An Efficient Quantum Algorithm 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 N P -hard with a brute-force complexity of O(N N) or O(N 2N) for...
| Autores: | , , , , |
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| Tipo de recurso: | artículo |
| Estado: | Versión publicada |
| Fecha de publicación: | 2025 |
| País: | España |
| Institución: | Consejo Superior de Investigaciones Científicas (CSIC) |
| Repositorio: | DIGITAL.CSIC. Repositorio Institucional del CSIC |
| OAI Identifier: | oai:digital.csic.es:10261/401701 |
| Acceso en línea: | http://hdl.handle.net/10261/401701 https://api.elsevier.com/content/abstract/scopus_id/105005323156 |
| Access Level: | acceso abierto |
| Palabra clave: | Traveling salesman problem Efficient quantum algorithm Hamiltonian cycle problem |
| Sumario: | 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 N P -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)κ/ϵ + 1/ϵ3), where the errors ϵ are O(1/poly(N)), and κ is the not-too-large condition number of the matrix encoding all Hamiltonian cycles. |
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