Undergrad Is A→E→D→F a valid solution for Dijkstra's algorithm?

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The solution A→E→D→F is valid for Dijkstra's algorithm, as both A→B→D and A→E→D have equal distances of 9 when reaching point D. The choice of A→B→D in the video is primarily due to it being evaluated first, rather than any inherent superiority. The alternative path A→E→D does not provide a better cost, which is why it may not have been highlighted. The discussion emphasizes that both paths are acceptable solutions within the algorithm's framework. Ultimately, the decision to favor one path over the other does not affect the correctness of the algorithm's outcome.
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In this video the solution is: A→B→D→F
but the paths A→B→D and A→E→D both have a distance of 9, so when you get to D you can either have [B,9] or [E,9] where the first entry is the predecessor and the second entry is the 'cost' on that path so far, or here the distance.
is the solution A→E→D→F also correct or is there some reason that he chose the path A→B→D→F?
 
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Erenjaeger said:


In this video the solution is: A→B→D→F
but the paths A→B→D and A→E→D both have a distance of 9, so when you get to D you can either have [B,9] or [E,9] where the first entry is the predecessor and the second entry is the 'cost' on that path so far, or here the distance.
is the solution A→E→D→F also correct or is there some reason that he chose the path A→B→D→F?


Hi Erenjaeger! :)

Yes, A→E→D→F is also correct.
The only reason to pick A→B→D is just because it was evaluated earlier, and A→E→D does not improve on it.
It seems that this decision was skipped in the explanation in the video.
 
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