where will xe move me, dunno let's find out lol

P1740
i'm guessing t-trash
best board btw

Have you watched any interesting math videos lately? I've been slowly going through a series of lectures on algebraic geometry. Slowly mostly because I find myself pausing a lot to work on the problems he's talking about. I haven't gotten very far yet, but so far it's pretty interesting.

https://www.youtube.com/playlist?list=PL8yHsr3EFj53j51FG6wCbQKjBgpjKa5PX

>youtube
you have to go back

yt-dlp works fine over Tor. I can't control where other people host their videos.

P1442
I thought this was neat. You know how a polynomial is determined up to a normalizing constant by the locations and multiplicities of its zeros in ℂ because it can be factored into linear factors?

P(x) = a(x-r₁)(x-r₂)⋯(x-r_n)

Well, what about a polynomial in two variables restricted to the circle x²+y²=1? Turns out it's still true.

As an example, consider the functions

f(x,y) = (5x-3)(5x-4)
g(x,y) = (5x+5y-7)(5x-5y-7)
h(x,y) = (7x+y-5)(7x-y-5)
k(x,y) = (5x-3)(5x-4) + x² + y² - 1

The curves f(x,y)=0, g(x,y)=0, h(x,y)=0, and k(x,y)=0 all intersect the circle at the four points (3/5,±4/5) and (4/5,±3/5). And when restricted to the circle, these functions differ only by a multiplicative constant. Second pic is the graph of f(cosθ,sinθ), g(cosθ,sinθ)/2, h(cosθ,sinθ)/2, and k(cosθ,sinθ), all forming the same curve.

I constructed k the way I did because it was the easiest way I could think of to make an ellipse pass through those same four points. But its form makes it obvious what's going on. It's the same for the rest of them.

f(x,y) = 25x²-35x+12
g(x,y) = 2(25x²-35x+12) - 25(x²+y²-1)
h(x,y) = 2(25x²-35x+12) - (x²+y²-1)
k(x,y) = (25x²-35x+12) + (x²+y²-1)

Any polynomial whose zeros on the circle are exactly those four points with multiplicity one will follow this same pattern. And it doesn't just work for circles, it works for any algebraic curve in ℂ² with only one component. This can be shown using Bézout's theorem using the same trick the guy mentions in the third video. Let f(x,y)=0 and g(x,y)=0 have the same intersections with a curve z(x,y)=0; then we can pick another point on the curve and choose λ to make f(x,y)+λg(x,y)=0 there. Since Bézout's theorem determines the number of intersections, this is impossible unless the curves f(x,y)+λg(x,y)=0 and z(x,y)=0 have a component in common.

I started from algebra and up through Calculus II with Professor Leonard. Was pretty great.

down again lol

Works for me

It's fixed now but they were breaking half the Internet again for a while.
https://www.cloudflarestatus.com/incidents/xvs51y9qs9dj

Now back to your regularly scheduled breaking privacy, making sites less secure, and fucking with Tor users or anyone who doesn't want to run JS garbage, as usual.

>run the entire internet on one server
>when it breaks, most of the internet goes down
why?

What's going on and why are they tearing apart the internet?
When will this end?

P1603
Because it's free. Free DDoS protection, free CDN caching, free for them to harvest tons of information.

http://zfswakltk3lb2idwrzh6zof2lktxjgms2ntgywfrb5c2f4jh5xu3oead.onion/

I am testing how solvable my CAPTCHA is. Please try to solve it.

My PGP public key:
http://zfswakltk3lb2idwrzh6zof2lktxjgms2ntgywfrb5c2f4jh5xu3oead.onion/pubkey.txt

Signed:
https://pastebin.com/N8P9wPzy

A potential issue I can see right away is how few possible answers there are. A bot could solve this by just spamming enough random guesses. Google's ReCaptcha shouldn't be imitated. It's very weak as a captcha, and mainly operates by banning IP addresses.

Yes, a bot can also solve this CAPTCHA by a fixed answer as well as random ones. In practical users, I must send 2 or more CAPTCHAs to users in order to increase n of answers. The expectation value of this CAPTCHA is equal to its number of selectable answers. In this test, I set it to 120. Easy to solve for bots. And also humans? :D

P1653
I would prefer having to solve a slightly more difficult CAPTCHA then being forced to repeatedly wait for the CAPTCHA to load.

Draw two lines g and h. Then draw three points A, B, C on line g and 3 points a, b, c on line h. Let X be the intersection of lines Ab and Ba, Y of Ac and Ca, and Z of Bc and Cb. Pappus's theorem says points X, Y and Z are collinear. Can you see why?

P1541
X is 0+0j. Without loss of generality we can put Y at 1+0j.
A=d cis e, R contains d>0, 0<=e<2pi
b=f cis e, R contains f<0
B=g cis h, R contains g>0, 0<=h<2pi
a=k cis h, R contains h<0
And then the problem actually begins, I will try it later.

Problems like this bring me back to high school math olympiads :)

The elementary proof I know of was using menelaus' theorem, see here: https://www.cut-the-knot.org/pythagoras/Pappus.shtml

There's a similar proof of its generalization, pascal's theorem, also using menelaus, also on the same website.

I swear to god I remember learning some kind of super elegant proof of this from Dušan Djukić (Yugoslavian IMO coach I used to know). I think it used some kind of circle theorem, like Miquel point or radical axes or something. But with no cyclic quads in sight...it could very well have just been the menelaus proof.

Anyway, >>P1540 is probably the better way to go about it, I just like elementary solutions to geometry puzzles for sentimental reasons.

P1548
I don't remember your fancy math names, but a circle idea would be to show that the points where the circles described by
AYC aYc XBZ XbZ
touch form a line.

P1549
To stay on theme you could show that they cannot possibly lie on the perimeter of the same circle ;p

P1548
That's pretty cool. Personally I found Pappus' original proof using cross ratios a little easier to grasp the essence of. But the concepts seem pretty similar; I wonder if there's some relation between cross ratios and Menelaus' theorem besides both being useful to prove Pappus.

This page about applying Pappus' theorem to games looks very interesting and it's a shame the Java applet doesn't work in browsers anymore. Maybe I'll have to get it working outside the browser and play with it.
https://www.cut-the-knot.org/Curriculum/Games/MixedStrategies.shtml

nanochan is down and there is no updated information in the nanochan bunker... what gives?

P929
Why would he seethe?
He is incapable of feeling emotions and any traits of humanity.
High chances he is here and patiently but eventually will control this imageboard as well.

Once I posted about hikariabove, for a few minutes, the real nanochan came back. I think he realized he got caught, but decided to continue attacking it in case people found out.

I thought about why people like him exist here, and realized it's better that people like him stay and get absolute control which they seek at imageboards so the real world would be safe from them.

P880
What's this from?
limit = (tonumber(FORM["postlimit"]) and tonumber(FORM["postlimit"]) <= 128)
and FORM["postlimit"] or limit;
oops, that's embarrassing.

P1451
I saw the chatlog posted on picochan.

P1451 nvm found it http://iliurdpsynwcmtxmsjuz4jdx3f7gfotaeeni5x5lveqknoakne4edjyd.onion/

Twinnano be like

no idea

>P60

is that Seattle?

you're in seatle

Why do so many corporations use Fedora and RHEL? What's so special about them?

Red Hat also provides companies with something to blame if their computers go offline. You don't want to be the one that's responsible for taking the blame right? Nobody got fired for choosing IBM!

Yeah and IBM owns Red Hat these days

RH has sales representatives. That's it.
They bribe corporate decision makers into buying their wares with little gifts, positive attention, and kickbacks. That's how you get sales in the corporate world, and the only reason companies like Red Hat and Microsoft are successful.

RHEL because their support is highly regarded. Fedora I can only assume is used by corporations because of its relation to RHEL.

P1406
Ah, the good Oracle strategy.

Feeling tired and alone and kinda wanna play renpy games, anyone got good suggestions?
Also the browse file button don't work for me so yea

Why not play Aquaplus games, their engine is GPL-licensed

aristocunts

Can you cover all the squares with dominos? Each domino covers two adjacent squares, and no part of the domino is allowed to go outside of the squares.

Right ;p

P1368
I can weaken this lemma by a lot I think.

So far, this is the strategy I have:

1. If there are tiles only connected to one other tile, we know it and the other tile must be covered by a domino, and so we can remove it. It makes sense to remove these first. Removing them may create new ones, so continue until there are no more.

2. This is a generalization of P1368. If two regions are only connected to each other at one line segment between tiles, then we can deduce whether a domino must be placed across that line segment by counting the number of tiles in each region. If both are odd, a domino must be placed there, and if both are even, a domino must not be placed there. Either way, we can now treat the two parts as if they were disconnected regions.

3. Look at each connected component separately. If any can be proven impossible to cover, the whole thing is impossible.

4. A lot of regions can be shown impossible by covering them with a checkerboard pattern and testing whether the number of white squares equals the number of black squares.

Still, at this point there will remain impossible regions we haven't proven impossible yet, and I don't know any better way to show them impossible except a search. Maybe we can get some mileage out of generalizing (2) more.

P1322
>It turns out for this one there's a slicker way to prove this one impossible, which I think you're pretty close to seeing based on the stuff you've said.
Now for my own idiot moment. When I wrote this I thought that this board was one that if covered in a checkerboard pattern, the number of white and black squares would differ. But I never bothered to check. I had been thinking of a similar puzzle I had seen posted.

Might as well link to the original thread:
https://warosu.org/sci/thread/5652006

This is a doodle I made of the Gaussian primes.
Have you made any math-related art you'd like to share?

found this on sushi

P1232
I think at some point you can try too hard to avoid saying the word parallelepiped

doodling with basic fractals
turtle graphics works well for this

Are there any database engines that implement something like SQL views, but which instead of being recalculated when you query them, get recalculated whenever one of their inputs changes?

P1213
Didn't notice the license.

>I don't believe in databases
What's your preferred way of achieving ACID?

P1214
Right now I am not doing anything performance bound, so I can use my own trivial programs with an emphasis on brevity **inb4 I have just praised common lisp for concision**.
In contrast, when performance has been limiting my solutions have been definitionally nonportable in ways you can imagine.
An attempt at generic benchmarking of the former is pointless, and the latter is generally dumb, hence some of my bile.

Now I have actually read P336 and in this case I would just do exactly what you said you don't want to and conventional wisdom is against and instead of adding a database dependency and a using-the-database-in-a-customized-way dependency I would have just written the trivial program, and if it was too slow made it faster, and proved reliability properties that warranted it (related to makin' it faster).

P1215
P1216
ACID is about correctness rather than performance.

>proved reliability properties that warranted it
An efficient database with formal proof of ACIDity is another thing that would be very nice to have.

P1218
A historical note is that there was a common lisp metaobject protocol based persistent object protocol/database based on the pre-postgres Berkeley DB (which was ACID). Lots of modern things like it are around. My dear acl2 doesn't have a formal notion of CLOS which is a hassle.

Can you color this map with the least number of colors possible? Each pair of regions that share a boundary must be a different color. Regions that only touch at a corner don't have to have different colors.

I think
1. Connected components
2. Take dual by connecting connected components nodes over edges, using something like a thick edge matched filter to agree on shared edges.
3. Choose a corner as a starting place, the dual is a tree with that root. Give it a color.
4. At each level of the tree, permute colors already used such that the maximum number of previously used colors are present
5. Add new colors for any uncolored nodes at that level.
<
I don't have a proof of 4. It is based on an intuition because the third level of the tree gives 0 fucks about the first level of said tree.

I didn't understand what you mean by 4. How are you assigning colors to the level? Iterating through possibilities? What is being permuted? Are you changing colors already assigned?

I should probably note that this problem in general is known to be NP-complete. Can be done in polynomial time if you know 4 is the minimum, though.
https://en.wikipedia.org/wiki/Graph_coloring#Computational_complexity

P1196
Probably because I am saying something trite. Do you agree that after (2) if I pick any starting point, it describes a tree where each level of the tree is the minimum number of edges to traverse to get to this region from the starting point.
And then I just said the word permute in the incredibly dumb sense of just trying every color in every allowable position. I thought I was getting something out of doing this separately at each level of the tree, oh I said something both dumb and wrong didn't I. Er.. I'm... Shitposting...
I guess OP was obliquely pointing out that the four color theorem was the first theorem to be proved by a mechanical theorem prover in such a way that a human could not meaningfully verify the theorem (too big).

P1201
>I guess OP was obliquely pointing out that the four color theorem was the first theorem to be proved by a mechanical theorem prover in such a way that a human could not meaningfully verify the theorem (too big).
I was just looking through my old folder full of stuff from 4chan /sci/ and remembered at one point people were posting maps and searching for colorings by hand. But yeah, the 4-color theorem is an example of something that's been proven mechanically but for which the proof has too many cases for a human to check.

How efficiently and simply can you implement this function?
Bonus: prove your solution correct.

P1107
Is that the only greedy way to do it? Would greedily swapping identical characters within cycles work?

Your sets representation solves it pretty much. We just need to disprove the existence of local optima.
Proof of a unique optimum (fail :():
From any permutations we have a set of possible moves (swaps) of identical characters to other permutations through composition. A swap will change the total cycle length by either +1 or -1. Obviously the global optimum will have all possible identical character swaps being +1. Let's choose the {QP|Q permutes identical characters of s2} form.
The identical characters are elements in various positions in the cycles of P.
Let each set of identical characters that can be swapped be Xi, containing swappable positions Xij.
Xij are unique for all i,j.
Lemma: Optimizing swaps of elements from each Xi independently optimizes Q
It doesn't?
In your counterexample there are no repeated characters within a cycle in either solution.

Maybe extending the moves to multiple swaps would work. Greedy but with some lookahead. Maybe swaps from different Xi's or something. It would work given enough swaps to represent all of Q. But what is the minimum for convexity?
In your counterexample there are extra possibilities? Like: ABFG AHICDEB. Do these change things?
If we can reduce something hard to this problem then accepting a brute force search of Q is probably fine.

Oh wait?
Proof of a unique optimum if there is only one set of identical characters X:
Assume you have Q and Q' with nswaps(QP)<nswaps(Q'P).
Q has more cycles than Q'.
Every cycle of P that Q or Q' can affect contains at least one element of X. We ignore the rest.
There are more such cycles with elements of X in Q than Q'.
Therefore, there exists a cycle in Q' with multiple elements of X. Therefore, there is a swap move from Q that decreases nswaps. Therefore, there is a unique optimum and the greedy swap algorithm works where there is only 1 set of identical characters.

This indicates that having moves be combinations of swaps from different Xi's might be convex for the general case.

P1125
Correction again...
>swap move from Q
swap move from Q'

P1125
Also "unique optimum" means unique optimal value not a unique Q...

I added two functions to get this:
ACL2 !>(get-swap-set "abcd")
((2 3) (1 3) (0 3) (1 2) (0 2) (0 1))
ACL2 !>(rank-swaps "abcd" "dabc" (get-swap-set "abcd") nil)
(1 0 1 1 0 1)
ACL2 !>(get-swap-set "abba")
((2 3) (1 3) (0 3) (1 2) (0 2) (0 1))
ACL2 !>(rank-swaps "abba" "baab" (get-swap-set "abba") nil)
(2 2 0 0 2 2)
ACL2 !>(rank-swaps "abba" "baba" (get-swap-set "abba") nil)
(2 0 0 0 0 -2)
ACL2 !>(rank-swaps "abcd" "acbd" (get-swap-set "abcd") nil)
(-1 -1 2 -2 -1 -1)
So I need to prove a lemma that only positively ranked swaps should ever be done and higher ranked moves are better, recursively break ties for multiple rank 1s (in the absence of a rank 2), and recursively collect the leftmost(choooosing) top ranked swaps.
Functions:
(defun get-swap-set (string)
(add-square-idxes nil (eta nil (length string)) (length string)))

(defun rank-swaps (string goal swaps ranks)
(if (zp (acl2-count swaps)) (reverse ranks)
(rank-swaps string goal (cdr swaps)
(append ranks
(let ((char-a (char string (car (car swaps))))
(char-b (char string (cadr (car swaps))))
(goal-a (char goal (car (car swaps))))
(goal-b (char goal (cadr (car swaps)))))
(list
(+ (if (equal char-a goal-a) -1 0)
(if (equal char-a goal-b) +1 0)
(if (equal char-b goal-b) -1 0)
(if (equal char-b goal-a) +1 0))))))))

#'add-square-idxes is in a text file above. I guess I also probably need to formalize those -1, 0, +1s by replacing them with the difference in number of matching characters in the string and goal (which is what they are, informally).
P1106
The game is proving a theorem your method is correct from the ground up, and probably then prooooooving your method is efficient or something.

Oops, c/(reverse ranks)/ranks/
I "reversed" it twice.

To what extent is hypnosis real/fake?
Have you ever been hypnotized?

As real as you believe it to be. You just have to submit yourself.
Not me, because I have racing thoughts in my head which makes it hard to submit to specific sensations and thoughts that aid in being hypnotised, but I've seen people get hypnotized with music and chanting and enter into a trance state where they lose themselves.

>One study suggests that about 10 percent of the population is highly hypnotizable. Although it's possible that the rest of the population could be hypnotized, they're less likely to be receptive to the practice.

Going to post random things I come across ITT that gets my attention.

Happened 2 days ago. Not sure if the admin would appreciate the livestream footage being posted here, but the guy was able to reload his automatic like John Wick.

Twitter rant that goes for pages about Fumos and how they're bad and must be treated like Funko Pops (No idea what it means), angry about how expensive they are.
https://twitter.com/BakedOuma/status/1525613812770123776?cxt=HHwWgMC96b_riKwqAAAA

>The eyes are emphasized to be as big as possible, while the hands are really small. It is meant to give off a child like appearance to it, no matter how much fumo Twitter tells you otherwise.
>Children don’t play the Touhou series. This is specifically designed in mind for terminally online people who are addicted to lolicon material.
>Age of consent (in Japan) is only two years older than an actual third world shithole
Some casual transphobia thrown in there as well.

Tactical ruffles reload.

P763
partial phase starting soon

These mechanical computers they used to build are pretty cool. I especially like the multiplier.

A network of radio telescopes managed to image the black hole at the center of our galaxy:
https://eventhorizontelescope.org/blog/astronomers-reveal-first-image-black-hole-heart-our-galaxy

ok