Since there is always a pointer to the minimum distance value node in the heap meaning we get an extremely efficient time frame for many actions. We then can use a circular linked list to connect the children nodes. Each node contains a pointer to its parent and one of its children. The reason why we are able to have O(1) for many operations as a apposed to in a binary heap is the nature of the fibonacci. For a fibonacci heap here are the time complexities of the same common operations as we saw in binary heap: Operation A fibonacci heap is a lazy data structure, meaning that common operations carried out on the heap are very fast due to the structure being less managed whilst its performing actions. Our final implementation of Dijkstra's is our most efficient. Inner loop operation time complexities can be found in table shown above Case 3: Fibonacci Heap + Priority Queue Total O(ElogV) where decrease key happens the least amount of times this means we will have a complexity of O(ElogV) still, just will the number of logV operations reduced. Our best case is identical to our worst case however it is when the number of key operations are the smallest. Total O(EVlog(E/V)logV) where decrease key happens O(log(E/V) times Best Case Time Complexity We know this since our inner loop still takes O(V + E) times, the only difference is the amount of decrease key operations is bounded by O(Vlog(E/V)) meaning that there is a constant on the calculations. The average running time in this case will be O(EVlog(E/V)logV). Total O(ElogV) where decrease key happens the most amount of times Average Case Time Complexity Where we have the largest number of decrease key operations (which take logV). If checking_dist V this simplifies to O(ElogV).

Implementation in Python from collections import defaultdictĭef bin_heap_dijkstras(times : list], x : int, y : int) -> int:ĭistance_costs = Ĭurrent_dist, current = heapq.heappopp(min_dist)Ĭhecking_dist = current_dist + graph Here are common heap operations and their time complexities using binary heap: Operation If not, repeat steps 3-5 until all nodes have been visitedĬost = unvisited + graph.Once all vertices have been added to visited set, the algorithm has complete.Once all neighbours have been checked, mark the current node as unvisited and place it to the visited set, this set will never be rechecked.For the current node (i), find all of its neighbours and calculate their costs by adding the edge weights connecting current node and neighbour (j) to the current nodes cost (i -> j + i).Assign each vertex a distance cost, 0 for the root node and infinity for the rest, set root node as current node.Create a set with all vertices as unvisted called unvisited set.Space complexity will be O(V) where V is number of vertices in graph, it is worse case scenario if it is a complete graph and every edge has to be visited. This will bring our total time complexity to O(V^2) where is the number of vertices in the graph. This means that our initial time complexity will be O(n) for this search. In this approach, using an unsorted array, we search through all vertices to find the closest within the graph. This is our simplest implementation as well as the least efficient. The algorithm works for the following scenarios:

In this article we will be analysing the time and space complexities in different use cases and seeing how we can improve it. It is one of the most popular pathfinding algorithms due to its diverse range of applications. Prerequisite: Dijkstra's Algorithm, Binary Heap, Fibonacci Heap Introduction to Dijkstra's Algorithmĭijkstra's Algorithm is a pathfinding algorithm, used to find the shortest path between the vertices of a graph. Case 3: Fibonacci Heap + Priority Queue.In this article, we have explored the Time & Space Complexity of Dijkstra's Algorithm including 3 different variants like naive implementation, Binary Heap + Priority Queue and Fibonacci Heap + Priority Queue.

Priority queue time complexity