找到一组二维点之间的最有效路径（以最短距离表示）

Question

I have a set of 2D points stored in a dictionary and i need to find the most efficient path to sampling all points (red traingles) in term of the shortest distance from a start-end point (yellow circle).

dict_points = OrderedDict([(0, (0.5129102892466411, 1.2791525891782567)), (1, (1.8571436538551014, 1.3979619805011203)), (2, (2.796472292985357, 1.3021773853592946)), (3, (2.2054745567697513, 0.5231652951626251)), (4, (1.1209493135130593, 0.8220950186969501)), (5, (0.16416153316980153, 0.7241249969879273))])

where the key is the ID of the point

My strategy is very simple. I use all points sequence possible (720 for 6 points) and i compute the euclidean distance point-by-point starting and ending from the start-end point (yellow point). The sequence with the shortest total distance is the most efficient.

The problem of this approach is that get very slow for a large number of points

import math
import itertools

base = (2.596, 2.196)

def segments(poly):
    """A sequence of (x,y) numeric coordinates pairs """
    return zip(poly, poly[1:] + [poly[0]])


 def best_path(dict_points, base=None):
    sequence_min_distance = None
    l = dict_points.keys()
    gen = itertools.permutations(l)
    min_dist = float('+inf')
    for index, i in enumerate(gen):
        seq = gen.next()
        seq_list = [dict_points[s] for s in seq]
        if base:
            seq_list.insert(0, base)
        points_paired = segments(seq_list)
        tot_dist = 0
        for points in points_paired:
            dist = math.hypot(points[1][0] - points[0][0], points[1][1] - points[0][1])
            tot_dist += dist
        if tot_dist <= min_dist:
            sequence_min_distance = seq
            min_dist = min(min_dist, tot_dist)
    return sequence_min_distance

best_seq = best_path(dict_points)
(5, 4, 3, 2, 1, 0)

Answer 1

您还可以查看项目tsp-solver https://github.com/dmishin/tsp-solver