Let's continue the habit of commenting “understood” if you got the entire video. Please give it a like too, you don't 😞 Do follow me on Instagram: striver_79
Thank u. Note: The Dijkstra's algorithm implemented in G-32 can handle any negative edges graph EXCEPT the following cases: 1- Directed graph having any negative cycle (cycle with sum < 0) 2- Undirected graph having any negative edge because the edge in undirected graph is by itself a cycle
Your thought is right, my doubt is why not follow Dijkstras algorithm implemented in G-32 for directed graphs with no negative cycles. The time complexity is even less than Bellmanford algorithm. What is the use of Bellman ford algorithm?
Thanks Striver. Trust me, even in my paid course, they just simply explained the working of Dijkstra and Bellman without going into such detail. U r the best teacher.
@@dharmvir2330 So why did you take the paid course? Is there any advantage you can think of that the paid course is giving better than Striver(just curious to know)?
Improve Time Complexity by exponential with just minor observation: Put int count =0; After the first loop & increment the value of count by 1 when the dist array will get updated and at the end of the second loop, if the value of count is not increased then directly return dist array. if no update in dist array happened that means we already calculate dist array, no need to do further iteration, In the worst case it does not have any impact, but for rest, it decreases TC a lot. It Reduce the number of iteration in Best & Average cases.
Is this case even possible? in a graph like this : a-> b-> c . if we have found smaller distance for a->b , we will surely find shorter distance for b->c in next iteration. Let me know if you think differently.
Things i figured out : 1. If you know a path to reach a node you can figure out the paths to reach nodes adjacent to it. 2. why repeat n-1 times, every time you run through the edges you find a path to reach the node and the cost to reach it, if its better you update, incase the cost doesnt update you have been able to exhaust the minimum cost path.
2 din baad DAA ki paper hai and Bellman Ford aayega exam me, soch hi rha tha ki striver agr next video yhi upload kr de fir to mjaa hi aa jaye and aaj subh dekha BOOM!!
Question: why do we need N-1 iterations? Reason: Because we first of all set the source distance out ot all the N edges, now we have N-1 edges, to fill their distances w.r.t source, we need at max N-1 iterations for each Node.
We can also fit the negetive cycle check in the for loop with extending its range to V and checking if its the Vth iteration in relaxtion without writing repeated code. Also the best and worst cases can be improved by keeping a count of how many relaxations done in each iteration which signifies that if at any point no relaxation can be done then no further iteration is required bcoz there will be no change in distances further. Here's how its done: vector bellman_ford(int V, vector& edges, int S) { vector dist(V, 1e8); dist[S] = 0; for(int i = 1; i
one thing should also be mentioned that if a graph on N nodes have cycle, then their is path exist having more than N - 1 edges from first to last node. By the way best explanation on TH-cam👌
Dijsktra's code which striver has taught works for negative edges , it just not works for Negative edge cycles. So all in all , it would work for DAG with positive / negative weights.
Of course, because Dijkstra doesn't work for negative edges in UNdirected grapsh: What if you have 0 - 1 with -1 weight? It will give TLE. But in directed, 0->1 with -1, 1 can never go back to 0 so the loop will not work.
23:34 - If someone is wondering like me as to why the edges are passed by reference using & If you don't use a reference, the function would create a copy of the edges vector every time the function is called, which can be very costly in terms of memory and time, especially for large graphs
I just wanted to play around with dijkstra and I know this in not an optimal solution and thought of doing the negative cycle part with dijsktra only. My thought process is very simple with djkstra also you could solve the problem the only issue was negative cycle. Now a cycle in itself is not an issue but a cycle with reducing distance does. According to the condition of dijkstra it would enter the if block if we get a decreasing distance and this will only happen in negative cycle or in linear path. To check that we can store paths for each node. Next time we visit that node we check do we already have the node itself in the path that means we are in a cycle or else there is a different path that led us to the node. Code: vector bellman_ford(int V, vector& edges, int S) { // Code here vector adj(V); for(int i=0;i
I guess for the question why we need to do n-1 relaxations to each node is that ,, suppose we have 3 nodes directly connected a node and we have relaxed those nodes,, now in typical dijkstra algo,, the node with smallest value of relaxation will never get relaxed again, but if there are negative weights,, then there is a chance that the the node with smallest value of relaxation will get relaxed again. Since each node at max gets connected to n-1 nodes, so thus n-1 relaxations.
Presense of negative edge does not always result incorrect result ...that could be easily proved in the second example if there is a path say from 4 to 3 with -1 edge weight ....but if the negative edge cycle change the previous path length which we got after n-1 iteration then it cause problem.
If we will take nodes in a increasing order then that time I think we can find the distance in 2-3 iteration as well we don't need to do n-1 iterations
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Let's continue the habit of commenting “understood” if you got the entire video. Please give it a like too, you don't 😞
Do follow me on Instagram: striver_79
understood
Understood
Understood
Striver you are amazing, you need to worry, your supporters will keep on increasing, infact next gen of engineers will be brought up "only by you" 😄✨
understood
Thank u.
Note: The Dijkstra's algorithm implemented in G-32 can handle any negative edges graph EXCEPT the following cases:
1- Directed graph having any negative cycle (cycle with sum < 0)
2- Undirected graph having any negative edge because the edge in undirected graph is by itself a cycle
What if graph is disconnected and negative cycle isnt reachable from source then your first point is false.
Your thought is right, my doubt is why not follow Dijkstras algorithm implemented in G-32 for directed graphs with no negative cycles. The time complexity is even less than Bellmanford algorithm. What is the use of Bellman ford algorithm?
@@Mercer80 I think we can apply a quick visited check? Like we did in the first lectures?
Thanks Striver. Trust me, even in my paid course, they just simply explained the working of Dijkstra and Bellman without going into such detail. U r the best teacher.
True , i am also here from a paid course , Someone believes it or not These explanations are way better than in a paid course.
@@dharmvir2330 So why did you take the paid course? Is there any advantage you can think of that the paid course is giving better than Striver(just curious to know)?
Improve Time Complexity by exponential with just minor observation:
Put int count =0; After the first loop & increment the value of count by 1 when the dist array will get updated and at the end of the second loop, if the value of count is not increased then directly return dist array. if no update in dist array happened that means we already calculate dist array, no need to do further iteration, In the worst case it does not have any impact, but for rest, it decreases TC a lot. It Reduce the number of iteration in Best & Average cases.
this should be pinned
I did the same thing.
exactly!
Is this case even possible? in a graph like this : a-> b-> c . if we have found smaller distance for a->b , we will surely find shorter distance for b->c in next iteration. Let me know if you think differently.
@@nalingoyal881 Bhai you would find smoller distance for c in the same iteration.
Thanks
Bhai mrko b bej dai
This guy got superpower. Can be cast as a Marvel hero "The Explainer" .
And we all guys as watchers 😂
nice one
@@samarthagarwal6965 already taken by the "Watcher"
I think Striver is already a good enough superhero name
Wahi yr his teachings skills with so much of patiencee
You explained it really well, If I was to trace this myself I would have sat for an entire day.
OMG! too much hype about bellman ford algorithm and this is what it is? WOW! you made it so simple. Thanks a ton striver!
Things i figured out :
1. If you know a path to reach a node you can figure out the paths to reach nodes adjacent to it.
2. why repeat n-1 times, every time you run through the edges you find a path to reach the node and the cost to reach it, if its better you update, incase the cost doesnt update you have been able to exhaust the minimum cost path.
when u said "yes u r correct", my confidence became infinity❤
2 din baad DAA ki paper hai and Bellman Ford aayega exam me, soch hi rha tha ki striver agr next video yhi upload kr de fir to mjaa hi aa jaye and aaj subh dekha BOOM!!
Question: why do we need N-1 iterations?
Reason: Because we first of all set the source distance out ot all the N edges, now we have N-1 edges, to fill their distances w.r.t source, we need at max N-1 iterations for each Node.
understood, I dont know why i was afraid of this algo in explaining. You made it a butter.
Have watched multiple videos, but got the understanding from this explaination. Thanks
Shortly, if N nodes, the node at the farthest level will be at
Bhaiya nind se uth ke breakfast mai apka video khaneka Maza hi kuch Alag h…….
Thank you so much bhaiya ….❤
We can also fit the negetive cycle check in the for loop with extending its range to V and checking if its the Vth iteration in relaxtion without writing repeated code. Also the best and worst cases can be improved by keeping a count of how many relaxations done in each iteration which signifies that if at any point no relaxation can be done then no further iteration is required bcoz there will be no change in distances further. Here's how its done:
vector bellman_ford(int V, vector& edges, int S) {
vector dist(V, 1e8);
dist[S] = 0;
for(int i = 1; i
One of the best playlist of Graph on youtube bhaia you deserve more
one thing should also be mentioned that if a graph on N nodes have cycle, then their is path exist having more than N - 1 edges from first to last node.
By the way best explanation on TH-cam👌
Dijsktra's code which striver has taught works for negative edges , it just not works for Negative edge cycles. So all in all , it would work for DAG with positive / negative weights.
Of course, because Dijkstra doesn't work for negative edges in UNdirected grapsh: What if you have 0 - 1 with -1 weight? It will give TLE. But in directed, 0->1 with -1, 1 can never go back to 0 so the loop will not work.
@@tasneemayham974 That's what I said. It works for DAG.
thanks striver, you are the real gem
Imp when to apply - > when have -ve cycles, idea-> relax all the edges v-1 times , tc->(O(VE))
SUBSCRIBED FROM FIRST RECURSION LIST VIDEO, SIRE!!!! UNDERRRSSTOOODDDD
start the relaxation loop from i=1 to i
the thought process is insane
Best explaination of this algo till date !!
Thank You So Much for this wonderful video...........🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻
Top notch explanation as usual. I would have included an update flag to pre-empt unnecessary iterations.
Amazing content sir !! ..... if I get a job will be because of you.🙂
Lggyi
we need to relax each edge for n - 1 times but in the code we are running loop for
if we assume it 0 index based then V-2 is right and if we take it 1 based than simply run it for 1 less as V-1
for(int i = 0; i < N; i++) : runs for N times ( 0 to N-1)
for(int i = 0; i < N-1; i++) : runs for N-1 times ( 0 to N-2) - which is required
Thanks for putting such kind of effort for us.
Habibi ye ek number bideo bana di tumne toh ...baut baut danyawaad
thank you so much for the clean and crisp explanation.
The fact that you explained N-1 is why you are the GOAT. Please make a paid course and we will pay
But the one thing should also be mentioned that if a graph on N nodes have cycle then their is path exist having edges more than N - 1.
Understood. Great explanation for the intuition.
takeaway: having n-1 iterations, helps bellman ford to avoid infinite loop. It won't again try to find a better distance.
Simple in every cycle worst can be 1 updation, so we need to update n - 1 updation as we already have 0 or src updated manually
Understood! Super amazing explanation as always, thank you very much!!
You got so much energy, bro!
Maja aa gaya
Itna khatarnak explanation bro 🎉
Thanks Striver for these wonderful lectures. Understood.
23:34 -
If someone is wondering like me as to why the edges are passed by reference using &
If you don't use a reference, the function would create a copy of the edges vector every time the function is called, which can be very costly in terms of memory and time, especially for large graphs
I just wanted to play around with dijkstra and I know this in not an optimal solution and thought of doing the negative cycle part with dijsktra only. My thought process is very simple with djkstra also you could solve the problem the only issue was negative cycle. Now a cycle in itself is not an issue but a cycle with reducing distance does. According to the condition of dijkstra it would enter the if block if we get a decreasing distance and this will only happen in negative cycle or in linear path. To check that we can store paths for each node. Next time we visit that node we check do we already have the node itself in the path that means we are in a cycle or else there is a different path that led us to the node.
Code:
vector bellman_ford(int V, vector& edges, int S) {
// Code here
vector adj(V);
for(int i=0;i
understood, thanks for the intuition part
Understood bhaiya 🙏❤️
Beautiful Explanation . Loved your content keep going 100%
Thank you very much. You are a genius.
Had no idea it was this easy, damn. Obviously now that i know the logic, i don't even need to remember it.
I guess for the question why we need to do n-1 relaxations to each node is that ,, suppose we have 3 nodes directly connected a node and we have relaxed those nodes,, now in typical dijkstra algo,, the node with smallest value of relaxation will never get relaxed again, but if there are negative weights,, then there is a chance that the the node with smallest value of relaxation will get relaxed again. Since each node at max gets connected to n-1 nodes, so thus n-1 relaxations.
Wow! very well explained, completely Understood
bhaiya , in the for loop terminating condition should be “ i < V “ for n-1 iterations
very nice explanation
@ 16:00 : explained why it has n-1 iterations
tq striver bhaiyya
intution was just 🔥🔥🔥🔥🔥🔥🔥🔥
5:28 - 22:00
20:11 activated SUPER SONIC MODE 🔥
Thank you, Striver!
Understood
great explanation 🔥🔥🔥🔥
bahi kya kar raha h
lots of love and respect🙌
greate explaination and with great energy while explaining that make people more creative affracting getting more..💖
Presense of negative edge does not always result incorrect result ...that could be easily proved in the second example if there is a path say from 4 to 3 with -1 edge weight ....but if the negative edge cycle change the previous path length which we got after n-1 iteration then it cause problem.
If we will take nodes in a increasing order then that time I think we can find the distance in 2-3 iteration as well we don't need to do n-1 iterations
Amazing Explanation!!🔥🔥
Understood😉 bhaiya
striver bhai is goated
Understood!! :)
Thank you! 🙏🏻😊
understood, It was so Awesome.
Master piece !
Understood !! Amazing as always
Well explained bhai!
great explanation thank you so much and please continue😘😍
Very well explained.
watching it at 4 a.m. and when u say , I got a better guy, it really hurts :)🤣🤣
Good, well explained.
Understood, thank you bhaiya
In Case of Negative weight cycles, Dijkstra don't run forever. It produces wrong answer
Thanks for the intution
Bahi abhi graph ki kitni video he
Aur bachi
He
Big fan 😍🙏
Thank you bhaiya
Thank you sir 🙏
Super explanation😀
Amazing , very well explained 🔥🔥
The best explanation doesn't exist.. Meanwhile Striver : Hold my beer 😵
Thank you so much 🙏🙏
Understood👍👍
Thanks a lot
Understood Sir
Understood Sir!
thank you bhaiyya , please can you make a playlist on examples on segment trees from codeforces
understood too
understood striver
Solid explanation man! Thanks!
Understood sir 🙂
at 17.25 Edges are 5 i think u r referring to the no of nodes not edges
clear explanation
baad m distance array ka sum nikal k check kr skte h n negative cycle ka
for(int i=0;i
Understood Raj bro
undershood sir
In for loop i
Understood :)
Sep'2, 2023 11:38 am