Travelling salesman problem

Symmetric TSP with four cities

TSP can be modelled as an undirected weighted graph, such that cities are the graph's vertices, paths are the graph's edges, and a path's distance is the edge's weight. It is a minimization problem starting and finishing at a specified vertex after having visited each other vertex exactly once. Often, the model is a complete graph (i.e., each pair of vertices is connected by an edge). If no path exists between two cities, adding a sufficiently long edge will complete the graph without affecting the optimal tour.

In the symmetric TSP, the distance between two cities is the same in each opposite direction, forming an undirected graph. This symmetry halves the number of possible solutions. In the asymmetric TSP, paths may not exist in both directions or the distances might be different, forming a directed graph. Traffic collisions, one-way streets, and airfares for cities with different departure and arrival fees are examples of how this symmetry could break down.

• An equivalent formulation in terms of graph theory is: Given a complete weighted graph (where the vertices would represent the cities, the edges would represent the roads, and the weights would be the cost or distance of that road), find a Hamiltonian cycle with the least weight.
• The requirement of returning to the starting city does not change the computational complexity of the problem, see Hamiltonian path problem.
• Another related problem is the bottleneck travelling salesman problem (bottleneck TSP): Find a Hamiltonian cycle in a weighted graph with the minimal weight of the weightiest edge. For example, avoiding narrow streets with big buses.[14] The problem is of considerable practical importance, apart from evident transportation and logistics areas. A classic example is in printed circuit manufacturing: scheduling of a route of the drill machine to drill holes in a PCB. In robotic machining or drilling applications, the "cities" are parts to machine or holes (of different sizes) to drill, and the "cost of travel" includes time for retooling the robot (single machine job sequencing problem).[15]
• The generalized travelling salesman problem, also known as the "travelling politician problem", deals with "states" that have (one or more) "cities" and the salesman has to visit exactly one "city" from each "state". One application is encountered in ordering a solution to the cutting stock problem in order to minimize knife changes. Another is concerned with drilling in semiconductor manufacturing, see e.g., U.S. Patent 7,054,798. Noon and Bean demonstrated that the generalized travelling salesman problem can be transformed into a standard TSP with the same number of cities, but a modified distance matrix.
• The sequential ordering problem deals with the problem of visiting a set of cities where precedence relations between the cities exist.
• A common interview question at Google is how to route data among data processing nodes; routes vary by time to transfer the data, but nodes also differ by their computing power and storage, compounding the problem of where to send data.
• The travelling purchaser problem deals with a purchaser who is charged with purchasing a set of products. He can purchase these products in several cities, but at different prices and not all cities offer the same products. The objective is to find a route between a subset of the cities that minimizes total cost (travel cost + purchasing cost) and enables the purchase of all required products.

In addition to the ${\displaystyle x_{ij}}$ variables as above, there is for each ${\displaystyle i=2,\ldots ,n}$ a dummy variable ${\displaystyle u_{i}}$ that keeps track of the order in which the cities are visited, counting from city ${\displaystyle 1}$; the interpretation is that ${\displaystyle u_{i}<u_{j}}$ implies city ${\displaystyle i}$ is visited before city ${\displaystyle j}$. For a given tour (as encoded into values of the ${\displaystyle x_{ij}}$ variables), one may find satisfying values for the ${\displaystyle u_{i}}$ variables by making ${\displaystyle u_{i}}$ equal to the number of edges along that tour, when going from city ${\displaystyle 1}$ to city ${\displaystyle i}$.

Because linear programming favours non-strict inequalities (${\displaystyle \geq }$) over strict (${\displaystyle >}$), we would like to impose constraints to the effect that

${\displaystyle u_{j}\geq u_{i}+1}$ if ${\displaystyle x_{ij}=1}$.

Merely requiring ${\displaystyle u_{j}\geq u_{i}+x_{ij}}$ would not achieve that, because this also requires ${\displaystyle u_{j}\geq u_{i}}$ when ${\displaystyle x_{ij}=0}$, which is not correct. Instead MTZ use the ${\displaystyle (n-1)(n-2)}$ linear constraints

${\displaystyle u_{j}+(n-2)\geq u_{i}+(n-1)x_{ij}}$ for all distinct ${\displaystyle i,j\in \{2,\dotsc ,n\}}$

where the constant term ${\displaystyle n-2}$ provides sufficient slack that ${\displaystyle x_{ij}=0}$ does not impose a relation between ${\displaystyle u_{j}}$ and ${\displaystyle u_{i}}$.

The way that the ${\displaystyle u_{i}}$ variables then enforce that a single tour visits all cities is that they increase by (at least) ${\displaystyle 1}$ for each step along a tour, with a decrease only allowed where the tour passes through city ${\displaystyle 1}$. That constraint would be violated by every tour which does not pass through city ${\displaystyle 1}$, so the only way to satisfy it is that the tour passing city ${\displaystyle 1}$ also passes through all other cities.

The MTZ formulation of TSP is thus the following integer linear programming problem:

${\displaystyle {\begin{aligned}\min \sum _{i=1}^{n}\sum _{j\neq i,j=1}^{n}c_{ij}x_{ij}&\colon &&\\x_{ij}\in {}&\{0,1\}&&i,j=1,\ldots ,n;\\u_{i}\in {}&\mathbf {Z} &&i=2,\ldots ,n;\\\sum _{i=1,i\neq j}^{n}x_{ij}={}&1&&j=1,\ldots ,n;\\\sum _{j=1,j\neq i}^{n}x_{ij}={}&1&&i=1,\ldots ,n;\\u_{i}-u_{j}+(n-1)x_{ij}\leq {}&n-2&&2\leq i\neq j\leq n;\\2\leq u_{i}\leq {}&n&&2\leq i\leq n.\end{aligned}}}$

The first set of equalities requires that each city is arrived at from exactly one other city, and the second set of equalities requires that from each city there is a departure to exactly one other city. The last constraints enforce that there is only a single tour covering all cities, and not two or more disjointed tours that only collectively cover all cities. To prove this, it is shown below (1) that every feasible solution contains only one closed sequence of cities, and (2) that for every single tour covering all cities, there are values for the dummy variables ${\displaystyle u_{i}}$ that satisfy the constraints.

To prove that every feasible solution contains only one closed sequence of cities, it suffices to show that every subtour in a feasible solution passes through city 1 (noting that the equalities ensure there can only be one such tour). For if we sum all the inequalities corresponding to ${\displaystyle x_{ij}=1}$ for any subtour of k steps not passing through city 1, we obtain:

${\displaystyle (n-1)k\leq (n-2)k,}$

which is a contradiction.

It now must be shown that for every single tour covering all cities, there are values for the dummy variables ${\displaystyle u_{i}}$ that satisfy the constraints.

Without loss of generality, define the tour as originating (and ending) at city 1. Choose ${\displaystyle u_{i}=t}$ if city ${\displaystyle i}$ is visited in step ${\displaystyle t(i,t=2,3,\ldots ,n)}$. Then

${\displaystyle u_{i}-u_{j}\leq n-2,}$

since ${\displaystyle u_{i}}$ can be no greater than ${\displaystyle n}$ and ${\displaystyle u_{j}}$ can be no less than 2; hence the constraints are satisfied whenever ${\displaystyle x_{ij}=0.}$ For ${\displaystyle x_{ij}=1}$, we have:

${\displaystyle u_{i}-u_{j}+(n-1)x_{ij}=(t)-(t+1)+n-1=n-2,}$

satisfying the constraint.

Label the cities with the numbers 1, …, n and define:

${\displaystyle x_{ij}={\begin{cases}1&{\text{the path goes from city }}i{\text{ to city }}j\\0&{\text{otherwise}}\end{cases}}}$

Take ${\displaystyle c_{ij}>0}$ to be the distance from city i to city j. Then TSP can be written as the following integer linear programming problem:

${\displaystyle {\begin{aligned}\min &\sum _{i=1}^{n}\sum _{j\neq i,j=1}^{n}c_{ij}x_{ij}\colon &&\\&\sum _{i=1,i\neq j}^{n}x_{ij}=1&&j=1,\ldots ,n;\\&\sum _{j=1,j\neq i}^{n}x_{ij}=1&&i=1,\ldots ,n;\\&\sum _{i\in Q}{\sum _{j\neq i,j\in Q}{x_{ij}}}\leq |Q|-1&&\forall Q\subsetneq \{1,\ldots ,n\},|Q|\geq 2\\\end{aligned}}}$

The last constraint of the DFJ formulation—called a subtour elimination constraint—ensures no proper subset Q can form a sub-tour, so the solution returned is a single tour and not the union of smaller tours. Because this leads to an exponential number of possible constraints, in practice it is solved with row generation.[22]

The most direct solution would be to try all permutations (ordered combinations) and see which one is cheapest (using brute-force search). The running time for this approach lies within a polynomial factor of ${\displaystyle O(n!)}$, the factorial of the number of cities, so this solution becomes impractical even for only 20 cities.

One of the earliest applications of dynamic programming is the Held–Karp algorithm, which solves the problem in time ${\displaystyle O(n^{2}2^{n})}$.[23] This bound has also been reached by Exclusion-Inclusion in an attempt preceding the dynamic programming approach.

Solution to a symmetric TSP with 7 cities using brute force search. Note: Number of permutations: (7−1)!/2 = 360

Improving these time bounds seems to be difficult. For example, it has not been determined whether a classical exact algorithm for TSP that runs in time ${\displaystyle O(1.9999^{n})}$ exists.[24] The currently best quantum exact algorithm for TSP due to Ambainis et al. runs in time ${\displaystyle O(1.728^{n})}$.[25]

Other approaches include:

• Various branch-and-bound algorithms, which can be used to process TSPs containing 40–60 cities.

Solution of a TSP with 7 cities using a simple Branch and bound algorithm. Note: The number of permutations is much less than Brute force search

• Progressive improvement algorithms which use techniques reminiscent of linear programming. Works well for up to 200 cities.
• Implementations of branch-and-bound and problem-specific cut generation (branch-and-cut[26]);[27] this is the method of choice for solving large instances. This approach holds the current record, solving an instance with 85,900 cities, see Applegate et al. (2006).

An exact solution for 15,112 German towns from TSPLIB was found in 2001 using the cutting-plane method proposed by George Dantzig, Ray Fulkerson, and Selmer M. Johnson in 1954, based on linear programming. The computations were performed on a network of 110 processors located at Rice University and Princeton University. The total computation time was equivalent to 22.6 years on a single 500 MHz Alpha processor. In May 2004, the travelling salesman problem of visiting all 24,978 towns in Sweden was solved: a tour of length approximately 72,500 kilometres was found and it was proven that no shorter tour exists.[28] In March 2005, the travelling salesman problem of visiting all 33,810 points in a circuit board was solved using Concorde TSP Solver: a tour of length 66,048,945 units was found and it was proven that no shorter tour exists. The computation took approximately 15.7 CPU-years (Cook et al. 2006). In April 2006 an instance with 85,900 points was solved using Concorde TSP Solver, taking over 136 CPU-years, see Applegate et al. (2006).

Various heuristics and approximation algorithms, which quickly yield good solutions, have been devised. These include the Multi-fragment algorithm. Modern methods can find solutions for extremely large problems (millions of cities) within a reasonable time which are with a high probability just 2–3% away from the optimal solution.[13]

Several categories of heuristics are recognized.

Nearest Neighbour algorithm for a TSP with 7 cities. The solution changes as the starting point is changed

The nearest neighbour (NN) algorithm (a greedy algorithm) lets the salesman choose the nearest unvisited city as his next move. This algorithm quickly yields an effectively short route. For N cities randomly distributed on a plane, the algorithm on average yields a path 25% longer than the shortest possible path.[29] However, there exist many specially arranged city distributions which make the NN algorithm give the worst route.[30] This is true for both asymmetric and symmetric TSPs.[31] Rosenkrantz et al.[32] showed that the NN algorithm has the approximation factor ${\displaystyle \Theta (\log |V|)}$ for instances satisfying the triangle inequality. A variation of NN algorithm, called nearest fragment (NF) operator, which connects a group (fragment) of nearest unvisited cities, can find shorter routes with successive iterations.[33] The NF operator can also be applied on an initial solution obtained by NN algorithm for further improvement in an elitist model, where only better solutions are accepted.

The bitonic tour of a set of points is the minimum-perimeter monotone polygon that has the points as its vertices; it can be computed efficiently by dynamic programming.

Another constructive heuristic, Match Twice and Stitch (MTS), performs two sequential matchings, where the second matching is executed after deleting all the edges of the first matching, to yield a set of cycles. The cycles are then stitched to produce the final tour.[34]

Creating a matching

Using a shortcut heuristic on the graph created by the matching above

The algorithm of Christofides and Serdyukov follows a similar outline but combines the minimum spanning tree with a solution of another problem, minimum-weight perfect matching. This gives a TSP tour which is at most 1.5 times the optimal. It was one of the first approximation algorithms, and was in part responsible for drawing attention to approximation algorithms as a practical approach to intractable problems. As a matter of fact, the term "algorithm" was not commonly extended to approximation algorithms until later; the Christofides algorithm was initially referred to as the Christofides heuristic.[9]

This algorithm looks at things differently by using a result from graph theory which helps improve on the lower bound of the TSP which originated from doubling the cost of the minimum spanning tree. Given an Eulerian graph we can find an Eulerian tour in ${\displaystyle O(n)}$ time.[5] So if we had an Eulerian graph with cities from a TSP as vertices then we can easily see that we could use such a method for finding an Eulerian tour to find a TSP solution. By triangular inequality we know that the TSP tour can be no longer than the Eulerian tour and as such we have a lower bound for the TSP. Such a method is described below.

1. Find a minimum spanning tree for the problem
2. Create duplicates for every edge to create an Eulerian graph
3. Find an Eulerian tour for this graph
4. Convert to TSP: if a city is visited twice, create a shortcut from the city before this in the tour to the one after this.

To improve the lower bound, a better way of creating an Eulerian graph is needed. By triangular inequality, the best Eulerian graph must have the same cost as the best travelling salesman tour, hence finding optimal Eulerian graphs is at least as hard as TSP. One way of doing this is by minimum weight matching using algorithms of ${\displaystyle O(n^{3})}$.[5]

Making a graph into an Eulerian graph starts with the minimum spanning tree. Then all the vertices of odd order must be made even. So a matching for the odd degree vertices must be added which increases the order of every odd degree vertex by one.[5] This leaves us with a graph where every vertex is of even order which is thus Eulerian. Adapting the above method gives the algorithm of Christofides and Serdyukov.

1. Find a minimum spanning tree for the problem
2. Create a matching for the problem with the set of cities of odd order.
3. Find an Eulerian tour for this graph
4. Convert to TSP using shortcuts.

An example of a 2-opt iteration

The pairwise exchange or 2-opt technique involves iteratively removing two edges and replacing these with two different edges that reconnect the fragments created by edge removal into a new and shorter tour. Similarly, the 3-opt technique removes 3 edges and reconnects them to form a shorter tour. These are special cases of the k-opt method. The label Lin–Kernighan is an often heard misnomer for 2-opt. Lin–Kernighan is actually the more general k-opt method.

For Euclidean instances, 2-opt heuristics give on average solutions that are about 5% better than Christofides' algorithm. If we start with an initial solution made with a greedy algorithm, the average number of moves greatly decreases again and is ${\displaystyle O(n)}$. For random starts however, the average number of moves is ${\displaystyle O(n\log(n))}$. However whilst in order this is a small increase in size, the initial number of moves for small problems is 10 times as big for a random start compared to one made from a greedy heuristic. This is because such 2-opt heuristics exploit 'bad' parts of a solution such as crossings. These types of heuristics are often used within Vehicle routing problem heuristics to reoptimize route solutions.[29]

The Lin–Kernighan heuristic is a special case of the V-opt or variable-opt technique. It involves the following steps:

1. Given a tour, delete k mutually disjoint edges.
2. Reassemble the remaining fragments into a tour, leaving no disjoint subtours (that is, don't connect a fragment's endpoints together). This in effect simplifies the TSP under consideration into a much simpler problem.
3. Each fragment endpoint can be connected to 2k − 2 other possibilities: of 2k total fragment endpoints available, the two endpoints of the fragment under consideration are disallowed. Such a constrained 2k-city TSP can then be solved with brute force methods to find the least-cost recombination of the original fragments.

The most popular of the k-opt methods are 3-opt, as introduced by Shen Lin of Bell Labs in 1965. A special case of 3-opt is where the edges are not disjoint (two of the edges are adjacent to one another). In practice, it is often possible to achieve substantial improvement over 2-opt without the combinatorial cost of the general 3-opt by restricting the 3-changes to this special subset where two of the removed edges are adjacent. This so-called two-and-a-half-opt typically falls roughly midway between 2-opt and 3-opt, both in terms of the quality of tours achieved and the time required to achieve those tours.

The variable-opt method is related to, and a generalization of the k-opt method. Whereas the k-opt methods remove a fixed number (k) of edges from the original tour, the variable-opt methods do not fix the size of the edge set to remove. Instead, they grow the set as the search process continues. The best-known method in this family is the Lin–Kernighan method (mentioned above as a misnomer for 2-opt). Shen Lin and Brian Kernighan first published their method in 1972, and it was the most reliable heuristic for solving travelling salesman problems for nearly two decades. More advanced variable-opt methods were developed at Bell Labs in the late 1980s by David Johnson and his research team. These methods (sometimes called Lin–Kernighan–Johnson) build on the Lin–Kernighan method, adding ideas from tabu search and evolutionary computing. The basic Lin–Kernighan technique gives results that are guaranteed to be at least 3-opt. The Lin–Kernighan–Johnson methods compute a Lin–Kernighan tour, and then perturb the tour by what has been described as a mutation that removes at least four edges and reconnects the tour in a different way, then V-opting the new tour. The mutation is often enough to move the tour from the local minimum identified by Lin–Kernighan. V-opt methods are widely considered the most powerful heuristics for the problem, and are able to address special cases, such as the Hamilton Cycle Problem and other non-metric TSPs that other heuristics fail on. For many years Lin–Kernighan–Johnson had identified optimal solutions for all TSPs where an optimal solution was known and had identified the best-known solutions for all other TSPs on which the method had been tried.

Optimized Markov chain algorithms which use local searching heuristic sub-algorithms can find a route extremely close to the optimal route for 700 to 800 cities.

TSP is a touchstone for many general heuristics devised for combinatorial optimization such as genetic algorithms, simulated annealing, tabu search, ant colony optimization, river formation dynamics (see swarm intelligence) and the cross entropy method.

Start with a sub-tour such as the convex hull, insert other vertexes. [35]

Artificial intelligence researcher Marco Dorigo described in 1993 a method of heuristically generating "good solutions" to the TSP using a simulation of an ant colony called ACS (ant colony system).[36] It models behaviour observed in real ants to find short paths between food sources and their nest, an emergent behaviour resulting from each ant's preference to follow trail pheromones deposited by other ants.

ACS sends out a large number of virtual ant agents to explore many possible routes on the map. Each ant probabilistically chooses the next city to visit based on a heuristic combining the distance to the city and the amount of virtual pheromone deposited on the edge to the city. The ants explore, depositing pheromone on each edge that they cross, until they have all completed a tour. At this point the ant which completed the shortest tour deposits virtual pheromone along its complete tour route (global trail updating). The amount of pheromone deposited is inversely proportional to the tour length: the shorter the tour, the more it deposits.

1) An ant chooses a path among all possible paths and lays a pheromone trail on it. 2) All the ants are travelling on different paths, laying a trail of pheromones proportional to the quality of the solution. 3) Each edge of the best path is more reinforced than others. 4) Evaporation ensures that the bad solutions disappear. The map is a work of Yves Aubry .

Ant colony optimization algorithm for a TSP with 7 cities: Red and thick lines in the pheromone map indicate presence of more pheromone

In the metric TSP, also known as delta-TSP or Δ-TSP, the intercity distances satisfy the triangle inequality.

A very natural restriction of the TSP is to require that the distances between cities form a metric to satisfy the triangle inequality; that is the direct connection from A to B is never farther than the route via intermediate C:

${\displaystyle d_{AB}\leq d_{AC}+d_{CB}}$.

The edge spans then build a metric on the set of vertices. When the cities are viewed as points in the plane, many natural distance functions are metrics, and so many natural instances of TSP satisfy this constraint.

The following are some examples of metric TSPs for various metrics.

• In the Euclidean TSP (see below) the distance between two cities is the Euclidean distance between the corresponding points.
• In the rectilinear TSP the distance between two cities is the sum of the absolute values of the differences of their x- and y-coordinates. This metric is often called the Manhattan distance or city-block metric.
• In the maximum metric, the distance between two points is the maximum of the absolute values of differences of their x- and y-coordinates.

The last two metrics appear, for example, in routing a machine that drills a given set of holes in a printed circuit board. The Manhattan metric corresponds to a machine that adjusts first one co-ordinate, and then the other, so the time to move to a new point is the sum of both movements. The maximum metric corresponds to a machine that adjusts both co-ordinates simultaneously, so the time to move to a new point is the slower of the two movements.

In its definition, the TSP does not allow cities to be visited twice, but many applications do not need this constraint. In such cases, a symmetric, non-metric instance can be reduced to a metric one. This replaces the original graph with a complete graph in which the inter-city distance ${\displaystyle d_{AB}}$ is replaced by the shortest path length between A and B in the original graph.

For points in the Euclidean plane, the optimal solution to the travelling salesman problem forms a simple polygon through all of the points, a polygonalization of the points.[37] Any non-optimal solution with crossings can be made into a shorter solution without crossings by local optimizations. The Euclidean distance obeys the triangle inequality, so the Euclidean TSP forms a special case of metric TSP. However, even when the input points have integer coordinates, their distances generally take the form of square roots, and the length of a tour is a sum of radicals, making it difficult to perform the symbolic computation needed to perform exact comparisons of the lengths of different tours.

Like the general TSP, the exact Euclidean TSP is NP-hard, but the issue with sums of radicals is an obstacle to proving that its decision version is in NP, and therefore NP-complete. A discretized version of the problem with distances rounded to integers is NP-complete.[38] With rational coordinates and the actual Euclidean metric, Euclidean TSP is known to be in the Counting Hierarchy,[39] a subclass of PSPACE. With arbitrary real coordinates, Euclidean TSP cannot be in such classes, since there are uncountably many possible inputs. Despite these complications, Euclidean TSP is much easier than the general metric case for approximation.[40] For example, the minimum spanning tree of the graph associated with an instance of the Euclidean TSP is a Euclidean minimum spanning tree, and so can be computed in expected O (n log n) time for n points (considerably less than the number of edges). This enables the simple 2-approximation algorithm for TSP with triangle inequality above to operate more quickly.

In general, for any c > 0, where d is the number of dimensions in the Euclidean space, there is a polynomial-time algorithm that finds a tour of length at most (1 + 1/c) times the optimal for geometric instances of TSP in

${\displaystyle O\left(n(\log n)^{(O(c{\sqrt {d}}))^{d-1}}\right),}$

time; this is called a polynomial-time approximation scheme (PTAS).[41] Sanjeev Arora and Joseph S. B. Mitchell were awarded the Gödel Prize in 2010 for their concurrent discovery of a PTAS for the Euclidean TSP.

In practice, simpler heuristics with weaker guarantees continue to be used.

In most cases, the distance between two nodes in the TSP network is the same in both directions. The case where the distance from A to B is not equal to the distance from B to A is called asymmetric TSP. A practical application of an asymmetric TSP is route optimization using street-level routing (which is made asymmetric by one-way streets, slip-roads, motorways, etc.).

Solving an asymmetric TSP graph can be somewhat complex. The following is a 3×3 matrix containing all possible path weights between the nodes A, B and C. One option is to turn an asymmetric matrix of size N into a symmetric matrix of size 2N.[42]

Asymmetric path weights

	A	B	C
A		1	2
B	6		3
C	5	4

To double the size, each of the nodes in the graph is duplicated, creating a second ghost node, linked to the original node with a "ghost" edge of very low (possibly negative) weight, here denoted −w. (Alternatively, the ghost edges have weight 0, and weight w is added to all other edges.) The original 3×3 matrix shown above is visible in the bottom left and the transpose of the original in the top-right. Both copies of the matrix have had their diagonals replaced by the low-cost hop paths, represented by −w. In the new graph, no edge directly links original nodes and no edge directly links ghost nodes.

Symmetric path weights

	A	B	C	A′	B′	C′
A				−w	6	5
B				1	−w	4
C				2	3	−w
A′	−w	1	2
B′	6	−w	3
C′	5	4	−w

The weight −w of the "ghost" edges linking the ghost nodes to the corresponding original nodes must be low enough to ensure that all ghost edges must belong to any optimal symmetric TSP solution on the new graph (w=0 is not always low enough). As a consequence, in the optimal symmetric tour, each original node appears next to its ghost node (e.g. a possible path is ${\displaystyle \mathrm {A\to A'\to C\to C'\to B\to B'\to A} }$) and by merging the original and ghost nodes again we get an (optimal) solution of the original asymmetric problem (in our example, ${\displaystyle \mathrm {A\to C\to B\to A} }$).

There is an analogous problem in geometric measure theory which asks the following: under what conditions may a subset E of Euclidean space be contained in a rectifiable curve (that is, when is there a curve with finite length that visits every point in E)? This problem is known as the analyst's travelling salesman problem.

Suppose ${\displaystyle X_{1},\ldots ,X_{n}}$ are ${\displaystyle n}$ independent random variables with uniform distribution in the square ${\displaystyle [0,1]^{2}}$, and let ${\displaystyle L_{n}^{\ast }}$ be the shortest path length (i.e. TSP solution) for this set of points, according to the usual Euclidean distance. It is known[8] that, almost surely,

${\displaystyle {\frac {L_{n}^{*}}{\sqrt {n}}}\rightarrow \beta \qquad {\text{when }}n\to \infty ,}$

where ${\displaystyle \beta }$ is a positive constant that is not known explicitly. Since ${\displaystyle L_{n}^{*}\leq 2{\sqrt {n}}+2}$ (see below), it follows from bounded convergence theorem that ${\displaystyle \beta =\lim _{n\to \infty }\mathbb {E} [L_{n}^{*}]/{\sqrt {n}}}$, hence lower and upper bounds on ${\displaystyle \beta }$ follow from bounds on ${\displaystyle \mathbb {E} [L_{n}^{*}]}$.

The almost sure limit ${\displaystyle {\frac {L_{n}^{*}}{\sqrt {n}}}\rightarrow \beta }$ as ${\displaystyle n\to \infty }$ may not exist if the independent locations ${\displaystyle X_{1},\ldots ,X_{n}}$ are replaced with observations from a stationary ergodic process with uniform marginals.[43]

• One has ${\displaystyle L^{*}\leq 2{\sqrt {n}}+2}$, and therefore ${\displaystyle \beta \leq 2}$, by using a naive path which visits monotonically the points inside each of ${\displaystyle {\sqrt {n}}}$ slices of width ${\displaystyle 1/{\sqrt {n}}}$ in the square.
• Few[44] proved ${\displaystyle L_{n}^{*}\leq {\sqrt {2n}}+1.75}$, hence ${\displaystyle \beta \leq {\sqrt {2}}}$, later improved by Karloff (1987): ${\displaystyle \beta \leq 0.984{\sqrt {2}}}$.
• Fietcher[45] showed an upper bound of ${\displaystyle \beta \leq 0.73\dots }$.

• By observing that ${\displaystyle \mathbb {E} [L_{n}^{*}]}$ is greater than ${\displaystyle n}$ times the distance between ${\displaystyle X_{0}}$ and the closest point ${\displaystyle X_{i}\neq X_{0}}$, one gets (after a short computation)

${\displaystyle \mathbb {E} [L_{n}^{*}]\geq {\tfrac {1}{2}}{\sqrt {n}}.}$

• A better lower bound is obtained[8] by observing that ${\displaystyle \mathbb {E} [L_{n}^{*}]}$ is greater than ${\displaystyle {\tfrac {1}{2}}n}$ times the sum of the distances between ${\displaystyle X_{0}}$ and the closest and second closest points ${\displaystyle X_{i},X_{j}\neq X_{0}}$, which gives

${\displaystyle \mathbb {E} [L_{n}^{*}]\geq \left({\tfrac {1}{4}}+{\tfrac {3}{8}}\right){\sqrt {n}}={\tfrac {5}{8}}{\sqrt {n}},}$

• The currently[46] best lower bound is

${\displaystyle \mathbb {E} [L_{n}^{*}]\geq ({\tfrac {5}{8}}+{\tfrac {19}{5184}}){\sqrt {n}},}$

• Held and Karp[47] gave a polynomial-time algorithm that provides numerical lower bounds for ${\displaystyle L_{n}^{*}}$, and thus for ${\displaystyle \beta (\simeq L_{n}^{*}/{\sqrt {n}})}$ which seem to be good up to more or less 1%.[48] In particular, David S. Johnson[49] obtained a lower bound by computer experiment:

${\displaystyle L_{n}^{*}\gtrsim 0.7080{\sqrt {n}}+0.522,}$

where 0.522 comes from the points near square boundary which have fewer neighbours, and Christine L. Valenzuela and Antonia J. Jones[50] obtained the following other numerical lower bound:

${\displaystyle L_{n}^{*}\gtrsim 0.7078{\sqrt {n}}+0.551}$.

In the general case, finding a shortest travelling salesman tour is NPO-complete.[51] If the distance measure is a metric (and thus symmetric), the problem becomes APX-complete[52] and the algorithm of Christofides and Serdyukov approximates it within 1.5.[53][54][9]

If the distances are restricted to 1 and 2 (but still are a metric) the approximation ratio becomes 8/7.[55] In the asymmetric case with triangle inequality, up until recently only logarithmic performance guarantees were known.[56] In 2018, a constant factor approximation was developed by Svensson, Tarnawski and Végh.[57] The best current algorithm, by Traub and Vygen, achieves performance ratio of ${\displaystyle 22+\varepsilon }$.[58] The best known inapproximability bound is 75/74.[59]

The corresponding maximization problem of finding the longest travelling salesman tour is approximable within 63/38.[60] If the distance function is symmetric, the longest tour can be approximated within 4/3 by a deterministic algorithm[61] and within ${\displaystyle {\tfrac {1}{25}}(33+\varepsilon )}$ by a randomized algorithm.[62]