Lecture 15: Single-Source Shortest Paths Problem

[@SparshPaliwal]

Everyone please stop saying the lecture is worthless. This lecture is a Primer for shortest path algorithms. Understanding negative weights, negative loops was pretty important. Also learning about all different problems like exponential number of vertices. Surely if you had gone through the topic already might be useless to you, so you can skip it. But if you are going through shortest path algos for the first time, its a good lecture.

[@ryanc7840]

Right. Without this one, subsequent lectures are hard to grasp at least regarding the notations. I concur that this one is a good and necessary primer.

[@zvezdalion]

This professor's lecture is so structured! He really stresses the important things. So easy to get the big picture, but also gives you the often-overlooked details (e.g., it's not the neg. weight edges, but the *cycles* that are a problem; Dijkstra is essentially *linear*)

[@sergeykholkhunov1888]

02:30 motivation, Dijkstra, Bellman-Ford 08:25 definitions (path, weight of path, shortest-path problem) 11:10 about time complexity 14:21 weighted graphs 17:56 example 23:55 predecessor relationship definition 26:40 negative weights 32:00 negative weight cycles 37:30 general structure of shortest-path algorithms, relaxation 45:02 example (problems with exponential number of paths) 50:35 optimal substructure (shortest path's subpaths, triangle inequality)

[@md.sayeedrahman2553]

loved it! personally i feel attached to this professor! his lecture is so amazing!!!

[@Qladstone]

If each edge represents a multiplier, and you want to minimize the product of the multipliers along a path; that is equivalent to minimizing the sum of the logarithms of those multipliers. Logarithms have a range from -inf to +inf, and so include both negative, zero, and positive values. One commonly cited example is where the multipliers represent currency conversion rates; where each vertex represents one currency, and edges represent conversion rates between currencies.

[@pengliu9987]

Great example of negative weight graph. Thanks for sharing.

[@anunaybagga2948]

i understood each and everything sir said and hence i don't get the hate comments out here.

[@mohitsingh-nf5vl]

Can u explain me time complexity plz??

[@zeronothinghere9334]

@@mohitsingh-nf5vl Look at the first video in the playlist. This is the 15th video in the series after all.

[@user-td8zb9xo9k]

happy to see indian professor here

[@loganwishartcraig]

This lecture was definitely fine.

[@aerronpro1133]

Wow this lec is genius, generalises whole shortest path in a lec

[@Tintak_hatpin]

I like the way he says "you could imagine"

[@MrKingoverall]

Operations Research Major here: LOVE YOU MAN !! :)

[@SzechSauce]

Isn’t there a mistake at 48:20? How can you relax the edge from v4 to v5 to then have d(v5) be 12? That would mean v4 can be reached in 12, and v5 can also be reached in 12? That can’t be correct...? Really great lecture otherwise, quite surprised at all the negative comments on here...

[@marvinmurphy4330]

neg weights? 28:55 - What about graphs being represented as electric circuits? Some branches have passive components like resistors that cause a potential drop and some branches have active components like batteries that cause a potential gain. Intersections or splits in the circuit are vertices. The connections between splits are edges and the weights are the voltage gain (+ve) or drop (-ve) of that connection.

[@surjayanghosh5872]

thats an interesting obeservation! ..... but how would we use the different paths/ because in electrical circuits, current (I) is gonna flow in all the branches, i mean we cant tell it to use only a certain path. So, can you think of any way it could be used? but good analogy nonetheless.

[@iamparitosh]

Also the total potential gain of any cycle is zero here

[@lockersrandom6161]

Thank YOU MIT.

[@eepassion9720]

29:36 you will see Victor giving a example about negative weight. We would know more about MIT TA work. That is so different in my college. I don't even know who is my TA in the entire of my college.

[@cskwillreturn2473]

At 12:43, did he say O(E^2) possible weights? I think it should be V^2.

[@moongumoongu]

yeah, jotted that down and got really confused later. if there are |E| edges, the number of possible unique weights is |E|... and E=O(V^2), so yeah, i think he meant to say roughly V^2 possible weights.

[@bigdildo]

agreee

[@csl1384]

Glad to see I am not the only one who thought this.

[@ShivamPatel-ww1up]

my doubt exactly

[@neuron8186]

i don't understand the hate both teacher are best and you haters are nothing more than a keyboard warrior how did you get the audacity to hate people providing free education

[@josh7910]

48:20 I think he got the 13 ->12 from v5 wrong

[@fabischn90]

It's perfectly correct. After relaxing the edge (v4, v6) he relaxed edge (v5, v6)

[@PK-en1bm]

***** I am thinking about relaxation of the edge (v4, v5)??

[@fabischn90]

Prasanjit Khuntia You would understand, if you just had continued watching for another 2 minutes. Keep calm :)

[@JustinLiang0]

Yeah odd... I am not sure how he can "relax" the edge like that because nothing seems to satisfy d[v] > d[u] + w(u,v) at that point.

[@botelhorui]

Justin Liang Yes, its a mistake

[@ChristopherPelnar]

Negative weights are like hitting wormholes in space-time.

[@isbestlizard]

why are cycles even a thing? if they add positive path length you'd obviously never take them, if they add negative path length you'd obviously go around around them forever so surely it makes no sense to even have a graph with cycles when asking the question 'what's the shortest path'.. without adding extra criteria like 'without retracing the same edge' or 'without visiting the same node more than x times'

[@SanghoBose5]

Travelling upstream in river be considered as negative weight path.

[@yashjain6945]

What does he mean by at least gets credit for it 4:27

[@hoboishjoeishme]

45:23 sounds like someone's squeezing out a little fart

[@FactVault5584]

lol

[@swappybird]

I came for the this exact comment.

[@mo_pak_tech]

legend

[@fredwu6812]

What are you here for? Listening to farting or lectures?

[@RalyJoey]

48:22 how did the instructor obtain 12 for the second last vertex of this directed graph?

[@fredwu6812]

that was a mistake.

[@anunaybagga2948]

Fred Wu no man it is not a mistake watch it again

[@tu-ningting2867]

Anunay Bagga It's been 2 years. Hope you now see it is a mistake. If not, watch it again

[@이택영-l9h]

His lecture is very meticulous

[@yettodecideahandle]

Srini is very good and seem to explain thing simple, Eric makes the same lectures complicated.

[@elliott8175]

Erik is the boss. I love his intuitive explanations. He gets straight to the ideas in a really down-to-earth way: "Dynamic programming is really just careful brute force". If you get DP, you realise how other teachers make it out to be much more complicated than it is.

[@gouravgoel2974]

this is gold

[@Mrimadeyoufall]

So helpful

[@rithwikteja2168]

I need to find victor now

[@brendawilliams8062]

Negative takes so much work with squares and cubes. People don’t want what appears to be infinities of work. They want a genesis of general and fast tracked work.

[@yangyinjedi]

I like the other professor a little better I feel like his lectures are more random and less structured...I felt like this professor was just teaching by the book...But they're are both great sup lectures

[@davidporterrealestate]

41:20 49:00

[@mayonesacosmica]

how about circuits where there are multiple paths w/ positive or negative voltage differentials for a motivation? if so, I can send you my address for a cushion... ;)

[@Joshuaposada]

Wouldn't you just use the magnitude? I'm not really sure, but that's something to think about

[@isbestlizard]

why would anyone imagine the weight defined for an edge WOULD affect the calculation speed of an algorithm? that'd be like thinking the 'weight' of an object would affect the time it took to sort it. 'this is a really heavy object so it takes longer to add it to the list' wot no 'this is a really long path so it takes longer to calculate its length' WOT NO WHY WOULD YOU THINK THAT :V

[@adhishmalviya2408]

49:30 the random relaxing of edges can be exponential and dependent on order of weights

[@nikhilrajput5030]

Can anyone help me why E=O(V^2)??

[@abhaikollara5811]

Number of possible edges between n vertices is nC2 i.e the number of ways to pick 2 vertices from a given set of n vertices. Unless I screwed up, replacing r=2 in the equation for nC2 should derive into (V^2 - 1)/2 which is O(V^2).

[@nikhilrajput5030]

@@abhaikollara5811 Thanks man!

[@legendarylea560]

@@abhaikollara5811 I think it's not about nC2 (combinations) but nP2 (permutations), so nPr -> n^r -> n^2. Am I wrong?

[@cvprasanthkumar]

I think it is because every Vertex can have atmost V Edges(Unless we are talking about Multi Graph). So you have V Vertex with each having V Edges, so overall edges are V^2

[@ShabbirAhmed-ug8sv]

22:25

[@douzigege]

I can feel the pain. The reason why sometimes he is not delivering the lectures is probably the fact that Dr. Devadas is not a mathematician like Erik.

[@ankushm3t]

Nonsense

[@saicharanmarrivada5077]

He has higher h-index compared to Erik

[@csl1384]

14:05 If Dijkstra is "basically" linear time [O(VlogV + E)], then Bellman-Ford [O(VE)] should be "basically" quadratic time, not cubic time.

[@philodev874]

No, the professor explains how E’s asymptotic upper bound is V^2(imagine each vertex containing an edge to every other vertex) therefore V * E = V * V^2 = V^3 (asymptotic upper bound)

[@csl1384]

@@philodev874 agreed, thanks

[@philodev874]

c sl no problem!

[@rohangupta3486]

how he wrote E=O(V^2) ?????????????????

[@cskwillreturn2473]

Assume two vertices v1 and v2. The total number of possible edges would be 4 which is 2^2 ( 2= total number of vertices). All possible edges v1 -> v2, v2 -> v1, v1->v1, v2->v2. They all add up to 4. But this is only correct if the graph is directed. If undirected, I think it should be v(v-1)/2.

[@kp2718]

imagine an adjacency matrix

[@legendarylea560]

@IndiaRockLovers I think it's not about nC2 (combinations) but nP2 (permutations), so nPr -> n^r -> n^2. Am I wrong?

[@legendarylea560]

@IndiaRockLovers Perfect, thx

[@mystmuffin3600]

it's about how many (u,v) pairs can be formed. Directed edges would yield nP2 edges given the graph is simple (u cannot be equal to v) and nC2 edges for undirected as (u,v) and (v,u) would only be counted once

[@tchappyha4034]

11:55 very dangerous!

[@jshellenberger7876]

#POW

[@abdelrhmansamir1426]

I know it's too late i guess. but can anyone explain why does the asymptotic relation between edges and vertices is v^2? don't know if i said it correctly as well just didn't get that point.

[@vamshidharreddy2580]

maximum number of edges possible is of the order v^2. which is same as maximum number of handshakes among 'v' people.

[@shubhankitsingh5196]

nC2

[@thehvhnk]

discrete math? Imagine you pick two random nodes in a bunch of nodes so when total of nodes is N therefore total of number of ways is NC2 ~ n^2

[@determinator2]

In a complete graph, the number of edges is V choose 2 (The total number of ways by which we can pair two vertices), which equals V*(V-1)/2. This is of order V^2. Hence, E = O(V^2)

[@connectication]

30:32, sooo awkward

[@ravitejapothala7114]

Mm,kumms

[@rj-nj3uk]

He did not taught dijkstra properly.

[@ChrisLeeX]

While I'm a big fan of this series and I like the lecturer, this particular lecture is terrible. Just skip it.

[@roylee3196]

+Chris Lee Yes you're right.

[@satadhi]

Hi can you be more specific please ? cause i don't see anything wrong in this , maybe i am missing something ?

[@xinli6243]

I like Eric.

[@kp2718]

You get to appreciate the quality of a lecture after watching something like this. Where's Eric??? I think this guy forgot whats' the point of lectures, so he just rambles.

[@piyushsharma6811]

Disappointed.

[@chenzhiye633]

waste of time=,=

[@jsridhar72]

Just unberable. Wasted precious time. You wont gain anything by watching this. Just skip it.