4:25. It does matter! The smaller group becomes part of the greater group, otherwise the worst case runtime would be different.

@TheDemeaN

4 жыл бұрын

yas it does matter but I think he was saying in the context of find the mst in the graph, like imagine an exam exercise where you need to draw the mst

@JanacMeena

3 жыл бұрын

Merging a smaller group into a larger group would require fewer operations, but it does not affect the worst-case runtime. Please let me know if I'm wrong!

@lordquaggan

2 жыл бұрын

@@JanacMeena It actually does! Weighted union keeps tree depth below log(n), even in the worst case (can prove by induction), so find is O(log(n)), while without weighted union, tree depth can be n in the worst case, leading to O(n) find. Applied to Kruskal's algorithm, this is the difference between an overall O(mlog(n)) and O(mn) complexity.

@SAJID-zs2gf

8 ай бұрын

@@lordquaggan during each find operation you can update the parent of children and attach children directly to it's top-most parent (root node), then in that case tree-depth won't reach O(n), since at every call to find(), the children node will be attached to root node of the whole group.

Such a neat and crisp explanation. People like you are making our lives simpler. Thanks a bunch !!! Keep it up.

I saw your Prim's explanation using PriorityQueue first. Kruskal's using PriorityQueue and Union Find was a piece of cake thence. Your explanations are great and so is your code. Thanks a lot

The color grouping is really intuitive. Thank you for the helpful video.

Great video! Finally I understood this algorithm. I had been trying to understand it for days until I watched your video

You have really worked hard in designing these colourful ppts/video. Thank you very much.

I've been trying to understand it for a long time. finally understood. thanks a lot!

@WilliamFiset-videos

7 жыл бұрын

That's really encouraging to hear! :)

Your channel is so underrated. Love it!

This was absolutely freaking great. I love this stuff, u explained it so well. Thank you for reminding me why I love what I do. Keep up the great work!

Great illustration! This is a great series!

Your explanation made it super simple... great work

You made my life so much easier!

i love u u explained it better than my textbook and professor regarding applications of mfset so I understand disjoint set better!

Fantastic simulation my friend - keep up

Sehr gut erklärt. Vielen Dank :)

This is amazing! Thank you so much for that, really!

Thank you!!!!!By far the best explanation!!!

Thank you, this is very helpful!

Thanks for the video, love your channel! Why did you stop after the pair "B to C"? how do I know when to stop iterating over the list on left?

very adaptive to my brain. thanks !

Excellent explaination! :)

This is brilliant, thank you!

greatexplanation ,you made it easy

You rock dude, thanks for the video

Great job, keep it up!

I swear you are the king of graph theory

nicely explained. thanks.

nice explanation, thanks!

Thanks a lot William 😁😁

Well presented.

Great explanation

nice explanation thanks

Very interesting, thanks ^^

This Algorithm is crazy!! I find beautiful how people come with these solutions. This is a beautiful application of Data Structures to simplify hard problems to solve. Just waw! Oh and Thanks William for bringing the quality content as usual!

@chaitanyak.n.4768

3 ай бұрын

Its just greedy search.

Just found your channel while studying for my algorithms exam. Cannot thank you enough for making these great videos! You are f@#$!ing awesome!

@WilliamFiset-videos

7 жыл бұрын

GL on your exam!

how do you assign the edge weights?

Is this algorithm work when graph is directed?

Great vid

hello，Willliam can you teach me how to do the ppt, I am very interested

I am your fan, bro!

Can you please help me understand how are giving weight to a junction.

mind blowing

very good

Why B to C instead of G to I or H to C is it because of node size?

Basically, sort the edge then run Union Find

i don't understand the logic behind each number being assigned to which path. Any kind of help is aprecciated.

at 5:20 you say we've found the minimum spanning tree, but how did you know that they were all connected at the point? Does Kruskals algorithm keep track of group size at each root node?

@kylewilliams9429

8 ай бұрын

The would all have the same 'root' parent. That's how you know the algorithm is complete. When every vertice has the same parent (belongs to the same 'group')

@gradientO

4 ай бұрын

You can keep track of group count. At the start, each node is a group, so it'll be the node count. Decrease the count on union. Stop immediately when the group size is one instead of going through remaining edges (which won't be added anyway)

Great illustration. Just wanted to add that union-find is an algorithm and not a data structure. Graph is a data structure.

@riteshrastogi5388

4 жыл бұрын

No its not. Read the first line ( written in black color ) of this article : cp-algorithms.com/data_structures/disjoint_set_union.html

Thanks

are the edge weights arbitrarily assigned?

@WilliamFiset-videos

4 жыл бұрын

A weighted graph will come with predetermined edge weights.

I'm still not sure where "size" comes into play

@WilliamFiset-videos

6 жыл бұрын

Whether you merge the smaller group into the larger group or vice versa doesn't matter. It's just a good heuristic to use for efficiency of the Union find if you merge the smaller group into the larger one.

ehrenmann

done

!!!!!!!!!! WAIT , The title of your video is very misleading . It should say something like "Application of Union Find data structure (Kruskal's Algorithm)" . Cause when you say adding "C and J" will create a cycle ,that is where we need to know how union Find Algo works . else video was awesome

@williamvanderscheer4327

5 жыл бұрын

I know this is obviously very late, but for others reading this, just in case... It has nothing to do with the union find algorithm, but with Kruskal's algorithm, which tries to find a minimum spanning tree. A minimum spanning tree is the minimum set of weighted edges needed to connect all nodes. If we select edges that form cycles, then we have by default not found a minimum spanning tree. It doesn't even have anything to do with a cycle itself, it's simply because since that node was already "spanned" before (aka already in the group) we did not have to actually use that edge to visit it, and we needlessly incurred the cost by travelling over that vertex, which violates the invariant of the minimum spanning tree algorithm. A scenario that happens to be equivalent to creating cycles. (because we are reaching a node in the group, from a node that is also in that group => cycle)

All these "Algorithms", are recipes that a six year old might use when introduced to such problems with no prior knowledge. Very simple !

Thanks