f-k has weight 10. j-g has weight 19. The weight text is oriented with the edge.

It is true that every vertex cover is a dominating set...but a graph can have a dominating set that is smaller than the smallest door takes cover. So, if we ask the question "is there a dominating set of size 1" in a triangle, the answer is yes, any vertex will do. But the smallest vertex has size 2, and knowing that there is a size 2 vertex cover doesn't help us to know the size of the smallest dominating set. This example gets much worse when you scale it: a clique with n vertices has a dominating set of size 1, but its smallest vertex cover is size n -1.

I think it just makes for cleaner code. Suppose that you are trying to find a minimum path length, and the edge weight represents a distance, with most distances being relatively small numbers. If you plug in a dummy value like the maximum possible integer (which might be 2 billion, but I will call infinity), then you can say "if a < b" instead of "if (a !=null && b != null && a < b)".

It seems reasonable to not have that analysis be part of this video, but maybe I should link to an explanation. But, because log (product) = sum (log), taking the top half of the factorial terms (each at least n/2), we can see lg n! Is at least (n/2) (lg (n/2)) = (n/2) ((lg n) - 1), where the lower bound is easier to see.

@@AlgorithmswithAttitude I see :) In my opinion, if the explanation fits into 1-2 lines, why not include it in the slideshow, since the result may not be clear to everyone? On the other hand, one could also argue that since the proof isn't that hard, it could be left as an exercise.

substitution method is also called inductive method

@@aakritipangeni8791 but my university substitution method is called where we iterate the t(n-1) ,t(n-2)... And put these to get nth t(n) ...they call iterative method substitution here ....like they are substituting t(n-1) in t(n) then t(n-2) and so on

@@vini_jr_07_20 yes that's called backward substitution method or iterative method too.

@@aakritipangeni8791 where are you from and from which college?

​@@aakritipangeni8791 From which method should i solve a recurrence if they just mention substitution

Definitions are kind of boring, but without a shared vocabulary, it is hard to discuss the topic. Also, I haven't actually monetized these videos...and if I did, they might only buy me a burger per year anyway.

Is there any specific questions that I could answer?

Basically, yes, though to make it precise, the wording is a bit tricky. If a set of vertices in a graph forms a clique, that set of vertices is an independent set in the complement graph. And, if a set of vertices is a vertex cover in a graph, the complement of those vertices is an independent set in the same graph.

No, the problem could be so hard that it is not NP.

A problem is NP-Complete if it is NP hard & has a non-deterministic polynomial time algorithm.

Wait, did I say it wasn't worth it. I don't think I did. I didn't read it, so that doesn't seem like something I would say. But I do feel like you can get a pretty good understanding of quicksort in under 300 pages.

It's the hair, right? I want to redo the whole playlist, but haven't made the time. Hopefully within the next few months...

@@AlgorithmswithAttitude you are singlehandedly saving my average (tentative). thank you for all this.

It Still takes additional study. You have to physically work it out. Just to have an Excellent teacher. Yes. Thankyou

That section describes how, for everything EXCEPT back edges, for edge u->v, v finishes first. Because DFS in a Directed ACYCLIC Graph will have no back edges, we don't need to worry about the fact that for back edge u->v, u would return first.