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04:29
This is interesting: the opposite of "Explicit is better than implicit" is "Convention over configuration"
in other words, when a language/framework/system has a decision to be made, you have two choices:
1. Present that choice to the user of the system, forcing them to choose
2. Choose the conventional option, hiding away the option (but still allowing for its configuration if they know how/can find it)
 
4 hours later…
08:57
@NathanMerrill @El'endiaStarman You recently talked about intersections, look at this
 
3 hours later…
11:55
> a(n) = A000217(n) for n=1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 86, 89, ...; checked up to 10^7. - Michel Marcus, Sep 15 2015 (checked up to 9*10^8. - Kenny Lau, Aug 01 2017)
@orlp @flawr any idea?
btw I've checked it to 2e9 now
I've checked it to 2^32 ~ 4e9 now
the size of unsigned long on my computer
12:18
@LeakyNun what is the question? :)
@flawr just the quote above
I'm trying to extend the list
I've checked to 2^32 now, and still haven't found any terms beyond 89
I still don't get what you're asking:)
the two sequences seem to race past each other twice in each 10^n - 10^(n+1) interval
@flawr to find solutions for a(n) = A000217(n)
i.e. the reversed range sum = the normal range sum
12:32
O_o you're a user on OEIS
@Mr.Xcoder I contributed to many of these sequences
basically nobody came up with an efficient algorithm except brute force
and come up with a good algorithm I did
my algorithm doesn't quite work if |r| >= 1 though
13:26
intuitively it seems like we should have A062918 < A000127 for sufficiently large n, due to trailing zeros. I'm surprised we still get equalities as high as n = 89.
1 hour ago, by Leaky Nun
the two sequences seem to race past each other twice in each 10^n - 10^(n+1) interval
So potential equalities is 2 for every interval
one around 1.1 x 10^n and one around 8.9 x 10^n
if you understand what I mean
oh I see, because the headstart of the reverse sequence after reaching a new 10^n is bigger than the loss it accumulates from trailing zeros
14:04
@MartinEnder so do you have any idea?
(cc @orlp) also, how is the g.f. of this sequence generated?
14:35
@flawr ...that looked fantastic until I saw that there were still lighted connections
why did they have to ruin it all with a stop light
14:58
apparently in dubai
@NathanMerrill ^
oooh, that is pretty
this is basically the most complicated we get in utah:
we call it the spaghetti bowl. And its feels a lot more complicated when you are driving through it than from the top-down view
haha=) fitting name=)
@NathanMerrill Louisville KY has the "Spaghetti Junction".
if you use reddit, post it to /r/intersectionporn/
15:07
oh oh oh!
@Americans: I have the impression this guy from WOP sounds like someone who wants to impersonate an american. Is this a certain accent or does that sound like an "average" american?
I don't really notice accents any more
I listen to enough europeans that I really can't tell the difference between them and american accents (unless the accents are really strong)
in other words, I don't hear an accent, but maybe it's just me :P
15:36
Just had a look at the lightning map, it seems switzerland is doomed.
@NathanMerrill Hm I'm really curious what others find then =)
16:09
@LeakyNun can you clarify a bit
exactly what you want to know
@orlp we have this problem:
> Number of real integers in n-th generation of tree T(2i) defined as follows.
Let T* be the infinite tree with root 0 generated by these rules: if p is in T*, then p+1 is in T* and x*p is in T*. Let g(n) be the set of nodes in the n-th generation, so that g(0) = {0}, g(1) = {1}, g(2) = {2,x}, g(3) = {3,2x,x+1,x^2}, etc. Let T(r) be the tree obtained by substituting r for x.
And we found out this g.f.:
> G.f.: 1/(1 - x/(1 - x^2/(1 - x^5/(1 - x^12/(1 - x^29/(1 - x^70/(1 -...- x^Pell(n)/(1 -...))))))))
Question: how did we know that the g.f. describes the problem?
shouldn't g(1) also contain 0?
since 0*x = 0
@orlp that's the only exception :)
just ignore it
the rule starts at g(1)
(please ping me next time)
@LeakyNun also is this a puzzle or do you also not know the answer?
either way I'll look at it in a bit, eating
@orlp I don't know the answer
16:21
@LeakyNun Gah, no LaTeX?
@SimplyBeautifulArt $\dfrac1{1 - \dfrac{x}{1 - \dfrac{x^2}{1 - \dfrac{x^5}{1 - \dfrac{\cdots}{1 - \dfrac{x^{\operatorname{Pell}(n)}}{1 - \cdots}}}}}}$
Hm, and what are we trying to prove?
I believe you meant to have x^Pell(n)
@SimplyBeautifulArt done
@SimplyBeautifulArt that the g.f. solves the problem
RIP, well I don't quite understand the problem at hand, so good luck
@SimplyBeautifulArt g(1) = {1}, g(2) = {2,x}, g(3) = {3,2x,x+1,x^2}
basically you perform "add one" and "times x" to the elements in the sets to get the next set
and then at the end, replace "x" with "2i" in g(n)
the number of real integers in g(n) is a(n)
16:26
Hm, I see...
@SimplyBeautifulArt any idea?
17:29
Oh yes, and would anyone like to support my large function challenge proposal?
0
A: Sandbox for Proposed Challenges

Simply Beautiful ArtCan you surpass Γ0? Can you make a function that grows faster than Feferman–Schütte ordinal (a.k.a. Γ0 or φ(1,0,0)) level in the fast growing hierarchy? For those unfamiliar with what all the above gibberish means, I recommend watching Giroux Studios' video series on the fast growing hierarchy....

@SimplyBeautifulArt this room isn't about PPCG. We do like math around here, so if you wanted to discuss the math or stuff behind it, that's awesome, but for "supporting a PPCG challenge", this really isn't the place
Ah, okay
Well, if anyone would like to take a look and tell me if I should make a guide for the math behind the stuff?
that said, I can't speak much about Γ0, as I've never really gotten into busy beaver stuff
The only busy beaver thing about it is you want to get as large as possible
Would it be fine if I tried to sketch a quick explanation of the math behind the problem in here?
but like, trying to find a faster function should be fairly trivial, right? Like, if I have two functions, f(x) = g(x) and h(x) = g(g(x)), assuming g(x) is strictly increasing, then h(x) >= f(x) for all x, right?
that's why "finding a faster function" is always trivial from a mathematical perspective.
unless I'm missing something
17:45
Haha, you think that it's that easy?
Consider you want to reach x^x
what do you mean by "reach"?
And you are able to construct g(x) = x^2
You want to make a function that grows faster than x^x as x → ∞
why am I only limited to x^2?
or is Γ0 impossible to program?
It's trivial to note that g(g(...g(x)...)) = x^(2^n) << x^x as x becomes very large
Nah, but its just an analogy. If you can only construct so much, then repeatedly nesting will only take you so far
@NathanMerrill a busy beaver isn't just any fast growing function
17:48
Also, I gave an example Ruby program that surpasses Γ0
it's the fastest growing function for a turing machine with a fixed number of states
def f(n)n.class==Array end;def g(a,b,x=-a)c,d,e=b;x<9?(f(b)?g(a*a,g(a,b,9),x):x<8?g(a,(z=-a;a.times{z=[z,z,z]};z),x+1):a):(h=f(c)?[g(a,c,9),d,e]:[c>9?c-1:a,d,e];c==9?e:f(d)?[h,g(a,d,9),h]:(d==9?h:[h,d>9?d-1:a,h]))end
In a bit over 200 bytes
@LeakyNun I don't know how to approach this as the tree portion is unclear to me in terms of growth
hard to keep track of duplicates
alright
Yeah, I was thinking much the same @orlp . You also can't say certain terms will never result in real numbers eventually :-/
17:49
ok, so you've already got a function that grows faster than Γ0. It's still about the bytes, because if you nest that function in itself, it'll grow even faster
like it's hard to manipulate a generating function
if you don't know what you're trying to reach
@NathanMerrill Sure, but you have to make that function that grows faster than Γ0. If you have a function that doesn't grow fast enough, repeated nesting itself probably won't take it anywhere near Γ0
oh! it's a code-golf challenge
Also, 512 bytes is max code length
its not about writing the fastest-growing function, its about writing one that grows faster than Γ0 with the fewest bytes
@SimplyBeautifulArt there's no reason to have a maximum code length if its code-golf
@LeakyNun do you have any recurrences or other identities for a(n)?
Really? Hm, okay
Thanks @NathanMerrill
@orlp I don't...
But yeah... my point was that there are functions "unreachable" by simple repeated nestings of other functions.
The example being the comparison between x^x and x^2
17:55
if f(x) = g(g(x)), and h(x) = f(f(x)), then is h(x) = g(g(g(g(x))))?
Yeah. h(x) = f(g(g(x))) = f(y) =g(g(y)) = g(g(g(g(x))))
@LeakyNun a geometric interpretation is that we start with one point at (1, 0)
and at every step we duplicate the points, one goes 1 step to the right, and the other rotates 90 degrees counter clockwise around the origin and scales 2 times
@LeakyNun Did you ask about this in the main math chat?
@SimplyBeautifulArt yes
@NathanMerrill Still think this is trivial?
18:13
@SimplyBeautifulArt I stand by my point of "making a faster-growing function" is trivial. It's always been about the bytes IMO
like, making X grow faster than Y isn't trivial, but coming up with any function that is faster than Y is
@NathanMerrill If you believe so, I recommend trying to make a function that grows faster than f_ω^ω (n) in the fast growing hierarchy, regardless of byte limitations.
f_ω^ω (f_ω^ω (n))?
Without using f_ω^ω (n) ofc
i.e. construct it from X, where X is... idk, f(n) = n^n
Then its very much not so trivial.
Ofc, you probably don't know what f_ω^ω even means
Would you like me to explain it?
18:19
eh, I'm ok.
xD Oh c'mon, just real quick
I'll start at f_0 and hopefully you'll see why your plan ain't gonna work
Well, anyways, your plan is already built into and surpassed in the fast growing hierarchy... by a long shot
By definition, applying fixed amounts of iterations of f_α will result in a function that grows slower than f_α+1
And brute force applying this technique would then be able to overcome limit ordinals
you mean f_α(x+1), right?
I'd expect that of any fast-growing function.
Its just shorthand notation, and no, I meant f_α (n) and f_α+1 (n)
Let (g)^k (n) = g(g(g(...g(n)...)))
what does +1 mean?
is it a family of functions?
Then for every k and α, (f_α)^k (n) << f_(α+1) (n)
The +1 is a successor ordinal thing
and f_α is a specific function
18:25
then what about the function f_(α+x)(x)?
That is easily dominated by f_(α+ω+1)(x)
therefore, I stand by my point: it's always trivial to make a bigger function
its basically a bunch of mathematicians saying "My infinity is bigger"
@NathanMerrill That's not what I'm arguing about. I agree its trivial to make a faster growing function, but I do not agree that you can take any function and trivially construct from it a function that grows faster than some other function
oh, I agree. I don't think that's the case
18:29
like, log(x) is one
you aren't going to get faster than log(x) by nesting more log(x)s
That's a completely different problem
that just means log(x) is sublinear to begin with, so why would you want to do anything with it?
@SimplyBeautifulArt Yeah, that's not what Nathan's saying. He's saying that if the challenge is to construct a function f(x) that grows faster than a given g(x), then setting f(x) = g(g(x)) will grow faster than g(x) (assuming that g(x) is strictly increasing).
@El'endiaStarman Yes, I know, but the problem is that you are not given g(x) in my proposed challenge, you have to construct everything from scratch more or less.
^^ this. I agree that if you ban g(x), then it's no longer trivial
but your challenge isn't about either of those: its about bytes
because you've already provided a function that grows faster
so all people need to do is write your function shorter, or come up with a different (but still fast enough) one that is shorter
Well, I do still think getting there is half the journey. I don't think anyone can look at my code and understand anything from it.
18:32
I can agree with that :)
If I were to make this into a real thing, I should remove my example code?
no, I think it's fine
you've provided a high bar that people can at least reproduce. They'll get out-golfed, though
wtf with that flagged chat message lol
19:00
Lol. Not exactly related to this chat, but meta.math has an interesting post.
24
Q: Recent influx of new users with female "glamour" pics for profile

ChrisI don't mean to 1) complain about women being present on the site or 2) suggest that women could not both be interested in glamour and mathematics, but I've perceived a curious uptick in profiles like this in the past couple of months: profiles with female usernames and glamour- or modeling-type ...

@SimplyBeautifulArt we aren't a strictly on-topic chat here :) We just have a couple of things that are strictly off-topic :P
that said, that's crazy
@LeakyNun do you have a moment?
@orlp ?
put えぐ into google translate and translate it from japanese -> english
19:13
@orlp and then?
@LeakyNun check what happens if you repeat えぐ 2 times, 3 times, etc
@orlp very interesting.
hi, first time trying the chat on SE
Regret = return returning?
19:15
just wondering which is the best chat room for Javascript
@HappySad how'd you land here?
@NathanMerrill not really the right way to dig into it...
google translate just builds statistic models without knowing grammar/vocabulary/etc
so invalid inputs like えぐ*n are just UB
and it just returns the most probable output
19:17
@LeakyNun oh, I assumed it was actually correct japanese :)
@NathanMerrill :-P
n1 Nathan tx u!
@HappySad what does n1 mean? no one? none? I recommend full words on stackexchange
I think it means "nice", but I have no idea how you derive it from "n1".
19:23
this challenge made me think: Are there boards that are impossible to get into?
there are some trivial ones, like a board filled all with 2s
@NathanMerrill well, boards with 2 at the 4 corners
why is that impossible?
the last swipe cannot be up
that seems quite possible to me
rotate the board by 90 deg and repeat the same argument, 4 times
19:25
we do allow for the randomly placed tile to be anywhere
@NathanMerrill yes, but only one at a time
which one of the bottom two tiles is the random tile, if the last swipe is up?
who says there wasn't already a full row?
oh, I meant the board with 2 at the 4 corners
sorry
oh, only 2s
yeah, I can agree with that
that sounds like a rather fun challenge: detect if the board is reachable

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