This question already has an answer here:

In a forward search algorithm, what do you do if both items are equal?

The forward search algorithm stems off of Dijkstra algorithm

Forward Search Algorithm

```
1. Initialize the Confirmed list with an entry for myself; this entry has
a cost of 0.
2. For the node just added to the Confirmed list in the previous step,
call it node Next and select its LSP.
3. For each neighbor (Neighbor) of Next, calculate the cost (Cost) to
reach this Neighbor as the sum of the cost from myself to Next and
from Next to Neighbor.
(a) If Neighbor is currently on neither the Confirmed nor the
Tentative list, then add (Neighbor, Cost, NextHop) to the
Tentative list, where NextHop is the direction I go to reach Next.
(b) If Neighbor is currently on the Tentative list, and the Cost is less
than the currently listed cost for Neighbor, then replace the
current entry with (Neighbor, Cost, NextHop), where NextHop
is the direction I go to reach Next.
4. If the Tentative list is empty, stop. Otherwise, pick the entry from
the Tentative list with the lowest cost, move it to the Confirmed list,
and return to step 2.
```

At the last step, I am unsure what to do when there is an occurrence with 2 entries with the same lowest cost. Do I move both to the Confirmed list? or do I choose the entry that stayed in the tentative list the longest?

Answer:

If there are two entries with the same lowest cost, you can choose either - it doesn't matter. For example, if your tentative list looks like this:

```
10 4 9 7 7 3 2 11 5 2
```

and you were finding the minimum cost by using a sorted list, it would look like this:

```
2 2 3 4 5 7 7 9 10 11
```

But it wouldn't matter whether you pick the first 2 or the second 2.

If you choose the "wrong" one, and it turns out that the one chosen leads to a longer path, the algorithm will correct itself as it goes on.

Edit:

This answer is actually super-similar to one in another related question which may well be a dupe. In the spirit of this meta-post, I won't delete the answer, but thought I'd add more detail so it's not just the same information again.

Obviously, the concern here is that by selecting one node over another, you "miss out" on a path that is still there. However, since the forward search algorithm you're quoting is derived from Dijkstra's, it's subject to the same rules. One of the concerns you might have about Dijkstra's is, what happens if I choose the lowest cost immediately, but there's actually a different path that is ultimately, if not immediately shorter?

For example, you might have a node of cost 1 and a node of cost 2. You choose the node of cost 1, but actually, in the long run, it's cheaper to choose the node of cost 2.

It's the same question here really, except both nodes have the same cost. However, if they all have positive weights, you can prove that Dijkstra's algorithm will always find the shortest path. There's a proof of that here.

Therefore, it's fine to select either node of minimum weight. The algorithm gets there eventually.

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