Question

有人可以给我一个2-opt算法的代码样本，用于旅行商问题。现在我使用最近邻居找到路径，但这种方法远非完美，经过一些研究后我发现2-opt算法可以将该路径纠正到可接受的水平。我发现了一些示例应用程序，但没有源代码。

Answer 1

所以我感到无聊并写下来。 看起来就像它有效，但我还没有彻底测试过它。它假设三角不等式，所有边都存在，那种东西。它的工作方式很像我概述的答案。它打印每次迭代;最后一个是2优化的。

我相信它可以通过无数种方式得到改善。

using System;
using System.Collections.Generic;
using System.Linq;


namespace TSP
{
    internal static class Program
    {
        private static void Main(string[] args)
        {
            //create an initial tour out of nearest neighbors
            var stops = Enumerable.Range(1, 10)
                                  .Select(i => new Stop(new City(i)))
                                  .NearestNeighbors()
                                  .ToList();

            //create next pointers between them
            stops.Connect(true);

            //wrap in a tour object
            Tour startingTour = new Tour(stops);

            //the actual algorithm
            while (true)
            {
                Console.WriteLine(startingTour);
                var newTour = startingTour.GenerateMutations()
                                          .MinBy(tour => tour.Cost());
                if (newTour.Cost() < startingTour.Cost()) startingTour = newTour;
                else break;
            }

            Console.ReadLine();
        }


        private class City
        {
            private static Random rand = new Random();


            public City(int cityName)
            {
                X = rand.NextDouble() * 100;
                Y = rand.NextDouble() * 100;
                CityName = cityName;
            }


            public double X { get; private set; }

            public double Y { get; private set; }

            public int CityName { get; private set; }
        }


        private class Stop
        {
            public Stop(City city)
            {
                City = city;
            }


            public Stop Next { get; set; }

            public City City { get; set; }


            public Stop Clone()
            {
                return new Stop(City);
            }


            public static double Distance(Stop first, Stop other)
            {
                return Math.Sqrt(
                    Math.Pow(first.City.X - other.City.X, 2) +
                    Math.Pow(first.City.Y - other.City.Y, 2));
            }


            //list of nodes, including this one, that we can get to
            public IEnumerable<Stop> CanGetTo()
            {
                var current = this;
                while (true)
                {
                    yield return current;
                    current = current.Next;
                    if (current == this) break;
                }
            }


            public override bool Equals(object obj)
            {
                return City == ((Stop)obj).City;
            }


            public override int GetHashCode()
            {
                return City.GetHashCode();
            }


            public override string ToString()
            {
                return City.CityName.ToString();
            }
        }


        private class Tour
        {
            public Tour(IEnumerable<Stop> stops)
            {
                Anchor = stops.First();
            }


            //the set of tours we can make with 2-opt out of this one
            public IEnumerable<Tour> GenerateMutations()
            {
                for (Stop stop = Anchor; stop.Next != Anchor; stop = stop.Next)
                {
                    //skip the next one, since you can't swap with that
                    Stop current = stop.Next.Next;
                    while (current != Anchor)
                    {
                        yield return CloneWithSwap(stop.City, current.City);
                        current = current.Next;
                    }
                }
            }


            public Stop Anchor { get; set; }


            public Tour CloneWithSwap(City firstCity, City secondCity)
            {
                Stop firstFrom = null, secondFrom = null;
                var stops = UnconnectedClones();
                stops.Connect(true);

                foreach (Stop stop in stops)
                {
                    if (stop.City == firstCity) firstFrom = stop;

                    if (stop.City == secondCity) secondFrom = stop;
                }

                //the swap part
                var firstTo = firstFrom.Next;
                var secondTo = secondFrom.Next;

                //reverse all of the links between the swaps
                firstTo.CanGetTo()
                       .TakeWhile(stop => stop != secondTo)
                       .Reverse()
                       .Connect(false);

                firstTo.Next = secondTo;
                firstFrom.Next = secondFrom;

                var tour = new Tour(stops);
                return tour;
            }


            public IList<Stop> UnconnectedClones()
            {
                return Cycle().Select(stop => stop.Clone()).ToList();
            }


            public double Cost()
            {
                return Cycle().Aggregate(
                    0.0,
                    (sum, stop) =>
                    sum + Stop.Distance(stop, stop.Next));
            }


            private IEnumerable<Stop> Cycle()
            {
                return Anchor.CanGetTo();
            }


            public override string ToString()
            {
                string path = String.Join(
                    "->",
                    Cycle().Select(stop => stop.ToString()).ToArray());
                return String.Format("Cost: {0}, Path:{1}", Cost(), path);
            }

        }


        //take an ordered list of nodes and set their next properties
        private static void Connect(this IEnumerable<Stop> stops, bool loop)
        {
            Stop prev = null, first = null;
            foreach (var stop in stops)
            {
                if (first == null) first = stop;
                if (prev != null) prev.Next = stop;
                prev = stop;
            }

            if (loop)
            {
                prev.Next = first;
            }
        }


        //T with the smallest func(T)
        private static T MinBy<T, TComparable>(
            this IEnumerable<T> xs,
            Func<T, TComparable> func)
            where TComparable : IComparable<TComparable>
        {
            return xs.DefaultIfEmpty().Aggregate(
                (maxSoFar, elem) =>
                func(elem).CompareTo(func(maxSoFar)) > 0 ? maxSoFar : elem);
        }


        //return an ordered nearest neighbor set
        private static IEnumerable<Stop> NearestNeighbors(this IEnumerable<Stop> stops)
        {
            var stopsLeft = stops.ToList();
            for (var stop = stopsLeft.First();
                 stop != null;
                 stop = stopsLeft.MinBy(s => Stop.Distance(stop, s)))
            {
                stopsLeft.Remove(stop);
                yield return stop;
            }
        }
    }
}

Answer 2

嗯，你对TSP的解决方案总是远非完美。没有代码，但这里是如何进行2-Opt。这还不错：

你需要一个名为Stop的类，它有一个Next，Prev和City属性，可能还有一个Stops属性，只返回包含Next和Prev的数组。
当你将它们链接在一起时，我们称之为巡回赛。 Tour有一个Stop属性（任何一个站点都可以），还有一个AllStops属性，它的getter只是走了一段路并返回它们
您需要一种方法来进行巡视并返回其成本。我们称之为Tour.Cost（）。
你需要Tour.Clone（），它只是走了一段路并单独克隆它们
您需要一种方法来生成一组切换两个边的游览。打电话给这个Tour.PossibleMutations（）
从您的NN解决方案开始
在其上调用PossibleMutations（）
对所有这些人调用Cost（）并取结果最低的那个
重复直至费用不降低

Answer 3

如果问题是欧几里德距离并且您希望算法产生的解决方案的成本在最佳值的3/2之内，那么您需要Christofides算法。 ACO和GA没有保证成本。

旅行商问题，2-opt算法c＃实现

3 个答案: