优化旅行推销员算法（时间旅行者算法）

Question

我尝试优化一个简单的 Python 算法，以近似解决旅行商问题：

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="\r")

        #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()

算法通过 3 个简单的步骤运行：

1/ 第一步我从点集合中随机取3个点连接起来作为初始路线。

2/ 然后每下一步，我从剩下的点集合中随机取一个点。并尝试找到我拥有的路线中最近的路段并将其连接到它。

3/ 我不断重复步骤 2/ 直到剩下的点集为空。

这是算法如何解决一组 120 个点的 gif 图像：TimeTravelerAlgorithm.gif

我将其命名为“时间旅行者”，因为它的运作方式类似于贪婪的推销员算法。但是，贪婪的推销员不是前往现在最近的新城市，而是穿越到过去他已经去过的最近的城市，然后去参观那个新城市，然后继续他的正常路线。

时间旅行者开始了3个城市的路线，旅行者在他的过去每一步都添加一个新城市，直到他到达一个他访问了所有城市并返回他的家乡的现在。

该算法可以快速为少量点提供体面的解决方案。以下是每组数的执行时间，均在 2.6GHz 双核 Intel Core i5 处理器 Macbook 上制作：

• 大约 0.03 秒内 120 个点
• 360 点大约在 0.23 秒内
• 大约 10 秒内 2520 个点
• 3分钟左右10000分
• 大约 5 小时内获得 100 000 个积分 (Solution Map)

该算法远未得到优化，因为在某些情况下它会给出次优的交叉路由。这一切都是用纯python制作的。也许使用 numpy 或一些高级库甚至 GPU 可以加速程序。

我希望得到您的评论并帮助我们优化它。对于可能非常大的点集（从 100 万到 1000 亿点），我尝试在没有交叉路径的情况下近似求解。

PS：我的算法和代码是开放的。来自互联网的人，请随时在您拥有的任何项目或任何研究中使用它。

Answer 1

我通过在每次插入时添加双链表和 2-opt 来改进算法：

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="\r")
            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，该解决方案可以快速获得结果，无需交叉路线。

使用 PyPy，它可以在 30 分钟内解决 100,000 个点且没有交叉路线。

现在我正在努力实现 KD-tree 来解决大型集合问题。

Answer 2

感谢 cmets。我使用对象、集合和链表重新实现了算法。我还从距离函数中删除了平方根。现在代码看起来更干净了：

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

#Distance between two point
class Node:
    def __init__(self, dataval=None):
        self.dataval = dataval
        self.nextval = None

class TSP_TimeTraveler():
    def __init__(self, dataval=None):
        self.count = 0
        self.position = None
        self.length = 0

    def get_position():
        return self.position

    def next_city():
        self.position = self.position.nextval
        return self.position

    #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
            self.position.nextval = node
            self.length = 2*distance(self.position.dataval,node.dataval)
        else : 

            #Creating the traveler
            traveler = self.position

            c = traveler.dataval #current position
            n = traveler.nextval.dataval #next position

            #Calculating the length of adding the city to the path
            Min_L = self.length-distance(c,n)+distance(c,node.dataval)+distance(node.dataval,n)
            Min_Node = traveler

            traveler = traveler.nextval

            while traveler != self.position :
                c = traveler.dataval #current position
                n = traveler.nextval.dataval #next position

                #Calculating the length of adding the city to the path
                L = self.length-distance(c,n)+distance(c,node.dataval)+distance(node.dataval,n)

                #Searching the path to the of city with minimum length
                if L < Min_L :
                    Min_L = L
                    Min_Node = traveler

                traveler = traveler.nextval


            #Adding the city to the minimum path
            node.nextval = Min_Node.nextval
            Min_Node.nextval = node
            self.length = Min_L

        #Incrementing the number of city in the route
        self.count = self.count + 1

    #Get the list of the route
    def getRoute(self):
        result = []

        traveler = self.position
        result.append(traveler.dataval)

        traveler = traveler.nextval

        while traveler != self.position :
            result.append(traveler.dataval)
            traveler = traveler.nextval

        result.append(traveler.dataval)

        return result

    def Solve(self, Set_points):
        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="\r")
            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()

        print("Points left : ", len(points),' Done              ',)
        print("Execution time : ", delta, "secs")

        return result

#######################
#Testing the Algorithm#
#######################

#Size of the set
size = 120

#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)

#Plot the solution
plt.scatter(*zip(*points),s=5)
plt.plot(*zip(*route))
plt.axis('scaled')
plt.show()

使用 PyPy 而不是普通的 python，它运行得更快：

• 120 大约 0.03 秒
• 360 大约 0.05 秒
• 2520 大约 0.22 秒
• 10 000 在大约 2 秒内
• 100 000 约 7 分钟

5小时前的10万件案子，现在7分钟解决。

接下来，我将尝试使用双链表和 KD-tree 实现 2-opt。所以它可以解决没有交叉的大集合。