traveling salesman problem, 2-opt algorithm c# implementation
Can someone give me a code sample of 2-opt algorithm for traveling salesman problem. For now im using nearest neighbour to find the path but this method is far from perfect, and after some research i found 2-opt algorithm that would correct that path to the acceptable level. I found some samp开发者_JAVA百科le apps but without source code.
So I got bored and wrote it. It looks like it works, but I haven't tested it very thoroughly. It assumes triangle inequality, all edges exist, that sort of thing. It works largely like the answer I outlined. It prints each iteration; the last one is the 2-optimized one.
I'm sure it can be improved in a zillion ways.
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;
}
}
}
}
Well, your solution to TSP is always going to be far from perfect. No code, but here's how to go about 2-Opt. It's not too bad:
- You need a class called Stop that has a Next, Prev, and City property, and probably a Stops property that just returns the array containing Next and Prev.
- When you link them together, we'll call that a Tour. Tour has a Stop property (any of the stops will do), and an AllStops property, whose getter just walks the stops and returns them
- You need a method that takes a tour and returns its cost. Let's call that Tour.Cost().
- You need Tour.Clone(), which just walks the stops and clones them individually
- You need a method that generates the set of tours with two edges switched. Call this Tour.PossibleMutations()
- Start with your NN solution
- Call PossibleMutations() on it
- Call Cost() on all of them and take the one with the lowest result
- Repeat until the cost doesn't go down
If the problem is euclidian distance and you want the cost of the solution produced by the algorithm is within 3/2 of the optimum then you want the Christofides algorithm. ACO and GA don't have a guaranteed cost.
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