Simulated Annealing algorithm to solve Travelling Salesmen Problem in Python
License: MIT License
Using simulated annealing metaheuristic to solve the travelling salesman problem, and visualizing the results.
Starts by using a greedy algorithm (nearest neighbour) to build an initial solution.
A simple implementation which provides decent results.
An example of the resulting route on a TSP with 100 nodes.
The fitness (objective value) through iterations.
References
Kirkpatrick et al. 1983: "Optimization by Simulated Annealing"
batch_anneal是为了多次从初始化状态运行模拟退火，相当于寻找time次中的最优解吗
What you are doing below is calculating an Adjacency matrix with distances, for this case with just 45 nodes, it is good, what about the case if this is a graph with 500,000 nodes, any ideas?
def to_dist_matrix(self, coords):
"""
Returns nxn nested list from a list of length n
Used as distance matrix: mat[i][j] is the distance between node i and j
'coords' has the structure [[x1,y1],...[xn,yn]]
"""
n = len(coords)
mat = [[self.dist(coords[i], coords[j]) for i in range(n)] for j in range(n)]
return mat
Thanks for the code for 2D geometry. Is there a simple way to extend your code for 3D?
According to the paper you quoted, you shouldn't use the absolute value of the energy value in the p_accept function
return math.exp(-abs(candidate_fitness - self.cur_fitness) / self.T)
Rather use:
return math.exp(-(candidate_fitness - self.cur_fitness) / self.T)
Thanks , for this informative repo. While using the coords.txt file for the node coordinates, I found the Euclidean distance function refers the x- coordinates of a node with indices 0 and the y-coordinates by indices 1. I think they should be respectively 1 and 2 because the 0 indices are the node numbers. I separately tested with the modified indices and got the distance equal to that calculated manually. Please correct me if my understanding is wrong? Below is the snippet of the dist function with modified indices:
def dist(self, node_0, node_1):
"""
Euclidean distance between two nodes.
"""
coord_0, coord_1 = self.coords[node_0], self.coords[node_1]
return math.sqrt((coord_0[1] - coord_1[1]) ** 2 + (coord_0[2] - coord_1[2]) ** 2)
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