优化旅行推销员算法（时间旅行者算法）

Posted 2023-02-19

【中文标题】优化旅行推销员算法（时间旅行者算法）

【英文标题】：Optimizing a Traveling Salesman Algorithm (Time Traveler Algorithm)

【发布时间】：2021-01-19 20:02:20

【问题描述】：

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。

【问题讨论】：

您可以对距离平方进行比较，避免计算 sqrt()。

那里有一个有点相似的算法，但我不记得它叫什么了。性能优化将使用一种方法，使您可以在O(log h) 而不是O(h) 中确定数据中最近的点，其中h 是解决方案中的当前点。可能是KD树或其他东西。还实施 2- 或 3-opt 以摆脱十字架。

算法的时间复杂度是多少，实现又是怎样的？如果您的实现速度较慢，那么您可能会遇到数据结构问题以及您对它们的操作时间复杂度的假设（例如，插入列表是 O(n)，来自 wiki.python.org/moin/TimeComplexity）。

如果您还不了解它们，我建议您针对“标准”TSP 基准测试数据案例运行您的算法，看看它们是如何做的。 math.uwaterloo.ca/tsp/data/index.html（这里是一个带有 TSPLIB 格式定义的 PDF comopt.ifi.uni-heidelberg.de/software/TSPLIB95/tsp95.pdf）

嗯，试试math.uwaterloo.ca/tsp/world/zitour.html。 VLSI 难以优化，因为它们在某些情况下可能是等距的（它们基本上是电路板），因此如果您不处理 AB 与 BC 距离相等的情况，您可能会在该点集上遇到问题。建议在其他基准实例上也对其进行测试。看看你哪里出错了。我可以看到算法是如何出错的，但 16% 的近似值还不错。如果您将 2-opt 提升为 3-opt，可能会有所帮助。我会实施并看看它是否有很大帮助。

【参考方案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。

【参考方案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。