@bobisabuilder If I recall correctly, no you do not have to start at A, like GreenfordMaths says, it will not affect the result, because the algorithm will always choose the lowest weighted edge connecting the current node to the rest of the graph. This is acutally the idea behind the proof for this algorithm.
@bobisabuilder i'm guessing you're doing this for University or college right? They'll normally tell you to start from written as starting from U{...}. In this case, written as U{A}.
very easy to understand visual representation - good work for the entire collection of
videos on graphs!
theObjectClass 11 months ago
Thank you very much !
Very nice video (good explanation, good visualization)
Can you also post a video for Bellman-Ford and also Walkerson?
SuchFrau 1 year ago
Basically, you can just go from letter to letter and draw a line from the shortest path? (Unless it makes a complete circuit of course).
This is much easier than Dijkstra
linkinpark9sc 1 year ago
@linkinpark9sc Though I see the rule to this one in comparison to Kruskal.
Kruskal allows you to add lines which are possible to connect but for Prims they must connect in order (From shortest number of the current node)?
They both produce practically the same result though?
linkinpark9sc 1 year ago
thx man ,i totally understand ,thax alot for the example
akhil1987jain 1 year ago
thanks again
L30TM 1 year ago
do you have to start from A?
or do you chose where to start?
or do they tell you where to start?
thanks
bobisabuilder 2 years ago
@bobisabuilder To be honest, I can't remember.
At the end of the day, unless you're told where to start, it doesn't affect the result.
GreenfordMaths 2 years ago
@GreenfordMaths
It does affect the result if there are multiple equally good solutions - a different start might prefer a different solution
If the edges are totally sorted by preference, there is always exactly one solution, independent on the starting point.
dvoraj20 1 year ago
Yes. Prim's Algorithm assumes you are given an initial starting point.
linuxarcher 1 year ago
@bobisabuilder If I recall correctly, no you do not have to start at A, like GreenfordMaths says, it will not affect the result, because the algorithm will always choose the lowest weighted edge connecting the current node to the rest of the graph. This is acutally the idea behind the proof for this algorithm.
comochinganconesto 1 year ago
@bobisabuilder You can choose any vertice you want
tekhiun 1 year ago
@bobisabuilder i'm guessing you're doing this for University or college right? They'll normally tell you to start from written as starting from U{...}. In this case, written as U{A}.
skezza 1 year ago
Brilliant! more videos like that. Its so simple and well-said. And You have great voice, its easy to understand if ur non-english guy.
Good job=]
Tuvitob 2 years ago
awesome
sudhanshu2340 2 years ago
Simple and easy to understand! Thank you!
mastterR7 2 years ago
Thank you mate
martinz66 2 years ago
thanks mate, very very useful.
I have an exam today, and thanks to your videos, I'll be OK with Kruskals and Prims :)
THANK YOU.
Ciwan2 2 years ago
Thanx pal
thablackmarkit 2 years ago