// creating a child node is O(n^2) for both time and space complexity
public DistMatrix(int toNode, DistMatrix parent)
{
// take in the index from the parent
this.maxIndex = parent.maxIndex;
// add the cost of moving from the previous node to the current node
this.b = parent.b + parent.m[parent.currNode * maxIndex + toNode];
// create a new m matrix and copy over all the values
this.m = new double[maxIndex * maxIndex];
for (int i = 0; i < m.Length; i++)
{
this.m[i] = parent.m[i];
}
// create a new visited list and copy those values
this.visited = new List <int>();
for (int i = 0; i < parent.visited.Count; i++)
{
this.visited.Add(parent.visited[i]);
}
// update the visited list
this.visited.Add(toNode);
this.toNode = toNode;
this.currNode = toNode;
this.fromNode = parent.currNode;
// mark infinity on the column for the node going to (the child)
markToNode();
// mark infinities on the row for the node coming from (the parent)
markFromNode();
// mark the awkward diagonal cell in the matrix as inifinity
markFromToNode();
// make sure there are still 0 values in row/column
updateMatrix();
}
/// <summary>
/// performs a Branch and Bound search of the state space of partial tours
/// stops when time limit expires and uses BSSF as solution
/// </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>
// This algorithm implements a greedy approach, 10,000 random attempts, and the branch and bound approach with pruning of leaf
// nodes that have values grater than the current BSSF.
// O = greedy + 10,000*random + branch and bound
// = O(n^3 + 10000*n + (2^n)*(n^2) = O((n^2)*(2^n))
public string[] bBSolveProblem()
{
// initialize the variables
int count = 0;
string[] results = new string[3];
int[] perm = new int[Cities.Length];
Route = new ArrayList();
Stopwatch timer = new Stopwatch();
int maxStates = 0;
int totalStates = 0;
int statesPruned = 0;
// start the timer
timer.Start();
// start BSSF as positive infinity;
double currSol = Double.PositiveInfinity;
// see if the BSSF updates with the greedy algorithm
// Finding greedy solution = O(n^3)
double greedySol = Convert.ToDouble(greedySolveProblem()[0]);
if (greedySol < currSol)
{
currSol = greedySol;
count += 1;
}
// attempt updating the BSSF with 10,000 random solutions
// finding random solution = O(n)
for (int cycleRand = 0; cycleRand < 10000; cycleRand++)
{
double tempSol = Convert.ToInt32(defaultSolveProblem()[0]);
if (tempSol < currSol)
{
currSol = tempSol;
count += 1;
}
}
// create the List of nodes
List <DistMatrix> children = new List <DistMatrix>();
// initialize the first node in the list (root node)
children.Add(new DistMatrix(Cities));
// while there are nodes in the node list
while (children.Count > 0)
{
// check if the timer has passed 60 seconds
timer.Stop();
long time = timer.ElapsedMilliseconds / 1000;
if (time >= 60)
{
// if it has passed 60 seconds, break out of the while loop
break;
}
timer.Start();
// Find the next node - O(n)
if (maxStates < children.Count)
{
maxStates = children.Count;
}
// collect the variables to find the best child node
DistMatrix currNode = children[0];
double best = children[0].b;
int numVisited = children[0].visited.Count;
// for each child node, check if it is optimal
for (int i = 0; i < children.Count; i++)
{
double value = children[i].b;
int visited = children[i].visited.Count;
double diffValue = value - best;
int diffVisit = visited - numVisited;
// used this little algorithm to try and balance breadth and depth
if (value < best * (1 + (diffVisit / .75)))
{
currNode = children[i];
best = children[i].b;
numVisited = children[i].visited.Count;
}
}
// remove that node from the list of possiblities
children.RemoveAt(children.IndexOf(currNode));
// if the child node has not reached the end
if (!currNode.finished())
{
// get the list of nodes that haven't been visited
List <int> nextChildren;
nextChildren = currNode.returnNodesLeft();
// create a child object for each of those nodes - O(n^2) happening 2^n times
// for the worst case scenario where each leaf must be reached in the tree.
// This gives a O((n^2)*(2^n))
for (int i = 0; i < nextChildren.Count; i++)
{
DistMatrix child = new DistMatrix(nextChildren[i], currNode);
children.Add(new DistMatrix(nextChildren[i], currNode));
totalStates += 1;
}
}
// if the node is a leaf, check if it's better than the current BSSF
//O(n) to check each one, with O(n) to check if any need removing after
// an update on the BSSF.
else
{
if (currNode.b < currSol)
{
// if it is better, replace the current BSSF
ArrayList tempRoute = buildRoute(currNode.visited);
TSPSolution test = new TSPSolution(tempRoute);
if (test.costOfRoute() < currSol)
{
Route.Clear();
Route = buildRoute(currNode.visited);
bssf = null;
bssf = new TSPSolution(Route);
currSol = costOfBssf();
count += 1;
}
// filter any children that has a b greater than the current bssf.
// remove those children in hopes of speeding up the algorithm
}
}
for (int i = 0; i < children.Count; i++)
{
if (children[i].b > currSol)
{
children.RemoveAt(i);
statesPruned += 1;
i--;
}
}
}
// Stop the time and report the results.
timer.Stop();
results[COST] = costOfBssf().ToString(); // load results array
results[TIME] = timer.Elapsed.ToString();
results[COUNT] = count.ToString();
Console.WriteLine("Max States: {0}, Total States: {1}, Total Pruned: {2}", maxStates, totalStates, statesPruned);
return(results);
}