travelling salesman problem belongs to which of the class

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Travelling salesman problem explained

• Published on April 26, 2024
• by Oguzhan Uyar
• Last updated: 1 week 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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Traveling Salesperson Problem

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A salesperson needs to visit a set of cities to sell their goods. They know how many cities they need to go to and the distances between each city. In what order should the salesperson visit each city exactly once so that they minimize their travel time and so that they end their journey in their city of origin?

The traveling salesperson problem is an extremely old problem in computer science that is an extension of the Hamiltonian Circuit Problem . It has important implications in complexity theory and the P versus NP problem because it is an NP-Complete problem . This means that a solution to this problem cannot be found in polynomial time (it takes superpolynomial time to compute an answer). In other words, as the number of vertices increases linearly, the computation time to solve the problem increases exponentially.

The following image is a simple example of a network of cities connected by edges of a specific distance. The origin city is also marked.

Network of cities

Here is the solution for that network, it has a distance traveled of only 14. Any other path that the salesman can takes will result in a path length that is more than 14.

Relationship to Graphs

Special kinds of tsp, importance for p vs np, applications.

The traveling salesperson problem can be modeled as a graph . Specifically, it is typical a directed, weighted graph. Each city acts as a vertex and each path between cities is an edge. Instead of distances, each edge has a weight associated with it. In this model, the goal of the traveling salesperson problem can be defined as finding a path that visits every vertex, returns to the original vertex, and minimizes total weight.

To that end, many graph algorithms can be used on this model. Search algorithms like breadth-first search (BFS) , depth-first search (DFS) , and Dijkstra's shortest path algorithm can certainly be used, however, they do not take into consideration that fact that every vertex must be visited.

The Traveling Salesperson Problem (TSP), an NP-Complete problem, is notoriously complicated to solve. That is because the greedy approach is so computational intensive. The greedy approach to solving this problem would be to try every single possible path and see which one is the fastest. Try this conceptual question to see if you have a grasp for how hard it is to solve.

For a fully connected map with \(n\) cities, how many total paths are possible for the traveling salesperson? Show Answer There are (n-1)! total paths the salesperson can take. The computation needed to solve this problem in this way grows far too quickly to be a reasonable solution. If this map has only 5 cities, there are \(4!\), or 24, paths. However, if the size of this map is increased to 20 cities, there will be \(1.22 \cdot 10^{17}\) paths!

The greedy approach to TSP would go like this:

• Find all possible paths.
• Find the cost of every paths.
• Choose the path with the lowest cost.

Another version of a greedy approach might be: At every step in the algorithm, choose the best possible path. This version might go a little quicker, but it's not guaranteed to find the best answer, or an answer at all since it might hit a dead end.

For NP-Hard problems (a subset of NP-Complete problems) like TSP, exact solutions can only be implemented in a reasonable amount of time for small input sizes (maps with few cities). Otherwise, the best approach we can do is provide a heuristic to help the problem move forward in an optimal way. However, these approaches cannot be proven to be optimal because they always have some sort of downside.

Small input sizes

As described, in a previous section , the greedy approach to this problem has a complexity of \(O(n!)\). However, there are some approaches that decrease this computation time.

The Held-Karp Algorithm is one of the earliest applications of dynamic programming . Its complexity is much lower than the greedy approach at \(O(n^2 2^n)\). Basically what this algorithm says is that every sub path along an optimal path is itself an optimal path. So, computing an optimal path is the same as computing many smaller subpaths and adding them together.

Heuristics are a way of ranking possible next steps in an algorithm in the hopes of cutting down computation time for the entire algorithm. They are often a tradeoff of some attribute - such as completeness, accuracy, or precision - in favor of speed. Heuristics exist for the traveling salesperson problem as well.

The most simple heuristic for this problem is the greedy heuristic. This heuristic simply says, at each step of the network traversal, choose the best next step. In other words, always choose the closest city that you have not yet visited. This heuristic seems like a good one because it is simple and intuitive, and it is even used in practice sometimes, however there are heuristics that are proven to be more effective.

Christofides algorithm is another heuristic. It produces at most 1.5 times the optimal weight for TSP. This algorithm involves finding a minimum spanning tree for the network. Next, it creates matchings for the cities of an odd degree (meaning they have an odd number of edges coming out of them), calculates an eulerian path , and converts back to a TSP path.

Even though it is typically impossible to optimally solve TSP problems, there are cases of TSP problems that can be solved if certain conditions hold.

The metric-TSP is an instance of TSP that satisfies this condition: The distance from city A to city B is less than or equal to the distance from city A to city C plus the distance from city C to city B. Or,

\[distance_{AB} \leq distance_{AC} + distance_{CB}\]

This is a condition that holds in the real world, but it can't always be expected to hold for every TSP problem. But, with this inequality in place, the approximated path will be no more than twice the optimal path. Even better, we can bound the solution to a \(3/2\) approximation by using Christofide's Algorithm .

The euclidean-TSP has an even stricter constraint on the TSP input. It states that all cities' edges in the network must obey euclidean distances . Recent advances have shown that approximation algorithms using euclidean minimum spanning trees have reduced the runtime of euclidean-TSP, even though they are also NP-hard. In practice, though, simpler heuristics are still used.

The P versus NP problem is one of the leading questions in modern computer science. It asks whether or not every problem whose solution can be verified in polynomial time by a computer can also be solved in polynomial time by a computer. TSP, for example, cannot be solved in polynomial time (at least that's what is currently theorized). However, TSP can be solved in polynomial time when it is phrased like this: Given a graph and an integer, x, decide if there is a path of length x or less than x . It's easy to see that given a proposed answer to this question, it is simple to check if it is less than or equal to x.

The traveling salesperson problem, like other problems that are NP-Complete, are very important to this debate. That is because if a polynomial time solution can be found to this problems, then \(P = NP\). As it stands, most scientists believe that \(P \ne NP\).

The traveling salesperson problem has many applications. The obvious ones are in the transportation space. Planning delivery routes or flight patterns, for example, would benefit immensly from breakthroughs is this problem or in the P versus NP problem .

However, this same logic can be applied to many facets of planning as well. In robotics, for instance, planning the order in which to drill holes in a circuit board is a complex task due to the sheer number of holes that must be drawn.

The best and most important application of TSP, however, comes from the fact that it is an NP-Complete problem. That means that its practical applications amount to the applications of any problem that is NP-Complete. So, if there are significant breakthroughs for TSP, that means that those exact same breakthrough can be applied to any problem in the NP-Complete class.

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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.

Elastic Net Method for the Travelling Salesman Problem

Given the positions of N cities, what is the shortest closed tour in which each city can be visited once?

All exact methods known for determining an optimal route require a computing effort that increases exponentially with number of cities, so in practice exact solutions can be attempted only on problems involving a few hundred cities or less. The travelling salesman problem belongs to the large class of nondeterministic polynomial time complete problems.

The elastic net method has been applied to the travelling salesman problem . The essence of the method is:

Using an iterative procedure, a circular closed path is gradually elongated non-uniformly until it eventually passes sufficiently near to all the cities to define a tour.

A tour can be viewed as a mapping from a circle to the plane so that each city in the plane is mapped to by some point on the circle. We consider mappings from a circular path of points to the plane in which neighboring points on the circle are mapped as close as possible on the plane. This is a special case of the general problem of best preserving neighborhood relations when mapping between different geometrical spaces.

By this means, each city becomes associated with a particular section on the path. The tightness of the association is determined by how the force contributed from a city depends on a distance, and the nature of this dependence changes as the algorithm progresses. Initially all cities have roughly equal influence on each point on the path. Subsequently, longer distance become less favored, and each city gradually becomes more specific for the points on the path closest to it.

If you were using a Java-enabled browser, you would see Travelling Salesman Problem demonstration instead of this paragraph.

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On inverse traveling salesman problems

• Research Paper
• Published: 08 January 2012
• Volume 10 , pages 193–209, ( 2012 )

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• Yerim Chung 1 &
• Marc Demange 2  

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The inverse traveling salesman problem belongs to the class of “inverse combinatorial optimization” problems. In an inverse combinatorial optimization problem, we are given a feasible solution for an instance of a particular combinatorial optimization problem, and the task is to adjust the instance parameters as little as possible so that the given solution becomes optimal in the new instance. In this paper, we consider a variant of the inverse traveling salesman problem, denoted by I TSP W,A , by taking into account a set W of admissible weight systems and a specific algorithm. We are given an edge-weighted complete graph (an instance of TSP), a Hamiltonian tour (a feasible solution of TSP) and a specific algorithm solving TSP. Then, I TSP W,A , is the problem to find a new weight system in W which minimizes the difference from the original weight system so that the given tour can be selected by the algorithm as a solution. We consider the cases \({W \in \{\mathbb{R}^{+m}, \{1, 2\}^m , \Delta\}}\) where Δ denotes the set of edge weight systems satisfying the triangular inequality and m is the number of edges. As for algorithms, we consider a local search algorithm 2-opt , a greedy algorithm closest neighbor and any optimal algorithm. We devise both complexity and approximation results. We also deal with the inverse traveling salesman problem on a line for which we modify the positions of vertices instead of edge weights. We handle the cases \({W \in \{\mathbb{R}^{+n}, \mathbb{N}^n\}}\) where n is the number of vertices.

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Ahuja R, Orlin J (2000) A faster algorithm for the inverse spanning tree problem. J Algorithms 34(1): 177–193

Article   Google Scholar  

Ahuja R, Orlin J (2001) Inverse optimization. Oper Res 49(5): 771–783

Angelov S, Harb B, Kannan S, Wang LS (2006) Weighted isotonic regression under the l1 norm. In: Proceedings of the 17th annual ACM-SIAM symposium on discrete algorithms, SODA, pp 783–791

Ausiello G, Crescenzi P, Gambosi G, Kann V, Marchetti-Spaccame-La A, Protasi M (1999) Complexity and approximation: Combinatorial optimization problems and their approximability properties. Springer, Berlin

Google Scholar  

Burton D, Toint P (1992) On an instance of the inverse shortest paths problem. Math Programm 53: 45–61

Burton D, Toint P (1994) On the use of an inverse shortest paths algorithm for recovering linearly correlated costs. Math Programm 63: 1–22

Carr SC, Lovejoy WS (1997) The inverse newsvendor problem: choosing an optimal demand portfolio for capacitated resources. Technical report, Department of Industrial and Operations Engineering, University of Michigan, Ann Arbor, MI

Chakravarti N (1989) Isotonic median regression; a linear programming approach. Math Oper Res 14(2): 303–308

Chung Y, Culus JF, Demange M (2008) Inverse booking problems. In: Proceedings of the 2d workshop on algorithms and computation, WALCOM 2008, Lecture Notes in Computer Science, vol 4921, pp 180–187

Chung Y, Demange M (2008) The 0-1 inverse maximum stable set problem. Discret Appl Math 156(13): 2501–2516

Ciura E, Deaconu A (2007) Inverse minimum flow problem. J Appl Math Comput 23(1-2): 193–203

Demange M, Ekim T, de Werra D (2009) A tutorial on the use of graph coloring for some problems in robotics. Eur J Oper Res 192(1): 41–55

Demange M, Monnot J (2010) An introduction to inverse combinatorial problems. In: Paradigms of combinatorial optimization (problems and new approaches). ISTE-WILEY, London-Hoboken (UK-USA), Vangelis Th., Paschos, pp 547–586

Dembo R, Merkoulovitch L, Rosen D (1998) Images from a portfolio. Algorithmics research working paper, Algorithmics, Inc., Canada

Dial B (1997) Minimum-revenue congestion pricing, Part 1: a fast algorithm for the single-origin case. Technical report, The Volpe National Transportation Systems Center, Kendall Square, Cambridge, MA

Dinur I, Safra S (2005) On the hardness of approximating minimum vertex cover. Ann Math 162(1): 439–485

Garey MR, Johnson DS (1979) Computers and intractability—a guide to the theory of NP-completeness. Freeman, San Francisco

Heuberger C (2004) Inverse combinatorial optimization: a survey on problems, methods, and results. J Comb Optim 8(3): 329–361

Karakostas G (2005) A better approximation ratio for the vertex cover problem. In: Proceedings of the 32nd international colloquium on automata, languages and programming, ICALP, Lecture Notes in Computer Science, vol 3580, pp 1043–1050

Khanna S, Motwani R, Sudan M, Vazirani V (1998) On syntactic versus computational views of approximability. SIAM J Comput 28(1): 164–191

Nolet G (1987) Seismic tomography. Reidel, Dordrecht

Papadimitriou C (1994) Computational complexity. Addison-Wesley, New York

Papadimitriou CH, Steiglitz K (1976) Some complexity results for the traveling salesman problem. In: STOC’76, pp 1–9

Robertson T, Wright F (1980) Algorithms in order restricted statistical inference and the cauchy mean value property. Ann Stat 8: 645–651

Tarantola A (1987) Inverse problem theory: methods for data fitting and model parameter estimation. Elsevier, Amsterdam

Xu S, Zhang J (1995) An inverse problem of the weighted shortest path problem. Jpn J Ind Appl Math 12(1): 47–59

Ye Y (1991) An O ( n 3 L ) potential reduction algorithm for linear programming. Math Programm 50: 239–258

Zhang J, Liu Z (2002) A general model of some inverse combinatorial optimization problems and its solution method under l-norm. J Comb Optim 6(2): 207–227

Zhang J, Yang X, Cai M (1999) The complexity analysis of the inverse center location problem. J Glob Optim 15(2): 213–218

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Chung, Y., Demange, M. On inverse traveling salesman problems. 4OR-Q J Oper Res 10 , 193–209 (2012). https://doi.org/10.1007/s10288-011-0194-4

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• Inverse problem
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