Mathematics

7:00 PM

if i can count the number of connected graphs by subtracting disconnected graphs, where disconnected graphs are just segregated connected graphs unto themselves, how do I set this up combinatorically?

i keep double(/triple/quad/quint../)counting stuff

Peter Tamaroff

@TobiasKildetoft Dude.

shobon

@user74918, inclusion exclusion?

user74918

7:15 PM

@shobon That is what I thought too but I couldn't set it up properly

shobon

@user74918, I dont know what the actual problem is just guessing from "i keep double(/triple/quad/quint../)counting stuff"

@user74918, you just want to count the number of connected graphs? Why did yo mention bipartite

user74918

i am counting the number of connected graphs if you know how many vertices in set A and set B

shobon

ah

so there's exactly 1 if |A|=2 and |B|=1?

user74918

right

shobon

ok this seems interesting

user74918

7:17 PM

exactly 1 if either |A| or |B| is 1 actually

CBenni

real quick, is there a theorem that shows a sufficient requirement for a subset of $\ell^2$ to be compact?

user74918

so if A=2 and B=2 there are 5 connected graphs

shobon

@user74918, wow I have no idea how to count this!

I'm trying to solve |A| arbitrary |B|=2

we need both B's to share at least one vertex of A

user74918

in a connected graph, all vertices are visited

shobon

given that, you can sort of consider both B elements 'the same' count the graphs on that then 'split them'

but to do that you need to count the number of edges into |B|-1

user19161

7:21 PM

Oh yeah @skull would be good if you changed the name of the fake JL, thanks!

shobon

so maye you can count this by strengthening what you're counting to include the number of edges?

user74918

right

right now i have a way to count the number of graphs (disconnected and connected alike) if you know A and B

shobon

do you get what I mean though? does it seem doable

user74918

so i am trying to subtract back out the disconnected graphs

shobon

isnt that just 2^{|A||B|}?

since there's at most that many edges

or are you factoring out permutations of A and B?

user74918

7:23 PM

that counts graphs where some vertices are not actually reached by any path

shobon

did you get what I mean earlier

user74918

sort of, i'm not very math literate when it comes to jargon or wording

shobon

@user74918, for example

user74918

the B=2 case I believe it just 3^A-2^A (has a closed form)

shobon

I just saw a problem with it

but let me explain anyway

@user74918, |A|=3, |B|=2, both vertices of b must connect to one of the 3 vertices of |A| then you can ignore that one and 'merge' the two vertices and you have |A|=2,|B|=1 situation where there is exactly 1 graph with 2 edges going into B - so there's 3 ways you could split it - 9 graphs in total [but this is wrong]

user74918

7:26 PM

there is a recursive solution if that is what you're implying (I just can't derive it yet)

shobon

you already have a solution?

user74918

yes but i don't understand it at all

so i am trying to figure out a new one

shobon

this problem is really difficult though

so what was your idea starting with counting all graphs that have a path through every vertex?

Theorem

hi everyone

skullpatrol

hi

user74918

7:28 PM

let T represent the total number of graphs where there is an edge touching every vertex (so no vertex is just "left out cold," I don't know the right math jargon for this)

the connected graphs is just T - disconnected graphs

disconnected graphs are just partitions of connected graphs

shobon

@user74918, I see

so possibly writing the number of disconnected graphs as a sum over connected graphs?

then hopefully you'd get a recurrence relation

user74918

the problem is that i can't just loop over 1<=i<A, 1<=j<=B and then say "ok, we have a partition that connects to i vertices on the left and j vertices on the right, then count how many ways the rest could be arranged" because this multicounts stuff down the road

shobon

@user74918, sorry to switch to something totally different but.. what about counting the total bitpartite graphs with |A| edges (suppose |A|>|B|), then |A|+1, then |A|+2 up to |A||B|

the edge cases seem easy (|B|p(|A|-|B|) and 1)

user74918

what does this total do?

what approach does it give for A=3 B=2 rather

3, 4, 5, 6

user19161

Hey @anon I am beginning to think that you are Marvis, lol.

user19161

7:34 PM

By the way, I saw that Gone is really gone and Math Gems has been suspended.

Peter Tamaroff

@JasperLoy Dude.

shobon

@user74918, I was just throwing out a different approach but there's a difficulty with this way too: when the number |A|+k gets large you can't freely add edges

Peter Tamaroff

This is alright, yes? @JasperLoy

user74918

@JasperLoy I thought BD=MG=G

Peter Tamaroff

@JasperLoy Bill is Math Gems, pho sho.

user74918

7:35 PM

@shobon ah ok I wasn't sure what it was doing though

user19161

@PeterTamaroff +1

user19161

@PeterTamaroff Indeed.

Peter Tamaroff

@JasperLoy Holy monkey. +6 already.

shobon

@user74918, really not seeing any way to approach this problem that looks promising

user74918

yeah it is tough

shobon

7:39 PM

@user74918, what about starting with the number of 1 component graphs then trying to 'invert' that to count bipartite ones

since we know a graph of N vertices can be split into total bipartite ones for each solution of |A|+|B|=N

user74918

does that matter if A and B have labeled vertices?

shobon

I dont know

user19161

@PeterTamaroff Good for you. Just remember me when you win the Fields medal.

shobon

@user74918, that way wont work either

user74918

any way to use inclusion exclusion for the approach mentioned earlier?

shobon

7:43 PM

I dont see how

user74918

i think it can work but i just couldnt set it up properly

skullpatrol

@JasperLoy What made you decide to come back?

user19161

@skullpatrol Nothing, just felt like it. I might delete the new account anytime, lol.

skullpatrol

@JasperLoy Lonely?

user19161

@skullpatrol Nope. I am used to being alone.

anon

7:49 PM

the internet equivalent of not being afraid of death.

skullpatrol

@JasperLoy Bored?

user19161

@anon The great anon!

user19161

@skull You and I really confused Marvis! He asked if we were the same person...

Peter Tamaroff

@anon Dude.

One thinks the Jacobian matrix is crazy, and then comes the Hessian.

user19161

@peter Your original answer was fine! Why did you add in more stuff?

Peter Tamaroff

7:57 PM

@JasperLoy Because the guy asked.

Vrouvrou

Hi

skullpatrol

Hi

Vrouvrou

who can help me plase

Peter Tamaroff

@JasperLoy He wanted to know if it was possible to define $f,g$ outside of $1,0$,$0,1$. I showed it cannot be defined.

Vrouvrou

Question about theorem 3.2 from Morse theory by Milnor

@skullpatrol how are you ?

skullpatrol

7:59 PM

@Vrouvrou Fine thanks, how are you?

Vrouvrou

i'm fine

user74918

I don't think anon is Marvis

user19161

@user74918 Haha, are you anon?

user74918

no

i'm Batman.

user19161

@user74918 Are you Jonas?

user74918

8:06 PM

►

user19161

I see. Jonas is Batman. I am Superman.

user74918

►

more mistaken identity in the Superman universe. Are you sure?

Vrouvrou

0

Q: Question about theorem 3.2 from Morse theory by Milnor

THe demonstration of the theorem 3.2 in the book Morse theory by Milnor is given in the special case whene the manifold is the Torus , My question is : can i prove it in the case where the manifold is a manifold with dimension 1 ? Please Thank you

Will Jagy

@OldJohn Jose Mourinho, currently in contract talks with Chelsea, put some public pressure on the club by announcing that, on the 10th of May, he would be at the Chinese arch in Manchester's Chinatown at 7:30 PM, wearing a white carnation. Neither Sir Alex Ferguson nor Roberto Mancini had any comment.

skullpatrol

Are you going to watch the FA Cup?

Will Jagy

8:21 PM

Maybe. Usually depends on time of day. Difference of about 8 hours.

amWhy

Hello @Will

Hello, @skull

:-D

Will Jagy

@amWhy, hello

skullpatrol

Hello, @amWhy
:D

amWhy

Hope I haven't interrupted anything ;-)

Will Jagy

@amWhy, no. These are top soccer teams in Europe being mentioned. The ongoing news is how poorly Real Madrid and Barcelona did against German clubs Bayern Munich and Borussia Dortmund. Just one of those things,

Granted, not as important as having several Jasper Loy's available.

amWhy

8:28 PM

@WillJagy hahahaha! I don't follow soccer much, but then, I don't follow many sports...

Will Jagy

@amWhy, meanwhile, Old John is a firm supporter of Manchester United, who have won the top English league this season. European soccer seasons are long, roughly August to May.

amWhy

@WillJagy Wow, that is a long season. On average, how often are games played?

I mean, american football: teams generally play once a week. Baseball, much more frequently than weekly...basketball seems in between those two...

Will Jagy

@amWhy, roughly one per week. England is mostly Saturday and Sunday, some Mondays. Germany like Fridays. About 20 teams, you play any opponent twice, home and away, so the season has 38 rounds. Same for Spain and Italy, I think Germany may have fewer, maybe 18 teams in the top league. Note the 3-starred comment by robjohn on the right, from about 18 hours ago.

amWhy

@WillJagy Yes, indeed, I saw that comment! It's just refreshing on this site that there is an international emphasis, in contrast to being entirely US-centric...

skullpatrol

They say the FA Cup is the oldest cup in all of sports.

Will Jagy

8:38 PM

@amWhy, I think I need to take credit for that. I embrace all nations, all ethnic backgrounds. I embody these in my own person.

amWhy

@skullpatrol Really? Forgive my ignorance, but what does FA stand for?

skullpatrol

Football Association.

amWhy

@WillJagy hoorah!

@skullpatrol Of course...dah! Why I didn't see that, is beyond me...

user19161

Functional analysis.

amWhy

@JasperLoy that's my Jasper!! hahahahaha!

skullpatrol

8:40 PM

The word "soccer" comes from As soc iation

amWhy

@skullpatrol I do know enough to know that American "Football" is a misnomer...i.e., that Football, internationally, means "soccer"

Or I should say, soccer means Football, and American "Football" means sanctioned violence :-)

skullpatrol

American Football came from Rugby.

amWhy

Ruby? Ohhh Rugby...!! Yes, that would make sense.

Old John

Hmm - in my neck of the woods, FA stands for something completely different :)

user19161

@OldJohn Hey! I am back with a new account!

Will Jagy

8:46 PM

@OldJohn, note for you about 20 minutes ago here

Old John

@JasperLoy So I see - but are you the REAL Jasper Loy? (a proof might be an email from your account!)

@WillJagy What coincidence - I am meeting someone at the same place and time!

user19161

@OldJohn Ah OK, I will email the SE guys soon with my new address...

Will Jagy

@OldJohn, yes, I thought that was odd.

user19161

The old one has been deleted!

Old John

@WillJagy It seems that Chelsea have nearly clinched the match with Basel - even without Mourinho's leadership

Will Jagy

8:51 PM

@OldJohn, currently 3-1, aggregate 5-2, 1 minute plus stoppage to go, that's about it

skullpatrol

@WillJagy @OldJohn Do you guys find this interesting?

Old John

@skullpatrol mildly interesting - but not something I would spend much time thinking about

Git Gud

Should that really be asked on English SE?

skullpatrol

@OldJohn Would you agree that there most certainly is value in having an unambiguous scheme for rendering equations to speech?

Will Jagy

Benfica Fenerbahce still close.

Old John

8:59 PM

@skullpatrol I'm not convinced there is a great deal of value in it - maybe a little, but after teaching maths for most of a lifetime, I have never felt that it was a big problem

skullpatrol

@OldJohn How about to record textbooks for the blind?

Old John

@skullpatrol sure

Transcript for

$Mathematics$ Mathematics