Background: A finite projective plane has point-line duality, and affine planes lack it. You can get a projective plane by adding a "line at infinity," and this is reversible.
Years ago I stumbled across this in constructing a counterexample to a graph conjecture, and I never found out if it was something people had realized or used before:
We can also recover point-line duality from an affine plane by just deleting one of the parallel classes of lines. <cont...>
"...now lastly, set \(x=10\), ..."
#proofinatoot that if a finite poset has a unique maximal \(x\), then \(x\) is maximum.
If not, there is a \(y_1||x\). \(y_1\) is not maximal, so there is \(y_2>y_1\); we cannot have \(y_2<x\), else transitivity would give \(y_1<x\), and we cannot have \(y_2>x\) because \(x\) is maximal, so \(y_2||x\). Continuing, we build a chain \(y_1<y_2<\dotsb\) (with \(y_i||x\) for all \(i\)), contradicting finiteness.
(This proof also suggests a construction of an infinite poset without the property.)
You MUST play AI Dungeon 2, a text adventure game run by a neural net.
@nickwalton00 built it using @OpenAI's huge GPT-2-1.5B model, and it will respond reasonably to just about anything you try. Such as eating the moon.
"Can you throw the bones for me?" the young woman asked.
"What is your concern?" said the witch.
"I don't fit."
A common concern among the young, but... The witch threw the bones and read them. "You must change, to stay who you are."
"Who I... " The young man blinked. "Oh!"
#MicroFiction #TootFic #SmallStories
Anyone else on mathstodon doing #AdventOfCode? If there's enough of us, would it be worth setting up a private leaderboard?
I've been playing "binary sudoku". You have a 2nx2n grid, with some 0s and 1s and you have to fill it in according to some rules:
* each row and column has equal number of 0s and 1s
* no row or column has a run of three 0s or 1s
* no pair of rows is the same
* no pair of columns is the same
conjecture: there must be a pair of rows (or columns) that are "inverse" i.e. a pair of strings s t where s[i]=/=t[i] for all i.
I can see a boring case analysis proof but is there an "at a glance" proof?
Bechdelgrams illustrate of whether a movie passes the Bechdel test: https://boingboing.net/2019/11/24/bechdelgrams-are-beautiful.html
A nice use of color to highlight the information you're looking for in a social network: Here, the network consists of interactions between characters in a film, and the women and conversations not about men are given distinctive colors to show the test criteria: does the film have at least two named female characters, who speak to each other, about something other than men?
New blog post: Reconfiguring 3-colorings, https://11011110.github.io/blog/2019/11/25/reconfiguring-3-colorings.html
Great insight, does your website fit on a floppy disk? 💾
Mine doesn't (yet), not sure how to cut back on size on @GoHugoIO@twitter.com w/ academic theme
What is a p value? And what's wrong with them? @david_colquhoun explained all in issue 02! http://chalkdustmagazine.com/features/the-perils-of-p-values/
Probing some mathematical minds here: how novel is this way of solving a general quadratic equation \(x^2 + Bx + C = 0\)? Kinda looks like a corollary of "completing the square", or am I mistaken?
3blue1brown on numberphile, with a dart game that connects to (spoiler, but it's in the title too) higher dimensions:
longread, mathematical, web
"My God, It’s Full of Dots!" Wonderful long essay at bit-player with well placed (and paced) interactive components (source cod and all)
(Also a ridiculous commenter.)
Costume victory: One daughter is the Lunar Lander, one an astronaut. This makes me giddy. I hope their candy sample return containers overflow.
I'm sure many of you have seen this, it's a couple of years old - there's a sort of lattice of quadrilaterals (a square is a rectangle is a parallelogram etc) but it's a bit ugly if you stick to standard named ones. Here a "kitoid" is introduced to make a really beautiful diagram
But it got me thinking, is there any well defined sense in which these categories are exhaustive (for convex quads)? And if so how many are there for n-gons
In his latest alphabet entry "Platonic" (https://nebusresearch.wordpress.com/2019/10/24/my-2019-mathematics-a-to-z-platonic/) @nebusj writes that "it was obvious" that eight equal charged particles free to move on a sphere would space themselves out into the vertices of a cube. But sadly, in this case, the obvious is false. According to https://en.wikipedia.org/wiki/Thomson_problem they form a square antiprism instead.
Combinatorist (esp. graph theorist) turned Data Scientist
A Mastodon instance for maths people. The kind of people who make \(\pi z^2 \times a\) jokes.
\) for inline LaTeX, and
\] for display mode.