public City[] Solve(City[] cities)
{
MatrixNode root = new MatrixNode(cities);
double lowB = root.reduce();
MatrixNode head = root;
while (!head.isComplete()) // while there are still unseen branches
{
head = head.Expand();
}
List <int> temp = head.FinishPath();
City[] path = new City[cities.Count];
int i = 0;
for (int item in temp)
{
path[i++] = cities[item];
}
return(path);
}
// The creation from a previous node is O(n^2) for the matrix copy
public MatrixNode(MatrixNode parent)
{
n = parent.n;
reduction = parent.reduction;
matrix = new double[n, n];
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
matrix[i, j] = parent.matrix[i, j];
}
}
path = new List <int>();
foreach (int item in parent.path)
{
path.Add(item);
}
unvisited = new List <int>();
foreach (int item in parent.unvisited)
{
unvisited.Add(item);
}
}
// Expands the node by building all of it's children
// and returning the list
// This is O(n^3) for the reduction of possibly n children
// since the reduction costs O(n^2)
// The sort is probably O(nlogn) although I don't know what algorithm
// the default c# function uses. This is dropped in the final big-O.
public List <MatrixNode> Expand(double bound)
{
List <MatrixNode> children = new List <MatrixNode>();
MatrixNode child;
double temp;
foreach (int item in unvisited)
{
child = new MatrixNode(this);
temp = child.Reduce(item);
if (temp < bound)
{
children.Add(child);
}
else
{
pruned++;
}
total_count++;
}
children.Sort(delegate(MatrixNode x, MatrixNode y)
{
if (x.reduction < y.reduction)
{
return(-1);
}
else if (x.reduction > y.reduction)
{
return(1);
}
else
{
return(0);
}
});
return(children);
}
/// <summary>
/// performs a Branch and Bound search of the state space of partial tours
/// stops when time limit expires and uses BSSF as solution
/// Time complexity is O(2^n * n^2) because it has to expand 2^n possible
/// MatrixNode's which take approximately O(n^2) time to Expand.
/// I wrote that the expand function is worst case O(n^3) but that is
/// only if there are n cities left unvisited, which is not going to be
/// the case most of the time. So on average it will turn out to be closer
/// to O(n^2).
/// The space complexity is O(n^2 * 2^n) for 2^n nodes and n^2 for each node.
/// </summary>
/// <returns>results array for GUI that contains three ints: cost of solution, time spent to find solution, number of solutions found during search (not counting initial BSSF estimate)</returns>
public string[] bBSolveProblem()
{
string[] results = new string[3];
int count = 0, max_stack = 0;
Random rnd = new Random();
Stopwatch timer = new Stopwatch();
timer.Start();
// My solution for branch and bound: Iain
MatrixNode.total_count = 0;
MatrixNode.pruned = 0;
MatrixNode root = new MatrixNode(Cities);
double lowB = root.Reduce(), upperBound = Double.PositiveInfinity;
List <MatrixNode> problems = new List <MatrixNode>();
problems.Add(root);
MatrixNode head;
List <MatrixNode> children;
while (problems.Count > 0) // while there are still unseen branches
{
if (timer.ElapsedMilliseconds > time_limit)
{
break;
}
if (problems.Count > max_stack)
{
max_stack = problems.Count;
}
head = problems[0];
problems.RemoveAt(0);
if (head.Reduction < upperBound)
{
if (head.isCompletePath())
{
List <int> temp = head.FinishPath();
if (head.Reduction < upperBound && temp != null && Cities[temp[temp.Count - 1]].costToGetTo(Cities[temp[0]]) != Double.PositiveInfinity)
{
upperBound = head.Reduction;
Route = new ArrayList();
foreach (int item in temp)
{
Route.Add(Cities[item]);
}
bssf = new TSPSolution(Route);
count++;
}
}
else
{
children = head.Expand(upperBound);
children.AddRange(problems);
problems = children;
}
}
else
{
MatrixNode.pruned++;
}
}
// End of my solution
timer.Stop();
results[COST] = costOfBssf().ToString(); // load results array
results[TIME] = timer.Elapsed.ToString();
results[COUNT] = count.ToString();
Console.Write("max_stack: ");
Console.WriteLine(max_stack);
Console.Write("total_count: ");
Console.WriteLine(MatrixNode.total_count);
Console.Write("pruned: ");
Console.WriteLine(MatrixNode.pruned);
return(results);
}
