Traveling Salesperson Problem

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.

Solution

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:

1. Find all possible paths.
2. Find the cost of every paths.
3. 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

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.