Question

Ok, I posted this question because of this exercise:

Can we modify Dijkstra’s algorithm to solve the single-source longest path problem by changing minimum to maximum? If so, then prove your algorithm correct. If not, then provide a counterexample.

For this exercise or all things related to Dijkstra’s algorithm, I assume there are no negative weights in the graph. Otherwise, it makes not much sense, as even for shortest path problem, Dijkstra can’t work properly if negative edge exists.

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Ok, my intuition answered it for me:

Yes, I think it can be modified.

I just

1. initialise distance array to MININT
2. change distance[w] > distance[v]+weight to distance[w] < distance[v]+weight

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Then I did some research to verify my answer. I found this post:

Longest path between from a source to certain nodes in a DAG

First I thought my answer was wrong because of the post above. But I found that maybe the answer in the post above is wrong. It mixed up The Single-Source Longest Path Problem with The Longest Path Problem.

Also in wiki of Bellman–Ford algorithm, it said correctly :

The Bellman–Ford algorithm computes single-source shortest paths in a weighted digraph. For graphs with only non-negative edge weights, the faster Dijkstra’s algorithm also solves the problem. Thus, Bellman–Ford is used primarily for graphs with negative edge weights.

So I think my answer is correct, right?
Dijkstra can really be The Single-Source Longest Path Problem and my modifications are also correct, right?

Answer 1

No, we cannot¹ – or at the very least, no polynomial reduction/modification is known – longest path problem is NP-Hard, while dijkstra runs in polynomial time!

If we can find a modfication to dijsktra to answer longest-path problem in polynomial time, we can derive P=NP

If not, then provide a counterexample.

This is very bad task. The counter example can provide a specific modification is wrong, while there could be a different modification that is OK.

The truth is we do not know if longest-path problem is solveable in polynomial time or not, but the general assumption is – it is not.

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regarding just changing the relaxation step:

        A
       / \
      1   2
     /     \
    B<--1---C
edges are (A,B),(A,C),(C,B)

dijkstra from A will first pick B, and then B is never reachable – because it is out of the set of distances.

At the very least, one will have also to change min heap into max heap, but it will have a different counter-example why it fails.

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(1) probably, maybe if P=NP it is possible, but it is very unlikely.