Confirmation of Dijsktra application explanation

Question

Please can someone confirm that my description of the application of Dijkstra's algorithm is correct for this graph?

    1   3
 A -- B -- C
  \      /
 2 \    / 1
    \  /
     D

Step   Current Node

                       B        C        D
1      A               1(A)     -        2(A)   
2      B               1(A)     4(B)     2(A)   
3      D               1(A)     3(D)     2(A)   
4      C               1(A)     3(D)     2(A)   

Description:

1. Tentative distances to B and D determined. C not yet visible.
2. Shortest distance from previous step taken. No new information on shortest path to B. Tentative distance to C now known. No new information on D.
3. We choose to backtrack to A, and then move to D because the distance to D (2) is smaller than that to C (4). We update the tentative distance to C because via D it is shorter than via B.
4. We choose to move to C directly from D because it is shorter (3) than backtracking and going via A (4). Moving to C gives us no new shorter distances. All nodes visited, so end.

Answer 1

It's more traditional to describe dijkstra with the open and closed sets:

You start with an empty closed set and an open set containing $ \{ (A,0) \}$

You select the node with the lowest cost and remove it from the open set, then you expand the neighbours and add them to the open set $ \{ \{B,1\} \{D,2\} \}$ or update the elements in the open set and Add the selected node to the closed set

step1:
open = $ \{ (B,1, A), (D,2, A) \}$
closed = $\{(A)\}$

step2:
open = $ \{ (C,4, B), (D,2, A) \}$
closed = $\{(A),(B,A)\}$

step3:
open = $ \{ (C,3, D) \}$
closed = $\{(A),(B,A),(D,A)\}$

There is no actual sense of "backtracking" because you are considering all the open nodes in parallel.

Many other search algorithms work in the same way with the only difference in how a node is selected from the open set. Breadth-first takes the oldest node in the set, depth-first takes the youngest, Dijkstra takes the node with the lowest cost, A* takes the node with the lowest cost+heuristic.