! 2 = ( 4 − 1 ) ! 2 = 6 2 = 3 ( n − 1 ) ! 2 = ( 4 − 1 ) ! 2 = 6 2 = 3 possible weights.
• Yes, they are all distinct cycles and there are 6 of them.
• Cycles 1 and 6 have the same weight, Cycles 2 and 4 have the same weight, and Cycles 3 and 5 have the same weight, because these pairs follow the same route through the graph but in reverse.

TIP! When listing the possible cycles, ignore the vertex where the cycle begins and ends and focus on the ways to arrange the letters that represent the vertices in the middle. Using a systematic approach is best; for example, if you must arrange the letters M, O, and P, first list all those arrangements beginning with M, then beginning with O, and then beginning with P, as we did in Example 12.42.

Your Turn 12.43

The brute force method.

The method we have been using to find a Hamilton cycle of least weight in a complete graph is a brute force algorithm, so it is called the brute force method . The steps in the brute force method are:

Step 1: Calculate the number of distinct Hamilton cycles and the number of possible weights.

Step 2: List all possible Hamilton cycles.

Step 3: Find the weight of each cycle.

Step 4: Identify the Hamilton cycle of lowest weight.

Example 12.44

Applying the brute force method.

On the next assignment, the air force officer must leave from Travis Air Force base, visit Beal, Edwards, and Vandenberg Air Force bases each exactly once and return to Travis Air Force base. There is no need to visit Los Angeles Air Force base. Use Figure 12.221 to find the shortest route.

Step 1: Since there are 4 vertices, there will be ( 4 − 1 ) ! = 3 ! = 6 ( 4 − 1 ) ! = 3 ! = 6 cycles, but half of them will be the reverse of the others; so, there will be ( 4 − 1 ) ! 2 = 6 2 = 3 ( 4 − 1 ) ! 2 = 6 2 = 3 possible distances.

Step 2: List all the Hamilton cycles in the subgraph of the graph in Figure 12.222.

To find the 6 cycles, focus on the three vertices in the middle, B, E, and V . The arrangements of these vertices are BEV, BVE, EBV, EVB, VBE , and VEB . These would correspond to the 6 cycles:

1: T → B → E → V → T

2: T → B → V → E → T

3: T → E → B → V → T

4: T → E → V → B → T

5: T → V → B → E → T

6: T → V → E → B → T

Step 3: Find the weight of each path. You can reduce your work by observing the cycles that are reverses of each other.

1: 84 + 410 + 207 + 396 = 1097 84 + 410 + 207 + 396 = 1097

2: 84 + 396 + 207 + 370 = 1071 84 + 396 + 207 + 370 = 1071

3: 370 + 410 + 396 + 396 = 1572 370 + 410 + 396 + 396 = 1572

4: Reverse of cycle 2, 1071

5: Reverse of cycle 3, 1572

6: Reverse of cycle 1, 1097

Step 4: Identify a Hamilton cycle of least weight.

The second path, T → B → V → E → T , and its reverse, T → E → V → B → T , have the least weight. The solution is that the officer should travel from Travis Air Force base to Beal Air Force Base, to Vandenberg Air Force base, to Edwards Air Force base, and return to Travis Air Force base, or the same route in reverse.

Your Turn 12.44

Now suppose that the officer needed a cycle that visited all 5 of the Air Force bases in Figure 12.221. There would be ( 5 − 1 ) ! = 4 ! = 4 × 3 × 2 × 1 = 24 ( 5 − 1 ) ! = 4 ! = 4 × 3 × 2 × 1 = 24 different arrangements of vertices and ( 5 − 1 ) ! 2 = 4 ! 2 = 24 2 = 12 ( 5 − 1 ) ! 2 = 4 ! 2 = 24 2 = 12 distances to compare using the brute force method. If you consider 10 Air Force bases, there would be ( 10 − 1 ) ! = 9 ! = 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 362 , 880 ( 10 − 1 ) ! = 9 ! = 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 362 , 880 different arrangements and ( 10 − 1 ) ! 2 = 9 ! 2 = 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 2 = 181 , 440 ( 10 − 1 ) ! 2 = 9 ! 2 = 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 2 = 181 , 440 distances to consider. There must be another way!

The Nearest Neighbor Method

When the brute force method is impractical for solving a traveling salesperson problem, an alternative is a greedy algorithm known as the nearest neighbor method , which always visit the closest or least costly place first. This method finds a Hamilton cycle of relatively low weight in a complete graph in which, at each phase, the next vertex is chosen by comparing the edges between the current vertex and the remaining vertices to find the lowest weight. Since the nearest neighbor method is a greedy algorithm, it usually doesn’t give the best solution, but it usually gives a solution that is "good enough." Most importantly, the number of steps will be the number of vertices. That’s right! A problem with 10 vertices requires 10 steps, not 362,880. Let’s look at an example to see how it works.

Suppose that a candidate for governor wants to hold rallies around the state. They plan to leave their home in city A , visit cities B, C, D, E , and F each once, and return home. The airfare between cities is indicated in the graph in Figure 12.223.

Let’s help the candidate keep costs of travel down by applying the nearest neighbor method to find a Hamilton cycle that has a reasonably low weight. Begin by marking starting vertex as V 1 Figure 12.224. The lowest of these is $100, which is the edge between A and F .

Mark F as V 2 Figure 12.225. The lowest of these is $150, which is the edge between F and D .

Mark D as V 3 Figure 12.226. The lowest of these is $120, which is the edge between D and B .

So, mark B as V 4 Figure 12.227. The lower amount is $160, which is the edge between B and E .

Now you can mark E as V 5 Figure 12.228. Make a note of the weight of the edge from E to C , which is $180, and from C back to A , which is $210.

The Hamilton cycle we found is A → F → D → B → E → C → A . The weight of the circuit is $ 100 + $ 150 + $ 120 + $ 160 + $ 180 + $ 210 = $ 920 $ 100 + $ 150 + $ 120 + $ 160 + $ 180 + $ 210 = $ 920 . This may or may not be the route with the lowest cost, but there is a good chance it is very close since the weights are most of the lowest weights on the graph and we found it in six steps instead of finding 120 different Hamilton cycles and calculating 60 weights. Let’s summarize the procedure that we used.

Step 1: Select the starting vertex and label V 1 V 1 for "visited 1st." Identify the edge of lowest weight between V 1 V 1 and the remaining vertices.

Step 2: Label the vertex at the end of the edge of lowest weight that you found in previous step as V n V n where the subscript n indicates the order the vertex is visited. Identify the edge of lowest weight between V n V n and the vertices that remain to be visited.

Step 3: If vertices remain that have not been visited, repeat Step 2. Otherwise, a Hamilton cycle of low weight is V 1 → V 2 → ⋯ → V n → V 1 V 1 → V 2 → ⋯ → V n → V 1 .

Example 12.45

Using the nearest neighbor method.

Suppose that the candidate for governor wants to hold rallies around the state but time before the election is very limited. They would like to leave their home in city A , visit cities B , C , D , E , and F each once, and return home. The airfare between cities is not as important as the time of travel, which is indicated in Figure 12.229. Use the nearest neighbor method to find a route with relatively low travel time. What is the total travel time of the route that you found?

Step 1: Label vertex A as V 1 V 1 . The edge of lowest weight between A and the remaining vertices is 85 min between A and D .

Step 2: Label vertex D as V 2 V 2 . The edge of lowest weight between D and the vertices that remain to be visited, B, C, E , and F , is 70 min between D and F .

Repeat Step 2: Label vertex F as V 3 V 3 . The edge of lowest weight between F and the vertices that remain to be visited, B, C, and E , is 75 min between F and C .

Repeat Step 2: Label vertex C as V 4 V 4 . The edge of lowest weight between C and the vertices that remain to be visited, B and E , is 100 min between C and B .

Repeat Step 2: Label vertex B as V 5 V 5 . The only vertex that remains to be visited is E . The weight of the edge between B and E is 95 min.

Step 3: A Hamilton cycle of low weight is A → D → F → C → B → E → A . So, a route of relatively low travel time is A to D to F to C to B to E and back to A . The total travel time of this route is: 85 min + 70 min + 75 min + 100 min + 95 min + 90 min = 515 min or 8 hrs 35 min 85 min + 70 min + 75 min + 100 min + 95 min + 90 min = 515 min or 8 hrs 35 min

Your Turn 12.45

Check your understanding, section 12.9 exercises.

Trains Moscow to Elektrostal: Times, Prices and Tickets

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Moscow to Elektrostal by train

The journey from Moscow to Elektrostal by train is 32.44 mi and takes 2 hr 7 min. There are 71 connections per day, with the first departure at 12:15 AM and the last at 11:46 PM. It is possible to travel from Moscow to Elektrostal by train for as little as or as much as . The best price for this journey is .

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Virail's search tool will provide you with the options you need when you want to go from Moscow to Elektrostal. All you need to do is enter the dates of your planned journey, and let us take care of everything else. Our engine does the hard work, searching through thousands of routes offered by our trusted travel partners to show you options for traveling by train, bus, plane, or carpool. You can filter the results to suit your needs. There are a number of filtering options, including price, one-way or round trip, departure or arrival time, duration of journey, or number of connections. Soon you'll find the best choice for your journey. When you're ready, Virail will transfer you to the provider's website to complete the booking. No matter where you're going, get there with Virail.

How can I find the cheapest train tickets to get from Moscow to Elektrostal?

Prices will vary when you travel from Moscow to Elektrostal. On average, though, you'll pay about for a train ticket. You can find train tickets for prices as low as , but it may require some flexibility with your travel plans. If you're looking for a low price, you may need to prepare to spend more time in transit. You can also often find cheaper train tickets at particular times of day, or on certain days of the week. Of course, ticket prices often change during the year, too; expect to pay more in peak season. For the lowest prices, it's usually best to make your reservation in advance. Be careful, though, as many providers do not offer refunds or exchanges on their cheapest train tickets. Unfortunately, no price was found for your trip from Moscow to Elektrostal. Selecting a new departure or arrival city, without dramatically changing your itinerary could help you find price results. Prices will vary when you travel from Moscow to Elektrostal. On average, though, you'll pay about for a train ticket. If you're looking for a low price, you may need to prepare to spend more time in transit. You can also often find cheaper train tickets at particular times of day, or on certain days of the week. Of course, ticket prices often change during the year, too; expect to pay more in peak season. For the lowest prices, it's usually best to make your reservation in advance. Be careful, though, as many providers do not offer refunds or exchanges on their cheapest train tickets.

How long does it take to get from Moscow to Elektrostal by train?

The journey between Moscow and Elektrostal by train is approximately 32.44 mi. It will take you more or less 2 hr 7 min to complete this journey. This average figure does not take into account any delays that might arise on your route in exceptional circumstances. If you are planning to make a connection or operating on a tight schedule, give yourself plenty of time. The distance between Moscow and Elektrostal is around 32.44 mi. Depending on the exact route and provider you travel with, your journey time can vary. On average, this journey will take approximately 2 hr 7 min. However, the fastest routes between Moscow and Elektrostal take 1 hr 3 min. If a fast journey is a priority for you when traveling, look out for express services that may get you there faster. Some flexibility may be necessary when booking. Often, these services only leave at particular times of day - or even on certain days of the week. You may also find a faster journey by taking an indirect route and connecting in another station along the way.

How many journeys from Moscow to Elektrostal are there every day?

On average, there are 71 daily departures from Moscow to Elektrostal. However, there may be more or less on different days. Providers' timetables can change on certain days of the week or public holidays, and many also vary at particular times of year. Some providers change their schedules during the summer season, for example. At very busy times, there may be up to departures each day. The providers that travel along this route include , and each operates according to their own specific schedules. As a traveler, you may prefer a direct journey, or you may not mind making changes and connections. If you have heavy suitcases, a direct journey could be best; otherwise, you might be able to save money and enjoy more flexibility by making a change along the way. Every day, there are an average of 18 departures from Moscow which travel directly to Elektrostal. There are 53 journeys with one change or more. Unfortunately, no connection was found for your trip from Moscow to Elektrostal. Selecting a new departure or arrival city, without dramatically changing your itinerary could help you find connections.

Book in advance and save

If you're looking for the best deal for your trip from Moscow to Elektrostal, booking train tickets in advance is a great way to save money, but keep in mind that advance tickets are usually not available until 3 months before your travel date.

Stay flexible with your travel time and explore off-peak journeys

Planning your trips around off-peak travel times not only means that you'll be able to avoid the crowds, but can also end up saving you money. Being flexible with your schedule and considering alternative routes or times will significantly impact the amount of money you spend on getting from Moscow to Elektrostal.

Always check special offers

Checking on the latest deals can help save a lot of money, making it worth taking the time to browse and compare prices. So make sure you get the best deal on your ticket and take advantage of special fares for children, youth and seniors as well as discounts for groups.

Unlock the potential of slower trains or connecting trains

If you're planning a trip with some flexible time, why not opt for the scenic route? Taking slower trains or connecting trains that make more stops may save you money on your ticket – definitely worth considering if it fits in your schedule.

Best time to book cheap train tickets from Moscow to Elektrostal

The cheapest Moscow - Elektrostal train tickets can be found for as low as $35.01 if you’re lucky, or $54.00 on average. The most expensive ticket can cost as much as $77.49.

Find the best day to travel to Elektrostal by train

When travelling to Elektrostal by train, if you want to avoid crowds you can check how frequently our customers are travelling in the next 30-days using the graph below. On average, the peak hours to travel are between 6:30am and 9am in the morning, or between 4pm and 7pm in the evening. Please keep this in mind when travelling to your point of departure as you may need some extra time to arrive, particularly in big cities!

Moscow to Elektrostal CO2 Emissions by Train

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