优化旅行商算法(时间旅行者算法)
问题描述
我试着优化我做的一个简单的python算法，它近似地解决了旅行商问题：
import math
import random
import matplotlib.pyplot as plt
import datetime
#Distance between two point
def distance(point1, point2):
return math.sqrt((point2[0]-point1[0])**2+(point2[1]-point1[1])**2)
#TSP TimeTraveler Algorithm
def TSP_TimeTraveler(Set_Points):
print("Solving TSP")
#For calculating execution time
time_start = datetime.datetime.now()
#Copy the set points
points = Set_Points.copy()
route = []
#Take 3 points at random
route.append(points.pop(random.randint(0,len(points)-1)))
route.insert(0,points.pop(random.randint(0,len(points)-1)))
route.insert(1,points.pop(random.randint(0,len(points)-1)))
#Calulating the initial route length
Length = distance(route[0],route[1]) + distance(route[1],route[-1]) + distance(route[-1],route[0])
#Time Traveler Algorithm
while len(points)>0 :
print("Points left : ", len(points),' ', end="")
#Take a random point from the Set
point = points.pop(random.randint(0,len(points)-1))
###############################################################################################################
#### Finding the closest route segment by calculation all lengths posibilities and finding the minimum one ####
###############################################################################################################
Set_Lengths = []
for i in range(1,len(route)):
#Set of Lengths when the point is on each route segment except the last one
L = Length - distance(route[i-1],route[i]) + distance(route[i-1],point) + distance(point, route[i])
Set_Lengths.append((i,L))
#Adding the last length when the point is on the last segement
L = Length - distance(route[-1],route[0]) + distance(route[-1],point) + distance(point, route[0])
Set_Lengths.append((0,L))
###############################################################################################################
###############################################################################################################
#Sorting the set of lengths
Set_Lengths.sort(key=lambda k: k[1])
#Inserting the point on the minimum length segment
route.insert(Set_Lengths[0][0], point)
#Updating the new route length
Length = Set_Lengths[0][1]
#Connecting the start point with the finish point
route.append(route[0])
#For calculating execution time
time_end = datetime.datetime.now()
delta = (time_end-time_start).total_seconds()
print("Points left : ", len(points),' Done ',)
print("Execution time : ", delta, "secs")
return route
#######################
#Testing the Algorithm#
#######################
#Size of the set
size = 2520
#Generating a set of random 2D points
points = []
for i in range(size):
points.append([random.uniform(0, 100),random.uniform(0, 100)])
#Solve TSP
route = TSP_TimeTraveler(points)
#Plot the solution
plt.scatter(*zip(*points),s=5)
plt.plot(*zip(*route))
plt.axis('scaled')
plt.show()
该算法分3个简单步骤运行：
1/第一步，我从点集中随机抽取3个点，并将它们连接起来作为初始路线。
2/然后，在下一步，我从剩下的一组点中随机取一个点。并尝试查找我拥有的最接近的路线段并将其连接到它。
3/i不断重复步骤2/，直到剩余的点集为空。
以下是该算法如何求解一组120个点的gif：TimeTravelerAlgorithm.gif
我将其命名为Time Traveler&是因为它的运行方式类似于贪婪的推销员算法。但是，贪婪的推销员不是去现在最近的新城市旅行，而是回到过去，回到他已经访问过的最近的城市，去拜访那个新城市，然后继续他的正常路线。时间旅行者开始了一条3个城市的路线，旅行者在过去的每一步中都会添加一个新的城市，直到他到达一个现在，他走遍了所有的城市，然后回到了他的家乡城市。
该算法对于较小的点集快速地给出像样的解。以下是每组的执行时间，均采用2.6 GHz双核Intel Corei5处理器Macbook：
- 在0.03秒左右获得120分
- 约0.23秒内360分
- 10秒左右2520分
- 3分钟左右10000分
- 5小时左右100000分(Solution Map)
该算法远未得到优化，因为在某些情况下，它给出的交叉路径不是最优的。它都是用纯蟒蛇做的。也许使用Numpy或一些高级库，甚至使用GPU可以提高程序的速度。
我需要您的审查和有关如何优化它的帮助。我尝试在没有交叉路线的情况下近似求解可能非常大的一组点(从100万到1000亿点)。
PS：我的算法和代码是开放的。来自互联网的人们，可以在任何项目或您的任何研究中使用它。解决方案
我改进了算法，在每次插入时添加了双向链表和2-opt：
import math
import random
import datetime
import matplotlib.pyplot as plt
#Distance between two point
def distance(point1, point2):
return (point2[0]-point1[0])**2 + (point2[1]-point1[1])**2
#Intersection between two segments
def intersects(p1, q1, p2, q2):
def on_segment(p, q, r):
if r[0] <= max(p[0], q[0]) and r[0] >= min(p[0], q[0]) and r[1] <= max(p[1], q[1]) and r[1] >= min(p[1], q[1]):
return True
return False
def orientation(p, q, r):
val = ((q[1] - p[1]) * (r[0] - q[0])) - ((q[0] - p[0]) * (r[1] - q[1]))
if val == 0 : return 0
return 1 if val > 0 else -1
o1 = orientation(p1, q1, p2)
o2 = orientation(p1, q1, q2)
o3 = orientation(p2, q2, p1)
o4 = orientation(p2, q2, q1)
if o1 != o2 and o3 != o4:
return True
if o1 == 0 and on_segment(p1, q1, p2) : return True
if o2 == 0 and on_segment(p1, q1, q2) : return True
if o3 == 0 and on_segment(p2, q2, p1) : return True
if o4 == 0 and on_segment(p2, q2, q1) : return True
return False
#Distance Double Linked Node
class Node:
def __init__(self, dataval=None):
self.dataval = dataval
self.prevval = None
self.nextval = None
class TSP_TimeTraveler():
def __init__(self):
self.count = 0
self.position = None
self.length = 0
self.traveler = None
self.travelert_past = None
self.is_2opt = True
def get_position():
return self.position
def traveler_init(self):
self.traveler = self.position
self.travelert_past = self.position.prevval
return self.traveler
def traveler_next(self):
if self.traveler.nextval != self.travelert_past:
self.travelert_past = self.traveler
self.traveler = self.traveler.nextval
return self.traveler, False
else :
self.travelert_past = self.traveler
self.traveler = self.traveler.prevval
return self.traveler, True
#adding a city to the current route with Time Traveler Algorithm :
def add_city(self, point):
node = Node(point)
if self.count <=0 :
self.position = node
elif self.count == 1 :
node.nextval = self.position
node.prevval = node
self.position.nextval = node
self.position.prevval = self.position
self.length = 2*distance(self.position.dataval,node.dataval)
elif self.count == 2 :
node.nextval = self.position.nextval
node.prevval = self.position
self.position.nextval.prevval = node
self.position.nextval = node
self.length = 2*distance(self.position.dataval,node.dataval)
else :
#Creating the traveler
traveler = self.traveler_init()
c = traveler #current position
prev = False #inverse link
n, prev = self.traveler_next()
#Calculating the length of adding the city to the path
Min_prev = prev
Min_L = self.length-distance(c.dataval,n.dataval)+distance(c.dataval,node.dataval)+distance(node.dataval,n.dataval)
Min_Node = c
traveler = n
while traveler != self.position :
c = n #current position
n, prev = self.traveler_next()
#Calculating the length of adding the city to the path
L = self.length-distance(c.dataval,n.dataval)+distance(c.dataval,node.dataval)+distance(node.dataval,n.dataval)
#Searching the path to the of city with minimum length
if L < Min_L :
Min_prev = prev
Min_L = L
Min_Node = c
traveler = n
if Min_prev :
Min_Next_Node = Min_Node.prevval
else :
Min_Next_Node = Min_Node.nextval
node.nextval = Min_Next_Node
node.prevval = Min_Node
if Min_prev :
Min_Node.prevval = node
else :
Min_Node.nextval = node
if Min_Next_Node.nextval == Min_Node:
Min_Next_Node.nextval = node
else :
Min_Next_Node.prevval = node
self.length = Min_L
#2-OP
if self.is_2opt == True :
self._2opt(Min_Node, node, Min_Next_Node)
#Incrementing the number of city in the route
self.count = self.count + 1
#apply the 2opt to a-b-c
def _2opt(self, a, b, c):
traveler = self.traveler_init()
c1 = a
c2 = b
n1 = b
n2 = c
c = traveler #current position
t_prev = False
n, t_prev = self.traveler_next()
traveler = n
while traveler != self.position :
cross = False
if (c.dataval != c1.dataval and c.dataval != c2.dataval and n.dataval != c1.dataval and n.dataval != c2.dataval) and intersects(c.dataval, n.dataval, c1.dataval, c2.dataval):
self._2optswap(c,n,c1,c2)
cross = True
a = n
n = c1
c2 = a
if (c.dataval != n1.dataval and c.dataval != n2.dataval and n.dataval != n1.dataval and n.dataval != n2.dataval) and intersects(c.dataval, n.dataval, n1.dataval, n2.dataval):
self._2optswap(c,n,n1,n2)
cross = True
a = n
n = n1
n2 = a
if cross:
return
c = n #current position
n, t_prev = self.traveler_next()
traveler = n
#swap between the crossed segment a-b and c-d
def _2optswap(self, a, b, c, d):
if a.nextval == b :
a.nextval = c
else :
a.prevval = c
if b.prevval == a :
b.prevval = d
else :
b.nextval = d
if c.nextval == d :
c.nextval = a
else :
c.prevval = a
if d.prevval == c :
d.prevval = b
else :
d.nextval = b
self.length = self.length - distance(a.dataval,b.dataval) - distance(c.dataval,d.dataval) + distance(a.dataval,c.dataval) + distance(b.dataval,d.dataval)
#Get the list of the route
def getRoute(self):
result = []
traveler = self.traveler_init()
result.append(traveler.dataval)
traveler, prev = self.traveler_next()
while traveler != self.position :
result.append(traveler.dataval)
traveler, prev = self.traveler_next()
result.append(traveler.dataval)
return result
def Solve(self, Set_points, with_2opt = True):
print("Solving TSP")
#For calculating execution time
time_start = datetime.datetime.now()
#Copy the set points list
points = Set_points.copy()
#Transform the list into set
points = set(tuple(i) for i in points)
#Add
while len(points)>0 :
print("Points left : ", len(points),' ', end="")
point = points.pop()
self.add_city(point)
result = self.getRoute()
#For calculating execution time
time_end = datetime.datetime.now()
delta = (time_end-time_start).total_seconds()
L=0
for i in range(len(result)-1):
L = L + math.sqrt((result[i-1][0]-result[i][0])**2 + (result[i-1][1]-result[i][1])**2)
print("Points left : ", len(points),' Done ',)
print("Execution time : ", delta, "secs")
print("Average time per point : ", 1000*delta/len(Set_points), "msecs")
print("Length : ", L)
return result
#######################
#Testing the Algorithm#
#######################
#Size of the set
size = 1000
#Generating a set of random 2D points
points = []
for i in range(size):
points.append((random.uniform(0, 100),random.uniform(0, 100)))
#Solve TSP
TSP = TSP_TimeTraveler()
route = TSP.Solve(points, with_2opt = True)
plt.scatter(*zip(*route), s=5)
plt.plot(*zip(*route))
plt.axis('scaled')
plt.show()
现在，该解决方案无需交叉路线即可快速得出结果。
使用PyPy，它可以在30分钟内求解100,000个点，没有交叉路线。
现在我正在研究实现KD-tree来解决大型集合的问题。
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