Prim Graph

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Category:

Education

Tags:

• prim
• algorithm
• d1
• maths

[bobisabuilder]

do you have to start from A?

or do you chose where to start?

or do they tell you where to start?

thanks

[GreenfordMaths]

@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.

[theObjectClass]

very easy to understand visual representation - good work for the entire collection of

videos on graphs!

[dvoraj20]

@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.

[SuchFrau]

Thank you very much !

Very nice video (good explanation, good visualization)

Can you also post a video for Bellman-Ford and also Walkerson?

[skezza]

@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}.

[linkinpark9sc]

@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]

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

[akhil1987jain]

thx man ,i totally understand ,thax alot for the example

[tekhiun]

@bobisabuilder You can choose any vertice  you want

[comochinganconesto]

@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.

[L30TM]

thanks again