01:52 graph requirements on the Bellman-Ford algorithm 04:28 review of generic shortest-path algorithm 06:40 first problem with generic shortest-path algorithm 07:45 second problem with generic shortest-path algorithm 11:29 Bellman-Ford 15:54 complexity of Bellman-Ford 18:02 correctness of Bellman-Ford 23:24 proof of the theorem (relaxation part) 30:22 proof of the corollary (check part) 33:35 questions 39:20 shortest-path vs longest-path
Amazing lecture. Some remarks: 1) In the proof of correctness (at 24:28) you don't need to pick a shortest path with the minimum number of edges. Any shortest path will do (always assuming that the graph has no negative cycles). 2) In the example (at 35:00) the professor affirms that you fix at the beginning an ordering of the edges and at each pass you relax the edges following that order. This is not strictly necessary for Bellman-Ford to work, as you can see from the proof of correctness. At each pass you can use any ordering of the edges. 3) This is more of a philosophical remark, connected to point 2), on how one could think about Bellman-Ford. Once one figures out that relaxing edges is the way to go to find shortest paths you need to choose in which order to relax them. This is nontrivial because the relax operation is non-commutative: if you first do relax(a,b) and then relax(b,c) you end up with different distances than if you do relax(b,c) and then relax(a,b). Bellman-Ford comes up with an ordering of the operations relax(a_i, b_i) such that if (s,v_1,v_2,..,,v_k) is a shortest path from s to v_k then the subsequence relax(s,v_1), relax(v_1,v_2), ..., relax(v_{k-1}, v_k) will appear for sure (in this order!), and this is enough.
who are there watching these beautiful lectures in 2022 also? Because i feel it was much better way of explaining the things previously as compare to now where all the things are explained in ppts.
You gotta love 42:40 when he catches an honest mistake, especially if you were able to see it first. The professor is brilliant, no doubt, but he is also human.
46:05 the reason why Bellman-Ford can not find the reachable shortest "simple" path when there exist any negative cycles : you potentially go through negative cycle before you reach the destination, hence |V-1| passes might not be enough, which makes this NP-hard. Also, finding the longest path is equivalent to finding the shortest simple path.
while negative cycles cause shortest path to be undefined, positive cycles do the same for longest path. so in a sense, if you negate everything and run B-F undefined result is still correct, isn't it?
Yeah, that's what I thought too. I think he may have been referring to trying to find the longest *simple* path from s to v (i.e. not including cycles) which would be 6 in that example, but because of the cycle, would cause BF to return undefined.
Toan Ngo The Bellman ford algorithm is explained in that interval. Since the video is named "Bellman Ford", this is what you searched for, and if you only want to understand how does it work, those 4 minutes are enough to understand it.
Alin Radulescu those 4 minutes don't really show how it works - it's just the algorithm. Well, at least I won't truly understand B-F just by looking at the algorithm.
what if you do 3 passes of E passes? 1 - first O(VE) pass, 2 - check pass (marking bad edges), 3 - another O(VE) pass skipping bad edges. I think that would give the simple shortest paths
26:22 how does only one relaxation give shortest path between two adjacent vertices. there could be v0->v2->v1 path which is shorter than v0->v1. thanks in advance
I think in the case you mentioned it does give the shortest path between two adjacent vertices, those adjacent vertices being v0 and v2. It does not necessarily need to give the shortest path between all adjacent vertices but since every edge is explored, at least one will be the shortest path.
In first place, I think he should have say the first iteration of the for loop. Because each iteration makes |E| relaxations. So the second relaxation will be in the first iteration of the loop. But anyways, you have to run the entire for loop even if you wanna get the shortest path between 2 adjacent nodes. But at the end you'll figure out that if the shortest path between was v0->v1 then it was given in the first iteration of the for loop. If the shortest path was v0->v2->v1 then it was given in the second iteration. But you wouldn't know until you run the entire algorithm. It is a bit tricky because basically, after 5 iterations you will have all the shortest paths consisting of 6 nodes, and you will also have paths of 6 nodes that are not the shortest, but you won't know which one are which until you finish.
it does not .it gives only current shortest path which will be updated once we relax edges coming out of v2 which wi;; be the shortest path of v0 to v1
because we already know that v0 -> v1 -> .... -> vk is a shortest path, that means all sub paths are also shortest paths, so v0 -> v1 is a shortest path from v0 to v1, there could be no shorter paths (otherwise v0 -> v1 -> .... -> vk won't be a shortest path from v0 to vk).
any weighted general graph containing (-)ve weights can be solved using bellman-ford where as other algorithms are all specific case related which is outside the scope of 6.006
This guy doesn't explain why after relaxing each edge E, V-1 times, why we're guaranteed to have found the shortest path assuming no negative cycles. That seems to be a big point. At 28:49, he writes on the board, that after the second pass we get delta of (S, V2) because in second pass we relax (V1,V2)..But we relaxed (V1, V2) in all passes since in all passes we relax each edge, according to the algorithm. Kinda frustrating
This lecture has a much better explanation. The first 50 minutes goes over it: th-cam.com/video/XIAQRlNkJAw/w-d-xo.html Shai Simonson is great. He's not currently an MIT lecturer, but I believe he gave that lecture when he was working at MIT. Go figure.
@@kevinryzack5712 Thanks for the share; I had the same thought about that portion and was going to rewatch the video/read CLRS to see if I missed something (which I probably will do), but glad to see another lecture!
@@kevinryzack5712 Had the same doubt. Went through a lot of resources and explanations, but couldn't find a better and intuitive explanation than this. Thanks for sharing.
+Ashish Yadav There could be various situations. Consider this, if you travel from city A to city B you earn 50 rupees (+50), but if you travel from city B to city C you spend 20 rupees (-20).
Pro Dev bellman ford is kindda dynamic programming right? unlike dijkstra's greedy criterion, bellman ford relaxes all the edges and gauges shortest distance then... what is the other solution anyway??
on this video this solution is effcient implementation for Save Space there is other solution the Original ;) th-cam.com/video/OQ5jsbhAv_M/w-d-xo.html see min 49
With digraphs, I think it's just n*(n-1) since every node can be connected to every other node, resulting in two edges per node pair. (since E(u, v) != E(v, u) in a digraph). Think you're right for undirected graphs though. Still O(v^2) either way.
@@loganwishartcraig That's not important. The point of analyzing complexity is understanding "in terms of what and how is your algorithm's time growing". For example, in Bellman-Ford, you are doing a check pass also, but you don't count the total complexity as O(VE + E), which technically is O((V+1)E), you say O(VE).
We're talking about the American Educational system here... they can never have enough money. Maybe only american hospitals are more obsessed with money than private colleges.
They're asking for donations for their open courseware, which is harder to get funding from the college administration. Why would the administration fund giving away the information for free? Open education is academia's last weapon against the commodification of information.
many things are not expalined in this video lecturer is giving short dialogs as if allpeople in the world is genius so please expalin every thing if you put something in youtube
Pathological selection of edge is pretty vague but It's clear that there is an exponential number of paths, where length of each path is at least 1 (or there is at least one edge). In particular, the number of paths from 1 to n-1 is 2^((n-1)/2) and we have this many choices of d[n-1]. In the worst case, it will try all of them and relax d[n-1] + (n-1, n)
Some of us prefer Eric (if saying the the "brown dude" is racist, isn't saying the "white dude" racist too, here is a question for you to meditate on), can't make all people happy all of the time.
see in ur comment box if every one can understand yoyr video no one will ever ask the question...bdw i had understand this from another book of indian mtech author...
![[LIVE] : ONE ลุมพินี 88 | คู่เอก "ป้อมเพชร vs อัสลามจอน"](http://i.ytimg.com/vi/1ZqTVVbMgbo/mqdefault.jpg)
"Polynomial time is great, exponential time is bad, and infinite time gets you fired". Love it!
At Google, quadratic time is bad.
If he hasn't convinced everyone already, there's further proof of what a great prof he is.
01:52 graph requirements on the Bellman-Ford algorithm
04:28 review of generic shortest-path algorithm
06:40 first problem with generic shortest-path algorithm
07:45 second problem with generic shortest-path algorithm
11:29 Bellman-Ford
15:54 complexity of Bellman-Ford
18:02 correctness of Bellman-Ford
23:24 proof of the theorem (relaxation part)
30:22 proof of the corollary (check part)
33:35 questions
39:20 shortest-path vs longest-path
Amazing lecture. Some remarks:
1) In the proof of correctness (at 24:28) you don't need to pick a shortest path with the minimum number of edges. Any shortest path will do (always assuming that the graph has no negative cycles).
2) In the example (at 35:00) the professor affirms that you fix at the beginning an ordering of the edges and at each pass you relax the edges following that order. This is not strictly necessary for Bellman-Ford to work, as you can see from the proof of correctness. At each pass you can use any ordering of the edges.
3) This is more of a philosophical remark, connected to point 2), on how one could think about Bellman-Ford. Once one figures out that relaxing edges is the way to go to find shortest paths you need to choose in which order to relax them. This is nontrivial because the relax operation is non-commutative: if you first do relax(a,b) and then relax(b,c) you end up with different distances than if you do relax(b,c) and then relax(a,b). Bellman-Ford comes up with an ordering of the operations relax(a_i, b_i) such that if (s,v_1,v_2,..,,v_k) is a shortest path from s to v_k then the subsequence relax(s,v_1), relax(v_1,v_2), ..., relax(v_{k-1}, v_k) will appear for sure (in this order!), and this is enough.
yeah bruh
who are there watching these beautiful lectures in 2022 also? Because i feel it was much better way of explaining the things previously as compare to now where all the things are explained in ppts.
it feels like im in class, learning and at the same time reading and writing comments , like talking with classmates , loved it !!
You gotta love 42:40 when he catches an honest mistake, especially if you were able to see it first. The professor is brilliant, no doubt, but he is also human.
why are you so cute?
I love the part after the correction when the prof asked the other student, "and why did you say undefined?". "That's wrong." hahahah
"5 minute university" sugestion is the point I like most in the lecture. Great, man!!
It is fantastic how he runs a Bellman-Ford at 43:20.
Two key moments:
42:40 Finding his mistake
and
43:20 Running Bellman Ford
I love this teacher!
correctness proof of bellman ford algorithm starts at 18:08
free acapella concert!!!
I wanted to go, but had to study for the quiz he mentioned in class....
JK. I don't go to MIT :(
Clean explanations, easy to understand, thank you very much.
46:05 the reason why Bellman-Ford can not find the reachable shortest "simple" path when there exist any negative cycles : you potentially go through negative cycle before you reach the destination, hence |V-1| passes might not be enough, which makes this NP-hard. Also, finding the longest path is equivalent to finding the shortest simple path.
while negative cycles cause shortest path to be undefined, positive cycles do the same for longest path. so in a sense, if you negate everything and run B-F undefined result is still correct, isn't it?
Yeah, that's what I thought too. I think he may have been referring to trying to find the longest *simple* path from s to v (i.e. not including cycles) which would be 6 in that example, but because of the cycle, would cause BF to return undefined.
great lectures, help me a lot.
You came here for 11:33 - 15:20
um why?
Toan Ngo The Bellman ford algorithm is explained in that interval. Since the video is named "Bellman Ford", this is what you searched for, and if you only want to understand how does it work, those 4 minutes are enough to understand it.
Alin Radulescu those 4 minutes don't really show how it works - it's just the algorithm. Well, at least I won't truly understand B-F just by looking at the algorithm.
Not all heroes wear a cap
hahahah
In worst case every pair of vertices is connected by an edge. So the number of edges will be (n choose 2), which is n*(n-1)/2. That's about n^2 edges.
Can you please explain how would there be an exponential number of edges in the generic shortest path algorithm?
I got the same question, why exponential?
I've watched most of the lectures so far and this one is pretty bad. Recitation is definitely needed here. Viktor where are you!!
I don't understand what he is proving?
what if you do 3 passes of E passes? 1 - first O(VE) pass, 2 - check pass (marking bad edges), 3 - another O(VE) pass skipping bad edges. I think that would give the simple shortest paths
#besten #Episode Full Service#Vielen Vielen dank#bestimmt Liebe grüße und zu Videos#so#mit#
@10:59 How does dijkstra turn into exponential time?
see @ 6:51 , he says for a graph if edges are selected wrongly and then complexcity in terms of number of vertices.
Think about coin change problem, how similar they are.
This doctor has mistaken concept of single source shortest path to shortest simple path.
26:22 how does only one relaxation give shortest path between two adjacent vertices.
there could be v0->v2->v1 path which is shorter than v0->v1.
thanks in advance
I think in the case you mentioned it does give the shortest path between two adjacent vertices, those adjacent vertices being v0 and v2.
It does not necessarily need to give the shortest path between all adjacent vertices but since every edge is explored, at least one will be the shortest path.
In first place, I think he should have say the first iteration of the for loop. Because each iteration makes |E| relaxations. So the second relaxation will be in the first iteration of the loop.
But anyways, you have to run the entire for loop even if you wanna get the shortest path between 2 adjacent nodes.
But at the end you'll figure out that if the shortest path between was v0->v1 then it was given in the first iteration of the for loop.
If the shortest path was v0->v2->v1 then it was given in the second iteration.
But you wouldn't know until you run the entire algorithm.
It is a bit tricky because basically, after 5 iterations you will have all the shortest paths consisting of 6 nodes, and you will also have paths of 6 nodes that are not the shortest, but you won't know which one are which until you finish.
it does not .it gives only current shortest path which will be updated once we relax edges coming out of v2 which wi;; be the shortest path of v0 to v1
because we already know that v0 -> v1 -> .... -> vk is a shortest path, that means all sub paths are also shortest paths, so v0 -> v1 is a shortest path from v0 to v1, there could be no shorter paths (otherwise v0 -> v1 -> .... -> vk won't be a shortest path from v0 to vk).
excellent .....
perfect video - thanks for sharing ;-)
K
at 4:44 , *throws hat in the air*
Are there other algorithms which could produce negative cycles?
any weighted general graph containing (-)ve weights can be solved using bellman-ford where as other algorithms are all specific case related which is outside the scope of 6.006
This guy doesn't explain why after relaxing each edge E, V-1 times, why we're guaranteed to have found the shortest path assuming no negative cycles. That seems to be a big point. At 28:49, he writes on the board, that after the second pass we get delta of (S, V2) because in second pass we relax (V1,V2)..But we relaxed (V1, V2) in all passes since in all passes we relax each edge, according to the algorithm.
Kinda frustrating
This lecture has a much better explanation. The first 50 minutes goes over it:
th-cam.com/video/XIAQRlNkJAw/w-d-xo.html
Shai Simonson is great. He's not currently an MIT lecturer, but I believe he gave that lecture when he was working at MIT. Go figure.
@@kevinryzack5712 Thanks for the share; I had the same thought about that portion and was going to rewatch the video/read CLRS to see if I missed something (which I probably will do), but glad to see another lecture!
@@kevinryzack5712 Had the same doubt. Went through a lot of resources and explanations, but couldn't find a better and intuitive explanation than this. Thanks for sharing.
This is too edgy for me
Ha! I get it. It's a bit graphic for my tastes, but a good joke nonetheless.
What does -ve stand for?
+Alexander List Negative
I don't think people should get to have trivial algoirthm named after themselves
this is my algorithm for calculating the sum of a collection of numbers.. for i in collection: sum += i PUT ME IN TEXTBOOK
@Lizard King its not about triviality, its about novelty. It was new, so it was recognized. Yours isn't new.
can anyone tell me why negative weights are used?
+Ashish Yadav There could be various situations. Consider this, if you travel from city A to city B you earn 50 rupees (+50), but if you travel from city B to city C you spend 20 rupees (-20).
An example would be traveling by boat up river ...
@@icaryslittle6370 Nice example
i dont really understand the loop, where he loops v-1 times? can someone please help.
question .. why is the geeric algorithm's complexity O(2^n/2) ...somebody please help :)
R. S. Joshi ahh! bad bad typo :P
thank you sir :) the proof makes the point so concrete!
Suneet Singh is this the Efficient implementation because i think there is other solution with DP ?
Pro Dev bellman ford is kindda dynamic programming right? unlike dijkstra's greedy criterion, bellman ford relaxes all the edges and gauges shortest distance then...
what is the other solution anyway??
on this video this solution is effcient implementation for Save Space there is other solution the Original ;)
th-cam.com/video/OQ5jsbhAv_M/w-d-xo.html see min 49
True ... ?
i need to learn math
42:40 that was funny :D
im sorry, but why is E order V^2?
In worst case every pair of vertices is connected by an edge. So the number of edges will be (n choose 2), which is n*(n-1)/2. That's about n^2 edges.
With digraphs, I think it's just n*(n-1) since every node can be connected to every other node, resulting in two edges per node pair. (since E(u, v) != E(v, u) in a digraph). Think you're right for undirected graphs though. Still O(v^2) either way.
@@loganwishartcraig That's not important. The point of analyzing complexity is understanding "in terms of what and how is your algorithm's time growing". For example, in Bellman-Ford, you are doing a check pass also, but you don't count the total complexity as O(VE + E), which technically is O((V+1)E), you say O(VE).
Think of adjacency matrix.
So, what exactly is difference between Dijkstra and Bellman-Ford?
+Liang Luo different assumptions about the graph, different complexity, and of course, different algorithm.
+Liang Luo Dijkstra won't work with negative edges.
+Liang Luo the difference is that belman ford can solve a cycle where you have negative values
MIT has an endowment of $13.182 billion, I wonder why it needs donations lol.
LMAO right??! that's what I always wonder whenever I hear that bit...
you need donations? for realsies now???
We're talking about the American Educational system here... they can never have enough money. Maybe only american hospitals are more obsessed with money than private colleges.
They're asking for donations for their open courseware, which is harder to get funding from the college administration. Why would the administration fund giving away the information for free? Open education is academia's last weapon against the commodification of information.
many things are not expalined in this video lecturer is giving short dialogs as if allpeople in the world is genius so please expalin every thing if you put something in youtube
Could any one please explain why is the exponential time? 7:03 Thanks
+Pengqian Y see lecture 15
Pathological selection of edge is pretty vague but It's clear that there is an exponential number of paths, where length of each path is at least 1 (or there is at least one edge). In particular, the number of paths from 1 to n-1 is 2^((n-1)/2) and we have this many choices of d[n-1]. In the worst case, it will try all of them and relax d[n-1] + (n-1, n)
i wish this guy taught all the videos. don't understand the white dude
Some of us prefer Eric (if saying the the "brown dude" is racist, isn't saying the "white dude" racist too, here is a question for you to meditate on), can't make all people happy all of the time.
see in ur comment box if every one can understand yoyr video no one will ever ask the question...bdw i had understand this from another book of indian mtech author...
+Broken Heart first of you grammar is terrible and secondly you are not paying them for this lecture so stop talking trash, if you have doubts ask
@@calyxwakefield3523 exactly,he is like i want it free ,i will never donate, but will give my precious advice ...SOAB!