Find transitive closure of the given graph. It is the Reachability matrix.
Пікірлер: 60
@AnnoyedUnicorn5 жыл бұрын
I'm a French student the day before my exam, and thank you very much for this video ;)
@coconut10103 жыл бұрын
Pareil c est fou, ya peu de contenu en français sur ça
@AnnoyedUnicorn3 жыл бұрын
Oh I failed it btw, but still ^^
@aniketbhoite71683 жыл бұрын
@@AnnoyedUnicorn ohhh you still came back after a year .....😅
@subhranil_mukherjee5 жыл бұрын
sir i don't think in a directed graph each vertex is obviously reachable to itself. i think what it means is "starting from a vertex v, can we get back to v following any path in the graph?". Let me know your thoughts.
@sanjeewanigunasekara43045 жыл бұрын
I think you are correct.
@AnnoyedUnicorn5 жыл бұрын
With the methode I learned with my math professor, we addition to the identity Matrix, so obviously, he is right ;)
@matthewstruble88814 жыл бұрын
of course not, but that's not what the graph is saying. think, if we knew that [2,1] is a path, then of course [2,1] -> [1,1] is also a valid way to reach node 1. if the matrix is not made this way the algorithm will fail
@missakasms Жыл бұрын
Yes u are correct i guess
@cicakmuhamed Жыл бұрын
You're right. Otherwise, the diagonal from left-to-right would always be filled with 1's in the transitive closure and that would not convey much useful information. Also, in the video, you cannot reach A starting from A (there exists no path, at 1:10) but he marked it as 1 in the transitive closure, which means he actually meant that "it is obvious every vertex can reach itself".
@craigruchman70073 жыл бұрын
My text book mentions transitive closure but did not give an example. Thanks so much for the clear example. Question: Closure is when we perform said operations on ALL vertices without adding any new elements to our set. Yes? Also, the edges don’t alow paths in both directions. So this makes it non-commutative. If they were bi-directional, it would be commutative - yes?
@Tyokok5 жыл бұрын
thanks for the video! quick Question: is diagonal always 1? meaning any vertex can reach itself? it doesn't need a self-circle edge? i thought without self-circle edge self reach=0.
@vibranteye43693 жыл бұрын
I learned it this way as well...
@Tyokok3 жыл бұрын
@@vibranteye4369 what you mean? can you make it clear please? So it should 0 or 1, and the meaning of each? Thanks a lot!
@manognadevineni89012 жыл бұрын
Noo here if we start from say a.. we will get back to a ; So here we'll take diagonal as 1.. i.e. if we do have a closed path for any vertex we can take its diagonal one as 1 or else 0 ..
@Tyokok2 жыл бұрын
@@manognadevineni8901 so you saying regardless we have self-circle edge, we always set diagonal to 1, like convention ?
@manognadevineni89012 жыл бұрын
@@Tyokok if a node has a self circle then its quite good it will be having 1 in its diagonal...but here look into the question if we start from 1 we will get back to 1.. that means a path exsits .. so we will take it as 1 .
@jongmagee4 жыл бұрын
Big help again, thank you!
@dulamshiva53403 жыл бұрын
sir can you please explain how A will reach to A again in the adjacent matrix...i feel their is a flaw in solution
@revanth64763 жыл бұрын
we can also use bfs right..?
@pg92275 жыл бұрын
sir your videos on graph theory is good and no doubt everybody can easily understand. sir i am requesting you please make videos on graph coloring ,vertex coloring,np completeness and linear programming in graph theory. I am eagerly waiting for response. Thanking you sir
@mayurwarpe27635 жыл бұрын
Great sir...you have clear my doubt Please make video on DFS algorithm
@joepaul94755 жыл бұрын
thank you. very useful
@tomsonthomas76583 жыл бұрын
In a digraph D, prove that a vertex y is reachable from a vertex x if and only if D contains an (x, y)− path. 2. Show that a digraph D is strong if every vertex of V (D) is reachable from every vertex other vertex of V (D). 3. Show that a digraph D is strong if and only if it has a closed Hamilton walk. 4. Develop a strong digraph D = (V, A) and extract possible separator from D.
@abdultamim1428 Жыл бұрын
Thank you, sir; it was great. Please send me some material about(Transit function and betweenness) if you have any!
@SOURABHGOLCHHABCE5 жыл бұрын
sir can u upload more questions on dynamic programming for interview purpose and sir upload daily one algorithm please sir i request you you should daily upload one algorithm please sir
@madanagopal55853 жыл бұрын
Today my xm this is video is very helpful ❤️
@balasrinivas75175 жыл бұрын
please teach particle swarm algorithm
@tekbssync57273 жыл бұрын
please provide the code for this Qs.
@Shree_krishna7720 күн бұрын
Thank you sir ❤❤❤❤
@hashcodez757 Жыл бұрын
thanks alot sir!!
@arifulislamhabib29372 жыл бұрын
Thank You.
@JananiAnbarasan5 жыл бұрын
Sir, please explain TSP!
@rohansodhi471320 күн бұрын
There is no path from A to A So how did you write the path from A to A?
@gangireddyakula63954 жыл бұрын
Ossm sir I'm easily understand to see ur videos 🥰TQ u so much sir
@tayebdz85812 ай бұрын
thank you
@ManiTeluguGamer232 жыл бұрын
Tqu so much sir
@mt03adventures5 жыл бұрын
Ty
@Rexxxxaaa122 жыл бұрын
i think the diagonals should be 0 as there is no selfloop
@masakkali99964 жыл бұрын
Is the transitive closure of this graph is correct??
@pavlos.konstantinidis Жыл бұрын
nope
@tazelhossan28384 жыл бұрын
Wrong teaching. There is no path for node a. Please check it Again.
@bwbs7410 Жыл бұрын
this is wrong. First, this isn't how the algorithm to obtain the transitive matrix is preformed. Second, a path from A to A signifies a looped edge. Thus, all values on the main diagonal [top left corner to bottom right corner] should all be zero because this graph has no looping edges.
@Anime_Animation_Illustration5 ай бұрын
You are talking about Adjacent matrix 😅
@jethanpaul8184 ай бұрын
you're confusing between adjacent matrix representation of a graph and transitive closure , get a book cormen to be specific and see the algorithm on transitive closure maybe you can understand that he is correct.
@samudralashanmukrao21854 жыл бұрын
How can reach form a to a 🤔
@shubhamshejaval85263 жыл бұрын
Same doubt bro
@Anime_Animation_Illustration5 ай бұрын
Because it is same node re
@tauchainagoras60442 жыл бұрын
check this out.. shortest code ever
@setYourHandleHttp4044 жыл бұрын
7 minutes for only describing a matrix!!! I thought you are also describing the Floyd-Warshall algorithm.
@venkyg6745 Жыл бұрын
A to C there is no path ..
@52_sopansharma525 жыл бұрын
galat padha rh ho
@nitinsingh96045 жыл бұрын
Jake khud padh le pahele
@gokuljs8375 жыл бұрын
Teach properly
@ossamabinhassanietstudent34184 жыл бұрын
Abey chutiye.... Loop kahan dikhaaya gaya hai Harr vertex Pai