优化旅行商算法(时间旅行者算法) [英] Optimizing a Traveling Salesman Algorithm (Time Traveler Algorithm)

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。

推荐答案

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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