Why does Dijkstra's Algorithm use a heap (priority queue)?

Question

I have tried using Djikstra\'s Algorithm on a cyclic weighted graph without using a priority queue (heap) and it worked.

Wikipedia states that the original implementatio

Answer 1

Like Moataz Elmasry said the best you can expect is O(|E| + |V|.|logV|) with a fib queue. At least when it comes to big oh values.

The idea behind it is, for every vertex(node) you are currently working on, you already found the shortest path to. If the vertex isn't the smallest one, (distance + edge weight) that isn't necessarily true. This is what allows you to stop the algorithm as soon as you have expanded(?) every vertex that is reachable from your initial vertex. If you aren't expanding the smallest vertex, you aren't guaranteed to be finding the shortest path, thus you would have to test every single path, not just one. So instead of having to go through every edge in just one path, you go through every edge in every path.

Your estimate for O(E + V) is probably correct, the path and cost you determined on the other hand, are incorrect. If I'm not mistaken the path would only be the shortest if by any chance the first edge you travel from every vertex just happens to be the smallest one.

So Dijkstra's shortest path algorithm without a queue with priority is just Dijkstra's path algorithm ;)