当这个常规队列版本也正确时，为什么 Dijkstra 的算法需要优先级队列？

Question

I have read the following but please take a look at my code below.

Why Dijkstra's Algorithm uses heap (priority queue)?

I have two versions of dijkstra, one good version with PQueue, and one bad version with regular linked list queue.

public static void computeDijkstra(Vertex source) {
    source.minDistance = 0.;
    Queue<Vertex> vertexQueue = new PriorityQueue<Vertex>();
    // Queue<Vertex> vertexQueue = new LinkedList<Vertex>();
    vertexQueue.add(source);

    while (!vertexQueue.isEmpty()) {
        Vertex fromVertex = vertexQueue.poll();

        if (fromVertex.neighbors != null) {
            for (Edge currentEdge : fromVertex.neighbors) {
                Vertex toVertex = currentEdge.target;
                if (currentEdge.weight + fromVertex.minDistance < toVertex.minDistance) {
                    toVertex.minDistance = currentEdge.weight + fromVertex.minDistance;
                    toVertex.previous = fromVertex;
                    vertexQueue.add(toVertex);
                }
            }
        }
    }
}

public static void computeDijkstraBad(Vertex source) {
    source.minDistance = 0.;
    // Queue<Vertex> vertexQueue = new PriorityQueue<Vertex>();
    Queue<Vertex> vertexQueue = new LinkedList<Vertex>();
    vertexQueue.add(source);

    while (!vertexQueue.isEmpty()) {
        Vertex fromVertex = vertexQueue.poll();

        if (fromVertex.neighbors != null) {
            for (Edge currentEdge : fromVertex.neighbors) {
                Vertex toVertex = currentEdge.target;
                if (currentEdge.weight + fromVertex.minDistance < toVertex.minDistance) {
                    toVertex.minDistance = currentEdge.weight + fromVertex.minDistance;
                    toVertex.previous = fromVertex;
                    vertexQueue.add(toVertex);
                }
            }
        }
    }
}

I also have graph creation with a text file like below

0, 1, 2, 3, 4, 5, 6 // vertices
0, 6 // from and to vertex
1, (2-5, 0-4, 4-6) // from vertex 1 it will have edge to 2 with weight 5 ...
0, (4-3, 3-7)
4, (2-11, 3-8)
3, (2-2, 5-5)
2, (6-2, 5-10)
5, (6-3)

Both the implementation renders the following [0, 3, 2, 6] which is the shortest path indeed from 0 to 6!

Now we know, if a Simple BFS is used to find shortest distance with positive integers, there will be cases where it will not find the minimum path. So, can someone give me a counter example for which my Bad implementation will fail to print the right path for a graph. Feel free to give me the answer in the graph format (the sample text file format) that I used.

So far all the graphs I have had, both implementations rendered the right result. This shouldn't happen because the bad implementation is runtime (E+V) and we know we can't find shortest path without at least E log V.

Another example,

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
0, 10
0, (1-9, 2-10, 3-11)
1, (4-1, 5-7)
2, (4-4, 5-3, 6-5)
3, (5-1, 6-4)
4, (7-9, 8-14, 5-3)
5, (7-4, 8-5, 9-9, 6-2)
6, (8-2, 9-2)
7, (10-3)
8, (10-2)
9, (10-5)

Both implementations renders [0, 3, 5, 6, 8, 10], which is the correct shortest path from 0-10.

Answer 1

I believe that the algorithm you've given is correct, but that it isn't as efficient as Dijkstra's algorithm.

At a high level, your algorithm works by finding an "active" node (one whose distance has been lowered), then scanning the outgoing edges to activate all adjacent nodes that need their distance updated. Notice that the same node can be made active multiple times - in fact, it's possible that a node will be activated once per time its candidate distance drops, which can happen potentially many times in a run of the algorithm. Additionally, the algorithm you have implemented might put the same node into the queue multiple times if the candidate distance drops multiple times, so it's possible that all dequeues except the first will be unnecessary. Overall, I'd expect this would result in a pretty big runtime hit for large graphs.

In a sense, the algorithm you've implemented is a shortest paths algorithm, but it's not Dijkstra's algorithm. The main difference is that Dijkstra's algorithm uses a priority queue to ensure that every node is dequeued and processed once and exactly once, leading to much higher efficiency.

So I guess the best answer I can give is "your algorithm isn't an implementation of Dijkstra's algorithm, and the reason Dijkstra's algorithm uses a priority queue is to avoid recomputing distances multiple times in the way that your algorithm does."

Answer 2

Your algorithm will find the right result but what your approach is doing is that it kills off the efficiency of Dijkstra's approach.

Example:

Consider 3 nodes named AB C.

A->C :7 
A->B :2
B->C :3

In your bad approach, You'll first set the shortest path from A to C as 7, and then, as you traverse, you will revise it to 5 (A->BC)

In Dijkstra's approach, A->C will not be traversed at all because, when using a min-heap, A->B will be traversed first, B will be marked as "traversed", and then B->C will be traversed, and, C will be marked as "traversed". Now, since C is already marked as "traversed", the path A->C (of length 7) will never be checked.

Therefore, as you can see, in your bad approach, you will be reaching C 2 times (A->C & A->B->C), while using Dijkstra's approach, you will go to C only once.

This example should prove that you will have fewer iterations with Dijkstra's algorithm.

Answer 3

Read the others' answers, they are right, and the summary is the bad implement method is correct, but the complexity of the owner's claim(O(E+V)) is not Right. One other thing that no one else has mentioned, and I'll add it here.

This poor implementation also corresponds to an algorithm, which is actually a derivative of BFS, formally known as SPFA . Check out the SPFA algorithm . When the author published this algorithm back in 1994, he claimed it has a better performance than Dijkstra with O(E) complexity, which is Wrong.

It became very popular among ACM students. Due to its simplicity and ease to implement. As for Performance, Dijkstra is preferred.

similar reply ref to this post

Answer 4

All of the answers are very well written, so I won't be adding much explanation. I am adding an example that I came across in a comment made by Striver in one of his videos of the graph playlist. Hope it will help to see the issue clearly.

If we don't use Priority Queue while implementing Dijkstra's Algo for an undirected graph with arbitrary edge weights then we might have to visit the same node, again and again, it will not be able to account for arbitrary edge weights.

Take the below example and dry run it by taking a Priority queue and by taking a simple queue, you will see that you have to visit node 6 twice for the normal queue dry run. So, in the case of much larger graphs, this will result in very poor performance.

Edit-- 5 to 6 node edge weight would be 1 and not 10. I will upload the correct image soon.