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////////////////////////////////////////////// Dijktra's Algorithm ////////////////////////////////////////////////////////
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/**
* This function will implement Dijkstra's Algorithm to find the shortest path to all the nodes from the source node
* Time Complexity: O((|V| + |E|) log|V|) with a heap as it iterates over all the nodes and all the edges and uses a queue to go
* through them where the heap queue has a worst case of log|V|. Whereas if the queue was implemented with an array, the complexity
* would be O((|V|^2) since the queue has a worst case of |V| and |E| is upper bounded by |V|^2 and so |V|^2 dominates.
* Space Complexity: O(|V|) as it creates arrays as big as the number of nodes in the graph
*/
private List<int> Dijkstras(ref PriorityQueue queue, bool isArray)
{
// Create Queue to track order of points
queue.makeQueue(points.Count);
// Set up prev node list
List<int> prev = new List<int>();
List<double> dist = new List<double>();
for (int i = 0; i < points.Count; i++)
{
prev.Add(-1);
dist.Add(double.MaxValue);
}
// Initilize the start node distance to 0
dist[startNodeIndex] = 0;
// Update Priority Queue to reflect change in start point distance
if (isArray)
queue.insert(startNodeIndex, 0);
else
queue.insert(ref dist, startNodeIndex);
// Iterate while the queue is not empty
while (!queue.isEmpty())
{
// Grab the next min cost Point
int indexOfMin;
if (isArray)
indexOfMin = queue.deleteMin();
else
indexOfMin = queue.deleteMin(ref dist);
PointF u = points[indexOfMin];
// For all edges coming out of u
foreach (int targetIndex in adjacencyList[indexOfMin])
{
PointF target = points[targetIndex];
double newDist = dist[indexOfMin] + computeDistance(u, target);
if (dist[targetIndex] > newDist)
{
prev[targetIndex] = indexOfMin;
dist[targetIndex] = newDist;
if (isArray)
queue.decreaseKey(targetIndex, newDist);
else
queue.decreaseKey(ref dist, targetIndex);
}
}
}
return prev;
}
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////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
////////////////////////////////////////////// Dijktra's Algorithm ////////////////////////////////////////////////////////
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
/**
* This function will implement Dijkstra's Algorithm to find the shortest path to all the nodes from the source node
* Time Complexity: O((|V| + |E|) log|V|) with a heap as it iterates over all the nodes and all the edges and uses a queue to go
* through them where the heap queue has a worst case of log|V|. Whereas if the queue was implemented with an array, the complexity
* would be O((|V|^2) since the queue has a worst case of |V| and |E| is upper bounded by |V|^2 and so |V|^2 dominates.
* Space Complexity: O(|V|) as it creates arrays as big as the number of nodes in the graph
*/
private List <int> Dijkstras(ref PriorityQueue queue, bool isArray)
{
// Create Queue to track order of points
queue.makeQueue(points.Count);
// Set up prev node list
List <int> prev = new List <int>();
List <double> dist = new List <double>();
for (int i = 0; i < points.Count; i++)
{
prev.Add(-1);
dist.Add(double.MaxValue);
}
// Initilize the start node distance to 0
dist[startNodeIndex] = 0;
// Update Priority Queue to reflect change in start point distance
if (isArray)
{
queue.insert(startNodeIndex, 0);
}
else
{
queue.insert(ref dist, startNodeIndex);
}
// Iterate while the queue is not empty
while (!queue.isEmpty())
{
// Grab the next min cost Point
int indexOfMin;
if (isArray)
{
indexOfMin = queue.deleteMin();
}
else
{
indexOfMin = queue.deleteMin(ref dist);
}
PointF u = points[indexOfMin];
// For all edges coming out of u
foreach (int targetIndex in adjacencyList[indexOfMin])
{
PointF target = points[targetIndex];
double newDist = dist[indexOfMin] + computeDistance(u, target);
if (dist[targetIndex] > newDist)
{
prev[targetIndex] = indexOfMin;
dist[targetIndex] = newDist;
if (isArray)
{
queue.decreaseKey(targetIndex, newDist);
}
else
{
queue.decreaseKey(ref dist, targetIndex);
}
}
}
}
return(prev);
}
public void CheckInsertPop()
{
PriorityQueue pq = new PriorityQueue(2,10);
node n = pq.insert(10);
Assert.AreEqual(n, pq.pop());
}
//-----------------------------------------------------------------------------//
//-------------------------- Dijkstra's Algorithm -----------------------------//
//-----------------------------------------------------------------------------//
/**
* This Dijkstra's is implemented based off of the pseudocode found in the text.
* Time complexity for the Binary Heap Priority queue is O(|V| * |logV|)
* because the implementation requires all the nodes to be visited but the
* update methods require |log V| time. The array priority queue is O(|V|^2)
* because it has to visit all the nodes on implementation and has to itterate
* through the nodes in order to delete and update the queue. The space complexity
* for both is O(|V|) because each queue stores all of the nodes generated
**/
private List <int> dijkstrasAlgorithm(PriorityQueue queue, bool isArray)
{
queue.makeQueue(points.Count);
List <int> previous = new List <int>();
List <double> distances = new List <double>();
// Sets all distance values to "infinity"
// sets previous as 'undefined'
for (int i = 0; i < points.Count; i++)
{
previous.Add(-1);
distances.Add(double.MaxValue);
}
// distance to start node
distances[startNodeIndex] = 0;
// checks which priority queue will be used
if (isArray)
{
queue.insert(startNodeIndex, 0);
}
else
{
queue.insert(ref distances, startNodeIndex);
}
// main loop
// loops until there arent any nodes with a permanent distance to the end node
while (!queue.isEmpty())
{
int minIndex;
if (isArray)
{
minIndex = queue.deleteMin();
}
else
{
minIndex = queue.deleteMin(ref distances);
}
PointF u = points[minIndex];
// finds the best path amongst its neighbors, if it exists
foreach (int index in adjacencyList[minIndex])
{
PointF alt = points[index];
double altDistance = distances[minIndex] + distanceBetween(u, alt);
if (altDistance < distances[index])
{
previous[index] = minIndex;
distances[index] = altDistance;
if (isArray)
{
queue.decreaseKey(index, altDistance);
}
else
{
queue.decreaseKey(ref distances, index);
}
}
}
}
//queue.printQueue();
// gives the path of the nodes
return(previous);
}
