[FIXED] How to update key of a relaxed vertex in Dijkstra's algorithm?

Issue

Just like it was asked here, I fail to understand how we can find the index of a relaxed vertex in the heap.

Programming style-wise, the heap is a black box that abstracts away the details of a priority queue. Now if we need to maintain a hash table that maps vertex keys to corresponding indices in the heap array, that would need to be done in heap implementation, right?

But most standard heaps don't provide a hash table that does such mapping.

Another way to deal with this whole problem is to add the relaxed vertices to the heap regardless of anything. When we extract the minimum we'll get the best one. To prevent the same vertex being extracted multiple times, we can mark it visited.

So my exact question is, what is the typical way (in the industry) of dealing with this problem?

What are the pros and cons compared what the methods I mentioned?

Solution

You are asking some very valid questions but unfortunately they are kind of vague so we won't be able to give you a 100% solid "industry standard" answer. However, I'll try to go over your points anyway:

Programming style-wise, the heap is a black box that abstracts away the details of a priority queue

Technically, a priority queue is the abstract interface (insert elements with a priority, extract the lowest priority element) and a heap is a concrete implementation (array-based heap, binomial heap, fibonacci heap, etc).

What I'm trying to say is that using an array is only one particular way to implement a priority queue.

Now if we need to maintain a hash table that maps vertex keys to corresponding indices in the heap array, that would need to be done in heap implementation, right?

Yes, because everytime you move an element inside the array you will need to update the index in the hash table.

But most standard heaps don't provide a hash table that does such mapping.

Yes. This can be very annoying.

Another way to deal with this whole problem is to add the relaxed vertices to the heap regardless of anything.

I guess that could work but I dont think I ever saw anyone do that. The whole point of using a heap here is to increase performance and by adding redundant elements to the heap you kind of go against that. Sure, you preserve the "black-boxness" of the priority queue but I don't know if that is worth it. Additionally, there could be a chance that the extra pop_heap operations could negatively affect your asymptoptic complexity but I'd have to do the math to check.

what is the typical way (in the industry) of dealing with this problem?

First of all, ask yourself if you can get away with using a dumb array instead of a priority queue. Sure, finding the minimum element in now O(N) instead of O(log n) but the implementation is the simplest (an advantage on its own). Additionally, using an array will be just as efficient if your graph is dense and even if your graph is sparse it might be efficient enough depending on how big your graph is.

If you really need a priority queue, then you are going to have to find one that has a decreaseKey operation implemented. If you can't find one, I would say its not that bad to implement it yourself - it might be less trouble than trying to find an existing implementation and then trying to fit it in with the rest of your code.

Finally, I would not recommend using the really fancy heap data structures (such as fibonacci heaps). While these often show up in textbooks as a way to get optimal asymptotics, in practice they have terrible constant factors and these constant factors are significant when compared with something that is logarithmic.

Answered By - hugomg

Answer Checked By - Robin (PHPFixing Admin)