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I have been practicing problems on graphs and shortest paths and I encountered a problem that I'm struggling to understand.

Can you give me any tips and/or can you confirm that I got the general concept of the problem right?

In a given undirected weighted graph $G=(V,E)$ and an integer $k$ and edges with positive weights, the starting point is node $v_0$. You have to visit every node from $v_0, v_1, v_2, \dots, v_{k-1}$. If you visit any of those $k$ nodes you are allowed to create a portal in them. You can only make portals in those $k$ nodes. From one portal you can teleport to any other without any cost to the path.

You have to create a function double shortestPath(int k) that returns a shortest path to visit all $k$ nodes using teleportation system (if necessary).

I sketched this graph and tried to understand the problem here. Lets say we have to visit nodes 0,1 and 2.

Edge-weighted graph

Edited (Please check the reply to see better example of the problem)

My approach is this: If I take this route $0 \to 5$ (cost 1) then $5 \to 4$ (cost 2), $4 \to 2$ (cost 3) and then jump from $2 \to 0$ and go $0 \to 1$ (cost 9). This is the shortest path using teleportations.

Can someone tell me if I understood this problem correctly and maybe give some tips on how to implement the algorithm?

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  • $\begingroup$ I retract my statement about this being equivalent to the Steiner Tree problem. A simple counter-example is the claw with the leaves as terminals. $\endgroup$ – Pål GD Dec 17 '20 at 22:28
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    $\begingroup$ I think perhaps the greedy works? Start with $T' = \{v_0\}$. While $|T'|<k$, pick a terminal not in $T'$ that is of shortest distance from $T'$ and add it to $T'$. Sum up the distances travelled in total. $\endgroup$ – Pål GD Dec 17 '20 at 22:52
  • $\begingroup$ I sometimes overthink. That sounds smooth and good. I will check it on few cases and then code it. Will leave a feedback! Thank you for the effort, really. $\endgroup$ – Exzone Dec 17 '20 at 23:46
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You can do this in polynomial time with a greedy algorithm by always traveling to the nearest un-visited terminal, and dropping a portal in every terminal.

The crucial observations are these:

  1. Since it's free to drop a portal and use one, you always put a portal in each of the terminals.
  2. There is no point traveling anywhere but to a terminal.
  3. When you move to a terminal, you always travel from the closest portal.

Let $G = (V, E, w)$ be the input graph and $T \subseteq V$ be your $k$ terminals, with $v_0 \in T$.

Start with $T' = \{v_0\}$. While $|T'|<k$, pick a terminal not in $T'$ that is of shortest distance from $T'$ and add it to $T'$. Sum up the distances traveled in total.

Notice that you can preprocess your graph in $O(t^2 (n+m))$ time to get a new (complete) graph $G' = (T, E', w')$ with only vertices from $T$ and where the edge $t_1t_2$ has $w'(t_1, t_2) = \text{dist}_G(t_1, t_2)$.

Edit: This answer is updated. The previous answer was based on Steiner Tree.

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Reply to @Pål GD:

Lets try and apply Steiner algorithm to this graph where $b,e$ and $g$ are terminals. Starting point will be node $b$. Closest of nodes $e$ and $g$ to the node $b$ is node $e$ with cost of $6+7=13$. So, for now, we have a subgraph with nodes $b \rightarrow c \rightarrow e $ . Finally we need $g$ in our subgraph that will make a minimum spanning tree which would cover terminal nodes. The closest node in our graph to the node $g$ is $c$ with the cost $15$. Thus our final subgraph will look like this:

enter image description here

And this subgraph indeed covers all the terminal nodes with the cost of 28. But in the exercise with teleportation we are allowed to make a portal in nodes $b,e$ and $g$ after we visit them. In this case, our starting point is $b$ and we make a portal in it. Next node we visit is $c$ with the cost $6$, after that we go to the node $e$ with the cost $6+7=13$ and now we're allowed to make a portal in node $e$. So in this case it's better to teleport back to the node $b$ and go to the node $f$ via $c$ than going back from $e$ to $c$ and then to $f$ with the cost of $10$. Cost to $f$ in this case would be additional $9$ to current $13$ and we have total cost for now of $22$. Now we visit $d$ and finally $g$ with additional cost of $12$ and the final cost would be $34$ with our teleportation system.

With the Steiners algorithm I could use the subgraph to traverse it with portals (still have no effective idea of changing Steiners algorithm to work like this). But the problem is, if we take a look at the graph we see that the best possible subgraph would be made of nodes $b,a,d,g,c$ and $e$. Lets explain why.

We're starting from the node $b$ and traverse all to the node $g$ via nodes $a$ and $d$ with the cost of $16$. We have now two portals: one in $g$ and oen in $b$ so we're allowed to teleport back to $b$ with the cost of $0$. Then we can traverse to the node $e$ via nodes $c$ and $e$ with the cost of $13$ and the final cost would be $29$ which is better than $34$ of the previous case.

This would apply in the case I understood Steiner algorithm correctly and would be pleased if You could confirm it.

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