A brief explanation of the count to infinity problem, with an easy to follow example of the mechanics. Video created using the Manim engine (see 3blue1brown channel). ===== Music: www.newgrounds.... Hourglass, by Noisysundae
Very Very and Very Good and straight Forward without any diversions...very well explained...i know that i used "Very" so many times..but you deserve that man!!
Please make more videos like these. I can bet your channel will grow fast. Nice and simple explanation. Also, the video style using Manim and the music seem to be really soothing. Keep up good work!
HEY SIR thank you for this video this was very simple and straightforward. Maybe in the next video you can talk about the solutions for this problem such as split horizon and poison reverse. love you
at 3:04 when router "B" sends a table to router "A" saying that it would help router "A" to reach router "C" by cost 4 in that case since router "A" (even though the connection is lost) assumes it already has a path c with cost 3(thats what the table of router A tells) which is less that than one offered by router "B" why will it update the entry of path A-->C to 5 . since it already know a better path (which however it is unaware that such a path might not exist though) any body please explain me this ?
This makes sense if we assume the good news travels faster than the bad ones, but why is that bad news travels slower?? okay Is it like both the possibilities are possible if the contrary one happens then all is good. but if the above mention one happens the we do run into a slow convergence problem?
Can this problem be overcome if router B checks how A reaches to C and if it realizes that if A reaches to C via B (itself), it doesn't update the entry and marks it with "- -", then the count to infinity problem won't occur?
That would definitely work. But keep in mind that if you scale this problem, say that you have 5000 routers, this would mean tables with a lot more information. And what if a package is already sent, but in between sending and reaching one of the connections breaks down?
@@thenerds9134 i went to search somemore. It seems possible and this is what "split horizon" method does. But if there is a loop with 3 or more nodes, the method womt work
But after link breaks b will first send its vector to A because B's routing table is changed.but you showed B is recived table from A and according to that B changes
I don't get it. Why doesn't C just send it's table to all neighbors after it updates its C entry? Seems like a trivial problem to solve, which must mean I'm misunderstanding something
So - where's the problem with that? Neither A nor B can reach C, but think the other one can. Therefore, they keep incrementing the cost. Since they can't reach C anyway, why does that matter? How is a high cost value for an impossible connection a problem?
The fun fact is each router sends its distance vector only so they don't know they are taking any node to the lost node via them only and that is where the problem arises.
