Rahul Narain @narain@mathstodon.xyz
- Homepage
- http://rahul.narain.name/
Assistant professor of computer science at IIT Delhi.
I'm just here for the TeX.
Joined Dec 2017
@christianp Curious if you found anything yet. I'd be tempted to explicitly tell the users up front that their headings must start at h3, and reject any descriptions that include h1's or h2's.
Roxanne is what they call Dwayne Johnson in Japan
Rahul Narain
boosted
The pseudorhomicuboctahedron can never be an Archimedean solid.
Boost if you agree
Rahul Narain
boosted
Animation of the minimum-weight matchings of increasingly many points of two colors in a unit square: https://twitter.com/thienan496/status/1446021847292669953
Because the color densities fluctuate, the matching develops regions of many parallel long edges transporting excess density from one place to another. This is reflected mathematically in the fact that the expected length is \(\Theta(\sqrt{n\log n})\) compared to \(\Theta(\sqrt{n})\) for non-bipartite matching; see https://doi.org/10.1007/BF02579135
Render onto Caesar the texture map that is Caesar's
@tpfto Figuring out the symmetries of this shape was extremely confusing. Great job!
Rahul Narain
boosted
Hanlon's razor for identical incorrect exam answers: never attribute to collusion that which is adequately explained by synchronised stupidity.
Rick Astley will let you borrow any of his Pixar DVDs, except one
@christianp It's the condition number of the matrix \(\begin{bmatrix}a&\\&b\end{bmatrix}\), but I agree with "aspect ratio" which I was going to independently suggest
@tpfto Is there any power-of-two (quasi?)periodicity in the source sequence? Or am I just hearing an auditory illusion caused by the cultural bias towards 4/4 time?
@tpfto @11011110 After some numerical experimentation I've found that it's easy to get specified curvatures at the endpoints, but it's very much not easy to get monotonic curvature. In fact now I'm pretty sure it's impossible: if the curvature decreases monotonically from κ to κ/√2 then the curve has to lie "outside" the osculating circle of radius 1/κ and "inside" the circle of radius √2/κ, but it seems there's no way to choose κ so that both endpoints lie outside/inside each other's circles.
The existence of pronouns implies that all other nouns are merely amateur nouns
Rahul Narain
boosted
Because of Euler we know that for polyhedra without holes we have
V+F=E+2. As a result we can't have a polyhedron made entirely of
hexagons, and need at least 12 pentagons or an equivalent set of
shapes. Etc.
But things change if we allow intersecting faces. Clearly we have
a lot of scope for variations, but we can always retain limits to
make things reasonable.
But ...
Can we make a "polyhedron" consisting only of hexagonal "faces"?
It's because of this video. https://youtu.be/J0mUVY9fLlw
Brains are weird. I spent half a minute today trying to figure out why the phrase "go to the dentist" reminded me of Star Wars.
Rahul Narain
boosted
The four points, two distances problem: https://www.theguardian.com/science/2019/oct/21/can-you-solve-it-the-four-points-two-distances-problem
Can you find all of the ways of arranging four distinct points in the plane so that they form only two distances? The link is not a spoiler but it has a separate link to the solution. "Nearly everyone misses at least one" says Peter Winkler; can you guess the one I missed?
Because I just spent 15 minutes trying to remember how to find out what variables a CMake find_package call sets:
cmake --help-module Find<name>
Goddammit. Why is this information so hard to find?
@christianp Use the decomposition of n² into 1+3+5+..., which appears naturally in both lattices starting from a vertex.
- Homepage
- http://rahul.narain.name/
Assistant professor of computer science at IIT Delhi.
I'm just here for the TeX.
Joined Dec 2017
