@@bhaktiarandolwala7681 On the second iteration, we are only considering 2 hops from root node, so node containing 7 is unreachable until we have 3 hops.
Why can't we make all edges positive by adding 2 (or more than 2) to each of the edges and then apply Dijkstra's alg. This would (I think) make the alg work even for negative edged graphs.
Indian lecturers always give me hope in my study
Thanks
very detailed explanation , helped a lot in my project
Thank U Sir
💐
Awesome! Super easy to understand when going by path length for each iteration. Thank you for this explanation.
Thanks
Why in 2nd iteration there is no change in 7?? Plz explaination
@@bhaktiarandolwala7681 On the second iteration, we are only considering 2 hops from root node, so node containing 7 is unreachable until we have 3 hops.
@@phishyp hey thankyou
2nd iteration why is cost of 2 not 5-2-2? path: 1-4-3-2.
Edit: oh i think its because it can only have max 2 edges.
Nice expaination
It helped me in my internal😄
clear explanation
Thank you so much! very helpful
Merci ❤️🙏
Why can't we make all edges positive by adding 2 (or more than 2) to each of the edges and then apply Dijkstra's alg. This would (I think) make the alg work even for negative edged graphs.
you can google counter example for that
Nice thinking
y to add extra overhead?
Single source and bellman comes in greedy algorithm!
not easy to understand, acutually
Aww horible. Explain it slowly, like einsten said you must be able explain to kids or else you don't really understand it yourself
very very refulgent illustration..
worst explained