Exactly like class attendance. One of my teachers at the university told it was the communicating vases principle : less people in the auditorium, more people drinking at the terraces
This is another fine lecture on Computer Science number one topic algorithms. Matching Problems use different algorithms to generate a solution that makes sense.
The students with the feet on top of the chair in front of them, come on guys, someone else will rest their head on it someday! Also, it feels disrespectful to the instructor. Like have some sense of decorum.
In the last proof , his assumption was that there exist a worser matching for G, in which the girl G has a boy B' married to him and B' is worser than G's match in TMA .but by Thm 4 , the boy B who is married to G in TMA has his favorite match in TMA , but in this matching B doesnt like his mate (compared to TMA) .thus B , G form a rogue couple . So no such stable matching in which G has worser mate exist
THM 1: Each boy has N girls on his list. So if the algorithm terminates in n^2 + 1 days doesn't that mean that each boy made N cross outs. In other words didn't each boy get denied by each girl and thus have no more options left? Wouldn't the maximum number of steps therefor be (n-1)^2 +1
+Adriaan de Clercq The proof of N^2 + 1 doesn't necessarily intend to come close to the actual # of days it would take. It's just an upper bound created using basic reasoning. I'm sure you could prove something stronger like what you're saying.
I was thinking to put the same comment actually. Yes, it is true. In the original proof of Gale and Shapley 1962, it says that the algorithm ends at most in (n-1)^2+1 stages.
There are recitations. Sorry we don't have videos of those but we do have written notes for them. See the Recitations section of the course on MIT OpenCourseWare at: ocw.mit.edu/6-042JF10.
How does we insure that the solution came out of TMA is most fair among all the stable matching possible out of preference list. and how we calculate realm of possibility for each man and woman.
Lee_Land_Y he proposes that there is a worse mate for G than the mate chosen for G using TMA and then goes and shows that his proposal is a contradiction thus there is no worse mate for G in any type of matching than the mate G is given using TMA.
Moral of the story:- Regardless of whether you are a boy or a girl, just confess to your beloved.... Who knows you might end up getting your optimal partner🙂 And of course, #𝙟𝙪𝙨𝙩𝙞𝙘𝙚𝙁𝙤𝙧𝙈𝙚𝙧𝙜𝙖𝙩𝙞𝙤𝙙
Karan Singh if Brad ended up with Jen, nd BB ended up with Ang, we'd be in a situation where Brad and Ang like each other better than their respective spouses, they would "cheat" on their spouses
I know i'm late to the party but at the end, when he proves Thm 3, he uses 2 cases to base his proof upon. In this particular example, it is kind of obvious that there is no other possible cases but imagine a different example where this is not the case. Don't I have to somehow ALSO prove that there is no other possible cases? And so, no matter how trivial, don't I still have to prove it in this example too?
In Theorem 3, the two cases are the following: [Case 1: G rejected B] [Case 2: G did not reject B] By the law of excluded middle, Case 1 and Case 2 are: mutually exclusive, AND mutually exhaustive - hence they supply all possible scenarios. (The law of excluded middle is a logical axiom which says that any proposition can only be true or false - not both, also not neither.)
In the end we proved that a girl cannot be worse off then the suitor provided by the TMA algo, how does that prove that she always marries her pessimal mate ?...is there no possiblity that she does better ? If so how to prove that?
There is a possibility. Don't take this lecture's example as simply girls and boys but as aggressors in suitors and waiting suitors. If we switch how the professor matched in 4 days(girl waiting in the balcony) to boys waiting in balcony we will get solution in 1 day without rogue couple. So the thing really is that the aggressor in the case is optimal and the waiting one is pessimal.
I have the same question. We only proved it cannot be worse than pessimistic match. But by definition of Pessimistic match, there cannot be any match that's worse than it. So I don't even see a point in his last proof. He must prove that there is no better match than pessimal partner.
@@karthicks2518 we proved that there is no stable matching where G has a worse partner compared to the partner she has in TMA, hence TMA is the pessimal matching for girls
Mathematically, there is no reason to split people into two discreet groups to solve the marriage problem. Other resources probably don't need to be split into two exclusive groups to find stable solutions either. If you make it so higher weights are better, and allow weights -1 and 0 (values -1 and lower all have the same effect), and make the graph n-node complete, you can run the (nearly) same algorithm and it finds stable pairs without needing to know gender. That is: heterosexuals will never be paired with same-sex partners, homosexuals will never be paired with opposite sex partners, bisexuals will be paired stabily, and people for whom attraction is less important than having any partner at all will potentially get paired with zeros to create stabil pairs, and people who would rather be alone than be paired with a zero, will be alone. Just never pair any resource with something that it weighted at -1. This could also be expressed with all positive weights and a >threshhold or
Suppose a boy B did not get partnered with his optimal mate in TMA for purposes of contradiction. This means he has the possibility of being partnered with a girl higher up in his set of ranked potential partners {1, 2, ... , n}, say girl x. Girl x then has to choose boy B as opposed to boy A that she got partnered with in TMA, so girl x's set has to change from from {... , A, B, ...} to {... , B, A, ...}. The set of ideal mates does not change, therefore, by contradiction, we have proved that B is pared with his optimal mate. (Not ''the proof'', just an attempt, may not be valid)
What a weird lecture: completely random topics, no formal theory of anything written on the black board, just a bunch of sample math problems thrown away to the students...
this is a most typical lectures series !! .... this course is for essential maths for computer science if you know the topic already you can skip it... you ask about formal theory her let me answer you - the main theory in lecture 6,7 is a "graph theory" .... the professor start with some definitions (which is the most importent in maths) about graphs .. and then start to say some common associated problems with "graphs theory" such as "Coloring" (in lecture 6) and "matching problem" (in this lecture) which have a several of applications like dating apps , networking , exams scheduling and many more mentioned in the lecture 6,7 ... and in my opinion he choose a clever and interesting examples to state those problems !!
this teacher is really awesome.
Exactly.
having this guy switch on and off lectures with the other guy is like giving and taking away cake from a fat kid
@@somemathkid2889 very true!
@@somemathkid2889 Yeah, I really wish there's a version of him teaching all the lessons.
It feels orgasmic to get an MIT lecture free of cost. The world is a good place now.
I like how the views go down exponentially later into the semester.
zipf law :D
Exactly like class attendance. One of my teachers at the university told it was the communicating vases principle : less people in the auditorium, more people drinking at the terraces
I wish I could give more than one like for lectures from this professor.
he's a complete teacher
Blockbuster professor and blockbuster lecture.A zillion thanks to MIT for making it free!
"On Tuesday, we talked about sex and today we're gonna talk about marriage." Aren't we moving too fast?
Nice one.
LOL
The other way round is better.
religious people: hold my beer
26:08 Well, that was smooth...
This is another fine lecture on Computer Science number one topic algorithms. Matching Problems use different algorithms to generate a solution that makes sense.
Lies again? MRT MIT
01:02:28 Student on the left is NOT buying that
rofl
He did that in the last one as well lol.
I'm desperate for the recitation videos.
Fast forward to 2016 >> Angelina don't seem to really like Brad .
No angelina lesbian
awesome choice of sociological parallel. very clever and applicable indeed.
The students with the feet on top of the chair in front of them, come on guys, someone else will rest their head on it someday!
Also, it feels disrespectful to the instructor. Like have some sense of decorum.
lol "doing their 6.042 hw together at night"
This is like he spied on me :D
Been in this exact situation during my student years.
In the video, at 48:19, Thm1 says TMA terminates in
Where did you get the course note from?
Shiva kumar Gupta ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-042j-mathematics-for-computer-science-fall-2010/readings/MIT6_042JF10_chap05.pdf
In the last proof , his assumption was that there exist a worser matching for G, in which the girl G has a boy B' married to him and B' is worser than G's match in TMA .but by Thm 4 , the boy B who is married to G in TMA has his favorite match in TMA , but in this matching B doesnt like his mate (compared to TMA) .thus B , G form a rogue couple .
So no such stable matching in which G has worser mate exist
Hence the contradiction!
48:32
I think there is a stronger statement as it will terminate in
Just Amazing
THM 1: Each boy has N girls on his list. So if the algorithm terminates in n^2 + 1 days doesn't that mean that each boy made N cross outs. In other words didn't each boy get denied by each girl and thus have no more options left? Wouldn't the maximum number of steps therefor be (n-1)^2 +1
+Adriaan de Clercq The proof of N^2 + 1 doesn't necessarily intend to come close to the actual # of days it would take. It's just an upper bound created using basic reasoning. I'm sure you could prove something stronger like what you're saying.
damn I have the same profile picture!!!! love for pi
I was thinking to put the same comment actually. Yes, it is true. In the original proof of Gale and Shapley 1962, it says that the algorithm ends at most in (n-1)^2+1 stages.
great teacher
lovely lecture...
At 24:55 he says "tomorrow at recitation..." does he mean the lecture by Dr. Marten van Dijk, because he just seems to like recite from his notes ?
There are recitations. Sorry we don't have videos of those but we do have written notes for them. See the Recitations section of the course on MIT OpenCourseWare at: ocw.mit.edu/6-042JF10.
Bro are U doing CS course as well right now
He really looks like tim cook in some shots.
How does we insure that the solution came out of TMA is most fair among all the stable matching possible out of preference list. and how we calculate realm of possibility for each man and woman.
Watching this video in 2017 and looking at the Brad Jennifer Billy Bob Angelina example, pretty ironic.
Amazing work!!
Is guy on min 4.15 on blue screen installing and old windows or what
excellent lecture, thank you
I don’t see how proof of theorem 5 proves that girls always get the pessimal mate. Can someone clarify? Thanks. This lecturer is awesome
Meaning how it proves that girls never get someone who isn’t neither pessimal not optimal
Lee_Land_Y he proposes that there is a worse mate for G than the mate chosen for G using TMA and then goes and shows that his proposal is a contradiction thus there is no worse mate for G in any type of matching than the mate G is given using TMA.
@15:55 Sadly Brad and Angelina weren't happy with each other ...
Brad never got matched with her
Moral of the story:-
Regardless of whether you are a boy or a girl, just confess to your beloved.... Who knows you might end up getting your optimal partner🙂
And of course, #𝙟𝙪𝙨𝙩𝙞𝙘𝙚𝙁𝙤𝙧𝙈𝙚𝙧𝙜𝙖𝙩𝙞𝙤𝙙
I dont really understand the first part. How is (Brad-Ang and BB-Jen) not a rogue couple, but (Brad-Jen and BB-Ang) is?
Karan Singh if Brad ended up with Jen, nd BB ended up with Ang, we'd be in a situation where Brad and Ang like each other better than their respective spouses, they would "cheat" on their spouses
I know i'm late to the party but at the end, when he proves Thm 3, he uses 2 cases to base his proof upon. In this particular example, it is kind of obvious that there is no other possible cases but imagine a different example where this is not the case. Don't I have to somehow ALSO prove that there is no other possible cases? And so, no matter how trivial, don't I still have to prove it in this example too?
In Theorem 3, the two cases are the following:
[Case 1: G rejected B] [Case 2: G did not reject B]
By the law of excluded middle, Case 1 and Case 2 are: mutually exclusive, AND mutually exhaustive - hence they supply all possible scenarios.
(The law of excluded middle is a logical axiom which says that any proposition can only be true or false - not both, also not neither.)
Inidans are everywhere flexing knowledge 😂
In the end we proved that a girl cannot be worse off then the suitor provided by the TMA algo, how does that prove that she always marries her pessimal mate ?...is there no possiblity that she does better ? If so how to prove that?
There is a possibility. Don't take this lecture's example as simply girls and boys but as aggressors in suitors and waiting suitors. If we switch how the professor matched in 4 days(girl waiting in the balcony) to boys waiting in balcony we will get solution in 1 day without rogue couple. So the thing really is that the aggressor in the case is optimal and the waiting one is pessimal.
In the last proof, Do we need to consider the other case in which some girls have a better choice than the pessimal partner?
I have the same question. We only proved it cannot be worse than pessimistic match. But by definition of Pessimistic match, there cannot be any match that's worse than it. So I don't even see a point in his last proof. He must prove that there is no better match than pessimal partner.
@@karthicks2518 we proved that there is no stable matching where G has a worse partner compared to the partner she has in TMA, hence TMA is the pessimal matching for girls
All the dislikes are from Mergatoid.
Matching algorithm reminds me of a show in Netflix called single inferno.
1:09:49 when you rarely find an Asian struggling. On the left.
LOL. Maybe he did't pre-view the lecture note?
Seems he was sleeping from doing homework or studying in small hours that day.
made my day..
when proving that invariant at around 57 min why was case 2 necessary?
Lovely lecture, such simplicity, and also poor little mergatoid :'(
suppose if we try to match boys with girls taking girls as serenaders can we solve the problem ? can any anyone help me out here
Mathematically, there is no reason to split people into two discreet groups to solve the marriage problem. Other resources probably don't need to be split into two exclusive groups to find stable solutions either.
If you make it so higher weights are better, and allow weights -1 and 0 (values -1 and lower all have the same effect), and make the graph n-node complete, you can run the (nearly) same algorithm and it finds stable pairs without needing to know gender. That is: heterosexuals will never be paired with same-sex partners, homosexuals will never be paired with opposite sex partners, bisexuals will be paired stabily, and people for whom attraction is less important than having any partner at all will potentially get paired with zeros to create stabil pairs, and people who would rather be alone than be paired with a zero, will be alone. Just never pair any resource with something that it weighted at -1.
This could also be expressed with all positive weights and a >threshhold or
can anyone pls provide the proof of thm_4?
Suppose a boy B did not get partnered with his optimal mate in TMA for purposes of contradiction. This means he has the possibility of being partnered with a girl higher up in his set of ranked potential partners {1, 2, ... , n}, say girl x. Girl x then has to choose boy B as opposed to boy A that she got partnered with in TMA, so girl x's set has to change from from {... , A, B, ...} to {... , B, A, ...}. The set of ideal mates does not change, therefore, by contradiction, we have proved that B is pared with his optimal mate.
(Not ''the proof'', just an attempt, may not be valid)
@@lovaaaa2451 this is invalid, nobody says this match has to be done by TMA, so I don't see why you mean by the girl's set has to change.
Isn't it Murgatroyd?
bravo
thanks! Do you have this lection as text?
i have a feeling i might be mergatoid
Thanks :)
can anyone pls tell me what TMA stands for?
The Marriage Algorithm
serhan kars The Mating Algorithm
Turnaround Management Association. He isn't referring to the mating algorithm.
Mergatoyd not Mergatoid ;-) -- Heavens to Mergatoyd
Poor brad :(
У нас проходят тоже самое за пол пары, что в мит за 7 лекций)
короче не значит лучше
This guarantees that all of you will no grasp of what is being taught
I would not pass this class
The girl in front is using steno machine!
But they anyways share well formatted digital notes
Hey MIT, Can you keep the camera close to the board? No one wants to see the student asking a question.
What a weird lecture: completely random topics, no formal theory of anything written on the black board, just a bunch of sample math problems thrown away to the students...
this is a most typical lectures series !! .... this course is for essential maths for computer science if you know the topic already you can skip it... you ask about formal theory her let me answer you
- the main theory in lecture 6,7 is a "graph theory" .... the professor start with some definitions (which is the most importent in maths) about graphs .. and then start to say some common associated problems with "graphs theory" such as "Coloring" (in lecture 6) and "matching problem" (in this lecture) which have a several of applications like dating apps , networking , exams scheduling and many more mentioned in the lecture 6,7 ... and in my opinion he choose a clever and interesting examples to state those problems !!
damn you sound so insecure