dijkstra | Dijkstra shortest path problem solved in Go

Your understanding is correct. However, there are some issues that could be improved.

Firstly, you usually want to measure only one algorithm at a time. If you measure each algorithm separately you can compare the two and get more accurate data. Also, I would stick to measuring it on the same graph multiple times and then average the time to get one more accurate (you can use copy.deepcopy() to make duplicates of the graph). Then do this on each of the graphs you have. Since each algorithm may be more efficient on different types of graphs (you would lose this information completely if measured as you propose).

Secondly, since timeit can repeat the measured operation multiple times but the setup is done only once before any measurement starts (see the docs) you would measure the algorithm on already traversed (or sorted as you put it) graphs. Which you correctly point out.

As a solution, I suggest manually timing the runs using the time.perf_counter_ns(). The code for measuring the bellmanFord algorithm could look as such:

Your spin loop while (data->locked[i]); is a data race; you don't hold the lock while reading it data->locked[i], and so another thread could take the lock and write to that same variable while you are reading it. In fact, you rely on that happening. But this is undefined behavior.

Immediate practical consequences are that the compiler can delete the test (since in the absence of a data race, data->locked[i] could not change between iterations), or delete the loop altogether (since it's now an infinite loop, and nontrivial infinite loops are UB). Of course other undesired outcomes are also possible.

So you have to hold the mutex while testing the flag. If it's false, you should then hold the mutex until you set it true and do your other work; otherwise there is a race where another thread could get it first. If it's true, then drop the mutex, wait a little while, take it again, and retry.

(How long is a "little while", and what work you choose to do in between, are probably things you should test. Depending on what kind of fairness algorithms your pthread implementation uses, you might run into situations where take_forks succeeds in retaking the lock even if put_forks is also waiting to lock it.)

Of course, in a "real" program, you wouldn't do it this way in the first place; you'd use a condition variable.

Move the definition of printPath outside of the body of dijkstraSSSP and, to give it access to pred, change it to accept two parameters instead of one. The additional parameter should have type int * and should point to the first element of pred.

In the call, pass pred for that new parameter. The array pred will be automatically converted to a pointer to its first element.

The Wiki article doesn't specify, but that code will only solve the 'loopy' version of k-shortest-paths, where paths are not required to be simple.

The simple path version of the problem is harder: you'll want to look at something like Yen's algorithm, which does clever filtering to avoid repeated points when generating paths. Yen's algorithm can use Dijkstra's algorithm as a subroutine, but any other shortest-path algorithm can also be used instead.

There is no obvious way to modify Dijkstra's algorithm to solve the k-shortest-simple-paths problem. You'd need to track the paths in the priority queue (which is already done in your posted code), but there's an exponential upper bound on the number of times each vertex can be explored.

Here, if count[u] <= K puts an upper bound of K+1 on the number of times a vertex can be explored, which works for the non-simple path case. On the other hand, a direct modification of Dijkstra's algorithm for simple paths would, in the worst case, require you to explore a node once for each of the 2^(V-1) possibilities of which nodes had been previously visited (or possibly a slightly smaller exponential).

Your somewhat odd comparator...

Perhaps you can try this

1. Despite the test, this implementation of Dijkstra may put Ω(E) items in the priority queue. This will cost Ω(E log E) with every comparison-based priority queue.

2. Why not E log V? Well, assuming a connected, simple, nontrivial graph, we have Θ(E log V) = Θ(E log E) since log (V−1) ≤ log E < log V² = 2 log V.

3. The O(E + V log V)-time implementations of Dijkstra's algorithm depend on a(n amortized) constant-time DecreaseKey operation, avoiding multiple entries for an individual vertex. The implementation in this question will likely be faster in practice on sparse graphs, however.

probably just try something like this

Since BFS is guaranteed to return an optimal path on unweighted graphs, and you've created the unweighted equivalent of your original graph, you'll be guaranteed to get the shortest path.

What you lose by doing this over Dijkstra's algorithm is runtime optimality. Now the runtime of your algorithm is dependent on the edge weights, whereas Dijkstra's is only dependent on the number of edges.

This sort of thought experiment is a great way to understand how Dijkstra's algorithm works (eg. how would you modify your algorithm to not require creating a new graph? Or not take 100 steps for an edge with weight 100?). In fact this is probably how Dijkstra discovered the algorithm to begin with.