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main.py
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main.py
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import re
import numpy
from functools import reduce
import sys
import getopt
alfa = 2
beta = 5
sigm = 3
ro = 0.8
th = 80
iterations = 1000
ants = 22
def getData(fileName):
f = open(fileName, "r")
content = f.read()
optimalValue = re.search("Optimal value: (\d+)", content, re.MULTILINE)
if(optimalValue != None):
optimalValue = optimalValue.group(1)
else:
optimalValue = re.search("Best value: (\d+)", content, re.MULTILINE)
if(optimalValue != None):
optimalValue = optimalValue.group(1)
capacity = re.search("^CAPACITY : (\d+)$", content, re.MULTILINE).group(1)
graph = re.findall(r"^(\d+) (\d+) (\d+)$", content, re.MULTILINE)
demand = re.findall(r"^(\d+) (\d+)$", content, re.MULTILINE)
graph = {int(a): (int(b), int(c)) for a, b, c in graph}
demand = {int(a): int(b) for a, b in demand}
capacity = int(capacity)
optimalValue = int(optimalValue)
return capacity, graph, demand, optimalValue
def generateGraph():
# TODO: 输入格式太复杂了
capacityLimit, graph, demand, optimalValue = getData(fileName)
vertices = list(graph.keys())
vertices.remove(1)
edges = {(min(a, b), max(a, b)): numpy.sqrt((graph[a][0] - graph[b][0]) ** 2 + (graph[a][1] - graph[b][1]) ** 2) for
a in graph.keys() for b in graph.keys()}
feromones = {(min(a, b), max(a, b)): 1 for a in graph.keys() for b in graph.keys() if a != b}
return vertices, edges, capacityLimit, demand, feromones, optimalValue
def solutionOfOneAnt(vertices, edges, capacityLimit, demand, feromones):
solution = list()
while len(vertices) != 0:
path = list()
city = numpy.random.choice(vertices)
capacity = capacityLimit - demand[city]
path.append(city)
vertices.remove(city)
while len(vertices) != 0:
# 轮盘赌算法
probabilities = list(map(lambda x: ((feromones[(min(x, city), max(x, city))]) ** alfa) * (
(1 / edges[(min(x, city), max(x, city))]) ** beta), vertices))
# 概率归一化
probabilities = probabilities / numpy.sum(probabilities)
# 根据概率选出城市
_city = numpy.random.choice(vertices, p=probabilities)
if capacity >= demand[_city]:
# 如果装载量 >= 需求，直接卸货，寻找下一个城市
capacity = capacity - demand[_city]
path.append(_city)
vertices.remove(_city)
else:
# 如果装载量 < 需求，进行拆分，保证去下一个城市+回城的距离最小
min_distance = float("inf")
for i in vertices:
# 下一个城市+回城的距离
dis = edges[(min(i, city), max(i, city))] + edges[(min(i, 1), max(i, 1))]
if dis < min_distance:
_city = i
min_distance = dis
path.append(_city)
if capacity >= demand[_city]:
# 如果装载量 >= 需求，直接卸货，寻找下一个城市
capacity = capacity - demand[_city]
vertices.remove(_city)
else:
# 如果装载量 < 需求，直接卸货，跳出当前路程
demand[_city] = demand[_city] - capacity
break
solution.append(path)
return solution
def rateSolution(solution, edges):
s = 0
for i in solution:
a = 1
for j in i:
b = j
s = s + edges[(min(a, b), max(a, b))]
a = b
b = 1
s = s + edges[(min(a, b), max(a, b))]
return s
def updateFeromone(feromones, solutions, bestSolution):
Lavg = reduce(lambda x, y: x + y, (i[1] for i in solutions)) / len(solutions)
feromones = {k: (ro + th / Lavg) * v for (k, v) in feromones.items()}
solutions.sort(key=lambda x: x[1])
if (bestSolution != None):
if (solutions[0][1] < bestSolution[1]):
bestSolution = solutions[0]
for path in bestSolution[0]:
for i in range(len(path) - 1):
feromones[(min(path[i], path[i + 1]), max(path[i], path[i + 1]))] = sigm / bestSolution[1] + feromones[
(min(path[i], path[i + 1]), max(path[i], path[i + 1]))]
else:
bestSolution = solutions[0]
for l in range(sigm):
paths = solutions[l][0]
L = solutions[l][1]
for path in paths:
for i in range(len(path) - 1):
feromones[(min(path[i], path[i + 1]), max(path[i], path[i + 1]))] = (sigm - (l + 1) / L ** (l + 1)) + \
feromones[(
min(path[i], path[i + 1]),
max(path[i], path[i + 1]))]
return bestSolution
def main():
bestSolution = None
vertices, edges, capacityLimit, demand, feromones, optimalValue = generateGraph()
for i in range(iterations):
solutions = list()
for _ in range(ants):
solution = solutionOfOneAnt(vertices.copy(), edges, capacityLimit, demand.copy(), feromones)
solutions.append((solution, rateSolution(solution, edges)))
bestSolution = updateFeromone(feromones, solutions, bestSolution)
print(str(i) + "/" + str(iterations) + ": " + str(bestSolution[1]), end='\r', flush=True)
return bestSolution
if __name__ == "__main__":
argv = sys.argv[1:]
try:
opts, args = getopt.getopt(argv, "f:a:b:s:r:t:i:n:", ["fileName=",
"alpha=", "beta=", "sigma=", "rho=", "theta=",
"iterations=", "numberOfAnts="])
except getopt.GetoptError:
print("""use: python main.py
-f <fileName>
-a <alpha>
-b <beta>
-s <sigma>
-r <rho>
-t <theta>
-i <iterations>
-n <numberOfAnts>
Default values:
fileName: E-n22-k4.txt
alpha: 80
beta: 5
sigma: 3
rho: 0.8
theta: 80
iterations: 1000
number of ants: 22""")
sys.exit(2)
for opt, arg in opts:
if (opt in ("-a", "--alpha")):
alfa = float(arg)
elif (opt in ("-b", "--beta")):
beta = float(arg)
elif (opt in ("-s", "--sigma")):
sigm = float(arg)
elif (opt in ("-r", "--rho")):
ro = float(arg)
elif (opt in ("-t", "--theta")):
th = float(arg)
elif (opt in ("-f", "--fileName", "--file")):
fileName = str(arg)
elif (opt in ("-i", "--iterations")):
iterations = int(arg)
elif (opt in ("-n", "--numberOfAnts")):
ants = int(arg)
print("file name:\t" + str(fileName) +
"\nalpha:\t" + str(alfa) +
"\nbeta:\t" + str(beta) +
"\nsigma:\t" + str(sigm) +
"\nrho:\t" + str(ro) +
"\ntheta:\t" + str(th) +
"\niterations:\t" + str(iterations) +
"\nnumber of ants:\t" + str(ants))
solution = main()
print("Solution: " + str(solution))