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07:02
their method seems slow
hmm ok
(i didnt read it as im a bit busy)
any reason you think a polynomial time algorithm can be trivially constructed
(i dare not say at all as p may equal np)
well, even their method calculates APL (in polynomial time), performs a random rewiring(sort of random), checks to see if graph is connected still (in polynomial time), and then recalculates APL (again in polynomial time) to see if APL has decreased( or increased)
and on and on until it arrives at a reasonable APL
so you'd have to proof that the number of times you'd need to do this process isn't polynomial
prove*
APL?
average path length
07:19
ah
well you forget
what im saying is that one should find the minimum number of edges to be toggled such that the result has the desired APL
finding that set is what might be NP
@Eric turns out there are a lot of REUs that I think I'd be into, it's p sick tbh
@Daminark REU program?
yeah there's not that many that im super into
if you're thinking about the REU program feel free to ask me about it
i was a participant last year
err this year
the summer
XD
I don't know to which you refer, since there are quite a few and I don't know where you are :P
07:24
finding a minimal set isn't what i'm interested in; i just want a "generic" graph on n nodes with e edges with a given average path length; these type of problems aren't often(or ever) NP as far as I know
if you generate a random ( connect) erdos-renyi or gilbert graph the average path length tends to be very small
connected*
@Salt yeah and thats great
now you have to adjust it to have the apl you desire
that is what im trying to convey is the hard part
if you want a random graph connection then you want to alter that random choice as little as possible
hence why i say the set minimal alterations needed to change the apl
by all means you can write a thing to find any such set of connections
still not easy though
hi chat
@Daminark Hello
@Eric yeah I've found 6 that could be good contenders
dami
I kinna need your help with something
Hey @Kasmir, how's it going?
07:31
do you have herstein book ?
ofc you do you are the one who recommended it to me :D
I don't have a pdf on me right now but it's very easily Googleable
@Daminark id prefer not to say which one as that might make me identifiable
@Typhon fair
I think that should work @Kasmir?
however if you say which ones you are interested in i can at least give advice
i did look through them a while back out of curiosity
Hi guys! What are we talking about here?
07:33
and there was a few math reu things being tossed around
@quartzfun fun stuff
@Daminark wow this has more stuff in it than third edition
@Kasmir there are two books, one is "Topics in Algebra", one is "Abstract Algebra"
The former has a lot more than the latter, so maybe that's where the discrepancy comes from?
hmm i got the AA one =p
but those topics what i need now
ill do some Reading on modules also later
but is that the last edition of topics in algebra?
You guys must be knowing Algebra series of videos by 3blue1brown right? They are fun and awesome.
are they good quartzfun ?
@quartzfun like are they Worth watching em ?
07:37
Yes, they definitely are.
Check one of the videos out.
@Daminark dami ! my question was on the proof of a'Na is contained in N, for all a in G, forces a'Na =N
what he did was
a (a'Na)a' in contained in aNa' ... and there was a chain of that type
did not really get it
@quartzfun never heard of thrm
@quartzfun I'm vaguely familiar with 3blue1brown however I am not particularly fond of videos so I haven't watched any
@Daminark how dare you
07:39
He's videos are very insightful and interesting.
you should be executed
What??
leaky is being leaky -.-
@KasmirKhaan the proof is really easy
a'Na $\subseteq$ N
i.e. N $\subseteq$ aNa'
however, aNa' $\subseteq$ N (by substituting a as a')
I get that part
but the way the book manipulated the symbols
07:41
therefore aNa' = N = a'Na
was not very clear that we can do that
i know leaky, we show one is contained in the other , however
there's nothing wrong with it
if we had an infinite Group , this wont be true
there are cases when a'Na is contained in N for all a in G
and N is not equal to a'Na
only for infinite cases tho
so that proof is not well written
you're wrong
@Daminark you hate all video media
thats unsettling
XD
07:43
I mean that's but a result of my general hate of media
"a'Na is contained in N for all a in G" implies "a'Na = N for all a in G"
and I just proved it
consider GL_2(R)
matrices of the form ( a b ) first row
( 0 1 ) second row
they form a subgroup and they have the property that a' Ha is contained in H without being equal to H
$\begin{bmatrix}3&2\\4&3\end{bmatrix}^{-1} \begin{bmatrix}1&1\\0&1\end{bmatrix} \begin{bmatrix}3&2\\4&3\end{bmatrix} = \begin{bmatrix}13&9\\-16&-11\end{bmatrix}$
a in Q / {0} , b in Q
are you doubting whether 1 is rational?
3
07:53
inb4 1 is irrational
conjugate by A = ( 2 , 0 )
( 0 ,1 ) second row
why should I conjugate by that element?
that is the counter example
read my statement again
N = (1 z )
( 0 1 ) second row
z in Z ( integer)
iv tested it, it works
08:04
what if I told you
15 mins ago, by Leaky Nun
$\begin{bmatrix}3&2\\4&3\end{bmatrix}^{-1} \begin{bmatrix}1&1\\0&1\end{bmatrix} \begin{bmatrix}3&2\\4&3\end{bmatrix} = \begin{bmatrix}13&9\\-16&-11\end{bmatrix}$
that is not what i asked =p
"they form a subgroup and they have the property that a' Ha is contained in H without being equal to H" was claimed though
ill write it more properly
but on phone now
and a counterexample was provided
08:24
@KasmirKhaan just accept the answer and stop pestering people
@Typhon we had this problem in our class , am sure am not wrong
ill try to find it
i mean i saw it also on some questions asked here
in that case you likely misquoted the question
@Daminark rarely makes mistakes
XD
I meant @LeakyNun
08:44
if a^n = b^n in a group G
what are the steps to show that a =b
@Jacksoja depends on whether that actually is true
@Typhon it is not?
i can neither confirm nor deny that statements truthfulness
idk
why did you answer?
what do you mean?
08:46
you can't help me , you dont know , why did you answer?
you entered chat
we are here
therefore i replied
do you own this chat?
thanks bye
geez
rude
@Typhon he has a Point tho
@Typhon he asked a question and your reply is "idk" that is rude
@Jacksoja $(-1)^2=1^2$ in $\Bbb Q$
08:53
@AlessandroCodenotti so in proving that the map f(a) = a^n is injective
starting from f(a) = f(b) , ie, a^n = b^n
how does one shows injectivity ?
it isn't injective
It isn't injective without some assumtpions on your group
that depends on the group
if the group is abelian
it might still not be injective
08:54
oh well that won't help
Morning @Alessandro
yes Z is abelian but
(-1) ^4 = 1^4
The key property here is that the group needs to be torsion-free
which says that if we have $a^n=1$, then $a=1$
okay, what would one answers if asked to show that this homorphism is injective or not
08:56
More generally if $f,g$ are elements of your group with the same finite order $f^n=g^n=1$
"it depends on the group, but generally it is wrong: [insert example]"
what is a torsion mathein ? :D
i Heard that Word before
in the context of groups, a torsion element is just an element of finite order
It is more interesting in the context of modules
hmm thanks
like Z,+
we have only 1 and -1 that are torsions :D
oups i mean
Z,*
09:00
Z,* is not a group
hmm what was i thinking then -.-'
R,*
yes thats it :D
R/{0}
yes, that's right
:)
that makes me wonder why you said it needs to be torsion free
dont you mean all elements has to be torsions?
in jack's question
we we're wondering if the map $G \to G$ $x\mapsto x^n$ is injective
And if you have 2 distinct elements both of order $n$ it's not
09:03
a^n =b^n ==> (ab')^n =e ( G is abelian)
If you have any element of order $n$ where $1<n<\infty$, then it's not
because then $x^n=1=1^n$
aha
so we need the criteria of torsion free :D
that way to guratee that only e^n =1
and hence ab' =e
a=b
did not Think of this ._.
the discussion was usefull thanks
Hello
Please some help
Is it 34
I calculated limit of first term at 1-
i found 34
and i calculated the second term i found
-2+k
09:24
@KasmirKhaan no his question was how to compute a or b when a^n = b^n and my point was that they first need to examine whether it is even true or not.
it likely isnt in general
and i was merely saying i dont know enough about group theory to actually say one way or the other
considering this is a chat room and not someone's question post it is completely normal to respond to someone when they enter a room that you were already in and having a discussion with. It wasnt like I wasnt willing to talk about group theory. I just i dont know. That doesnt mean someone cant think about or discuss it
09:40
@Typhon then you should have said that , i thought you were being rude
@MatheinBoulomenos mathein here? :D
10:10
Fun fact: It is difficult to caught me trolling, because frequently when the target being trolled missed the point of being trolled, their responses are so amusing that they generate interesting questions, which caused my next reply to turn to genuine mode and go forward on that question
Or in short: I can switch erratically between trolling and genuine, thus there is no way to find out
and most importantly, my trolls often end up triggering interesting investigation on whatever maths they are asking
11:04
@LeakyNun leaky, i Think i told you the problem wrong
@LeakyNun about gNg'
@Secret This is a strange thing to brag about?
lol, I am not sane except when discussing maths
too much feeling of loneliness on the internet
Feel less lonely by studying browns cohomology of groups with the lads :D.
11:06
face to face interaction is something no VR can substitute
@Secret In what sense?
the social bonding, the facial expressions, the gestures, the environment
and also the hugs, kisses and running about
Oh, sure. I thought we were only talking in the math sense :P.
well, maths lads has their own sense of socialisation in the form of discussion of various algebraic groups, jokes made by chaining real analysis results together and some mumbles about topology
I used to make those back in my old uni where I met most of my maths friends
You've discussed algebraic groups?
Or do you mean abstract groups?
11:10
probably the latter, those tied to symmetry groups and isometries
Sure, the latter. The former is a group equipped with the Zariski topology (whichever level of sophistication you are equipped with)
hi @Narcissusjewel
@AlessandroCodenotti if you have one then you have another
Not necessarily
@LeakyNun found the example i was talking about :D
11:15
Some elements are their own inverses sometimes
guys in proving the Z(G) is normal subgroup
let x be in Z(G), g in G, gxg^-1 =x
@AlessandroCodenotti but then x^2 = 1^2
is that the step needed?
feels way to easy unless am missing something
Nah, that's correct
@KasmirKhaan so the right statement is “gNg’ subset N for all g” implies “gNg’ = N for all g”, and the wrong statement is “gNg’ subset N implies gNg’ = N” for all g
11:47
@KasmirKhaan Hello.
@Daminark Proof of five lemma is 7/10
Proof of snake lemma is 8/10 however
It's not as fun as the schneck
prove it without diagram chasing though
no
why would you do that
because i’m a horrible person
i mean the diagram chase gives the most insight
because you see what the shneck map does
Technically you don't really need the snake lemma to prove the homology long exact sequence
12:02
I mean five
Oh. I don't know of a diagram chase-free proof of that dude
its all in that yellow boi
(that i havent read)
is that a big book of category theory
category and sheaves
i see
I'd be interested to see a diagram-free proof of the five lemma. I don't think anybody ever needs the middle isomorphism explicitly.
12:11
inb4 my machine blows up
calculating degree assortativity and average path length of a million node graph
i killed ipython
rip
12:36
@LeakyNun thanks for the help !
Generally matrix is written as $A[m][n]$.
However, what does it mean when the matrix is written as $A[m...p][n...q]$?
where $m,n, p,q \in I$
12:51
is it a submatrix?
no
well, my other guess would just be explicitly stating the size of the matrix, but then i'd expect m=n=1 rather than some random int
is there some context?
13:12
@Salt it appeared in a question: ""A 2D array defined as $A[4...7, -1..3]$ requires 2 bytes of storage space for each element....."
Not sure what it actually means, but understood how to use it in calculations..using the above info^
i see, seems they just want you to use a A[4][5] matrix...this way of specifying arrays makes me think compsci
Hi guys. I'm trying to learn how to do proof by induction and I just gave it a shot.
Could anybody of you confirm this is true or not (small example)
https://i.imgur.com/LCH1MNe.jpg
your induction step is false
try to derive f(k+1) from your assumption
don't assume f(k)=f(k+1), it doesn't
Thanks!
I will take a look at it immediatley
np
13:56
anyone know average path length of random acyclic graphs? i think it's around sqr(N) but i'm failing to find any literature
duh, dunno why i didn't just look for random trees
turns out it's approx 1/2Pi(N)^(1/2)
rip latex
$\approx (1/2\pi N)^{1/2}$
this means I know an extremely inefficient way of computing $\pi$
14:44
@Abcd this means that the rows are indexed from m to p and columns are indexed n to q (typically inclusive, but could exclude the upper limit too).
in your case it's inclusive.
@ParthKohli is that only in theory or is it allowed in programming too?
python implements this, yeah.
java?
@ParthKohli does it?
@LeakyNun isn't slicing kinda like that?
14:52
@ParthKohli right
but it doesn't have negative indices...
well yeah, in that case it acts in a circular manner
we've met before, I think :)
have we?
15:37
[Random]
15:55
Since art and body language are syntax free and does not depend much on natural language, perhaps they might be expressive enough to uniquely represent all uncountably many reals
16:10
0
Q: Calculate conditional probability for the Poisson Process

JeffI need to calculate $P[N_{s}=k||N_{u},\,u \geq t]\,(s \leq t)$ for the Poisson process. However, I have been instructed to use the following example in order to do so: Example: The Poisson process $[N_{t}:t \geq 0]$ has independent increments. Suppose that $0 \leq t_{1} \leq \cdots \leq t_{k}...

16:49
1
Q: Computing the Kernel of an Epimorphism

user193319 Let $R$ be a commutative ring with identity, $I$ an ideal in $R$, $\pi : R \to R/I$ the canonical projection, and let $S$ be some multiplicative subset of $R$. Define $\theta : S^{-1}R \to (\pi(S))^{-1} (R/I)$ by $\theta (r/s) = \pi(r)/\pi(s)$. Prove that $\theta$ is an epimorphism with kernel...

@salt, with $N$ the number of vertices?
17:09
Does someone know about percolation ?
@Blue does
@LeakyNun Just in the off chance you might be interested, I posted that problem I was asking about earlier and got two answers: math.stackexchange.com/questions/2575362/… . Thank you again for your help earlier.
Anonymous
What's up?
Anonymous
@Lucas I know a teeny weeny bit
Hello. Are there links between groups and percolation?
or algebra in general
17:16
0
Q: Does finding this conditional expectation boil down to finding a conditional probability?

ALannisterFor a random variable $T \geq 0$ with distribution function $F$, a real number $t > 0$ and $\mathcal{G}= \{ \emptyset, \{ T > t \}, \, \{ T \leq t \}, \, \Omega \}$, I need to evaluate the conditional expectation of $E|T-t||\mathcal{G}$, where $\mathcal{G}$ is my sub-$\sigma$-algebra. Now, I as...

uuuggghhh the index for Munkres's Topology is awful
Anonymous
@Lucas The mathematics involved in percolation is mostly probability, statistics, topology. Maybe group theory is used somewhere (perhaps where crystal percolation is involved). I don't know enough to give a knowledgeable answer. Perhaps you could ask on Math SE main site
missing "hausdorff", "klein bottle", "knot"
there's actually no J, or K section at all, it goes form I to L in the index
to wikipedia
in my course evaluation forms, they asked what supplemental sources I used the most. I put in math.se and wikipedia
@Blue Ok thank you. Anyway it's just to have a preview. Briefly, how is this related to the topology, if you know ?
1
Q: Computing the Kernel of an Epimorphism

user193319 Let $R$ be a commutative ring with identity, $I$ an ideal in $R$, $\pi : R \to R/I$ the canonical projection, and let $S$ be some multiplicative subset of $R$. Define $\theta : S^{-1}R \to (\pi(S))^{-1} (R/I)$ by $\theta (r/s) = \pi(r)/\pi(s)$. Prove that $\theta$ is an epimorphism with kernel...

17:22
also missing $T_0$, $T_1$, etc
I need help with this. I think Santana might be confused by what I am doing.
anyone have a good authoritative introductory topology book they can recommend with a better index?
Anonymous
@Lucas Check this and this
@Blue It seems it is linked to group via Cayley graphs.
Thanks
Anonymous
Algebra can be applicable almost everywhere :P
Anonymous
17:25
Welcome
OK :)
17:41
what's this notation? Thm: The number of topologies on a finite set with $p+n$ elements, with $p$ prime, satisfies
$a_{p+n} \equiv (n) \pmod p$
what's $(n)$?
oeis.org/A000798 in "comments"
@Lucas salut, t'es en prépa ?
18:26
@Gabriel Salut. Oui :)
Bientôt les écrits, tu vises quoi ?
Ca va, il reste 4 mois :) . J'aimerai bien une ENS
@Lucas tu passes en MP info ?
Non physique
MP physique
bon courage !
18:31
Merci :)
18:44
@GFauxPas You're not reproducing that quite right, I think: It should be a(p+n) == A265042(n) mod p
and when you look up that sequence entry, you find that it's defined as: the unique number k such that T(p + n) == k mod p for all primes p, where T(n) = A000798(n) = number of topologies on n points.
...which basically means that the previous statement is just the definition of A265042. so it's not telling you anything beyond "the number of topologies on p+n points, mod prime p, is the same for every n."
The fact that this works for all primes is presumably the result of Kizman that's cited.
i.e. this preprint: arxiv.org/abs/1503.08359
ah silly me, thanks
Can someone tell me the obvious argument for the parity of the last determinant in this answer ? math.stackexchange.com/a/2574991/66096
put in another way, why $$\det\begin{pmatrix}
1 & 0 & 0 & 0 \\
1 & 1 & 1 & 1 \\
1 & 1 & 0 & 1 \\
0 & 1 & 0 & 0 \\
\end{pmatrix}\equiv 1\pmod{2}$$ ?
@GabrielRomon Well, the determinant is in fact $1$
without computing it explicitly of course
Jack wrote it like it's obvious

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