Talk:Dijkstra's algorithm

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Any video of animation available?

Latest comment: 9 years ago1 comment1 person in discussion

That animation is killing me. I so much want to carefully look at it's calculations, but they flash up there for just a half second. It is maddening.

That animation does not work as described. The calculated cost as shown in the animation is 20, while the correct one based on the described procedure is 28 (or 26 if bidirectional [1]) [1]: http://rosettacode.org/wiki/Dijkstra's_algorithm#Java — Preceding unsigned comment added by Ckorakidis (talk • contribs) 19:31, 17 October 2015 (UTC)Reply

Terminological Dissonance?

The terminology in the caption for the animation is not in full agreement with the description of the algorithm: (a) the algorithm description does not mention a "tentative set" - it does mention "tentative distances", and it mentions an "unvisited set"; (b) the algorithm description mentions no "heuristic", but the caption to the animation says that the algorithm uses a "heuristic that is identically zero". At first it seemed to me that the caption of the animation intended "tentative set" to mean that which is meant by "unvisited set" in the algorithm description, but as I thought about it, I am unsure; on the other hand, it seems clear to me that even if that is not the intent of the caption, it does follow that the "tentative set" is identical to the "unvisited set" in that example. Matt Insall 13:15, 5 August 2017 (UTC) — Preceding unsigned comment added by Espresso-hound (talk • contribs)

Description section of artice

Latest comment: 9 years ago1 comment1 person in discussion

In the second paragraph after the note, it says "This is done by determining the sum of the distance between an unvisited intersection and the value of the current intersection". There should be two intersections following between, but there is only one intersection since value is not an intersection.

The phrase, "This is done by determining the sum of the distance between an unvisited intersection and the value of the current intersection" should be changed to read, "This is done by adding the value of the current intersection to the distance between the current intersection and an unvisited intersection". RHB100 (talk) 00:34, 12 January 2016 (UTC)Reply

Pronunciation of the inventor's name

Latest comment: 9 years ago1 comment1 person in discussion

Could we add the correct phonetical expression for the inventors name? Anyone researching for this topic will have trouble to pronounce the name correctly. — Preceding unsigned comment added by 2A02:8109:80C0:7EC:E8F6:F2B8:8896:A526 (talk) 10:33, 22 April 2016 (UTC)Reply

Description: "Pencil arrows" vs. "parents"

Latest comment: 8 years ago1 comment1 person in discussion

A paragraph in the Description advises "in pencil, mark the road with an arrow pointing to the relabeled intersection." The next paragraph talks about a visited node's "parent." I think these are talking about the same thing, but it's not very clear. Perhaps better wording might be "by following the nodes' parents (that is, traversing the arrows backward)", or perhaps in the first paragraph, "mark the road with an arrow pointing to the relabeled intersection (from 'parent' to 'child')" --Jackrepenning (talk) 18:54, 24 September 2016 (UTC)Reply

Complexity with Fibonacci heaps

Latest comment: 7 years ago1 comment1 person in discussion

A few users have been changing the complexity with Fibonacci heaps from ${\displaystyle O(|E|+|V|\log |V|)}$  to ${\displaystyle O((|E|+|V|)\log |V|)}$  without any justification. The former appears to be correct based on the reasoning in the article:

For any implementation of the vertex set Q, the running time is in

${\displaystyle O(|E|\cdot T_{\mathrm {dk} }+|V|\cdot T_{\mathrm {em} })}$ ,

where ${\displaystyle T_{\mathrm {dk} }}$  and ${\displaystyle T_{\mathrm {em} }}$  are the complexities of the decrease-key and extract-minimum operations in Q, respectively.

since ${\displaystyle T_{\mathrm {dk} }=\Theta (1)}$  for Fibonacci heaps. Hence I am reverting the complexity with Fibonacci heaps back to ${\displaystyle O(|E|+|V|\log |V|)}$ . -- Paddu (talk) 15:44, 27 December 2017 (UTC)Reply