I HONESTLY SPEND AN ENTIRE DAY LOOKING FOR A GOOD SOLUTION TO PRIMS AND I AM SO HAPPY I FOUND IT. THANK YOU SO MUCH FOR THIS . YOU GOT A NEW SUBSCRIBER
I thought I had my answer at about :20 into the video But I wanted to make sure so I watched until about 1:30 and my answer was confirmed About 1:30 plus the 2 minutes searching google to find your video or "read" 40 pages of death to maybe find the same conclusion I'll pick the 2 and a half minutes with you every time! thanks for this awesome video that cut the BS and got straight to business - no frills no BS new subscriber!
at 5:17 we can not select AH not only because it would create a cycle but because A and H are already discovered before...so there is no need in examining those 2 edges at that moment Great explanation though!! clear and to the point!!! love your videos on Kruskal's and Dijkstra as well!! :D
How would you do it if you can’t backtrack? You went from C to I and the C to F. If you had to continue from the last point you reached, how would you make it efficient?
Does it mean that we have to stop selecting edges until most edges that do not form a cycle will be selected? Then that would be the time the MST is already completed?
So is it safe to say, that this video finds the shortest solution one could travel to get to all locations? Meaning that every node must be reachable, but it's the most effecient way to reach all nodes? I'm strugling with this a bit because you could take the road from A to H and get there much faster that going way around.
I have a question b-c and a-h are basically ties because their distances are the same so why after choosing a-c did we not choose a-h ? But rather chose c-i ?
I have a question, why would i not be able to simply look at every node and assume that the cheapest path for that node was one that would be a part of the minimum spanning tree?
Thank you for the timely response, my goal isnt to find a minimum spanning tree but instead find the sum of the value of all paths that must be taken. so i was wondering if i could instead look at each node in the input independently and simply take the cheapest path for each node?