This article needs attention from an expert on the subject. The specific problem is: "This page is very poor in terms of contents. No pseudo-code, scarce references. This algorithm deserves a lot better description and treatment.".(September 2016)
3-opt analysis involves deleting 3 connections (or edges) in a network (or tour), to create 3 sub-tours. Then the 7 different ways of reconnecting the network are analysed to find the optimum one. This process is then repeated for a different set of 3 connections, until all possible combinations have been tried in a network. A single execution of 3-opt has a time complexity of . Iterated 3-opt has a higher time complexity.
This is the mechanism by which the 3-opt swap manipulates a given route:
def reverse_segment_if_better(tour, i, j, k): "If reversing tour[i:j] would make the tour shorter, then do it." # Given tour [...A-B...C-D...E-F...] A, B, C, D, E, F = tour[i-1], tour[i], tour[j-1], tour[j], tour[k-1], tour[k % len(tour)] d0 = distance(A,B) + distance(C,D) + distance(E,F) d1 = distance(A,C) + distance(B,D) + distance(E,F) d2 = distance(A,B) + distance(C,E) + distance(D,F) d3 = distance(A,D) + distance(E,B) + distance(C,F) d4 = distance(F,B) + distance(C,D) + distance(E,A) if d0 > d1: tour[i:j] = reversed(tour[i:j]) return -d0 + d1 elif d0 > d2: tour[j:k] = reversed(tour[j:k]) return -d0 + d2 elif d0 > d4: tour[i:k] = reversed(tour[i:k]) return -d0 + d4 elif d0 > d3: tmp = tour[j:k], tour[i:j] tour[i:k] = tmp return -d0 + d3 return 0
The principle is pretty simple. You compute, the original distance and you compute the cost of each modification. If you find a better cost, apply the modification and return (relative cost). This is the complete 3-opt swap making use of the above mechanism:
def three_opt(tour): "Iterative improvement based on 3 exchange." delta = 0 for (a,b,c) in all_segments(len(tour)): delta += reverse_segment_if_better(tour, a, b, c) if delta < 0: return three_opt(tour) return tour def all_segments(N): "Generate all segments combinations" return [(i, j, k) for i in range(N) for j in range(i+2, N) for k in range(j+2, N+(i>0))]
For the given tour, you generate all segments combinations and for each combinations, you try to improve the tour by reversing segments. While you find a better result, you restart the process.
- Munim, Ziaul Haque; Haralambides, Hercules (2018). "Competition and cooperation for intermodal container transhipment: A network optimization approach". Research in Transportation Business & Management. 26: 87–99. doi:10.1016/j.rtbm.2018.03.004.
- Blazinskas, Andrius; Misevicius, Alfonsas (2011). "Combining 2-OPT, 3-OPT and 4-OPT with K-SWAP-KICK perturbations for the traveling salesman problem" (PDF).
- F. BOCK (1965). An algorithm for solving traveling-salesman and related network optimization problems. unpublished manuscript associated with talk presented at the 14th ORSA National Meeting.
- S. LIN (1965). Computer solutions of the traveling salesman problem. Bell Syst. Tech. J. 44, 2245-2269. Available as PDF
- S. LIN AND B. W. KERNIGHAN (1973). An Effective Heuristic Algorithm for the Traveling-Salesman Problem. Operations Res. 21, 498-516. Available as PDF
- Local Search Heuristics. (n.d.) Retrieved June 16, 2008, from http://www.tmsk.uitm.edu.my/~naimah/csc751/slides/LS.pdf