Lucio López Lecube Glad it helped! Not really sure on the algorithm, but the premise can be found on the Articulation Points wiki. en.wikipedia.org/wiki/Biconnected_component
hi, thanks for the video, however there is something i don't get, you said that the root is an articulation point iff it has more than one child. But what if b and d were connected in your first example, a is no more an articulation point even it has more than one child. I hope you see my question i spent a lot of my time looking for an answer ...
Wow . That's really great explanation .
Very helpful, thank you! The practical examples make it easier to grasp the algorithm.
Thanks for the video. Is directed graph treated as an undirected graph in depth first search analysis?
Very helpful and wonderfully explained.
hello Michael
thank you so much for the amazing explanation
can we use this method with weighed graph
Your video was helpful to me. Thank you. That said, please get a better mic for future videos, the background noise is terrible.
Very informative. Thanks!
Hi michael! Thanks for the explanation!! very well.
Do you have the name of the algorithm you used?
Lucio López Lecube Glad it helped! Not really sure on the algorithm, but the premise can be found on the Articulation Points wiki. en.wikipedia.org/wiki/Biconnected_component
+Michael Mroczka
the name of this algorithm is Tarjan's algorithm
In example 2, did you miss vertex D, which is also an articulation point?
great video. Thanks!
Really good explanation! Thanks! ;)
Thanks. In Question 2 you explained, is B a child of G?
Thanks Michael.
hi, thanks for the video, however there is something i don't get, you said that the root is an articulation point iff it has more than one child.
But what if b and d were connected in your first example, a is no more an articulation point even it has more than one child.
I hope you see my question i spent a lot of my time looking for an answer ...
What if our root has more than one child buts that root belongs to a cycle.
Example a square obviously formed with 4 vertex.
last answer plz :)
+Kostas Giannakoulas I've found the following solution: AP = c, e, f
+Ewout Merckx b as well, no?
no, i agree with Ewout. It's C, E, F