When I studied law I had several sociology classes; the laugh I had because of 26:25 reminded me why I hated them, and turn myself to science. :) Great lesson, MIT, you should be awarded for spreading those classes online. This is really some true human behaviour. Thank you from someone living far away.
@@alskfaslkfjkl no, it doesn’t. Sure, one subgroup can have more promiscuous men on average but this has to be evened out by the same number of equally promiscuous women in some other subgroup (or across multiple subgroups) assuming the same # of men and women. And vice versa. So on average each subgroup must have an equal number of romantic relations for men and women. For minority and non-minority studying groups, this is even more so the case since studying with people outside of one’s subgroup (people in the same university or class) for a midterm makes less sense. So minority students must study with more non-minority students on average than the reverse because there are fewer of them. That’s it. And making any generalized conclusion based on an outlier subgroup would be the very definition of tendentious. Or if the subgroup refers to "minority" and "non-minority", it's actually the other way around. In "an absolute sense" each subgroup studies with the other exactly the same number of times. It only becomes "more than" when that number is taken relative to the subgroup sizes (the average).
@@alskfaslkfjkl Okay, so you just don't understand the point of this lecture, the question, or maybe just the math in general. Here is the Boston Globe's finding applied to your scenario: 1 out of 10 is 10%, true, however we are calculating the average number of non-minority students each of the minority students studies with and vice versa, not how many minority students have studied with non-minority students at all. That one student studied with 100 non-minority students. (100+0+0...)/10 = 10, i.e. *on average* each minority student studied with 10 non-minority students. The 100 non-minority students each studied with exactly 1 minority. (100)/100 = 1, i.e. on average each non-minority student studied with 1 minority student (obviously, since they all studied with exactly 1 minority student the average is also 1). 10 is greater than 1 (I can't prove this but trust me). Thus minority students studied with ten times more non-minority students on average than the other way around. So sorry, no, you're wrong. Minority students do indeed study with more non-minority students on average than the other way around. Now I said I don't think you understood the point of the lecture, question, or maybe simply don't understand the math, because your example would probably be a more useful statistic than the example in the lecture. Why? The reason minority students on average study with ten times more non-minority etc. etc. IS BECAUSE there are ten times more non-minority students than minority students. If just one minority student studied with just one non-minority student this would still be true. Mathematically the statement "minority students study with more non-minority students on average than the other way around" is equivalent to "there are more non-minority students than minority students". That's it. There's no social commentary, no hidden psychological phenomena, nothing. That's the point, the professor is saying that it's a very dumb question whose solution can have no meaningful implications besides "minority students are in the minority". Yet the Boston Globe still published this "explosive survey" to draw some sort of groundbreaking conclusion.
@@alskfaslkfjkl Wow, I really hope you’re a professor at a community college at best and not a prestigious university like MIT. And I’m not sure what to tell you, your “counterexample” doesn’t disprove anything. I literally wrote out all the math for you in my last comment, look at it and disprove me if you’re so confident about your obviously incorrect notions. Otherwise, I want you to rewatch the video and *think* about the mathematics, work it out for yourself, and take the time to understand why the professor brought up that headline. And why your “counterexample” is totally, 100% irrelevant. As I already said, taking the Boston Globe’s findings and applying it to your counterexample demonstrates that it only proves the Boston Globe’s headline. And I’ve done all I can to show why that is the case. What mathematics field do you teach and where? Normally I would not be so presumptuous as to think I am more capable at math or logic than a professor, but for you I make an exception. You clearly do not understand the point the professor and I are making and your blatant disregard for all the math and explanations I presented in my last comment reflects that. “Hold yourself to a higher standard than respected MIT professors”, yeah okay alksfaslkfjkla.
This is a solid quality of teaching by MIT professor Tom Leighton. The introduction to Graph Theory and Coloring is the backbone of Computer Science. There are many application of Graph Theory and Coloring in the world in which we live.
This session reminds me of junior high school mathematics. Too bad my college profs were not as fun and interesting as Prof. Leighton. This was a fun watch, even for an old guy like me. I bet he made the mistake at the end to make sure the kids are paying attention.
I have been watching videos on Graph Theory but this is the first time some one is providing an actual real life example and I now finally understand what the heck graphs are good for. Man, why are maths professors so bad? Instead of showing their ability, they need to focus on actually teaching aka transmitting real knowledge to students with real life examples. Thank you!
At around 52:19, the professor said that induction on d didn't work and would be a nightmare. But I can prove it by induction on d only. First, pick a the vertex whose deg is d + 1, removed it and all the edges containing it. If the remaining graph stil has a vertex whose deg is d + 1, then keep doing like above untill we get the final graph such that each degree is
I’m surprised how simple these lectures are compared to other lectures. Is it because the material is simple, or is it because the presentation is clear?
He just talks more than what he writes so it gives you enough time to keep up. Past 2 lessons were so much writting on blackboard that you have to pause a lot
I think at 1h5m the alternative proof suggested does work. The student is saying that as d+1 is the maximum chromatic number for the clique case, other cases can't be higher since any other case is just the clique with one or more edges removed.
Students were totally right when they said the worst case situation because this was as well how lecturer proved by adding one p(n+1) as it has d adjacents.
If the video would have been named ' MIT teachings on sex studies ' like the professors daughter pointed out it would easily have more then a million views. :D
The issue with the studies vs. the calculation is that the calculation is on a closed, self-contained set. The studies are not, because there are certainly partners who are outside of the study universe ... what impresses me about this video is how much material is covered in a relatively short time. I am an MIT grad from many many years back and it is a reminder of the fast pace and intensity of those days.
The term 'opposite-gender partners' is not the best choice for the example discussed in the lecture. The example is actually about comparing the average number of opposite-gender sexual relations, not the number of distinct opposite-gender sexual partners. The lecturer mentioned that the sum of the degrees of the F vertices represents the total number of opposite-gender partners for all females. However, this isn't entirely accurate. Imagine a scenario where all females have the same male partner. In that case, the total number of opposite-gender partners for all females would be just one, which doesn’t reflect the actual total number of edges in the graph. A more accurate description would be to say that the sum of the degrees of the F vertices represents the total number of sexual relations with opposite-gender partners for all females. This phrasing correctly accounts for the actual number of interactions (edges) in the graph, rather than just the number of distinct partners.
Stupid question: isn't the fixed wall behind the two mobile blackboards, a third blackboard as well? Because they always write on the two mobile ones only.
Yes(I went there). Sometimes, but not usually, profs would use that *forbidden 3rd back blackboard*. To answer your question, when they did that, then you couldn't follow lecture as good bcuz part of the board was blocked
There is only one problem with the "promiscuity calculation": The assumption is that all men and all women find at least one partner. However, everyone who understands biology/behavioural science should know that not everyone will. Moreover, there will be always more men than women who *cannot* find a partner. Therefore - in a realistic scenario - there will be fewer "male" nodes than "female" nodes in the graph. And, I guess, no one wants to admit what that implies...
There is no problem with his calculation even then, it's just that average is really useless indicator here. Manually calculate some extreme examples to convince yourself of this
Assignments with solutions : ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-042j-mathematics-for-computer-science-fall-2005/assignments/
The average for women is more evenly distributed, the distribution for men is heavily skewed, the average for the top 20% would be miles higher than the average of the bottom 80% which is close to 0. The way its set up, if you randomly picked a man and woman off the street, the chance that the man has more than the women is extremely low as theres an 80% chance the man will be in the bottom 80%. Thats the only useful observation you can make.
Curious that a question that would seem to be entirely empirical turns out to be a priori, given one empirical bit of information: the ratio 'number of men/number of women'.
I just ge a bit frustrated when I notice there are empty chairs in the room... If I had known I would have taken a flight from Sweden just to attend...
I like this approach to graph theory, but the population statistic cannot sufficiently analyze the social patterns studied in a real survey. The graph theory assumes the maximum amount of edges based on the maximum amount of nodes. A real survey is a statistical method for refining that graph theory, therefor the opinion of a professional statistician would be that the 2500 samples from the University of Chicago survey is a better estimate than universal graph theory. The graph theory would estimate a minimum average by virtue of the population of the universal set.
There are not any recitation videos but there are recitation notes: ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-042j-mathematics-for-computer-science-fall-2010/recitations/. Best wishes on your studies!
If the study is on the median instead of avarage number of partners then it's not that trivial. Also the surveys still tell a story. Men tend to overexaggerate the number of partners they had much more compared to women.
At first, we proved that the basic algorithm will use at most d + 1 colors on any graph. But then in the star graph, the basic algorithm could only color it with two colors no matter the order but according to the theorem we should be able to color it with d + 1 colors using the basic algorithm but we can't color it with more than two colors using basic algorithm. Isn't this a contradiction? Plz, help me understand.
True but misleading, one can draw completely lopsided graphs where M or W have way more more connections (increasing the number of "singles" on the other side) and still get the same average. The median however, will tell a completely different story.
at 1:05:19 of the video The student claims that Kn-complete graph is Worst Case. In fact, it is true. You give me any graph with n-nodes, and I say it is the subset of Kn-complete graph with n-nodes Because you could some edges from Kn-complete graph to form your own graph. Now let say Kn-complete graph require at max X colors to get legally colored, it means your graph has some X colors. Hence it is the proof to theorem. I think, it is better than Induction Method.
EDIT : Because you could REMOVE some edges from Kn-complete graph to form your own graph. Now let say Kn-complete graph require at max X colors to get legally colored, it means your graph REQUIRE AT MAX X colors, TOO. Hence it is the proof to theorem. I think, it is better than Induction Method.
Doesn't he ask the wrong question? Instead of finding the average number of edges per node, shouldn't he be comparing the percentage of nodes with an incidence of 1 in both groups. That would be a better modeling of promiscuity i believe.
I wonder what exactly do Vm and Vw cover. I think it's only correct if they are the number of men and women who was once in a relationship... it can't be the total number otherwise it's not correct. Taking an extreme example, if there are 100 men and 100 women, but only 50 women have ever entered into a relationship, then it should be the number 50 that is used here rather than 100. But how can we know this number?
Writing this for future people cause I was confused about this too. He is correct even in extreme examples, write out simple extreme example (with say 10 women and 10 men) and follow procedure he did, and calculate AVERAGE NUMBER OF DIFFERENT PARTNERS manually too, compare results, it checks out
Well, he is just drawing numbers. The exercise means nothing and it is unscientific to model life this naively with no argument whatsoever. He puts axioms that oversimplifies the problem, thus the result is (as he says) stupid. It's a good example for studying graphs though.
we have to consider degree of each node if d is in predicate, which is pretty hard whereas in case of nodes, it is pretty to prove the induction hypothesis
One major point of this lecture is Graph Coloring. Winston had an entire lecture on Graph Coloring in 6.034 that is extremely good: ruclips.net/video/dARl_gGrS4o/видео.html This is a bit strange as 6.042J is a pre-req for 6.034 . Why is same topic covered twice?
+Akshit Sharma thanks for the answer, I understand what's on your first statement, but I got a little confuse with your second statement: 1.0325x is female population and 1x is male population.
Roy lee it's not difficult. 1.0325 = 103.25% (decimal to percentage conversion) That means the number of opp. gender partners men have is 103.25% of that of women. That's the same as saying men have (100-103.25)% = 3.25% more opposite gender partners than those of females.
The promiscuous example is oversimplified and interpretation of the problem is incorrect. Here it is an example of the proper interpretation: Let it be a set of 2 males and a set of 11 females. One male has 10 females, the other male has only one female. So. There are 11 edges in total over 2 males that is an average of 5.5 edges per male, and an average of 1 edge per female. That is, promiscuity is 5.5 higher in the male side. And no promiscuity in the female side. Promiscuity is not the ratio of populations but the degree of the node. I believe that mathematics is an excellent tool to study nature but we need to be very careful how we translate natural phenomena into the language of math, otherwise we could get erroneous information.
But the assumption that every woman has dated and that every man has dated is flawed. If so, then the ratio of men to women dating would be tipped, and so greater numbers are possible
But the metric is average number of partners. To get the average you need to divide by the total number -- all men, not just all men who have had partners.
I know this was explained for the purpose of learning graphs but suppose only one man had sex with the entire population of women and the other men have never had sex. The average would've still be 1.03. Wouldn't some more meaningful indicators be the mode and the median in this case?
@@satishchandrachitrapu6471 I think he means that average does not take into consideration how the data is distributed, because it's just sum over frequency. On the other hand Median gives us number of relationships the person in the exact middle of the distribution would have. So that is a better indicator of how the distribution is. For example let's say values are - 1,3,2,1,3,4,99999,8,5,2 now the average would be - around 1000 and this would lead you to believe that on average everyone has 1000 relations... On the other hand median would be 3 which is a much more reliable metric in this scenario.
In that first example involving two opposite genders, you are assuming that everybody in the population performs that "act". So, I think it is important to note that. I write act because I think my comment got auto removed multiple times for using the original word.
When I studied law I had several sociology classes; the laugh I had because of 26:25 reminded me why I hated them, and turn myself to science. :)
Great lesson, MIT, you should be awarded for spreading those classes online. This is really some true human behaviour. Thank you from someone living far away.
@@alskfaslkfjkl no, it doesn’t. Sure, one subgroup can have more promiscuous men on average but this has to be evened out by the same number of equally promiscuous women in some other subgroup (or across multiple subgroups) assuming the same # of men and women. And vice versa. So on average each subgroup must have an equal number of romantic relations for men and women.
For minority and non-minority studying groups, this is even more so the case since studying with people outside of one’s subgroup (people in the same university or class) for a midterm makes less sense. So minority students must study with more non-minority students on average than the reverse because there are fewer of them. That’s it.
And making any generalized conclusion based on an outlier subgroup would be the very definition of tendentious.
Or if the subgroup refers to "minority" and "non-minority", it's actually the other way around. In "an absolute sense" each subgroup studies with the other exactly the same number of times. It only becomes "more than" when that number is taken relative to the subgroup sizes (the average).
@@alskfaslkfjkl Okay, so you just don't understand the point of this lecture, the question, or maybe just the math in general. Here is the Boston Globe's finding applied to your scenario:
1 out of 10 is 10%, true, however we are calculating the average number of non-minority students each of the minority students studies with and vice versa, not how many minority students have studied with non-minority students at all. That one student studied with 100 non-minority students. (100+0+0...)/10 = 10, i.e. *on average* each minority student studied with 10 non-minority students.
The 100 non-minority students each studied with exactly 1 minority. (100)/100 = 1, i.e. on average each non-minority student studied with 1 minority student (obviously, since they all studied with exactly 1 minority student the average is also 1).
10 is greater than 1 (I can't prove this but trust me). Thus minority students studied with ten times more non-minority students on average than the other way around.
So sorry, no, you're wrong. Minority students do indeed study with more non-minority students on average than the other way around. Now I said I don't think you understood the point of the lecture, question, or maybe simply don't understand the math, because your example would probably be a more useful statistic than the example in the lecture. Why? The reason minority students on average study with ten times more non-minority etc. etc. IS BECAUSE there are ten times more non-minority students than minority students. If just one minority student studied with just one non-minority student this would still be true. Mathematically the statement "minority students study with more non-minority students on average than the other way around" is equivalent to "there are more non-minority students than minority students".
That's it. There's no social commentary, no hidden psychological phenomena, nothing. That's the point, the professor is saying that it's a very dumb question whose solution can have no meaningful implications besides "minority students are in the minority". Yet the Boston Globe still published this "explosive survey" to draw some sort of groundbreaking conclusion.
@@alskfaslkfjkl Wow, I really hope you’re a professor at a community college at best and not a prestigious university like MIT. And I’m not sure what to tell you, your “counterexample” doesn’t disprove anything. I literally wrote out all the math for you in my last comment, look at it and disprove me if you’re so confident about your obviously incorrect notions. Otherwise, I want you to rewatch the video and *think* about the mathematics, work it out for yourself, and take the time to understand why the professor brought up that headline. And why your “counterexample” is totally, 100% irrelevant. As I already said, taking the Boston Globe’s findings and applying it to your counterexample demonstrates that it only proves the Boston Globe’s headline. And I’ve done all I can to show why that is the case.
What mathematics field do you teach and where? Normally I would not be so presumptuous as to think I am more capable at math or logic than a professor, but for you I make an exception. You clearly do not understand the point the professor and I are making and your blatant disregard for all the math and explanations I presented in my last comment reflects that.
“Hold yourself to a higher standard than respected MIT professors”, yeah okay alksfaslkfjkla.
@@alskfaslkfjkl No, *I* tried to help you understand by thoroughly showing and explaining the math. You did nothing of the sort.
This guy has to be the best mathematics lecturer in the world.
The professor is really good! Thanks MIT!
I have encountered a graph problem, that is not degree constrained, but there exist an matching, how to identify that for big graphs?please help
omg my mind was blown when the cardinality of the edges cancelled out
Dad: What are you learning?
Me: Who shagged who.
Dad: Oh okay! Just make sure you are not watching anything about sex.
That is an execellent teacher
If my graph theory course was this interesting, I wouldn't have skipped all but 6 of the lectures :D
I have encountered a graph problem, that is not degree constrained, but there exist an matching, how to identify that for big graphs?
@Susuya Juuzou I'm sorry if I'm wrong but isn't this an introduction to graphs and graph theory
Thank you MIT for sharing such quality teaching
This is a solid quality of teaching by MIT professor Tom Leighton. The introduction to Graph Theory and Coloring is the backbone of Computer Science. There are many application of Graph Theory and Coloring in the world in which we live.
This is really fun! I studied in many classes about graph theory but only this lesson gave me examples which are similar with real life.
Damn, He is the coolest Prof ever. I wish he is the Prof all through collage
This session reminds me of junior high school mathematics. Too bad my college profs were not as fun and interesting as Prof. Leighton. This was a fun watch, even for an old guy like me. I bet he made the mistake at the end to make sure the kids are paying attention.
WOW. Simply amazing from CEO of Akamai. He also linked these problems to what was solved at Akamai.
I watched the whole lecture and answered the questions as a real student
it's really fun to learn from MIT OCW
Graph coloring starts around 31:00
I have been watching videos on Graph Theory but this is the first time some one is providing an actual real life example and I now finally understand what the heck graphs are good for. Man, why are maths professors so bad? Instead of showing their ability, they need to focus on actually teaching aka transmitting real knowledge to students with real life examples. Thank you!
I have encountered a graph problem, that is not degree constrained, but there exist an matching, how to identify that for big graphs?please help
mathematics does not need real life examples
This is pure gold lying here for free!
I think the last two lectures by that other professor are put in just to remind us how good this one is.
At around 52:19, the professor said that induction on d didn't work and would be a nightmare. But I can prove it by induction on d only. First, pick a the vertex whose deg is d + 1, removed it and all the edges containing it. If the remaining graph stil has a vertex whose deg is d + 1, then keep doing like above untill we get the final graph such that each degree is
Great Course
Basic Graph Coloring Algorithm : 43:00
I’m surprised how simple these lectures are compared to other lectures. Is it because the material is simple, or is it because the presentation is clear?
He just talks more than what he writes so it gives you enough time to keep up. Past 2 lessons were so much writting on blackboard that you have to pause a lot
Given it is a introductory course on mathematic proofs, it is well explained; but yeah it is not very in depth considering the material.
The lecturer does more talking that writing, which is easy to keep up with.
You could even follow along with audio only.
Thank you Prof. Tom. GOD bless
mind blown at 23:21
World class teaching ! this is MIT
I think at 1h5m the alternative proof suggested does work. The student is saying that as d+1 is the maximum chromatic number for the clique case, other cases can't be higher since any other case is just the clique with one or more edges removed.
Other cases include graphs with more elements, too.
damn. guy does the university's job for them and saves the day. and having himself a stess-free christmas. a real chad
This lecture is really entertaining, spent some nice time.
Awesome learning experience, sir. Thank you.
Students were totally right when they said the worst case situation because this was as well how lecturer proved by adding one p(n+1) as it has d adjacents.
If the video would have been named ' MIT teachings on sex studies ' like the professors daughter pointed out it would easily have more then a million views. :D
Christoph G. Exactly
im not sure if pimping out science is the right idea, people are either interested in this stuff or they're not...
Thats called marketing brother!
Great professor! Learned a lot today!
I have became your fan! Thankyou very much for you wonderfull lecture
I'm amazed this guy hasn't been cancelled
I don't understand... why do you say that?
Why do you want him cancelled?
@@Nora-bv2el I don't want him to be cancelled, I'm just amazed someone wasn't "triggered" by his discussion of two genders
This was an awesome lecture
5 min into the lecture, I've had to grab my earphones 👀
I loved it, really useful lesson.
That's very interesting lecture, I really enjoyed watching.
Lies again? MRT MIT
That's excellent.... outstanding..
It'd be nice if the videos from this playlist had the subject in the title and not just in the description.
Now i learnt graph theory. Amazing lecture
Example : 11:00
Vera level!!!
Perhaps the Chicago University and ABC News studies looked at the median and not the mean?
The issue with the studies vs. the calculation is that the calculation is on a closed, self-contained set. The studies are not, because there are certainly partners who are outside of the study universe ... what impresses me about this video is how much material is covered in a relatively short time. I am an MIT grad from many many years back and it is a reminder of the fast pace and intensity of those days.
@@bobnewell7888 Ha, it's been so long, I don't even remember what my comment was about.
Loved this.
Awesome professor.
The term 'opposite-gender partners' is not the best choice for the example discussed in the lecture. The example is actually about comparing the average number of opposite-gender sexual relations, not the number of distinct opposite-gender sexual partners.
The lecturer mentioned that the sum of the degrees of the F vertices represents the total number of opposite-gender partners for all females. However, this isn't entirely accurate. Imagine a scenario where all females have the same male partner. In that case, the total number of opposite-gender partners for all females would be just one, which doesn’t reflect the actual total number of edges in the graph.
A more accurate description would be to say that the sum of the degrees of the F vertices represents the total number of sexual relations with opposite-gender partners for all females. This phrasing correctly accounts for the actual number of interactions (edges) in the graph, rather than just the number of distinct partners.
Thank you sir ...really awesome
Sociology PhD from down the street(a.k.a Harvard) xD
You'd be better off with a degree from a community tech school, it's still sociology...
John Waas What? :\
+xXxBladeStormxXx Just be nice, and humble. If you lost a character, you lost 8 bits.
that's negligent given I spend billions of bit on random crap. there are better motives out there.
Literally, just down the street :-)
Feels like I'm studying from Mobius. A great teacher indeed.
Ik this is 10 years late but this was a great lecture
Awesome lec 🔥🔥
Stupid question: isn't the fixed wall behind the two mobile blackboards, a third blackboard as well? Because they always write on the two mobile ones only.
Yes(I went there). Sometimes, but not usually, profs would use that *forbidden 3rd back blackboard*. To answer your question, when they did that, then you couldn't follow lecture as good bcuz part of the board was blocked
24:30 I think he meant to write Vw over Vm
but he wrote Vm twice
I love MIT
Excellent lecture..
teaches like a God
There is only one problem with the "promiscuity calculation":
The assumption is that all men and all women find at least one partner.
However, everyone who understands biology/behavioural science should know that not everyone will.
Moreover, there will be always more men than women who *cannot* find a partner.
Therefore - in a realistic scenario - there will be fewer "male" nodes than "female" nodes in the graph.
And, I guess, no one wants to admit what that implies...
There is no problem with his calculation even then, it's just that average is really useless indicator here. Manually calculate some extreme examples to convince yourself of this
That should not be a problem, The degree can then be 0 , and the thing is for every woman who finds a partner, there is a man, So the logic holds
Mind blowing
Where are lectures 1 to 5?
ruclips.net/p/PLB7540DEDD482705B
@@mitocw Thanks. I got to this video through a different playlist which started at lecture 6.
Amazing :o !
Wish he taught at my school, I'd live in office hours 😊😊😊
Assignments with solutions : ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-042j-mathematics-for-computer-science-fall-2005/assignments/
@Michael Nesler You're welcome :)
Much more interesting than some netflix tv series lmao
The average for women is more evenly distributed, the distribution for men is heavily skewed, the average for the top 20% would be miles higher than the average of the bottom 80% which is close to 0. The way its set up, if you randomly picked a man and woman off the street, the chance that the man has more than the women is extremely low as theres an 80% chance the man will be in the bottom 80%. Thats the only useful observation you can make.
Actually, it can't simply divide the number of nodes, because there can be nodes with zero degree, so they shold be excluded first.
Does not matter.
we are looking for the average
Curious that a question that would seem to be entirely empirical turns out to be a priori, given one empirical bit of information: the ratio 'number of men/number of women'.
27:01 because minority !
0:25 dude sitting in the gray shirt shares meme with his friend
mind.blow();
I just ge a bit frustrated when I notice there are empty chairs in the room... If I had known I would have taken a flight from Sweden just to attend...
This is pretty old, is this still up to date and everything?
Indeed it is. Graph theory actually changes very little. Just compare a 20 year old book, to a current one.
Fundamentals in maths rarely change.
pretty much. I use these this as supplementary material for my course and its in 2021.
Can I watch just graph theory lectures skipping the first 5
I like this approach to graph theory, but the population statistic cannot sufficiently analyze the social patterns studied in a real survey. The graph theory assumes the maximum amount of edges based on the maximum amount of nodes. A real survey is a statistical method for refining that graph theory, therefor the opinion of a professional statistician would be that the 2500 samples from the University of Chicago survey is a better estimate than universal graph theory. The graph theory would estimate a minimum average by virtue of the population of the universal set.
18:45 "In the US there's about 147.6 men" xD
Are there any recitation videos for this course?
There are not any recitation videos but there are recitation notes: ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-042j-mathematics-for-computer-science-fall-2010/recitations/. Best wishes on your studies!
If the study is on the median instead of avarage number of partners then it's not that trivial. Also the surveys still tell a story. Men tend to overexaggerate the number of partners they had much more compared to women.
At first, we proved that the basic algorithm will use at most d + 1 colors on any graph. But then in the star graph, the basic algorithm could only color it with two colors no matter the order but according to the theorem we should be able to color it with d + 1 colors using the basic algorithm but we can't color it with more than two colors using basic algorithm.
Isn't this a contradiction? Plz, help me understand.
it is atmost d+1, d+1>2 so the theorem still holds for this case.
The sex partner ratio still tricks me. Is the calculation involving average relationship telling the true stats story?
It would be a different story if we used median instead of average, and I think this is what the surveys are telling us.
True but misleading, one can draw completely lopsided graphs where M or W have way more more connections (increasing the number of "singles" on the other side) and still get the same average. The median however, will tell a completely different story.
at 1:05:19 of the video
The student claims that Kn-complete graph is Worst Case. In fact, it is true.
You give me any graph with n-nodes, and I say it is the subset of Kn-complete graph with n-nodes Because you could some edges from Kn-complete graph to form your own graph.
Now let say Kn-complete graph require at max X colors to get legally colored, it means your graph has some X colors.
Hence it is the proof to theorem. I think, it is better than Induction Method.
EDIT :
Because you could REMOVE some edges from Kn-complete graph to form your own graph.
Now let say Kn-complete graph require at max X colors to get legally colored, it means your graph REQUIRE AT MAX X colors, TOO.
Hence it is the proof to theorem. I think, it is better than Induction Method.
Then that's another theorem you have to prove first, and then you have to prove that this n-complete graph is the worst case.
if teacher == "This Great at Teaching":
blow(mind)
are np completeness solved or not yet ?
Unlikely, otherwise it will be such big news you'll hear about it.
i have no idea what is a quantum satellite ... how did you get this information ?
Bessalah Amar you will never find out because hazard77 was kidnapped by the chinese government because he released their secret .
@@hazzard77 how the hell do you know that.
@@thangible r/woooosh
1.07.55 at this time indian student is asking doubt # proud of it
I have encountered a graph problem, that is not degree constrained, but there exist an matching, how to identify that for big graphs?please help
@1:02 he says we are working with d+1, so I'm wondering where is this coming from?
Wouldn't single people be considered in the male and female population aka nodes with zero degree?
They wouldn't even be in the set of people who had heterosexual partners.
Doesn't he ask the wrong question? Instead of finding the average number of edges per node, shouldn't he be comparing the percentage of nodes with an incidence of 1 in both groups. That would be a better modeling of promiscuity i believe.
He is just doing what those research claimed to do.
He's talking out of his ass. He presents an awful model all around.
I wonder what exactly do Vm and Vw cover. I think it's only correct if they are the number of men and women who was once in a relationship... it can't be the total number otherwise it's not correct. Taking an extreme example, if there are 100 men and 100 women, but only 50 women have ever entered into a relationship, then it should be the number 50 that is used here rather than 100. But how can we know this number?
Writing this for future people cause I was confused about this too. He is correct even in extreme examples, write out simple extreme example (with say 10 women and 10 men) and follow procedure he did, and calculate AVERAGE NUMBER OF DIFFERENT PARTNERS manually too, compare results, it checks out
Well, he is just drawing numbers. The exercise means nothing and it is unscientific to model life this naively with no argument whatsoever. He puts axioms that oversimplifies the problem, thus the result is (as he says) stupid. It's a good example for studying graphs though.
thank you
I'm here because of Zach Star.
Zach Star?
Robotic cameras siting in the classroom.
52:12 can somone explain why should i prefer working on n instead of d, and is this generally a good approach? i didn't understand very well.
we have to consider degree of each node if d is in predicate, which is pretty hard whereas in case of nodes, it is pretty to prove the induction hypothesis
I have encountered a graph problem, that is not degree constrained, but there exist an matching, how to identify that for big graphs? Please help
One major point of this lecture is Graph Coloring. Winston had an entire lecture on Graph Coloring in 6.034 that is extremely good: ruclips.net/video/dARl_gGrS4o/видео.html
This is a bit strange as 6.042J is a pre-req for 6.034 . Why is same topic covered twice?
is your daughter at MIT now??
Haaha
How did he deduce 3% out of 1.0325?
+Roy lee (1.0325-1)/1=0.325=3.25%
1.0325x is female population and 1*x is male population.
+Akshit Sharma thanks for the answer, I understand what's on your first statement, but I got a little confuse with your second statement: 1.0325x is female population and 1x is male population.
+Roy lee 1.0325 is the ratio of female population to the male population
Roy, i think those lectures are a bit too advanced then for you))
Roy lee it's not difficult.
1.0325 = 103.25% (decimal to percentage conversion)
That means the number of opp. gender partners men have is 103.25% of that of women.
That's the same as saying men have (100-103.25)% = 3.25% more opposite gender partners than those of females.
The promiscuous example is oversimplified and interpretation of the problem is incorrect. Here it is an example of the proper interpretation: Let it be a set of 2 males and a set of 11 females. One male has 10 females, the other male has only one female. So. There are 11 edges in total over 2 males that is an average of 5.5 edges per male, and an average of 1 edge per female. That is, promiscuity is 5.5 higher in the male side. And no promiscuity in the female side. Promiscuity is not the ratio of populations but the degree of the node.
I believe that mathematics is an excellent tool to study nature but we need to be very careful how we translate natural phenomena into the language of math, otherwise we could get erroneous information.
You just explained what was done in the lecture. Same analysis and same figure (#females/#males = 11/2)
But the assumption that every woman has dated and that every man has dated is flawed. If so, then the ratio of men to women dating would be tipped, and so greater numbers are possible
he makes no such assumption
But the metric is average number of partners. To get the average you need to divide by the total number -- all men, not just all men who have had partners.
I have encountered a graph problem, that is not degree constrained, but there exist an matching, how to identify that for big graphs?please help
28:48
I know this was explained for the purpose of learning graphs but suppose only one man had sex with the entire population of women and the other men have never had sex. The average would've still be 1.03.
Wouldn't some more meaningful indicators be the mode and the median in this case?
I am not sure I understand what you mean. Could you elaborate with examples on how median or mode are more meaningful indicators?
@@satishchandrachitrapu6471 I think he means that average does not take into consideration how the data is distributed, because it's just sum over frequency. On the other hand Median gives us number of relationships the person in the exact middle of the distribution would have. So that is a better indicator of how the distribution is.
For example let's say values are - 1,3,2,1,3,4,99999,8,5,2
now the average would be - around 1000 and this would lead you to believe that on average everyone has 1000 relations...
On the other hand median would be 3 which is a much more reliable metric in this scenario.
The teacher presents an oversimplification that basically means nothing. Just a graph example.
49:18
In that first example involving two opposite genders, you are assuming that everybody in the population performs that "act". So, I think it is important to note that.
I write act because I think my comment got auto removed multiple times for using the original word.